Calculus 1 - Differentiation and Integration: Over 1,900 Solved Problems 0964995441

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Calculus I

Differentiation and Integration

Book Title: Calculus I - Differentiation and Integration Author: Dan Hamilton Editor: Linda Hamilton Cover design by: Kathleen Myers Copyright  2002 All rights reserved. Printed in the United States of America. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the author. Request for permission or further information should be addressed to Hamilton Education Guides via [email protected]. First published in the year 2002 Library of Congress Catalog Card Number 20022-81883 Library of Congress Cataloging-in-Publication Data ISBN 0-9649954-4-1

This book is dedicated to my wife and children for their support and understanding.

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Contents Introduction and Overview ....................................................................................................... iii Chapter 1 Sequences and Series Quick Reference to Chapter 1 Problems ........................................................................ 1 1.1 Sequences ........................................................................................................... 2 1.2 Series .................................................................................................................. 7 1.3 Arithmetic Sequences and Arithmetic Series..................................................... 13 1.4 Geometric Sequences and Geometric Series ..................................................... 20 1.5 Limits of Sequences and Series ......................................................................... 31 1.6 The Factorial Notation ....................................................................................... 42 Chapter 2 Differentiation (Part I) Quick Reference to Chapter 2 Problems ........................................................................ 53 2.1 The Difference Quotient Method ....................................................................... 54 2.2 Differentiation Rules Using the Prime Notation ................................................ 59 d Notation ..................................................... 71 2.3 Differentiation Rules Using the dx 2.4 2.5 2.6 2.7 2.8

The Chain Rule .................................................................................................. 82 Implicit Differentiation ...................................................................................... 97 The Derivative of Functions with Fractional Exponents ................................... 102 The Derivative of Radical Functions ................................................................. 109 Higher Order Derivatives ................................................................................... 124

Chapter 3 Differentiation (Part II) Quick Reference to Chapter 3 Problems ........................................................................ 139 3.1 Differentiation of Trigonometric Functions ....................................................... 140 3.2 Differentiation of Inverse Trigonometric Functions .......................................... 158 3.3 Differentiation of Logarithmic and Exponential Functions ............................... 166 3.4 Differentiation of Hyperbolic Functions ............................................................ 181 3.5 Differentiation of Inverse Hyperbolic Functions ............................................... 187 3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule ............................. 193 Chapter 4 Integration (Part I) Quick Reference to Chapter 4 Problems ........................................................................ 212 4.1 Integration Using the Basic Integration Formula ............................................... 213 4.2 Integration Using the Substitution Method ........................................................ 222 4.3 Integration of Trigonometric Functions ............................................................. 232 4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions........ 259 4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions 273

Hamilton Education Guides

i

Calculus I

Contents

Chapter 5 Integration (Part II) Quick Reference to Chapter 5 Problems ........................................................................ 286 5.1 Integration by Parts ............................................................................................ 287 5.2 Integration Using Trigonometric Substitution ................................................... 308 5.3 Integration by Partial Fractions .......................................................................... 320

5.4 Appendix

Case I - The Denominator Has Distinct Linear Factors 320 Case II - The Denominator Has Repeated Linear Factors 327 Case III - The Denominator Has Distinct Quadratic Factors 334 Case IV - The Denominator Has Repeated Quadratic Factors 344

Integration of Hyperbolic Functions .................................................................. 350

Exercise Solutions Chapter 1 Solutions .........................................................................................379 Chapter 2 Solutions .........................................................................................401 Chapter 3 Solutions .........................................................................................426 Chapter 4 Solutions .........................................................................................436 Chapter 5 Solutions .........................................................................................454

Index ...........................................................................................................................................478

Hamilton Education Guides

ii

Introduction and Overview Similar to the previous books published by the Hamilton Education Guides, the intent of this book is to build a strong foundation by increasing student confidence in solving mathematical problems. To achieve this objective, the author has diligently tried to address each subject in a clear, concise, and easy to understand step-by-step format. A great deal of effort has been made to ensure that the subjects presented in each chapter are explained simply, thoroughly, and adequately. It is the authors hope that this book can fulfill these objectives by building a solid foundation in pursuit of more advanced technical concepts. The scope of this book is intended for educational levels ranging from the 12th grade to adult. The book can also be used by students in home study programs, parents, teachers, special education programs, tutors, high schools, preparatory schools, and adult educational programs, including colleges and universities as a main text, a thorough reference, or a supplementary book. A thorough knowledge of algebraic concepts in subject areas such as linear equations and inequalities, fractional operations, exponents, radicals, polynomials, factorization, non-linear and quadratic equations is required. “Calculus I” is divided into five chapters. Sequences and series are introduced in Chapter 1. How to compute and find the limit of arithmetic and geometric sequences and series including expansion and simplification of factorial expressions is discussed in this chapter. Derivatives and its d notations are introduced in Chapter 2. In applicable differentiation rules using the Prime and dx

addition, use of the Chain rule in solving different types of equations, the implicit differentiation method, derivative of functions with fractional exponents, derivative of radical functions, including the steps for solving higher order equations is discussed in this chapter. Differentiation of trigonometric functions, exponential and logarithmic functions, hyperbolic functions, and inverse hyperbolic functions is discussed in Chapter 3. Furthermore, evaluation of expressions referred to as indeterminate forms using a general rule known as L’Hopital’s Rule is discussed in Chapter 3. The subject of integration is introduced in Chapter 4. Integration using basic integration formulas and methods such as the substitution method is discussed in this chapter. Additionally, integration of trigonometric functions, inverse trigonometric functions, exponential and logarithmic functions is addressed in Chapter 4. Other integration techniques such as integration by parts, integration using trigonometric substitution, and integration by partial fractions is introduced in Chapter 5. The steps in integrating hyperbolic functions is also discussed in this chapter. Finally, detailed solutions to the exercises are provided in the Appendix. Students are encouraged to solve each problem in the same detailed and step-by-step format as shown in the text. In keeping with our commitment of excellence in providing clear, easy to follow, and concise educational materials to our readers, I believe this book will again add value to the Hamilton Education Guides series for its clarity and special attention to detail. I hope readers of this book will find it valuable as both a learning tool and as a reference. Any comments or suggestions for improvement of this book will be appreciated. With best wishes, Dan Hamilton Hamilton Education Guides

iii

Calculus I

Quick Reference to Chapter 1 Problems

Chapter 1

Sequences and Series Quick Reference to Chapter 1 Problems 1.1

Sequences .................................................................................................................... 2 an =

1.2

2n + 1 −2n

n( n + 1) k ( k + 1) = ; bk = = ; sn = = 2 −1 k

Series ........................................................................................................................... 7 5

n

1  1 ∑  ai − bi  = ; 4  i =1  2

1.3

∑

i =1

( n − 1)2 ( n + 1) = ∑ n =0

15

∑ (3i−2) = ; i =1

15

∑ (5 j −1)

=

j =3

Geometric Sequences and Geometric Series............................................................ 20 10

∑ 3k − 2

=;

k =1

10

∑ ( −3)k −2

=;

k =1

6

 1 8 −  2 k =2 

∑

k +1

=

Limits of Sequences and Series ................................................................................. 31 lim n→∞

1.6

2i

4

=;

Arithmetic Sequences and Arithmetic Series .......................................................... 13

∑ ( 2i+1) = ;

1.5

( −1) i+1

i =1

20

1.4

2n

n2 + 5 =; n2

1  lim n→∞ 1 +  2 

−n

= ; limn→∞

n6 = 12n4 + 5

The Factorial Notation .............................................................................................. 42 2

( 2n − 2 )! 2 ( n!) ( n!) ( 4 − 2 )!8! =; =; = 11! ( 5 − 3)! 2 n ! n − 1 ! n + 1 ( )( ) ( )! ( n − 1)!

Hamilton Education Guides

1

Chapter 1 - Sequences and Series The objective of this chapter is to improve the student’s ability to solve problems involving sequences and series. Sequences and series are introduced in Sections 1.1 and 1.2. How to solve arithmetic sequences and arithmetic series are discussed in Section 1.3. Solutions to geometric sequences and geometric series are addressed in Section 1.4. The process of identifying convergence or divergence of a sequence or a series, for large values of n , is discussed in Section 1.5. Finally, the factorial notation and its use in expanding binomial expressions is addressed in Section 1.6. Each section is concluded by solving examples with practice problems to further enhance the student’s ability.

1.1

Sequences

A sequence is a function whose domain contains a set of positive integer terms such as (1, 2, 3, 4,) . Functions generate sequences. For example, the function s( n) = sn = n − 2 whose domain is (1, 2, 3, 4, 5, 6) generates the sequence s (1) =s1 = 1 − 2 =−1

s ( 2 ) = s2 = 2 − 2 = 0

s ( 3) = s3 = 3 − 2 = 1

s ( 4 ) = s4 = 4 − 2 = 2

s ( 5) = s5 = 5 − 2 = 3

s ( 6 ) = s6 = 6 − 2 = 4

where the first six terms of the sequence are ( s1 , s2 , s3 , s4 , s5 , s6 ) = ( −1, 0, 1, 2, 3, 4) . In general, a function f ( x ) whose domain is the set of positive integers (1, 2, 3,  , n) including a fixed value for n is called a finite sequence function. On the other hand, a function whose domain is the set of (1, 2, 3, ) is called an infinite sequence function. The elements of the range of a sequence function are called the terms of the sequence function. In some instances a sequence is given by presenting its first few terms, followed by its nth term, sn = s( n) , which is commonly referred to as the general term of a sequence. For example, the sequence 2

4, 3,

( n + 1) 16 25 , ,, 5 9 2n − 1

shows the first four terms and the general term of the sequence. In the

following examples we will learn how the various terms of a sequence are found: Example 1.1-1 List the first six terms of the given sequence. a. a n =

(− 3)n

b. bk =

n3

( −1) k k +1

c. d n =

1

5 n(2n − 1)

n

(−1) n

d. cn =   ⋅  2 n

Solutions: a. a1 = a3 =

a5 =

(− 3) 1 3

1

(− 3) 3 3

3

(− 3) 5 5

3

=

−3 = −3 1

a2 =

=

−27 27

a4 =

=

−243 125

= −1

Hamilton Education Guides

= −1.944

a6 =

(− 3) 2 2

3

(− 3) 4 4

3

(− 3) 6 6

3

=

9 = 1.125 8

=

81 64

=

729 216

= 1.265

= 3.375

2

Calculus I

b. b1 = b3 =

b5 =

c. d1 =

1.1 Sequences

(− 1) 1 1+1

(− 1) 3 3 +1

(− 1) 5 5 +1

=

−1 = −0.5 2

b2 =

=

−1 = −0.25 4

b4 =

=

−1 6

5 1 ⋅ (2 ⋅ 1 − 1)

=

= −0.167 5 2 −1

=

5 = 5 1

(− 1) 2

=

1 3

= 0.333

=

1 5

= 0.2

b6 =

(− 1) 6 =

1 7

= 0.143

d2 =

5 5 5 = = = 0.833 2 ⋅ (2 ⋅ 2 − 1) 2⋅3 6

2 +1

(− 1) 4 4 +1

6 +1

d3 =

5 5/ 1 = = = 0.333 3 ⋅ (2 ⋅ 3 − 1) 3 ⋅ 5/ 3

d4 =

5 5 5 = = = 0.178 4 ⋅ (2 ⋅ 4 − 1) 4⋅7 28

d5 =

5 5 ⋅ (2 ⋅ 5 − 1)

d6 =

5 6 ⋅ (2 ⋅ 6 − 1)

=

5/ 5/ ⋅ 9

=

1 9

= 0.111

1 1 1 1  1  (− 1) d. c1 =   ⋅ = ⋅ −1 = − = −0.5

2

2

1

3

3

5

5

 1  (− 1) c3 =   ⋅ 3 2  1  (− 1) c5 =   ⋅ 5 2

=

1 1 ⋅− 8 3

=

1 1 ⋅− 32 5

= −

1 24

= −

2

4

4

6

6

 1  (− 1) c4 =   ⋅ 4 2

= −0.042

1 160

2

 1  (− 1) c2 =   ⋅ 2 2

2

 1  (− 1) c6 =   ⋅ 6 2

= −0.006

=

5 6 ⋅11

=

5 66

= 0.076

=

1 1 ⋅ 4 2

=

1 1 ⋅ 16 4

=

1 64

=

1 1 ⋅ 64 6

=

1 384

=

1 8

= 0.125

= 0.016

= 0.003

Example 1.1-2 Find the indicated terms for the following sequences. a. Write the third and sixth terms of sn =

( −2) n+1 n3

b. Write the tenth term of a i = (i − 1)3 ⋅ 2 i−4

c. Write the third and fourth terms of a n = ( −1) n − 2

d. Write the seventh term of a k = (0.2) k −1

e. Write the third and twelfth terms of a i = ( − i)3

f. Write the eleventh term of sn = ( −1) n −1 ⋅ 2 n +1

Solutions: a. s3 =

(− 2) 3+1 3

3

=

(− 2) 4 27

=

16 27

= 0.593

s6 =

(− 2) 6+1 6

3

=

(− 2) 7 216

=

−128 216

= −0.593

b. a10 = (10 − 1) 3 ⋅ 210−4 = 9 3 ⋅ 2 6 = 729 ⋅ 64 = 4.6656 × 10 4 c. a3 = (− 1) 3−2 = (− 1) 1 = −1

Hamilton Education Guides

a4 = (− 1) 4− 2 = (− 1) 2 = 1

3

Calculus I

1.1 Sequences

d. a7 = (0.2) 7 −1 = (0.2) 6 = 0.000064 = 6.4 × 10 −5 e. a3 = (− 3) 3 = − 33 = −27

a12 = (− 12 ) 3 = − 12 3 = −1728

f. s11 = (− 1) 11−1 ⋅ 211+1 = (− 1) 10 ⋅ 212 = 1⋅ 212 = 4096 Example 1.1-3 Write s5 , s6 , s7 , and s15 for the following sequences. a. s n =

(n + 1) π 2

Solution: a. s5 = s7 =

b. s5 = s6 =

s7 =

s15 =

b. sn =

(5 + 1) π 2

6π 2

= 3π

s6 =

=

8π 2

= 4π

s15 =

2

=

5−1

2

26

=

2 6−1

27

=

2 7 −1 215 2

15−1

c. s n = 2n(n − 1) (n − 2)

2 n −1

=

(7 + 1) π 25

2n

=

25 2

2

(15 + 1) π 2

= =

7π 2

= 3.5π

16π 2

= 8π

; 2 5 ⋅ 2 −4 ; 2 5−4 ; 21 = 2

4

26

= 2 6 ⋅ 2 −5 = 2 6−5 = 21 = 2

25

27

= 2 7 ⋅ 2 −6 = 2 7 −6 = 21 = 2

26

215 2

(6 + 1) π

14

= 215 ⋅ 2 −14 = 215−14 = 21 = 2

c. s5 = 2 ⋅ 5 ⋅ (5 − 1) (5 − 2) = 10 ⋅ 4 ⋅ 3 = 120

s6 = 2 ⋅ 6 ⋅ (6 − 1)(6 − 2 ) = 12 ⋅ 5 ⋅ 4 = 240

s15 = 2 ⋅15 ⋅ (15 − 1)(15 − 2 ) = 30 ⋅14 ⋅13 = 5460

s7 = 2 ⋅ 7 ⋅ (7 − 1) (7 − 2) = 14 ⋅ 6 ⋅ 5 = 420

Example 1.1-4 Find the twelfth term of the following sequences: a. 0.5, 0.25, 0125 . ,,

1n +1

1 b. 8, 5.063, 4.214,  , 1 +  

2n

n+2

n

c. 4, − 12, 32,  ,

2 n ⋅ ( n + 1)

( −1) n+1

Solutions: a. s12 =

112+1 2

12

=

113 2

12

=

1 4096

= 0.000244 = 2.44 × 10 −4

12 + 2 14 14 1   13   12 + 1  14 b. s12 = 1 +  =   =   = 1.0833 = 3.066



12 

Hamilton Education Guides

 12 

 12 

4

Calculus I

c. s12 =

1.1 Sequences

212 ⋅ (12 + 1)

(− 1)

12+1

=

212 ⋅13

(− 1)

13

=

4096 ⋅13 −1

= −53248 = − 5.3248 × 10 4

Example 1.1-5 Given the general term of the sequence s( n) = sn = n( n − 2) + 5 , write its k th and k + 1 term. Solutions: a. To write the k th term of the sequence simply substitute k in place of n in the general term of the sequence, i.e., s ( k ) = sk = k (k − 2) + 5 = k 2 − 2k + 5 b. To write the k + 1 term of the sequence simply substitute k + 1 in place of n in the general term of the sequence, i.e., s ( k + 1) = sk +1 = ( k + 1) ( k + 1) − 2  + 5 = ( k + 1) k + 1 − 2 + 5 = ( k + 1)( k − 1) + 5 = k 2 − k/ + k/ − 1 + 5 = k 2 + 4 Example 1.1-6 For the given domain (1, 2, 3, 4) , write the first four terms of the following functions: b. s ( x= ) 3x − 5

a. f ( x ) = x 2 + 2 x + 1 2

c. g ( x ) = x

−1 d. h (= x) x + 1

3

Solutions: a. f (1) = f1 = 12 + ( 2 ⋅1) + 1 = 1 + 2 + 1 = 4 f ( 3)

= f3 = 32 + ( 2 ⋅ 3) + 1 = 9 + 6 + 1 = 16

f ( 2)

= f 2 = 22 + ( 2 ⋅ 2 ) + 1 = 4 + 4 + 1 = 9

f ( 4)

= f 4 = 42 + ( 2 ⋅ 4 ) + 1 = 16 + 8 + 1 = 25

Therefore, the first four terms of the sequence are ( f 1 , f 2 , f 3 , f 4 ) = ( 4, 9, 16, 25) b. s (1) = s1 = ( 3 ⋅1) − 5 = 3 − 5 = −2

s ( 2)

= s2 = ( 3 ⋅ 2 ) − 5 = 6 − 5 = 1

= s3 = ( 3 ⋅ 3) − 5 = 9 − 5 = 4

s ( 4)

= s4 = ( 3 ⋅ 4 ) − 5 = 12 − 5 = 7

s ( 3)

Therefore, the first four terms of the sequence are ( s1 , s2 , s3 , s4 ) = ( −2, 1, 4, 7) . 2 2 c. g (1) = g1 = ×1 = 3

g ( 3)

3

2 2 = g3 = × 3/ = = 2 3/

1

g ( 2)

4 2 = g2 = × 2 =

g ( 4)

8 2 = g4 = × 4 =

3

3

3

3

2 4 8 Therefore, the first four terms of the sequence are ( g1 , g 2 , g 3 , g 4 ) =  , , 2,    3 3

d. h (1) = h1 = 1−1 + 1 =

Hamilton Education Guides

1 +1 1

=

1+ 1 1

= 2

h ( 2)

3

3 1 1+ 2 = h2 = 2−1 + 1 = + 1 = = 2

2

2

5

Calculus I

h ( 3)

1.1 Sequences

= h3 = 3−1 + 1 =

1 3

+1

=

1+ 3 4

=

4 3

h ( 4)

5 1 1+ 4 = h4 = 4−1 + 1 = + 1 = = 4

4

4

3 4 5 2 3 4

Therefore, the first four terms of the sequence are ( h1 , h2 , h3 , h4 ) =  2, , ,    In the following section we will discuss series and identify its relation with sequences. Section 1.1 Practice Problems - Sequences 1. List the first four and tenth terms of the given sequences. a. a n =

2n + 1 −2n

b. bk =

k ( k + 1)

c. d n = 3 − ( −2) n

k2

n 1 ( −1) d. k n =  − 

 2

n +1

n+2

2. Write s3 , s4 , s5 , and s8 for the following sequences. a. sn =

n( n + 1)

b. sn = ( −1) n+1 2 n−2

2n −1

c. sn =

( −2) n+1( n − 2) 2n

3. Write the first five terms of the following sequences. 1    100 

i −2

a. a n = ( −1) n+1 ( n + 2)

b. a i = 3 

d. a n = (3n − 5) 2

e. uk = ar k −2 + 2

g. c j = j. y k =

j + j j +1

h. y n = 1 − 

k

k. y n =

2 k −1

1 9k

1 c. ci = 3  − 

i −1

2 f. bk = −3  

k −2

 5

1   n + 2

3

n +1

( k − 2)

i. uk = 1 − ( −1) k +1 l. cn =

n2 − 2 n +1

4. Given n ! read as “n factorial” which is defined as n ! = n (n − 1) (n − 2) (n − 3)  5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 , find a. The first eight terms of n ! . b. The first four terms of a n =

2n + 1 . n!

c. The tenth and twelfth terms of the c n =

1 + 3 n −1

(n !)2

.

d. The first, fifth, tenth, and fifteenth terms of y n =

n ! (n − 1) . 2+n!

5. Write the first three terms of the following sequences.

(2n − 3) (n + 1) (n − 4) n k +1 k ( k − 1) = ( −1)

a. c n = d. y k

2

Hamilton Education Guides

1   n − 2    n − 1  2 + n 

b. a n = 

n −1   2 + n

e. bn = n 2 

c. sn = ( −1) n+1 2 n+1 f. x a = (5 − a) a +1 2 a 6

Calculus I

1.2

1.2 Series

Series

Addition of the terms in any finite sequence result in having the sum of the sequence. The sum of the sequence is referred to as a series. k = 1, 2, 3, 4, 5, and 6

For example, the sequence

yk =

can be summed and expressed in the following way:

y1 + y 2 + y3 + y 4 + y5 + y 6

1

=

2

0

+

1

+

1

2

1 2

2

+

1 2

+

3

1 2

4

+

1 2

1 2

1 4

1 8

= 1+ + + +

5

1

2

k −1

for

1 1 = 1.9687 + 16 32

The sum of a sequence is generally shown by the Greek letter “ ∑ ” (sigma) which is also called

summation. Thus, using the sigma notation, the above example can be expressed in the following 6

1

∑ y k where y k = 2 k −1 .

way

Note that the variable i is referred to as the index of summation and

i =1

the integer range over which the summation occurs is referred to as the range of summation. The following are three properties of summation that students should be familiar with: n

∑ ( ai + bi ) = i =1

n

∑ i =1

kai n

∑k

n

∑ i =1

ai +

n

∑ bi i =1

n

= k ∑ ai i =1

= nk

i =1

These properties are used extensively in solving the sum of sequences over a specified range as shown in the following examples: Example 1.2-1

Given

n

∑ i =1

a i = 20

and

∑ bi = 40 , find the solution to the following problems i =1

using the summation properties. a.

n

∑ i =1

( 2a i + 3bi )

=

b.

n

∑ i =1

n

( a i − bi )

=

c.

n

∑ i =1

( −5a i + 2bi )

=

d.

n

1

1

∑  2 ai − 4 bi 

=

i =1

Solutions: n

n

n

n

n

a. ∑ ( 2ai + 3bi ) = ∑ 2ai + ∑ 3bi = 2∑ ai + 3∑ bi = ( 2 × 20 ) + ( 3 × 40 ) = i =1

n

=i 1 =i 1

n

n

b. ∑ ( ai − bi ) = ∑ ai − ∑ bi = i =1

n

20 − 40

=i 1 =i 1

n

n

c. ∑ ( −5ai + 2bi ) = ∑ −5ai + ∑ 2bi = i =1

=i 1 =i 1

Hamilton Education Guides

40 + 120

=i 1 =i 1

=

=

160

−20

n

n

−5∑ ai + 2∑ bi

=i 1 =i 1

= ( −5 × 20 ) + ( 2 × 40 ) =

−100 + 80

=

−20

7

Calculus I

1.2 Series

1

n

1

1

n

n

1 n 1 n ai − ∑ bi ∑ 2 i 1= 4i1 =

1

d. ∑  ai − bi  = ∑ ai − ∑ bi = 2 4 2 4 i =1





=i 1 =i 1

1   1   × 20  +  − × 40  2 4    

=

=

=

10 − 10

0

Example 1.2-2 Solve the following series: a. Find

6

∑

n =1

c. Find

a n where a n =

5

∑( j =0

e. Find

)

2n + 1 n

2

x j −1

where x j =

5

∑ ( y n − 2) n+1

where y n =

n =1

∑ x i where x i = (1 + i 2 ) (−2) i i =1

1 1+ j

2

7

b. Find

4

∑ ( uk ) 2 where uk = k + 1

d. Find

k =0

n 1+ n

f. Find

5

∑ ( ua + a) 2 where ua = a 2 − 1

a =0

Solutions: 6

a. ∑ an where an = n =1

2n + 1 n

=

3 5 7 9 11 13 + + + + + 1 2 3 4 5 6

=

a1 + a2 + a3 + a4 + a5 + a6

=

=

2 + 1 4 + 1 6 + 1 8 + 1 10 + 1 12 + 1 + + + + + 1 2 3 4 5 6

3 + 2.5 + 2.33 + 2.25 + 2.2 + 2.17

7

b. ∑ xi where xi =(1 + i 2 ) ( −2 )i = i =1

x1 + x2 + x3 + x4 + x5 + x6 + x7

=

14.45

= (1 + 12 ) ( −2 )1 + (1 + 22 ) ( −2 )2

+ (1 + 32 ) ( −2 ) + (1 + 42 ) ( −2 ) + (1 + 52 ) ( −2 ) + (1 + 62 ) ( −2 ) + (1 + 7 2 ) ( −2 ) 3

4

5

+ (17 ⋅ 16 ) + ( 26 ⋅ −32 ) + ( 37 ⋅ 64 ) + ( 50 ⋅ −128 ) 5

c. ∑ ( x j − 1) j =0

2

6

=

−4 + 20 − 80 + 272 − 832 + 2368 − 6400

2

2

=

−4656

1 where x j = = ( x0 − 1)2 + ( x1 − 1)2 + ( x2 − 1)2 + ( x3 − 1)2 + ( x4 − 1)2 + ( x5 − 1)2 1+ j 2

2

2

2

1   1   1   1   1   1   − 1 +  − 1 +  − 1 +  − 1 +  − 1 +  − 1 1   2   3   4   5   6 

=

= ( 2 ⋅ −2 ) + ( 5 ⋅ 4 ) + (10 ⋅ −8)

7

4  5 +  +  −  5  6

2

=

4

d. ∑ ( uk )2 where uk=

1 4 9 16 25 + + + + 4 9 16 25 36

k +1

k =0

=

2

=

2

2

 1  2  3 0 + −  + −  + −   2  3  4

0.25 + 0.4444 + 0.5625 + 0.64 + 0.6944

=

2

2.5913

= ( u0 )2 + ( u1 )2 + ( u2 )2 + ( u3 )2 + ( u4 )2 = ( 0 + 1)2 + (1 + 1)2 + ( 2 + 1)2

+ ( 3 + 1) + ( 4 + 1) = 12 + 22 + 32 + 42 + 52 = 1 + 4 + 9 + 16 + 25 = 55 2

2

Hamilton Education Guides

8

Calculus I

1.2 Series

n

5

e. ∑ ( yn − 2 )n +1 where yn = = ( y1 − 2 )1+1 + ( y2 − 2 )2 +1 + ( y3 − 2 )3+1 + ( y4 − 2 )4 +1 + ( y5 − 2 )5+1 1+ n n =1

2

3

4

5

 1   2   3   4   5  − 2 +  − 2 +  − 2 +  − 2 +  − 2  1+1  1+ 2  1+ 3  1+ 4  1+ 5 

=

+ ( −1.2 ) + ( −1.17 ) 5

=

6

=

2.25 − 2.35 + 2.44 − 2.49 + 2.56

6

= ( −1.5)2 + ( −1.33)3 + ( −1.25)4

2.41

5

2 2 2 2 2 2 a 2 − 1 = ( u0 + 0 ) + ( u1 + 1) + ( u2 + 2 ) + ( u3 + 3) + ( u4 + 4 ) + ( u5 + 5 ) f. ∑ ( ua + a )2 where ua = a =0

= ( −1 + 0 )2 + ( 0 + 1)2 + ( 3 + 2 )2 + (8 + 3)2 + (15 + 4 )2 + ( 24 + 5)2 = Example 1.2-3 Solve the following series. a.

5

∑

a=1

d.

a n (2a − 1)

=

5

∑

b.

a =1

4

∑ (n − 1)2 (n + 1) =

2a + 1 a

3

∑

e.

n=0

j =−3

=

1 + 1 + 25 + 121 + 361 + 841

=

c.

1350

( −1) i+1 = ∑ 5

i =1

2j = j+5

f.

2i

(1 − k ) k −1 = ∑ 5

k =1

k

Solutions: 5

a. ∑ a ( 2a − 1) = a =1

+28 + 45

=

95

2a + 1 a a =1

=

( 2 ⋅ 1) + 1 + ( 2 ⋅ 2 ) + 1 + ( 2 ⋅ 3) + 1 + ( 2 ⋅ 4 ) + 1 + ( 2 ⋅ 5) + 1

5

b. ∑

1

2

=

+2.33 + 2.25 + 2.2 5

c. ∑ i =1

( −1) 2i

i +1

=

( −1)

1 + 6 + 15

2 ⋅1

3

4

=

5

3 5 7 9 11 + + + + 1 2 3 4 5

=

3 + 2.5

12.28

1 +1

1 1 1 1 1 − + − + 2 4 6 8 10

=

=

1 ⋅ ( 2 ⋅ 1 − 1)  +  2 ⋅ ( 2 ⋅ 2 − 1)  + 3 ⋅ ( 2 ⋅ 3 − 1)  +  4 ⋅ ( 2 ⋅ 4 − 1)  + 5 ⋅ ( 2 ⋅ 5 − 1) 

+

=

( −1)

2 +1

2⋅2

+

( −1)

3 +1

2⋅3

+

( −1)

4 +1

2⋅4

+

( −1)

5 +1

=

2⋅5

0.5 − 0.25 + 0.167 − 0.125 + 0.1

=

( −1) 2

2

+

( −1) 4

3

+

( −1) 6

4

+

( −1) 8

5

+

( −1)

6

10

0.392

4

d. ∑ ( n − 1)2 ( n + 1) = ( 0 − 1)2 ( 0 + 1) + (1 − 1)2 (1 + 1) + ( 2 − 1)2 ( 2 + 1) + ( 3 − 1)2 ( 3 + 1) + ( 4 − 1)2 ( 4 + 1) n=0

= (1 ⋅ 1) + ( 0 ⋅ 2 ) + (1 ⋅ 3) + ( 4 ⋅ 4 ) + ( 9 ⋅ 5) =

Hamilton Education Guides

1 + 0 + 3 + 16 + 45

=

65

9

Calculus I

2j j+5

3

e. ∑

j = −3

+

23 8

5

f. ∑

1.2 Series

0.125 0.25 0.5 1 2 4 8 + + + + + + 2 3 4 5 6 7 8

=

(1 − k )

k

=

k

k =1

2−3 2−2 2−1 20 21 22 23 + + + + + + −3 + 5 −2 + 5 −1 + 5 0 + 5 1 + 5 2 + 5 3 + 5

=

+20.25 − 204.8

(1 − 1)

1

1

=

+

(1 − 2 )

2

2

+

(1 − 3)

3

3

+

=

=

2−3 2−2 2−1 20 21 22 + + + + + 2 3 4 5 6 7

0.0625 + 0.0833 + 0.125 + 0.2 + 0.3333 + 0.5714 + 1

(1 − 4 ) 4

4

+

(1 − 5) 5

5

=

0 1 8 81 1024 + − + − 1 2 3 4 5

=

=

2.3755

0.5 − 2.67

−186.72

Example 1.2-4 Prove that both sides of the following series are equal to one another. a. c.

n

n

i =1

i =1

∑ 2xi + ∑ 4 yi

n

= 2∑ ( x i + 2 y i )

b.

i =1

n

∑ a yi2 i =1

n

∑ a = na

d.

i =1

n

= a ∑ y i2 i =1

n

n

i =1

i =1

∑ ( xi + a) = ∑ xi + na

Solutions: n

n

a. ∑ 2 xi + ∑ 4 yi = ( 2 x1 + 2 x2 + 2 x3 +  + 2 xn ) + ( 4 y1 + 4 y2 + 4 y3 +  + 4 yn ) = ( 2 x1 + 4 y1 ) + ( 2 x2 + 4 y2 )

=i 1 =i 1

+ ( 2 x3 + 4 y3 ) +  + ( 2 xn + 4 yn )

= 2 ( x1 + 2 y1 ) + 2 ( x2 + 2 y2 ) + 2 ( x3 + 2 y3 ) +  + 2 ( xn + 2 yn )

n

= 2∑ ( x i + 2 y i ) i =1

n

b. ∑ a yi2 = i =1

n

c. ∑

a

i =1

=

ay12 + ay22 + ay32 + ay42 +  + ayn2

a + a + a + a ++ a  n terms

=

= a ( y12 + y22 + y32 + y42 +  + yn2 ) =

n

a ∑ yi2 i =1

na

n

d. ∑ ( xi + a ) = ( x1 + a ) + ( x2 + a ) + ( x3 + a ) +  + ( xn + a ) = ( x1 + x2 + x3 +  + xn ) + ( a + a + a +  + a ) i =1

n

= ∑ xi + na i =1

Example 1.2-5 Use the properties of summation to evaluate the following series. a.

6

∑

2k

=

i=1

Hamilton Education Guides

b.

7

∑ (4k − 3) = i =1

c.

4

∑ ( k 3 − 2k ) = k =1

10

Calculus I

d.

5

∑ (k

1.2 Series

2

+a

k =1

)=

4

∑

e.

k =1

Solutions: 6

6

i =1

i =1

a. ∑ 2k = 2 ∑ k = 7

=

2 ⋅ 6k

7

 2 2⋅−   3

c.

7

7

4

4

4

7

−(2 ⋅ 10)

=

=

100 − 20 5

5

5

55 + 5a

4

2

=



28k − 21

= (13 + 23 + 33 + 43 ) − 2 (1 + 2 + 3 + 4 ) =

1 + 8 + 27 + 64

5

k =1

= 5 (11 + a )

e. ∑ 2 ⋅  −  3 k =1

4

k 3 − 2∑ k

=

= ∑ k 2 + 5a = (12 + 22 + 32 + 42 + 52 ) + 5a = (1 + 4 + 9 + 16 + 25) + 5a

a

= k 1= k 1

=

∑ (2 k + k ) =

80

d. ∑ ( k 2 + a ) = ∑ k 2 + ∑ k =1

4 ⋅ 7k − 7 ⋅ 3

= k 1= k 1

= k 1= k 1

k =1

4

= ∑

+ ∑ −2k

3

f.

k =1

=i 1 =i 1

=i 1 =i 1

∑ (k 3 − 2k ) = ∑ k

=

5

12k

b. ∑ ( 4k − 3) = ∑ 4k + ∑ −3 = 4∑ k − ∑ 3 = i =1

k +1

k +1



2

4

= 2∑  −  3 k =1



k +1



=

 2  2  2 3  2  4  2 5  2 ⋅  −  +  −  +  −  +  −    3   3   3   3  

2 ⋅ ( 0.4444 − 0.2963 + 0.1975 − 0.1317 )

5

5

=

2 ⋅ 0.2139

=

=

 4 8 16 32  2⋅ − + −   9 27 81 243 

0.4278

5

f. ∑ ( 2k + k ) = ∑ 2k + ∑ k = ( 21 + 22 + 23 + 24 + 25 ) + (1 + 2 + 3 + +4 + 5) = ( 2 + 4 + 8 + 16 + 32 ) + 15 = k 1= k 1

k =1

= 62 + 15 =

77

Section 1.2 Practice Problems - Series 1. Given

n

∑ i =1

a.

a i = 10

and

n

∑ bi = 25 , find i =1

n

∑ ( 2ai + 4bi ) = i =1

b.

n

∑ ( −ai + bi ) =

c.

i =1

n

∑ (3ai + 5bi ) =

d.

n

1

1

∑  2 ai + 5 bi 

=

i =1

i =1

2. Evaluate each of the following series. a.

5

∑

k =1

y k where y k = 2 + k

Hamilton Education Guides

b.

6

∑

n=0

xn where xn =

1

(− 2)

n +1

c.

4

∑ x n where x n = (− 1) n+1

n =0

11

Calculus I

d.

1.2 Series

3

∑

u j where u j = j − 3 j 2

e.

y k + 2 where y = 2k − 3

h.

3

∑

k = −2

∑

y a where y = a + 2

f.

a =3

j = −3

g.

5

5

∑ i =0

5

x i where x i =

(− 1) i +1 2i

1

∑ (x m − 1) 2 where x m = m

m =1

3. Find the sum of the following series within the specified range. a.

3

∑

10 i

=

6

i = −3

d.

5

∑ (n 6

2

−n

)=

∑

5(a − 1) + 3

4

n +1 − n

∑

n =1

4

∑

n =1

n

6

∑ (−1)

e.

=

c.

=

m+1

5

 1 −  3 k =0 

∑

h.

n2 n +1

=

5

∑

k.

4

1

∑ 10 a

=

a =0

=

f.

m= 0

a =1

j.

2

n=0

n =1

g.

n −1

∑

b.

5k −1

5

∑

1 + ( −1)

k =0

k −1

=

k

2k

=

5

∑( j − 3 j2 ) =

i.

j =1

=

4

∑ (− 0.1) 2i−5

l.

=

i =1

k =1

4. Rewrite the following terms using the sigma notation. a.

1 1 1 1 1 1 + + + + + = 2 3 4 5 6 7 1 2

1 3

1 4

1 5

d. 1 + + + + +

1 = 6

Hamilton Education Guides

b.

1 2 3 4 5 6 + + + + + 2 3 4 5 6 7

1 2

2 3

3 4

4 5

e. 0 + + + + +

5 6

=

c. 2 + 4 + 8 + 16 + 32 + 64 =

=

f. 1 − + − + −

1 2

1 3

1 4

1 5

1 = 6

12

Calculus I

1.3

1.3 Arithmetic Sequences and Arithmetic Series

Arithmetic Sequences and Arithmetic Series

An arithmetic sequence is a sequence in which each term, after the first term, is obtained by adding a common number to the preceding term. The common number added to each term can be found by taking the common difference, denoted by d , of two successive terms. For example, the sequences 3, 6, 9, 12, 15,  and

5 7 3 1 , 1, , 2, , 3, ,  2 2 2 2

are arithmetic sequences because

the common difference that is added to each term in order to obtain the next term is 6 − 3 = 3 and 1 2

1 , respectively. Note that the n th term in both 2 1 and sn = n . Therefore, the two arithmetic sequences 2 1 7 5 3 1 , 1, , 2, , 3, ,  , n . 2 2 2 2 2 1−

=

examples can easily be stated as sn = 3n can be written as 3, 6, 9, 12, 15,  , 3n and

To obtain the n th term of an arithmetic sequence, a general form can be developed by letting sn and d be the n th term and the common difference of an arithmetic sequence. Thus, the first terms can be written as: s1 = a s2

= s1 + d

s3

= s2 + d = ( s1 + d ) + d = s1 + 2d

s4

= s3 + d = ( s1 + 2d ) + d = s1 + 3d

s5

= s4 + d = ( s1 + 3d ) + d = s1 + 4d

 sn

= sn−1 + d = [s1 + (n − 2) d ] + d = s1 + nd − 2d + d = s1 + nd − d = s1 + (n − 1) d

sn+1

where a and d are real numbers and n is a positive integer

= sn + d = [s1 + (n − 1) d ] + d = s1 + nd − d + d = s1 + nd

Thus, the n th and n + 1 term of an arithmetic sequence is equal to sn

= s1 + (n − 1) d

s n+1

= s1 + nd

( 1) (2)

In the following examples the above equations ( 1) and (2) are used in order to find several terms of arithmetic sequences. Example 1.3-1 Find the next five terms of the following arithmetic sequences. a. s1 = 5 , d = 3

b. s1 = −5 , d = 2

c. s1 = 20 , d = 0.4

Solutions: a. The n th term for an arithmetic sequence is equal to sn = s1 + (n − 1) d . Substituting s1 = 5 and d = 3 into the general arithmetic expression for n = 2, 3, 4, 5, and 6 we obtain s2 = s1 + (2 − 1) d = s1 + d = 5 + 3 = 8 s3 = s1 + (3 − 1) d = s1 + 2d = 5 + ( 2 × 3) = 5 + 6 = 11

Hamilton Education Guides

13

Calculus I

1.3 Arithmetic Sequences and Arithmetic Series

s4 = s1 + (4 − 1) d = s1 + 3d = 5 + ( 3 × 3) = 5 + 9 = 14 s5 = s1 + (5 − 1) d = s1 + 4d = 5 + ( 4 × 3) = 5 + 12 = 17 s6 = s1 + (6 − 1) d = s1 + 5d = 5 + ( 5 × 3) = 5 + 15 = 20

Thus, the first six terms of the arithmetic sequence are ( 5, 8, 11, 14, 17, 20 ) . b. Substituting s1 = −5 and d = 2 into the general arithmetic expression for n = 2, 3, 4, 5, and 6 we obtain s2 = s1 + (2 − 1) d = s1 + d = −5 + 2 = −3 s3 = s1 + (3 − 1) d = s1 + 2d = −5 + ( 2 × 2 ) = −5 + 4 = −1

s4 = s1 + (4 − 1) d = s1 + 3d = −5 + ( 3 × 2 ) = −5 + 6 = 1 s5 = s1 + (5 − 1) d = s1 + 4d = −5 + ( 4 × 2 ) = −5 + 8 = 3 s6 = s1 + (6 − 1) d = s1 + 5d = −5 + ( 5 × 2 ) = −5 + 10 = 5

Thus, the first six terms of the arithmetic sequence are ( −5, − 3, − 1, 1, 3, 5 ) . c. Substituting s1 = 20 and d = 0.4 into the general arithmetic expression for n = 2, 3, 4, 5, and 6 we obtain s2 = s1 + (2 − 1) d = s1 + d = 20 + 0.4 = 20.4 s3 = s1 + (3 − 1) d = s1 + 2d = 20 + ( 2 × 0.4 ) = 20 + 0.8 = 20.8

s4 = s1 + (4 − 1) d = s1 + 3d = 20 + ( 3 × 0.4 ) = 20 + 1.2 = 21.2 s5 = s1 + (5 − 1) d = s1 + 4d = 20 + ( 4 × 0.4 ) = 20 + 1.6 = 21.6 s6 = s1 + (6 − 1) d = s1 + 5d = 20 + ( 5 × 0.4 ) = 20 + 2 = 22

Thus, the first six terms of the arithmetic sequence are ( 20, 20.4, 20.8, 21.2, 21.6, 22 ) . Example 1.3-2 sequences.

Find the general term and the fiftieth term of the following arithmetic

a. s1 = 3 , d = 5

b. s1 = −2 , d = 4

c. s1 = 10 , d = −2.5

Solutions: a. The n th term for an arithmetic sequence is equal to sn = s1 + (n − 1) d . Substituting s1 = 3 and d = 5 into the general arithmetic expression we obtain

sn = 3 + (n − 1) 5 = 3 + 5n − 5 = 5n + ( 3 − 5 ) = 5n − 2 Hamilton Education Guides

14

Calculus I

1.3 Arithmetic Sequences and Arithmetic Series

substituting n = 50 into the general equation sn = 5n − 2 we find s50 = ( 5 × 50 ) − 2 = 250 − 2 = 248

b. Substituting s1 = −2 and d = 4 into the general arithmetic expression sn = s1 + (n − 1) d we obtain sn = − 2 + (n − 1) 4 = −2 + 4n − 4 = 4n + ( −2 − 4 ) = 4n − 6

substituting n = 50 into the general equation sn = 4n − 6 we find s50 = ( 4 × 50 ) − 6 = 200 − 6 = 194

c. Substituting s1 = 10 and d = −2.5 into the general arithmetic expression sn = s1 + (n − 1) d we obtain sn = 10 + ( n − 1) × −2.5 = 10 − 2.5n + 2.5 = −2.5n + (10 + 2.5) = −2.5n + 12.5 substituting n = 50 into the general equation sn = −2.5n + 12.5 we find s50 = ( −2.5 × 50 ) + 12.5 = −125 + 12.5 = −112.5

Example 1.3-3 Find the next four terms in each of the following arithmetic sequences. a. 6, 10, 

b. x, x + 2, 

c. 2 x + 1, 2 x + 5, 

d. x, x − 29, 

Solutions: a. The first term s1 and the common difference d are equal to s1 = 6 and d = 10 − 6 = 4 . Thus, using the general arithmetic equation sn = s1 + (n − 1) d or sn+1 = sn + d the next four terms are as follows: Let’s use sn+1 = sn + d . Then, s3 = s2 + d = 10 + 4 = 14

s4 = s3 + d = 14 + 4 = 18

s5 = s4 + d = 18 + 4 = 22

s6 = s5 + d = 22 + 4 = 26

b. The first term s1 and the common difference d are equal to s1 = x and d = x + 2 − x = 2 . Thus, s3 = s2 + d = ( x + 2 ) + 2 = x + 4

s4 = s3 + d = ( x + 4 ) + 2 = x + 6

s5 = s4 + d = ( x + 6 ) + 2 = x + 8

s6 = s5 + d = ( x + 8 ) + 2 = x + 10

c. The first term s1 and the common difference d are equal to s1 = 2 x + 1 and d = (2 x + 5) − (2 x + 1) . = 2 x + 5 − 2 x − 1 = 4 . Thus, s3 = s2 + d = ( 2 x + 1) + 4 = 2 x + 5

s4 = s3 + d = (2 x + 5) + 4 = 2 x + 9

s5 = s4 + d = ( 2 x + 9 ) + 4 = 2 x + 13

s6 = s5 + d = ( 2 x + 13) + 4 = 2 x + 17

d. The first term s1 and the common difference d are equal to s1 = x and d = ( x − 29) − x = −29 . Thus, s3 = s2 + d = ( x − 29 ) − 29 = x − 58 Hamilton Education Guides

s4 = s3 + d = ( x − 58 ) − 29 = x − 87 15

Calculus I

s5

1.3 Arithmetic Sequences and Arithmetic Series

= s4 + d = ( x − 87) − 29 = x − 116

= s5 + d = ( x − 116) − 29 = x − 145

s6

Example 1.3-4 The first term of an arithmetic sequence is −5 and the fourth term is 10 . Find the twentieth term. Solution: Since s1 = −5 and s4 = 10 we use the general formula sn = s1 + ( n − 1)d in order to solve for d . s4 = s1 + (4 − 1)d ; 10 = −5 + (4 − 1)d ; 10 = −5 + 3d ; 10 + 5 = 3d ; 15 = 3d ; d =

15 3

; d = 5 . Then,

s20 = s1 + (20 − 1)d ; s20 = −5 + 19d ; s20 = −5 + (19 × 5) ; s20 = −5 + 95 ; s20 = 90

Having learned about arithmetic sequences and the steps for finding the terms of an arithmetic sequence, we will next learn about arithmetic series and the steps for finding the sum of arithmetic series over a given range. Addition of the terms in an arithmetic sequence result in having an arithmetic series. To obtain the arithmetic series formula let sk = s1 + ( k − 1)d be an arithmetic sequence and denote the sum of the first n terms by =

Sn

n

∑ s1 + (k − 1)d k =1

then, S n = s1 + ( s1 + d ) +  + [ s1 + ( n − 2)d ] + [ s1 + ( n − 1)d ]

(a )

Let’s write the sum in reverse order and add the two series (a ) and (b ) together. Sn

= [ s1 + ( n − 1)d ] + [ s1 + ( n − 2)d ] +  + ( s1 + d ) + s1

Sn + Sn

(b )

= { s1 + [ s1 + ( n − 1)d ]} + {( s1 + d ) + [ s1 + ( n − 2)d ]} +  + {[ s1 + ( n − 2)d ] + ( s1 + d )} + {[ s1 + ( n − 1)d ] + s1}

2S n

= [ s1 + s1 + ( n − 1)d ] + [ s1 + d + s1 + ( n − 2)d ] +  + [ s1 + ( n − 2)d + s1 + d ] + [ s1 + ( n − 1)d + s1 ]

2S n

= [2s1 + ( n − 1)d ] + [ s1 + s1 + ( n − 2)d + d ] +  + [ s1 + s1 + ( n − 2)d + d ] + [ s1 + s1 + ( n − 1)d ]

2S n

= [2s1 + ( n − 1)d ] + [2s1 + nd − 2d + d ] +  + [2s1 + nd − 2d + d ] + [2s1 + ( n − 1)d ]

2S n

= [2s1 + ( n − 1)d ] + [2s1 + nd − d ] +  + [2s1 + nd − d ] + [2s1 + ( n − 1)d ]

2S n

= [2s1 + ( n − 1)d ] + [2s1 + ( n − 1)d ] +  + [2s1 + ( n − 1)d ] + [2s1 + ( n − 1)d ]

2S n

= n [2s1 + ( n − 1)d ] ; S n =

n[2 s1 + (n − 1)d ] 2

=

n 2s1 + ( n − 1)d 2

[

]

Therefore, the arithmetic series can be written in the following two forms:

Hamilton Education Guides

Sn

=

Sn

=

n

∑ s1 + ( k − 1)d

(1)

k =1

n 2s1 + ( n − 1) d 2

[

]

(2) 16

Calculus I

1.3 Arithmetic Sequences and Arithmetic Series

Note that equation (2) , similar to the n th term of the arithmetic sequence [s n = s1 + (n − 1) d ] , is given in terms of s1 , n , and d . In the following examples the above equations (1) and (2) are used in order to find the sum of arithmetic series. Example 1.3-5 Find the sum of the following arithmetic series. a.

20

∑ (2i + 1) =

15

∑ (3i − 2) =

b.

i =1

i =1

c.

15

∑ (5 j − 1) = j =3

Solutions: a. First - Write the first three terms of the arithmetic series in expanded form, i.e., 20

∑ ( 2i + 1) = ( 2 + 1) + ( 4 + 1) + ( 6 + 1) +  = i =1

3+ 5+ 7 +

Second - Identify the first term, s1 , the difference between the two terms, d , and n , i.e., s1 = 3 , d = 5 − 3 = 2 , and n = 20 . Third - Use the arithmetic series formula to obtain the sum of the twenty terms. Sn S 20

=

20 [2s1 + (20 − 1) d ] 2

=

n [2s1 + (n − 1) d ] 2

= 10[ 2 s1 + 19d ] = 10 ( 2 × 3) + (19 × 2 ) = 10[6 + 38] = 10 × 44 = 440

Note that prior to learning the arithmetic series formula the only method that we could use was by summing each term as shown below: 20

∑ ( 2i + 1) = ( 2 + 1) + ( 4 + 1) + ( 6 + 1) + (8 + 1) + (10 + 1) + (12 + 1) + (14 + 1) + (16 + 1) + (18 + 1) + ( 20 + 1) i =1

+ ( 22 + 1) + ( 24 + 1) + ( 26 + 1) + ( 28 + 1) + ( 30 + 1) + ( 32 + 1) + ( 34 + 1) + ( 36 + 1) + ( 38 + 1) + ( 40 + 1)

= 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 = 440 As you note, it is much easier to use the arithmetic series formula as opposed to the summation of each term which is fairly long and time consuming. b. First - Write the first three terms of the arithmetic series in expanded form, i.e., 15

∑ ( 3i − 2 ) = ( 3 − 2 ) + ( 6 − 2 ) + ( 9 − 2 ) +  = i =1

1+ 4 + 7 + 

Second - Identify the first term, s1 , the difference between the two terms, d , and n , i.e., s1 = 1 , d = 4 − 1 = 3 , and n = 15 . Third - Use the arithmetic series formula to obtain the sum of the fifteen terms. Sn

Hamilton Education Guides

=

n [ 2s1 + (n − 1) d ] 2

17

Calculus I

S15

=

1.3 Arithmetic Sequences and Arithmetic Series

15 [2s1 + (15 − 1) d ] 2

= 7.5[ 2 s1 + 14d ] = 7.5  ( 2 × 1) + (14 × 3)  = 7.5[ 2 + 42] = 7.5 × 44 = 330

or, we can obtain the answer by summing up the first fifteen terms of the series, i.e., 15

∑ ( 3i − 2 ) =

1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 + 25 + 28 + 31 + 34 + 37 + 40 + 43

i =1

= 330

c. First - Write the first three terms of the arithmetic series in expanded form, i.e., 15

∑ ( 5 j − 1) = (15 − 1) + ( 20 − 1) + ( 25 − 1) +  = j =3

14 + 19 + 24 + 

Second - Identify the first term, s1 , the difference between the two terms, d , and n , i.e., s1 = 14 , d = 19 − 14 = 5 , and n = 13 . Third - Use the arithmetic series formula to obtain the sum of the thirteen terms. Sn S15

=

13 [2s1 + (13 − 1) d ] 2

=

n [ 2s1 + (n − 1) d ] 2

= 6.5[ 2 s1 + 12d ] = 6.5  ( 2 × 14 ) + (12 × 5)  = 6.5[ 28 + 60] = 6.5 × 88 = 572

or, we can obtain the answer by summing up the first thirteen terms of the series, i.e., 15

∑ ( 5 j − 1) =

14 + 19 + 24 + 29 + 34 + 39 + 44 + 49 + 54 + 59 + 64 + 69 + 74

j =3

Example 1.3-6 sequences. a. s1 = 5 , d = 2

= 572

Given the first term s1 and d , find S 80 for each of the following arithmetic b. s1 = −10 , d = 3

Solutions: a. The n th term for an arithmetic series is equal to S n =

c. s1 = 500 , d = 25 n [2s1 + (n − 1) d ] . 2

Substituting s1 = 5

and d = 2 into the general arithmetic expression we obtain S80

=

80 [2s1 + (80 − 1) d ] 2

= 40[ 2 s1 + 79d ] = 40  ( 2 × 5) + ( 79 × 2 )  = 40[10 + 158] = 40 ×168 = 6720

b. Substituting s1 = −10 and d = 3 into S n = S80

=

80 [2s1 + (80 − 1) d ] 2

=

80 [2s1 + (80 − 1) d ] 2

Hamilton Education Guides

we obtain

= 40[ 2 s1 + 79d ] = 40  ( 2 × −10 ) + ( 79 × 3)  = 40[ −20 + 237] = 8680

c. Substituting s1 = 500 and d = 25 into sn = S80

n [2s1 + (n − 1) d ] 2

n [2s1 + (n − 1) d ] 2

we obtain

= 40[ 2 s1 + 79d ] = 40  ( 2 × 500 ) + ( 79 × 25)  = 40[1000 + 1975] = 119000

18

Calculus I

1.3 Arithmetic Sequences and Arithmetic Series

Example 1.3-7 Find the sum of the following sequences for the indicated values. a. S35 for the sequence −5, 3, 

b. S 200 for the sequence −10, 10, 

Solutions: a. The first term s1 and the common difference d are equal to s1 = −5 and d = 3 − (− 5) = 3 + 5 = 8 . Thus, using the general arithmetic series S n = S35

=

35 [2s1 + (35 − 1) d ] 2

n [2s1 + (n − 1) d ] 2

we obtain

= 17.5[ 2 s1 + 34d ] = 17.5  ( 2 × −5) + ( 34 × 8)  = 17.5[ −10 + 272] = 4585

b. The first term s1 and the common difference d are equal to s1 = −10 and d = 10 − (− 10) = 20 . Thus, using the general arithmetic series S n = S 200

200 [2s1 + (200 − 1) d ] 2

=

n [2s1 + (n − 1) d ] 2

we obtain

= 100[ 2 s1 + 199d ] = 100  ( 2 × −10 ) + (199 × 20 )  = 100[ −20 + 3980]

= 396000 Section 1.3 Practice Problems - Arithmetic Sequences and Arithmetic Series 1. Find the next seven terms of the following arithmetic sequences. a. s1 = 3 , d = 2

b. s1 = −3 , d = 2

c. s1 = 10 , d = 0.8

2. find the general term and the eighth term of the following arithmetic sequences. a. s1 = 3 , d = 4

b. s1 = −3 , d = 5

c. s1 = 8 , d = −12 .

3. find the next six terms in each of the following arithmetic sequences. a. 5, 8, 

b. x, x + 4, 

c. 3x + 1, 3x + 4, 

d. w, w − 10, 

4. Find the sum of the following arithmetic series. a.

20

∑ (2i − 4) =

b.

i =10

d.

15

∑ 3i = 10

∑

(3i + 4)

∑k =

c.

e.

10

∑ (i + 1) =

f.

i =1

=

i=4

h.

100

∑ (2k − 3) = k =1

k=1

i=1

g.

1000

15

∑ (2k − 1) =

k =5

13

∑ (3 j + 1) =

i.

18

∑ (4k − 3) =

k =7

j =5

5. The first term of an arithmetic sequence is 6 and the third term is 24 . Find the tenth term. 6. Given the first term s1 and d , find S 50 for each of the following arithmetic sequences. a. s1 = 2 , d = 5

b. s1 = −5 , d = 6

c. s1 = 30 , d = 10

7. Find the sum of the following sequences for the indicated values. a. S15 for the sequence −8, 6,  Hamilton Education Guides

b. S100 for the sequence −20, 20,  19

Calculus I

1.4

1.4 Geometric Sequences and Geometric Series

Geometric Sequences and Geometric Series

A geometric sequence is a sequence in which each term, after the first term, is obtained by multiplying the preceding term by a common multiplier. This common multiplier is also called the common ratio and is denoted by r . The common ratio r is obtained by division of two successive terms in a sequence. For example, the sequences 3, 6, 12, 24, 48,  and 1 1 1 1 1 , , , , ,  are geometric sequences because the common ratio that each term is multiplied 2 4 8 16 32 6 by in order to obtain the next term is equal to = 2 and 3

1 4 1 2

=

1× 2 4 ×1

=

2 1 = , respectively. 4 2

To obtain the n th term of a geometric sequence, a general form can be developed by letting sn and r be the n th term and the common multiplier (common ratio) of a geometric sequence. Thus, the first terms can be written as: s1

=a

s2

= s1r

s3

= s2 r = ( s1r ) ⋅ r = s1r 2

s4

= s3r = s1r 2 ⋅ r = s1r 3

s5

( ) = s r = (s r ) ⋅ r = s r 1

4



sn

where a is a real number and n is a positive integer

3

1

4

( ) = s r = (s r ) ⋅ r = s r

= sn−1r = s1r n−2 ⋅ r = s1r n−2+1 = s1r n−1

sn+1

n

1

n −1

1

n −1+1

= s1r n

Thus, the n th and n + 1 term of an arithmetic sequence is equal to sn

= s1 r n−1

( 1)

= s1 r n

(2)

s n+1

In the following examples the above equations ( 1) and (2) are used in order to find several terms of geometric sequences. Example 1.4-1 Find the next five terms of the following geometric sequences. a. s1 = 5 , r = 2 b. s1 = −3 , r = 3 c. s1 = 10 , r = 0.5 Solutions: a. The n th term for an geometric sequence is equal to sn = s1r n−1 . Substituting s1 = 5 and r = 2 into the general geometric expression for n = 2, 3, 4, 5, and 6 we obtain s2

= s1r 2−1 = s1r = 5 × 2 = 10

s3

= s1r 3−1 = s1r 2 = 5 × 22 = 5 × 4 = 20

s4

= s1r 4−1 = s1r 3 = 5 × 23 = 5 × 8 = 40

Hamilton Education Guides

20

Calculus I

1.4 Geometric Sequences and Geometric Series

s5

= s1r 5−1 = s1r 4 = 5 × 24 = 5 ×16 = 80

s6

= s1r 6−1 = s1r 5 = 5 × 25 = 5 × 32 = 160

Thus, the first six terms of the geometric sequence are ( 5, 10, 20, 40, 80, 160) . b. Substituting s1 = −3 and r = 3 into the general geometric expression for n = 2, 3, 4, 5, and 6 we obtain s2 = s1r 2−1 = s1r = −3 × 3 = −9 s3

= s1r 3−1 = s1r 2 = −3 × 32 = −3 × 9 = −27

s4

= s1r 4−1 = s1r 3 = −3 × 33 = −3 × 27 = −81

s5

= s1r 5−1 = s1r 4 = −3 × 34 = −3 × 81 = −243

s6

= s1r 6−1 = s1r 5 = −3 × 35 = −3 × 243 = −729

Thus, the first six terms of the geometric sequence are ( −3, − 9, − 27, − 81, − 243, − 729) . c. Substituting s1 = 10 and r = 0.5 into the general geometric expression for n = 2, 3, 4, 5, and 6 we obtain s2 = s1r 2−1 = s1r = 10 × 0.5 = 5 s3

= s1r 3−1 = s1r 2 = 10 × 0.52 = 10 × 0.25 = 2.5

s4

= s1r 4−1 = s1r 3 = 10 × 0.53 = 10 × 0.125 = 1.25

s5

= s1r 5−1 = s1r 4 = 10 × 0.54 = 10 × 0.0625 = 0.625

s6

= s1r 6−1 = s1r 5 = 10 × 0.55 = 10 × 0.03125 = 0.3125

Thus, the first six terms of the geometric sequence are (10, 5, 2.5, 1.25, 0.625, 0.3125) . Example 1.4-2 find the general term and the tenth term of the following geometric sequences. a. s1 = 3 , r = 12 .

b. s1 = −2 , r = 0.8

c. s1 = 10 , r = −0.5

Solutions: a. The n th term for a geometric sequence is equal to sn = s1r n−1 . Substituting s1 = 3 and r = 12 . into the general geometric expression we obtain sn

= 3 × r n −1 = 3×1.2n-1

substituting n = 10 into the general equation sn = 3 × 12 . n −1 we have s10

= 3 ×1.210−1 = 3 ×1.29 = 3 × 5.1598 = 15.479

b. Substituting s1 = −2 and r = 0.8 into the general geometric expression sn = s1r n−1 we obtain

Hamilton Education Guides

21

Calculus I

1.4 Geometric Sequences and Geometric Series

= −2 × r n −1 = −2× 0.8n-1

sn

substituting n = 10 into the general equation sn = −2 × 0.8 n−1 we have = −2 × 0.810−1 = −2 × 0.89 = −2 × 0.1342 = − 0.2684

s10

c. Substituting s1 = 10 and r = −0.5 into the general geometric expression sn = s1r n−1 we obtain n-1 = 10 × r n −1 = 10× ( −0.5 )

sn

substituting n = 10 into the general equation sn = 10 × ( −0.5) n −1 we have = 10 × ( −0.5)10−1 = 10 × ( −0.5)9 = 10 × ( −0.0019 ) = − 0.019

s10

Example 1.4-3 sequences.

Find the next four terms and the n th term in each of the following geometric

1 2

1 1 3 9

a. 1, , 

b. − , , 

c.

Solutions: a. The first term s1 and the common ratio r are equal to s1 = 1 and r =

1 2

1 1 x, − x,  2 4

1

1× 1 1 = 2 = = . 1 1 2 ×1 2

using the general geometric equation sn = s1r n−1 the next four terms are:

Thus,

1

s3

2 1 1 1 = s1r 2 = 1 ⋅   = 2 =

s4

3 1 1 1 = s1r 3 = 1 ⋅   = 3 =

s5

4 1 1 1 = s1r 4 = 1 ⋅   = 4 =

s6

5 1 1 1 = s1r 5 = 1 ⋅   = 5 =

2

4

2

2

16

2

2

2

8

2

32

2

1 1 1 1 1 Thus, the first six terms of the geometric sequence are  1, , , , ,  and the n th term is 

1 equal to sn = 1 ⋅  

n−1

2

=

n −1

1 2n −1

=

2 4 8 16 32 

1

2

n-1

b. The first term s1 and the common ratio r are equal to s1 = −

1

1 1 1× 3 and r = 91 = − = − . Thus, 3 3 9 ×1 −3

using the general geometric equation sn = s1r n−1 the next four terms are: s3

2 1 1 1 1 = s1r 2 = − ⋅  −  = − 3 = −

27

s4

3 1 1 1 1 = s1r 3 = − ⋅  −  = 4 =

s5

4 1 1 1 1 = s1r 4 = − ⋅  −  = − 5 = −

s6

5 1 1 1 1 = s1r 5 = − ⋅  −  = 6 =

3  3

3  3

3

3

243

3  3

3  3

3

3

81

729

1 1 1 1 1 1  Thus, the first six terms of the geometric sequence are  − , , − , , − ,  and the   3 9

Hamilton Education Guides

27 81

243 729

22

Calculus I

n th

1.4 Geometric Sequences and Geometric Series

1 1 term is equal to sn = − ⋅  − 

n−1

3  3

= −

(− 1) n−1 3 ⋅ 3 n −1

= −

(− 1) n−1 3n

1 2

1 2

c. The first term s1 and the common ratio r are equal to s1 = x and r = − . Thus, using the general geometric equation sn = s1r n−1 the next four terms are: s3

2 1 1 1 1 = s1r 2 = x ⋅  −  = 3 x = x

s4

3 1 1 1 1 = s1r 3 = x ⋅  −  = − 4 x = − x

s5

4 1 1 1 1 x = s1r 4 = x ⋅  −  = 5 x =

s6

5 1 1 1 1 = s1r 5 = x ⋅  −  = − 6 x = − x

2

 2

2

8

2

 2

32

2

2

2

 2

16

2

 2

64

2

1 1 1 1 1 1 Thus, the first six terms of the geometric sequence are  x, − x, x, − x, x, − x 2

4

8

16

32

64 

n −1 x (− 1) n −1 x (− 1) n −1 x (− 1) n −1 1 1 and the n th term is equal to sn = x ⋅  −  = = = n n −1 n −1+1

2

 2

2⋅2

2

2

Example 1.4-4 Given the following terms of a geometric sequence, find the common ratio r . a. s1 = 32 and s7 = Solutions:

1 2

b. s1 = 3 and s5 =

1 27

c. s1 = 5 and s8 = 1

1 into sn = s1r n−1 and solve for r . 2 1 1 1 1× 1 1 1 1 6 6 6 6 2 2 = r ; r = 5 ; r = 25 ; r 6 = ; = 32r ; ; r6 = 6 ; r = 5 2 32 2 2 2 2 2× 2

a. Substitute s1 = 32 and s7 = s 7 = s1 r 7 −1

1

1 b. Substitute s1 = 3 and s5 = into sn = s1r n−1 and solve for r . 27 s 5 = s1 r 5−1

;

1 = 3r 4 27

1 1 1 3 3 3 1× 1 3 1 1 ; 27 = r 4 ; r 4 = ; r4 = 3 ; r4 = 3 ; r4 = 4 ; r = 3 3 3 3 3 ×3 1

c. Substitute s1 = 5 and s8 = 1 into sn = s1r n−1 and solve for r . 1 = r7 5

; 1 = 5r 7 ;

s8 = s1 r 8−1

; r7 =

1 5

1

; r=7

5

Example 1.4-5 Write the first six terms and the n th term of the following geometric sequences. 1 a. sn =  

n −1

1 b. sn =    2

 3

n+2

1 c. sn =  − 

2n

1 d. sn =  − 

n +1

 2

 3

Solutions: 1−1

1 a. s1 =   3

=

1   3

0

= 1

Hamilton Education Guides

1 s2 =   3

2 −1

=

1

1   3

=

1 3

23

Calculus I

1.4 Geometric Sequences and Geometric Series 3−1

1 s3 =   3

5−1

1 s5 =   3

=

1   3

2

=

1   3

4

=

1 1 = 9 32

1 s4 =   3

=

1 1 = 81 34

1 s6 =   3

4 −1

6 −1

3

=

1   3

=

1   3

5

=

1 1 = 27 33

=

1 1 = 243 35

1  1 1 1 1 Thus, the first six terms of the geometric sequence are  1, , , , , .   3 9 27 81 243

1+ 2

1 b. s1 =   2

3+ 2

1 s3 =   2

5+ 2

1 s5 =   2

3

=

1   2

=

1   2

=

1   2

5

7

=

1 1 3 = 8 2

1 s2 =   2

2+ 2

=

1 1 5 = 32 2

1 s4 =   2

4+ 2

=

1 27

=

1 128

1 s6 =   2

6+ 2

=

1   2

4

=

1   2

6

=

1   2

8

=

1 1 4 = 16 2

=

1 1 6 = 64 2

=

1 28

1 256

=

1 1 1 1 1 1  Thus, the first six terms of the geometric sequence are  , , , , , .   8 16 32 64 128 256

c. s1 =

 1 −   3

2

 1 s3 =  −   3

6

10

 1 s5 =  −   3

=

1 1 2 = 9.0×100 3

=

1 36

=

=

1

3

4

8

1 7.29×102

=

10

 1 s2 =  −   3  1 s4 =  −   3

12

1 5.9×104

 1 s6 =  −   3

=

1 1 4 = 8.1×101 3

=

1 38

=

=

1 6.56×103

1 1 12 = 5.31×105 3

  1 1 1 1 1 1 , , , , ,   9.0 × 10 0 8.1 × 101 7.29 × 10 2 6.56 × 10 3 5.9 × 10 4 5.31 × 10 5 

Thus, the six terms are  d. s1 =

 1 −   2

2

 1 s3 =  −   2

4

 1 s5 =  −   2

6

3

=

1 1 2 = 4 2

 1 s2 =  −   2

=

1 1 4 = 16 2

 1 s4 =  −   2

=

1 26

1 64

 1 s6 =  −   2

=

5

7

.

1 1 = − 3 = −

2

8

1 1 = − 5 = −

2

32

1 1 = − 7 = −

2

128

1 1 1 1 1 1  Thus, the first six terms of the geometric sequence are  , − , , − , , − .   4

8 16

32 64

128

Having learned about geometric sequences and the steps for finding the terms of a geometric sequence, we will next learn about geometric series and the steps for finding the sum of geometric series over a given range.

Hamilton Education Guides

24

Calculus I

1.4 Geometric Sequences and Geometric Series

Similar to arithmetic series, addition of the terms in a geometric sequence result in having a geometric series. To obtain the geometric series formula let sk = s1r k −1 be a geometric sequence and denote the sum of the first n terms by Sn

n

∑ s1r k −1

=

k =1

then, S n = s1 + s1r + s1r 2 +  + s1r n −2 + s1r n −1

(a )

Let’s multiply both sides of the equation (a ) by r and subtract (b ) from (a ) . Sn ⋅ r rS n

= s1 ⋅ r + s1r ⋅ r + s1r 2 ⋅ r +  + s1r n−2 ⋅ r + s1r n−1 ⋅ r

(b )

= s1r + s1r 2 + s1r 3 +  + s1r n−1 + s1r n

( S (1 − r ) = ( s + s r + s r

) ( + s r +  + s r ) + (− s r − s r − s r S (1 − r ) = s + ( s r − s r ) + ( s r − s r ) + ( s r − s r ) +  + ( s r s (1 − r ) s −s r ; S = S (1 − r ) = s − s r ; S = S n − rS n

)

= s1 + s1r + s1r 2 + s1r 3 +  + s1r n −1 − s1r + s1r 2 + s1r 3 +  + s1r n −1 + s1r n 1

n

1

n

1

n

1

1

1

1

1

n

2

1

3

2

1

n

1

1

2

1

1

n −1

1

1

3

1

n

3

2

1

1

3

−  − s1r n −1 − s1r n

)

)

− s1r n −1 − s1r n

n −1

n

1

n

1− r

1

1− r

r ≠1

Therefore, the geometric series can be written in the following two forms: Sn

Sn

n

∑ s1 r k −1

=

k =1

(

s1 1 − r n

=

( 1)

)

(2)

r ≠1

1− r

)

(

Note that equation (2) , similar to the n th term of a geometric sequence sn = s1r n−1 , is given in terms of s1 , n , and r . A third alternative way of expressing the geometric series is by substituting s1r n with its

(

)

equivalent value s1r n = s1r n−1 ⋅ r = r s1r n−1 = rsn which result in having Sn

=

s1 − s1r n 1− r

=

(

s1 − r s1 r n −1 1− r

) = s − r (s r ) = s 1

1

1− r

n −1

− rs n 1− r

(3)

1

where the geometric series is given in terms of s1 , sn (the geometric sequence), and r . In the following examples we will use the above equations ( 1) , (2) , and (3) in order to find the sum of geometric series. Example 1.4-6 Evaluate the sum of the following geometric series. a.

10

∑

3k − 2

=

k =1

Hamilton Education Guides

b.

10

∑ (−3) k −2 = k =1

c.

6

 1 8 −  2 k =2 

∑

k +1

=

25

Calculus I

1.4 Geometric Sequences and Geometric Series

Solutions: a. First - Write the first few terms of the geometric series in expanded form, i.e., 10

∑ 3k − 2

= 31−2 + 32−2 + 33−2 + 34−2 +  = 3−1 + 30 + 31 + 32 +  = 3−1 + 1 + 3 + 6 + 

k =1

Second - Identify the first term, s1 , the common ratio between the two terms, r , and n , i.e., s1 = 3−1 , r =

1

=

3 −1

1

1× 3 = 3, = 1 = 1 1 1× 1

1 3

and n = 10 .

3

Third - Use the geometric series formula to obtain the sum of the ten terms. Sn

S10

(

1 1 − 310 3 = 1− 3

)

=

(

s1 1 − r n

)

1− r

1 59048 (1 − 59049) − − 59048 59048 × 1 59048 3 3 3 = = = = = = 9841.333 −2 −2 6 3× 2 − 12

Note that prior to learning the geometric series formula the only method that we could use was by summing each term as shown below: 10

∑ 3k − 2

= 3−1 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 = 3−1 + 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187

k =1

+6561 = 3−1 + 9841 = 0.333 + 9841 = 9841.333

As you note, it is much easier to use the geometric series formula as opposed to the summation of each term which is somewhat long and time consuming. b. First - Write the first few terms of the geometric series in expanded form, i.e., 10

∑ (−3) k −2

= ( −3)1−2 + ( −3) 2−2 + ( −3)3−2 + ( −3) 4−2 +  = ( −3) −1 + ( −3) 0 + ( −3)1 + ( −3) 2 + 

k =1

= −3−1 + 1 − 3 + 9 +  Second - Identify the first term, s1 , the common ratio between the two terms, r , and n , i.e., s1 = −3−1 , r =

1 − 3 −1

=

1

1

1× 3 =− 1 =− = −3 , 1 1 1× 1 −3 3

and n = 10 .

Third - Use the arithmetic series formula to obtain the sum of the ten terms. Sn =

S10

=

−

[

1 10 1 − ( −3) 3 1 − ( −3)

] = − 13 (1 − 59049) = 59048 3 = 4

4

(

s1 1 − r n

)

1− r 59048 3 4

=

59048 × 1 3× 4

=

59048 12

= 4920.666

1

or, we can obtain the answer by summing up the first ten terms of the series, i.e., Hamilton Education Guides

26

Calculus I

1.4 Geometric Sequences and Geometric Series

10

∑ (−3) k −2

= ( −3) −1 + ( −3) 0 + ( −3)1 + ( −3) 2 + ( −3)3 + ( −3) 4 + ( −3)5 + ( −3) 6 + ( −3) 7 + ( −3)8

k =1

= −3−1 + 1 − 3 + 9 − 27 + 81 − 243 + 729 − 2187 + 6561 = −3−1 + 4921 = −0.333 + 4921 = 4920.666 c. First - Write the first few terms of the geometric series in expanded form, i.e., 6

 1 8 −  2 k =2 

∑

k +1

 1 = 8 − 

2 +1

 2

6

 1 + 8 −  +   2

1 8

 1 + 8 −   2

= 8× − + 8×

3+1

 1 + 8 −   2

4 +1

 1 + 8 −   2

5+1

3

 2

 2

8 8 8 8 − + + 8 16 32 64

1 1 1 − 8× + 8× +  16 32 64

4

 1  1  1 = 8 −  + 8 −  + 8 − 

+

 2

1 2

= − +

5

1 4

1 8

= −1 + − + + 

Second - Identify the first term, s1 , the common ratio between the two terms, r , and n , i.e., s1 = −1 , r =

1 2

1

1 1× 1 =− , =− 2 =− 1 2 2 ×1 −1

and n = 5 .

1

Third - Use the geometric series formula to obtain the sum of the five terms. =

Sn

S5

=

( )5  ( )

− 1 ⋅ 1 − − 12  1 − − 12

=

− 1 + 15   2  1 + 12

=

(

1 − 1+ 32 3 2

)

=

(

s1 1 − r n

)

1− r

− (1+ 0.03125) 1.5

= −

1.03125 1.5

= −0.6875

or, we can obtain the answer by summing up the first five terms of the series, i.e., 6

 1 8 −  2 k =2 

∑

k +1

=

 1  3  1  4  1  5  1  6  1  7  8 ⋅  −  +  −  +  −  +  −  +  −    2   2   2  2  2  

1

1

1

1

1

  = 8 ⋅ − + − + −  8 16 32 64 128  

= 8 ⋅ (− 0.125 + 0.0625 − 0.03125 + 0.01563 − 0.00781) = 8 × −0.08593 = −0.6875 Example 1.4-7 Given the first term s1 and r , find S10 for each of the following geometric sequences. a. s1 = 5 , r = 2 b. s1 = −10 , r = 3 c. s1 = 50 , r = −2 Solutions: th

a. The n term for a geometric series is equal to S n = into the general geometric expression we obtain S10

=

(

5 1 − 210 1− 2

)

=

5 (1 − 1024) −1

1− r

) . Substituting s

1

=5

and r = 2

= 5115

b. Substituting s1 = −10 and r = 3 into S n =

Hamilton Education Guides

(

s1 1 − r n

(

s1 1 − r n 1− r

) we obtain 27

Calculus I

S10

1.4 Geometric Sequences and Geometric Series

(

−10 1 − 310

=

)

=

1− 3

−10 (1 − 59049) −2

= 5 × −59048 = −295240

c. Substituting s1 = 50 and r = −2 into S n = S10

[

10

50 1 − ( −2)

=

1 − ( −2)

(

s1 1 − r n 1− r

) we obtain

] = 50[1 − 1024] = 50 × −1023 = −17050 1+ 2

3

Example 1.4-8 Find the x and y values to the following problems. a. x if

6

∑

ix

= 20

b. x and y if

∑ (ix + 2 y) =

30 and

5

∑ (ix + 5) = 15 i =1

d. x and y if

6

∑ ( x + iy) =

50

5

∑ (ix + 2 y) = 10 i=2

i =1

i=1

c. x if

5

8

∑ ( x + iy) = 24

and

i=4

i =3

Solutions: a. Expanding

6

∑ ix

= 20 we obtain x + 2 x + 3x + 4 x + 5x + 6 x = 20 ; 21x = 20 ; x =

i=1

b. Expanding

20 ; x = 0.952 21

5

∑ (ix + 2 y) = 30 we obtain ( x + 2 y) + (2x + 2 y) + (3x + 2 y) + (4x + 2 y) + (5x + 2 y) = 30 i =1

; (x + 2 x + 3x + 4 x + 5 x ) + (2 y + 2 y + 2 y + 2 y + 2 y ) = 30 ; 15x + 10 y = 30 Expanding

5

∑ (ix + 2 y) = 10 we obtain (2x + 2 y) + (3x + 2 y) + (4 x + 2 y) + (5x + 2 y) = 10

; 14 x + 8 y = 10

i=2

The two linear equations with two unknowns x and y are solved using the substitution method to obtain x = −7 and y = 13.5 c. Expanding

5

∑ (ix + 5) = 15 we obtain ( x + 5) + (2x + 5) + (3x + 5) + (4x + 5) + (5x + 5)

; 15x + 25 = 15

i =1

10 ; x = − ; x = −0.667 15

d. Expanding

6

∑ ( x + iy) = 50 we obtain ( x + 3 y) + ( x + 4 y) + ( x + 5 y) + ( x + 6 y) = 50

; 4 x + 18 y = 50

i =3

Expanding

8

∑ ( x + iy) = 24 we obtain ( x + 4 y) + ( x + 5 y) + ( x + 6 y) + ( x + 7 y) + ( x + 8 y) = 24

; 5x + 30 y = 24

i=4

The two linear equations with two unknowns x and y are solved using the substitution method to obtain x = 35.6 and y = −5.133

Hamilton Education Guides

28

Calculus I

1.4 Geometric Sequences and Geometric Series

Example 1.4-9 Find the value of x for the following geometric sequences. b. 2 −1 , 2 −1 x, 2 −3

a. 2, 4 x, 16 .

c. 5, 5x, 125

Solutions: a. Since the common ratio r of a geometric sequence is defined as the ratio of the ( n + 1) st term to the n th term, we can use this principal to solve for x , i.e., r

4 x 16 = 2 4x

=

therefore 4 x × 4 x = 16 × 2 ; 16 x 2 = 32 ; x 2 =

32 ; x2 = 2 ; x = ± 2 16

b. Using the common ratio principal we can solve for x in the following way: r

2 −1 x

=

2

−1

2 −3

=

2

−1

x

therefore 2 −1 x × 2 −1 x = 2 −3 × 2 −1 ; 2 −2 x 2 = 2 −4 ; x 2 =

; x 2 = 2 −4+ 2 ; x 2 = 2 −2 ; x 2 =

1 2

2

; x=±

2 −4 2

; x 2 = 2 −4 ⋅ 2 2

−2

1 2

c. Using the common ratio principal we can solve for x in the following way: r

5 x 125 = 5 5x

=

therefore 5x × 5x = 125 × 5 ; 25x 2 = 625 ; x 2 =

625 ; x 2 = 25 ; x = ±5 25

Section 1.4 Practice Problems - Geometric Sequences and Geometric Series 1. Find the next four terms of the following geometric sequences. a. s1 = 3 , r = 0.5

b. s1 = −5 , r = 2

c. s1 = 5 , r = 0.75

2. Find the eighth and the general term of the following geometric sequences. a. s1 = 2 , r = 3

b. s1 = −4 , r = 12 .

c. s1 = 4 , r = −2.5

3. Find the next six terms and the n th term in each of the following geometric sequences. 1 4

1 1 2 4

a. 1, , 

b. − , , 

c.

1 p, − 3 p,  3

4. Given the following terms of a geometric sequence, find the common ratio r . a. s1 = 25 and s4 =

1 5

b. s1 = 4 and s5 =

1 64

c. s1 = 3 and s8 = 1

5. Write the first five terms of the following geometric sequences. 1 a. sn =  − 

2 n −1

 3

1 b. sn =  

2n+2

1 c. sn =  − 

2 n −3

1 d. sn =  −   2

 5

 3

n

6. Evaluate the sum of the following geometric series. a.

6

∑

3 k −1

=

k =1

Hamilton Education Guides

b.

10

∑ (− 2)k −3 = k =3

c.

8

∑

j =4

 1 4 −   2

j +1

=

29

Calculus I

d.

1.4 Geometric Sequences and Geometric Series

4

∑ (− 2)m−3 =

e.

5

∑

4m

=

∑ (− 3)n−4 =

f.

n =5

m =1

g.

10

h.

m =1

4

∑

j =1

5

∑ (− 3)k −1

=

k =1

3j = 27

i.

6

1 6  2 k =3  

∑

k +1

=

7. Given the first term s1 and r , find S 8 for each of the following geometric sequences. a. s1 = 3 , r = 3

b. s1 = −8 , r = 0.5

c. s1 = 2 , r = −2.5

8. Solve for x and y . a.

7

∑ (ix + 2) =

30

i =3

Hamilton Education Guides

b.

4

∑ (ix + y ) = i =1

20

and

6

∑ (ix + y ) = 10 i=2

30

Calculus I

1.5

1.5 Limits of Sequences and Series

Limits of Sequences and Series

A sequence s1 , s2 , s3 , s4 ,  , sn ,  is said to converge to the constant K , lim n→∞ s n = K

if and only if, for a large value of n , the absolute value of the difference between the n th term 1  5 10 17  , ,  , 1 + 2  ,   4 9 16 n 

and the constant K is very small.

For example, the sequence 2, ,

 

converges to 1 . This is because, the absolute value of the difference between 1 +

1  , n2 

for large

and 1 is very small. On the other hand, the sequence s1 , s2 , s3 , s4 ,  , sn ,  is said to diverge, if and only if, for a large value of n , the sequence approaches to infinity (∞) . For example, the n,

sequence 4, 8, 16, 32,  , 2 n+1 ,  does not converge. This is because, as n increases, the n th term increases without bound, i.e., it approaches to infinity. In the following examples we will learn how to identify a convergent or a divergent sequence: Example 1.5-1 State which of the following sequences are convergent. 5 10 17 n2 + 1 , ,, , 8 27 64 n3

a. 1, 2, 3, 4, 5,  , n,  =

b. 2, ,

8 26 80 3n − 1 , ,  , n −1 ,  3 9 27 3

c. 2, ,

d.

=

f. 6, , ,

9 19 33 1 , ,, 2+ 2 , 4 9 16 n

g.

1 1 1 1 1 , , , ,, , 2 3 4 5 n +1

1 1 1 1 1 , , , ,  , n +1 ,  16 64 256 1024 4

=

e. 3, ,

7 8 9 n+5 ,, 2 , 4 9 16 n

=

= =

=

h. 3, 6, 9, 12, 15,  , 3n,  =

Solutions: In solving this class of problems write the n th term and observe if it converges or diverges as n approaches to infinity. a. The sequence 1, 2, 3, 4, 5,  , n,  continues to increase. lim n→∞ n = ∞ which is undefined. Hence, the sequence diverges or is divergent. b. lim n→∞

n2 + 1 n3

 n2

= lim n→∞ 

 n3

+

1 1 1  1 1  = lim n →∞  +  = + 3 = 0+0 = 0 3 3  n ∞ n ∞ n 

The sequence converges to 0 c. lim n →∞

3n − 1 3n −1  

= lim n→∞  3 −

 3n

= lim n→∞ 

 3n−1

1 

3

n −1 

= 3−

Hamilton Education Guides

1

−

∞−1

3

 3n 3−( n−1)  3n/ − n/ +1 1  1  1     = lim n→∞  − lim − = →∞ n  1 3n−1  3n−1  3n−1   1 

= 3−

1 ∞

3

= 3−

1 ∞

= 3− 0 = 3

The sequence converges to 3

31

Calculus I

d. lim n→∞

1.5 Limits of Sequences and Series

1

4

n +1

 

e. lim n→∞  2 + f. lim n→∞

n+5 n

2

1

=

4

1   n2 

1

=

∞ +1

= 2+

4

∞

=

1 ∞

The sequence converges to 0

= 0

= 2+

1 = 2+0 = 2 ∞

5  n + 2 2 n n 

= lim n→∞  +

1 ∞

2

= lim n →∞ 

1 n

The sequence converges to 2 5  n2 

=

1 5 + 2 ∞ ∞

=

1 5 + ∞ ∞

= 0+0 = 0

The sequence converges to 0 g. lim n→∞

1 n +1

1 ∞ +1

=

=

1 ∞

= 0 The sequence converges to 0

h. The sequence 3, 6, 9, 12, 15,  , 3n,  continues to increase. lim n→∞ 3n = 3⋅ ∞ = ∞ which is undefined. Hence, the sequence diverges or is divergent. Example 1.5-2 State which of the following geometric sequences are convergent. a.

1 1 1 1 1 , , , ,, n , 3 9 27 81 3

b. 2, 4, 8, 16, 32,  , 2 n ,  =

=

1 1 1  1 , , ,  , 10 ⋅    10  10 100 1000

c. 1, − 1, 1, − 1,  , ( −1) n+1 ,  =

d. 10, 1,

n e. 0.2, 0.02, 0.0002,  , 2(01 . ) , =

f. 1, ,

g. −2, 4, − 8, 16, − 32,  , ( −1) n 2 n ,  =

h. 27, 3, ,

Solutions:

4 16 64 256  4 , , ,,    3 3 9 27 81 1 1  1 ,  , 27 ⋅    9 3 27

1 1 1 1 1 , ,, n , 3 9 27 81 3 1

a. The sequence , ,

value of the difference between

3n

n −1

n −1

n −1

,

,

,

=

=

=

converges to 0 since, for large value of n , the absolute and 0 is very small.

b. The sequence 2, 4, 8, 16, 32,  , 2 n ,  diverges since, as n increases, the n th term increases without bound. c. The sequence 1, − 1, 1, − 1,  , ( −1) n+1 ,  diverges since, as n increases, the n th term oscillates back and forth from +1 to −1 . d. The sequence 10, 1,

1 1 1  1 , , ,  , 10 ⋅    10  10 100 1000

n −1

,

converges to 0 since, for large value of n ,

1 the absolute value of the difference between 10 ⋅    10 

n −1

and 0 is very small.

n e. The sequence 0.2, 0.02, 0.0002,  , 2(01 . ) ,  converges to 0 since, for large value of n , the

Hamilton Education Guides

32

Calculus I

1.5 Limits of Sequences and Series

n absolute value of the difference between 2(01 . ) and 0 is very small.

4 16 64 256  4 , , ,,    3 3 9 27 81

f. The sequence 1, ,

n −1

,

diverges since, as n increases, the n th term

increases without bound. g. The sequence −2, 4, − 8, 16, − 32,  , ( −1) n 2 n ,  diverges since, as n increases, the n th term oscillates back and forth from a large positive number to a large negative number. 1 1  1 ,  , 27 ⋅    9 3 27

h. The sequence 27, 3, ,

n −1

,

converges to 0 since, for large value of n , the

1 absolute value of the difference between 27 ⋅    9

n −1

and 0 is very small.

Example 1.5-3 Discuss the limiting behavior of the following sequences as n approaches ∞ . a.

1

n2

=

e. (1) −n i.

b. 1 −

1 = n

1

n2

1 f. 1 +  

5n + 10 = n

−n

2

2n − 1

j.

=

2n

c. =

n+5 n2

g. 2 +

1

n2

=

d.

=

n2 + 5 n2

=

h. 100n =

=

Solutions: a. lim n →∞

1 n

1

b. lim n→∞ 1 − c. lim n →∞

=

2

n

2

∞

2

1 = 0 ∞

=

= 1−

1 ∞

2

= 1−

converges to 0 1 = 1− 0 = 1 ∞

converges to 1

1 5  n 1 5 5 1 5 = lim n→∞  2//=1 + 2  = lim n→∞  + 2  = + 2 = + = 0 + 0 = 0

n+5 n

1

2

n

n 

n

n 

∞

∞

∞

∞

converges to 0 d. lim n→∞

n2 + 5 n

2

 n2

5  5 5 + 2  = lim n→∞ 1 + 2  = 1 + 2 = 1 + 0 = 1  n  n  ∞ n

= lim n→∞ 

2

converges to 1 e. lim n→∞ (1) − n

1 1 1 1 1  1 1 = lim n →∞  n ⋅  = ∞ ⋅ = ⋅ = 1 ⋅ 0 = 0 1 n n 1 ∞ 1 ∞

 1 f. lim n →∞ 1 +  

2

−n

 3 = lim n →∞  

Hamilton Education Guides

 2

−n

 1    15 . n

−n . ) = lim n →∞  = lim n→∞ (15

converges to 0 =

1 15 .

∞

=

1 = 0 ∞

converges to 0

33

Calculus I

1.5 Limits of Sequences and Series

1

g. lim n→∞ 2 +

n

2

= 2+

1 ∞

= 2+

2

1 ∞

converges to 2

= 2+0 = 2

h. lim n→∞ 100n = 100 ⋅ ∞ = ∞

diverges

5n + 10 10  5n 10   10  = lim n →∞  / +  = lim n →∞  5 +  = 5 + = 5 + 0 = 5     n ∞ n/ n n

i. lim n→∞

converges to 5 2n − 1

j. lim n→∞

2

n

 2/ n/

= lim n→∞  

2/ n/

−

1 1   1  1  = lim n→∞ 1 − = 1− 0 = 1  = 1− ∞ = 1− n n  2  ∞ 2  2

converges to 1 Note that an easier way of finding the answer to sequences as n → ∞ is by rewriting the sequence in its “almost equivalent” form. This approach is only applicable to cases where n is approaching to infinity. For example, lim n →∞ n + 8 is almost the same as lim n→∞ n . (This is because ∞ + 8 is the same as ∞ . Addition of the number eight to a very large number such as infinity does not significantly change the final answer.) Let’s use this approach to solve few of the above problems. lim n→∞

lim n →∞

lim n→∞

lim n→∞

n+5 n

≈ lim n→∞

2

n2 + 5 n

2

5n + 10 n 2n − 1 2

n

n n

n2

≈ lim n→∞

n

≈ lim n→∞ ≈ lim n→∞

= lim n→∞

2

2

1 n

= lim n→∞

1 ∞

=

= 0

which is the same answer as in 1.5-3c.

1 = 1 1

which is the same answer as in 1.5-3d.

5n 5 = lim n→∞ = 5 1 n

2n 2

n

= lim n→∞

which is the same answer as in 1.5-3i.

1 = 1 1

which is the same answer as in 1.5-3j.

Example 1.5-4 State whether or not the following sequences are convergent or divergent as n approaches infinity. If the sequence does converge, find its limit. a. e.

n2 n −1

=

8n 2n + 1

b.

=

f.

n

1 i.   =  

j.

4

m.

n6 4

12n + 5

=

Hamilton Education Guides

n.

5n

=

n2 + 1

105 n = 1+ n n3 + 1

n3 + n + 1 n 4 + 3n 5

n +3

c.

8n

2 n + 105

=

g. 10 n = =

=

k. o.

25 −

h. 1 n

n2 4

d.

8n + 1

= =

l. p.

2n

8 n + 105 0.5 n

n2 + 1 3n + 1 3n

=

= =

1 1 = − n n+3

34

Calculus I

1.5 Limits of Sequences and Series

q. (0.5) n =

r. (0.5) −n =

s.

Solutions: n2 n −1

a. lim n→∞

≈ lim n →∞

5n

b. lim n→∞

n +1 8n

c. lim n→∞

n

2 + 10

5n n

≈ lim n→∞

5

=

n n2 = lim n→∞ = lim n→∞ n = ∞ 1 n

≈ lim n→∞

2

n −1 2 n

8n 2

= lim n →∞

2

n

2

n

=

1 2

∞

n

8 + 10

5

≈ lim n→∞

2n 8

2n

= lim n→∞

n

2

3n

= lim n→∞

1 = 0 ∞ 8n

e. lim n→∞

≈ lim n→∞

n

2 +1

10 5

8n 2

= lim n→∞

n

2 3n 2

n

1

∞2

≈

0.5 n

1

2

3n − n

= lim n →∞

1

2 2n

= lim n→∞

10 5 n 2 10 5 10 5 10 5 lim lim = lim n→∞ = = n → ∞ n → ∞ 1 1− 1 −1 n n 2 n2 n⋅n 2

converges to 0

≈ lim n→∞

2

n +1 n

1 1 i. lim n →∞   =      

n3 + 1

3

n + n +1

k. lim n→∞ 25 −

∞

4

4

j. lim n→∞

= lim n→∞

1

10 5 n lim n→∞ n

g. lim n→∞ 10 n = 10 ∞ = ∞ h. lim n →∞

⋅2

−n

= lim n→∞ 23n ⋅ 2 − n = lim n →∞ 23n − n = lim n→∞ 2 2 n

10 5 = 0 ∞

=

1

diverges

105 n lim n→∞ 1+ n

=

2

3n

converges to 0

= 2∞ = ∞ f.

converges to 5

diverges 2n

=

=

= lim n→∞ 23n ⋅ 2 − n = lim n →∞ 23n − n = lim n→∞ 2 2 n

= 2∞ = ∞ d. lim n→∞

n +1

diverges

5n = lim n→∞ 5 = 5 n

2 3n

= lim n→∞

2 n

t.

1 n

diverges 0.5 n n

Hamilton Education Guides

2

0.5∞ ∞

2

=

0 = 0 ∞

= 0.25∞ = 0

≈ lim n →∞ ≈

=

25 −

1 ∞

n3 n

3

=

converges to 0 1 = 1 1

converges to 1

= 52 = 5

converges to 5

= lim n→∞ 25 − 0

converges to 0

35

Calculus I

1.5 Limits of Sequences and Series

3n + 1

l. lim n→∞

3

m. lim n→∞ n. lim n→∞

o. lim n→∞

n

≈ lim n→∞

n6

n + 3n

3

n +3

n6

n2

n

4

converges to 1

n 6 n −4 = lim n→∞ n 2 = ∞ 2 = ∞ 1

diverges = lim n→∞

5

n2

≈ lim n →∞

8n 4 + 1

= lim n→∞

n4

n

≈ lim n →∞

5

1 = 1 1

= lim n→∞

n

≈ lim n→∞

12n 4 + 5 4

3n

1

n n

= lim n→∞

8n 4

= lim n →∞

5 −4

1 1 = = 0 n ∞

converges to 0 n2 2n

2

2

= lim n→∞

1 2 2

=

1 2 2

converges to 1 n

 p. lim n→∞  −

1   n + 3

0 1 1 1 − 1 ≈ lim n→∞  −  = lim n →∞  = 0  = n

 n 

n

∞

q. lim n→∞ (0.5) n = (0.5) ∞ = 0 r. lim n→∞ (0.5) − n =

s. lim n→∞ t. lim n→∞

n −1 2 n 2 n n +1

1

(0.5)

=

n

1 2 2

converges to 0 converges to 0

1

(0.5)

∞

1 0

=

diverges

= ∞

≈ lim n→∞

n/ 2 n/

= lim n→∞

1 1 = 2 2

converges to

≈ lim n→∞

2 n/ n/

= lim n→∞

2 = 2 1

converges to 2

1 2

Infinite Geometric Series: In section 1.4 we stated that the geometric series can be written in the following two forms: Sn

Sn

=

n

∑ s1r k−1 k =1

=

(

s1 1 − r n

)

1− r

(1) where r ≠ 1

(2)

In equation (2) , let’s consider the criteria where r is less than one and n is considerably large. n

Then, under these conditions r approaches to zero and S n = S∞

=

∞

∑ s1r n =1

n −1

= lim n→∞

(

s1 1 − r n 1− r

) = s (1 − 0) = 1

1− r

s1 1− r

(

s1 1 − r n 1− r

) reduces to S

n

=

s1 , 1− r

i.e.,

.

Thus, the sum of an infinite geometric series as n → ∞ and if r 1 is equal to

Hamilton Education Guides

36

Calculus I

1.5 Limits of Sequences and Series

=

S∞

∞

∑

n=0

s1 r n

=

∞

∑ s1r n−1

s1 1− r

=

n=1

(3)

.

Example 1.5-5 Find the sum of the following geometric series. a.

∞

1

∑ 2  4 

j

=

b.

∞

1

j

=

c.

18

∑ 10 k +1

=

e.

∞

2

∑  3 

j

=

f.

j =0

k =1

∞

5

∑ 5 3

k −1

=

k =1

j =0

j =0

d.

∞

∑ 5 − 2 

∞

 1

∑  − 8 

j

=

j =0

Solutions: 1 4

a. s1 = 2 and r = . Since r 1 therefore, we can use equation (3) to obtain the sum ∞

∑

j =0

 1 2   4

j

=

b. s1 = 5 and ∞

∑

j =0

 1 5 −   2

2

1 4 1 r=− . 2

j

1−

=

=

2 4 −1 4

2 3 4

=

2 8 2×4 = 1 = = 3 3 1× 3 4

Since r 1 therefore, we can use equation (3) to obtain the sum

5  1 1− −   2

=

5 1+

1 2

=

5 2 +1 2

=

5 3 2

5 10 5× 2 = 1 = = 3 3 1× 3 2

5 3

c. s1 = 5 and r = . Since r 〉 1 therefore, the geometric series d.

∞

18

∑ 10 k +1 k =1

=

∞

18

∑ 10 2 ⋅10 k −1 k =1

=

∞

∑

k =1

18  1    100  10 k −1 

=

∞

∑

k =1

18  1    100  10 

∞

5

∑ 5 3

k −1

has no finite sum.

k =1

k −1

18 1 and r = . Since r 1 therefore, we can use equation (3) to obtain the sum 100 10 18 18 18 k −1 ∞ 1 180 18 × 10 18  1  = 100 = 100 = 100 = = =   1 10 − 1 9 5 100 × 9 900 100  10  1− k =1 10 10 10 2 e. s1 = 1 and r = . Since r 1 therefore, we can use equation (3) to obtain the sum 3 1 j ∞ 1× 3 3 1 1 1 2 = = = 1 = = = 3   = 3 − 2 2 1 1 1×1 1 3 1− j =0   3 3 3 3 1 f. s1 = 1 and r = − . Since r 1 therefore, we can use equation (3) to obtain the sum 8 1 j ∞ 8 1× 8 1 1 1 1  1 = = = = 1 = = −  = 8 + 1 1 9 9 9 1× 9 8  1 1+ j =0  1− −  8  8 8 8 8 s1 =

∑

∑

∑

Hamilton Education Guides

37

Calculus I

1.5 Limits of Sequences and Series

Example 1.5-6 Find the sum of the following infinite geometric series. 1 2

1 4

1 3

a. 2 − 1 + − +  = 1 9

c. 1 + +

b. − + 1 − 3 + 9 +  = 1 2

1 1 + + = 81 729

1 4

d. 1 + + +

1 + = 16

Solutions: 1 2

1 4

1 2

a. Given 2 − 1 + − +  , s1 = 2 and r = − . Since r 1 therefore, we can use equation (3) to obtain the sum, i.e., 1 1 2 −1+ − +  2 4

=

2  1 1− −   2

=

2

1 1+ 2

=

2 2 +1 2

=

2 3 2

2 2×2 4 1 = = = = 1.333 3 1× 3 3 2

Note that in example 1.4-6c the answer to the same problem when n = 5 was 1375 . However, . as n → ∞ the answer approaches to 1333 . . 1 3

b. Given − + 1 − 3 + 9 +  , s1 = −

1 3

and r =

1 3

1 = −3 . − 13

Since r = − 3 = 3 is greater than one therefore,

the geometric sequence − + 1 − 3 + 9 +  has no finite sum. 1 9

c. Given 1 + +

1

1 1 1 + +  , s1 = 1 and r = 9 = . Since r 81 729 1 9

1 therefore, we can use equation

(3) to obtain the sum, i.e., 1 1 1 1+ + + + = 9 81 729 1 2

1

1 1− 9

=

1 1 + , s1 = 1 4 16

d. Given 1 + + +

(3) to obtain the sum, i.e.,

1 9 −1 9

=

and r =

1 8 9 1 2

1

=

=

1 1 8 9

1 . 2

=

9 1× 9 = 8 1× 8

Since r 1 therefore, we can use equation

1 1 1 1 1× 2 2 1 1 1 + = 1+ + + = = = 1 = = = 2 2 −1 1 1 1 2 4 16 1×1 1 1− 2 2 2 2

Repeating Decimals:

An application of infinite geometric series is in representation of repeating decimals as the quotient of two integers. For example, 0131313 .  and 0.666 66  are repeating decimals. The bar above the repeating numbers denotes that the numbers appearing under it are repeated endlessly. The following examples show the steps as to how repeating decimals are converted to fractional forms:

Hamilton Education Guides

38

Calculus I

1.5 Limits of Sequences and Series

Example 1.5-7 Write the following repeating decimals as the quotient of two positive integers. a. 0131313  = .

b. 5.510510510  =

c. 012451245 . =

Solutions: a. First - write the decimal number 0131313  in its equivalent form of .  = 013 . 0131313 . + 0.0013 + 0.000013 + 

Second - Since this is a geometric series, write the first term in the series and its ratio, i.e., s1 = 013 .

and r =

0.0013 = 0.01 . . 013

Third - Since the ratio r is less than one, use the infinite geometric series equation s∞ =

s1 1− r

to obtain the sum of the infinite series 013 . + 0.0013 + 0.000013 +  , i.e., s∞ =

s1 1− r

; s∞ =

13 . 013 13 013 . ; s∞ = ; s∞ = thus 0.131313 = 99 1 − 0.01 99 0.99

b. First - Consider the decimal portion of the number 5.510510510  and write it in its equivalent form of 0.510510510 = 0.510 + 0.000510 + 0.000000510 +  Second - Since this is a geometric series, write the first term in the series and its ratio, i.e., s1 = 0.510

and r =

0.000510 = 0.001 . 0.510

Third - Since the ratio r is less than one, use the infinite geometric series equation s∞ =

s1 1− r

to obtain the sum of the infinite series 0.510 + 0.000510 + 0.000000510 +  , i.e., s∞ =

s1 1− r

; s∞ =

510 0.510 510 0.510 ; s∞ = ; s∞ = thus 5.510510510 = 5 999 999 1 − 0.001 0.999

c. First - write the decimal number 012451245 .  in its equivalent form of 012451245 .  = 01245 . + 0.00001245 + 

Second - Since this is a geometric series, write the first term in the series and its ratio, i.e., s1 = 01245 .

and r =

0.00001245 = 0.0001 . 01245 .

Third - Since the ratio r is less than one, use the infinite geometric series equation s∞ =

s1 1− r

to obtain the sum of the infinite series 01245 . + 0.00001245 +  , i.e., s∞ =

s1 1− r

; s∞ =

1245 1245 . 01245 01245 . ; s∞ = ; s∞ = thus 0.12451245 = 9999 9999 0.9999 1 − 0.0001

Hamilton Education Guides

39

Calculus I

1.5 Limits of Sequences and Series

Section 1.5 Practice Problems - Limits of Sequences and Series 1. State which of the following sequences are convergent. 3 2

5 2

a. 1, , 2, , 3,  ,

3 8 15 n2 −1 ,, , = 2 3 4 n

n +1 , = 2

b. 0, , ,

c. 4, 8, 16, 32,  , 2 n+1 ,  = 1 2 3 n −1 ,, 2 , 4 9 16 n

e. 0, , ,

=

d.

1 1 1 1 1 , , , ,  , n +1 ,  16 64 256 1024 4

f.

1 1 1 1  1 , , , ,,    5 25 125 625 3125

n +1

=

,

=

2. State which of the following geometric sequences are convergent. a.

1 1 1 1 1 , , , ,, n , 4 16 64 256 4

b. −5, 25, − 125, 625, − 3125,  , ( −5) n ,  =

=

c. 2, − 2, 2, − 2,  , 2( −1) n+1 ,  =

1 1 1 1 d. 1, , , ,  ,  

e. −9, 27, − 81, 243,  , ( −1) n 3n+1 ,  =

f. 1, , ,

n −1

 2

2 4 8

=

,

1 1 1 1  1 , ,,    3 3 9 27 81

n −1

,

=

3. State whether or not the following sequences converges or diverges as n → ∞ . If it does converge, find the limit. a.

n2

=

n3 − 4

b.

e.

( n + 2) 2 = 2

f.

i.

n +1 = n −1

j.

n

m. 100 q.

−

1 n

=

1 −1 n +1

n.

5n + 1 n2 + 1

2n

n3 − 1 3 n 3

c.

5 n+1

=

d.

n 2 + 2n

g.

=

k. 10 n =

4

n +1

=

h.

1

=

o.

r. (0.25) −n =

=

25 n

=

2n + 1 n

=

s.

n + 100 n 3 − 10 n +1 n +1

l.

5 n + 25 125 n 5

n2 + 1

=

=

( n − 1) 2 = (1 − n)(1 + n) 1

=

p.

=

100 n n2 + 3 n

t.

n +1

= +2

=

4. Find the sum of the following geometric series. a.

∞

∑

j =0

d.

∞

 1 3   8 5

j

∑ 100 k +1

=

b.

∞

∑

j =0

=

k =1

Hamilton Education Guides

e.

∞

 1 3 −   4

1   3 j =0  

∑

j

j

=

=

c.

∞

∑

k =1

f.

∞

 3 3   2

 1 −  5 j =0 

∑

k −1

j

=

=

40

Calculus I

1.5 Limits of Sequences and Series

5. Find the sum of the following infinite geometric series. 1 5

a. 5 − 1 + − 1 6

c. 1 + +

1 + = 25

1 2

b. − + 2 − 8 + 32 +  =

1 1 + + = 36 216

d. 1 +

1 1 1 + + + = 10 100 1000

6. Write the following repeating decimals as the quotient of two positive integers. a. 0.666666 =

Hamilton Education Guides

b. 3.027027 027  =

c. 0111111 . =

41

Calculus I

1.6

1.6 The Factorial Notation

The Factorial Notation

As was stated earlier, the product of several consecutive positive integers is usually written using a special symbol n! , read as “ n factorial,” which is defined by n! = n ( n − 1)( n − 2)( n − 3)( n − 4)  ( 4)( 3)( 2)(1)

For example, 1! (read as “one factorial”) through 10! (read as “ten factorial”) are written in their equivalent form as: 1!

= 1

6!

= 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 720

2! = 2 ⋅ 1 = 2

7! = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 5,040

3! = 3 ⋅ 2 ⋅ 1 = 6

8! = 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 40,320

4! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24

9! = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 362,880

5! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120

10! = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 3,628,800

Note that, for n 〉 1 , since n! = n ( n − 1)( n − 2)( n − 3)  4 ⋅ 3 ⋅ 2 ⋅1 and ( n − 1)! = ( n − 1)( n − 2)( n − 3)  4 ⋅ 3 ⋅ 2 ⋅ 1 we can rewrite the recursive n! relationship in the following way: n! = n ( n − 1)( n − 2)( n − 3)( n − 4)  4 ⋅ 3 ⋅ 2 ⋅ 1 = n ( n − 1) !    ( n − 1)!

n ( n − 1) ! = n ( n − 1)( n − 2)( n − 3)( n − 4)  4 ⋅ 3 ⋅ 2 ⋅ 1 = n ( n − 1)( n − 2) !  ( n − 2) !

n ( n − 1)( n − 2)! = n ( n − 1)( n − 2)( n − 3)( n − 4)  4 ⋅ 3 ⋅ 2 ⋅ 1 = n ( n − 1)( n − 2)( n − 3) !  ( n − 3) !

n ( n − 1)( n − 2)( n − 3)! = n ( n − 1)( n − 2)( n − 3)( n − 4)  4 ⋅ 3 ⋅ 2 ⋅ 1 = n ( n − 1)( n − 2)( n − 3)( n − 4) !    n − 4 ! ( )

For example, 7 ! = 7 ⋅ 6 ! = 7 ⋅ 6 ⋅ 5! = 7 ⋅ 6 ⋅ 5 ⋅ 4! = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3! = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ! 10! = 10 ⋅ 9 ! = 10 ⋅ 9 ⋅ 8! = 10 ⋅ 9 ⋅ 8 ⋅ 7! = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6! = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ! 35! = 35 ⋅ 34 ! = 35 ⋅ 34 ⋅ 33! = 35 ⋅ 34 ⋅ 33 ⋅ 32! = 35 ⋅ 34 ⋅ 33 ⋅ 32 ⋅ 31! = 35 ⋅ 34 ⋅ 33 ⋅ 32 ⋅ 31 ⋅ 30 !

( n + 4)! = ( n + 4)( n + 3)! = ( n + 4)( n + 3)( n + 2)! = ( n + 4)( n + 3)( n + 2)( n + 1) ! ( n + 8)! = ( n + 8)( n + 7)! = ( n + 8)( n + 7)( n + 6)! = ( n + 8)( n + 7)( n + 6)( n + 5) !

(2n + 2)! = (2n + 2)(2n + 1) ! = (2n + 2)(2n + 1)(2n)! = (2n + 2)(2n + 1)(2n)(2n − 1)! = (2n + 2)(2n + 1)(2n) (2n − 1)(2n − 2)! = ( 2n + 2)( 2n + 1)( 2n)( 2n − 1)( 2n − 2)( 2n − 3) ! Hamilton Education Guides

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Calculus I

1.6 The Factorial Notation

(2n + 5)! = (2n + 5)(2n + 4)! = (2n + 5)(2n + 4)(2n + 3)! = (2n + 5)(2n + 4)(2n + 3)(2n + 2)! = (2n + 5)(2n + 4)

(2n + 3)(2n + 2)(2n + 1)! = ( 2n + 5)( 2n + 4)( 2n + 3)( 2n + 2)( 2n + 1)( 2n) ! The above principal can be used to prove that 0 ! = 1 . Since 1! = 1 ⋅ (1 − 1) ! = 1 ⋅ 0! in order for equality on both sides of the equation to hold true 0! must be equal to one. Hence, we state that 0! = 1 . Example 1.6-1 Expand and simplify the following factorial expressions. b. (6 − 2)! =

a. 13! = e.

10! = 5! 4 !

f.

8! 3! (8 − 2)!

=

c.

8! = 4!

g.

11! 8 ! 16 !

d. =

h.

12 ! = 11!

( 4 − 2) ! 8 ! = 11! (5 − 3) !

Solutions: a. 13! = 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 6,227,020,800 b. (6 − 2)! = 4! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24 c.

8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4! 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4/ ! 8! = = = 8 ⋅ 7 ⋅ 6 ⋅ 5 = 1680 4! 4/ ! 4!

d.

/ /! 12 ⋅ 11! 12 ⋅ 11 12 ! = = = 12 / /! 11! 11 11!

e.

10! 5! 4 !

f.

8! 3! (8 − 2)!

4 28 8! 8 ⋅ 7 ⋅ 6/ ! 8⋅7 4⋅7 8/⋅ 7 = = = = = = 3 3! 6 ! 3! 6/ ! 3 3 ⋅ 2/ ⋅ 1 3!

g.

11! 8 ! 16 !

11! 8 ! 16 ⋅ 15 ⋅ 14 ⋅ 13 ⋅ 12 ⋅ 11!

h.

(4 − 2)! 8! = 11! (5 − 3)!

=

=

10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5! 5! 4 !

2/ ! 8! 11! 2/ !

=

=

10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5/ ! 5/ ! 4 !

=

=

10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 4!

/ / ! 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 11 / /! 16 ⋅ 15 ⋅ 14 ⋅ 13 ⋅ 12 ⋅ 11

=

=

2 10 ⋅ 9 ⋅ 8/⋅ 7 ⋅ 6/ 4/ ⋅ 3/ ⋅ 2/ ⋅ 1

8/ ⋅ 7/ ⋅ 6/ ⋅ 5/ ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 / /⋅ 13 ⋅ 12 // / /⋅ 15 / /⋅ 14 16 2 2 3 2

=

=

10 ⋅ 9 ⋅ 2 ⋅ 7 1

= 1260

1 4/ ⋅ 3/ ⋅ 2/ ⋅ 1 = 13 2/ ⋅ 3/ ⋅ 2/ ⋅ 13 ⋅ 2/

1 8/ ! 8! 8! = = = 990 11 ⋅ 10 ⋅ 9 ⋅ 8 ! 11 ⋅ 10 ⋅ 9 ⋅ 8/ ! 11!

Example 1.6-2 Write the following products in factorial form. a. 4 ⋅ 3 ⋅ 2 ⋅1 =

b. 10 ⋅ 11 ⋅ 12 ⋅ 13 =

c. 20 ⋅ 21 ⋅ 22 ⋅ 23 ⋅ 24 =

d. 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 =

e. 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7 =

f. 20 =

g. 15 ⋅14 ⋅13 ⋅12 ⋅11 =

h. 8 =

i. 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 =

Hamilton Education Guides

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Calculus I

1.6 The Factorial Notation

Solutions: a. 4 ⋅ 3 ⋅ 2 ⋅1 = 4! d. 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 =

b. 10 ⋅ 11 ⋅ 12 ⋅ 13 = 9! 4!

g. 15 ⋅14 ⋅13 ⋅12 ⋅11 =

e. 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7 = 15 ! 10 !

h. 8 =

13 ! 9!

7! 2!

8! 7!

c. 20 ⋅ 21 ⋅ 22 ⋅ 23 ⋅ 24 = f. 20 =

24 ! 19 !

20 ! 19 !

i. 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 = 6!

Example 1.6-3 Expand the following factorial expressions. a. n! =

b. ( n − 1)! =

c. (n − 3)! =

d. ( n − 5)! =

e. ( n − 8)! =

f. (2n)! =

g. (2n − 1)! =

h. (3n − 5)! =

i. (2n + 1)! =

j. (n + 1)! =

k. (n + 3)! =

l. (n + 8)! =

m. (2n + 2)! =

n. (2n + 5)! =

Solutions: a. n! = ( n)( n − 1)( n − 2)( n − 3)( n − 4)( n − 5)( n − 6)( n − 7)( n − 8)( n − 9)  4 ⋅ 3 ⋅ 2 ⋅ 1 b. ( n − 1)! = ( n − 1)( n − 2)( n − 3)( n − 4)( n − 5)( n − 6)( n − 7)( n − 8)( n − 9)  4 ⋅ 3 ⋅ 2 ⋅ 1 c. ( n − 3)! = ( n − 3)( n − 4)( n − 5)( n − 6)( n − 7)( n − 8)( n − 9)  4 ⋅ 3 ⋅ 2 ⋅ 1 d. ( n − 5)! = ( n − 5)( n − 6)( n − 7)( n − 8)( n − 9)( n − 10)( n − 11)  4 ⋅ 3 ⋅ 2 ⋅ 1 e. ( n − 8)! = ( n − 8)( n − 9)( n − 10)( n − 11)  4 ⋅ 3 ⋅ 2 ⋅ 1 f. (2n)! = ( 2n)( 2n − 1)( 2n − 2)( 2n − 3)( 2n − 4)  4 ⋅ 3 ⋅ 2 ⋅ 1 g. (2n − 1)! = ( 2n − 1)( 2n − 2)( 2n − 3)( 2n − 4)  4 ⋅ 3 ⋅ 2 ⋅ 1 h. (3n − 5)! = ( 3n − 5)( 3n − 6)( 3n − 7)( 3n − 8)( 3n − 9)( 3n − 10)( 3n − 11)  4 ⋅ 3 ⋅ 2 ⋅ 1 i. (2n + 1)! = ( 2n + 1)( 2n)( 2n − 1)( 2n − 2)( 2n − 3)( 2n − 4)  4 ⋅ 3 ⋅ 2 ⋅ 1 j. ( n + 1)! = ( n + 1)( n)( n − 1)( n − 2)( n − 3)( n − 4)( n − 5)( n − 6)( n − 7)( n − 8)  4 ⋅ 3 ⋅ 2 ⋅ 1 k. ( n + 3)! = ( n + 3)( n + 2)( n + 1)( n)( n − 1)( n − 2)( n − 3)( n − 4)( n − 5)( n − 6)( n − 7)( n − 8)  4 ⋅ 3 ⋅ 2 ⋅ 1 l. ( n + 8)! = ( n + 8)( n + 7)( n + 6)( n + 5)( n + 4)( n + 3)( n + 2)( n + 1)( n)( n − 1)( n − 2)( n − 3)  4 ⋅ 3 ⋅ 2 ⋅ 1

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Calculus I

1.6 The Factorial Notation

m. (2n + 2)! = ( 2n + 2)( 2n + 1)( 2n)( 2n − 1)( 2n − 2)( 2n − 3)( 2n − 4)  4 ⋅ 3 ⋅ 2 ⋅ 1 n. (2n + 5)! = ( 2n + 5)( 2n + 4)( 2n + 3)( 2n + 2)( 2n + 1)( 2n)( 2n − 1)( 2n − 2)( 2n − 3)( 2n − 4)  4 ⋅ 3 ⋅ 2 ⋅ 1 Example 1.6-4 Expand and simplify the following factorial expressions. a.

( n − 1)! = ( n − 3)!

b.

( n + 2)! = n!

2 n !) ( f. = ( n + 1)! ( n − 1)!

(3n + 1)! (3n − 1)! = e. (3n)! (3n − 3)!

c.

( n + 3)! = ( n − 1)!

g.

(2n − 2)! 2( n !) = (2n)! ( n − 1)!

d.

( n + 2)( n + 2)! = ( n + 3)!

Solutions: a. b.

1 ( n − 1)! ( n/ − 1/)! ( n − 1)! = = = ( n − 3)! ( n − 3)( n − 2)( n − 1)! ( n − 3)( n − 2)( n/ − 1/)! ( n − 3)( n − 2)

( n + 2)! n!

=

( n + 2)( n + 1) n ! n!

=

( n + 2)( n + 1) n/ ! n/ !

= ( n + 2)( n + 1)

c.

( n + 3)! ( n + 3)( n + 2)( n + 1)( n)( n − 1)! ( n + 3)( n + 2)( n + 1)( n)( n/ − 1/)! = = = ( n + 3)( n + 2)( n + 1) n ( n − 1)! ( n/ − 1/)! ( n − 1)!

d.

( n + 2)( n + 2)! ( n + 2)( n + 2)! ( n + 2)( n/ + 2/ )! = = = ( n + 3)( n + 2)! ( n + 3)( n/ + 2/ )! ( n + 3)!

e.

(3n + 1)! (3n − 1)! [(3n + 1)(3n)!] [(3n − 1)(3n − 2)(3n − 3)!] [(3n + 1)(3/ n/ )!] [(3n − 1)(3n − 2)(3/ n/ − 3/ )!] = = (3n)! (3n − 3)! (3/ n/ )! (3/ n/ − 3/ )! (3n)! (3n − 3)! =

(3n + 1) (3n − 1)(3n − 2) 1

n+2 n+3

= ( 3n + 1) ( 3n − 1)( 3n − 2)

f.

n ( n/ − 1/) ! ( n !) 2 ( n !)( n/ !) ( n !)( n !) n! = = = = = ( n + 1)! ( n − 1)! [( n + 1)( n)!] ( n − 1)! [( n + 1)( n/ )!] ( n − 1)! ( n + 1)( n − 1)! ( n + 1)( n/ − 1/)!

g.

(2n − 2)! [2 n ( n − 1)!] (2/ n/ − 2/ )! [2 n ( n/ − 1/)!] (2n − 2)! 2( n !) 2n = = = (2n)! ( n − 1)! [(2n)(2n − 1)(2n − 2)!] ( n − 1)! [(2n)(2n − 1)(2/ n/ − 2/ )!] ( n/ − 1/)! (2n)(2n − 1) =

2/ n/ / n 2 ( / )(2n − 1)

=

n n+1

1 2n − 1  n r

The factorial notation is used in expansion of the   read as “the binomial coefficient n, r ” in the following way:  n n!  =  r  r ! ( n − r) !

Hamilton Education Guides

45

Calculus I

1.6 The Factorial Notation

Example 1.6-5 Write the following expressions in factorial notation form. Simplify the answer. a.   =

 6  4

b.   =

 6  2

c.   =

10 5 

g.   =

 8  4

h.   =

f.   = Solutions: a.

 6    4

b.

 6    2

=

=

6! 2! (6 − 2)!

=

6! 2! ⋅ 4!

c.   =

 5  0

5! 0 ! ( 5 − 0) !

=

5! 1 5/ ! = = = 1 1 ⋅ 5/ ! 0 !⋅ 5! 1

d.   =

 5  5

5! 5! (5 − 5) !

=

5/ ! 5! 1 = = = 1 5/ ! ⋅ 1 5! ⋅ 0 ! 1

e.

 7    5

=

7! 5! (7 − 5) !

f.

10   5 

=

10 ! 5! (10 − 5) !

g.

 8    4

=

8! 4 ! (8 − 4) !

h.   =

n! n ! ( n − n) !

10 ! 5! ⋅ 5!

=

 n  n

i. 

n    n − 1

=

3 6/⋅ 5 ⋅ 4/ ⋅ 3/ ⋅ 2/ ⋅ 1/ (2/ ⋅ 1) ⋅ (4/ ⋅ 3/ ⋅ 2/ ⋅ 1/)

10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 (5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1) ⋅ (5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1)

=

 7  5

e.   = n    n − 4

=

j. 

=

=

3⋅ 5 1

= 15

=

2 2 / /⋅ 9 ⋅ 8/⋅ 7 ⋅ 6/ ⋅ 5/ ⋅ 4/ ⋅ 3/ ⋅ 2/ ⋅ 1/ 10 (5/ ⋅ 4/ ⋅ 3/ ⋅ 2/ ⋅ 1) ⋅ (5/ ⋅ 4/ ⋅ 3/ ⋅ 2/ ⋅ 1/)

=

2⋅9⋅2⋅7 1

= 252

2 8! 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4! 7⋅ 2⋅5 8/ ⋅ 7 ⋅ 6/⋅ 5 ⋅ 4/ ! = = = = = 70 4! ⋅ 4! 4/ ! ⋅ 4/ ⋅ 3/ ⋅ 2/ ⋅ 1 1 4! ⋅ 4 ⋅ 3 ⋅ 2 ⋅1

=

n! 1 n! = / = = 1 n/ ! ⋅ 1 n ! 0! 1

n! ( n − 1)! n − ( n − 1) !

n    n − 4

=

n! ( n − 4) ! n − ( n − 4 ) !

=

 5  5

3 7! 7 ⋅ 6 ⋅ 5! 7⋅3 7 ⋅ 6/⋅ 5/ ! = = = = = 21 5! ⋅ 2 ! 5! ⋅ 2 ! 1 5/ ! ⋅ 2/ ⋅ 1

=

j. 

6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 (2 ⋅ 1) ⋅ (4 ⋅ 3 ⋅ 2 ⋅ 1)

=

n    n − 1

i. 

d.   =

3 6! 3⋅ 5 6/⋅ 5 ⋅ 4/ ⋅ 3/ ⋅ 2/ ⋅ 1/ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = = = = = 15 4! ⋅ 2! 1 (4 ⋅ 3 ⋅ 2 ⋅ 1) ⋅ (2 ⋅ 1) (4/ ⋅ 3/ ⋅ 2/ ⋅ 1/) ⋅ (2/ ⋅ 1)

6! 4 ! ( 6 − 4) !

 n  n

 5  0

[

]

[

]

=

n! ( n − 1)! ( n/ − n/ + 1)!

=

n ⋅ ( n − 1) ⋅ ( n − 2) ⋅ ( n − 3) ⋅ ( n/ − 4/ ) ! ( n/ − 4/ )! 4!

Hamilton Education Guides

n! ( n − 4)! ( n/ − n/ + 4)!

=

n! ( n −1)!1!

=

=

n! ( n − 4)! 4!

n ⋅ ( n − 1) ⋅ ( n − 2) ⋅ ( n − 3) 4 ⋅ 3 ⋅ 2 ⋅1

=

=

n ⋅ ( n − 1) !

( n − 1)!

=

=

n ⋅ ( n/ − 1/) ! ( n/ − 1/)!

= n

n ⋅ ( n − 1) ⋅ ( n − 2) ⋅ ( n − 3) ⋅ ( n − 4) !

( n − 4)! 4!

n ⋅ ( n − 1) ⋅ ( n − 2) ⋅ ( n − 3) 24

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Calculus I

1.6 The Factorial Notation

 n r

An application of the binomial coefficient n, r , i.e.,   is in its use for expansion of binomials of the form (a + b) n , where n is a positive integer. In general, the binomial equation of order n can be expanded in the following form:

( a + b) n =

 n n  n n −1  n  n  n − r +1 r −1  n n −2 2 +  +   bn b ++ b   a +  a b+  a a  0  r − 1  n  2 1 

(1)

note that the above equation is used for expanding binomial expressions that are raised to the second, third, fourth, or higher powers. For example, ( x − 2) 2 , ( x + 3)3 , ( x −1)5 , ( x − 3) 6 , (2 x + y ) 4 , y   2x −   2

5

(

)

4

, x − 3 , ( x − 2 y ) 6 , ( x − y )16 , etc. can all be expanded using the above equation. The

following examples show the steps as to how binomial coefficients are used in expanding binomial equations: Example 1.6-6 Expand the following binomial expressions. b. ( x + 2)5 =

a. ( x −1)3 =

c. ( x − 3) 4 =

Solutions: a. First - Identify the a, b, and n terms, i.e., a = x , b = −1 , n = 3 . Second - Use the general binomial expansion formula, i.e., equation (1) above to expand ( x −1)3 .

( x − 1)3 =  3  0

 3 3  3 3−1  3 3−2 2  3 3−3 3   x +   x ⋅ ( −1) +   x ( −1) +   x ( −1)  0 1  2  3  3 1

 3  2

 3  3

=   x 3 −   x 2 +   x −   ⋅1 = =

3! 3 3! 2 3! 3! x − x + x− 3! 3! 2! 2!

=

 3  0

 3 1

 3  2

 3  3

=   x3 −   x2 +   x −   x0

3! 3! 3! 3! x2 + x− x3 − 2 ! (3 − 2) ! 3! (3 − 3) ! 0 ! (3 − 0) ! 1! (3 − 1) !

3/ ! 3 3 ⋅ 2/ ! 2 3 ⋅ 2/ ! 3/ ! x − x + x− 3/ ! 2/ ! 2/ ! 3/ !

= x3 − 3 x2 + 3 x − 1

b. First - Identify the a, b, and n terms, i.e., a = x , b = 2 , n = 5 . Second - Expand ( x + 2)5 using equation (1) .

( x + 2) 5 =  5  0

 5 5  5 5−1  5 5−2 2  5 5−3 3  5 5−4 4  5 5−5 5 ⋅2 +   x ⋅2 +   x ⋅2 +   x ⋅2   x +   x ⋅2 +   x  0 1   2  3  4  5  5 1 

 5  2

 5  3

 5  4

 5  5

 5  0

 5 1 

 5  2

 5  3

=   x 5 + 2   x 4 + 4   x 3 + 8   x 2 + 16   x + 32   x 0 =   x 5 + 2   x 4 + 4   x 3 + 8   x 2  5  5 + 16   x + 32    4  5

=

5! 5! 5! 5! 5! x5 + 2 x4 + 4 x3 + 8 x 2 + 16 x 0 ! ( 5 − 0) ! 1! (5 − 1) ! 2 ! ( 5 − 2) ! 3! (5 − 3) ! 4 ! ( 5 − 4) !

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Calculus I

+ 32

+4

1.6 The Factorial Notation

5! 5! (5 − 5) !

=

5! 5! 5! 2 5! 3 5! 5 5! x + 32 x +8 x + 16 x + 2 x4 + 4 5! 4! 3! 2 ! 2 ! 3! 4! 5!

5/! 5 ⋅ 4/ ! 5 ⋅ 4 ⋅ 3/! 2 5 ⋅ 4 ⋅ 3/! 3 x +8 x + 16 x + 32 4/ ! 5/! 3/!2! 2!3/!

=

5/ ! 5 5 ⋅ 4/ ! 4 x +2 x 4/ ! 5/ !

= x 5 + 10 x 4 + 40 x 3 + 80 x 2 + 80 x + 32

c. First - Identify the a, b, and n terms, i.e., a = x , b = −3 , n = 4 . Second - Expand ( x − 3) 4 using equation (1) .

( x − 3) 4 =

 4 4  4 4−1  4 2  4 3  4 4 ⋅ ( −3) +   x 4−2 ( −3) +   x 4−3 ( −3) +   x 4−4 ( −3)   x +  x  0 1   2  3  4

 4  4  4 + 9  x 2 − 27  x + 81  x 0  4 3  2

 4 1 

4! 4! 4! 4! x4 − 3 x3 + 9 x 2 − 27 x 0 ! ( 4 − 0) ! 1! (4 − 1) ! 2 ! (4 − 2) ! 3! (4 − 3) !

4! 4 4! 4! 2 4! 4! 4/ ! 4 ⋅ 3/ ! 3 4 ⋅ 3 ⋅ 2/ ! 2 x +9 x x − 3 x3 + 9 x − 27 x + 81 = x4 − 3 4/ ! 3/ ! 2! 2/ ! 4! 3! 2! 2! 3! 4! 0!

+ 81

4! 4! (4 − 4)!

−27

4 ⋅ 3/ ! 4/ ! x + 81 3/ ! 4/ ! 0 !

=

=

 4  0

=   x4 − 3   x3

= x 4 − 12 x 3 + 54 x 2 − 108 x + 81

Example 1.6-7 Use the general equation for binomial expansion to solve the following exponential numbers to the nearest hundredth. a. (0.83) 6 =

8 b. (105 . ) =

10 c. (121 . ) =

Solutions: 6 a. First - Write the exponential expression in the form of (0.83) 6 = (1 − 017 . ) .

Second - Identify the a, b, and n terms, i.e., a = 1 , b = −017 . , n=6. 6 Third - Use the general binomial expansion formula, i.e., equation (1) above to expand (1 − 017 . ) . 6 . ) = (1 − 017

 6 6 +  ( −017 . )  6

+ 0.0289

=

 6 6  6 5  6 2  6 3  6 4  6 5 . ) +   14 ⋅ ( −017 . ) +   13 ⋅ ( −017 . ) +   12 ⋅ ( −017 . ) +   1 ⋅ ( −017 . )   1 +   1 ⋅ ( −017 0 1 2 3 4 5              6  0

 6 1 

 6  2

 6  3

.   + 0.0289   − 0.0049   +  = =   − 017

6! 6! − 0.0049 + 2!(6 − 2 )! 3!(6 − 3)!

=

6! 6! − 017 . 1! (6 − 1) ! 0! (6 − 0) !

6! 6! 6! 6! − 017 . + 0.0289 − 0.0049 + 6! 5! 2! 4! 3! 3!

6 ⋅ 5 ⋅ 4 ⋅ 3/ ! 6/ ! 6 ⋅ 5/ ! 6 ⋅ 5 ⋅ 4/ ! . +  = 1 − 102 + 0.0289 − 0.0049 − 017 . + 0.4335 − 0.098 +  ≈ 0.3155 3! 3/ ! 6/ ! 5/ ! 2! 4/ !

Hamilton Education Guides

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Calculus I

1.6 The Factorial Notation

6 Therefore, (1 − 017 . ) , to the nearest hundredth, is equal to 0.32 8 8 b. First - Write the exponential expression in the form of (105 . ) = (1 + 0.05) .

Second - Identify the a, b, and n terms, i.e., a = 1 , b = 0.05 , n = 8 . Third - Use the general binomial expansion formula, i.e., equation (1) above to expand (1 + 0.05)8 .

(1 + 0.05)8 =  8  0

 8 6  8 8  8 7 5 4  8 3  8 2 2  8 3   1 +   1 ⋅ (0.05) +   1 ⋅ (0.05) +   1 ⋅ (0.05) +   1 ⋅ (0.05) +   1 ⋅ (0.05) +   5  4  3  2 1   0

 8 1 

 8  2

 8  3

=   + 0.05   + 0.0025   + 0.000125   +  = +0.000125

=

8! + 3! (8 − 3)!

=

8! 8! 8! + 0.05 + 0.0025 0! (8 − 0) ! 1! (8 − 1) ! 2! (8 − 2) !

8! 8! 8! 8! + 0.05 + 0.0025 + 0.000125 + 8! 7! 2! 6! 3! 5!

8 ⋅ 7 ⋅ 6/ ! 8 ⋅ 7 ⋅ 6 ⋅ 5/ ! 8 ⋅ 7/ ! 8/ ! +  = 1 + 0.4 + 0.07 + 0.007 +  ≈ 1477 + 0.000125 + 0.0025 + 0.05 . 3! 5/ ! 2! 6/ ! 7/ ! 8/ !

Therefore, (1 + 0.05)8 , to the nearest hundredth, is equal to 1.48 5 5 c. First - Write the exponential expression in the form of (121 . ) = (1 + 0.21) .

Second - Identify the a, b, and n terms, i.e., a = 1 , b = 0.21 , n = 5 . Third - Use the general binomial expansion formula, i.e., equation (1) above to expand (1 + 0.21)5 .

(1 + 0.21)5 = 5

 5 3  5 5  5 4 2  5 2 3  5 4  5 5   1 +   1 ⋅ (0.21) +   1 ⋅ (0.21) +   1 ⋅ (0.21) +   1 ⋅ (0.21) +   ⋅ (0.21)  5  4  2  3 1   0

 5

5

 5

 5

5

5! 5! 5! + 0.21 + 0.0441 =   + 0.21 + 0.0441  + 0.00926  + 0.0019 + 0.0004  = 0 ! ( 5 − 0) ! 1! (5 − 1) ! 2! (5 − 2) !  5  4  3  2 1  0 + 0.00926

=

5! 5! 5! + 0.0019 + 0.0004 3! (5 − 3)! 5! (5 − 4 )! 5! (5 − 5)!

=

5! 5! 5! 5! 5! 5! + 0.21 + 0.0441 + 0.00926 + 0.0019 + 0.0004 5! 4! 2!3! 3!2! 4!1! 5!

5/! 5 ⋅ 4/ ! 5 ⋅ 4 ⋅ 3/! 5 ⋅ 4/ ! 5/! 5 ⋅ 4 ⋅ 3/! + 0.21 + 0.0441 + 0.00926 + 0.0019 + 0.0004 4/ ! 5/! 4/ ! 2!3/! 3/!2! 5/!

= 1+ 1.05+ 0.441 + 0.0926 + 0.0095 + 0.0004

= 2.594 Therefore, (1 + 0.21)5 , to the nearest hundredth, is equal to 2.59 Note that in equation (1) the r th term in a binomial expansion is given by

Hamilton Education Guides

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Calculus I

1.6 The Factorial Notation

 n  n − r +1 r −1 b  a  r − 1

n!

=

( r − 1) ! ( n − r + 1) !

(2)

a n − r +1 b r −1

this implies that we can use the above equation to find any specific term of a binomial. For example, the sixth term of ( x − 3)8 is equal to  8 3 5   x ( −3)  5

=

8! 5 x 3 ( −3) 5! (8 − 5) !

=

8! 3 8 ⋅ 7 ⋅ 6/ ⋅ 5/ ! 3 x ⋅ ( −243) = −243 ⋅ x = −13,608 x 3 5!3! 5/ ! ⋅ 3/ ⋅ 2/ ⋅ 1

and the fourth term of ( x − 3) 4 is equal to  4 4−4+1 3 −3) (   x  3

=

4! 3 x( −3) 3! (4 − 3) !

=

4! 4 ⋅ 3/ ! x ⋅ ( −27) = −27 ⋅ x = −108 x 3!1! 3/ !

Example 1.6-8 Find the stated term of the following binomial expressions. a. The sixth term of ( x + 2)10

b. The eighth term of ( x − y )12

c. The fifth term of ( w − a )13 =

d. The tenth term of ( x + 1) 20

Solutions: a. First - Identify the a, b, r and n terms, i.e., a = x , b = 2 , r = 6 , and n = 10 . Second - Use equation (2) above to find the sixth term of ( x + 2)10 . 10! x10−6+1 ⋅ 2 6−1 (6 − 1)! (10 − 6 + 1)!

=

10 ! 5 5 x 2 5! 5!

=

10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5/ ! 32 x 5 5! 5/ !

=

32 / / ⋅ 9/⋅ 8/⋅ 7 ⋅ 6 10 32 x 5 5/ ⋅ 4/ ⋅ 3/ ⋅ 2/ ⋅ 1

=

6⋅7⋅6 32 x 5 1

= 8064 x 5 b. First - Identify the a, b, r and n terms, i.e., a = x , b = − y , r = 8 , and n = 12 . Second - Use equation (2) above to find the eighth term of ( x − y )12 . 12! x12−8+1 ⋅ b 8−1 (8 − 1)! (12 − 8 + 1)!

=

/ / ⋅9⋅8 5 7 / / ⋅ 11 ⋅ 10 12 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7/ ! 5 7 12! 5 7 x y x (− y) = − x y = − 5/ ⋅ 4/ ⋅ 3/ ⋅ 2/ ⋅ 1 7 ! 5! 7/ ! 5!

= − (11 ⋅ 9 ⋅ 8)x 5 y 7 = −792 x 5 y 7 c. First - Identify the a, b, r and n terms, i.e., a = w , b = − a , r = 5 , and n = 13 . Second - Use equation (2) above to find the fifth term of ( w − a )13 . 13! 5−1 w13−5+1 ( − a ) 5 − 1 ! 13 − 5 + 1 ! ( )( )

=

5 / / ⋅ 11 ⋅ 10 // 9 4 13! 9 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9/ ! 9 4 13 ⋅ 12 4 w a = w (−a) = w a = 4 ! 9/ ! 4! 9! 4/ ⋅ 3/ ⋅ 2/ ⋅ 1

13 ⋅ 11 ⋅ 5 9 4 w a = 715 w 9 a 4 1

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Calculus I

1.6 The Factorial Notation

d. First - Identify the a, b, r and n terms, i.e., a = x , b = 1 , r = 10 , and n = 20 . Second - Use equation (2) above to find the tenth term of ( x + 1) 20 . 20! x 20−10+1 ⋅110−1 (10 − 1)! (20 − 10 + 1)!

=

20! 11 9 x ⋅1 9 !11!

=

/ / ! 11 20 ⋅ 19 ⋅ 18 ⋅ 17 ⋅ 16 ⋅ 15 ⋅ 14 ⋅ 13 ⋅ 12 ⋅ 11 x / /! 9 !11

2 / / ⋅ 17 ⋅ 16 / /⋅ 15 ⋅ 14 / / ⋅ 13 ⋅ 12 11 / / ⋅ 19 ⋅ 18 19 ⋅ 17 ⋅ 2 ⋅ 15 ⋅ 13 ⋅ 12 11 20 x = 167,960 x 11 x = = 9 9 ⋅ 8/ ⋅ 7/ ⋅ 6/ ⋅ 5/ ⋅ 4/ ⋅ 3/ ⋅ 2/ ⋅ 1

Section 1.6 Practice Problems - The Factorial Notation 1. Expand and simplify the following factorial expressions. b. (10 − 3)! =

a. 11! = e.

15! = 8! 4!

f.

10 ! 4 ! (10 − 2) !

=

c.

12 ! = 5!

g.

12 ! 6 ! 14 !

=

d.

14 ! = 10 !

h.

(7 − 3)!9 ! = 12!(7 − 2)!

2. Write the following products in factorial form. a. 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 =

b. 10 ⋅ 11 ⋅ 12 ⋅ 13 ⋅ 14 ⋅ 15 =

c. 22 ⋅ 23 ⋅ 24 ⋅ 25 =

d. 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 =

e. 4 ⋅ 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 =

f. 35 =

3. Expand the following factorial expressions. a. 5( n!) =

b. ( n − 7)! =

c. ( n + 10)! =

d. (5n − 5)! =

e. (2n − 8) ! =

f. (2n + 6)! =

g. (2n − 5)! =

h. (3n + 3)! =

4. Expand and simplify the following factorial expressions. a.

(n − 2)! = (n − 4)!

b.

( n + 4)! =

c.

( n + 5)! = ( n − 2) !

e.

(3n)! (3n − 2)! = (3n + 1)! (3n − 4)!

f.

( n − 1)! = ( n + 2)! ( n !) 2

g.

(2n − 3)! 2( n !) = (2n)! ( n − 2)!

n!

d.

( n − 1)( n + 1)! = ( n + 2)!

5. Write the following expressions in factorial notation form. Simplify the answer.  5  3

b.   =

 5 1 

g. 

a.   = f.   =

10 6 

c.   =

 8  0

n   =  n − 5

h. 

 2n    2n − 1

 8  8

 6  3

d.   = =

 3n    3n − 3

i. 

e.   = =

n    n − 6

j. 

=

6. Expand the following binomial expressions. a. ( x − 2) 4 =

Hamilton Education Guides

b. (u + 2) 7 =

c. ( y − 3)5 =

51

Calculus I

1.6 The Factorial Notation

7. Use the general equation for binomial expansion to solve the following exponential numbers to the nearest hundredth. a. (0.95)5 =

b. (2.25) 7 =

c. (1.05)4 =

8. Find the stated term of the following binomial expressions. a. The eighth term of ( x + 3)12

b. The ninth term of (x − y )10

c. The seventh term of (u − 2a )11

d. The twelfth term of ( x − 1)18

Hamilton Education Guides

52

Calculus I

Quick Reference to Chapter 2 Problems

Chapter 2

Differentiation (Part I) Quick Reference to Chapter 2 Problems 2.1

The Difference Quotient Method .............................................................................. 54 lim h→0

2.2

1 x + h +1 + x +1

1

(θ + 1)

3

3

x+3

h

=

 2 + x3    = ;  x +1 

f ( x) =

2 x 2 + 3x + 1 x2 +1

=

d Notation .......................................................... 71 dx

[

)] = ;

(

u2  d  u   = + du  1 − u 1 + u 

2

  t3 + t 2  h( t ) =  4   t −1

=;

(

 f ( x) =  x 3 + 2x 

)

2

 − x2 

4

=

= ; xy + x 2 y 2 + y 3 = 10 x = ; 3x 3 y 3 + 2 y 2 = y + 1 =

The Derivative of Functions with Fractional Exponents........................................ 102

( )

1

2 3

+ (2 x + 1)

3 5

=;

y = ( x + 1)

1 2

( x + 3) 2

1 3

1

=;

y=

x 3 (x − 1)

=

1 + x2

The Derivative of Radical Functions ........................................................................ 109 r (θ= ) θ2 +

2.8

(x + h ) + 3 −

Implicit Differentiation .............................................................................................. 97

y= x

2.7

2 x

d ( x + 1) x 2 − 3 dx

=;

=;

x 2 y 2 + y = 3y 3 − 1

2.6

= ; lim h→0

The Chain Rule .......................................................................................................... 82  1  f ( x ) =  2 + x x 

2.5

f ( x) =

=;

Differentiation Rules Using the d  2x   5x + 2  dx  x + 1

2.4

4(x + h )2 − 4 x 2 h

Differentiation Rules Using the Prime Notation ..................................................... 59 r (θ ) = θ 2 +

2.3

= ; lim h→0

1

θ3 +1

(θ + 1) = ; r(θ ) = 3

θ 2 +1

= ; h(t ) =

3 2

t +1 t3

=

Higher Order Derivatives.......................................................................................... 124 f (u) =

Hamilton Education Guides

u3 − 1 u +1

= ; r (θ ) = θ 2 +

1

(θ + 1) 3

=;

f (t) =

t 3 + t 2 +1 10

=

53

Chapter 2 - Differentiation The objective of this chapter is to improve the student’s ability to solve problems involving derivatives. In Section 2.1 derivatives are computed using the Difference Quotient method. Various differentiation rules using the prime and d notation are introduced in Sections 2.2 and dx

2.3, respectively. The use of the Chain Rule in finding the derivative of functions that are being added, subtracted, multiplied, and divided are addressed in Section 2.4. Implicit differentiation is discussed in Section 2.5. Finding the derivative of exponential and radical expressions is addressed in Sections 2.6 and 2.7, respectively. Finally, computation of second, third, fourth, or higher order derivatives are discussed in Section 2.8. Each section is concluded by solving examples with practice problems to further enhance the student’s ability.

2.1

The Difference Quotient Method

In this section students learn how to differentiate functions using the difference quotient equation as the limit h approaches zero. The expression

f ( x + h) − f ( x ) h

is referred to as the difference

quotient equation. A function f ( x ) is said to be differentiable at x if and only if lim h→0 f ( x + h) − f ( x ) h

exists. If the limit exists, then the result is referred to as the derivative of f ( x ) at

which is denoted by f ′( x ) . It should be noted that this approach is rather long and time consuming and is merely presented in order to show the usefulness of the differentiation rules which are addressed in the subsequent sections. The following examples show the steps in finding derivatives of functions using the Difference Quotient method: x

Example 2.1-1: Use the Difference Quotient method to find the derivative of the following functions. a. f ( x ) = 3x + 1

b. f ( x ) = 4 x 2

c. f ( x ) = x + 3

d. f ( x ) = (2 x − 7) 2

e. f ( x ) = − x + 1

f. f ( x ) =

Solutions:

1 x +1

a. To find the derivative of the function f (x ) = 3x + 1 First - Substitute f (x + h ) = 3(x + h ) + 1 = 3x + 3h + 1 and f (x ) = 3x + 1 into the difference quotient equation, i.e.,

f (x + h ) − f (x ) h

=

(3x + 3h + 1) − (3x + 1)

3h/ 3/ x + 3h + 1/ − 3/ x/ − 1/ = / = = 3 h

h

h/

Second - Compute f ′(x ) as the lim h→0 in the difference quotient equation, i.e., f ′( x )

= lim h→0

f (x + h ) − f (x ) h

= lim h→0 3 = 3

b. To find the derivative of the function f (x ) = 4x 2

(

)

First - Substitute f (x + h ) = 4(x + h )2 = 4 x 2 + h 2 + 2 xh = 4 x 2 + 4h 2 + 8 xh and f (x ) = 4x 2 into the Hamilton Education Guides

54

Calculus I

2.1 The Difference Quotient Method

f (x + h ) − f (x ) h

difference quotient equation, i.e., =

4h/ (h + 2 x ) h/

4(h + 2 x ) 1

=

=

( 4x

2

)

+ 4h 2 + 8 xh − 4 x 2 h

=

4/ x/ 2/ + 4h 2 + 8 xh − 4 x/ 2/ h

= 4h + 8 x

Second - Compute f ′(x ) as the lim h→0 in the difference quotient equation, i.e., f ′( x )

f (x + h ) − f (x ) h

= lim h→0

= lim h→0 (4h + 8 x ) = (4 ⋅ 0) + 8 x = 0 + 8 x = 8 x

c. To find the derivative of the function f ( x ) = x + 3 First - Substitute f (x + h ) = (x + h ) + 3 = f (x + h ) − f (x ) h

equation, i.e.,

x+h+3

and f (x ) = x + 3 into the difference quotient

x+h+3 − x+3 h

=

To remove the radical from the numerator multiply both the numerator and the denominator by x + h + 3 + x + 3 to obtain the following: f (x + h ) − f (x ) h

=

x+h+3 − x+3 h

=

x+h+3 − x+3 x+h+3 + x+3 ⋅ h x+h+3 + x+3

1

=

x+h+3 + x+3

Second - Compute f ′(x ) as the lim h→0 in the difference quotient equation, i.e., f ′( x )

= lim h→0

f (x + h ) − f (x ) h

= lim h→0

1 x+h+3 + x+3

=

1 x+3 + x+3

=

1 2 x+3

d. To find the derivative of the function f (x ) = (2 x − 7 )2 First - Substitute f (x + h ) = [ 2 (x + h ) − 7 ] 2 = 4 (x + h ) 2 + 49 − 28 (x + h ) = 4 x 2 + 4h 2 + 8 xh + 49 − 28 x − 28h and f (x ) = (2 x − 7 ) 2 = 4 x 2 + 49 − 28 x into the difference quotient equation, i.e., = =

(4 x

2

) (

+ 4h 2 + 8 xh + 49 − 28 x − 28h − 4 x 2 + 49 − 28 x h

4h 2 + 8 xh − 28h h

=

h/ (4h + 8 x − 28) h/

)=

f (x + h ) − f (x ) h

4/ x/ 2/ + 4h 2 + 8 xh + 4/ 9/ − 2/ 8/ x/ − 28h − 4/ x/ 2/ − 4/ 9/ + 2/ 8/ x/ h

= 4h + 8 x − 28

Second - Compute f ′(x ) as the lim h→0 in the difference quotient equation, i.e., f ′( x )

= lim h→0

f (x + h ) − f (x ) h

= lim h→0 (4h + 8 x − 28) = (4 ⋅ 0) + 8 x − 28 = 0 + 8 x − 28 = 8 x − 28

e. To find the derivative of the function f (x ) = − x + 1 First - Substitute f (x + h ) = − (x + h ) + 1 = − x + h + 1 and f (x ) = − x + 1 into the difference Hamilton Education Guides

55

Calculus I

2.1 The Difference Quotient Method

quotient equation, i.e.,

f (x + h ) − f (x ) h

x + h +1 + x +1 h

= −

To remove the radical from the numerator multiply both the numerator and the denominator by x + h + 1 − x + 1 to obtain the following: f (x + h ) − f (x ) h

x + h +1 + x +1 h

= −

x + h +1 + x +1 x + h +1 − x +1 ⋅ h x + h +1 − x +1

= −

1

= −

x + h +1 + x +1

Second - Compute f ′(x ) as the lim h→0 in the difference quotient equation, i.e., f ′( x )

= lim h→0

f (x + h ) − f (x ) h

= − lim h→0

1 x + h +1 + x +1

f. To find the derivative of the function f (x ) = 1

First - Substitute f (x + h ) =

f (x + h ) − f (x ) h

i.e.,

x + h +1

1

=

x + h +1 h

−

= −

1

= −

x +1 + x +1

1 2 x +1

1 x +1

and f (x ) =

1 x +1

into the difference quotient equation,

1 x +1

=

x +1 − x + h +1 h x +1 ⋅ x + h +1

To remove the radical from the numerator multiply both the numerator and the denominator by x + 1 + x + h + 1 to obtain the following: f (x + h ) − f (x ) h

=

=

x +1 − x + h +1 h x +1 ⋅ x + h +1

⋅

x +1 + x + h +1 x +1 + x + h +1

=

(x + 1) − (x + h + 1) h (x + 1) x + h + 1 + h (x + h + 1)

x/ + 1/ − x/ − h − 1/ −h/ = h (x + 1) x + h + 1 + h (x + h + 1) x + 1 h/ (x + 1) x + h + 1 + (x + h + 1) x + 1

[

]

=

(x + 1)

x +1 −1

x + h + 1 + (x + h + 1) x + 1

Second - Compute f ′(x ) as the lim h→0 in the difference quotient equation, i.e., f ′( x )

= −

= lim h→0

f (x + h ) − f (x ) h

= lim h→0

(x + 1)

−1

x + h + 1 + (x + h + 1) x + 1

=

(x + 1)

−1

x + 1 + (x + 1)

x +1

1

2 ( x + 1)

x +1

Example 2.1-2: Given the derivative of the functions in example 2.1-1, find: a. f ′(2)

b. f ′(3)

c. f ′(1)

d. f ′(0)

e. f ′(15)

f. f ′(0)

Solutions: a. Given f ′(x ) = 3 then, f ′(2) = 3 Hamilton Education Guides

56

Calculus I

2.1 The Difference Quotient Method

Note that since the derivative is constant f ′(x ) is independent of the x value. f ′(2) can also f (x + h ) − f (x ) h

be calculated directly by using f ′(x ) = lim h→0

[ 3 (2 + h ) + 1] − (6 + 1)

and by replacing x with 2 , i.e., f ′(2) = lim h→0 = lim h→0

3h/ h/

= lim h→0

h

[ 3 (x + h ) + 1] − (3x + 1) h

= lim h→0

6 + 3h + 1 − 7 h

= lim h→0 3 = 3

b. Given f ′(x ) = 8 x then, f ′(3) = 8⋅ 3 = 24 4 (x + h ) 2 − 4 x 2 f (x + h ) − f (x ) = lim h→0 h h

f ′(3) can also be calculated directly by using f ′(x ) = lim h→0

and by replacing x with 3 , i.e., f ′(3) = lim h→0

= lim h→0

(

)

4 9 + h 2 + 6h − 36 4 ( 3 + h) 2 − 4 ⋅ 32 = lim h→0 h h

h/ (4h + 24 ) 36 + 4h 2 + 24h − 36 = lim h→0 = lim h→0 4h + 24 = 0 + 24 = 24 h h/

c. Given f ′(x ) =

1 2 x+3

then, f ′( 1) =

1

=

2 1+ 3

1 2 4

=

1 2⋅2

=

1 4

Again, f ′( 1) can also be calculated directly by using the equation f ′(x ) = lim h→0 = lim h→0 = lim h→0

=

(x + h ) + 3 −

x+3

h

h+4 − 4 h

1 0+4 +2

=

1 4 +2

and by replacing x with 1 , i.e., f ′( 1) = lim h→0 h+4 − 4 h+4 + 4 ⋅ h h+4 + 4

= lim h→0

=

1 2+2

=

= lim h→0

h

(

h h+4 +2

)

f (x + h ) − f (x ) h

(1 + h ) + 3 −

1+ 3

h

= lim h→0

1 h+4 +2

1 4

d. Given f ′(x ) = 8 x − 28 then, f ′(0) = (8 × 0) − 28 = −28 e. Given f ′(x ) = − f. Given f ′(x ) = −

1 2 x +1

then, f ′(15) = −

1 2

(x + 1)3

1 2 15 + 1

then, f ′(0) = −

= −

1 2 (0 + 1)3

1 2 16

= −

= −

1 2 ⋅1

1 2⋅4

= −

= −

1 8

1 2

In problems 2.1-2 d, e, and f students may want to practice finding f ′(x ) for the specific values of

Hamilton Education Guides

57

Calculus I

x

2.1 The Difference Quotient Method

by using the general equation f ′(x ) = lim h→0

above stated solutions.

f (x + h ) − f (x ) . h

The answers should agree with the

Finally, it should be noted that every differentiable function is continuous. However, not every continuous function is differentiable. The proof of this statement is beyond the scope of this book and can be found in a calculus book. In the following section we will learn simpler methods of finding derivative of functions using various differentiation rules. Section 2.1 Practice Problems – The Difference Quotient Method 1. Find the derivative of the following functions by using the Difference Quotient method. a. f (x ) = x 2 − 1

b. f (x ) = x 3 + 2 x − 1

c. f (x ) =

x x −1

e. f (x ) = 20 x 2 − 3

f. f (x ) = x 3

g. f (x ) =

10 x−5

d. f (x ) = − h. f (x ) =

1 x2

ax + b cx

2. Compute f ′(x ) for the specified values by using the difference quotient equation as the lim h→0 . a. f (x ) = x 3 at x = 1

b. f (x ) = 1+ 2 x at x = 0

c. f (x ) = x 3 + 1 at x = −1

d. f (x ) = x 2 (x + 2) at x = 2

e. f (x ) = x −2 + x −1 + 1 at x = 1

f. f (x ) = x + 2 at x = 10

Hamilton Education Guides

58

Calculus I

2.2

2.2 Differentiation Rules Using the Prime Notation

Differentiation Rules Using the Prime Notation

In the previous section we differentiated several functions by writing the difference quotient equation and taking the limit as h approaches to zero. This process, however - as was mentioned earlier, is rather long and time consuming. Instead, we can establish some general rules that make the calculation of derivatives simpler. These rules are as follows: Rule No. 1 - The derivative of a constant function is equal to zero, i.e., if f (x ) = k ,

then f ′( x ) = 0

For example, the derivative of the functions f ( x ) = 10 , g( x ) = −100 , and s( x ) = 250 is equal to f ′( x ) = 0 , g ′( x ) = 0 , and s ′( x ) = 0 , respectively. Rule No. 2 - The derivative of the identity function is equal to one, i.e., if f ( x ) = x ,

then f ′( x ) = 1

For example, the derivative of the functions f ( x ) = 5x , g( x ) = −10 x , and s( x ) = 5x is equal to f ′(x ) = 5 ⋅1 = 5 , g ′(x ) = −10 ⋅1 = −10 ,

and s ′(x ) = 5 ⋅1 = 5 , respectively.

Rule No. 3A - The derivative of the function f ( x ) = x n is equal to f ′( x ) = n x n −1 , where n is a positive integer. For example, the derivative of the functions f ( x ) = x , f ( x ) = x 2 , f ( x ) = x 3 , and f ( x ) = x 4 is equal to f ′( x ) = 1 ⋅ x1−1 = x 0 = 1 , f ′( x ) = 2 ⋅ x 2−1 = 2 x1 = 2 x , f ′( x ) = 3 ⋅ x 3−1 = 3x 2 , and f ′(x ) = 4 ⋅ x 4−1 = 4x 3 , respectively. Rule No. 3B - The derivative of the function f ( x ) = x n is equal to f ′( x ) = n x n −1 , where n is a negative integer. For example, the derivative of the functions f ( x ) = x −1 , f ( x ) = x −2 , f ( x ) = x −3 , f ( x ) = x −4 , and −

1

f ( x) = x 8

is equal to f ′( x ) = −1 ⋅ x −1−1 = − x −2 , f ′( x ) = −2 ⋅ x −2−1 = −2 x −3 , f ′(x ) = − 3 ⋅ x −3−1 = −3x −4 , 1 8

and f ′( x ) = − ⋅ x

1 − −1 8

9

1 − =− x 8, 8

respectively.

Note that this rule can also be used to obtain the derivative of functions that are in the form of 1 f ( x) = n x

by rewriting the function in its equivalent form of f ( x ) = x − n . For example, the

derivative of the functions f (x ) = is equal to f ′(x ) = −8 ⋅ x

−8−1

f ′( x ) = −1 ⋅ x = −8 x

−9

−1−1

= −x

2 1 1 = −2 x −3 , = x − 2 , f (x ) = − = x −1 , f (x ) = 3 2 x x x −2

,

f ′( x ) = −2 ⋅ x

−2−1

= −2 x

−3

,

and f (x ) =

f ′( x ) = ( −2 ⋅ −3) ⋅ x

−3−1

= 6x

1 x

8

−4

= x −8

, and

, respectively.

Rule No. 4 - If the function f ( x ) is differentiable at x , then a constant k multiplied by f ( x ) is also differentiable at x , i.e., ( k f ) ′ x = k f ′( x )

Hamilton Education Guides

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Calculus I

2.2 Differentiation Rules Using the Prime Notation

Note that this rule is referred to as the scalar rule. For example, the derivative of the functions f ( x ) = 5x , g( x ) = −10 x , and s( x ) = 5x is equal to f ′( x ) = 5 , g ′( x ) = −10 , and s ′( x ) = 5 , respectively. Rule No. 5 - If the function f ( x ) and g( x ) are differentiable at x , then their sum is also differentiable at x , i.e.,

( f + g )′ (x ) =

f ′( x ) + g ′( x )

In other words, the derivative of the sum of two differentiable functions, ( f + g )' (x ) , is equal to the derivative of the first function, f ' (x ) , plus the derivative of the second function, g ' (x ) . Note that this rule is referred to as the summation rule.

)

(

For example, the derivative of the functions h( x ) = (5x − 3) + 2 x 2 − 1 and s( x ) = 6 x 3 + (3x + 2) is        f ( x)

f ( x)

g( x )

    g( x )

equal to h ′( x ) = f ′( x ) + g ′( x ) = 5 + 4 x and s ′( x ) = f ′( x ) + g ′( x ) = 18 x 2 + 3 . Rule No. 6 - If the function f ( x ) and g( x ) are differentiable at x , then their product is also differentiable at x , i.e., ( f ⋅ g )′ (x ) = f ′( x ) g( x ) + g ′( x ) f ( x ) In other words, the derivative of the product of two differentiable functions, ( f ⋅ g )' (x ) , is equal to the derivative of the first function multiplied by the second function, f ' (x ) ⋅ g (x ) , plus the derivative of the second function multiplied by the first function, g ' (x ) ⋅ f (x ) . Note that this rule is referred to as the product rule.

(

)

For example, the derivative of the functions f ( x ) = (3x − 5)(6 x + 1) and g( x ) = −10 x 2 5x 3 − 2 is equal to f ′( x ) = [ 3 ⋅ (6 x + 1) ] + [ 6 ⋅ (3x − 5) ] = 18 x + 3 + 18 x − 5 = (18 x + 18 x ) + (− 5 + 3) = 36 x − 2 and

[

)] [

(

]

g ′( x ) = − 20 x ⋅ 5 x 3 − 2 + 15 x 2 ⋅ −10 x 2 = −100 x 4 + 40 x − 150 x 4 = −250 x 4 + 40 x

Rule No. 7 - Using the rules 1 , 4 , 5 , and 6 we can write the formula for differentiating polynomials, i.e., if f ( x ) = a n x n + a n−1 x n−1 + a n−2 x n−2 +  + a3 x 3 + a2 x 2 + a1 x1 + a0 , then f ′( x )

= n a n x n−1 + ( n − 1) a n−1 x n−2 + ( n − 2) a n−2 x n−3 +  + 3a3 x 2 + 2a2 x + a1

For example, the derivative of the polynomials f ( x ) = 6 x 4 + 5x 3 − 3 , g( x ) = 2 x 5 − 3x 2 − 4 x , and h( x ) =

1 2 x − 2x + 5 3

is equal to f ′( x ) = (6 ⋅ 4) x 4−1 + (5 ⋅ 3) x 3−1 = 24 x 3 + 15x 2 , g ′( x ) = (5 ⋅ 2) x 4 − (2 ⋅ 3) x − 4

2 1 = 10x 4 − 6 x − 4 , and h′( x ) =  2 ⋅  x − 2 = x − 2 , respectively. 

3

3

Rule No. 8 - If the function f ( x ) and g( x ) are differentiable at x , then their quotient is also differentiable at x , i.e.,

Hamilton Education Guides

60

Calculus I

2.2 Differentiation Rules Using the Prime Notation

′  f   ( x)  g

f ′(x )g (x ) − g ′(x ) f (x )

=

[ g (x ) ] 2

'

 f    ( x ) , g

In other words, the derivative of the quotient of two differentiable functions,

is equal to

the derivative of the function in the numerator multiplied by the function in the denominator, f ' ( x ) ⋅ g ( x ) , minus the derivative of the function in the denominator multiplied by the function in the numerator, g ' (x ) ⋅ f (x ) , all divided by the square of the denominator, [ g (x ) ] 2 . Note that this rule is referred to as the quotient rule. For example, the derivative of the functions f ( x ) =

[ 3 ⋅ (1 + x ) ] − [ 1⋅ (1 + 3x ) ] = = (1 + x )2 =

6 x 4 + 6 x − 9 x 4 − 15x 2

( x + 1) 3

2

=

3 + 3x − 1 − 3x

(1 + x ) 2

=

1 + 3x 1+ x

and g( x ) =

3x 2 + 5 x3 +1

is equal to f ′(x )

[ 6x ⋅ (x + 1) ]− [ 3x ⋅ (3x + 5) ] and g ′(x ) = (x + 1) 3

2

(1 + x ) 2

2

3

2

2

−3x 4 − 9 x 2

( x + 1) 3

2

In the following examples the above rules are used in order to find the derivative of various functions: Example 2.2-1: Differentiate the following functions. a. f ( x ) = 5 + x

b. f ( x ) = x 3 + 3x − 1

d. f ( x ) = 10 x 3 + 5x 2 + 5

e. f (x ) = 3 x 2 + 2 x

g. f ( x ) = ax 2 + b

h. f ( x ) =

1 1 − x x3

j. f ( x ) =

m. f ( x ) =

x4 1− x

p. f ( x ) =

2 x

k.

 2 + x3     x +1 

v. f ( x ) =

x 

)

x5 x4 x3 x2 − + − 10 4 6 4

f ( x ) =x −5 + 3 x −3 − 2 x −1 + 10

x 4 + 10

n. f ( x ) =

2

x +1

(

)

q. f ( x ) = x 2 + 1 ( x + 5)

1  1  s. f (x ) = 1 +  1 − 3  

(

x 

2 x 2 + 3x + 1 x2 +1

 3x 2 + 5   ( x + 1)  x 

t. f ( x ) =  w. f ( x ) =

ax 2 + bx + c ax 2 − b

Solutions:

c. f ( x ) = 2 − x f. f ( x ) = ax 3 + bx 2 + c i. f ( x ) = l. f ( x ) = o. f ( x ) =

10 x2

x2 + 2 x4

3+ x

x3 − 5

r. f ( x ) = ( x + 1)( x + 2) u. f ( x ) = x. f ( x ) =

1+ x2 1− x2 3− x 1 −5 x

a. Given f ( x )= 5 + x then f ′( x ) = 0 + x1−1 = x0 = 1

Hamilton Education Guides

61

Calculus I

2.2 Differentiation Rules Using the Prime Notation

b. Given f ( x ) = x3 + 3x − 1 then f ′ ( x ) = 3x3−1 + 3 ⋅ x1−1 − 0 = 3x 2 + 3 ⋅ x0 = 3x 2 + 3 ⋅1 = 3 x 2 + 3 c. Given f ( x )= 2 − x then f ′ ( x ) = 0 − x1−1 = − x0 = −1 d. Given f ( x ) = 10 x3 + 5 x 2 + 5 then f ′ ( x ) = 10 ⋅ 3x3−1 + 5 ⋅ 2 x 2−1 + 0 = 30 x 2 + 10 x e. Given f = ( x ) 3( x 2 + 2 x ) then f ′ ( x ) = 3( 2 x 2−1 + 2 ⋅ x1−1 ) = 3 ( 2 x + 2 ⋅ x 0 ) = 3( 2 x + 2 ⋅ 1) = 6 x + 6 f. Given f ( x ) = ax3 + bx 2 + c then f ′ ( x ) = a ⋅ 3x3−1 + b ⋅ 2 x 2−1 + 0 = 3ax 2 + 2bx x ) ax 2 + b then f ′ ( x ) = a ⋅ 2 x 2−1 + 0 = 2ax g. Given f (= x5 x 4 x3 x 2 1 1 1 1 h. Given f ( x ) = − + − then f ′ ( x ) = ⋅ 5 x5−1 − ⋅ 4 x 4−1 + ⋅ 3x3−1 − ⋅ 2 x 2−1 10

=

=

4

10

x4 x3 x2 x + − − 2 1 2 2

or, using the quotient rule we obtain

6

5 4 4 3 3 2 2 x − x + x − x 10 4 6 4

f ′( x )

=

4

=

4

6

4

(5x5−1 ⋅10) − ( 0 ⋅ x5 ) − ( 4 x4−1 ⋅ 4) − ( 0 ⋅ x4 ) + (3x3−1 ⋅ 6) − ( 0 ⋅ x3 ) − ( 2 x2−1 ⋅ 4) − ( 0 ⋅ x2 ) 102

42

50 x 4 16 x3 18 x 2 8 x − + − 100 16 36 16

62

42

x4 x3 x2 x − − + 2 1 2 2

=

(

) ( ( )

)

0 ⋅ x 2 − 2 x 2−1 ⋅10 0 − ( 2 x ⋅10 ) 20 10 −20x i. Given f ( x ) = 2 then f ′ ( x ) = = = 4/ =3/ = − 3 2 4 x x x x x2 3 3−1 1 3 ( 0 ⋅ x ) − (1 ⋅1) − ( 0 ⋅ x ) − ( 3x ⋅1) 1 1 1 3x 2/ − 2+ 4 − + j. Given f ( x )= − 3 then f ′ ( x ) = = 2 2 2 6/ =4 =

x

x

( x3 )

x

x

x

x

x

k. Given f ( x ) =x −5 + 3x −3 − 2 x −1 + 10 then f ′ ( x ) = −5 x −5−1 + 3 ⋅ −3x −3−1 − 2 ⋅ −1x −1−1 + 0 = −5 x −6 − 9 x −4 + 2 x −2

1 3 2 or, we can rewrite f ( x ) as f ( x ) = 5 + 3 − + 10 and then find f ′( x ) x

using the quotient rule. f ′( x )

=

( 0 ⋅ x ) − ( 5 x ⋅1) + ( 0 ⋅ x ) − ( 3x ⋅ 3) − ( 0 ⋅ x ) − (1 ⋅ 2 ) + 0 = x (x ) (x ) 5

4

5 2

Hamilton Education Guides

3

2

3 2

2

−

5x 4 x10

x

x

−

9x 2 x6

+

2 x2

5 x 4/ 9 x 2/ 2 − + 2 = − 10 / / =6 6/ =4 x

x

x

62

Calculus I

2.2 Differentiation Rules Using the Prime Notation

5 9 2 = − 6 − 4 + 2 = −5 x −6 − 9 x −4 + 2 x −2 x

x

x

x2 + 2 l. Given f ( x ) = 4 then f ′ ( x ) = x

=

2 x5 − 4 x5 − 8 x3 x8

m. Given

=

=

x4 f ( x) = 1− x

4 x3 − 3 x 4

(1 − x )

2

−2 x5 − 8 x3 x8

then

[ ( 2x (

−2 x3/ x 2 + 4

=

8/ =5

(1 − x )

4 x5 + 4 x3 − 2 x5 − 20 x

( x2 + 1)

2

p.

x3 − 5 − 9 x 2 − 3 x3 2

Given f ( x ) =

(

)]

[ 2x ⋅ x ]− [ 4x ⋅ ( x = 4

[ 4x f ′( x ) =

4 −1

(

x

5

+2

)]

][

⋅ (1 − x ) − (0 − 1) ⋅ x 4

)

]

=

(1 − x ) 2

4 x3 (1 − x ) + x 4

(1 − x )

2

=

4 x3 − 4 x 4 + x 4

(1 − x )

2

2

[ ( 4x

=

=

) ] [ ( 2x (x + 1)

)(

+ 0 x 2 +1 −

( x + 1)

2

=

(

2 −1

2 x x 4 + 2 x 2 − 10

(x

2

+1

)

[ (0 + 1) ( x − 5) ]− [ ( 3x ( x − 5) 3

2

3−1

3

)(

+ 0 x 4 + 10

2

2

2 x 5 + 4 x 3 − 20 x 2

4 −1

)]

=

(

) ( ( x + 1)

4 x 3 x 2 + 1 − 2 x x 4 + 10

)

− 0 ( 3 + x)

2

]

=

( x3 − 5 ) − 3 x 2 ( 3 + x ) 2 ( x3 − 5 )

− 2x 3 − 9x 2 − 5

(x

3

−5

)

2

(

)

(

3−1  ( x + 1)  − (1 + 0 ) 2 + x3  ( 0 ⋅ x ) − (1 ⋅ 2 )   2 + x3    0 + 3x 2  2 + x3  +     then f ′ ( x ) =   2   x  x + 1  x2    x +1   ( x + 1)

(

 3x 2 ( x + 1)  − 2 + x3    2  ( x + 1) 

)   

)  2     x  

3 3 2 2  2 + x  2  2 x + 3x − 2  2  2 + x3  2  3x3 + 3x 2 − x3 − 2  − + +   = − 2  =  2  2 2  x  x + 1  x  x  x + 1  x  ( x + 1) ( x + 1)   

q. Given f ( x ) =( x 2 + 1) ( x + 5) then f ′ ( x ) =

Hamilton Education Guides

2

2

)



=

2

x8

− 2 x2 + 4

=

3

4 x3 − 3 x4

=

( x3 − 5 )

)

x

3+ x o. Given f ( x ) = 3 then f ′ ( x ) = x −5

=

) ][ (x )

+ 0 ⋅ x 4 − 4 x 4−1 ⋅ x 2 + 2 4 2

x 4 + 10 n. Given f ( x ) = 2 then f ′ ( x ) = x +1

=

2 −1

(

)

(

)

 2 x 2−1 + 0 ( x + 5 )  + (1 + 0 ) x 2 + 1     

= 2 x ( x + 5) + ( x 2 + 1)

63

)

Calculus I

2.2 Differentiation Rules Using the Prime Notation

= 2 x 2 + 10 x + x 2 + 1 = 3 x 2 + 10 x + 1 A second method would be to multiply the binomial terms by one another, using the FOIL method, and then taking the derivative of f (x ) as follows: f (x )

(

)

2

= x 2 + 1 (x + 5) = x 3 + 5 x 2 + x + 5 then f ′ ( x ) = 3x 3−1 + (5 ⋅ 2) x 2−1 + x1−1 + 0 = 3 x + 10 x + 1

r. Given f ( x ) =( x + 1)( x + 2 ) then f ′ ( x ) = (1 + 0 )( x + 2 )  + (1 + 0 )( x + 1)  = ( x + 2 ) + ( x + 1) = 2 x + 3 An alternative way is by multiplying the terms in parenthesis together and then taking the derivative of the product, i.e., f ( x)

= ( x + 1)( x + 2 ) = x 2 + 2 x + x + 2 then f ′ ( x ) = 2 x 2−1 + 2 ⋅1 + 1 + 0 = 2 x + 2 + 1 = 2 x + 3

1   1 s. Given f (x ) = 1 +  1 − 3  then f ′ ( x ) = x



x 

 ( 0 ⋅ x ) − (1 ⋅ 1)  1 − 1  + 0 − ( 0 ⋅ x3 ) − ( 3x3−1 ⋅ 1)  1 + 1  0 +     x2 x3   x6 x     

1 1 3x 2 1 1 1 3x 2 3x 2 1 1 3x 2/ 3x 2/ = − 2 1 − 3  + 6 1 +  = − 2 + 5 + 6 + 7 = − 2 + 5 +=6/ 4 + 7/ =5 x 

x 

x

x 

x

x

x

x

x

x

x

x

4 3 1 1 1 3 3 = − 2 + 5+ 4 + 5 = 5+ 4 − 2 x

x

x

x

x

x

x

A perhaps simpler way is to write f (x ) in the following form: f ′( x )



1 

= 1 +  1 −  x

[



(

1   x3 

(

)(

(

)]

= 1 + x −1 1 − x −3

)] [

= − x − 2 ⋅ 1 − x −3 + 3x −4 ⋅ 1 + x −1 = 4 x −5 + 3x −4 − x −2 =

= f ( x) t. Given

−1−1

)(



−3

)

−3−1

−1

= − x −2 + x −5 + 3x −4 + 3x −5

then

2−1

)][ (

+ 0 x − 1⋅ 3x 2 + 5 x2

) ]  (x + 1) + 1⋅  3x   

  

+ 5   x  

2

 3 x 2 − 5    3 x2 + 5  x + 1 ( ) =  +   2 x      x 

[ ( 0 + 2 x )(1 − x ) ]− [ ( 0 − 2 x )(1 + x ) ] = (1 − x ) 2 −1

f ′( x )

[ ( 3 ⋅ 2x

 

  2  6 x 2 − 3 x 2 − 5   (x + 1) +  3 x + 5   x    x2    

Hamilton Education Guides

(

[ ( 0 − x )⋅ (1 − x ) ]+ [ ( 0 + 3x )⋅ (1 + x ) ]

= − x −2 + x −2−3 + 3x −4 + 3x −4−1

 then f ′ ( x ) = 

= 

u. Given

then f ′ ( x ) =

4 3 1 + − x5 x4 x2

 3x 2 + 5    ( x + 1) a  x  

1 + x2 f ( x) = 1 − x2

)

2

2 −1

2 2

2

=

(

)

(

2 x 1 − x2 + 2 x 1 + x2

(1 − x2 )

2

64

)

Calculus I

=

2.2 Differentiation Rules Using the Prime Notation

2x − 2x 3 + 2x + 2x 3 4

=

2

x − 2x + 1

(2 x + 2 x ) − 2/ x 3/ + 2/ x 3/

(

1− x

=

)

2 2

[ ( 4x

2 x 2 + 3x + 1 v. Given f ( x ) = then f ′ ( x ) = 2

2−1

4x

(1 − x )

2 2

) ] [ ( 2x (x + 1)

)(

+ 3 + 0 x 2 +1 −

x +1

[ (4x + 3) (x + 1) ]− [ 2x ( 2x = (x + 1) 2

2

)]

+ 3x + 1

2

2

ax 2 + bx + c a then f ′ ( x ) = w. Given f ( x ) = 2

4/ x/ 3/ + 4 x + 3 x 2 + 3 − 4/ x/ 3/ − 6 x 2 − 2 x

(x + 1) 2

[ ( 2ax

2−1

2

− abx 2 − 2abx − 2acx − b 2

( ax

x. Given

2

−b

)

2

3− x f ( x) = a 1 −5 x

+ bx + c

)]

2

2

=

2

=

1 3 1 − + 2 − +5 x x x 2 1   − 5 x 

( ax

2

−b

( ax

( ax

2

−b

)

+1

)

)

2

)]

2

2

−b

)

2

2

  1     ( 0 ⋅ x ) − (1 ⋅1)   (3 − x )  −1 ⋅  x − 5   −   2      x    1   − 5 x  

=

x/ 5 x 2 − 2 x + 3 5 x2 − 2 x + 3 5 x3 − 2 x 2 + 3 x = = 25 x 2 − 10 x + 1 25 x3 − 10 x 2 + x x/ 25 x 2 − 10 x + 1

=

)

3 2 − +5 x2 x 1 10 − + 25 x2 x

=

2

3x − 2 x 2 5 + 1 x3 x − 10 x 2 25 + 1 x3

=

(

(

2

− abx 2 − 2a (b + c ) x − b 2

then f ′ ( x ) =

(

(x

2/ a/ 2/ x/ 3/ − 2abx + abx 2 − b 2 − 2/ a/ 2/ x/ 3/ − 2abx 2 − 2acx

=

1 3− x − + 2 +5 x x 2 1  − 5   x 

=

− 3x 2 + 2x + 3

+ b ax 2 − b − 2ax 2−1 ax 2 + bx + c

ax − b

[ (2ax + b) (ax − b) ] − [ 2ax ( ax = ( ax − b)

=

2

)] [

)(

)]

)(

+ 0 2 x 2 + 3x + 1

2

2

=

2−1

=

=

 1  1  − + 5  + 2 (3 − x )  x  x 1   − 5 x  

2

3 x − 2 x 2 + 5 x3 x3 x − 10 x 2 + 25 x3 x3

)

Example 2.2-2: Find f ′( 0) , f ′( 1) , and f ′(−2) for the following functions. a. f (x ) = (x + 5) x 2

b. f ( x ) = 3x 2 + 1

d. f ( x ) = x −1 ( x + 2)

e. f ( x ) = x 3 + 2 x +

Hamilton Education Guides

c. f ( x ) = x −5 − 2 x −4 − 3x −2 + 1 1 x

f. f ( x ) = x 2 ( x + 1)

65

Calculus I

2.2 Differentiation Rules Using the Prime Notation

g. f ( x ) =

x2 + 4

(

3x + 1

)

i. f (x ) = x 2 + 3 (x − 1)

h. f ( x ) = x 5 − 2 x 2 + 3x + 10

2

Solutions: a. Given f (x ) = (x + 5) x 2 , then f ′( x )

= (1 + 0 ) ⋅ x 2  +  2 x 2−1 ⋅ ( x + 5)  = x 2 + 2 x (x + 5) = x 2 + 2 x 2 + 10 x = 3x 2 + 10 x and

f ′( 0)

= ( 3 ⋅ 02 ) + (10 ⋅ 0 ) = 0

f ′ (1)

= ( 3 ⋅12 ) + (10 ⋅1) = 3 + 10 = 13

f ′ ( −2 )

= 3 ⋅ ( −2 )2  + (10 ⋅ −2 ) = ( 3 ⋅ 4 ) − 20 = 12 − 20 = −8  

b. Given f ( x) = 3x 2 + 1 , then f ′( x )

= (3 ⋅ 2) x 2−1 + 0 = 6x and

f ′( 0)

= 6⋅0 = 0

f ′ (1)

= 6 ⋅1 = 6

f ′ ( −2 )

= 6 ⋅ −2 = −12

c. Given f ( x ) = x −5 − 2 x −4 − 3x −2 + 1 , then f ′( x )

5 8 6 = − 5 x −5−1 + (− 2 ⋅ −4) x −4−1 + (− 3 ⋅ −2) x −2−1 + 0 = −5 x −6 + 8 x −5 + 6 x −3 = − 6 + 5 + 3 and

f ′( 0)

5 8 6 5 8 6 = − 6+ 5+ 3 = − + +

f ′ (1)

5 8 6 5 8 6 = − 6+ 5+ 3 = − + + = 9

x

x

0

1

f ′ ( −2 )

0

1

0

0

1 1

1

0

0

f ′(0)

x

is undefined due to division by zero

1

5 8 6 5 8 6 + + − + + = − = −0.078 − 0.25 − 0.75 = −1.078 5 6 3 = 64 − 32 − 8 ( −2 ) ( −2 ) ( −2 )

d. Given f ( x ) = x −1 ( x + 2) , then f ′( x )

1 2 1 2 x+2 1 =  − x −1−1 ( x + 2 )  + (1 ⋅ x −1 ) =  − x −2 ( x + 2 )  + x −1 =  − 2  + = − − 2 + = − 2 and

f ′( 0)

2 2 = − 2 = −

f ′ (1)

2 2 = − 2 = − = −2



0

1

0

f ′(0)

x



x

x

x

x

x

is undefined due to division by zero

1

Hamilton Education Guides

66

Calculus I

2.2 Differentiation Rules Using the Prime Notation

f ′ ( −2 )

1 2 2 − = − = − 2 = 2 4 ( −2 )

1 x

e. Given f ( x ) = x 3 + 2 x + = x 3 + 2 x + x −1 , then f ′( x )

= 3x3−1 + 2 x1−1 − x −1−1 = 3x 2 + 2 − x −2 and

1 1 1 f ′ ( 0 ) = 3x 2 + 2 − x −2 = 3x 2 + 2 − 2 = 3 ⋅ 02 + 2 − 2 = 2 − 0 x 0 f ′ (1)

f ′(0) is undefined due to division by zero

1 1 1 = 3x 2 + 2 − x −2 = 3x 2 + 2 − 2 = 3 ⋅12 + 2 − 2 = 3 + 2 − = 3 + 2 − 1 = 4

1

x

f ′ ( −2 )

1

1 1 3 ⋅ 4 + 2 − = 12 + 2 − 0.25 = 13.75 = 3x 2 + 2 − x −2 = 3 ⋅ ( −2 )2 + 2 − 2 = 4 ( −2 )

f. Given f ( x ) = x 2 ( x + 1) , then f ′( x )

=  2 x 2−1 ( x + 1)  + (1 ⋅ x 2 ) =  2 x ( x + 1)  + x 2 = 2 x 2 + 2 x + x 2 = 3x 2 + 2 x and

f ′( 0)

= ( 3 ⋅ 02 ) + ( 2 ⋅ 0 ) = 0

f ′ (1)

= ( 3 ⋅12 ) + ( 2 ⋅1) = 3 + 2 = 5

f ′ ( −2 )

= 3 ⋅ ( −2 )2  + ( 2 ⋅ −2 ) = ( 3 ⋅ 4 ) − 4 = 12 − 4 = 8  

g. Given f ( x ) =

(

3x 2 + 1

)

, then

( 2 (3x2 + 1)

)

 3x 2 + 1 ⋅ 2 x  −  x 2 + 4 ⋅ 6 x     

f ′( x )

=

f ′( 0)

= −

f ′ (1)

= −

f ′ ( −2 )

x2 + 4

22 ⋅ 0

(3 ⋅ 02 + 1) 22 ⋅1

(3 ⋅12 + 1)

= −

= −

2

2

0

( 0 + 1)

2

=

( 6 x3 + 2 x ) − ( 6 x3 + 24 x ) 2 (3x2 + 1)

=

6 x3 + 2 x − 6 x3 − 24 x

(3x2 + 1)

2

= −

22 x

(3x2 + 1)

2

0 0 = − 2 = − = 0

1

1

22 22 22 − 2 = − = − = −1.375 2 = 16 4 ( 3 + 1)

22 ⋅ −2 2

3 ⋅ ( −2 ) + 1  

Hamilton Education Guides

2

−44 44 44 0.26 = − 2 = 2 = 169 = 13 12 + 1 ( )

67

Calculus I

2.2 Differentiation Rules Using the Prime Notation

h. Given f ( x ) = x 5 − 2 x 2 + 3x + 10 , then f ′( x )

= 5 x5−1 − 2 ⋅ 2 x 2−1 + 3 + 0 = 5 x 4 − 4 x + 3

f ′( 0)

= 5 ⋅ 04 − 4 ⋅ 0 + 3 = 0 − 0 + 3 = 3

f ′ (1)

= 5 ⋅14 − 4 ⋅1 + 3 = 5 − 4 + 3 = 4

f ′ ( −2 )

= 5 ⋅ ( −2 )4 + ( −4 ⋅ −2 ) + 3 = ( 5 ⋅ 16 ) + 8 + 3 = 80 + 11 = 91

(

)

i. Given f (x ) = x 2 + 3 (x − 1) , then

[ (

)]

f ′( x )

= [ 2 x ⋅ (x − 1) ] + 1 ⋅ x 2 + 3

f ′( 0)

= 3 ⋅ 02 − 2 ⋅ 0 + 3 = 0 − 0 + 3 = 3

f ′ (1)

= 3 ⋅12 − 2 ⋅1 + 3 = 3 − 2 + 3 = 4

f ′ ( −2 )

= 2 x 2 − 2 x + x 2 + 3 = 3x 2 − 2 x + 3

= 3 ⋅ ( −2 )2 + ( −2 ⋅ −2 ) + 3 = 3 ⋅ 4 + 4 + 3 = 12 + 7 = 19

Example 2.2-3: Given g( x ) =

1 + 1 and h( x ) = x , find f (x ) , f ′( x ) and f ′(0) . x x h( x )

a. f ( x ) = x g( x )

b. f ( x ) = 2 x 2 − 5h( x )

c. f ( x ) = g( x ) +

d. h( x ) = 3x f ( x )

e. h( x ) = 1 − f ( x )

f. 3h( x ) = 2 x f ( x ) − 1

Solutions: a. Given g( x ) =

1 + 1 and f ( x ) = x g( x ) , then x

1  f ( x ) = x g (x ) = x ⋅  + 1 = 1 + x therefore f ′ ( x ) = 1 and f ′ ( 0 ) = 1 x 

b. Given h( x ) = x and f ( x ) = 2 x 2 − 5h( x ) , then 2

f ( x ) = 2 x 2 − 5h(x ) = 2 x 2 − 5 ⋅ x = 2 x − 5 x therefore f ′ ( x ) = (2 ⋅ 2 ) x 2−1 − 5 = 4 x − 5 and f ′( 0)

= (4 ⋅ 0) − 5 = −5

c. Given g( x ) = f ( x)

= g (x ) +

1 + 1 , h( x ) = x , x x h( x )

=

and f ( x ) = g( x ) +

1  x  + 1 + x  x

=

1   + 1 + 1 x 

=

x , h( x )

then

1 +2 x

= x

−1

+ 2 therefore f ′ ( x ) = − x −1−1 + 0

1 1 1 = − x −2 = − 2 and f ′ ( 0 ) = − 2 = − which is undefined due to division by zero.

x

Hamilton Education Guides

0

0

68

Calculus I

2.2 Differentiation Rules Using the Prime Notation

d. Given h( x ) = x and h( x ) = 3x f ( x ) , then f ( x) =

h( x ) 3x

x 3x

=

=

1 3

therefore, f ′ ( x ) = 0 and f ′ ( 0 ) = 0

e. Given h( x ) = x and h( x ) = 1 − f ( x ) , then f ( x ) = 1 − h(x ) = 1 − x therefore, f ′ ( x ) = −1 and f ′ ( 0 ) = −1

f. Given h( x ) = x and 3h( x ) = 2 x f ( x ) − 1 , then f ( x) =

3h(x ) + 1 2x

=

3x + 1 2x

therefore, f ′ ( x ) =

[ 2 x ⋅ 3] −  2 ⋅ ( 3x + 1)  6/ x − 6/ x − 2 2 = / 2/ = − 2 2 4x 4x ( 2x )

1 1 1 = − 2 and f ′ ( 0 ) = − 2 = − which is undefined due to division by zero.

2x

2⋅0

0

Section 2.2 Practice Problems - Differentiation Rules Using the Prime Notation 1. Find the derivative of the following functions. Compare your answers with Practice Problem number 1 in Section 2.1. a. f (x ) = x 2 − 1

b. f (x ) = x 3 + 2 x − 1

c. f (x ) =

x x −1

e. f (x ) = 20 x 2 − 3

f. f (x ) = x 3

g. f (x ) =

10

2. Differentiate the following functions: a. f (x ) = x 2 + 10 x + 1

(

d. f (x ) = 2 x 5 + 10 x 4 + 5 x

(

)(

g. f (x ) = x 3 + 1 x 2 − 5 j. f (x ) =

)

)

(

e. f (x ) = a 2 x 3 + b 2 x + c 2

f. f (x ) = x 2 ( x − 1) + 3x

(

)

 2x 3 + 5   p. f (x ) = x 2 − 1   

)

x 5 + 2x 2 −1 3x 2

n. f (x ) = (x + 1) ⋅

x



x



x2 ax + b f (x ) = cx

c. f (x ) = 3x 4 − 2 x 2 + 5

k. f (x ) =

1 m. f (x ) = x 2  2 + 

h.

1

b. f (x ) = x 8 + 3x 2 − 1

h. f (x ) = 3x 2 + x − 1 (x − 1)

x3 +1 x

x−5

d. f (x ) = −

q. f (x ) =

2x x −1

3x 4 + x 2 + 2 x −1

(

)

i. f (x ) = x x 3 + 5 x 2 − 4 x l. f (x ) = o. f (x ) =

x2 (x − 1) + 3x

x 3 + 3x − 1 x4

r. f (x ) = x −1 +

1 x −2

3. Compute f ′(x ) at the specified value of x . Compare your answers with the practice problem number 2 in Section 2.1. a. f (x ) = x 3 at x = 1

b. f (x ) = 1+ 2 x at x = 0

c. f (x ) = x 3 + 1 at x = −1

d. f (x ) = x 2 (x + 2) at x = 2

e. f (x ) = x −2 + x −1 + 1 at x = 1

f. f (x ) = x + 2 at x = 10

Hamilton Education Guides

69

Calculus I

2.2 Differentiation Rules Using the Prime Notation

4. Find f ′(0) and f ′(2) for the following functions:

(

)

(

)

a. f (x ) = x 3 − 3x 2 + 5

b. f (x ) = x 3 + 1 (x − 1)

c. f (x ) = x x 2 + 1

d. f (x ) = 2 x 5 + 10 x 4 − 4 x

e. f (x ) = 2 x −2 − 3x −1 + 5 x

f. f (x ) = x −2 x 5 − x 3 + x

g. f (x ) =

x 1+ x 2

1 x

h. f (x ) = + x 3

i. f (x ) =

(

)

ax 2 + bx cx − d

5. Given f (x ) = x 2 + 1 and g (x ) = 2 x − 5 , find h(x ) and h ′(x ) . a. h(x ) = x 3 f (x )

b. f (x ) = 3 + h(x )

c. 2 g (x ) = h(x ) − 1

d. 3h(x ) = 2 x g (x ) − 1

e. 3 [ f (x ) ] 2 − 2h(x ) = 1

f. h(x ) = g (x ) ⋅ 3 f (x )

g. 3h(x ) − f (x ) = 0

h. 2 g (x ) + h(x ) = f (x )

i. f (x ) = x 3 + 5 x 2 + h(x )

j. h(x ) =

k. h(x ) = 2 f (x ) + g (x )

l. [ h(x ) ] 2 − f (x ) = 10

m.

n.

x3 +1 − f (x ) x 2 g (x ) f (x ) = h( x )

Hamilton Education Guides

3 f (x ) 1 = h( x ) x

o. f (x ) =

1 h( x ) + 4

70

Calculus I

2.3

2.3 Differentiation Rules Using the

Differentiation Rules Using the

d dx

d Notation dx

Notation

In the previous section the prime notation was used as a means to show the derivative of a function. For example, derivative of the functions y = f ( x ) = x 2 + 3x + 1 was represented as y ′ = f ′(x ) = 2 x + 3 . However, derivatives can also be represented by what is referred to as the “double-d” notation. For example, the derivative of the function y = f (x ) = x 2 + 3x + 1 can be shown as form:

dy d = f (x ) = 2 x + 3 . dx dx

Following are the differentiation rules in the double-d notation

Rule No. 1 - The derivative of a constant function is equal to zero, i.e., if f ( x ) = k ,

then

d f ( x) = 0 dx

Rule No. 2 - The derivative of the identity function is equal to one, i.e., if f ( x ) = x ,

then

d f ( x) = 1 dx

Rule No. 3 - The derivative of the function f ( x ) = x n is equal to positive or negative integer.

d f ( x ) = n x n −1 , dx

where n is a

Rule No. 4 (scalar rule) - If the function f ( x ) is differentiable at x , then a constant k multiplied by f ( x ) is also differentiable at x , i.e., d d ( k f ) x = k  dx f ( x ) dx  

[

]

Rule No. 5 (summation rule) - If the function f ( x ) and g( x ) are differentiable at x , then their sum is also differentiable at x , i.e., d dx

[( f + g ) x ] =

d d g( x ) f ( x) + dx dx

Rule No. 6 (product rule) - If the function f ( x ) and g( x ) are differentiable at x , then their product is also differentiable at x , i.e., d   dx ( f ⋅ g ) (x )  

d d =  f ( x ) g( x ) +  g( x ) f ( x )  dx



 dx



Rule No. 7 - Using the rules 1 , 4 , 5 , and 6 we can write the formula for differentiating polynomials, i.e., if f ( x ) = a n x n + a n−1 x n−1 + a n−2 x n−2 +  + a3 x 3 + a2 x 2 + a1 x1 + a0 then, d f ( x ) = n a n x n −1 + ( n − 1) a n −1 x n −2 + ( n − 2) a n −2 x n −3 +  + 3a3 x 2 + 2a 2 x + a1 dx

Rule No. 8 (quotient rule) - If the function f ( x ) and g( x ) are differentiable at x , then their quotient is also differentiable at x , i.e.,

Hamilton Education Guides

71

Calculus I

2.3 Differentiation Rules Using the

 d  f      ( x)  dx  g  

=

d Notation dx

d  d   dx f ( x ) g( x ) −  dx g( x ) f ( x )    

[ g( x )]2

Note 1 - depending on the letter used to express the terms of a function, the double-d notation of a derivative is then shown as

( where a is equal to the letter used in the left hand side of the equation) . ( where b is equal to the letter used in the right hand side of the equation)

da db

For example, •

if the function y is represented by f (x ) , i.e., y = f ( x ) = x 2 + 2 x , then its derivative is shown as

(

)

dy d d 2 f ( x) = x + 2x = 2x + 2 . = dx dx dx

•

if the function y is represented by f (t ) , i.e., y = f (t ) = t 3 + 2t 2 + 4 , then its derivative is shown as

•

(

)

dy d d 3 = f (t) = t + 2t 2 + 4 = 3t 2 + 4t . dt dt dt

if the function u is represented by f (v ) , i.e., u = f ( v ) = v 3 + 3v , then its derivative is shown as

(

)

d du d = v 3 + 3v = 3v 2 + 3 . f (v ) = dv dv dv

•

if the function p is represented by f (r ) , i.e., p = f ( r ) = 2r 3 − 2r 2 − 3 , then its derivative is shown as

•

)

if the function y is represented by f (z ) , i.e., y = f ( z ) = z 5 + 3z 2 + 1 , then its derivative is shown as

•

(

dp d d f (r) = = 2r 3 − 2r 2 − 3 = 3r 2 − 4r . dr dr dr

(

)

dy d d 5 = z + 3z 2 + 1 = 5z 4 + 6z . f ( z) = dz dz dz

if the function v is represented by f (x ) , i.e., v = f ( x ) = x 8 + 4 , then its derivative is shown as

(

)

dv d d 8 f ( x) = x + 4 = 8 x 7 , etc. = dx dx dx

In the following examples the above rules are used in order to find the derivative of various functions: Example 2.3-1: Find

dy for the following functions. dx

a. y = x 3 − 2 x 2 + 5 d. y =

3x 2 1+ x

b. y = 4 x 5 − 3x 2 − 1 e. y = 5x +

(

g. y = (x + 1) x 2 − 3

)

x − 3   5 

k. y = x 2 

Solutions: dy dx

=

d 3 x − 2 x2 + 5 dx

(

x +1

h. y = 5 x (x + 1)

j. y = x (x + 1) (x − 2)

a.

2x

2

)

Hamilton Education Guides

=

d 3 d d x + −2 x 2 + ( 5) dx dx dx

( )

(

)

c. y = x 2 +

(

1 x

)

f. y = x 3 x 2 + 1 i. y = 5 +

1− x x

l. y = (x + 1) (x − 1)−2

= 3x 2 + ( −2 ⋅ 2 ) x + 0 = 3 x 2 − 4 x

72

Calculus I

b.

dy dx

=

d 4 x5 − 3 x 2 − 1 dx

c.

dy dx

=

d  2 1 x +  dx  x

d.

dy dx

=

d  3x 2    dx  1 + x 

e.

dy dx

=

2x  d  5x + 2  dx  x +1

) = dxd ( 4 x5 ) + dxd ( −3x2 ) + dxd ( −1)

(

(

(

d x 2 + x −1 dx

=

(1 + x )

=

g.

( x2 + 1)

2

dy dx

=

d  3 2 x x + 1   dx 

)

=

dy dx

=

d  3 2 x x + 1   dx 

=

dy dx

=

d  ( x + 1) x 2 − 3  dx 

(

)

( )

( )

)

(

2

d d  2x  (5x ) +  2  dx dx  x + 1 

)

= (4 ⋅ 5) x 4 + (− 3 ⋅ 2) x + 0 = 20 x 4 − 6 x

d 2 d −1 x + x dx dx

=

( ) ( )

=

(

)

d   2   2 d (1 + x ) dx 3x  −  3x dx (1 + x )     

 x2 + 1 ⋅ 2 − [ 2 x ⋅ 2 x] 

 = 5+ 

f.

d Notation dx

2.3 Differentiation Rules Using the

=

[ ( 1 + x )⋅ 6 x] − [ ( 3x 2 )⋅1] 3 x2 + 6 x = 2 (1 + x ) ( 1 + x) 2 

=

( x2 + 1) dxd ( 2 x ) − 2 x dxd ( x2 + 1) 2 ( x2 + 1)

−2 x 2 + 2 2 x2 + 2 − 4 x2 = 5+ = 5+ 2 2

( x 2 + 1)

( x2 + 1)

) ( )

d

5 3 x +x ) = ( dx

=

x

 d (5x ) +  dx

d 3   3 d 2  2   x + 1 dx x  +  x dx x + 1     

(

1 = 2x + ( − x −2 ) = 2x − x −2 = 2x − 2

(

)

d 5 d 3 x + x dx dx

= ( x 2 + 1) ⋅ 3x 2 + x3 ⋅ 2 x = 5 x 4 + 3 x 2 or, 4

= 5 x 5−1 + 3x 3−1 = 5 x + 3 x

d d 2  2     x − 3 dx ( x + 1)  + ( x + 1) dx x − 3     

(

)

(

)

=

(

2

)

 x 2 − 3 ⋅1 + ( x + 1) ⋅ 2 x     

= x 2 − 3 + 2 x 2 + 2 x = 3 x 2 + 2 x − 3 or,

h.

i.

(

)=

dy dx

=

dy dx

=

d d 5 x ( x + 1) = ( x + 1) ( 5 x )  + ( 5 x ) ( x + 1)  = ( x + 1) ⋅ 5 + [5 x ⋅1] = 5 x + 5 + 5 x = 10 x + 5 or, dx dx     dx

dy dx

=

d d 2 5x + 5x) = 5 x 2 + 5 x = (5 ⋅ 2 ) x 2−1 + 5 x1−1 = 10 x + 5 5 x ( x + 1) = ( dx dx dx dx

=

d  1− x  5+ dx  x 

dy dx

d  ( x + 1) x 2 − 3  dx 

(

)

=

d x 3 − 3x + x 2 − 3 dx

d 3 d 2 d (− 3x ) + d (− 3) x + x + dx dx dx dx

= 3 x2 + 2 x − 3

d

d

d

=

Hamilton Education Guides

d d 1− x  ( 5) +  dx dx  x 

=

d  d     x dx (1 − x )  − (1 − x ) dx ( x )  d    ( 5) +  dx x2

= 0+

[ x ⋅ −1] − (1 − x ) ⋅1 x2

73

Calculus I

j.

dy dx

2.3 Differentiation Rules Using the

=

1 −x −1+ x = − 2 2 x x

=

d [x (x + 1) (x − 2)] dx

=

d  2 x + x ( x − 2 )   dx 

(

)

d 2 x +x dx

(

= ( x − 2 ) 

d Notation dx

) + ( x2 + x ) dxd ( x − 2)

= ( x − 2 ) ⋅ ( 2 x + 1)  + ( x 2 + x ) ⋅1 = 2 x 2 + x − 4 x − 2 + x 2 + x = 3 x 2 − 2 x − 2 or, dy dx

k.

d  x ( x + 1)( x − 2 )  dx 

=

d d 3 d (− 2 x ) x + − x2 + dx dx dx

dy dx

dy dx

dy dx

=

( )

] = dxd ( x

3

− 2x 2 + x 2 − 2x

) = dxd ( x

=

 2 x2 − 6 x   2 1   +x ⋅     5 5  

=

d  2  x − 3  x dx   5 

=

(

=

2 x2 − 6 x x2 + 5 5

d  x3 − 3 x 2     dx  5 

) (

)

2 2    3  5 3x − 6 x  −  x − 3x ⋅ 0  2

5

d  −2 ( x + 1)( x − 1)  dx 

2 x2 − 6 x + x2 5

=

=

3x 2 − 6 x 5

) (

(

(

)

2

5

or,

)

=

(

5/ 3x 2 − 6 x 2/ =1

5

d d −2 ( x + 1)  + ( x + 1) ( x − 1)  dx dx    1

d  x + 1  dx  ( x − 1)2 

3 x ( x − 2) 5

52

5 3x 2 − 6 x − 0

( x − 1)

=

 d 3  2   3 2 d 5 dx x − 3x  −  x − 3x dx ( 5 )     

=

=

= ( x − 1)−2 

)

( ) ⋅  1⋅ 552− 0 

( ) ( )

=

− x 2 − 2x

3

= 3x 3−1 + −2 x 2−1 − 2 x1−1 = 3x 2 − 2 x − 2 x 0 = 3 x 2 − 2 x − 2

 x −3   2  x − 3  d 2   2 d  x − 3  d  2  x − 3  x + x x =  =   ⋅ 2x +  x  dx   5  dx  5    5     5  dx  

= ( x − 1)−2 − 2 ( x + 1)( x − 1)−3 =

dy dx

)

=

= l.

[(

d x 2 + x (x − 2 ) dx

=

2

−

2 ( x + 1)

( x − 1) 3

)

=

3x 2 − 6 x 5

=

3 x ( x − 2) 5

= ( x − 1)−2 ⋅1 + ( x + 1) ⋅ −2 ( x − 1)−3     

or,

d 2 d 2    ( x − 1) dx ( x + 1)  − ( x + 1) dx ( x − 1)     

[ (x −1) ⋅1] 2

=

d  −2 ( x + 1)( x − 1)  dx 

−

[ (x + 1)⋅ 2 (x − 1) ] 2 ( x + 1) 1 (x − 1)2 − 2(x + 1) (x − 1) (x − 1)2/ − 2(x/ + 1/ ) (x − 1) − = = = 2 4 4/ = 2 4/ =3 4 ( x − 1) ( x − 1) 3 (x − 1) (x − 1) (x − 1) (x − 1)

Hamilton Education Guides

=





=

( x − 1)

4

=

(x − 1)4

74

Calculus I

2.3 Differentiation Rules Using the

d Notation dx

Example 2.3-2: Find the derivative of the following functions.

(

)

(

)

a.

d 3x 2 + 5 x − 1 = dx

b.

d 8 x 4 + 3x 2 + x = dx

d.

d  2t 2 + 3t + 1   = dt  t3 

e.

g.

d dt

j.

d  3  1   w + 1   dw   w 

m.

 3  − 2t   t    4 

(

=

)

=

[( )

]

d 3 2 t t + 1 ( 3t − 1) = dt

[(

]

)

c.

d u3 + 5 (u + 1) = du

d 1 − t 2 (1 + t ) + t = dt

]

f.

d  t 2 + 1   = dt  t 2 − 1

h.

d  u u2    = + du  1 − u 1 + u 

i.

d  s 3 + 3s 2 + 1   =  ds  s3 

k.

d  2x    = dx  1 + 2 x 

l.

d  s2   =  ds  1+ s 2 

n.

d  4u 3 + 2    du  u 2 

o.

d dx

[( )

=

 x3   2   x +1 

=

Solutions: a.

d 3x 2 + 5 x − 1 dx

b.

d 8 x 4 + 3x 2 + x dx

c.

d  3 u + 5 ( u + 1)   du 

) = dxd (3x2 ) + dxd (5x ) + dxd ( −1)

(

(

) = dxd (8x4 ) + dxd (3x2 ) + dxd ( x ) = (8 ⋅ 4) x4−1 + (3 ⋅ 2) x2−1 + x1−1 =

)

(

= ( 3 ⋅ 2 ) x 2−1 + 5 x1−1 + 0 = 6 x + 5 x0 = 6 x + 5

= ( u + 1) 

d 3 d   u + 5  +  u3 + 5 ( u + 1)  du du   

(

) (

)

32 x 3 + 6 x + 1

= ( u + 1) ⋅ 3u 2  + ( u3 + 5) ⋅1

= 3u3 + 3u 2 + u3 + 5 = 4u3 + 3u2 + 5 d.

d  2t 2 + 3t + 1     dt  t3 

=

( 4t

)(

=

t

(

6

t 2/ 2t 2 + 6t + 3 t

6/ = 4

d  1 − t 2 (1 + t ) + t   dt 

(

) (

+ 3t 3 − 6t 4 + 9t 3 + 3t 2 t

= − e.

4

=

(

)

) ( )

d 3  3d   2 2 t dt 2t + 3t + 1  −  2t + 3t + 1 dt t     

) = − 2t d dt

=

2

)=

6

Hamilton Education Guides

(

)

t

6

(

− t 2 2t 2 + 6t + 3 4t 4 + 3t 3 − 6t 4 − 9t 3 − 3t 2 −2t 4 − 6t 3 − 3t 2 = = t6 t6 t6

)

+ 6t + 3 t

4

[ (1 − t ) (1 + t ) ]+ dtd t =  (1 + t ) dtd (1 − t ) + ( 1 − t ) dtd ( 1 + t )  + 1 2

2

{(1 + t ) ⋅ −2t  + (1 − t ) ⋅1} + 1 = ( −2t − 2t 2

=

t 3 ⋅ ( 4t + 3)  −  2t 2 + 3t + 1 ⋅ 3t 2     

2

)

+1− t2 +1

2

= −2t − 2t 2 + 1 − t 2 + 1 = −3t 2 − 2t + 2

75

Calculus I

f.

(

) (

) ( ( )

) (

d dt

2t 3 − 2t − 2t 3 − 2t

(t 2 − 1)

 3  2t  t  −    4 

= −

=

2

=

− 4t

( t 2 − 1)

( t 2 − 1)

 − 2t  d 3   3 d  − 2t   t  + t     4  dt   dt  4 

6t 3  3  [4 ⋅ −2] − [(− 2t ) ⋅ 0]  + t   16 4   

h.

 3  2t  t  −    4 

=

d  2t 4 − dt  4

= −

  d d  4 (− 2t ) − (− 2t ) (4)   − 2t  2   3  dt dt   ⋅ 3t  + t  2   4    4      

=

6t 3  3  − 8  + t   4   16 

  / 4  = d  − 2t   dt  4/   2

= −

    

= −

1 d 4 t 2 dt

3 6t 3 t 3 − 12t 3 − 4t 3 16t 3 − = = − = −2t 4 2 8 8

d  3  2t  t  −  dt   4  1 2

as follows:

= − ⋅ 4t 4−1 = −

3 4t 3 = −2t 2

d   d d 2  2 d     (1 − u ) du u  − u du (1 − u )  (1 + u ) du u  − u du (1 + u )  d  u u2  d  u  d  u2          + + +    =  = 2 2 du  1 − u 1 + u  du  1 − u  du  1 + u  (1 − u ) (1 + u )

[(1 − u )⋅1] − [u ⋅ −1] + [(1 + u )⋅ 2u ] − [ u = (1 − u )2 (1 + u )2 i.

(t 2 − 1)

2

A second way of solving this problem is to simplify d dt

d Notation dx

3 3  ( t 2 − 1) ⋅ 2t  −  ( t 2 + 1) ⋅ 2t      = ( 2t − 2t ) − ( 2t + 2t ) 2 2

)

d 2 d 2  2   2   t − 1 dt t + 1  −  t + 1 dt t − 1  d  t2 +1    =   =  2 dt  t 2 − 1  2 t −1

=

g.

2.3 Differentiation Rules Using the

d  s3 + 3s 2 + 1     ds  s3 

=

( 3s

5

+ 6 s 4 − 3s 5 + 9 s 4 + 3s 2 s

(

)

j.

d  3  1   w + 1   dw   w 

k.

d  2x  dx  1 + 2 x 

=

6

=

=

1− u + u

(1 − u )

2u + u 2 − u 2

(1 + u )

)

5

=

(1 + 2 x )

2

2

=

1

(1 − u )

2

+

2u

(1 + u )

2

 s 3 ⋅ ( 3s 2 + 6 s )  −  ( s 3 + 3s 2 + 1) ⋅ 3s 2     =  6 s

(

)

(

−3s 2 s 2 + 1 − 3 s2 +1 −3s 4 − 3s 2 + 6s 4 − 3s5 − 9s 4 − 3s 2 = = = s6 s6 s 6/ =4 s4 d w3−1 + w−1 dw

(

d d     (1 + 2 x ) dx ( 2 x )  − ( 2 x ) dx (1 + 2 x )     

Hamilton Education Guides

2

+

s6

) = 3s

d  3 w + 1 w−1   dw 

(

]

⋅1

) − ( s3 + 3s2 + 1) dsd s3 

(

 3 d 3 2  s ds s + 3s + 1 

=

)(

2

=

)

=

d 2 d −1 w + w dw dw

)

1 = 2w − w−2 = 2w − 2 w

[(1 + 2 x )⋅ 2] − [2 x ⋅ 2] 2 2 + 4x − 4x = 2 2 = 2 (1 + 2x ) (1 + 2 x ) (1 + 2 x )

76

Calculus I

l.

m.

2.3 Differentiation Rules Using the

d  s2    ds  1 + s 2 

=

(

)

(1 + s2 )

d3 2 t t + 1 ( 3t − 1)   dt 

(

(

)

 2 d 2  2 d 2   1 + s ds s  −  s ds 1 + s     

)

2

d 5 3 t + t ( 3t − 1)   dt 

(

=

)

) 2 (1 + s2 )

(

 1 + s 2 ⋅ 2s  −  s 2 ⋅ 2s     

=

= ( 3t − 1) 

d 5 3 t +t dt

(

=

2s + 2s3 − 2s3

(1 + s2 )

2

=

d Notation dx 2s

(1 + s 2 )

2

) + (t 5 + t 3 ) dtd (3t − 1)

= ( 3t − 1) ⋅ ( 5t 4 + 3t 2 )  + ( t 5 + t 3 ) ⋅ 3 = (15t 5 + 9t 3 − 5t 4 − 3t 2 ) + ( 3t 5 + 3t 3 ) = 18t − 5t + 12t − 3t 5

4

3

2

Another way of solving this problem is by multiplication of the binomial terms using the FOIL method prior to taking the derivative of the function as follows. d3 2 t t + 1 ( 3t − 1)   dt 

(

n.

)

d  4u 3 + 2    du  u 2 

=

=

[ (

)]

d 3 3 2 t 3t − t + 3t − 1 dt

=

(

) (

=

(

d 3t 6 − t 5 + 3t 4 − t 3 dt

)

d 2  2 d   3 3 u du 4u + 2  −  4u + 2 du u     

=

u4

(

)

(

u4

)

)

(

(

(

=

(u 2 ⋅12u 2 ) − ( 4u3 + 2) ⋅ 2u

4u/ u 3 − 1 4 u3 − 1  u3  u 3 −1  1 4u 4 − 4u = = = 4 3  = 4 3 − 3 4 / =3 3 4 u u u u u  u 

)

d 3  3 d  2  x +1 x − x x 2 +1  3    du   du d x   o. =   = 2 dx  x 2 +1  x 2 +1

x 4 + 3x 2

( x + 1) 2

2

=

x

2

(x

(x

2

2

+3

+1

)

2

)

) = 18t

  = 4  1 − 1   u3  

[ ( x + 1)⋅ 3x ]− ( x ( x + 1) 2

2

2

3

⋅ 2x

)

5

3

2

12u 4 − 8u 4 − 4u u4

=   

=

2

4

− 5 t + 12 t − 3 t

3x 4 + 3x 2 − 2 x 4

( x + 1) 2

2

)

Example 2.3-3: Find the derivative of the following functions at the specified value. a.

d (t + 1)(t − 2) + 3t dt

[

]

 d  x −1 c.   dx  ( x + 1)(2 x + 1) 

e.

d  s 2 + 3s    ds  s 2 + 1 

d  u 2 + 1   du  u3 − 1 

at t = 1

b.

at x = 0

  3 d  x + 1 ( x − 1)  d.  dx  2x 2  

)

(

at s = −1

f.

at u = 2

d  z 2 + 3z − 5    z dz  

at x = 2

at z = 2

Solutions: a.

d ( t + 1)( t − 2 ) + 3t  dt 

=

Hamilton Education Guides

d d ( t + 1)( t − 2 )  + ( 3t ) dt  dt

d dt

d dt

= ( t − 2 ) ( t + 1) + ( t + 1) ( t − 2 )  + 



d ( 3t ) dt

77

Calculus I

2.3 Differentiation Rules Using the

d Notation dx

= ( t − 2 ) ⋅1 + ( t + 1) ⋅1 + 3 = ( t − 2 ) + ( t + 1) + 3 = t − 2 + t + 1 + 3 = 2t + 2 d ( t + 1)( t − 2 ) + 3t  dt 

Therefore, at t = 1

b.

d  u2 + 1    du  u 3 − 1 

=

=

) (

(

 d  x −1   dx  ( x + 1)( 2 x + 1) 

=

=

(

−u 4 − 3u 2 − 2u

=

2

Therefore, at u = 2

c.

 ( u 3 − 1) ⋅ 2u  −  ( u 2 + 1) ⋅ 3u 2     =  2

) − (u 2 + 1) dud (u3 − 1) 2 (u3 − 1)

d 2  3  u − 1 du u + 1 

2u 4 − 2u − 3u 4 − 3u 2

(u3 − 1)

= 2t + 2 = ( 2 ⋅1) + 2 = 2 + 2 = 4

(u3 − 1)

( (

d  x −1  dx  2 x 2 + 3x + 1 

)

 2 x 2 + 3x + 1 ⋅1 − ( x − 1) ⋅ ( 4 x + 3)     

( 2 x2 + 3x + 1)

2

2 x 2 + 3x + 1 − 4 x 2 + x + 3 2

Therefore, at x = 0

3

u u + 3u + 2

( u3 − 1)

u u 3 + 3u + 2 d  u2 + 1   3  = − 2 du  u − 1  u 3 −1

=

( 2 x2 + 3x + 1)

= −

2

(

=

=

=

( u3 − 1)

)

)

(

2

= −

)

2 ⋅  23 + ( 3 ⋅ 2 ) + 2  

( 23 − 1)



2

)

= −

32 2 ⋅16 − = −0.653 2 = 49 7

(

)

d d     2 2  2 x + 3x + 1 dx ( x − 1)  − ( x − 1) dx 2 x + 3x + 1     

( 2 x2 + 3x + 1)

2

( 2 x2 + 3x + 1) − ( 4 x2 + 3x − 4 x − 3) 2 ( 2 x2 + 3x + 1)

=

( 2 x2 + 3x + 1) − ( 4 x2 − x − 3) 2 ( 2 x2 + 3x + 1)

−2 x 2 + 4 x + 4

( 2 x2 + 3x + 1)

 d  x −1   dx  ( x + 1)( 2 x + 1) 

2

=

−2 x 2 + 4 x + 4

( 2x

2

)

+ 3x + 1

( ) (

2

=

( −2 ⋅ 02 ) + ( 4 ⋅ 0) + 4 2  2 ⋅ 02 + [3 ⋅ 0] + 1 ) ( 

=

) ( ( )

4 1

= 4

) ( )

d    4 2 d 4 3 3 2   3   2 x dx x − x + x − 1  −  x − x + x − 1 dx 2 x  d  x + 1 ( x − 1)  d  x 4 − x3 + x − 1      d.  = 2 2 2  = dx  dx  2 x 2 2 x    2x  

(

(

)

) (

)

 2 x 2 4 x3 − 3 x 2 + 1  −  x 4 − x3 + x − 1 ⋅ 4 x     

4 x4

Hamilton Education Guides

=

( 8 x5 − 6 x 4 + 2 x 2 ) − ( 4 x5 − 4 x 4 + 4 x 2 − 4 x ) 4 x4

78

Calculus I

2.3 Differentiation Rules Using the

(

d Notation dx

)

2 x/ 2 x 4 − x3 − x + 2 8 x5 − 6 x 4 + 2 x 2 − 4 x5 + 4 x 4 − 4 x 2 + 4 x 4 x5 − 2 x 4 − 2 x 2 + 4 x 2 x 4 − x3 − x + 2 = = = 4 4 / =3 4 4x 4x 2 x3 4x

(

)



e.

d  s 2 + 3s    ds  s 2 + 1 

=

f.

(

) ( 2 ( s2 + 1) =

d  s 2 + 3s    s = −1 ds  s 2 + 1 

d  z 2 + 3z − 5     dz  z 

=

) (

( 2s3 + 3s2 + 2s + 3) − ( 2s3 + 6s2 ) 2 ( s2 + 1)

Therefore, at

=

)

(

)



) (

)

d 2 d 2  2   2   s + 1 ds s + 3s  −  s + 3s ds s + 1     

=

(

 3  2 ⋅ 24 − 23 − 2 + 2 d  x + 1 ( x − 1)  24 2 x 4 − x3 − x + 2 = = = = 1.5 2 3 3  16 dx  2 x 2 x 2 ⋅ 2  

Therefore, at x = 2

−3s 2 + 2s + 3

=

(s

2

) (

(

( s2 + 1)

2s3 + 3s 2 + 2s + 3 − 2s3 − 6s 2

( s2 + 1) )

+1

2

)

2

=

z

=

)

( s2 + 1)

( )

 −12 + 1  

=

2

−3s 2 + 2s + 3 2

 −3 ⋅ ( −1)2  + ( 2 ⋅ −1) + 3  

d  d 2   2   z dz z + 3z − 5  −  z + 3z − 5 dz ( z )      2

(

 s 2 + 1 ( 2s + 3)  −  s 2 + 3s ⋅ 2s     

=

2

=

1 −3 − 2 + 3 = − 2 22

(

)

 z ⋅ ( 2 z + 3)  −  z 2 + 3z − 5 ⋅1  

z

2

z2 + 5 2 z 2 + 3z − z 2 − 3z + 5 = z2 z2 d  z 2 + 3z − 5     dz  z 

Therefore, at z = 2

=

9 22 + 5 z2 + 5 = = = 2.25 2 2 4 z 2

Example 2.3-4: Given the functions below, find their derivatives at the specified value. x 2 + 2x − 1

a.

dy dx

c.

dv , given v = x 3 + 1 3x 2 + 5 dx

, given y =

(

x3

)(

u2 dv , given v = at u = 4 du 1− u du 3x d. , given u = at x = 2 dx ( x − 1) 2

at x = 2

)

b.

at x = 5

Solutions: a.

dy dx

=

d  x2 + 2 x − 1     dx  x3 

Hamilton Education Guides

=

(

) − ( x2 + 2 x − 1) dxd ( x3 ) 2 ( x3 )

 3 d 2  x dx x + 2 x − 1 

=

(

)( )

 x3 ( 2 x + 2 )  −  x 2 + 2 x − 1 3 x 2     

x

6

79

Calculus I

=

2.3 Differentiation Rules Using the

(

2 x 4 + 2 x3 − 3 x 4 + 6 x3 − 3 x 2 x

6

b.

=

d  u2    du  1 − u 

dv dx

=

=

(1 − u ) dv du

)(

d  u2    du  1 − u 

=

d  3 x + 1 3x 2 + 5   dx 

(

[(1 − u )⋅ 2u ] − [u 2 ⋅ −1] = = (1 − u )2

( )

d 2   2 d   (1 − u ) du u  − u du (1 − u )     

Therefore, at u = 4 c.

2 x 4 + 2 x3 − 3 x 4 − 6 x3 + 3 x 2 − x 4 − 4 x3 + 3 x 2 = 6 x6 x

=

dy −36 − x 4 − 4 x3 + 3 x 2 −24 − 4 ⋅ 23 + 3 ⋅ 22 d  x2 + 2 x − 1    = = = = = −0.56 6 6 3   dx 64 dx  x 2 x 

Therefore, at x = 2

dv du

)

d Notation dx

)

2

−u 2 + 2u

=

(1 − u )

2

=

− (4 )2 + (2 ⋅ 4 )

(1 − 4)

2

−16 + 8

=

(− 3)

2

2u − 2u 2 + u 2

(1 − u ) −8 9

=

2

=

−u 2 + 2u

(1 − u )

2

= −0.889

dv dv = ( 3x 2 + 5) ( x3 + 1) + ( x3 + 1) ( 3x 2 + 5) = ( 3x 2 + 5) ⋅ 3x 2 + ( x3 + 1) ⋅ 6 x dx

dx

= 9 x 4 + 15 x 2 + 6 x 4 + 6 x = 15 x 4 + 15 x 2 + 6 x dv dx

Thus, at x = 5

d.

du dx

=

=

d  3x  dx  ( x − 1)2  

(



=

=

d  x2 + 2 x − 1     dx  x3 

d 2 d 2    ( x − 1) du ( 3x )  − 3x du ( x − 1)     

( x − 1)

)

3 x 2 − 2x +1 − 6x 2 + 6x

(x − 1) 4 du dx

Therefore, at x = 2

= 15 x 4 + 15 x 2 + 6 x = 15 ⋅ 54 + 15 ⋅ 52 + 6 ⋅ 5 = 9375 + 375 + 30 = 9780

=

=

( x − 1) d  3x  dx  ( x − 1)2  



=

4

3x 2 − 6 x + 3 − 6 x 2 + 6 x 4

=

=

− 3x 2 + 3

(x − 1)

4

( x − 1)2 ⋅ 3 − 3x ⋅ 2 ( x − 1)     

( x − 1)

dy dx

( x − 1)4

(x − 1)4

=

(− 3 ⋅ 2 )+ 3 2

(2 − 1)

4

= d dx

−12 + 3 1

= −

9 1

= −9

Notation

for the following functions:

a. y = x 5 + 3x 2 + 1 d. y =

3 ( x − 1)2 − 6 x ( x − 1)

− 3x 2 + 3

Section 2.3 - Differentiation Rules Using the 1. Find

4

=

b. y = 3x 2 + 5

x2 1− x3

(

g. y = x 3 x 2 + 5 x − 2

e. y = 4 x 2 +

)

Hamilton Education Guides

1 x −1

h. y = x 2 (x + 3)(x − 1)

c. y = x 3 − f. y =

1 x

x 2 + 2x x3 +1

i. y = 5 x −

1 x3

80

Calculus I

2.3 Differentiation Rules Using the

(x − 1)(x + 3)

j. y =

x

x −1    3 

k. y = x

2

x  x −3   1+ x   5 

n. y = x 3 1 +

ax 2 + bx + c bx

q. y =

m. y =  p. y =



d Notation dx

l. y = x 2 (x + 3)−1

1   x −1 

1 2x −1  o. y =   x  3x + 1 

x3 − 2

r. y =

4

x −3

5x

(1 + x ) 2

2. Find the derivative of the following functions: a.

d 3t 2 + 5t dt

)=

b.

d 6 x 3 + 5x − 2 dx

d.

d  t 2 + 2t  = dt  5 

e.

d  s 3 + 3s − 1  =  ds  s2 

g.

d 2 t (t + 1) t 2 − 3 dt

h.

d (x + 1) x 2 + 5 dx

j.

d  3r 3 − 2r 2 + 1   dr  r 

k.

d  3s 2 1 − 3 ds  s + 1 s 2

(

[

)] =

(

=

(

[

)=

(

(

)=

c.

d u 3 + 2u 2 + 5 du

f.

d  3 w 2  = w + dw  1 + w 

)] =

i.

d  u 2 u  = − du  1 − u 1 + u 

=

l.

d  u 3 u + 1  − du  1 − u u 2 

   

=

3. Find the derivative of the following functions at the specified value. b.

d dx

[ ( x + 1) ( x − 1) ]

d  t 2 + 1  at t = −1 dt  t − 1 

e.

d du

 u3     ( u + 1) 2 

[ ( v + 1)v ] at

h.

d  x 3  at x = 0 dx  x 2 +1 

(

)

a.

d 3 x + 3x 2 + 1 dx

d. g.

d dv

2

at x = 2

3

v = −2

2

at x = 1

at u = 1

[

] at s = 0  w ( w + 1)   at w = 2

c.

d 3s 2 (s − 1) ds

f.

d dw 

i.

d  3  u 2  u  du   1 − u   

2

3w 2



at u = 0

4. Given the functions below, find their derivatives at the specified value.

(

)

a.

ds , given s = t 2 − 1 + (3t + 2) 2 dt

c.

dw , given w = x 2 + 1 dx

(

Hamilton Education Guides

)

2

+ 3x

at t = 2

b.

t 3 + 3t 2 + 1 dy , given y = dt 2t

at x = −1

d.

dy , given y = x 2 x 3 + 2 x + 1 dx

(

at t = 1

)

2

+ 3x

at x = 0

81

Calculus I

2.4

2.4 The Chain Rule

The Chain Rule

The chain rule is used for finding the derivative of the composition of functions. In general, the chain rule for two and three differentiable functions are defined in the following way: a. The chain rule for two differentiable functions f ( x ) and g( x ) is defined as:

( f  g )′ (x ) =

d { f [g (x )]} dx

= f ′[ g( x ) ] ⋅ g ′( x )

b. The chain rule for three differentiable functions f ( x ) , g( x ) and h( x ) is defined as:

( f  g  h) ′ ( x ) =

d [ f {g [h(x )]}] dx

= f ′{g [h(x )]}⋅g ′[h(x )]⋅ h ′(x )

The derivative of four or higher differentiable functions using the chain rule involves addition of additional link(s) to the chain. Note that the pattern in finding the derivative of higher order functions is similar to obtaining the derivative of two or three functions, given that they are differentiable. One of the most common applications of the chain rule is in taking the derivative of functions that are raised to a power. In general, the derivative of a function to the power of n is defined as: d n n−1 d f ( x) = n f ( x) ⋅ f ( x) = n f ( x) dx dx

[

]

[

]

[

]n−1 ⋅ f ′( x)

[ ]

n

which means, the derivative of a function raised to an exponent, f ' (x ) , is equal to the exponent times the function raised to the exponent reduced by one, n [ f (x )] n −1 , multiplied by the derivative of the function, f ' (x ) , i.e., n [ f (x )] n −1 ⋅ f ' (x ) . Note that the key in using the chain rule is that we always take the derivative of the functions by working our way from outside toward inside. The following examples show in detail the use of chain rule in differentiating different types of functions. Students are encouraged to spend adequate time working these examples. Example 2.4-1: Find the derivative of the following functions. (It is not necessary to simplify the answer to its lowest level. The objective is to learn how to differentiate using the chain rule.)

(

d. f ( x ) = x 3 − x 5

(

g. f ( x ) = x + x 3 j.

 1  f (t ) =   1+ t2 

(

)

)

 1  + x  x2 

8

e. f ( x ) = 

 1   1+ x2 

4

h. f ( x ) = 

3

m. g(u) = u3 + 3u 2

(

b. f ( x ) = (1 + x ) 6

a. f ( x ) = (3 − 5x ) −2

 θ2   k. r(θ ) =  1+ θ 

)

3

Hamilton Education Guides

 t3

c. f ( x ) = 1 + 2 x 2 3



2

x

 a x + b  cx −d

 r2 + r  l. p(r ) =   1+ r  2

(



o. s(t ) =  1 + t 3 

3

3

i. f ( x ) = 

3

 + t 2   t 4 −1 

n. h( t ) = 

1 f. f ( x ) = 1 + 

)

)

2

3

−1  3

 

82

Calculus I

2.4 The Chain Rule

 x2

p. f ( x ) = 

 3

−

2x   5 

( 

)

2

( 

)

3



−1

 θ   1+ θ 2 

q. r(θ ) =  

s. f ( x ) =  x 3 + 2 x − x 2  

4



 t3   1+ t2 

r. f (t ) = 

(  

t. f ( x ) =  x −1 + x −3



v. f ( x ) =  1 + 2 x 2 − x −2 

−1

5

(  

w. f ( x ) =  2 − x −1



Solutions:

)

)

2

−3

 + x 

3

(

u. f ( x ) = 2 − x −1

 + 2x 3  

(

4

)

−3

−1

)

a. Given f ( x=) ( 3 − 5 x )−2 then f ′ ( x ) = − 2 ( 3 − 5 x )−2−1 ⋅ 0 − 5 x1−1 = − 2 ( 3 − 5 x )−3 ⋅ (− 5) = 10 ( 3 − 5 x )−3

(

)

b. Given f ( x )= (1 + x )6 then f ′ ( x ) = 6 ( 1 + x ) 6−1 ⋅ 0 + x1−1 = 6 ( 1 + x ) 5 ⋅1 = 6 ( 1 + x ) 5

(

3

c. Given f ( x )= (1 + 2 x 2 ) then f ′ ( x ) = 3 1 + 2 x 2

) ⋅ ( 0 + 4 x ) = 3 (1 + 2 x ) 3−1

(

8

x ) ( x3 − x5 ) then f ′ ( x ) = 8 x 3 − x 5 d. Given f (= 3

 1   2 + x x 

x) e. Given f (=

) ⋅ ( 3x 8−1

3−1

(

3

= ( x −2 + x ) then f ′ ( x ) = 3 x −2 + x

3

2 2

2 −1

− 5 x 5−1

) ⋅ (− 2 x 3−1

(

3

1 f. Given f ( x=) 1 +  = f ( x )= (1 + x −1 ) then f ′ ( x ) = 3 1 + x −1 

x

(

4

g. Given f ( x=) ( x + x3 ) then f ′ ( x ) = ( f  g )′ (x ) = 4 x + x 3 1 h. Given f ( x ) = 

   1 + x2 

=

2 ⋅ −2 x

(1 + x )⋅ (1 + x ) 2

2 2

2

=

  2  1+ x  

1

then f ′ ( x ) = 2  −4 x

(1 + x )

2 1+ 2

=

−4 x

(1 + x 2 )

2 −1

(

1

derivative of f ( x ) . Hence, f ′ ( x ) = − 2 1 + x 2

Hamilton Education Guides

− 2 −1

3

(

= 12 x 1 + 2 x 2

− x5

) ( 3x 7

) = 3 (x

+1

−2

2

− 5x 4

+x

)

2

)

) (− 2 x 2

−3

+1

) ⋅ ( 0 − x ) = −3 x (1 + x−1 ) a 3−1

2

−2

−1−1

) (1 + 3 x ) 3

2

)]

  2  0 ⋅ 1 + x − [2 x ⋅1] ⋅  2  1+ x 2 

(

)

−2 x  2   ⋅  1 + x2  1 + x2 2

=

(

)

3

  2  1+ x  

A second method is to rewrite f (x ) = 

[ (

) = 8( x

⋅ 4x

)

2

= (1 + x 2 )

−2 −1

 

⋅ ( 0 + 2x)

−1 

 

2

= (1 + x 2 )

(

= − 2 1+ x 2

)

−2

−3

and then take the

⋅ ( 2x)

=

−4 x

(1 + x 2 )

3

83

)

Calculus I

2.4 The Chain Rule

2

2 −1

 ax + b 

ax+b  i. Given f ( x ) =   then f ′ ( x ) = 2  cx−d   cx − d  



 a x + b   (acx − ad ) − (acx + bc )    (cx − d )2

  = 2   c x − d  

=

2 (a x + b )(− ad − bc )

(c x − d )(cx − d )

=

2

 a x + b   acx − ad − acx − bc   

/ //   / / / = 2  (cx − d )2  c x − d  

− 2 (a x + b )(ad + bc )

(c x − d )

( cx − d )

3−1

1+ t 

3

2

−2t

⋅

(1 + t 2 ) (1 + t 2 )

2

3 ⋅ −2t

=

=

(1 + t ) ⋅ (1 + t ) 2 2

2 2



(1 + t )

(

derivative of f ( t ) . Hence, f ′ ( t ) = − 3 1 + t 2

k. Given

=

3θ 4

(1 + θ ) (1 + θ ) 2

l. Given

=

⋅

θ 2 + 2θ 2

3

   

=

3

 r2 + r  p(r ) =    1+ r   

 r2 +r   3  (1+ r)  

2

then

r ′ (θ )

(

=

3θ 4 ⋅ θ 2 + 2θ

(1 + θ ) ⋅ (1 + θ ) 2

 θ2 3  1+θ 

)

2

then p′ ( r ) =

 2r + 2r 2 + 1 + r − r 2 − r      ( 1 + r )2

Hamilton Education Guides

=

   

3

−3−1

= (1 + t 2 ) ⋅ ( 0 + 2t )

−1  3

 

1

(

= (1 + t 2 )

(

= − 3 1+ t 2

)

−4

−3

(1 + θ )

3−1

 r (1/ + r/ )  = 3   (1/ + r/ ) 

=

)

(1 + θ )

=

2

  1+θ   



3θ 4+1 (θ + 2 ) 4

and then take the

2t

[ ] = 3  θ

2+ 2

  2 

4

 [2θ ⋅ (1 + θ )] − 1 ⋅θ 2 ⋅  (1 + θ )2

3θ 4 ⋅θ (θ + 2 )

 r2 +r   3  1+ r   

(1 + t 2 )



2   − 2t   1 + t 2   1 + t 2



= 3 

)

− 6t

=



)

3−1

)

(

2 2+ 2

  2  1+ t 

 θ2 r (θ ) =  1+θ 

(

3

  0 ⋅ 1 + t 2 − 2t ⋅1  ⋅ 2   1+ t 2  

−6t

1

A second method is to rewrite f (t ) = 

 a x + b   − ad − bc 

  = 2    c x − d   (cx − d )2 

−2 ( ax + b ) ( ad + bc )

=

1+ 2

3  1  1  j. Given f ( t ) =  2  then f ′ ( t ) = 3  2  1+ t 

=

 [a ⋅ (cx − d )] − [c ⋅ (ax + b )] ⋅  (cx − d )2  

=

[(

− 6t

(1 + t 2 ) 2

4

 2θ + 2θ 2 − θ 2     (1 + θ )2 

(θ + 2) 4 (1 + θ )

3θ

5

2  [( 2r + 1) ( 1 + r )] − 1 ⋅ r + r ⋅  ( 1 + r )2

)] 

2 2

 

 (1/ + r/ )2/  r + 2r + 1  2  = 3r 2   = 3r 2 ⋅1 = 3r  (1/ + r/ )2/  ( 1 + r )2 

84

Calculus I

2.4 The Chain Rule 3

 r2 +r     1+ r   

A simpler way is to note that p(r ) =

=

(

3

u ) ( u 3 + 3u 2 ) then g ′ ( u ) = 3 u 3 + 3u 2 m. Given g (=

n.

 t3  h (t )  4 + t2  Given=  t −1   

 t3

then

h′ ( t )

  3t 6 − 3t 2 − 4t 6

 + 2t   

= 2  4 + t 2      t −1

( t − 1)

2

( 

)

(

−4

9t 2 = −

4

−1 

o. Given s(t ) =  1 + t 3 = −9t 2 (1 + t 3 )

x) p. Given f (=

q. Given

=

r. Given

=

4t 9

−2

⋅

3  = 1 + t

 x2 2 x  −    3 5  

⋅

−1

1−θ 2

(1 + θ 2 )

 t3  f (t ) =   1+ t2   

3

3

(1 + t 3 )

 θ  r (θ ) =   1+θ 2 

 θ  −  1+θ 2 

2

3t 2 + t 4

(1 + t 2 ) (1 + t 2 )

2

4

=

Hamilton Education Guides

2

−1

)

−3

) ⋅ [ 3u 3−1

  t3 +t2  2  t 4 −1   

=

2 −1

3

 r (1/ + r/ )     (1/ + r/ ) 

=

+ ( 3 ⋅ 2) u

2

[(

= r 3 then p′ ( r ) = 3r 2

] = 3(u

3

) ( 3u 2

+ 3u 2

) ][ ( ( )

2

+ 6u

)

)]

 + 2t   

  − t 6 − 3t 2

 t3

3

  4 3 3 2   t − 1 ⋅ 3t − t ⋅ 4t − 0 + 2t  ⋅ 2   t 4 −1

= 2  4 + t 2     t4 −1  t −1

(

(

then s ′ ( t ) = − 3 1 + t 3

)

)

2

−3−1

(

⋅ 0 + 3t 2

) = −3(1 + t ) 3

−4

⋅ 3t 2

4

 2



 3

5 

x 2x then f ′ ( x ) = −  − 

 θ  −  1+θ 2 

then r ′ (θ ) =

=

 r (1 + r )     (1 + r ) 

2

1+θ 2  1−θ 2 −  ⋅ 2  θ    1+θ 2

(

then f ′ ( t ) =

4t 9

3

⋅

(

)

 t3  4  1+ t2   

t2 3 + t2

(1 + t 2 ) (1 + t 2 )

2

)

=

−1−1

−1−1 1 ⋅

 ⋅  

2 2 ⋅ x −  5 3

=

2/

θ2

(

4 −1 

1−θ 2

⋅

)

(1 + θ 2 )

(

4t 9+ 2 3 + t 2

(1 + t )

2 3+ 2

 3

5 

=

 θ  −  1+θ 2 

)=

)

(

1−θ = − 2

2/

2 2 3  3t ⋅ 1 + t − 2t ⋅ t  ⋅  2   1+ t2  

(



(1 + θ 2 ) − 2θ ⋅θ  2 (1 + θ 2 ) 

(1 + θ 2 ) −

θ

(1 + t 2 )

5

2

2 2  x−  5 3

=

−2 

 1 + θ 2 − 2θ 2   2   1+θ 2   

(

)

θ2 − 1 θ2

 3  t 3   3t 2 + 3t 4 − 2t 4  4   2 1+ t2       1+ t2  

=

4t 11 3 + t 2

−2

 2

x 2x = −  − 

(

)

)

85

Calculus I

2.4 The Chain Rule

2

(

4

s. Given f ( x ) = ( x3 + 2 x ) − x 2  then f ′ ( x ) = 4  x 3 + 2 x    

(

 = 4  x 3 + 2x 

)



3

2

 − x2 ⋅ 

{[ 2( x

2

3

)

(

)

2

u. Given f ( x=) ( 2 − x −1 ) = −

(

3

x 2 2 − x −1

)

2

 + x 

(

−1

+ x −3

) (− x

−2

)

3

then f ′ ( x ) = − 3 2 − x −1

3



4

(

5

)

−3−1

)

2

 + x 

(

⋅ 0 + x −2

(

  − x − 2  12 x 1 + 2 x 2   −3

−1 3 w. Given f ( x ) = ( 2 − x ) + 2 x 





−3





−1



= −  ( 2 − x −1 ) + 2 x 3  

2−1

2

)

  + 2  − 2 x  

3−1

(

 −1 −3  2 x + x  

) (− x 2 −1

−2

)

  − 3 x − 4  + 1  

) = −3(2 − x )

−1 −4 − 2

x

=

−3x −2

( 2 − x−1 )

4

4



(

) ⋅ ( 3x

)] }

 2 2 −2 v. Given f ( x ) = (1 + 2 x ) − x  then f ′ ( x ) = 5  1 + 2 x     = 5  1 + 2x 2 

(

 ⋅  2 x 3 + 2 x 

− 3 x −4 + 1

(

−3

4−1

)] }

)(

3

{[2 ( x

 − x2  

+ 2x 3x 2 + 2 − 2x

   t. Given f ( x ) = ( x −1 + x −3 ) + x  then f ′ ( x ) = 3  x −1 + x −3   

 = 3  x −1 + x − 3 

2



−2



)

2

)

3

 − x −2  

5−1

(

 ⋅ 3 1 + 2 x 2 

)

3−1

 ⋅ 4 x + 2 x − 2−1  

 + 2 x −3  

(

 then f ′ ( x ) = −  2 − x −1 

)

−3

 + 2 x3  

−1−1

(

)

−3−1 − 2   ⋅ − 3 2 − x −1 ⋅ x + 6 x2   

−4   −2 2 − x −1 + 6 x2   −3 x  

(

)

Example 2.4-2: Find the derivative at x = 0 , x = −1 , and x = 1 in example 2.4-1 for problems a - g. Solutions: a. Given f ′( x) = 10 ( 3 − 5x ) −3 , then f ′( 0)

f ′ ( −1)

= 10 3 − ( 5 ⋅ 0 ) 

−3

= 10 3 − ( 5 ⋅ −1) 

Hamilton Education Guides

10 1 = 10[3 − 0]−3 = 10 ⋅3−3 = 10 ⋅ 3 = = 0.37 27

3

−3

10 1 = 10[3 + 5]−3 = 10 ⋅8−3 = 10 ⋅ 3 = = 0.019

8

512

86

Calculus I

f ′ (1)

2.4 The Chain Rule

= 10[ 3 − ( 5 ⋅ 1)]

−3

= 10[ 3 − 5]

−3

= 10 ⋅ ( −2) −3 = 10 ⋅

1 ( −2)

=

3

10 −8

10 = − = −1.25 8

b. Given f ′( x ) = 6 (1 + x ) 5 , then = 6 (1 + 0 )5 = 6 ⋅15 = 6 ⋅1 = 6

f ′( 0)

f ′ ( −1) f ′ (1)

= 6 (1 − 1)5 = 6 ⋅ 05 = 6 ⋅ 0 = 0

= 6 (1 + 1)5 = 6 ⋅ 25 = 6 ⋅ 32 = 192

( ) f ′ ( 0 ) = ( 12 ⋅ 0 ) [ 1 + ( 2 ⋅ 0 ) ]

2

c. Given f ′( x ) = 12 x 1 + 2 x 2 , then 2

2

f ′ ( −1)

[

= ( 12 ⋅ −1) 1 + 2 ⋅ (− 1)2

= 0 ⋅ ( 1 + 0 ) 2 = 0 ⋅ 12 = 0 ⋅ 1 = 0

]

2

= − 12 ( 1 + 2) 2 = − 12⋅ 3 2 = −12⋅ 9 = −108

[ ( ) ] = 12 (1 + 2) = 12⋅ 3 = 12⋅ 9 = 108 d. Given f ′( x ) = 8 ( x − x ) ( 3x − 5 x ) , then f ′( 0) = 8 ( 0 − 0 ) [ 3 ⋅ 0 − 5 ⋅ 0 ] = 8 ⋅ 0 ⋅ 0 = 8 ⋅ 0 ⋅ 0 = 0 f ′ (1)

2

= ( 12 ⋅1) 1 + 2 ⋅12 3

3

f ′ ( −1) f ′ (1)

5 7

[

= 8 (− 1)3 − (− 1)5

(

= 8 13 − 15

) [ 3 ⋅1 7

5 7

2

2

4

] [ 3 ⋅ (− 1)

2

(

7

− 5 ⋅ 14

e. Given f ′( x ) = 3 x −2 + x  1

 + 0  2 0 

= 3 

f ′( 0)

2

2

)( 2

2

4

7

2

− 5 ⋅ (− 1)4

] = 8 ( 1 − 1)

7

] = 8 [− 1 + 1]

7

( 3 ⋅1 − 5 ⋅1) =

[ 3 ⋅1 − 5 ⋅1] = 8 ⋅ 07 ( 3 − 5 )

8 ⋅ 07 ⋅ −2

= 8 ⋅ 0 ⋅ −2 = 0

= 8 ⋅ 0 ⋅ −2 = 0

2

)

 1   2  + x  − + 1 , then −2 x −3 + 1 = 3  x2   x3  2

 2   1   2  − + 1 = 3  + 0   − + 1  3 0 0      0 

f ′ (0)

is undefined due to division by zero.

2

f ′ ( −1)

f ′ (1)

=

 1    2 3 − 1  − + 1 = 3 (1 − 1) 2 (2 + 1) = 3 ⋅ 02 ⋅ 3 = 0  (− 1) 2   (− 1) 3      2

 1   2  = 3  2 + 1  − 3 + 1 = 3 (1 + 1) 2 (− 2 + 1) = 3 ⋅ 22 ⋅ −1 = 3 ⋅ 4 ⋅ −1 = −12 1   1 

(

f. Given f ′( x ) = −3x −2 1 + x −1

Hamilton Education Guides

)

2

= −

(

3 1+ x x

)

−1 2

2

= −

 1 3 1 +   x x2

2

, then

87

Calculus I

2.4 The Chain Rule

= −

f ′( 0)

f ′ ( −1)

f ′ (1)

 1 3 1 +   0

1   3 1 +  1 − 

1

= −

= −

2

(

f ′ ( −1) f ′ (1)

= −

3⋅ 02 0 = − = 0 1 1

2

3⋅ 2 = − = − 1

3

3⋅ 4 1

= −12

2

2

[

(

3 (1 − 1) 2 1

3 (1 + 1) 2 1

3

= 4 − 1 + (− 1)3

= 4 1 + 13

is undefined due to division by zero.

) (1 + 3x ), then ) (1 + 3 ⋅ 0 ) = 4 ⋅ 0 ⋅ 1 = 4 ⋅ 0 ⋅ 1 = 0 (

= 4 0 + 03

f ′ (0)

2

g. Given f ′( x ) = 4 x + x 3 f ′( 0)

2

2

(− 1)2

 1 3 1 +   1

1  3 1 +  0  − 0

=

02

= −

= −

2

3

] [1 + 3 ⋅ (− 1) ] 3

= 4 ⋅ (− 1 − 1) 3 ⋅ ( 1 + 3) = 4 ⋅ (− 2) 3 ⋅ 4 = 4 ⋅ (−8) ⋅ 4 = −128

2

) (1 + 3 ⋅1 ) = 4 ⋅ (1 + 1) ⋅ (1 + 3) = 4 ⋅ 2 ⋅ 4 = 4 ⋅ 8 ⋅ 4 = 128 3

3

2

3

Example 2.4-3: Use the chain rule to differentiate the following functions. Do not simplify the answer to its lowest term.

(

)

5

a.

d 2 x +3 dx

d.

 2 d  t +3  dt  t − 1 

g.

d  2 u +4 du 

j.

d  x + 2   dx  x − 2 

m.

=

(

)

(

)( 6

2

=

)

 u −1   3

Solutions:

(

)

5

a.

d 2 x +3 dx

b.

3 d  2  x + 5 + 1  dx  

(

e.

3  3 d  θ + 2θ    dθ  (θ + 1) 2   

)

4

=

(

4

)

5−1

=

=

)(

)

(

k.

3 d  ( 2 x + 5)    dx  (1 + x ) 2   

n.

3 4 d  3 2x + 1 x 2 + 1   dx  

(

(

) = 5(x

)

4 −1

d x2 + 3 dx

3 = 4  x 2 + 5 + 1  

Hamilton Education Guides

)

(

=

= 5 x2 + 3

(

4

(

2

=

)(

)

+3

4

)

⋅ 2x

f.

4 d  2 2 r r +3   dr  

(

)

2 d  (1 − x )    dx  x 3 + 2 x   

8

=

=

=

=

=

= 10 x ( x 2 + 3 )

3 d  2  x + 5 + 1  dx  

)

3 d  2  u + 1 (u + 5)  du  

l.

)

(

c.

d  x 3 + 3x    i. dx  1 − x 2 

3 d  3 h. t + 2t 2 + 1 t 2 + t + 1  =  dt  

=

1  d  2 x + 3  dx  x + 5

)

3 d  2  x + 5 + 1  dx  

5

   

(

b.

(

4

)

3

3

(

 = 4  x 2 + 5 + 1 ⋅ 3 x 2 + 5   

)

3−1

(

)

d  x 2 + 5 + 0 dx 

88

Calculus I

2.4 The Chain Rule

(

)

3

3

c.

(

)

(

)

(

)

3 d  2  u + 1 (u + 5)  du  

d.

 2 d  t +3  dt  t − 1 

=

e.

   

(

)

(

3

3

)

(

=

(

)

5

( 

−  θ 3 + 2θ

3

)

( t − 1) 4

)

(

 ⋅ 2t  − t 2 + 3 

=

5

(

=

 d ( t − 1)  dt 

=

(

)

(

2 −1

⋅

4

 d (θ + 1)  dθ 

) ( 3θ 2

2

2

)

4

(

)

(

)

3

=

(

) (

) (

)

4

3

4 4 −1 d  2   2  2 ⋅ r2 + 3   r + 3 ⋅ 2r  +  r ⋅ 4 r + 3 dr    

(

=

4

)

3

(

3



)

(

(

)

3

6  2 2 +  u + 4 ⋅ 3u   

3

(

6

2

) (

)

6

2

(

3

3

)

(

)

)

) (u − 1) = (u − 1) dud (u + 4)  + (u + 4) dud (u − 1) = ( u − 1)⋅ 6 ( u 6

(

 r 2 + 3 + 4r 2   

(

d  2 u +4 du 

)

( θ + 1) 4

  = 2r ( r 2 + 3) +  4r 2 ( r 2 + 3) ⋅ 2r  = 2r ( r 2 + 3) + 8r 3 ( r 2 + 3) = 2r ( r 2 + 3 )

g.

(

)

3−1 d   2 3 θ 3 + 2θ  ⋅ ( θ + 1) ⋅ 3 θ + 2θ d θ  



)

(



)

2

4

)

4

3

( θ + 1)

4 d 4  2   d 2 r 2  + r 2 r +3   r +3 dr   dr  

(

2

5

(θ + 1)

  + 2  −  2 θ 3 + 2θ  

( θ + 1)

)

2 3    3  2 3 2 (θ + 1) ⋅ 3 θ + 2θ ⋅ 3θ + 2  −  θ + 2θ ⋅ 2 (θ + 1) ⋅1    

=

)

+5

)

) (

(

)

( t − 1)

) (

( t − 1)

(

2

5−1 d 5     2 ⋅ t 2 + 3  −  t 2 + 3 ⋅1 ( t − 1) ⋅ 5 t + 3 dt    

4

(

)

2

=

3

  = 24 x  x 2 + 5 + 1  

= 6u ( u + 5 ) ( u 2 + 1) + ( u 2 + 1)

3

10t ( t − 1) t 2 + 3 − t 2 + 3

(θ + 1)

 2 3  3 ( θ + 1) θ + 2θ

)

2

(

(θ + 1)

(

)

3  3 d  2 d 2 θ 3 + 2θ  −  θ 3 + 2θ (θ + 1)  (θ + 1) d θ d θ    

⋅ 2 (θ + 1)

4 d  2 2 r r +3   dr  

5

)  − (t 2 + 3)

2

(

(



2

)

)

(

)

(

( t − 1)2

3  3 d  θ + 2θ    dθ  (θ + 1) 2   

=



(

)

(x

3  3 d 3−1 d      d 2 u2 + 1  u + 1  +  u2 + 1 ( u + 5)  = ( u + 5) ⋅ 3 u 2 + 1 du du du       

= ( u + 5 )

 d 2 ( t − 1) dt t + 3 

 2 ( t − 1) ⋅ 5 t + 3



f.

=

3



(

2

= (u + 5) ⋅ 3 u 2 + 1 ⋅ 2u   + u 2 + 1   

5

)

)



3   2 +  u + 1 ⋅1  

(

(

= 4  x 2 + 5 + 1 ⋅ 3 x 2 + 5 ⋅ 2 x  = 4  x 2 + 5 + 1 ⋅ 6 x x 2 + 5      

)(

2

+4

)

)

6 −1

⋅

(

)

d  u2 + 4  du 

(

)

5 6 5 6 =  u 3 − 1 ⋅ 6 u 2 + 4 ⋅ 2u  + 3u 2 u 2 + 4  = 12u u 3 − 1 u 2 + 4  + 3u 2 u 2 + 4         

Hamilton Education Guides

89

Calculus I

2.4 The Chain Rule

(

= 3u u 2 + 4 h.

) [ 4 ( u − 1) + u ( u 5

3

(

)(

(

) (

)

d  3 t + 2t 2 + 1 t 2 + t + 1 dt 

2

+4

)]

(

3

) dtd ( t

 2  =  t + t + 1 

) (

3

) (

) (

)

d 2   + 2t 2 + 1  +  t 3 + 2t 2 + 1 t + t +1 dt  

3

) (

)

(

 

)

) (

(

)

3

3 3−1 d 3   =  t 2 + t + 1 ⋅ 3t 2 + 4t  +  t 3 + 2t 2 + 1 ⋅ 3 t 2 + t + 1 ⋅ t 2 + t + 1  =  t 2 + t + 1 ⋅ 3t 2 + 4t  dt      

(

) (

(

)

2  3  2 2 +  t + 2t + 1 ⋅ 3 t + t + 1 ⋅ (2t + 1)  

d  x 3 + 3x    i. dx  1 − x 2 

=

j.

8

=

 x 3 + 3x   8  1− x 2   

7

[(

d  x + 2   dx  x − 2 

2

 x−2

=

−

l.

d dx

)] [( ( )

 x+2 2   x−2

2 −1

d  x+2 ⋅  dx  x − 2 

( x − 2 ) ⋅1 − ( x + 2 ) ⋅1

( x − 2)

3 d  ( 2 x + 5)    dx  (1 + x ) 2   

−

d  x 3 + 3 x  ⋅  dx  1 − x 2 

=

) {[ t ( t 2

(

)

2

 x + 2   2 ⋅  x−2

]

=

( x − 2)

( x − 2)

(1 + x )4

[(

)( 3 x

) (

)

)] [ ( (1 − x )

2

+ 3 + 2x 2 x 2 + 3

)]

2 2

2

−4 x+2 ⋅ 2 − x  ( x − 2 )2 

 = 2

4

2

) ( ( )

]}

d d ( x + 2 )  − ( x + 2 ) ( x − 2 )  dx dx   

x + 2  x/ − 2 − x/ − 2 ⋅  x − 2  ( x − 2 )2

[( 1 + x) =

) (

7

 = 2

(1 + x )

)

1− x2  x 3 + 3x   ⋅ = 8  2   1− x 

2 d 3  3 d 2  (1 + x ) dx ( 2 x + 5 )  − ( 2 x + 5 ) dx (1 + x )     

 3 2 −1 d (1 + x ) (2 x + 5) ⋅ 2(1 + x ) dx  

[ 2 ( 2 x + 5)



=

][ (

)

+ t + 1 ( 3 t + 4 ) + 3 t 3 + 2 t 2 + 1 ( 2 t + 1)

2

d d     7 1− x 2  1− x 2 x 3 + 3x  −  x 3 + 3x  x 3 + 3 x   dx dx     ⋅ 8 2  1− x 2    1− x 2

1 − x 2 ⋅ 3 x 2 + 3 − x 3 + 3 x ⋅ (− 2 x )  x 3 + 3x   ⋅ 8 2  1− x 2    1− x 2

x+2  = 2  ⋅

k.

)(

8−1

= t2 + t +1

= −

8( x + 2 )

( x − 2)

3

 2 3−1 d ( 2 x + 5) ( 1 + x ) ⋅ 3 ( 2 x + 5 ) dx  

=

( 1 + x) 4

] [ ( 2 x + 5)

⋅ 3 ( 2 x + 5) 2 ⋅ 2 −

3

⋅ 2 ( 1 + x ) ⋅1

]

( 1 + x) 4

[6 ( 1 + x) =

( 2 x + 5) 2 ] ( 1 + x) 4 2

2 ( 1 + x )] 2( 2 x + 5) ( x − 2) 2(1/ + x/ ) (2 x + 5)2 [3(1 + x ) − (2 x + 5) ] 2(2 x + 5)2 (3 + 3 x − 2 x − 5) = = = 3 (1 + x )4/ =3 ( 1 + x) 4 (1 + x )3 (1 + x )

 (1 − x ) 2     x 3 + 2x   

3

=

(

)

(

)

d 2  2 d  3  3  x + 2 x dx (1 − x )  − (1 − x ) dx x + 2 x     

Hamilton Education Guides

( x3 + 2 x )

2

=

2 −1 d  3  ⋅ (1 − x )   x + 2 x ⋅ 2 (1 − x ) dx  

(

)

( x3 + 2 x )

2

90

Calculus I

2.4 The Chain Rule

 ( 1 − x )2 ⋅ ( 3 x 2 + 2 )   − 2 x 3 + 2 x (1 − x )  − (1 − x ) 2 3 x 2 + 2   2 ( x3 + 2 x ) ( 1 − x ) ⋅ −1 −  ( 1 − x ) 2 ( 3 x 2 + 2 )  ) ( )        (   = = − 2 2 2

( x3 + 2 x )

( x3 + 2 x )

1  d  2 x + 3  dx  x + 5

m.

=

n.

4

1   4  x2 + 3  x +5 

=

(

d  1  ⋅  x2 + 3  dx  x +5

)

(

 d   d 3  3 3  x + 5 dx (1)  − 1 ⋅ dx x + 5 1    2    4 x + 3  ⋅ 2x + 2 x +5  3  x +5  

(

(

)(

)

) (

)

3 4 d  3 2x + 1 x 2 + 1   dx  

( 



4

=  x 2 + 1 ⋅ 3 2 x3 + 1

+

4 −1

3−1

=

⋅

( x3 + 2 x )

)

)     

3

=

 d  1  1  d 2 x +  3 4  x 2 +  ⋅   3 dx dx  x + 5  x +5  

=

 3 1   3 x2   2 4 x + 3  2 x − 2 x +5   x3 + 5   

( )

(

)

4 d 3  3 d 4  2 2 x3 + 1  +  2 x3 + 1 x2 + 1   x +1 dx dx    

) (

(

)

) (

(

)

(

) ⋅ 3(2 x + 1)

3 4 −1 d     d ⋅ 2 x3 + 1  +  2 x3 + 1 ⋅ 4 x 2 + 1 x2 + 1  =  x 2 + 1 dx dx    

(

) (

) (

)

(

)

4

3

2

 ⋅ 6x 2  

( 2 x3 + 1)3 ⋅ 4 ( x 2 + 1)3 ⋅ 2 x  = 18 x 2 ( x 2 + 1)4 ( 2 x 3 + 1)2  + 8 x ( 2 x 3 + 1)3 ( x 2 + 1)3           

In some instances students are asked to find the derivative of a function y , where y is a function of u and u is a function of x . We can solve this class of problems using one of two methods. The first method, and perhaps the easiest one, is performed by substituting u into the y equation and taking the derivative of y with respect to x . The second method is to find the derivative of y

by using the equation

dy dy du = ⋅ dx du dx

. This method is most often used in calculus books and can

be time consuming. For example, let’s find the derivative of the function y = u 2 + 1 where u = x + 1 using each of these methods. First Method: Given the function y = u 2 + 1 where u = x + 1 , substitute u with x + 1 in the function y and simplify, i.e., y = u 2 + 1 = ( x + 1)2 + 1 = x 2 + 2 x + 1 + 1 = x 2 + 2 x + 2 . Next, take the derivative of y with respect to x , i.e.,

dy dx

=

(

dy 2 x + 2x + 2 dx

) = 2x

2 −1

+ 2 x1−1 + 0

= 2x + 2 .

Second Method: Given the function y = u 2 + 1 where u = x + 1 , find 2u 2−1 + 0

= 2u and

du dx

= x1−1 + 0 = x 0 = 1 . Next, substitute

dy dy dy du to find the derivative = ⋅ dx du dx dx dy obtain = 2u = 2 (x + 1) = 2 x + 2 . dx

Hamilton Education Guides

, i.e.,

dy dx

dy du

dy du

and

and

du dx

du dx

, i.e.,

dy du

=

in the equation

= 2u ⋅1 = 2u . Substituting u = x + 1 in place of u we

91

Calculus I

2.4 The Chain Rule

The second method is generally beyond the scope of this book, therefore the first method is used in order to solve this class of problems. Examples 2.4-4 and 2.4-5 below provide additional examples as to how these types of problems are solved. Example 2.4-4: Find a. y = c. y = e. y =

1 1+ u u 1+ u

2

1+ u

dy dx

given:

and u = 3x + 1

b. y = u 2 + 2u + 1 and u = 3x + 2

and u = 5x + 1

d. y = u 2 + 1 and u =

(

Solutions: a. Given y = dy dx

=

1 1+ u

and u = 3x + 1 , then y =

d   d   ( 3x + 2 ) dx (1)  − 1 ⋅ dx ( 3x + 2 )     

( 3x + 2 )

x2 −1

)

f. y = u3 + 1 and u = x 2 + 1

and u = x 2 + 1

u3

x +1

2

=

1 1 + ( 3x + 1) 0−3

( 3x + 2 )

2

=

= −

1 3x + 2 3

( 3 x + 2)

−1

and

2

b. Given y = u 2 + 2u + 1 and u = 3x + 2 , then y = ( 3x + 2) 2 + 2( 3x + 2) + 1 = ( 3x + 2) 2 + 6 x + 5 and dy dx

=

d d d 2 ( 3x + 2 ) + 6 x + 5 dx dx dx

c. Given y = dy dx

=

=

u 1+ u

2

dx

and u = 5x + 1 , then y =

(5x + 1) and 2 1 + ( 5x + 1)

d  2 d 2      1 + ( 5 x + 1)  dx ( 5 x + 1)  − ( 5 x + 1) ⋅ dx 1 + ( 5 x + 1)       1 + ( 5 x + 1)2   

[

][

5 1 + ( 5 x + 1) 2 − 10 ( 5 x + 1) 2

[1 + ( 5x + 1) ]

2 2

d. Given y = u 2 + 1 and u = dy dx

d = 2 ( 3x + 2 )2−1 ⋅ ( 3x + 2 ) + 6 + 0 = 2 ( 3x + 2 ) ⋅ 3 + 6 = 18 ( x + 1)

x +1 x2 −1

=

=

2

)

=

Hamilton Education Guides

)

2 2

2

(

=

5 − 5 (5 x + 1) 2

[1 + ( 5 x + 1) ]

2 2

and

)

(

)

d d  2 x −1 ( x + 1)  − ( x + 1) ⋅ x 2 − 1   x + 1   dx dx    2 2  ⋅ 2 2  x −1 x −1

 2     x + 1   x − 1 ⋅1 − ( x + 1) ⋅ 2 x  ⋅  2  x2 − 1  x2 − 1

(

2 2

[1 + (5x + 1) ]

 x +1   +1  x 2 − 1

=

=

2

5 + 5 (5 x + 1) 2 − 10 (5 x + 1) 2

, then y = 

  x + 1 2−1 d  x + 1   2  ⋅  2  + 0  dx  x − 1     x2 − 1   

(

]

2

{[ 1 + ( 5x + 1) ]⋅ 5 }− [( 5x + 1)⋅ 2 ( 5x + 1)⋅ 5 ] [1 + ( 5x + 1) ]

(

=

)

   x + 1   x2 − 1 − 2 x2 − 2 x  ⋅    2  x 2 − 1    x2 − 1  

2

(

)





x + 1  x2 + 2 x + 1  = 2  2  ⋅  − 2   x − 1   x 2 − 1  

(

)



92

Calculus I

2.4 The Chain Rule 2

2

( x2 − 1)

( x2 − 1)( x2 − 1)

( x + 1)( x + 1) x + 1 ( x + 1) −2 = −2  2  ⋅ 2 =  x −1

1+ u

e. Given y =

dy dx

2

and u = x + 1 , then y =

u3

3  x +1 ( x + 1) −2  2 = −2 3 =

2

( x2 − 1)

(

=

(

)

( x + 1) 2

)

) (

) ( 6 ( x2 + 1)

3  2   2 2 2 2x x + 1  −  x + 2 ⋅ 3 x + 1 ⋅ 2x     

(

=

)

)

( ) ( 6 ( x2 + 1)

(

3

2

)

f. Given y = u + 1 and u = x + 1

−1

, then

)

1+ x 2 + 1

3 d 3  2   d 2 x2 + 2  −  x2 + 2 ⋅ x +1   x +1 ⋅ dx dx    

(

   x −1

=

3

x2 + 2

( x + 1) 2

(

=

(

(

3

3 2 d  2   2  2 x2 + 1   x + 1 ⋅ 2x −  x + 2 ⋅ 3 x + 1 dx    

)

 2 2 x x + 1 

=

3

)

 y =  x2 +1 

)

(

3

) ( 6 ( x2 + 1)

(

)(

 2 2  − 6 x x + 2 x + 1  

( x 2 + 1)

−1  3

 +1 

6

(

)

= x2 +1

−3

)

)

(

)

2

 

+1

−3−1 d −4 −4   6x 2 2 2 ⋅ ( x + 1) + 0 = −3( x + 1) ⋅ 2 x = −6 x ( x + 1) = − = −3( x 2 + 1) 4  2 dx   ( x + 1)

dy dx

Example 2.4-5: Find y ′ given that: 3u 2 and u = x 2 1+ u 1 − 5u d. y = 2 and u = 1 − x u

a. y = 3u3 − 1 and u = x 2 + 1 c. y = e. y = g. y =

b. y =

u +1 and u = x 2 + 3 u −1 u

and u = 2 x − 1

2

u +1 2u

(u − 1)

2

f. y = u +

1 and u = (2 x − 1) 4 4

h. y = u 4 − 1 and u =

and u = x + 2

1+ x 1− x

Solutions:

(

)

3

a. Given y = 3u3 − 1 and u = x 2 + 1 , then y = 3 x 2 + 1 − 1 and y′

(

b. Given y = y′

)

= 3⋅3 x 2 +1

=

(

3−1

⋅ 2x − 0

(

)

= 9 x 2 + 1 ⋅ 2 x = 18 x ( x 2 + 1) 2

2

3u 2 3x 4 and u = x 2 , then y = and 1+ u 1+ x 2

)( ) 2 (1 + x2 )

 3 ⋅ 4 x 4−1 ⋅ 1 + x 2  −  2 x ⋅ 3x 4     

Hamilton Education Guides

=

( ) 2 (1 + x2 )

12 x3 1 + x 2 − 6 x5

=

12 x3 + 6 x5

(1 + x2 )

2

=

(

6 x3 2 + x2

(1 + x 2 )

2

)

93

Calculus I

2.4 The Chain Rule

c. Given

( x + 3) + 1 = x + 4 and and u = x + 3 , then y = ( x + 3) − 1 x + 2 [ 2x ( x + 2) ]− [2x ( x + 4) ] ⋅ ( x + 2 ) ]− [ 2 x ⋅ ( x + 4) ] = ( x + 2) ( x + 2)

u +1 y= u −1

[ 2x y′ =

2 −1

d. Given y =

2

1 − 5u

2

2

2

2

2−1

1 − 5(1 − x )

(1 − x )

⋅ −1 ⋅ ( 5 x − 4)

]

) (

(

(1 − x )

e. Given y =

u

4

u +1

=

=

2 ⋅ ( 2 x − 1) + 1 −  2 ( 2 x − 1)    ( 2 x − 1)2 + 1   2

2 ( 2 x − 1) + 2 − 4 ( 2 x − 1) ( 2 x − 1)2 + 1  

f. Given y = u + y′

2

=

=

−4 x

( x 2 + 2)

(1 − x ) 2

(1 − x ) 2x − 1

4

−5 x 2 + 8 x − 3

=

(1 − x )

4

and

( 2 x − 1) 2 + 1

2

⋅ 2 ⋅ ( 2 x − 1) 

=

2 − 2 ( 2 x − 1)

2

( 2 x − 1)2 + 1  

2

2 ⋅ ( 2 x − 1) + 1 −  4 ( 2 x − 1) ⋅ ( 2 x − 1)    ( 2 x − 1)2 + 1  

=

[ ] [ (2x − 1) + 1]

2 1 − (2 x − 1)2

=

2

2

2

−2

[ ( 2 x − 1)

[ ( 2 x − 1)

2

2

−1

+1

]

2

]

1 1 and u = ( 2 x − 1) 4 , then y = ( 2 x − 1) 4 + and 4 4

2u

(u − 1)

2

and u = x + 2 , then y =

[ 2 ⋅ ( x + 1) ]− [ 2 ( x + 1) = 2

( x + 1)

=

2 −1

4

2 x 2 + 4 x + 2 − 4 x 2 − 12 x − 8

( x + 1)

4

Hamilton Education Guides

]

⋅ ( 2 x + 4)

=

2( x + 2)

( x + 2 − 1)

=

( x + 1)

2

=

2x + 4

( x + 1) 2

2

2 ( x + 1) − 2 ( x + 1)( 2 x + 4 )

−2 x 2 − 8 x − 6 4

2

5x − 4

=

2



2

( x2 + 2 )

2

3 = 4 ( 2 x − 1)4−1 ⋅ 2 + 0 = 4 ( 2 x − 1)3 ⋅ 2 = 8 ( 2 x − 1)

g. Given y =

y′

2

2 −1

(1 − x )

2

5 x 2 − 10 x + 5 − 10 x 2 + 18 x − 8

=

and u = 2 x − 1 , then y =

2

2

y′

)

1 − 5 + 5x

2 x3 + 4 x − 2 x3 − 8 x

5 (1 − x ) −  −2 (1 − x )( 5 x − 4 )  5 ( 1 − x ) 2 + 2 ( 1 − x ) ( 5 x − 4) = = 4 ( 1 − x) 4 (1 − x )

( 1 − x) 4

5 x2 − 2 x + 1 + 2 5x − 4 − 5x2 + 4 x

=

2

=

2

2

[ 5 ⋅ (1 − x) ]− [ 2 ( 1 − x) y′ = =

2

2

and u = 1 − x , then y =

2

2

2

2 −1

2

u

2

2

( x + 1)

=

(

4

− 2 x 2 + 4x + 3

( x + 1)

4

=

(

) (

2 x2 + 2 x + 1 − 2 2 x2 + 6 x + 4

( x + 1)

4

)

)

94

Calculus I

2.4 The Chain Rule

h. Given y = u 4 − 1 and u = y′

1+ x    1− x 

 = 4

=

4 −1

⋅

4

1+ x 1+ x  , then y =   − 1 and  1− x  1− x

[1⋅ (1 − x )] − [− 1⋅ (1 + x )] (1 − x )2

 (1 + x )3 2  ⋅ 4   (1 − x )3 (1 − x )2 

=

8 (1 + x )3

(1 − x )3+ 2

−0

=

3 3 1 + x  (1 − x ) + (1 + x ) 2 1+ x  ⋅ 4 = 4  =    ⋅ 2 2 1 − x 1− x    (1 − x ) (1 − x )

8 (1 + x )

(1 − x )

3

5

Section 2.4 Practice Problems - The Chain Rule 1. Find the derivative of the following functions. Do not simplify the answer to its lowest term.

(

)



1   x2 

a. y = x 2 + 2 d. y = 1 − 

(

3 2

x +1   3 

g. y = x 2 

(

3

)

4

(x + 1)−2 x

5

e. y = 2 x 3 +

1

1+ x f. y =  3   r 

2





3x 2

k.

 t2 y= 1+ t 2  

   

]

3

3

x 3

x3

o. y =

−1



(

2

4

i. y =  − 2 x 3 

)

−1

l. y = 1 + x −2

 1  + 3 x 1− x  1

n. y = 

3

)

c. y = x 3 − 1

h. y = x ( x + 1)2 + 2 x

j. y = x + 3x + 1 m. y =

(

−2

[

3

2

)

b. y = x 2 + 1

− x2

3

x +2

2. Find the derivative of the following functions at x = 0 , x = 1 , and x = −1 .

(

)

a. y = x 3 + 1

(

b. y = x 3 + 3x 2 − 1

)

2

d. y = x x + 1   2  x +1 

g. y = 

x

)

(

5

2

5

3

(

4

)

2

e. y = x + 2 x + 1

(

)

2

 x2 y= 1+ x 2 

   

f.

 x3 

1

3

h. y = x 2 + 1 ⋅

3

x   x  +1

c. y = 

3

2

 + 5x i. y =  x −1 

x2





3. Use the chain rule to differentiate the following functions. a.

d dt

 (t + 1)3   2   t 

(

=

)

2 d  3  d. x − 1 ( 2 x + 1) 3  =  dx  

Hamilton Education Guides

(

)

b.

 d  u 2 +1 du  3u 4 

e.

d ds

3

   

 3 1  s − 2  s + 6 

= 2

=

 ( 2 x + 1) 3     ( 1 − x ) 2 

c.

d dx

f.

3  d  t 2 −1  dt  t 2 + 1   

(

)

=

=

95

Calculus I

g.

2.4 The Chain Rule

2 3 1   d  2  u +1    du   u + 1   

(

)

=

h.

d dθ

 θ 2 +3    3  ( θ − 1) 

2

=

i.

 r7 d  dr  2  r + 2r

(

)

  3 

=

4. Given the following y functions in terms of u find y ′ . a. y = 2u 2 − 1 and u = x − 1 d. y = u 2 −

1 2

and y = x 4

Hamilton Education Guides

b. y =

u u −1

and u = x3

e. y = u 4 and u =

1 1− x 2

c. y = f. y =

u

1 + u2 u2

(u + 1)3

and u = x 2 + 1 and u = x − 1

96

Calculus I

2.5

2.5 Implicit Differentiation

Implicit Differentiation

In many instances an equation is explicitly represented in the form where y is the only term in the left-hand side of the equation. In these instances y ′ is obtained by applying the differentiation rules to the right hand side of the equation. However, for cases where y is not explicitly given, we must either first solve for y (if y can be factored) and then differentiate or use the implicit differentiation method. For example, to differentiate the equation x y = x 2 + y we can either solve for y by bringing the y terms to one side of the equation and then differentiate as follows: x y = x2 + y

y′

[2x =

2 −1

; x y − y = x 2 ; y( x − 1) = x 2 ; y =

][ ]

⋅ (x − 1) − 1 ⋅ x 2

(x − 1)2

x2 x −1

therefore, y ′ is equal to:

[2 x(x − 1)] − x 2 = = (x − 1)2

2x 2 − 2x − x 2

( x − 1) 2

=

2

x − 2x

( x − 1)

2

or, we can use the implicit differentiation method as shown below. x y = x2 + y

; (1⋅ y + y ′ ⋅ x ) = 2 x 2 −1 + y ′ ; y + y ′x = 2 x + y ′ ; y ′x − y ′ = 2 x − y ; y ′( x − 1) = 2 x − y ; y ′ =

Substituting y =

y′

=

2x − y x −1

=

x2 into the y ′ equation we obtain: x −1 x2 x −1 x −1

2x −

=

2 x( x − 1) − x 2

2 x( x − 1) − x 2

x −1 x −1

x −1 x −1 1

=

[2 x( x − 1) − x ]⋅1 = = 2

( x − 1) ⋅ ( x − 1)

2x − y x −1

2

x − 2x

( x − 1)

2

Note that the key in using the implicit differentiation method is that the chain rule must be applied each time we come across a term with y in it. Following are additional examples showing the two methods of differentiation when y is not explicitly given: Example 2.5-1: Given xy + x = y + 3 , find y ′ . Solution: First Method: Let’s solve for y by bringing the y terms to the left-hand side of the equation, i.e., xy + x = y + 3 ; y (x − 1) = − x + 3 ; y =

−x + 3 x −1

We can now solve for y ′ using the differentiation rule for division. y′

=

[− 1⋅ (x − 1)] − [1⋅ (− x + 3)] = (x − 1)2

− x/ + 1 + x/ − 3

(x − 1)

2

=

−2

( x − 1)

2

Second Method: Use the implicit differentiation method to solve for y ′ , i.e., given xy + x = y + 3

then, (1 ⋅ y + y ′ ⋅ x ) + 1 = y ′ + 0 ; y +1 = y ′ − y ′x ; y ′(1 − x ) = y + 1 ; y ′ =

Hamilton Education Guides

y +1 1− x 97

Calculus I

2.5 Implicit Differentiation

Substituting y =

y′

−x + 3 x −1

into the y ′ equation we obtain:

−x + 3 +1 x −1 1− x

−x + 3 1 + x −1 1 1− x

[(− x + 3)⋅1] + [1⋅ (x − 1)] (x − 1)⋅1 = =

=

y +1 y′ = 1− x

=

2 2 2 ⋅1 = = − = (x − 1)(x − 1) (x − 1)⋅ (1 − x ) (x − 1)⋅ −(x − 1)

=

=

− x/ + 3 + x/ − 1 x −1 1− x

1− x

−2

( x − 1)

=

2 x −1 1− x

=

2 x −1 1− x 1

2

Example 2.5-2: Given x 2 y + 5 = y + 2 x , find y ′ . Solution: First Method: Let’s solve for y by bringing the y terms to the left-hand side of the equation,

(

)

i.e., x 2 y − y = 2 x − 5 ; y x 2 − 1 = 2 x − 5 ; y =

2x − 5 x2 −1

We can now solve for y ′ using the differentiation rule for division.

[2 ⋅ (x − 1)]− [2x y′ = (x − 1) 2

2 −1

⋅ (2 x − 5)

]

=

2

2

2 x 2 − 2 − 4 x 2 + 10 x

(

)

x2 −1

2

=

2

−2 x + 10 x − 2

(x

2

−1

)

2

Second Method: Use the implicit differentiation method to solve for y ′ , i.e., given x 2 y + 5 = y + 2x

(

(

)

then, 2 x ⋅ y + y ′ ⋅ x 2 + 0 = y ′ + 2 x 1−1 ; 2 xy + y ′x 2 = y ′ + 2 ; y ′x 2 − y ′ = 2 − 2 xy

)

; y ′ x 2 − 1 = 2 x − 2 xy ; y ′ = Substituting y =

y′

=

=

2 − 2 xy x2 −1

=

2x − 5 x2 −1 2 − 2x ⋅

2 − 2 xy x2 −1

into the y ′ equation we obtain: 2x − 5

x2 −1 x2 −1

=

2−

−2 x 2 + 10 x − 2

−2 x 2 + 10 x − 2

x2 −1 x2 −1

x2 −1 x2 −1 1

=

(

) (

2 x 2 − 1 − 4 x 2 − 10 x

4 x 2 − 10 x x2 −1 x2 −1

x2 −1 x2 −1

=

)

2 x 2 − 2 − 4 x 2 + 10 x

=

x2 −1 x2 −1

(−2 x + 10x − 2) ⋅1 = −2 x + 10 x − 2 = ( x − 1) ⋅ ( x − 1) ( x − 1) 2

2

2

2

2

2

In the previous examples, to find y ′ we could either first solve for y and then differentiate or use the implicit differentiation rule. However, sometimes we can not simply solve for y by bringing the y terms to the left-hand side of the equation. In these instances, as is shown in the following examples, we can only use implicit differentiation in order to differentiate y .

Hamilton Education Guides

98

Calculus I

2.5 Implicit Differentiation

Example 2.5-3: Given x 2 y 2 + y = 3 y 3 − 1 , find y ′ = Solution:

dy dx

.

d 2 2 d x y += y 3 y3 − 1 dx dx

(

)

(

) ; ( 2 x ⋅ y2 + 2 y2−1y′ ⋅ x2 ) + y′ = (3 ⋅ 3) y3−1 ⋅ y′ − 0 ;

−2 xy 2 ; y ′ ( 2 x 2 y − 9 y 2 + 1) = −2 xy 2 ; y ′ = − ; 2 x2 y y′ − 9 y 2 y′ + y′ =

Example 2.5-4: Given xy + x 2 y 2 + y 3 = 10 x , find Solution: d d xy + x 2 y 2 + y3 =(10 x ) dx dx

(

)

dy dx

2 xy 2 + 2 yy ′x 2 + y ′ = 9 y2 y′

2 xy 2 2 x2 y − 9 y2 + 1

in terms of x and y .

10 ; ( y + x y′) + ( 2 x y 2 + 2 x2 y y′) + 3 y 2 y′ = 2

−2 x y − y + 10 ; x y ′ + +2 x 2 y y ′ + 3 y 2 y ′ = −2 x y 2 − y + 10 ; y ′ ( x + +2 x 2 y + 3 y 2 ) = −2 x y 2 − y + 10 ; y ′ = 2 2 x + 2x y + 3 y

Example 2.5-5: Given 3x 3 y 3 + 2 y 2 = y + 1 , find Solution: d d 3 x3 y 3 + 2 y 2 = ( y + 1) dx dx

(

)

dy dx

in terms of x and y .

y′ y ′ + 0 ; 9 x 2 y 3 + 9 x3 y 2 y ′ + 4 y y ′ = ; 3( 3 x 2 y 3 + 3 y 2 y ′x3 ) + ( 4 y y ′ ) =

−9 x 2 y3 ; y ′ ( 9 x3 y 2 + 4 y − 1) = −9 x 2 y3 ; y ′ = − ; 9 x3 y 2 y ′ + 4 y y ′ − y ′ =

Example 2.5-6: Given x y + x 3 y 3 = 5 , find Solution: d d x y + x3 y 3 = ( 5) dx dx

(

)

dy dx

9 x2 y3 9 x3 y2 + 4 y − 1

in terms of x and y .

0 0 ; y + x y ′ + 3 x 2 y 3 + 3 x3 y 2 y ′ = ; (1 ⋅ y + y ′ ⋅ x ) + ( 3x 2 ⋅ y3 + 3 y 2 y ′ ⋅ x3 ) =

3 x2 y3 + y −3x 2 y3 − y ; y ′ ( x + 3x3 y 2 ) = −3x 2 y3 − y ; y ′ = − ; x y ′ + 3 x3 y 2 y ′ = 3 2 x + 3x y

(

)

Example 2.5-7: Given 3x y + y = x 2 + y 2 , find Solution: d d 2 y) x + y2 ( 3x y + = dx dx

(

)

dy dx

in terms of x and y .

; 3( 1 ⋅ y + y ′ ⋅ x ) + y ′ = 2 x + 2 y y ′ ; 3 y + 3 x y ′ + y ′ = 2 x + 2 y y ′

2x − 3 y ; 3x y ′ + y ′ − 2 y y ′ = 2 x − 3 y ; y ′ ( 3x + 1 − 2 y )= 2 x − 3 y ; y ′ =

3x − 2 y + 1

Hamilton Education Guides

99

Calculus I

2.5 Implicit Differentiation

1 1 + = 10 x , x y2

Example 2.5-8: Given Solution: d  1 1  d (10 x ) + = dx  x y 2  dx

;

y′ =

x −2 + 10 − 2y

−3

find

dy dx

d −1 d x + y −2 = (10 x ) dx dx

)

(

; 1

in terms of x and y .

x −2 + 10 10 ; −2 y −3 y ′ = ; − x −2 − 2 y −3 y ′ =

1 + 10 x 2

+ 10

2 ; y′ = x −2

x2 −2

; y′ =

y3

;

y′ =

(

y 3 1 + 10 x 2 − 2x

y3

Example 2.5-9: Given xy 2 + yx 2 = x 2 , find Solution:

dy dx

)

;

2

y′ = −

3

 1  + 10   2 2 x 

y

in terms of x and y .

d d xy 2 + yx 2 =x 2 dx dx

(

)

2x ( ) ; (1⋅ y2 + 2 y y′ ⋅ x ) + ( y′ ⋅ x2 + 2 x ⋅ y ) =

2x ; y 2 + 2 x y y′ + x2 y′ + 2 x y =

− y2 − 2 x y + 2 x − y 2 − 2 x y + 2 x ; y′ ( x2 + 2 x y ) = − y2 − 2x y + 2x ; y′ = ; 2 x y y′ + x2 y′ = 2 x + 2x y

2 3

Example 2.5-10: Given y + x 3 y = y , find Solution:  d d  23  y + x3 y  = (y)   dx   dx

;

;

(

)



2 ; y ′ y 3

−1

3

 + x 3 − 1 = −3 x 2 y 

1

Example 2.5-11: Given xy + y 2 = y 8 , find Solution:

(

)

=′ ; (1 ⋅ y ) + ( y ′ ⋅ x )  + 2 yy

−7



1 1 − y ; y′ x + 2 y − y ; x y′ + 2 y y′ − y 8 y′ =  8

8



dy dx

−7  8= −y

 

Solution:

(

)

(

− 3x 2 y 2 − 13 y + x3 −1 3

−7

1 8 y y′ ; y + x y′ + 2 y y′ = 8

; y′ = − dy dx

y x + 2 y − 18 y

−7

8

in terms of x and y .

) ; (1⋅ y2 ) + ( 2 y y′ ⋅ x ) + y′ = 2 x + 0

2x − y ; y ′ ( 2 xy + 1) = 2 x − y 2 ; y ′ =

; y′ =

in terms of x and y .

1 18 −1 y ⋅ y′ 8

Example 2.5-12: Given xy 2 + y = x 2 + 3 , find d d 2 xy 2 + = y x +3 dx dx

in terms of x and y .

2 32 −1 2 −1 ⋅ y ′ + 3x 2 ⋅ y + y ′ ⋅ x 3 = y ′ ; y 3 y ′ + 3x 2 y + x 3 y ′ = y ′ y 3 3

2 − 13 y y ′ + x 3 y ′ − y ′ = −3 x 2 y 3

d d  1 xy + y 2 = y 8  dx dx  

dy dx

2 x ; 2 xy y ′ + y ′ = 2 x − y 2 ; y 2 + 2 xy y ′ + y ′ =

2

2 xy + 1

Hamilton Education Guides

100

Calculus I

2.5 Implicit Differentiation

Example 2.5-13: Given x 4 y 3 + y 2 = x + 4 , find Solution: d 4 3 d x y + y2 = ( x + 4) dx dx

(

)

;

(

) (

dy dx

in terms of x and y .

)

 4 x3 ⋅ y 3 + 3 y 2 y ′ ⋅ x 4  + 2 y y ′ = 1+ 0  

1 ; 4 x3 y 3 + 3 x 4 y 2 y ′ + 2 y y ′ =

1 − 4 x3 y3 1 − 4 x3 y 3 ; y ′ ( 3 x 4 y 2 + 2 y ) = 1 − 4 x3 y 3 ; y ′ = 4 2 ; 3x 4 y 2 y ′ + 2 y y ′ =

3x y + 2 y

Example 2.5-14: Given y 6 + x 3 y 5 + x 2 = 5 , find Solution: d 6 3 5 d y + x y + x 2 =( 5 ) dx dx

(

)

dy dx

in terms of x and y .

0 ; 6 y 5 y ′ + 3 x 2 y 5 + 5 x3 y 4 y ′ + 2 x = 0 ; 6 y5 y ′ + ( 3x 2 ⋅ y5 ) + ( 5 y 4 y ′ ⋅ x3 )  + 2 x =

3 x 2 y5 + 2 x −3x 2 y5 − 2 x ; y ′ ( 6 y5 + 5 x3 y 4 ) = −3x 2 y5 − 2 x ; y ′ = − 5 ; 6 y 5 y ′ + 5 x3 y 4 y ′ = 3 4 6 y + 5x y

Section 2.5 Practice Problems - Implicit Differentiation Use implicit differentiation method to solve the following functions. a. x 2 y + x = y

b. x y − 3x 2 + y = 0

c. x 2 y 2 + y = 3y 3

d. x y + y 3 = 5 x

e. 4 x 4 y 4 + 2 y 2 = y − 1

f. x y + x 2 y 2 − 10 = 0

g. x y 2 + y = x 2

h. x y 3 + x 3 y = x

i. y 2 + x 2 y = x

j. x 2 y + y 2 = y 4

k. x + y 2 = x 2 − 3

l. x 4 y 2 + y = −3

m. y 7 − x 2 y 4 − x = 8

n. (x + 3) 2 = y 2 − x

o. 3x 2 y 5 + y 2 = − x

1

Hamilton Education Guides

1

101

Calculus I

2.6

2.6 The Derivative of Functions with Fractional Exponents

The Derivative of Functions with Fractional Exponents

The derivative of a function f ( x ) with fractional exponent is obtained by applying the chain rule in the following way: d f ( x) dx

[

a

]b

=

a

a f ( x) b

a

] b −1 ⋅ dxd [ f ( x )]

[

For example, the derivative of f ( x ) = x b is equal to a

d b x dx

=

a

a b −1 d x ⋅ x b dx

=

a

a b −1 x ⋅1 b

=

a

a b −1 x b

=

a −b

a b x b

Note that the steps in finding the derivative of a function with fractional exponent is similar to finding the derivative of a function that is raised to a power as discussed in Section 2.4. The following examples illustrate how to obtain the derivative of exponential functions: Example 2.6-1: Find the derivative for the following exponential expressions.

( )

2

b. y = 3x 2

a. y = x 3

(

d. y = 3x 2 + 6 x

m.

(

3 5

)

2

1

)

1 4

2 7

(

i. y = ( x + 1) 2 x 2 + 3

1

( x + 3) 5 k. y = 2

x

)

f. y = 3x 2 + 8

h. y = x x 2 + 1 3

1 2

l. y =

x3

)

1 3

x3 2

(x + 1) 3

1

y = (x + 1) ⋅

Solutions:

(

3

1 3

2

(

c. y = 3x 3 + 2 x

e. y = (2 x + 1) 4

( ) + (2x + 1) ( x + 1) y=

g. y = x 2 j.

)

2 5

1 3

1

x7 2

a. Given y = x 3 then y ′ =

2 2 2 −1 2 −1 2 23 −1 2 2 −3 2 1 2 1 x = x 3 1 ⋅1 = x 3 = x 3 = ⋅ 1 = ⋅ 3 = 3 3 3 3 3 3 3 x 3 x x3

Note that the answer does not necessarily need to be in radical form. We can simply stop when y ′ =

2 − 13 x 3

. However, for review purposes only, the answer to some of the problems are

shown in radical form (see Sections 1.1 and 1.2 from Mastering Algebra – Advanced Level to review the subjects of exponents and radicals). 1

b. Given y = ( 3x 2 ) 3 then y ′ = = 2x

1

3

( 3x )

2 2

=

2x

Hamilton Education Guides

1 3

9 x4

=

( )

1 3x 2 3 2 x/ 3

x/ 9 x

1 −1 3

=

⋅ 6x

=

( )

6 x 3x 2 3

1 −1 3 1

( )

= 2 x 3x 2

1−3 3

( )

= 2 x 3x 2

−2 3

= 2x

1

( 3x ) 2

2 3

9x

102

2 3

Calculus I

2.6 The Derivative of Functions with Fractional Exponents

(

y = c. Given

=

=

)

) ( 9x

2

3

(

1 3x 3 + 2 x 4

(

= y d. Given

1 4

3x + 2 x 1− 4 4

2

)

(

12 3x 2 + 6 x 5

)

−3 5

(

1 3x 3 + 2 x 4

+2 =

3x + 6 x

)

2 5

12 ( x + 1)⋅ 5

( x + 1) =

3 1 ⋅ 2 4 2x + 1

=

=

(

)

−5

( )

g. Given y = x 2 =

=

7

1 3

12 x ⋅ 7

( 3x 3

−3 4

2

(

( 3x

1 2

+ 6x

)

)

+2

)

2 −1 5

=

3 5

(

1 3x 3 + 2 x 4

+2 =

=

(

9x 2 + 2

4 3x 3 + 2 x

2

+ 6x

3 1 6 (2 x + 1) 4 − 1 4

=

(

2 3x 2 + 8 7

1 2

+8

)

2 −1 7

⋅ 6x

12 x ⋅ 7 7

=

5 7

)

2

3

+ ( 2 x + 1) 5 = x 3 + ( 2 x + 1) 5

=

)

(

)

(

)

= x 2 +1

2 3

+

2 3

( 3x

(

=

12 x 3x 2 + 8 7

1

=

2

+8

)

5

(





)

2−3 3

1

(

)

i. Given y = ( x + 1) 2 x 2 + 3

Hamilton Education Guides

1 3

( 3x

4

)

2 −1 5 1

3

)

+ 2x

5

( 3x

5

)

3

( x + 1) =

12 ( x + 1)

=

3

+2

2

+ 6x

)

(

12 3x 2 + 6 x 5

)

2−5 5

( x + 1)

3

1 3 (2 x + 1) − 4 2

=

)

2 −1 7 1

(

)

2

(

)

= x 2 +1

2 3

+

1

2

7

2 −1 3

  

(

)

4x 2 2 x +1 3

1

( 3x

2

−1

3

(

(

2 5

+8

)

)

2−7 7

5

)

= x 2 +1

)

1

2 3

+

 

3

3 x

+

6 5 5 (2 x + 1)2

(

)

= x 2 +1

(

4x 2

)

3 x 2 +1

(x 3

 1

2

=

  ⋅ 2 x ⋅ x   

then y ′ =  ( x + 1) 2 −1 ⋅1 ⋅ x 2 + 3 3  +   2

12 x 3x 2 + 8 7

12 x

7

( x + 1)  3

  2

(

=

3 3 1 2 2 −1 6 2 32 −1 3 x + (2 x + 1) 5 −1 ⋅ 2 = x 3 1 + (2 x + 1) 5 − 1 3 5 3 5

then y ′ =

 then y ′ = 1 ⋅ x 2 + 1 3  +  

4x 2 2 x +1 3

4

3− 4 3 (2 x + 1) 4 2

3− 5 2 2 1 6 1 2 2 −3 3 6 2 −1 6 x + (2 x + 1) 5 = x 3 + (2 x + 1) − 5 = ⋅ 1 + ⋅ 3 3 5 (2 x + 1) 3 5 5 3 x

h. Given y = x x 2 + 1

3 4

2

9x 2 + 2

=

(

x +1

( 3x

1 −1 4 1

12 3x 2 + 6 x 5

( 6 x + 6) = 12 ⋅ 5 5

)

) ( 9x

2 2x + 1

2

12 x 3x 2 + 8 7

) ( 9x

)

2

3

4

y ( 3x 2 + 8) 7 then y ′ = f. Given=

=

1 −1 4

3 3 (2 x + 1) 4 −1 ⋅ 2 4

3

3 1 ⋅ 2 (2 x + 1) 14

) ( 9x

2 3x 2 + 6 x 5

then y ′ =

y ( 2 x + 1) 4 then y ′ = e. Given =

=

(

1 3x 3 + 2 x 4

then y ′ =

2

)

+3

1 −1 3

1 3

2 3

=

+

3

(

)

4x 2 2 x +1 3

(x

2

+1

)

2

+

2 −1 3 1

4x 2 3

1 ⋅ 2 x ⋅ ( x + 1) 2  

103

3

x2 +1

Calculus I

=

2.6 The Derivative of Functions with Fractional Exponents

(

1 −1 2  ( x + 1) 2 ⋅ x + 3 2 

(x =

2

+3

)

2 ( x + 1)

1 3

x2

=

(

2 x ( x + 1) 2

+

(

2

3 x +3

)

2

x +1

1 2

x

( x + 1) 2

x2 + 3

)

2

x3

2

1 2

x2 x2 + 1

2

=

( x + 1)

2

( x + 1) 3

m. Given y = (x + 1) ⋅

=

7

7 2

x

2

+3

1 −1 2

)

− 4

3x 2

(

1 1 x7

x

3

=

7

=

Hamilton Education Guides

x +1

x− 7

1

⋅

1

( x + 1) 2

1 2

  

(

)

x 2 − x 2 +1 x +1 x

=

2

(

)

3

1

=

x2

5 5 ( x + 3) 4 3

−

33 x + 1

(x + 1)4

)

=

−1

2

(

)

− x 2 +1

1 2

x2

x2 + 1 x2

=

−1 x2 x2 + 1

1 1 2 1 −4  2 −  x 3 (x + 3) 5  −  x 3 (x + 3)5    5   3 4

25 x + 3 33 x

x4

4

2x 3

(

2x 2 2 x +1 2

x3

2(x + 3) 5 3x 3

=

(x + 1) 3 −

  1 1  +  2 x ⋅ ( x + 1) 2 ⋅   3   x2 + 3

−1

2

=

2

then y ′ =

x 17

)

   ⋅ 2 x  ⋅ x  − 1 ⋅ x 2 + 1   

3 x 3−1 ⋅ ( x + 1) 23  −  2 ( x + 1) 23 −1 ⋅ x 3     3  

(x + 1)2

=

1 7 6

3

)

1 3

2

1

4

then y ′ =

2 x3

1 3 ( x +1) 3 4 3

7 x − x +1 ⋅

(x

x3

x3

(

=

4 x3

5(x + 3) 5

=

 1 ⋅ x2 + 3 2 

2  1 1 −1 1 2 2 −1  ( x + 3) 5 ⋅ x 3  −  x 3 ⋅ ( x + 3) 5   5   3 

x3

4

3 x 2 (x + 1) 3 −

3

)

x

x3

l. Given y =

3

− x2 + 1

2

1

  

x2

then y ′ =

x3 1 2 ( x + 3) 5 ⋅ − ⋅ 1 5 (x + 3) 54 3 x3

1 2

2x x + 1

  1 2   x +1   2

=

1 5

⋅ ( x + 1)

3

+

2 x +1

then y ′ =

2

x + 3)

3

=

2 3

)

− 2

(

− x 2 +1

1 2

x

)

(

1

( k. Given y =

=

(

  2x 2 x +3 +   3

1

1 2

j. Given y =

)

1 3

3x 2

3

=

( x + 1)2

−

=

3 3 x 2 ( x + 1) 23  −  2 x ( x + 1)− 13        3 4

( x + 1) 3

2x 3 33 x + 1

( x + 1) 3 x + 1

1    1 1 −1 1 ⋅ x 7  −  x 7 ⋅ (x + 1) 7     2

x7

=

1 x7

x + 1 − 76 − x 7 2

x7

1

x7 −

=

x +1 1 ⋅ 6 7 x7 2

x7

x +1 7

7 x6 x2 104

2 3

   

Calculus I

2.6 The Derivative of Functions with Fractional Exponents

d of the following functions. dx

Example 2.6-2: Find a.

d  − 23  x  dx  

d.

d 3x 2 + 4 x dx

b.

1 d (x + 1)− 4 dx

=

e.

−2 d x x 2 +1 3 dx

  =  

h.

d 2 x + 3x + 5 dx

)

)

=

k.

d dx

   

= − x

=

(

)

1 d  ( x + 1)− 2 g. dx  x 

j.

(

7 8

−1 d 2 x +5 6 dx

)

(

(

(

)

=

f.

d 3x 3 + 4 dx

=

i.

5 d  (x + 1)(x − 1) 4  =  dx  

l.

d  x 2 dx  ( x + 1)− 13 

1 4

  

=

(

Solutions: d  − 23 x dx 

b.

1 d (x + 1)− 4 dx

1 4

c.

3 2 x ⋅ 8

(x + 1) 3

(

)

7 8

7 3x 2 + 4 x 8

(

(

)

−2 3

7 8

−1

)

)

)

=

=

−2 d x x 2 +1 3 dx

= x 2 +1

)

−1

3

   

=

=

(

−

8

(

3x 2

7

)

⋅

)

8 x3 +1 8

(

7 3x 2 + 4 x 8

⋅ (6 x + 4 )

(

)

(

)

= x 2 +1

7 ⋅ 8

= −2 3

⋅

− 2 −1 2x 2 x +1 3 1 3

Hamilton Education Guides

3

8

1 −1 8 1

⋅ 3x 2

(

)

1−8 3 2 3 x x +1 8 8

=

=

(

)

−7 3 2 3 x x +1 8 8

(x + 1)

7

3

(

d 3x 2 + 4 x dx 6x + 4

(3x

2

+ 4x

)

1 8

=

(

) = 78 (3x

)

= x 2 +1

−2 3

+ 4x

)

7 −1 8 1

⋅ (6 x + 4 )

=

(

7 3x 2 + 4 x 8

)

7 −8 8

⋅ (6 x + 4 )

8

8 3 x2 + 4 x

)

−

2

7(6 x + 4 )

−2 d d 2 x + x⋅ x +1 3 dx dx

(

5

3 x2

8

7 −1 8

1 4

= − (x + 1)− 4

4 ( x + 1) 4 x + 1

) = 18 (x + 1)

=

−1− 4 4

1

= −

4 4 ( x + 1)5

(

1

(

1

1 4

1 1

= − (x + 1)− 4 − 1 ⋅1 = − (x + 1)

1 −1 1 3 d 3 x +1 8 ⋅ x +1 8 dx

=

d 3x 2 + 4 x dx

=

e.

)

1 4

d (x + 1) dx

1

= −

5 4

(x + 1)

1 d 3 x +1 8 dx

=

d.

1 4

= − (x + 1)− 4 −1 ⋅

1

(

=

−2 2 1 2 − 23 −1 d 2 − 2 −1 2 −2 −3 2 −5 −2 ⋅ ( x ) = − x 3 1 ⋅1 = − x 3 = − x 3 = − ⋅ 5 = = 3 3 3 3 3 dx 3 3 3 3x x 2 3 x5 x

a.

= − ⋅

1

d 3 x +1 8 dx

1 4

 3 2 x 5 x +1 

)

(

c.

=

(

(

)

)

= x2 + 1

= x 2 +1

2x 2 x +1 3

−2 −3 3

−2 3

(

⋅1 − x ⋅

)

(

)

− 2 −1 2 2 x +1 3 3

−2 3

−

(

)

−5 2x 2 x +1 3 3

105

Calculus I

=

f.

2.6 The Derivative of Functions with Fractional Exponents

1

(x + 1)

2 3

(

)

2

d 3x 3 + 4 dx

2x ⋅ 3

−

−1

(

h.

i.

(

)

−4 3

)

= − 3x 2 ⋅

3

(3x

2

2

− 1 −1

⋅

3

(

d 3x 3 + 4 dx

1 3

2x

−

+4

)

4 3

3

(x + 1)

5

2

) = − 13 (3x 3x 2

= −

3

(3x

3

+4

1

=

3

(x + 1)

3

+4

)

− 1 −1

= −

)

4

2

2

3 1

⋅ 9x 2

−

2x

(

) (x + 1) 3

3 x2 +1

= −

2

2

(

9x 2 3x 3 + 4 3

)

−1−3 3

3 x2

(3 x + 4) 3

3

3 x3 + 4

1  d     −1   −1 d − 1 −1 d −1    x ⋅ dx ( x + 1) 2  − ( x + 1) 2 ⋅ dx ( x )  x ⋅ − 2 ( x + 1) 2 ⋅ dx ( x + 1) − ( x + 1) 2 ⋅1      =  =  2 2  x x 

x2

(

(x + 1)

3

 x − 1 −1  −1 − 2 ( x + 1) 2 1 ⋅1 − ( x + 1) 2  

d 2 x + 3x + 5 dx

=

2

1 3

= − 3x 2 3x 3 + 4

=

(x + 1)

1

=

5 3

= − 3x 3 + 4

3

1 d  ( x + 1)− 2 g. dx  x 

1

)

1 4

=

)

(

−1− 2 1 x (x + 1) 2 − (x + 1)− 2 2 x2

(

1 −1 1 2 d 2 x + 3x + 5 x + 3x + 5 4 ⋅ 4 dx

=

)

(

−

3

− 1 2 x + 3 x + 5 4 (2 x + 3) 4

2x + 3 ⋅ 4

=

1

(x2 + 3x + 5)

3 4

=

−

) = 14 (x2 + 3x + 5) −

−3 1 x ( x + 1) 2 − ( x + 1)− 2 2 x2

1 1 4 1

⋅ (2 x + 3)

2x + 3 1 ⋅ 4 3 4 2 x + 3x + 5

=

)

(

=

=

(

1 2 x + 3x + 5 4

)

1− 4 4

⋅ (2 x + 3)

2x + 3 4 4  x 2 + 3 x + 5   

3

5 5 5 5 5 d  (x + 1)(x − 1) 4  = (x − 1) 4 d (x + 1) + (x + 1) d (x − 1) 4  = (x − 1) 4 ⋅1 + (x + 1)⋅ 5 (x − 1) 4 −1 ⋅ d (x − 1)  4 dx dx dx dx           5 1 5− 4  5 5 5 1  5 5 5   = (x − 1) 4 + (x + 1) ⋅ (x − 1) 4 − 1 ⋅1 = (x − 1) 4 + (x + 1) ⋅ (x − 1) 4  = ( x − 1)4 + ( x + 1)( x − 1)4

4



j.

(

)

−1 d 2 x +5 6 dx

x 3

= − ⋅

k.

d dx

(x

= −

1 2

+5

(

)

)

7 6



(

)

(

− 1 −1 d 1 2 x +5 6 ⋅ x2 + 5 6 dx

x 3 6

= − ⋅

(

(x

1 2

+5

4



= −

)

7

)

) = − 16 (x

(

2

+5

4



)

− 1 −1

6 1

⋅ 2x

= −

(

2x 2 x +5 6

)

−1− 6 6

= −

(

)

−7 x 2 x +5 6 3

x 2

3 x +5

)

6

x2 + 5

(

)

(

)

(

)

(

)

1 1 1 1 −1 1  2  3 2   d 53   53 d 2 d 2 3 3 −1   3 1 2 x  + x x +1 4 ⋅ x +1 4  =  x 2 +1 4 ⋅ x 5  + x 5 x +1  x 5 x +1 4  =  x +1 4 dx dx dx 5 4          

3  1 1 −1 1− 4  1 3 −1   3 3− 5  5  3 2  3 2 1 x x ⋅ 2 2  ⋅ x 2 +1 4  =  x + 1 4 ⋅ x 5 1  +  x 5 x + 1 4 1 ⋅ 2 x =  x + 1 4 ⋅ x 5  +   4  5     5   4  

(

)

Hamilton Education Guides

(

)

(

)

(

)

106

Calculus I

=

l.

2.6 The Derivative of Functions with Fractional Exponents

8  −2 3 1   3 x 5 ⋅ x2 + 1 4  +  x 5 ⋅ x2 + 1 − 4    5   2    

(

d  x 2 dx  ( x + 1)− 13 

)

(

  = d  dx 

)

1 1 d   x 2 ( x + 1) 13  2  2 d   = ( x + 1) 3 dx x  +  x dx ( x + 1) 3 

1 2 1 1 1 1  1 x2 1 d 1    = (x + 1) 3 ⋅ 2 x  +  x 2 ⋅ (x + 1) 3 −1 ⋅ (x + 1) = 2 x (x + 1) 3 +  x 2 ⋅ (x + 1) 3 − 1 ⋅1 = 2 x ( x + 1)3 + ( x + 1)− 3



3

 

dx



3



3



Example 2.6-3: Find y ′(1) in example 5.6-1 a through i . Solutions: In Example 2.6-1 we obtained the derivative of the exponential functions a through e to be equal to the following: a. Given y ′ =

2

then y ′(1) =

3

3 x

2x b. Given y ′ = 3

c. Given y ′ =

9x 4

9x 2 + 2

4

(3x

3

+ 2x

5

5

(3x

2

)

+ 6x

3

3 1

)

3

2 2x + 1

= 0.67 6

2 ⋅ −3 3

= −3

9 ⋅ 14

(3 ⋅1

4

3

5

2 2 ⋅1 + 1

2

=

9

3

)

+ 6 ⋅1

3

3 4

= −

0.33

=

)

+ 2 ⋅1

(3 ⋅1

3 4

9

12(1 + 1)

5

6

= −

9 ⋅ 12 + 2

then y ′(1) =

then y ′(1) =

4

2 3

=

then y ′(1) =

3

12(x + 1)

d. Given y ′ =

e. Given y ′ =

then y ′(1) =

2 3

2 2 +1

11

=

4 3

5

=

=

6 2.08

11 125

24 5

5 729

3 4

2 3

= −2.88

=

=

0.25

=

11 3.343

= 3.29

=

24 5 ⋅ 3.74

24 5 ⋅ 729 3

2⋅3

0.25

0.2

=

3 2 ⋅1.32

= 1.28

= 1.14

Section 2.6 Practice Problems - The Derivative of Functions with Fractional Exponents 1. Find the derivative of the following exponential expressions. 1

( ) y = (x ) − (3 x − 1) 1

d. y = 2 x 2 + 1 8 g.

j. y =

( ) y = (2 x + 3 x )

b. y = 4x 3

a. y = x 5

1

3 2

e. 1 3

x +1 2

x3

Hamilton Education Guides

1 2

3

1

h. y = x 2 (x + 1) 8 k.

(x + 1) y= 2

x2

1 2

1

c. y = (2 x + 1) 3 3 5

( ) y = (x + 1)

f. y = x 3 + 8 i.

l. y =

2 3

2 5

3

1

+x2

(x + 1)2 1

x3

107

Calculus I

2.6 The Derivative of Functions with Fractional Exponents

2. Use the

d dx

notation to find the derivative of the following exponential expressions.

2

a.

d  15 x dx 

d.

−1 d 3 x +1 4 dx

g.

d dx

j.

(

   

)

(

=

b. =

=

1 d  ( x − 1) 2  = dx  x 2     d  3 1 h. x ⋅ 1 dx   x 2 +1 2

e.

)( )  =

 3 2  x +1 x 

1 d (x − 1) 2 dx

1 3



1 1 d  (x − 1) 2 (x + 1) 3  =  dx  

Hamilton Education Guides

(

k.

d dx

(

)

)

1  3 2 x x +1 2   

  =  

(

)

c.

1 d 2 x +1 3 dx

f.

d 3 x + 2x dx

i.

 d  x5 dx  3  x + 1

l.

d dx

(

(

)

)

(

= 1 8

2 3

=   =  

)

 3 2 −1  x x +1 3  =  

108

Calculus 1

2.7

2.7 The Derivative of Radical Functions

The Derivative of Radical Functions

In this section finding the derivative of radical expressions and the steps as to how it is calculated is discussed. The derivative of radical functions is found by using the following steps: 3

x 3 as x 2 .

First - Write the radical expression in its equivalent fraction form, i.e., write

Second - Apply the differentiation rules to find the derivative of the exponential expression. Third – Change the answer from an expression with fractional exponent to an expression with radical expression (optional). The following examples show the steps in solving functions containing radical terms. Students who have difficulty with simplifying radical expressions may want to review radicals addressed in Chapter 5 of the “Mastering Algebra - An Introduction”. Example 2.7-1: Find the derivative for the following radical expressions. a. f ( x ) = x 3 + 1

b. f ( x ) = x 2 + 3x + 1

c. f ( x ) = 2 x 5 + 1

d. f (u) = 5 u3 + 3u

e. f (t ) = t 2 + t + 1

f. g( x ) = x 2 x 3 + x − 5

g. h( w) = 3 w 2 + 1

h. f ( z ) = 4 z 3 − z 2 + z

i. f ( x ) =

j. f ( x ) = m. g(u) =

x

k. r (θ ) =

x2 −1 u −1

n. h(t ) =

u +1

Solutions: a. Given f ( x ) = f ′( x ) =

(

(

)

θ +1

(

x +1

r3

l. p(r ) =

θ 2 +1 3 2

t +1

o. s(r ) =

t3

r3 −1 r2 −1 r −1

1

= x 3 + 1 2 , then

(

)

)

(

)

1 −1 −1 −1 1 3 3 1 3 3 x + 1 2 ⋅ 3x 2 = x 2 x 3 + 1 2 = x + 1 2 ⋅ 3 x 3−1 = 2 2 2 2

b. Given f ( x ) = f ′( x ) =

x3 +1

3

1 2

x 2 + 3x + 1

(

)

x2

(x + 1) 3

1 2

3 2

=

x2 x3 + 1

1

= x 2 + 3x + 1 2 , then

) (

)

(

)

1 −1 −1 1 2 1 2 x + 3 x + 1 2 ⋅ 2 x 2−1 + 3 x1−1 = x + 3 x + 1 2 ⋅ (2 x + 3) = 2 2

(

)

(

2x + 3

)

2 x 2 + 3x + 1

1 2

=

2x + 3 2 x 2 + 3x + 1

1

c. Given f ( x ) = 2 x 5 + 1 = 2 x 5 + 1 2 , then

(

) (

)

(

)

1 −1 −1 1 1 2 x 5 + 1 2 ⋅ 2 × 5 x 5−1 + 0 = 2 x 5 + 1 2 ⋅10 x 4 = f ′( x ) = 2 2

5

(

1/ 0/ x 4

)

2/ 2 x 5 + 1

1 2

=

5 x4 2 x5 + 1

3

d. Given f (u) = 5 u3 + 3u = u 5 + 3u , then f ′(u) =

3 −2 3 −2 3 53 −1 u + 3u 1−1 = u 5 + 3u 0 = u 5 + 3 = 5 5 5

Hamilton Education Guides

3 5u

2 5

+3 =

3 5

5 u2

+3

109

Calculus 1

2.7 The Derivative of Radical Functions 1

e. Given f (t ) = t 2 + t + 1 = t 2 + (t + 1) 2 , then 1 2

1 1 (t + 1) 2 −1 ⋅ t 1−1 2

f ′(t ) = 2t 2−1 +

1

1

= 2t + (t + 1)− 2 ⋅ t 0 = 2t +

(

)

2(t + 1)

1

= 2t +

1 2

2 t +1

1

f. Given g( x ) = x 2 x 3 + x − 5 = x 2 x 3 + x − 5 2 , then

(

)

(

1   1 g ′( x ) = 2 x 2−1 x 3 + x − 5 2  +  x 3 + x − 5    2

(

)

)

(

1

= 3x 2 + 1 x 2 = 2 x x 3 + x − 5 2 +

(

)

) ⋅ (3x 1 −1 2

3−1

(3x + 1) x 2 (x + x − 5) 2

2

3

)

(

 + x1−1 + 0  x 2 = 2 x x 3 + x − 5 

= 2 x x3 + x − 5 +

1 2

)

1 2

(

+

(

)

−1 1 3 x + x−5 2 2

)

x2 3 x2 + 1

2 x3 + x − 5

1

g. Given h( w) = 3 w 2 + 1 = w 2 + 1 3 , then h ′( w) =

(

) (

1 −1 1 2 w + 1 3 2 w 2−1 + 0 3

4 3

h. Given f ( z ) =

(

= z3 − z2

) (3z

2 −1

(

1 −1 4

1

3−1

− 2z

)

1 4

3

+z,

(z

)+ 1 =

− z2

)

= x 2 +1

x +1

− 1 −1

2

−1

2

x

x −1

2

1 2

1

=

−

(x − 1) 2

1 2

x 2 −1 −

1

(x2 − 1) (x2 − 1) 1 2

Hamilton Education Guides

(x − 1) ⋅ (x − 1) (x − 1) 1 2

2

x2

= −

=

2

4

3

3

(w

2

+1

2

− 2z

)

2

) + 1 = 3(3z − 2 z ) + 1 = 3z 4 (z 4(z − z ) 2

3 2 4

3

⋅ 2/ x =

−x

(x + 1) 2

(x − 1) = 1 2

4

−x

=

3 2

2

− 2z

3

−z

)

2 3

+1

2

( x + 1) 2

−x

=

3

(x + 1) 2

(x − 1)

1 +1 2

=

−

1

(x2 − 1)

3 2

(

(x − 1)

)

2

 1 −1 −  x 2 − 1 2 ⋅ 2/ x 2    2/

(x − 1) − x (x − 1) 1+1 2 2

2

=

x 2 −1

1 2

=

x 2 −1

− x2

1 2

2

1

2

−3

2

x 2 −1 2

) (3z

)

 − 0 ⋅ x 

2 −1

 



(x − 1)

=

1 −1 2

2

)

x2 +1

, then

1 2

2

1 2

2

f ′( x ) =

x

=

(x − 1)  1  1 ⋅ (x − 1)  −  (x − 1) (2 x  2  2

(

3 w2 +1

2w

=

2 3

, then

−3

2

2 −1

2

j. Given f ( x ) =

1 2

2

2w

4

(x + 1) 1 (2 x + 0) = − 12/ (x + 1) f ′( x ) = − (x + 1) 2 2

=

⋅ 2w

then

3

(

1

=

−2

2

z − z2 + z

1 3 z − z2 f ′( z ) = 4

i. Given f ( x ) =

) = 13 (w + 1)

= −

1 2

2

x 2 −1

1

(x − 1) 2

3

= −

2

1 2

−

x2

(x − 1) 2

x 2 −1

x/ 2/ − 1 − x/ 2/

=

(x − 1) 2

1 2

x 2 −1

1

(x − 1) 2

x2 −1

110

1 2

Calculus 1

2.7 The Derivative of Radical Functions

θ3 +1

k. Given r (θ ) =

r ′(θ ) =

=

=

(

(

(

)

3

2θ θ + 1

=

θ 2 +1

)

2 θ 2 +1

 3−1 3r 

+1

1 −1 2

2 −1

)

)(

(

 + 0 θ 3 +1  

1 2

r3

(

)

2 θ 2 +1

=

r3

1 2

3r 5

=

3−1

3r 5 − 6r 2

)(

)

2 r 3 −1 r 3 −1

1 2

6θ 4 + 6θ 2 − 2θ 4 − 2θ

=

(

) (θ

2 θ 2 +1

1 2

2

)

+1

)

=

(

)

=

2

2

3

(

1 2

(

−

)

2 r 3 −1

)

1 +1 2

(

3r 5

(

)

2 r 3 −1

(

)

r 3 −1

3r 2 r 3 − 2

=

(

)

1   2 3  −1 2 − 1 r 3 − 1 2 3r 2 ⋅ r 3 r r − 3 1        2 

=

r 3 −1

3r 2 r 3 − 2

(

3

)

2 r 3 −1

3 2

)=

( ) 2(r − 1)

6r 2 r 3 − 1 − 3r 5 1 2

3

=

(

)

2 r 3 −1

3

)=

(

(

(

3r 2 r 3 − 2

2 r 3 −1

)

)

2 r 3 −1

=

r 3 −1

3r 2 r 3 − 2

(

6r 5 − 6r 2 − 3r 5

1 2

1 2

r 3 −1

)

r 3 −1

1

u +1

1 −1 1  1 1 −1 1 1  2 (u − 1) 2 (u + 1) 2  −  2 (u + 1) 2 (u − 1) 2      u +1

  (u − 1) 12 −   2(u + 1) 12   u +1

   

4/

=

1 2

4/ (u + 1) (u − 1) (u + 1)

Hamilton Education Guides

1 2

(u − 1) 2 , then = 1 (u + 1) 2

u −1

 (u + 1) 12   2(u − 1) 12 

)

θ 2 +1

)

 −0  

3r 2 r 3 − 1

1 2

2 r 3 −1

r 3 −1

(

(

(

2 θ 2 +1

2

1 −1 2

3

−

)

θ 2 +1

3

 

1 2

(

 1 2 −1  − θ +1 2  2

, then

1 2

3

1 2

)

1 2

6θ 2 θ 2 + 1 − 2θ 4 − 2θ

3 2

2

(r − 1) (r − 1)  −  12 (r − 1) (3r 3

)

)

 2 2 3θ θ + 1 

) = θ (2θ + 3θ − 1) = θ (2θ + 3θ − 1) (θ + 1) θ + 1 (θ + 1) 2/ (θ + 1)

(

=

r3 −1

(

)

2/ θ 2θ 3 + 3θ − 1

=

1 +1 2

(

2θ θ 3 + 1

−

r 3 −1

m. Given g(u) =

=

) (2θ

2

θ 2 +1

4θ 4 + 6θ 2 − 2θ

(

)

3θ 2 θ 2 + 1

=

=

+1

, then

 

(

g ′( u) =

1 2

1 2

2

3r 2 r 3 − 1

=

(θ

2

θ 2 +1

l. Given p(r ) =

p ′( r ) =

)  1 + 0 )(θ + 1)  −  (θ  2 θ 2 +1

 3−1  3θ 

θ 3 +1

=

1 2

=

1 1  1 1 −1 −1  2 (u − 1) 2 (u + 1) 2  −  2 (u + 1) 2 (u − 1) 2      u +1

2(u + 1) − 2(u − 1)

=

1 2

4(u − 1) (u + 1) u +1

=

1 1 2

(u + 1) (u + 1) (u − 1)

4

2/ u/ + 2 − 2/ u/ + 2

1 2

1 2

1 2

4(u − 1) (u + 1) u +1

=

(u + 1)

1 2

=

1 2

1

4(u − 1) (u + 1) 2 u +1

1 u+1

u −1

111

Calculus 1

2.7 The Derivative of Radical Functions

n. Given h(t ) =

h ′( t ) =

3 2

t +1 t3

(

1 −1 3

t 2 −1

, then

3 2

)

(

)

(

)

1  2 1+ 3 2 −2  3 1 2  t 2 t +1 3  −  t 2 t +1 3    3   2

=

t3 5

2t 2 −

(

r2 −1

r 2 −1

=

r −1

1

(r − 1) 2

(

=

3r 2 − 4r + 1 1

2(r − 1) (r − 1) 2

=

)

(

)

3t 3 t 2 + 1

(

)   

)

=

2 3

=

2t 2 t − 3t 3

3

2t

(

(

5 2

)

3 t 2 +1

(

2 3

)

1 3 12 2 t t +1 3 −2 1

t3

)

9 t t 2 +1 2

(t

2

)

+1

2

, then

)

(

(

1+ 2 9 12 2 t t +1 3 3 2

5

)

 2r 2−1 − 0 (r − 1) 12  −  1 (r − 1) 12 −1 r 2 − 1      2  r −1

 2 2r (r − 1) 12  −  r − 1    1  2(r − 1) 2 = r −1

2 3

 1 1  −  3 t 2 t 2 +1 3     2    t3

2t 2 −

=

t3

o. Given s(r ) =

)

t3

2 3

2 3

2

=

)

( ) ( 3(t + 1)

1 3 12 2 t t +1 3 ⋅3 t 2 +1 2

=

(

)

1 1 2  3 3 1 2 −2  t + 1 3 ⋅ 2t  t 2 −  t 2 t + 1 3   3  2  

=

5   2t 2   3 t 2 + 1

(

)

(

)

1  3  3 3 −1 + 0  t 2 −  t 2 ⋅ t 2 +1 3    2 

t3

(

s ′( r ) =

2

) ⋅ (2t

1 2  t +1  3

1 3

(t + 1) =

=

 2r (r − 1) 12   1 

=

(

)

1  1   −1 2 2r (r − 1) 2  −  2 (r − 1) 2 r − 1  r −1

(

) 

  2  −  r −1 1     2(r − 1) 2 r −1

 

(

)

4r (r − 1) − r 2 − 1

=

1

2(r − 1) 2 r −1

4r 2 − 4r − r 2 + 1 1

=

2(r − 1) 2 r −1

3r 2 − 4r + 1

2(r − 1) r − 1

Example 2.7-2: Use the chain rule to differentiate the following radical expressions. a. d

  dx 

x−

  dx 

1   x 

b. d

 x3    dt  x + 1   

d. d

e. d

x + 2  3x + 1 

 3 x2

dx 

  dx 

c. d

+ x −2  

x 2  x 2 + 1 

f. d 5 x3 + 1 dx

Solutions: 



d  12 1  d  12 − 12  1 1 −1 1 − 1 −1 d 12 d − 12 d  1  x − 1 = x − x = x2 + x 2 x −x a. =  x −  =   2 2 dx dx dx  dx  dx  x  x 2  

Hamilton Education Guides

112

Calculus I

=

b.

2.7 The Derivative of Radical Functions

1 − 12 1 − 32 x + x 2 2

1

=

1

+

1 2x 2

d  x+2 2   dx  3 x + 1 

=

1

1

+

2 x

1  x+2 2   2  3x + 1 

=

 3/ x/ + 1 − 3/ x/ − 6  1   = = 1 1  2    x+2 2  x + 2  2  ( 3 x + 1) 2 2    3x + 1   3x + 1  1

= −

= −

c.

5 2 (x + 2)

2

(

1 2

x+2

(

)

5

)

2x x +1

1 2

=

)

( x + 1) 2

2

( x + 1)

=

1 2

2

(

)

1 −  x 2 +1  2/

−1

2

) ( x + 1) ( x + 1)

(

1 2

2

 ⋅ 2/ x  ⋅ x 2 

x3

1 2

2

=

1 2−

x 3 + 2x

( x + 1) 2

3

=

(x

(

x x2 + 2 2

+1

)

2x x

1  x+2    2  3x + 1 

=

2

     1

 5 (3 x + 1) 2 5  = − = − 1 1  2 (x + 2 ) 2 ( 3 x + 1) 2   x+2 2 2 2  ( 3 x + 1)  3x + 1 

 −5   ( 3 x + 1) 2 

1 2

 [ 1 ⋅ ( 3 x + 1) ] − [ 3 ⋅ ( x + 2 ) ]  ( 3x + 1) 2  

−1

( 3x + 1)

= −

2− 1

2

5 2 (x + 2)

1 2

( 3x + 1)

5

= −

4 −1 2

2 (x + 2 )

1 2

3

( 3x + 1) 2

5

1 2

(

)

d 2 d x − x2 x 2 +1 dx dx x 2 +1

=

(

)

2x x 2 +1

( ) ( x + 1)

1 2−

(

1 2

(

)

(

1  1 2−

=

x3 x 2 +1

x 2 +1

)

2x x 2 + 1

) ( 2x 1 −1 2

2  x +1 2

2 −1

x 2 +1

(

)

2x x 2 +1

−1

2

=

1 2−

x3

( x + 1) 2

2

1 2

x 2 +1

=

2x 3 + 2x − x 3

=

1 2

x 2 +1

x 3 + 2x

( x + 1) ( x + 1) ( x + 1) 2

1 2

2

2

1 +1 2

=

x 3 + 2x

( x + 1) 2

3 2

)

2

x +1

   d 32   32 d 3 32 − 1   32  3  ( ) ( ) ( ) x x x 1 + ⋅ x x + ⋅ − x + 1 1      −  x ⋅1   dx dx 2 d  x 3  d  x2          d. =  =  =   2 2 dt  x +1  dt  x +1  (x + 1) (x + 1)    

=

3 3 12 x (x + 1) − x 2 2 (x + 1) 2

=

Hamilton Education Guides

3 3 12 + 1 3 12 x + x −x2 2 2 (x + 1) 2

=

3 3 32 3 12 x + x −x2 2 2 (x + 1) 2

=

3  3 3 1  − 1 x 2 + x 2 2 2 

(x + 1) 2

=

)

 + 0  x2 

2x x 2 +1 − x 3

=

x 2 +1

1

x + 2 (3 x + 1) 3 x + 1

x 2 +1

2x x 2 +1

=

x

(

d  x+2  ⋅   dx  3 x + 1 

2 (x + 2)

2

2

2 x

+

5

= −

−1

= −

( 3x + 1) 3

d = dx

 d  x   dx  x 2 + 1  2

2

( 3x + 1) ( 3x + 1) 2

1

=

2 x3

1 −1

1

d  x+2    dx  3 x + 1 

=

3 2x 2

1 32 3 12 x + x 2 2 (x + 1) 2 113

Calculus I

2.7 The Derivative of Radical Functions

3

=

e.

3

1

x 2 + 3x 2

=

2 (x + 1)

x3 + 3 x

=

2

2 (x + 1)

x x +3 x

=

2

2 (x + 1)

=

2

x ( x + 3)

2 ( x + 1) 2

2−3  d 3 2 d  23 2 −1 d 23 d − 2 2 23 − 1 2 3 −2  − 2 −1  x + x− 2  = x + x 2 − x x x − 2 x − 3 = x 3 − 2 x −3 = =  x +x  =   dx  dx  3 dx dx 3 3  

=

f.

1

x 2 + 3x 2 2 (x + 1) 2

2

−

1 3x 3

d 5 3 x +1 dx

=

2 x

2

=

3

3

3 x

(

)

d x3 +1 dx

=

)

(

3x 2 x 3 + 1 5

2

−

x3 1 5

−4 5

=

(

3x 2

(

)

1 3 x +1 5

=

)

5 x3 +1

4 5

1 −1 5

⋅

(

3x 2

=

(x

5

5

) = 15 ( x + 1) ( 3x

d x3 +1 dx

3

+1

)

−4

3

5

3 −1

+0

) = 15 ( x + 1) 3

−4 5

⋅ 3x 2

4

Example 2.7-3: Use the chain rule to differentiate the following radical expressions. a. y = x 3 x 2 − 1

b. y = 3x 2 + x + 1

c. y = x 3 + x 2 + 1

d. y = 5 x 5 + x 2 − 1

e. y = x 2 − 1 ⋅ x + 1

f. y = x ( x + 1)3

h.

i. y = (x − 1) 3 3 x 5

g. y =

(x

2

)

+ 5x − 1

5

Solutions:

(

)

y = x2 3 x + 1

1

a. Given y = x 3 x 2 − 1 = x 3 x 2 − 1 2 , then y′



(

)

(

)

1

(



)

 1 2 x −1   2

= 3x 2 ⋅ x 2 − 1 2  +  1− 2   4 2 + x x −1 2   

1 −1 2

(

(

)

= 3x 2 x 2 − 1

)

  ⋅ 2 x ⋅ x 3  = 3 x 2 x 2 − 1   1 2

(

)

+ x 4 x 2 −1

−1

2

1 2

(

)

  2x 4 2 x −1 +   2

(

)

( x − 1) 2

)

 2 2  = 3 x x − 1  

x4

1

= 3x 2 x 2 − 1 2 +

(

1 −1 2 1

1 2

= 3x2 x2 − 1 +

1 2

   x4 x2 − 1

1

b. Given y = 3x 2 + x + 1 = 3x 2 + (x + 1) 2 , then y′

1 −1

1 2

= (3 ⋅ 2) x 2 − 1 + (x + 1) 2

1 2

= 6 x + (x + 1)

(

)

1− 2 2

1 2

1

= 6 x + (x + 1) − 2 = 6 x +

1 2 (x + 1)

1 2

= 6x +

1 2 x +1

1

c. Given y = x 3 + x 2 + 1 = x 3 + x 2 + 1 2 , then y′

=

(

)

1 3 x + x 2 +1 2

1 −1 2

Hamilton Education Guides

(

)

(

)

 3x 2 + 2 x  3  x + x 2 +1 ⋅ 3x 2 + 2 x =    2  

1 −1 2 1

(

)

 3x 2 + 2 x  3  x + x2 + 1 =   

2



1− 2 2

114

Calculus I

2.7 The Derivative of Radical Functions

(

)

 3x 2 + 2 x  3  x + x 2 +1 =   2





−1

2

 3x 2 + 2 x  ⋅ =   2



)

(



(x

1 3

)

+ x 2 +1

3x2 + 2x

=

1 2

2 x3 + x2 + 1

1

d. Given y = 5 x 5 + x 2 − 1 = x 5 + x 2 − 1 5 , then y′

=

(

)

1 5 x + x 2 −1 5

1 −1 5

(

(

)

 5x 4 + 2x  5  x + x 2 −1 =   5





(

)

)

 5x 4 + 2x  5  x + x 2 −1 ⋅ 5x 4 + 2 x =    5   −4 5

 5x 4 + 2 x  ⋅ =   5



(

)



1

(x

1 5

)

+ x 2 −1

5





5x 4 + 2x

=

4 5

(

)

 5x 4 + 2 x  5  x + x 2 −1 =  

1 −1 5 1

5

5

(x

5

+ x2 −1

)

1− 5 5

4

1

e. Given y = x 2 − 1 ⋅ x + 1 = x 2 − 1 2 ⋅ (x + 1) 2 , then y′

1

(

)

1 −1 2 ⋅ 2x ⋅

(

)

−1

=  x 2 −1 2 

= x x 2 −1 

2

1

1 −1

 

(x + 1) 2  + (x + 1) 2  

(

(

)

1  1 1 ⋅ (x + 1) 2  +  (x + 1) − 2 ⋅ x 2 − 1   2

  1  ⋅ x 2 − 1 ⋅ x + 1 +    2 x +1  x 2 −1 



1

= x ⋅  

)

1 2

1 2

  

⋅ x 2 −1

( x − 1)  2/  2/ x

  

1 −1 2 1





=  x ⋅ 

x x +1

=

2

= 

+

x2 −1

1

( x − 1) 2

1 2

(

)

1  1 1 1 ⋅ (x + 1) 2  +  (x + 1) 2 − 1 ⋅ x 2 − 1   2

  1 1 ⋅ (x + 1) 2  +    2 ( x + 1)

1 2

(

)

⋅ x 2 −1

1 2

1 2

   

x2 −1 2 x +1

1

f. Given y = x ( x + 1)3 = x 2 (x + 1) 3 , then y′

1

1 −1

=  x2 2

1   ⋅ (x + 1) 3  + 3 (x + 1) 2 ⋅ x 2    

 (x + 1) 3

= 

2



 (x + 1) 3

1 −1  1+ 3

1 2  (x + 1) x 2    

⋅x2

=  

2

x

−1

2

1   2  + 3 ( x + 1) x 2    

 (x + 1) 3 ⋅ 1  + 3 ( x + 1) 2 x 12  ( x + 1) 3 + 3 x ( x + 1) 2 =   =  1  

g. Given y = y′

=

 x2  

2

(

)

2

x + 5x − 1

(

)

5 2 x + 5x − 1 2

= h. Given

5 −1 2

5

(

y = x2 3 x + 1

)

5 =  x 2 + 5 x − 1   

⋅ ( 2 x + 5) =

5 ( 2 x + 5) ⋅ x 2 + 5x − 1 2

(

2 x



)

3 2

=

1 2

(

)

5 ( 2 x + 5) ⋅ x 2 + 5x − 1 2

5 ( 2 x + 5) ⋅ 2

5

= x 2 + 5 x − 1 2 , then

(

(x

)

2

)

+ 5x − 1

5 −1 2 1

3

=

=

5 ( 2 x + 5) ⋅ x 2 + 5x − 1 2

(

(

5 ( 2 x + 5) x 2 + 5 x − 1 2

)

)

5−2 2

x 2 + 5x − 1

1

= x 2 (x + 1) 3 , then

Hamilton Education Guides

115

  

Calculus I

2.7 The Derivative of Radical Functions

1  2 1 1  x 1  = 2 x ⋅ (x + 1) 3  +  (x + 1) 3 − 1 ⋅ x 2  = 2 x (x + 1) 3  +  (x + 1)

y′

 3



1

= 2 x (x + 1) 3 +



2 x2 (x + 1) − 3 3

1

= 2 x (x + 1) 3 +

( )

i. Given y = (x − 1) 3 3 x 5 = (x − 1) 3 x 5 5



 5  



3 = 3 x x 2 ( x − 1) 2 +

Example 2.7-4: Find a.

x+

d.

2x + 1 = y 2

g.

5 −1

= 3 (x − 1) 2 ⋅ x 3  +  x 3 3

y′

  3



1 3

x2 1 ⋅ 3 (x + 1) 23

1 −1 3 1



1

= 2 x (x + 1) 3 +

= 2x 3 x + 1 +

1− 3 x2 (x + 1) 3 3

x2 3 3 ( x + 1) 2

5

= (x − 1) 3 x 3 , then  5

5



 ⋅ (x − 1) 3  

= 3x 3 (x − 1) 2  +  x 3

5−3 3

2  53 2  5 3  3 x (x − 1)  +  x 3 (x − 1)     3



 





(x − 1) 3  =

53 2 x ( x − 1) 3 3

dy by implicit differentiation. dx 3

x2 + 5 y3 = 2

x +1 = y 3

f.

4

x2 y2 = x

x3 + y 3 = 2

i.

2

5 2

e. h.

7 x y 2 = 3x 2

c.

x2 + y2 = x

b.

y = 10

x 2 −1 = x y

Solutions: d dx

a.

(

)

x+ y =

d (10) dx

;

1 d  12 x + y2  = 0   dx  

1 −1 1 −1 1 12 − 1 d 1 1 −1 d x ⋅ (x ) + y 2 ⋅ ( y ) = 0 ; x 2 ⋅1 + y 2 ⋅ y ′ = 0 2 2 dx dx 2 2

;

1

1 1 − y y −y2 1 1 1 −1 1 − 1 1 y′ 1 y′ 1 1 ; y 2 y ′ = − x 2 ; ⋅ 1 = − ⋅ 1 ; 2/ y 2 ⋅ ⋅ 1 = 2/ y 2 ⋅ − ⋅ 1 ; y ′ = 1 ; y ′ = ; y′ = − x 2 2 / 2 2 2 2 2/ 2 x y x x2 y2 x2

(

d  2 d 2 (x ) ; d x 2 + y 2  x +y = dx dx   dx

b.

;

c.

2x + 2 y y′

(

2 x2 + y2

)

1 2

=1

(

) 2

1 2=1

; 2x + 2 y y′ = 2 x +

(

1 2 x + y2 2

;

)

1

y2 2

( ) +(y )

 d 3 2 5 3  d (2) ; d  x 2  x + y = dx  dx   dx

1 3

1

3 5

;

(

)

(

)

−1 1 2 d 2 d 2 x + y 2 2 (2 x + 2 y y ′) = 1 y  =1 ; x + 2 dx   dx

1 −1  2 

2 x2 + y2 y′ = 2y

)

1 2

− 2x

; y′ =

x2 + y2 − x y

3  2 2 −1 d 3 3 −1 d d  23 =0 ;  x + y 5  = 0 ; x 3 ⋅ (x ) + y 5 ⋅ ( y ) = 0 dx dx 3 5 dx    2

1 2 10 5 y 2 3y′ − 10 y 5 2 −1 3 −2 3 −2 − 2 − 13 −2 3 y ′ = −10 y 5 ; y ′ = y′ = − x 9 x = ; ; ; ; x 3 ⋅ 1 + y 5 ⋅ y′ = 0 ; y 5 y ′ = 1 2 1 5 3 3 5 93 x 9x 3 5 y 5 3x 3

Hamilton Education Guides

116

Calculus I

d dx

d.

(

2.7 The Derivative of Radical Functions

( );

)

1 d (2 x + 1) 2 = 2 y ⋅ d ( y ) dx dx

d 2 y dx

2x +1 =

; 2 y y ′ = x (2 x + 1)

−1

2

;

(

)

1 2 x +1 5

−4 5

⋅ 2x =

; y′ =

f.

(

2 − 13 ⋅ y′ y 3

2 xy 2 + 2 x 2 y y ′

5

(

)

4 x 2 y 2 −1

(

5

(x

)

2 x 2 y 2 −1

3 4

−

+1 4

)

1 4=1

)

( )

; y′ =

( )

d 7 2  d d 3x 2 ; xy 2  xy  = dx  dx  dx

g.

;

y 2 + 2 y y ′x 7

( )

6

xy 2 7

= 6x

3 4

(

)

−1

2

+

;

)

5

(

1 2 2 x y 4

;

) = 1 ; xy

(

2 x2 y2 −1 2

1 7

(

=0 ;

(

)

Hamilton Education Guides

)

−1

)

d  2 d  d (x y ) x 2 −1 ;  x −1  = dx dx   dx

(

)

1 2 x +1 5

−4

2 y′

;

1 3y 3

=

( )

d  2 23 − 1 d d 2 ⋅ (y)  dx x + dx (1) = 3 y dx  

1 −1  5 ⋅

2x

(

)

5 x 2 +1

4 5

(

)

4

1

; 10 x 2 + 1 5 y ′ = 6 x y 3

2

1 2

1 2 = 1⋅

(

;

)

+ x 2 y y′ = 2 x 2 y 2 −1

−

)

(

3 4

)

3

; x 2 y y ′ = 2 x 2 y 2 − 1 4 − xy 2

y x 1 −1 7

⋅

( )

d xy 2 = 6 x dx

; 2 y y ′x = 42 x xy 2

(

1 3 x + y3 2

)

;

6 7−

( ) (1 ⋅ y

1 xy 2 7

y 2 ; y′ =

−6 7

42 x

13 7

2

)

+ 2 y y′ ⋅ x = 6x 12

y 7 − y2 2x y

(

) (

)

−1 1 3 d 3 d 3 x + y 3 2 ⋅ 3x 2 + 3 y 2 y ′ = 0 x + y =0 ; 2 dx  dx 

(

d (y) dx

;

1 −1  2 ⋅

3y 2 y′ 3 x + y3 2

y + x⋅

) (

(

)

( )

6 xy 2 7

=0

(

−3 1 2 2 d 2 2  4 2 xy 2 + 2 x 2 y y ′ = 1 y x − 1 1 = x y ;   dx 4  

1 −1  4

( )

( )

(

3 4

)

1 xy 2 7

;

= 6x

; y 2 + 2 y y ′x = 6 x ⋅ 7

3y 2 y′ 3 x + y3 2

)

2

x y

d d  3 3 (2) ; d x 3 + y 3  x +y = dx dx   dx 3x 2 3 x + y3 2

1 1 (2 x + 1) − 2 ⋅ 2 x = 2 y ⋅ y ′ 2

)

2

(

x2 y

2 32 −1 d ⋅ (y) y 3 dx

(

4 x 2 y 2 −1

xy 2

=

2 − 13 2x 2 y y′ = x +1 3 5

2 xy 2 + x 2 y y ′

=1 ;

3 4

x2 y

(

1 5

3x3 y

; y′ =

4 5

;

(

; y′ =

i.

)

)

d x 2 +1 dx

;

d 4 2 2  d (x ) ; d x 2 y 2  x y = dx  dx  dx

;

h.

(

10 x 2 + 1

;

1

1

6x y 3

1− 2 2

x x (2 x + 1) − 2 x ; y′ = ; y′ = ; y′ = 1 2y 2 y 2x + 1 2 y (2 x + 1) 2

2 d 5 2  d  y 3   x +1 = dx   dx  

e.

1 2

; 6 x + (x + 1)

(

)

−1

2

=−

)

1 2 x −1 2

(

3x 2 3 x + y3 2

1 −1 2

)

−1

2

; y 2 y′ = − x 2 ; y ′ = −

(

)

⋅ 2x = y + x y ′ ; x x 2 −1

−1

2

= y + x y′

117

x2 y2

Calculus I

2.7 The Derivative of Radical Functions

(

2

)

(

−1

; y + x y′ = x x −1

)

2

; x y′ = x x −1

2

−1

2

−y

(

x x2 −1 y′ = x

;

)

−1

2

−y

Example 2.7-5: Compute the derivative for the following Radical expressions.

(

)=

a.

d 2x + x dx

d.

d  3 x+ x  dx  

g.

d  x +1   dx  x 

=

=

(

)=

b.

d dx

e.

d  5   x − 3 dx  

h.

d  1  = dx  x +1 

x +1

=

c.

d 3 2   x +1 = dx  

f.

d x3 x +1 dx

i.

d  x 2 dx  3 x

(

)=

  =  

Solutions:

(

)=

1 1 1 −1 1 −1 1 d d 12 2x + x = 2+ x2 = 2+ x 2 = 2+ 1 = 2+ 2 2 dx dx 2 x 2x 2

a.

d 2x + x dx

b.

d dx

c.

d 3 2 d  x 2 +1  x +1 = dx  dx 

)

1 3

d.

3 d  d  3 x+ x2 x+ x  = dx  dx  

   

e.

 5x x 5 d  5 5 32 5 52 − 1 d  52 d 52 d   x − 3 = x3 = x = 0 + x ( ) x + − 3 = =  x − 3 =   2 2 2 2 dx  dx  dx dx  

f.

d x3 x +1 dx

(

x +1

)=

1 d (x + 1) 2 dx

(

(

= 3x (x + 1) 2

=

g.

d d 2x + dx dx

)=

1 2

d  x +1   dx  x 

=

=

=

1 1 (x + 1) 2 − 1 ⋅1 2

=

=

(

)

1 2 x +1 3

1 1 (x + 1) − 2 2

=

1 −1 3

⋅ 2x

=

=

1 2 (x + 1)

(

)

2x 2 x +1 3

=

1 2

−2

=

3

1 2 x +1

(

2x

)

3 x 2 +1

3 3 1 3 3 −1 d d 32 x+ x = 1+ x 2 = 1+ x 2 = 1 + 2 2 2 dx dx

2 3

= 3

3

(x

2x 2

+1

)

2

x

1 1 −1 1 d 1  x 3 (x + 1) 12  2 3 1 3 3 d  = (x + 1) 2 dx x + x dx (x + 1) 2 = (x + 1) 2 ⋅ 3 x + x ⋅ 2 (x + 1) 2 

1 x3 (x + 1) − 2 + 2

6x 3 + 6x 2 + x 3 2 x +1

d dx

=

x

7x 3 + 6x 2 2 x +1

  d  x +1 dx  12   x 

Hamilton Education Guides

=

1 2

= 3x (x + 1) + 2

=

1

1

x3 1

2 (x + 1) 2

=

3 x 2 (x + 1) 2 ⋅ 2 (x + 1) 2 + x 3 1

2 (x + 1) 2

=

6 x 2 (x + 1) + x 3 2 x +1

x 2 (7 x + 6 ) 2 x +1

 12 d   d 12  x   x ⋅ (x + 1) − (x + 1) ⋅ dx dx     x

=

 12   1 1 −1   x ⋅1 − (x + 1) ⋅ x 2  2     x

118

Calculus I

1 x2

=

2.7 The Derivative of Radical Functions

1 x2

 1 −1 − (x + 1) ⋅ x 2  2   x

−

=

1

x 2 ⋅ 2 x 2 − (x + 1)

1 2x 2

1 2x 2

=

x

=

−

1 1 (x + 1) − 2 2 x +1

d  x 2 i. dx  3 x

=

 2   = d x  dx  13  x

x −1

2x − x −1

x −1 = 2 x =

2 x x

=

x

x

2x x

1 1 d   d  1 1 0 − (x + 1) 2 − 1  (x + 1) 2 ⋅ dx (1) − 1 ⋅ dx (x + 1) 2       2 =  = x +1 x +1 

d  1 d  1  d  1    = h. = dx  (x + 1) 12 dx  x +1  dx  x +1  

−

1

x +1

1 1

2 (x + 1) 2 (x + 1)

    

=

= −

1 2 (x + 1)

d  2 − 13  x ⋅x   dx  

1 2

= −

(x + 1)

1 2 (x + 1)

d 2 − 13 d x x = = dx dx

6−1 3

= −

3 2

1 2 5

d 3 x dx

=

(x + 1) 3 =

5

= −

5 3 −1 x 3

1

2 ( x + 1) x + 1 2

5 3 x 3

=

=

3

5 x2 3

Example 2.7-6: Evaluate the derivative of the following equations at the given values. a. y =

( x + 1) 2

3

at x = 2

b. y =

d. y = 3 2 x + 3 x 2 at x = 2

e. y =

5x + 1 2x + 1

3

6x + 1 x

2

c. y = x x 2 + 1 at x = 5

at x = 10

x + x +2

at x = 3

f. y =

at x = 3

h. y = x x 2 − 10 + 2 x at x = 5

2

at x = 0

k. y =

a. Given y =

( x + 1)

g. y = j. y =

x2 +1 1− x

x5 +1 2x

1 2

x +1

at x = −5

i. y = x 2 ( x + 1)3 at x = 1

at x = 2

l. y =

x +1 x3 +1

at x = 1

Solutions:

y′

=

2

3

) ( 2x

(

3 2 x +1 2

3 −1 2

(

)

3

= x 2 + 1 2 , then

2 −1

)

+0 =

(

)

3 2 x +1 2/

3−2 2

⋅ 2/ x

)

(

= 3x x 2 + 1

1 2

= 3x x 2 + 1

Substituting x = 2 in place of x in the y ′ equation we obtain the following value: y′

= (3 ⋅ 2) 2 2 + 1 =

b. Given y = y′

=

3

6 5

x + x +2

=

13.42 1

1

= x 3 + x 2 + 2 , then

1 −2 1 −1 1 13 − 1 1 12 − 1 x + x +0 = x 3 + x 2 = 3 2 3 2

1 2 3x 3

+

1 1 2x 2

=

1 3

3 x2

+

1 2 x

Substituting x = 10 in place of x in the y ′ equation we obtain the following value:

Hamilton Education Guides

119

Calculus I

y′

=

2.7 The Derivative of Radical Functions

1 2 3 ⋅10 3

+

1

=

1 2 ⋅10 2

1 1 + 3 ⋅ 4.634 2 ⋅ 3.162

(

)

= 0.072 + 0.158 = 0.23

1

c. Given y = x x 2 + 1 = x x 2 + 1 2 , then

(

)

(

)

1   1 y ′ = 1 ⋅ x 2 + 1 2  +  x 2 + 1   2 

(

)

2

x2

1 2

= x +1 +

(x + 1) 2

(

(

)

)

 ⋅ 2 x 2 − 1 + 0 ⋅ x = x 2 + 1 

(x + 1) (x + 1) (x + 1) 1 2

2

=

1 2

1 −1 2

2

1 2

+ x2

1 2

2

=

1 2

(

)

1 +  x 2 +1  2/

(x + 1) + x (x + 1) 1+1 2 2

2

1 2

2

2

−1

=

2

)

(

 ⋅ 2/ x 2  = x 2 + 1  x 2 +1+ x 2

(x + 1) 2

1 2

=

1 2

)

(

+ x 2 x 2 +1

2x 2 + 1 x 2 +1

Substituting x = 5 in place of x in the y ′ equation we obtain the following value: y′

=

2 ⋅ 52 +1

(5 + 1) 2

1 2

=

2 ⋅ 25 + 1

(25 + 1)

=

1 2

50 + 1 1 26 2

=

51 5.1

= 10

2

1

d. Given y = 3 2 x + 3 x 2 = (2 x ) 3 + x 3 , then y′

=

2 −1 1 2 −1 1 2 1 + (2 x ) 3 − 1 + 2 x 3 = 1 (2 x )− 3 + 2 x 3 = 1 2 + 21 = 3 3 3 3 3 3(2 x ) 3 3 x 3 33 (2 x )2 3 x

Substituting x = 2 in place of x in the y ′ equation we obtain the following value: y′

=

1 3 ⋅ (2 ⋅ 2)

e. Given y =

y′

=

2 3

+

2

=

1 3⋅ 2 3

6x + 1 x2

2 3⋅ 4 3

+

2 1 3⋅ 2 3

=

2 1 + 3 ⋅ 2.53 3 ⋅1.26

= 0.132 + 0.53 = 0.662

1

(6 x ) 2 + 1 , then = 2 x

1 −1 1  1   2   2 −1  ⋅ (6 x ) 2 + 1    (6 x ) 2 ⋅ 6 + 0 ⋅ x  − 2 x     2  

x4 1

=

1

=

1  1  2    −1   (6 x ) 2 ⋅ 6 ⋅ x  − 2 x ⋅ (6 x ) 2 + 1      2  

x4

1

3 x 2 (6 x ) − 2 − 2 x(6 x ) 2 − 2 x x4

Substituting x = 3 in place of x in the y ′ equation we obtain the following value: 1

y′

= =

1

3 ⋅ 3 2 ⋅ (6 ⋅ 3) − 2 − 2 ⋅ 3 ⋅ (6 ⋅ 3) 2 − 2 ⋅ 3 3

4

=

1

1

27 ⋅ (18) − 2 − 6 ⋅ (18) 2 − 6 81

=

(27 ⋅ 0.236) − (6 ⋅ 4.243) − 6 81

25.086 6.372 − 25.458 − 6 = − = −0.31 81 81

Hamilton Education Guides

120

−1

2

Calculus I

2.7 The Derivative of Radical Functions

1

f. Given y =

= =

1

, then

(x + 1)   1 ⋅ (2 x 0 ⋅ (x + 1)  −  (x + 1)  2  x2 +1 1 2

2

y′

=

1 2

2

1 −1 2

2

2−1

 



x 2 +1 −x

(x + 1)(x + 1) 2

2

=

1 2

−x

(

)

(x + 1) 2

−1

2

 ⋅ 2/ x  

=

x 2 +1

=

3 2

)

1 0 −  x 2 +1  2/

=

−x

=

1+ 1 x 2 +1 2

(

)

 + 0 ⋅1 

−x

(x + 1) 2

3

(

)

− x x 2 +1

−1

2

x 2 +1

−x

=

(x + 1) 2

x 2 +1

Substituting x = −5 in place of x in the y ′ equation we obtain the following value: y′

= −

−5

[(− 5) + 1] 2

g. Given y =

y′

=

5

=

3

(25 + 1)

3

5

=

=

17576

5 = 0.038 132.57

1

5x + 1 (5 x + 1) 2 , then = 2x + 1 2x + 1

(

)

(

)

1 1 −1 1  1−1 + 0 ⋅ (2 x + 1) −  2 x1 − 1 + 0 (5 x + 1) 2   2 (5 x + 1) 2 ⋅ 5 x    

(2 x + 1)2

=

2.5(5 x + 1) − 12 ⋅ (2 x + 1) − 2(5 x + 1) 12     

(2 x + 1)2

Substituting x = 3 in place of x in the y ′ equation we obtain the following value: y′

=

2.5(15 + 1) − 12 ⋅ (6 + 1) − 2(15 + 1) 12     

=

(6 + 1)2

(

)

17.5 ⋅ (16) − 12  − 2⋅ (16) 12      72

[17.5 ⋅ 0.25] − [2 ⋅ 4] =

=

49

−0.074

1

h. Given y = x x 2 − 10 + 2 x = x x 2 − 10 2 + 2 x , then y′



(

)

1

 

(

= 1 ⋅ x 2 − 10 2  + 0.5 x 2 − 10 

 

)

1 −1 2

(

)

(

 ⋅ 2 x 2 − 1 − 0 ⋅ x  + 2 x1 − 1 = x 2 − 10 

)

1 2

(

 +  x 2 x 2 − 10 

)

−1

2

 +2 

Substituting x = 5 in place of x in the y ′ equation we obtain the following value: y′

(

= 5 2 − 10

)

1 2

(

 + 5 2 5 2 − 10 

)

−1

2

 +2 

1



= 15 2 + 25 ⋅15 

−1

2

 +2 

. + [ 25 ⋅ 0.258] + 2 = 12.323 = 3873

3

i. Given y = x 2 ( x + 1)3 = x 2 (x + 1) 2 , then y′

(

)

3 3 3 1 3 3  = 2 x 2−1 ⋅ (x + 1) 2  +  (x + 1) 2 − 1 ⋅ x1−1 + 0 ⋅ x 2  = 2 x(x + 1) 2  +  x 3 (x + 1) 2    2   2  

Substituting x = 1 in place of x in the y ′ equation we obtain the following value: y′

3   3 1  3 1 3 . = 2(1 + 1) 2  +  (1 + 1) 2  =  2 ⋅ 2 2  +  ⋅ 2 2  = 5.657 + 2121 = 7.778 2   2 

Hamilton Education Guides



 



121

Calculus I

2.7 The Derivative of Radical Functions 1

x2 +1

 x 2 +1 2   , then =  1− x 2  1− x2  

j. Given y =

) ( ( )

(

1 −1 

)

1

1

− −  2 2  2  2  2 x − 2/ x 3/ + 2/ x 3/ + 2 x   2  2 4x  2x 1 − x + 2x x +1  = 1  x + 1    = 1  x +1 2     2  1 − x 2  2  1 − x 2  2 2 2 2 1− x 1− x 1− x 2    

1  x 2 + 1  2 y′ =  2  1 − x 2 

(

)

(

)

Substituting x = 0 in place of x in the y ′ equation we obtain the following value: 1  0 +1 y′ =   2  1− 0 

)

(

=

2

⋅

1 5  x +1  2/

4⋅0

1 − 12 ⋅1 ⋅ 0 2

=

(1 − 0)2

x5 +1 2x

k. Given y =

y′

−1

(x + 1) = 5

1 2

2x

1 −1 2 5x 4

= 0

, then

(

)

  ⋅ 2/ x  − 2 x1 − 1 ⋅ x 5 + 1  

1 2

(2 x )2

(

  

=

)

 5 5 5 x x + 1 

−1

2

(

)

  5  − 2 ⋅ x + 1  

1 2

  

4x 2

Substituting x = 2 in place of x in the y ′ equation we obtain the following value: y′

=

160(32 + 1) − 12  − 2 ⋅ (32 + 1) 12      4⋅4 x +1

l. Given y =

x +1 1 2



1 −1 2

3

2

 

(x + 1)

1 2

=

3 2 x (x + 1) 2 − 1

(x + 1) 3

(x + 1) 3

2

2 x 3 + 2 − 3x 3 − 3x 2

(

) (x + 1)

2 x3 +1

1 2

3

1

=

x3 +1

=

. 27.8 − 115 = 1.02 16

=

16

)

  + 0  (x + 1)   

x3 +1

3

=

1 2

3

[160 ⋅0.174] − [2 ⋅ 5.744]

, then

(x + 1)     1 (3x 1 ⋅ (x + 1)  −   (x + 1)    2 3

3

y′ =

x +1

=

=

1 2

−

(

)

x3 +1 − x 3 − 3x 2 + 2

(

) (x + 1) 1 2

3

3

=

1 2

1 2

=

=

)

(x + 1) 2(x + 1) − 3x 2(x + 1) 1 2

1 2

3

3

3

)

2 x3 +1

1 +1 2

=

(

1 2

)

2 x 3 + 1 − 3x 3 − 3x 2

− 3x 2

x3 +1

− x 3 − 3x 2 + 2

(

(

   1  −1 −   x 3 + 1 2 3 x 2  (x + 1)     2  x3 +1

3

3x 3 + 3x 2 2 x3 +1

2 x3 +1

(x + 1)

=

(

)

2 x3 +1

1 2

x3 +1

− x 3 − 3x 2 + 2

(

)

2 x3 +1

3 2

Substituting x = 1 in place of x in the y ′ equation we obtain the following value: y′

=

( ) 2(1 + 1)

− 13 − 3 ⋅12 + 2 3

3 2

Hamilton Education Guides

=

−1 − 3 + 2 3 2⋅2 2

=

−2/ 2/ ⋅ 2.828

=

−1 2.828

= −0.354

122

Calculus I

2.7 The Derivative of Radical Functions

Section 2.7 Practice Problems - The Derivative of Radical Functions 1. Find the derivative of the following radical expressions. Do not simplify the answer to its lowest term. b. y = x 3 + 3x − 5

a. y = x 2 + 1 d. y = g. y = 2. Use the

x +1 x

x2 + 3

f. y = x 3 + 3x 2

2

x −1 4

h. y =

x +1 d dx

x2

e. y =

c. y = x 2 + x − 1

x 3 −1

x3

i. y =

x

x2 x

notation to find the derivative of the following radical expressions.

a.

d  2 1  x +  x dx 

d.

d  x + 5  = dx  x 

=

b.

d  x  =  dx  x −1 

c.

d  x 3  = dx  x +1 

e.

d  3 x  = x +  dx  x 

f.

d  2 x 1+ dx  x3

c.

d (x y ) = d dx dx

  =  

3. Find the derivative of the following radical expressions. a.

d  3  d (x )  x + y= dx   dx

d.

d dx

g.

d d (2) x y2 + x = dx dx

(

)

y + x3 = 0

(

)

(

)

d (2) dx

( x)

b.

d dx

e.

d  4 d 2 (x )  x +y = dx   dx

f.

d dx

(

x +1 =

h.

d  3 d  x  + (x y ) = 0 dx   dx

i.

d dx

(

x + 3y =

x + y3 =

)

)

( )

d 3 y dx

d (y) dx

4. Evaluate the derivative of the following radical expressions for the specified values of x . a. y = 3x 3 + x 2 at x = 1 d. y =

x x 2 +1

at x = 1

Hamilton Education Guides

b. y = x 2 + 1 x at x = 0

(

)

c. y =

e. y = x 3 + 1 + 4 x 3 at x = 0

f. y =

x 2 −1 4x 2 x 2 +1 x3

at x = 2 at x = 3

123

Calculus I

2.8

2.8 Higher Order Derivatives

Higher Order Derivatives

If the function y = f ( x ) is differentiable, then we can form a new function y ′ = f ′( x ) which is referred to as the first derivative of y = f ( x ) . Consequently, if y ′ = f ′( x ) is differentiable, then we can form another new function y ′′ = f ′′( x ) called the second derivative of y = f ( x ) . This process of obtaining a new function can continue as long as we have differentiability. First, second, third, and higher order derivatives are denoted in various notations. In general, however, first, second, third, and nth derivatives are shown in the following forms: d dy f ( x) = D y = D f ( x) = dx dx

y′

= f ′( x ) =

y ′′

= f ′′( x ) =

y ′′′

= f ′′′( x ) =

d2y dx 2

=

d  dy  d d d2  f ( x ) = D 2 y = D 2 f ( x ) f ( x) =   = 2 dx  dx  dx  dx  dx

=

 d  d2 d  d2 y d3  =  f x f x = = D3 y = D3 f ( x ) ( ) ( )   2 2 3 dx  dx  dx  dx dx 

=

d  d n −1 y    dx  dx n −1 

d3y dx 3

 yn

dny

= f n ( x) =

dx n

=

dn dx n

f ( x)

=

 d  d n −1 n n  n −1 f ( x )  = D y = D f ( x ) dx  dx 

Students are encouraged to become familiar with these notations for finding the derivative of different functions. Note that the prime notation is not used beyond the third derivative. In general, the fourth or higher derivatives are shown as y 4 = f 4 ( x ) ; y 5 = f 5 ( x ) ; y 6 = f 6 ( x ) ;  ; yn = f

n

( x ) instead of

f ( x) = x 6 + x 3 + 1 , f 5 ( x ) = 720 x , f

6

then

y iv = f

iv

(x ) ;

(x ) ; y v i = f v i (x ) , etc. For example, given f ′′( x ) = 30 x 4 + 6 x , f ′′′( x ) = 120 x 3 + 6 , f 4 ( x ) = 360 x 2 ,

yv = f

f ′ ( x ) = 6 x 5 + 3x 2 ,

v

( x ) = 720 , and all derivatives higher than

are equal to zero. The following examples show in detail how higher order derivatives are obtained: 7

Example 2.8-1: Find the second derivative of the following functions. a. f ( x ) = 5x 8 − 3x 3 + 1 d. f (u) =

u3 − 1 u +1

(

)

)

−1

t 3 + t 2 +1 10

h. r (θ ) = θ 2 + k. p( r ) = r 2 −

x

1 x

f. h( x ) = a 2 + x 3

3

1

(θ + 1) 1 r

c. f ( x ) = x 2 +

(

1

e. g( x ) = x 2 +

g. f ( x ) = x 2 + 1 j. f ( t ) =

(

b. f ( x ) = x 3 x 2 + x + 5

3

(

2

)

i. s(r ) = r 2 r 2 + 1 l. f ( x ) =

)

3

x3 x +1

Solutions: a. Given f ( x ) = 5x 8 − 3x 3 + 1 , then f ′( x ) = ( 5 ⋅ 8) x 8−1 − ( 3 ⋅ 3) x 3−1 + 0 = 40 x 7 − 9 x 2 and

Hamilton Education Guides

124

Calculus I

f ′′( x )

2.8 Higher Order Derivatives

= ( 40 ⋅ 7) x 7 −1 − ( 9 ⋅ 2) x 2 −1 = 280 x 6 − 18 x

(

)

b. Given f ( x ) = x 3 x 2 + x + 5 , then f ′( x )

[

)] [(

(

) ] [ (

)] [

= 3x 3−1 ⋅ x 2 + x + 5 + 2 x 2−1 + 1x 1−1 + 0 ⋅ x 3 = 3x 2 x 2 + x + 5 + ( 2 x + 1) ⋅ x 3

]

= 3x 4 + 3x 3 + 15x 2 + 2 x 4 + x 3 = 5 x 4 + 4 x 3 + 15 x 2 and f ′′( x ) = ( 5 ⋅ 4) x 4 −1 + ( 4 ⋅ 3) x 3−1 + (15 ⋅ 2) x 2 −1 = 20 x 3 + 12 x 2 + 30 x 1 x

c. Given f ( x ) = x 2 + , then f ′( x ) = 2 x 2 −1 +

[0 ⋅ x] − (1⋅ 1) = 2

2x +

x

0−1 x

2

1

= 2x −

x2

A second way is to rewrite f ( x ) as f (x ) = x 2 + x −1 and find its derivative, i.e., f ′( x ) = 2 x 2 −1 − 1 ⋅ x −1−1 = 2 x − x −2 = 2 x − f ′′( x )

1 x2

and

= 2 x 1−1 + ( −1⋅ −2) x −2 −1 = 2 x 0 + 2 x −3 = 2 + 2 x −3 = 2 + u3 − 1 , then u +1

d. Given f (u) =

[ (3u f ′( x ) =

3−1

)]

)(

(u + 1)2

[ (2 ⋅ 3u f ′′( x ) = [ (6u =

] [(

)

− 0 ⋅ (u + 1) − u 1−1 + 0 ⋅ u 3 − 1

2

3−1

=

][

)

x3

(

) = 3u

3u 2 ⋅ ( u + 1) − u 3 − 1

( u + 1) 2

3

+ 3u 2 − u 3 + 1

( u + 1) 2

=

2u 3 + 3 u 2 + 1

( u + 1)2

)]

(

+ 3 ⋅ 2u 2−1 + 0 ⋅ (u + 1)2 − 2(u + 1)2−1 ⋅ 2u 3 + 3u 2 + 1

(u + 1)4

][

)

)]

(

+ 6u (u + 1)2 − 2(u + 1) 2u 3 + 3u 2 + 1

(u + 1)

e. Given g( x ) = x 2 + g ′( x ) = 2 x 2−1 +

2

1

x3

4

=

[ (6u

2

][

)

(

)]

+ 6u (u + 1)2 + (− 2u − 2 ) 2u 3 + 3u 2 + 1

(u + 1)

4

, then

(0 ⋅ x )− (3x 3

x

6

3−1

)

⋅1

= 2x +

−3x 2 x

6

= 2x −

3x 2/ x

6/ = 4

= 2x −

3 x4

A second way is to rewrite g( x ) as g (x ) = x 2 + x −3 and find its derivative, i.e.,

Hamilton Education Guides

125

Calculus I

2.8 Higher Order Derivatives

g ′( x ) = 2 x 2 −1 − 3 ⋅ x −1− 3 = 2 x − 3x −4 = 2 x −

3 x4

and 12

g ′′( x ) = 2 x1−1 + (− 3 ⋅ −4 )x −4−1 = 2 x 0 + 12 x −5 = 2 + 12 x −5 = 2 +

(

) ) ⋅ (0 + 3x ) = 2(a

x5

2

f. Given h( x ) = a 2 + x 3 , then

(

h ′( x ) = 2 a 2 + x 3

2 −1

3−1

(

2

)

(

)

+ x 3 ⋅ 3x 2 = 6 x 2 a 2 + x 3 = 6a 2 x 2 + 6 x 5 = 6 x 5 + 6a 2 x 2

)

h ′′( x ) = ( 6 ⋅ 5) x 5−1 + 6a 2 ⋅ 2 x 2 −1 = 30 x 4 + 12a 2 x

(

)

g. Given f ( x ) = x 2 + 1 f ′( x )

(



)

= −1⋅ x 2 + 1 

(

( 

)

h. Given r (θ ) = θ 2 + r ′ (θ )

= 2θ 2−1 +

r ′′(θ )

= 2θ 1−1 −

) = −( x + 1)

(

2

2 −1 +0  ⋅ 2x 

)



, then

−1−1 

 f ′′( x ) = −2 1 ⋅ x 2 + 1 

= −2 x 2 + 1

−1

−2

−2

(

)

 + −2 x 2 + 1 

(

)

− 4x x 2 + 1 1

(θ + 1) 3

−2 −1

−3 

 

(

)

(

)

= −2 x 2 + 1

−2

(

)

= −2x x 2 + 1

( 

)



= −2 x 2 + 1

)

(

+ 8x x 2 + 1

−2

−2

(

)

 + −2 x 2 + 1 

−3

 ⋅ 2 x  

−3

, then 3−1

3

(θ + 1)

= 2θ +

6

[ 0 ⋅ (θ + 1) ]− [2(θ + 1) 2

2 −1

⋅3

]

= 2−

(θ + 1)4

)

 ⋅ 2x 

 ⋅ 2 x 2 −1 + 0   

[ 0 ⋅ (θ + 1) ]− [3(θ + 1) ⋅1]

(

−2 

0 − 3 (θ + 1)

(θ + 1)

2

6

0 − 6 (θ + 1)

= 2θ −

= 2+

(θ + 1) 4

3 (θ + 1)

(θ + 1)

= 2θ −

6/ = 4

6 (θ/ + 1/ )

(θ + 1)

2/

4/ = 3

= 2+

3

(θ + 1) 2

and

6

(θ + 1) 3

3

i. Given s( r ) = r 2 r 2 + 1 , then

)

(

(

) ⋅ (2r

3   s ′(r ) = 2r 2 −1 ⋅ r 2 + 1  + 3 r 2 + 1   

(

)

(

3

)

= 2r r 2 + 1 + 6 r 3 r 2 + 1

(

)

(

2

2 −1

)

 + 0 ⋅r2 



(

)   (

)

3

2



= 2 r r 2 + 1  + 3 r 2 + 1 ⋅ 2 r  ⋅ r 2 



and

) ⋅ (2r

3   s ′′( r ) = 2 1 ⋅ r 2 + 1  + 3 r 2 + 1   

Hamilton Education Guides

3−1

3−1

2 −1

)

(

)

(

) ⋅ ( 2r

2     + 0  ⋅ r  + 63r 3−1 ⋅ r 2 + 1  + 2 r 2 + 1     

2 −1

2 −1

)

  + 0 ⋅r3  

126

Calculus I

2.8 Higher Order Derivatives

( 

)

(

)



)

( 

3



= 2  r 2 + 1 + 3 r 2 + 1

3−1

(

) [(

(

) ]

2     ⋅ 2r  ⋅ r  + 63r 2 r 2 + 1  + 2 r 2 + 1 ⋅ 2r ⋅ r 3      

)

) [ (

(

)]

3  2  2    = 2  r 2 + 1 + 6 r 2 r 2 + 1   + 6   3 r 2 r 2 + 1  + 4 r 4 r 2 + 1     



j. Given f (t ) = f ′( t )

[ (3t =

f ′′( t ) =

3−1

3





t + t +1 , then 10

)]

) ][ (

+ 2t 2−1 + 0 ⋅10 − 0 ⋅ t 3 + t 2 + 1

=

10 2

[ (3 ⋅ 2t



2

2 −1

) ][ (

+ 2t 1−1 ⋅10 − 0 ⋅ 3t 2 + 2t 10

k. Given p(r ) = r 2 −

)]

2

(

)

10 3t 2 + 2t − 0 100

10 ( 6t + 2) − 0

=

100

=

=

(

/ / 3t 2 + 2t 10 / / / = 10 100

/ / ( 6t + 2) 10 / / / = 10 100

=

)=

3 t 2 + 2t 10

2/ (3t + 1) 1/ 0/ = 5

=

and

3t + 1 5

1 which is equal to p(r ) = r 2 − r −1 , then r

(

)

p ′(r ) = 2r 2 −1 − −1 ⋅ r −1−1 = 2r + r −2 and p ′′( r ) = 2r 1−1 − 2r −2 −1 = 2r 0 − 2r −3 = 2 − 2r −3

l. Given f ( x ) =

[ 3x f ′( x ) =

3−1

x3 , then x +1

) ] = 3x

] [(

⋅ ( x + 1) − 1 ⋅ x 1−1 + 0 ⋅ x 3

{[(2 ⋅ 3) x f ′′( x ) =

( x + 1) 2 2

]

+ ( 3 ⋅ 2) x ⋅ ( x + 1)

2

( x + 1) − x 3 = ( x + 1) 2

} − {2( x + 1) ⋅ (2 x

3

+ 3x 2

( x + 1) 4

Example 2.8-2: Find

d3y dx 3

3x 3 + 3x 2 − x 3

( x + 1) 2

)} = [ (6 x

2

)

=

2x 3 + 3 x 2

and

( x + 1) 2

][

(

+ 6 x ( x + 1 ) 2 − 2( x + 1) 2 x 3 + 3 x 2

)]

( x + 1)4

for the following functions.

(

a. y = (1 − 5x )3 1 5

2

b. y = a − bx 2

1 4

d. y = x 5 + x 4 + x

e. y =

Solutions:

)

−2

ax 2 + b c

c. y = f. y =

x 2 + 3x + 1 x +1 x2 +1 x3

a. Given y = (1 − 5x )3 , then

)

(

y′

2 = 3(1 − 5x ) 3−1 ⋅ 0 − 5x 1−1 = 3 (1 − 5x ) 2 ⋅ ( −5) = −15 (1 − 5 x)

y ′′

= ( −15 ⋅ 2) (1 − 5x ) 2 −1 ⋅ 0 − 5x 1−1 = −30 (1 − 5x ) ⋅ ( −5) = 150 (1 − 5 x)

Hamilton Education Guides

(

)

127

Calculus I

y ′′′

2.8 Higher Order Derivatives

(

) )

= 150 0 − 5x 1−1 = 150 ⋅ ( −5) = −750

(

b. Given y = a − bx 2

(

)

−2

= −2 a − bx 2

y ′′

= ( 4b ⋅ 1) a − bx 2

(

 

(



= 4b a − bx 2 

y ′′′

)

)

)

−3−1

(



= 24b 2 x a − bx 2 

c. Given y =

[ (2x y′ = (2 x =

2

[ (2x y ′′ =

)

x 2 + 3x + 1 , x +1

2 −1

(

)

(

)

 2  + −3 a − bx  

−3 

−3−1

)

−3

  ⋅ ( 0 − 2bx )  ⋅ ( 4bx )   

(

)(

(

)

  ⋅ ( −2bx )  +  48b 2 x ⋅ a − bx 2  

−4 

−4



)

(

−3 

 2  + 6bx a − bx  

] [(

)

) (

(

)]

)(

)

( x + 1) 2

)

−4

 ⋅ ( 4bx )  

=

)

2 x 2 + 5x + 3 − x 2 − 3x − 1

( x + 1) 2

][

)

(

(x + 1)4 2

2

+ 2x + 2

2 x 3 + 6x 2 + 6x + 2 − 2 x 3 − 6x 2 − 8x − 4

( x + 1)

4

][

)] =

[2x =

x 2 + 2x + 2

( x + 1) 2

[ (2x + 2)(x + 1) ]− [2(x + 1) (x = 2

2

+ 2x + 2

)]

][

3

+ 6 x 2 + 6 x + 2 − 2 x 3 + 6 x 2 + 8x + 4

]

(x + 1)4

−2 x − 2

( x + 1)4

+ 0 ⋅ (x + 1)4 − 4(x + 1)4−1 ⋅ (− 2 x − 2 )

(x + 1)8

)]

=

(x + 1)4

(x + 1)4

Hamilton Education Guides

 ⋅ ( −2bx )  

[ (2 x + 3)⋅ (x + 1) ] − [1⋅ (x 2 + 3x + 1) ] = (x + 1)2

+ 2 x1−1 + 0 ⋅ (x + 1)2 − 2(x + 1)2−1 ⋅ x 2 + 2 x + 2

)

2 −4 −1

2 2

+ 3 x1−1 + 0 ⋅ (x + 1) − x1−1 + 0 ⋅ x 2 + 3 x + 1

1−1

(



= 4b a − bx 2

then

+ 2 x + 3x + 3 − x 2 + 3x + 1

[ (− 2x y ′′′ =

−3

) + (−96b x )(a − bx )

(x + 1)2

2 −1

)

−4 −5   2 2 + 192b 3 x 3 a − bx 2  + 48 b x a − bx    

[ (2x + 2) (x + 1) ]− [2(x + 1) (x = =

(

⋅ ( −2bx ) = 4 bx a − bx 2

−4   2 2 2  +  24 b x a − bx    

= −12b a − bx 2 

⋅ ( 0 − 2bx ) = −2 a − bx 2

−3 

(



(

−2 −1

y′

, then

]

=

[− 2 (x + 1) ]+ [ (x + 1) (8 x + 8) ] 4

3

( x + 1)8 128

Calculus I

2.8 Higher Order Derivatives

1 5

1 4

d. Given y = x 5 + x 4 + x , then =

y′ y ′′′

1 1 ⋅ 5x 5−1 + ⋅ 4 x 4 −1 + x 1−1 = x 4 + x 3 + 1 4 5

ax 2 + b , then c

=

[ (a ⋅ 2x

=

=

) ][ (

+ 0 ⋅ c − 0 ⋅ ax 2 + b

x2 +1

[ (2x

x3

2 −1

)]

) ][

)

− x 2/ x 2 + 3

= −

x 6/ = 4

= −

2 −1

)]

(

=

) ][ =

x8

)]

(

y ′′

(

x 3/ 2 x 2 + 12 x 8/ =5

) ][

−6 x 6 − 60 x 4

)=

(

x10

=

[2x ⋅ x ]− [3x ⋅ (x + 1) ] 3

2

2

=

2a c

and y ′′′ = 0

x6

2 x 4 − 3x 4 − 3x 2

=

x6

[2x ⋅ x ]− [4 x ⋅ (x + 3) ] = − 4

3

2

x8

+ 0 ⋅ x 5 − 5 x 5−1 ⋅ 2 x 2 + 12

x 10

2ax c

=

− x 4 − 3x 2 x6

x4

x8

2 −1

c

=

2/ =1

x2 + 3

+ 0 ⋅ x 4 − 4 x 4−1 ⋅ x 2 + 3

−2 x 5 − 12 x 3

[ (4x y ′′′ =

c

2acx /

=

2

, then

+ 0 ⋅ x 3 − 3 x 3−1 ⋅ x 2 + 1

(

( 2ax ⋅ c) − 0

=

2

x6

[ (2x y ′′ = −

=

2 −1

c

f. Given y = y′

= 4 x 4 −1 + 3x 3−1 + 0 = 4 x 3 + 3 x 2 and

= ( 4 ⋅ 3) x 3−1 + ( 3 ⋅ 2) x 2 −1 = 12x 2 + 6 x

e. Given y = y′

y ′′

)]

(

x

/ / =6 10

2 x 5 − 4 x 5 − 12 x 3 x8

2 x 2 + 12 x5

[ 4x ⋅ x ]− [5x ⋅ (2x = 5

4

2

+ 12

x10

) = − 6x

x 4/ −6 x 2 − 60

= −

2

+ 60

x

6

)]

=

4 x 6 − 10 x 6 − 60 x 4 x 10

Example 2.8-3: Find f ′′(0) and f ′′(1) for the following functions. a. f ( x ) = 6x 7 + 7 x 2 − 2 d. f ( x ) =

x3 +1 x

b. f (x ) = x 5 (x − 1)2 e. f ( x ) = x 3 −

g. f ( x ) = ( x − 1) −2

h. f ( x ) =

j. f ( x ) = (1 + 5x )3

k. f (x ) =

Hamilton Education Guides

1 x +1

( x + 1) 2 x

1+ x x

3

c. f ( x ) = x −

1 x

f. f ( x ) = (ax + b) 2

(

)

i. f ( x ) = ( x + 1) x 2 + 1 + 5 1 3

1 2

l. f ( x ) = x 3 + x 2 + x + 10

129

Calculus I

2.8 Higher Order Derivatives

Solutions: a. Given f ( x ) = 6 x 7 + 7 x 2 − 2 , then f ′( x ) = 6 ⋅ 7 x 7−1 + 7 ⋅ 2 x 2−1 − 0 = 42 x 6 + 14 x and f ′′( x ) = 42 ⋅ 6 x 6−1 + 14 ⋅ 1x 1−1 = 252 x 5 + 14

Therefore, f ′′(0) = 252 ⋅ 05 + 14 = 0 + 14 = 14 and f ′′(1) = 252 ⋅15 + 14 = 252 + 14 = 266 b. Given f ( x ) = x 5 ( x − 1) 2 , then

(

)

[

2

]

2

2

f ′( x ) = 5x 5−1 ⋅ 1 ⋅ ( x − 1) + 2( x − 1) ⋅ 1 ⋅ x 5 = 5x 4 ( x − 1) + 2 x 5 ( x − 1) = 5x 4 ( x − 1) + 2 x 6 − 2 x 5 and

[

]

2

2

f ′′( x ) = 5 ⋅ 4 x 4−1 ( x − 1) + 2( x − 1) ⋅ 5x 4 + 2 ⋅ 6 x 6−1 − 2 ⋅ 5x 5−1 = 20 x 3 ( x − 1) + 10 x 4 ( x − 1) + 12 x 5 − 10 x 4

(

)

= 20 x 3 x 2 − 2 x + 1 + 10 x 5 − 10 x 4 + 12 x 5 − 10 x 4 = 20 x 5 − 40 x 4 + 20 x 3 + 22 x 5 − 20 x 4 = 42 x 5 − 60 x 4 + 20 x 3 Therefore, f ′′(0)

= 42 ⋅ 0 5 − 60 ⋅ 0 4 + 20 ⋅ 0 3 = 0 and f ′′(1) = 42 ⋅15 − 60 ⋅14 + 20 ⋅13 = 42 − 60 + 20 = 2 1 x

c. Given f ( x ) = x − , then f ′( x ) = 1 −

0 ⋅ x − 1⋅1 x

= 1+

2

2

Therefore, f ′′(0) = −

0

1 x

[3x ⋅ x]− [1⋅ (x + 1) ] f ′( x ) = 2

3

=

x2

f ′′( x )

2

2 0

= −

3

0 ⋅ x − 2x ⋅1 x

4

= −

2x x

4

2

= −

which is not defined and f ′′(1) = −

x3

2

2 = −2 1

= −

3

1

x3 +1 , then x

d. Given f ( x ) =

[6x =

and f ′′( x ) = 0 +

2

][ (

3x 3 − x 3 − 1 x2

)]

⋅ x 2 − 2x ⋅ 2x 3 −1 x4

(

e. Given f ( x ) = x 3 −

)=

2 03 + 1

Therefore, f ′′(0) =

f ′( x ) = 3x 3−1 −

=

0

3

2 0

=

2x 3 − 1 x2

6x 4 − 4x 4 + 2x x4

=

and 2x 4 + 2x x4

=

(

) = 2(x + 1)

2 x/ x 3 + 1

3

x 4/ =3

x3

which is not defined and f ′′(1) =

(

)=

2 13 + 1 3

1

4 1

= 4

1 , then x +1

0 ⋅ ( x + 1) − 1 ⋅ 1

( x + 1)

Hamilton Education Guides

2

= 3x 2 +

1

( x + 1) 2

and

130

Calculus I

2.8 Higher Order Derivatives 2

0 ⋅ ( x + 1) − 2( x + 1)

f ′′( x ) = 3 ⋅ 2 x 2−1 +

f ′′(0) = 6 ⋅ 0 −

( x + 1)

2

(0 + 1)

= −

3

2−1

⋅1

= 6x −

4

2( x + 1)

( x + 1)

4

= 6x −

2

( x + 1)3

Therefore,

2 1 2 2 2 = −2 and f ′′(1) = 6 ⋅ 1 − = 6 − 3 = 6 − = 6 − = 5.75 3 1 4 8 2 (1 + 1)

f. Given f ( x ) = (ax + b) 2 , then f ′( x ) = 2(ax + b)

2−1

⋅ a = 2a(ax + b) = 2a 2 x + 2ab and f ′′( x ) = 2a 2 + 0 = 2a 2

Therefore, f ′′(0) = 2a 2 and f ′′(1) = 2a 2 Note that since f ′′( x ) is independent of x , f ′′( x ) is equal to 2a 2 for all values of x . g. Given f ( x ) = ( x − 1) −2 , then f ′( x ) = −2( x − 1) f ′′(0) = 6(0 − 1)

h. Given f ( x ) = f ′( x ) =

f ′′( x ) =

−2 −1

−4

⋅ 1 = −2( x − 1)

=

6

( −1)

=

4

−3

[

]

and f ′′( x ) = −2 ⋅ −3( x − 1) −3−1 ⋅ 1 = 6( x − 1) −4 Therefore,

6 6 6 = 6 and f ′′(1) = 6(1 − 1) −4 = 4 = which is undefined 0 1 0

( x + 1) 2 , then

[2 (x + 1)

x

2 −1

][

⋅1 ⋅ x − 1 ⋅ (x + 1)2 x2

[ 2x ⋅ x ]− [2x ⋅ (x − 1) ] 2

2

x

4

2

Therefore, f ′′(0) =

0

(

3

2 0

=

=

]

=

2 x ( x + 1) − ( x + 1) x

2x 3 − 2x 3 + 2x x

4

2

=

2

=

2x x

4

=

2x 2 + 2x − x 2 − 2x − 1 x

2

=

x2 −1 x2

and

2

x3

which is undefined and f ′′(1) =

2

13

=

2 = 2 1

)

i. Given f ( x ) = ( x + 1) x 2 + 1 + 5 = x 3 + x + x 2 + 1 + 5 = x 3 + x 2 + x + 6 , then f ′( x ) = 3x 2 + 2 x + 1 and f ′′( x ) = 6 x + 2

Therefore, f ′′(0) = 6 ⋅ 0 + 2 = 2 and f ′′(1) = 6 ⋅ 1 + 2 = 8 j. Given f ( x ) = (1 + 5x )3 , then f ′( x )

= 3(1 + 5x )3−1 ⋅ 5 = 15(1 + 5x ) 2 and f ′′( x ) = 0 ⋅ (1 + 5x ) 2 + 2(1 + 5x ) ⋅ 5 ⋅ 15 = 150(1 + 5x )

Therefore, f ′′(0) = 150(1 + 5 ⋅ 0) = 150 and f ′′(1) = 150(1 + 5 ⋅ 1) = 150(1 + 5) = 150 ⋅ 6 = 900

Hamilton Education Guides

131

Calculus I

2.8 Higher Order Derivatives

k. Given f ( x ) =

x3

, then

1 ⋅ x 3 − 3x 2 (1 + x )

f ′( x ) =

f ′′(x )

1+ x

x

=

6

x 3 − 3x 2 − 3x 3 x

[ − 2 ⋅ x ]− [4x (− 2x − 3) ] = 4

3

x

l. Given f ( x ) =

6 ⋅ 0 + 12 0

5

=

12 0

−2 x 3 − 3x 2 x

6

− 2 x 4 + 8 x 4 + 12 x 3 x

8

=

=

− x 2/ (2 x + 3) x

6/ = 4

6 x 4 + 12 x 3 x

8

=

which is not defined and f ′′(1) =

=

−2 x − 3 x4

6 x 3/ (x + 2 ) x

6 ⋅1 + 12 5

1

=

6 x + 12

18 1

= 18

8/ = 5

=

x5

1 3 1 2 x + x + x + 10 , then 3 2

3 3−1 2 2−1 + x + x 1−1 + 0 x 2 3

f ′( x ) =

=

8

Therefore, f ′′(0) =

6

=

= x 2 + x + 1 and f ′′( x ) = 2 x 2 −1 + x1−1 + 0 = 2 x + 1

Therefore, f ′′(0) = 2 ⋅ 0 + 1 = 1 and f ′′(1) = 2 ⋅ 1 + 1 = 3 dy d 2 y d3y , 2 , and 3 for the following functions. dx dx dx 1 a. y = x 4 + 5x 3 + 6 x 2 + 1 b. y = x + x

Example 2.8-4: Find

(

)

d. y = x 2 + 1 g. y = x −

−2

1 x

c. y = x( x + 1)3

e. y = x 3 + 3x 2 + 10

f. y =

h. y = ax 3 + bx

i. y =

1 1+ x x3 +1 x2

Solutions: a. Given y = x 4 + 5x 3 + 6 x 2 + 1 , then d 4 d d d dy x + 5 x 3 + 6 x 2 + 1 = 4 x 4−1 + (5 ⋅ 3) x 3−1 + (6 ⋅ 2) x 2−1 + 0 = 4 x 3 + 15 x 2 + 12 x = dx dx dx dx dx d2y dx

2

d3y dx

3

= 4

d d 3 d x + 15 x 2 + 12 x = (4 ⋅ 3) x 3 −1 + (15 ⋅ 2) x 2 −1 + (12 ⋅ 1) = 12 x 2 + 30 x + 12 dx dx dx

= 12

d 2 d d x + 30 x + 12 = (12 ⋅ 2) x 2 −1 + (30 ⋅ 1) x1−1 + 0 = 24 x 1 + 30 x 0 = 24 x + 30 dx dx dx

b. Given y = x +

1 x

which is the same as y = x + x −1 , then

(

)

dy d d −1 x+ x = 1 + −1 ⋅ x −1−1 = 1 − x −2 = dx dx dx d2y dx

2

=

(

)

d d −2 1− x = 0 − −2 ⋅ x −2−1 = 2 x −3 dx dx

Hamilton Education Guides

132

Calculus I

d3y dx

3

2.8 Higher Order Derivatives

= 2

d −3 x = (2 ⋅ −3) x −3 −1 = −6 x −4 dx

c. Given y = x( x + 1)3 , then dy d   d   = ( x + 1)3 x  +  x ( x + 1)3  = dx dx   dx   d2y dx

2

=

[( x + 1) ⋅1] + [ x ⋅ 3( x + 1) ⋅1] = ( x + 1) 3

3−1

[

3

+ 3 x( x + 1)

][

2

][

]

d d d ( x + 1)3 + 3( x + 1) 2 x  + 3x ( x + 1) 2  = 3( x + 1)3−1 ⋅ 1 + 3( x + 1) 2 ⋅ 1 + 3x ⋅ 2 ( x + 1) 2−1 ⋅ 1 dx dx dx    

2 = 3( x + 1) 2 + 3( x + 1) 2 + 6 x ( x + 1) = 6( x + 1) + 6 x ( x + 1)

d3y dx

3

= 6

d (x + 1)2 +  6(x + 1) d dx dx 

  d  x  + 6 x (x + 1)  dx   

=

[12(x + 1)

2 −1

]{

[

⋅1 + [6(x + 1) ⋅1] + 6 x ⋅(x + 1)1−1 ⋅1

]}

= 12(x + 1) + 6(x + 1) + 6 x(x + 1)0 = 18(x + 1) + 6 x = 18 x + 18 + 6 x = 24 x + 18 = 6(4 x + 3) d. Given y = ( x + 1) −2 , then

[

]

dy d d = ( x + 1) −2 = −2( x + 1) −2 −1 ⋅ ( x + 1) = −2( x + 1) −3 ⋅1 = −2 ( x + 1) −3 dx dx dx d2y dx

2

d3y dx

3

= −2 = 6

d ( x + 1) −3 dx

[

] dxd ( x + 1) = 6 ( x + 1)

−4

⋅ 1 = 6 ( x + 1)

[(6⋅ −4) ( x + 1) ] ⋅ dxd ( x + 1) = −24 ( x + 1)

−5

⋅ 1 = −24 ( x + 1)

= ( −2 ⋅ −3) ( x + 1) −3−1 ⋅

d ( x + 1) −4 = dx

−4 −1

−4

−5

e. Given y = x 3 + 3x 2 + 10 , then d dy d 3 d 3 d d x + x + 3 x 2 + 0 = 3x 3−1 + (3 ⋅ 2) x 2−1 = 3 x 2 + 6 x 3x 2 + 10 = = dx dx dx dx dx dx d2y dx

2

d3y dx

3

=

d d d d 3x 2 + 6 x = 3 x 2 + 6 x = (3 ⋅ 2) x 2 −1 + (6 ⋅ 1) = 6 x + 6 dx dx dx dx

=

d d d 6x + 6 = 6 x + 0 = 6 ⋅1 = 6 dx dx dx

f. Given y =

1 1+ x

, then

d   d   (1 + x ) dx 1 − 1 ⋅ dx (1 + x ) 0 − (1 ⋅ 1) 1 dy     = = = − 2 2 dx (1 + x ) (1 + x ) (1 + x ) 2

Hamilton Education Guides

133

Calculus I

2

d y dx

=

2

d3y dx

2.8 Higher Order Derivatives

=

3

 2 2 d   d (1 + x ) dx 1 − 1 ⋅ dx (1 + x )  −

(1 + x )

= −

4

d  3 d   3 (1 + x ) dx 2 − 2 ⋅ dx (1 + x )     

(1 + x )

(1 + x )

[

(1 + x )

]

4

0 − (2 ⋅ 3) (1 + x )3−1

=

6

[

0 − (1 ⋅ 2 ) (1 + x )2−1

]

=

= −

6

2(1/ + x/ )

(1 + x)

4/ = 3

6 (1 + x )

(1 + x)

2/

6/ = 4

=

2

(1 + x ) 3

= −

6

(1 + x ) 4

1 which is the same as y = x − x −1 , then x

g. Given y = x −

dy d d −1 x− x = 1 + x −1−1 = 1 + x −2 = dx dy dx d2y dx

=

2

d3y dx

d d −2 x 1+ dy dx

= −2

3

d −3 x dy

= 0 − 2 x −2−1 = −2 x −3

= ( −2 ⋅ −3) x −3−1 = 6 x −4

h. Given y = ax 3 + bx , then dy d d d 3 d bx = a ax 3 + x + b x = (a ⋅ 3) x 3−1 + (b ⋅ 1) x 1−1 = 3ax 2 + bx 0 = 3ax 2 + b = dx dx dx dx dx d2y dx

2

d3y dx

3

=

d d 3ax 2 + b dx dx

=

d d 6ax = 6a x = 6a ⋅ 1 = 6a dx dx

i. Given y =

2

d y dx

2

x2

(

d 2 x +0 dx

= (3a ⋅ 2) x 2−1 = 6ax

, then

)

(

)

d 2  2 d 3   3  x dx x + 1  −  x + 1 dx x     

dy = dx

=

x3 +1

= 3a

x 3x 4 − 2 x 4 − 2 x x4

(

= =

=

4

x 4 − 2x

=

=

x4

)

(

(

x/ x 3 − 2 x 4/ =3

)

d 3  3 d 3   3  x dx x − 2  −  x − 2 dx x      x 3x 5 − 3x 5 + 6 x 2 x

6

Hamilton Education Guides

6

=

6 x 2/ x

(

6/ = 4

=

)

d  d x 2  x 3 + 1 − x 3 + 1 ⋅ 2 x dx dx   x4

=

)

=

=

[

] (

)

x 2 3x 3−1 + 0 − 2 x x 3 + 1 x

4

x3 − 2 x3

(

)

d  d x3  x3 − 2  − x 3 − 2 ⋅ 3x 2 dx dx   x6

=

[

]

(

x 3 3 x 3−1 − 0 − 3 x 2 x 3 − 2 x

6

6

x4

134

)

Calculus I

d3y dx

3

2.8 Higher Order Derivatives

x4

=

( x ⋅ 0) − (6 ⋅ 4)x =

d 4 d 6−6 x dx dx x8

4

x

4−1

=

8

0 − 24 x 3 x

= −

8

24 x 3/ x

= −

8/ =5

24

x5

Example 2.8-5: Find y ′ and y ′′ for the following functions. Do not simplify the answer to its lowest term. b. xy + y 2 = 1

a. x 2 + y 2 = 2

d. x 3 y + y = 1

c. 1 + x 2 y 2 = x

Solutions: −2 x 2y

a. Given x 2 + y 2 = 2 , then y ′ is equal to 2 x 2−1 + 2 y ⋅ y ′ = 0 ; 2 x + 2 y y ′ = 0 ; 2 y y ′ = −2 x ; y ′ = ; y′ = −

(1⋅ y ) − ( y ′ ⋅ x ) y ′′ = − y − y ′ x x and y ′′ = − ; y y2 y2

b. Given xy + y 2 = 1 , then y ′ is equal to (1 ⋅ y + y ′ ⋅ x ) + 2 y ⋅ y ′ = 0 ; y + y ′ x + 2 y y ′ = 0 ; y ′ ( x + 2 y ) = − y

; y′ = −

[y ′ ⋅ (x + 2 y ) ] − [ (1 + 2 y ′)⋅ y ] y and y ′′ = − x + 2y ( x + 2 y )2

; y ′′ = −

(

c. Given 1 + x 2 y 2 = x , then y ′ is equal to 0 + 2 x ⋅ y 2 + 2 y y ′ ⋅ x 2

2

2

; 2 y y ′ x = 1 − 2x y ; y′ =

; y ′′ =

(

1 − 2 x y2

and

2x2 y

) (

)(

−4 x 2 y y 2 + 2 xyy ′ − 2 2 xy + x 2 y ′ 1 − 2 xy 2

)

( x + 2 y) 2

; 2x y 2 + 2 y y ′ x 2 = 1

=1

+ 2 yy ′ ⋅ x

xy ′ − y

) ] (2 x y ) }− {2(2 xy + x y ′)(1 − 2 xy )} (2 x y ) 2

2

2

2

2

)

4x y

(

= −

=

2

( x + 2 y)

; y ′′ = −

2

4 2

)

d. Given x 3 y + y = 1 ; y x 3 + 1 = 1 ; y =

y ′′

{[0 − 2(1⋅ y y ′′ =

xy ′ + 2 y y ′ − y − 2 y y ′

1 x3 +1

) [( ) (x + 1)

(

2  2 2 3 3 − 6 x ⋅ x + 1  − 2 x + 1 ⋅ 3 x ⋅ −3 x 4

3

(

6 x − x 3 − 1 + 3x 3

(x + 1) 3

3

Hamilton Education Guides

)

=

(

, then y ′ =

]

=

(

[0 ⋅ (x + 1) ]− [3x ⋅1 ] (x + 1) 3

2

2

3

)

2

(

3

4

− 3x 2

and

(x + 1)

2

3

) = 6 x(x + 1)[ − (x + 1)+ 3x ] (x + 1)

− 6 x x 3 + 1 + 18 x 4 x 3 + 1

(x + 1)

=

3

3

3

3

4

)

6x 2x 3 −1

(x + 1) 3

3

135

Calculus I

2.8 Higher Order Derivatives

Example 2.8-6: Find the first, second, and third derivative of the following functions. a. f (x ) = 6 x 7 + 7 x 2 − 2

b. f ( x ) = 3x 4 − 2 x 2 + 5x + 9

c. f ( x ) = x −4 + 3x −3 + x −2 + x

d. f ( x ) = x 7 + 6 x 5 + 8 x + 3x −3

e. f ( x ) =

3 x

3

+

g. f ( x ) = x 3 +

2 x

2

+

1 x

1

x3

f. f ( x ) = x (2 x + 1)3 h. f ( x ) = ( x + 1) 2 − x 3

Solutions: a. Given f ( x ) = 6 x 7 + 7 x 2 − 2 , then f ′( x ) = (6 ⋅ 7) x 7−1 + (7 ⋅ 2) x 2−1 − 0 = 42 x 6 + 14 x f ′′( x ) = (42 ⋅ 6) x 6−1 + 14 x 1−1 = 252 x 5 + 14 f ′′′( x ) = (252 ⋅ 5) x 5−1 + 0 = 1260 x 4

b. Given f ( x ) = 3x 4 − 2 x 2 + 5x + 9 , then f ′( x )

= (3 ⋅ 4) x 4−1 − (2 ⋅ 2) x 2−1 + 5x1−1 + 0 = 12 x 3 − 4 x + 5

f ′′( x ) = (12 ⋅ 3) x 3−1 − 4 x 1−1 = 36 x 2 − 4 f ′′′( x ) = (36 ⋅ 2) x 2−1 − 0 = 72x

c. Given f ( x ) = x −4 + 3x −3 + x −2 + x , then f ′( x )

= −4 x −4−1 + (3 ⋅ −3) x −3−1 − 2 x −2−1 + x1−1 = −4 x −5 − 9 x −4 − 2 x −3 + 1

f ′′( x ) = ( −4 ⋅ −5) x −5−1 + ( −9 ⋅ −4) x −4−1 + ( −2 ⋅ −3) x −3−1 + 0 = 20 x −6 + 36 x −5 + 6 x −4

f ′′′( x ) = (20 ⋅ −6) x −6−1 + (36 ⋅ −5) x −5−1 + (6 ⋅ −4) x −4−1 = −120 x −7 − 180 x −6 − 24 x −5

d. Given f ( x ) = x 7 + 6 x 5 + 8 x + 3x −3 , then f ′( x ) = 7 x 7−1 + (6 ⋅ 5) x 5−1 + 8 x 1−1 + (3 ⋅ −3) x −3−1 = 7 x 6 + 30 x 4 + 8 − 9 x −4 f ′′( x ) = (7 ⋅ 6) x 6−1 + (30 ⋅ 4) x 4−1 + 0 + ( −9 ⋅ −4) x −4−1 = 42 x 5 + 120 x 3 + 36 x −5 f ′′′( x )

= (42 ⋅ 5) x 5−1 + (120 ⋅ 3) x 3−1 + (36 ⋅ −5) x −5−1 = 210 x 4 + 360 x 2 − 180 x −6

Hamilton Education Guides

136

Calculus I

2.8 Higher Order Derivatives

e. Given f ( x ) =

3

2 1 + 2 + x x x 3

which is equal to f ( x ) = 3x −3 + 2 x −2 + x −1 , then

f ′( x ) = (3 ⋅ −3) x −3−1 + (2 ⋅ −2) x −2−1 − x −1−1 = −9 x −4 − 4 x −3 − x −2 f ′′( x ) = ( −9 ⋅ −4) x −4−1 + ( −4 ⋅ −3) x −3−1 + ( −1 ⋅ −2) x −2−1 = 36 x −5 + 12 x −4 + 2 x −3 f ′′′( x ) = (36 ⋅ −5) x −5−1 + (12 ⋅ −4) x −4−1 + (2 ⋅ −3) x −3−1 = −180 x −6 − 48 x −5 − 6 x −4

f. Given f ( x ) = x (2 x + 1)3 , then f ′( x )

]

][

[

3 2 = 1 ⋅ (2 x + 1)3 + 3(2 x + 1)3−1 ⋅ 2 ⋅ x = ( 2 x + 1) + 6 x ( 2 x + 1)

[

f ′′( x ) = 3(2 x + 1)

3−1

][

2

⋅ 2 + 6 ⋅ (2 x + 1) + 2 (2 x + 1)

2−1

⋅ 2 ⋅ 6x

] = 6(2x + 1)

2

2

+ 6(2 x + 1) + 24 x (2 x + 1)

2 = 12( 2 x + 1) + 24 x ( 2 x + 1)

f ′′′( x )

]

][

[

= (12 ⋅ 2)(2 x + 1) 2−1 ⋅ 2 + 24 ⋅ (2 x + 1) + (2 x + 1)1−1 ⋅ 2 ⋅ 24 x = 48 (2 x + 1) + [24 (2 x + 1) + 48 x ] = 72 ( 2 x + 1) + 48 x

g. Given f ( x ) = x 3 +

1

x3

which is equal to f ( x ) = x 3 + x −3 , then

f ′( x ) = 3x 3 −1 − 3x −3 −1 = 3 x 2 − 3 x −4 f ′′( x ) = (3 ⋅ 2) x 2−1 + ( −3 ⋅ −4) x −4−1 = 6 x + 12 x −5 f ′′′( x ) = 6 x 1−1 + (12 ⋅ −5) x −5−1 = 6 x 0 − 60 x −6 = 6 − 60 x −6

h. Given f ( x ) = ( x + 1) 2 − x 3 , then

[

f ′( x ) = 2( x + 1)

2−1

]

⋅ 1 − 3x 3−1

= 2( x + 1) − 3x 2 = 2 x + 2 − 3x 2 = −3 x 2 + 2 x + 2

f ′′( x ) = ( −3 ⋅ 2) x 2−1 + 2 x 1−1 + 0 = −6 x + 2 x 0 = −6 x + 2 f ′′′( x )

= −6 x1−1 + 0 = −6 x 0 = −6 Section 2.8 Practice Problems - Higher Order Derivatives

1. Find the second derivative of the following functions. a. y = x 3 + 3x 2 + 5 x − 1 Hamilton Education Guides

b. y = x 2 (x + 1)2

c. y = 3x 3 + 50 x 137

Calculus I

2.8 Higher Order Derivatives

d. y = x 5 + g. y = x 4 +

1 x

2

x 8 − 7 x 5 + 5x 10

e. y =

(

x3 − 5x 2 x +1

)

f. y = x 3 x 2 − 1 1

h. y = x 2 −

1 x +1

i. y =

b. y = x 2 +

1 x

c. y = 4 x 3 (x − 1)2

− 3x

x2

2. Find y ′′′ for the following functions. a. y = x 5 + 6 x 3 + 10 d. y =

x x +1

e. y = x 8 − 10 x 5 + 5 x − 10

f. y =

x −1 x2

+ 5x 3

3. Find f ′′(0) and f ′′(1) for the following functions. a. f (x ) = 6 x 5 + 3x 3 + 5

b. f (x ) = x 3 (x + 1)2

d. f (x ) = (x − 1)−3

e. f (x ) = (x − 1) x 2 + 1

Hamilton Education Guides

(

c. f (x ) = x + (x − 1)2

)

(

)

2

f. f (x ) = x 3 − 1 + 2 x

138

Calculus I

Quick Reference to Chapter 3 Problems

Chapter 3

Differentiation (Part II) Quick Reference to Chapter 3 Problems 3.1

Differentiation of Trigonometric Functions ............................................................ 140 d  tan 2 x  dx  cot x 

3.2

[

)]

(

(

)

(

=;

=;

)

d  3 x 2 dx  arc tan

  x 

=;

d  arc sin x  + 5x 3   x dx  

=

(

d cos 5 x −3 x e e dx

)

[

d ln x 3 + ln (csc x ) dx

=;

]

=

=;

(

d sinh x 2 + cosh x 2 dx

)

=;

d 1   sinh 3 x + 10 x  dx  6 

=

Differentiation of Inverse Hyperbolic Functions .................................................... 187

(

d sinh −1 x 2 + ln x 3 dx

3.6

=

Differentiation of Hyperbolic Functions .................................................................. 181 d sinh x 3 + cosh 3 x dx

3.5

d  sin 4 x    dx  x + 3 

Differentiation of Logarithmic and Exponential Functions .................................. 166 d e ln x tan x dx

3.4

=;

Differentiation of Inverse Trigonometric Functions .............................................. 158 d tan −1 x 2 + 3 dx

3.3

d  sin 3 x    dx  cos 2 x 

=;

)

=;

(

d tanh −1 2 x + ln e x dx

)

=;

(

d tanh −1 x 3 + 5 x dx

)

=

Evaluation of Indeterminate Forms Using L’Hopital’s Rule ................................ 193 lim x → 10

Hamilton Education Guides

e x − e10 x − 10

= ; lim x → 5

x 3 − 25 x x 3 − 125

= ; lim x →+ ∞

2 x 2 − ln x 3 x 2 + 3 ln x

=

139

Calculus I

3.1 Differentiation of Trigonometric Functions

Chapter 3 – Differentiation (Part II) The objective of this chapter is to improve the student’s ability to solve additional problems involving derivative of functions that was not addressed in the previous chapter. The derivative of trigonometric functions is addressed in Section 3.1. Finding the derivative of inverse trigonometric functions is discussed in Section 3.2. How to obtain the derivative of exponential and logarithmic expressions is addressed in Section 3.3. Differentiation of hyperbolic and inverse hyperbolic functions is addressed in Sections 3.4 and 3.5, respectively. Finally, evaluation of indeterminate forms of the type 00 , ∞∞ , 0 ⋅ ∞ , ∞ − ∞ , 1∞ , ∞ 0 , 0 0 using L’Hopital’s rule is discussed in Section 3.6. Each section is concluded by solving examples with practice problems to further enhance the student’s ability.

3.1

Differentiation of Trigonometric Functions

The differential formulas involving trigonometric functions are defined as: Table 3.1: Differentiation Formulas for Trigonometric Functions d du sin u = cos u ⋅ dx dx du d cos u = − sin u ⋅ dx dx du d tan u = sec 2 u ⋅ dx dx

d du cot u = − csc 2 u ⋅ dx dx du d sec u = sec u tan u ⋅ dx dx du d csc u = − csc u cot u ⋅ dx dx

In the following examples we will solve problems using the above differentiation formulas: Example 3.1-1: Find the derivative of the following trigonometric functions: b. y = sin 3x 4

c. y = cos x 2 + tan x 3

e. y = sin x 3 + cos x

f. y = cos 3 x

g. y = sec 5 x

h. y = sin 6 3 x 2

i. y = x 3 sin 2 3x

j. y = (x + 1) cot x 2

k. y = sin 5 x

l. y = csc 5 x

a. y = sin 8 x − cos 3x

(

d. y = tan 1 + 3x 2

)

Solutions: a. Given y = sin 8 x − cos 3x then b. Given y = sin 3x 4 then

dy dx

=

c. Given y = cos x 2 + tan x 3 then

Hamilton Education Guides

dy d (sin 8 x − cos 3x ) = cos 8 x ⋅ d 8 x + sin 3x ⋅ d 3x = 8 cos 8 x + 3 sin 3 x = dx dx dx dx

(

d sin 3 x 4 dx dy dx

=

(

) = cos 3x

4

⋅

d 3x 4 dx

d cos x 2 + tan x 3 dx

)

= cos 3x 4 ⋅12 x 4−1 = 12 x 3 cos 3 x 4

= − sin x 2 ⋅

d 2 d 3 x + sec 2 x 3 ⋅ x dx dx

140

Calculus I

3.1 Differentiation of Trigonometric Functions

= − sin x 2 ⋅ 2 x + sec 2 x 3 ⋅ 3 x 2 = − 2 x sin x 2 + 3 x 2 sec2 x 3

(

d. Given y = tan 1 + 3x 2

(

)

then

dy dx

(

d tan 1 + 3 x 2 dx

=

(

)

= sec 2 1 + 3x 2 ⋅ 6 x = 6 x sec 2 1 + 3 x 2 dy dx

e. Given y = sin x 3 + cos x then

=

(

) dxd (1 + 3x )

= sec 2 1 + 3x 2 ⋅

2

(

)

= sec 2 1 + 3x 2 ⋅ ( 0 + 6 x )

)

=

d   sin dx 

x 3 + cos x  

3 1 d   sin x 2 + cos x 2 dx 

=

   

3 3 3 1 1 1 1 1 −1 3 3 −1 d 32 d 12 d d x − sin x 2 ⋅ x = cos x 2 ⋅ x 2 − sin x 2 ⋅ x 2 sin x 2 + cos x 2 = cos x 2 ⋅ 2 2 dx dx dx dx

3

1

3 2

1

1 − 12 2

= cos x 2 ⋅ x 2 − sin x 2 ⋅ x

dy dx

f. Given y = cos 3 x then = 3 ( cos x ) ⋅ − sin x ⋅ 2

2

1 2

g. Given y = sec 5 x then

5 2 x

sec 4

h. Given y = sin 2

x

2

2

= sin

2 3

6

2 −1

= 6 sin 5 x 3 ⋅ cos x 3 ⋅ x 3

i. Given y = x 3 sin 2 3x then Hamilton Education Guides

=

(

)

2

(

=

d sec 5 dx

( x)

2 x3

=

(

(

= −

3 2 x

1 2

5

sec5

dy dx 2

=

) 2 ⋅ − sin

x⋅

cos 2

1 −1

=

2 x

d cos x dx

=

(

1 12 −1 x 2

x sin x

(

d sec x dx

5 sec 4 2

)

x ⋅ sec x tan x ⋅ x

−1

2

x tan x

2 d   sin x 3 dx 

2

   

6

=

2  6  sin x 3 

5

2  d  ⋅ sin x 3  dx 

2 2 2 − 13 12 = 3 sin 5 x 3 ⋅ cos x 3 = 3 3 x

= 6 sin 5 x 3 ⋅ cos x 3 ⋅ x dy dx

sin x 2

)

= 5 ( sec x ) 4 ⋅

)

d sec x 5 dx

1

1

(

2

= 5 sec 4 x ⋅ sec x tan x ⋅ x 2

2 x

then

=

x cos x 2 −

d 12 x = 3 cos x dx

)

x

3

3 2

=

= 3 ( cos x ) ⋅

3 d cos x dx

2 3 − 12 cos x ⋅ sin x 2

dy dx d dx

x

= 3 ( cos x ) ⋅ − sin x ⋅

= − x

2

x ⋅ sec x tan x

63

d cos 3 dx

=

−1

= 5 sec 4 x ⋅ sec x tan x ⋅

3 1 1 −1 3 12 x cos x 2 − x 2 sin x 2 2 2

=

( x)

d dx

= 3 ( cos x ) ⋅ − sin x ⋅ x

=

)

d x 3 sin 2 3 x dx

)

= sin 2 3x ⋅

2

2

= 6 sin 5 x 3 ⋅ cos x 3 ⋅ 4 3

x

2

d 23 x dx

2

sin 5 x 3 cos x 3

d 3 d x + x 3 ⋅ sin 2 3 x dx dx 141

Calculus I

3.1 Differentiation of Trigonometric Functions

= sin 2 3x ⋅ 3x 2 + x 3 ⋅

d (sin 3x )2 dx

= 3x 2 sin 2 3x + x 3 ⋅ 2 sin 3x ⋅

d sin 3 x dx

= 3x 2 sin 2 3x + x 3 ⋅ 2 sin 3x ⋅ cos 3x ⋅

d 3x dx

= 3x 2 sin 2 3x + x 3 ⋅ 2 sin 3x ⋅ cos 3x ⋅ 3 = 3 x 2 sin 2 3 x + 6 x 3 sin 3 x cos 3 x dy dx

j. Given y = (x + 1) cot x 2 then = cot x 2 ⋅1 + (x + 1) ⋅ − csc 2 x 2 ⋅

=

d 2 x dx

1 1 (sin 5 x )− 2 ⋅ cos 5 x ⋅ 5 2

=

2

1 d (sin 5 x ) 2 dx

5 cos 5 x ⋅ 2 (sin 5 x ) 12

=

5 cos 5 x ⋅ 2 sin 5 x

dy dx

=

1

1 1 (csc 5 x )− 2 ⋅ − csc 5 x cot 5 x ⋅ d 5 x 2 dx

= cot x 2

d (x + 1) + (x + 1) d cot x 2 dx dx

= cot x 2 − (x + 1) ⋅ csc 2 x 2 ⋅ 2 x = cot x 2 − 2 x ( x + 1) csc2 x 2 =

l. Given y = csc 5 x = (csc 5 x ) 2 then =

[ (x + 1) cot x ]

dy dx

1

k. Given y = sin 5 x = (sin 5 x ) 2 then =

d dx

=

1 d (csc 5 x ) 2 dx

=

1 1 (sin 5 x ) 2 −1 d sin 5 x dx 2

=

1 1 (csc 5 x ) 2 −1 d csc 5 x dx 2

1 1 (csc 5 x )− 2 ⋅ − csc 5 x cot 5 x ⋅ 5 2

=

1 1 (sin 5 x )− 2 ⋅ cos 5 x ⋅ d 5 x 2 dx

5 csc 5 x cot 5 x 2 (csc 5 x ) 12

= − ⋅

5 5 5 = − ⋅ csc 5 x ⋅ (csc 5 x ) − 2 ⋅ cot 5 x = − ⋅ (csc 5 x ) 1 − 2 ⋅ cot 5 x = − ⋅ (csc 5 x ) 2 ⋅ cot 5 x = − 1

1

2

1

2

2

5 csc 5 x cot 5 x 2

Example 3.1-2: Find the derivative of the following trigonometric functions: a. y =

csc x

b. y =

x2

d. y = sin x 5

(

)

(

csc x x2

then

dy dx

− x 2 csc x cot x − 2 x csc x x4

Hamilton Education Guides

=

=

)

h. y = tan 3 x 2 + 1

i. y = cos x 3

k. y = csc 10 x

l. y = csc x10

d  csc x    dx  x 2 

=

(x

2

x 4/ = 3

)(

d csc x − csc x ⋅ d x 2 ⋅ dx dx

− x/ csc x(x cot x + 2 )

)

f. y = tan x 2 + 1

(

3

j. y = cos 3 x

=

c. y = sin 5 x

e. y = sin 5 x

g. y = tan x 2 + 1

a. Given y =

sin x 2 x+3

)

x4

=

=

x 2 ⋅ − csc x cot x − csc x ⋅ 2 x x4

− csc x ( x cot x + 2) x3

142

Calculus I

3.1 Differentiation of Trigonometric Functions

sin x 2 y= x+3

b. Given

dy dx

then

d  sin x 2 dx  x + 3

=

   

=

[ (x + 3)⋅

][

d ( x + 3) sin x 2 − sin x 2 ⋅ dx

d dx

]

(x + 3)2

(x + 3)⋅ cos x 2 ⋅ dxd x 2 − sin x 2 ⋅1 (x + 3)⋅ cos x 2 ⋅ 2 x − sin x 2 = = = (x + 3)2 (x + 3)2

2 x ( x + 3) cos x 2 − sin x 2

( x + 3) 2

c. Given y = sin 5 x then

dy dx

=

d sin 5 x dx

= cos 5 x ⋅

d 5x dx

= cos 5 x ⋅ 5 = 5 cos 5 x

d. Given y = sin x 5 then

dy dx

=

d sin x 5 dx

= cos x 5 ⋅

d 5 x dx

= cos x 5 ⋅ 5 x 4 = 5 x 4 cos x 5

e. Given y = sin 5 x then

dy dx

=

d sin 5 x dx

=

d (sin x )5 dx

d sin x dx

= 5(sin x )4 ⋅

= 5 (sin x )4 ⋅ cos x ⋅

d x dx

= 5 (sin x )4 ⋅ cos x ⋅1 = 5 sin4 x cos x

(

)

(

)

dy dx

f. Given y = tan x 2 + 1 then then

dy dx

) (

)

(

(

) [ (

g. Given y = tan x 2 + 1

(

3

3

2

= sec 2 x 2 + 1 ⋅ 3 x 2 + 1 ⋅

(

)

)

(

)]

2

3

then

3

2

dy dx

2

)

3

= sec 2 x 2 + 1 ⋅

) = sec ( x + 1) ⋅ 3 ( x + 1)

d 2 x +1 dx

2

2

[ (

(

dy dx

=

2

2

d cos x 3 dx

j. Given y = cos 3 x = (cos x )3 then

dy dx

=

2

= − sin x 3 ⋅ d (cos x )3 dx

d 3 x dx

2

)

)

(

)

2

⋅ 2x

= 6x x 2 + 1

)]

= 3 tan x 2 + 1

3 d tan x 2 + 1 dx

=

(

= 2 x sec 2 x 2 + 1

3 d x 2 +1 dx

[ (

) dxd ( x + 1) = 3 tan ( x + 1)⋅ sec ( x + 1)⋅ 2 x

(

= 3 tan 2 x 2 + 1 ⋅ sec 2 x 2 + 1 ⋅ i. Given y = cos x 3 then

(

) dxd ( x + 1)

= sec 2 x 2 + 1 ⋅

3 d tan x 2 + 1 dx

=

h. Given y = tan 3 x 2 + 1 = tan x 2 + 1

(

(

)

d tan x 2 + 1 dx

=

2

(

sec 2 x 2 + 1

)]

2

⋅

)

3

(

)

d tan x 2 + 1 dx

(

)

(

= 6 x tan 2 x 2 + 1 sec 2 x 2 + 1

= − sin x 3 ⋅ 3 x 2 = − 3 x 2 sin x 3

= 3 (cos x )2 ⋅

d cos x dx

= 3 (cos x )2 ⋅ − sin x ⋅

d x dx

= 3 (cos x )2 ⋅ − sin x ⋅1 = − 3 sin x (cos x ) 2 k. Given y = csc 10 x then

Hamilton Education Guides

dy dx

=

d csc 10 x dx

= − csc 10 x cot 10 x ⋅

d 10 x dx

= − 10 csc 10 x cot 10 x

143

)

Calculus I

3.1 Differentiation of Trigonometric Functions

l. Given y = csc x10 then

dy dx

d csc x10 dx

=

= − csc x10 cot x10 ⋅

d 10 x dx

= − 10 x 9 csc x10 cot x10

Example 3.1-3: Find the derivative of the following trigonometric functions. a. y = csc10 x

b. y = sec 5 x 4

c. y = sec 4 5 x 4

d. y = sec x ⋅ cot x

e. y = sin 2 x ⋅ sec (x + 1)

f. y = sin x 2 + 1 ⋅ cos 2 x

g. y = sin 2 x + cos 2 x

h. y = 4 + cos 4 x

i. y = sin 2 x 2

j. y = sec 2 x − tan 2 x

k. y = cos 1 − x 2

(

Solutions: a. Given y = csc10 x then

dy dx

(

d csc10 x dx

=

) = 10 csc

9

(

)

x⋅

)

l. y = cos 2 (3x + 2) d csc x dx

= 10 csc 9 x ⋅ − csc x cot x ⋅

d x dx

= 10 csc 9 x ⋅ − csc x cot x ⋅1 = − 10 csc 10 x cot x b. Given y = sec 5 x 4 then

dy dx

=

(

d sec 5 x 4 dx

) = sec 5x

4

( )

d 5x4 dx

⋅ tan 5 x 4 ⋅

= sec 5 x 4 ⋅ tan 5 x 4 ⋅ (5 ⋅ 4)x 4−1

= sec 5 x 4 ⋅ tan 5 x 4 ⋅ 20 x 3 = 20 x 3 sec 5 x 4 tan 5 x 4

(

c. Given y = sec 4 5 x 4 = sec 5 x 4 = × tan 5 x 4 ⋅

then

dy dx

( ) = 4( sec 5x ) ⋅ sec 5x

=  cot x ⋅ sec x tan x ⋅ 

4

d 5x 4 dx

4 3

d. Given y = sec x ⋅ cot x then



)

dy dx

=

= 4

(

)

4 d sec 5 x 4 dx

⋅ tan 5 x 4 ⋅ 20 x 3

d (sec x ⋅ cot x ) dx

d   d  x  +  sec x ⋅ − csc 2 x ⋅ x dx   dx 

(

= sec (x + 1) ⋅ 2 sin x ⋅ cos x ⋅

Hamilton Education Guides

=

[

(

d sec 5 x 4 dx

(



=  cot x ⋅ 

)

4

) = 4(sec 5x ) ⋅ sec 5x 4 3

tan 5 x 4

d d    sec x  +  sec x ⋅ cot x  dx dx   

(

)

= (cot x ⋅ sec x tan x ⋅1) + sec x ⋅ − csc 2 x ⋅1

(

dy dx

3

= 80 x 3 sec 5 x 4

= cot x ⋅ sec x tan x + sec x ⋅ − csc 2 x = sec x tan x cot x − csc 2 x e. Given y = sin 2 x ⋅ sec (x + 1) then

)

= 4 sec 5 x 4 ⋅

d sin 2 x ⋅ sec (x + 1) dx

]

) = sec (x + 1) ⋅

d d x + sin 2 x ⋅ sec (x + 1) ⋅ tan (x + 1) ⋅ (x + 1) dx dx

d d sin 2 x + sin 2 x ⋅ sec (x + 1) dx dx

= sec (x + 1) ⋅ 2 sin x ⋅ cos x ⋅1

144

4

Calculus I

3.1 Differentiation of Trigonometric Functions

+ sin 2 x ⋅ sec (x + 1) tan (x + 1) ⋅1

= 2 sin x cos x sec (x + 1) + sin 2 x sec (x + 1) tan (x + 1)

= sin x sec ( x + 1) [ 2 cos x + sin x tan ( x + 1) ]

(

)

dy dx

f. Given y = sin x 2 + 1 ⋅ cos 2 x then

(

)

d  + sin x 2 + 1 ⋅ cos 2 dx 

[

 x 

2

)

= cos 2 x ⋅ cos x 2 + 1 ⋅ 2 x + sin x 2 + 1 ⋅ 2 cos x ⋅ − sin x g. Given y = sin 2 x + cos 2 x then = + 2 cos x ⋅

d cos x dx

dy dx

= 2 sin x ⋅ cos x ⋅

x⋅

(

)

d  sin x 2 + 1  dx 

]

2

(

)

(

= 2 x cos 2 x cos x 2 + 1 − 2 sin x cos x sin x 2 + 1

(

d sin 2 x + cos 2 x dx

=

2

) dxd ( x + 1)  + sin ( x + 1)⋅ 2 cos x ⋅ dxd cos x

(



] = cos

)

 = cos 2 x ⋅ cos x 2 + 1 ⋅

) ][ (

(

[ (

d sin x 2 + 1 ⋅ cos 2 x dx

=

)

=

d d x x + 2 cos x ⋅ − sin x ⋅ dx dx

d d sin 2 x + cos 2 x dx dx

= 2 sin x ⋅

)

d sin x dx

= 2 sin x ⋅ cos x ⋅1 + 2 cos x ⋅ − sin x ⋅1

= 2 sin x cos x − 2 sin x cos x = 0 A second way of solving this problem is by noting that sin 2 x + cos 2 x = 1 . Therefore, given y = sin 2 x + cos 2 x = 1

then

dy dx

=

(

d sin 2 x + cos 2 x dx 1

h. Given y = 4 + cos 4 x = (4 + cos 4 x ) 2 then =

1 1 (4 + cos 4 x )− 2 2

d d  cos 4 x  ⋅ 4 + dx dx  

=

dy dx

=

)=

d (1 ) dx

= 0

1 d (4 + cos 4 x ) 2 dx

1 1 (4 + cos 4 x )− 2 2

=

1 1 (4 + cos 4 x ) 2 −1 ⋅ d (4 + cos 4 x ) 2 dx

d   ⋅  0 − sin 4 x ⋅ 4x  dx  

=

1 1 (4 + cos 4 x )− 2 ⋅ (− 4 sin 4 x ) 2

2

1 1 2 sin 4 x 4/ sin 4 x ⋅ (4 + cos 4 x )− 2 = − 2 sin 4 x ⋅ (4 + cos 4 x )− 2 = − = − 1 2/ (4 + cos 4 x ) 2

i. Given y = sin 2 x 2 then

dy dx

=

(

d sin 2 x 2 dx

) = dxd ( sin x )

2 2

= 2 sin x 2 ⋅

d sin x 2 dx

= 2 sin x 2 ⋅ cos x 2 ⋅

d 2 x dx

= 2 sin x 2 ⋅ cos x 2 ⋅ 2 x = 4 x sin x 2 cos x 2

Hamilton Education Guides

145

Calculus I

3.1 Differentiation of Trigonometric Functions

j. Given y = sec 2 x − tan 2 x then = − 2 tan x ⋅

d tan x dx

dy dx

(

d sec 2 x − tan 2 x dx

=

= 2 sec x ⋅ sec x tan x ⋅

)

=

d d sec 2 x − tan 2 x dx dx

d d x − 2 tan x ⋅ sec 2 x ⋅ x dx dx

= 2 sec x ⋅

d sec x dx

= 2 sec 2 x ⋅ tan x ⋅1 − 2 tan x ⋅ sec 2 x ⋅1

= 2 sec 2 x tan x − sec 2 x 2 tan x = 0 A second way of solving this problem is by noting that sec 2 x − tan 2 x = 1 . Therefore, given y = sec 2 x − tan 2 x = 1

(

then

)

dy dx

(

d sec 2 x − tan 2 x dx

=

[ (

)]

then

dy dx

=

d cos 1 − x 2 dx

l. Given y = cos 2 (3x + 2) then

dy dx

=

d cos 2 (3 x + 2 ) dx

k. Given y = cos 1 − x 2

= 2 cos (3x + 2) ⋅ − sin (3x + 2) ⋅

d (3x + 2) dx

[

)

=

d (1 ) dx

(

= 0

]=

(

) dxd ( 1 − x )

= − sin 1 − x 2 ⋅

d [ cos (3x + 2) ]2 dx

= 2 x sin 1 − x 2

2

= 2 cos (3x + 2) ⋅

)

d cos (3 x + 2) dx

= 2 cos (3x + 2) ⋅ − sin (3x + 2) ⋅ 3 = − 6 cos (3 x + 2) sin (3 x + 2)

Note that since sin 2 α = 2 sin α cos α the above answer can also be written as: dy dx

= − 6 cos (3x + 2) sin (3x + 2) = − 3 ⋅ [2 sin (3x + 2) cos (3x + 2) ] = − 3 ⋅ [ sin 2 ⋅ (3x + 2) ] = − 3 sin (6 x + 4)

Example 3.1-4: Find the derivative of the following trigonometric functions: sin 2 x cos x

tan 2 x cot x

a. y = x 2 sin 3 2 x

b. y =

d. y = sec 2 10 x

e. y = cos 3 2 x

f. y = x 3 cot x 5

h. y = cos 2 ( 1 + 3x )

i. y = x 5 cot 3 x

g. y =

sin 2 x x

j. y = tan 4 x sin 5 x

k. y =

sin 3 x cos 2 x

c. y =

l. y =

sin 4 x x+3

Solutions: a.

dy dx

=

(

d x 2 sin 3 2 x dx

)

= sin 3 2 x ⋅

= 2 x sin 3 2 x + x 2 ⋅ 3 sin 2 2 x ⋅ cos 2 x ⋅

Hamilton Education Guides

d d 2 x + x 2 ⋅ sin 3 2 x dx dx d 2x dx

= sin 3 2 x ⋅ 2 x + x 2 ⋅ 3 sin 2 2 x ⋅

d sin 2 x dx

= 2 x sin 3 2 x + x 2 ⋅ 3 sin 2 2 x ⋅ cos 2 x ⋅ 2 = 2 x sin 3 2 x + 6 x 2

146

Calculus I

3.1 Differentiation of Trigonometric Functions

= 2 x sin 2 2 x ( sin 2 x + 3 x cos 2 x )

× sin 2 2 x cos 2 x

b.

dy dx

=

c.

dy dx

=

d.

dy dx

=

d  sin 2 x  dx  cos x 

=

dy dx

=

( cot x ⋅

= d dx

2 sin x cos 2 x + sin 3 x

)(

d cot x tan 2 x − tan 2 x ⋅ dx

2

2

x csc 2 x

2

cot x

(

d sec 2 10 x dx

=

d x dx

(

)

= 2 sec 10 x ⋅

=

)

( cos x ⋅ 2 sin x ⋅ dxd sin x )− ( sin 2 x ⋅ − sin x )

=

cos 2 x

(

sin x 2 cos 2 x + sin 2 x

=

cos 2 x

cot 2 x

( cot x ⋅ 2 tan x ⋅ sec x)+ tan =

)

cos 2 x

cos 2 x

d  tan 2 x  dx  cot x 

)(

d cos x sin 2 x − sin 2 x ⋅ dx

d dx

( cos x ⋅ 2 sin x ⋅ cos x ) + sin 3 x

× tan 10 x ⋅10 ⋅

e.

=

( cos x ⋅

)

cos 2 x

=

( cot x ⋅ 2 tan x ⋅ dxd tan x )− ( tan 2 x ⋅ − csc 2 x ) cot 2 x

(

tan x 2 cot x sec2 x + tan x csc2 x

)

2

cot x

d ( sec 10 x ) dx

= 2 sec 10 x ⋅ sec 10 x tan 10 x ⋅

d (10 x ) dx

= 2 sec 10 x ⋅ sec 10 x

= 2 sec 2 10 x tan 10 x ⋅10 = 20 sec 2 10 x tan 10 x

d cos 3 2 x dx

)

= 3 cos 2 2 x ⋅

d ( cos 2 x ) dx

= 3 cos 2 2 x ⋅ − sin 2 x ⋅

d 2x dx

= 3 cos 2 2 x ⋅ − sin 2 x ⋅ 2 ⋅

d x dx

= 3 cos 2 2 x ⋅ − sin 2 x ⋅ 2 = − 6 cos 2 2 x sin 2 x f.

dy dx

=

(

d x 3 cot x 5 dx

)

= cot x 5 ⋅

d 3 d x + x 3 ⋅ cot x 5 dx dx

= cot x 5 ⋅ 3x 2 + x 3 ⋅ − csc 2 x 5 ⋅

d 5 x dx

(

= 3x 2 cot x 5 − x 3 ⋅ csc 2 x 5 ⋅ 5 x 4 = 3x 2 cot x 5 − 5 x 7 csc 2 x 5 = x 2 3 cot x 5 − 5 x 5 csc 2 x 5

g.

dy dx

=

=

d  sin 2 x  dx  x 

=

( x⋅

( x ⋅ 2 sin x ⋅ cos x ) − sin 2 x x

2

Hamilton Education Guides

d dx

)(

d x sin 2 x − sin 2 x ⋅ dx

)

x2

=

2 x sin x cos x − sin 2 x x

2

=

)

( x ⋅ 2 sin x ⋅ dxd sin x )− ( sin 2 x ⋅1) x2

=

sin x (2 x cos x − sin x ) x2

147

Calculus I

h.

dy dx

=

3.1 Differentiation of Trigonometric Functions

[

d cos 2 ( 1 + 3 x ) dx

]

= 2 cos ( 1 + 3x ) ⋅ d d  1 + 3x  dx   dx

 = − 2 cos ( 1 + 3x ) ⋅ sin (1 + 3x ) ⋅ 

i.

dy dx

=

(

d x 5 cot 3 x dx

)

= cot 3 x ⋅

dy dx

=

d (tan 4 x sin 5 x ) dx

= tan 4 x ⋅

d x dx

= 2 cos ( 1 + 3x ) ⋅ − sin (1 + 3x ) ⋅

d ( 1 + 3x ) dx

= − 2 cos ( 1 + 3x ) ⋅ sin (1 + 3x ) ⋅ (0 + 3) = − 6 cos ( 1 + 3 x ) sin (1 + 3 x )

d d 5 x + x 5 ⋅ cot 3 x dx dx

= 5 x 4 cot 3 x + x 5 ⋅ 3 cot 2 x ⋅ − csc 2 x ⋅ j.

d cos ( 1 + 3 x ) dx

= cot 3 x ⋅ 5 x 4 + x 5 ⋅ 3 cot 2 x ⋅

d cot x dx

(

= 5 x 4 cot 3 x − x 5 ⋅ 3 cot 2 x ⋅ csc 2 x ⋅1 = x 4 cot 2 x 5 cot x − 3 x csc 2 x

d d sin 5 x + sin 5 x ⋅ tan 4 x dx dx

= tan 4 x ⋅ cos 5 x ⋅

d d 5 x + sin 5 x ⋅ sec 2 4 x ⋅ 4 x dx dx

= tan 4 x ⋅ cos 5 x ⋅ 5 + sin 5 x ⋅ sec 2 4 x ⋅ 4 = 5 tan 4 x cos 5 x + 4 sin 5 x sec2 4 x

k.

dy dx

=

l.

=

d  sin 3 x    dx  cos 2 x 

=

(cos 2 x ⋅ dxd sin 3x )− (sin 3x ⋅ dxd cos 2 x ) cos 2 2 x

( cos 2 x ⋅ cos 3x ⋅ 3) − ( sin 3x ⋅ − sin 2 x ⋅ 2) 2

cos 2 x

d  sin 4 x    dx  x + 3 

=

(cos 2 x ⋅ cos 3x ⋅ dxd 3x )− (sin 3x ⋅ − sin 2 x ⋅ dxd 2 x ) cos 2 2 x

3 cos 3 x cos 2 x + 2 sin 3 x sin 2 x

cos2 2 x

[ (x + 3)⋅ dxd sin 4 x]− [sin 4 x ⋅ dxd (x + 3) ]

dy dx

=

=

[ (x + 3)⋅ cos 4 x ⋅ 4] − sin 4 x 4 ( x + 3) cos 4 x − sin 4 x = ( x + 3) 2 (x + 3)2

=

=

(x + 3)2

=

[ (x + 3)⋅ cos 4 x ⋅ dxd 4 x]− [sin 4 x ⋅1 ] ( x + 3) 2

Example 3.1-5: Find the derivative of the following trigonometric functions: a. sin y = cot 3x

b. cos 3 y = tan 5 x

c. sin ( y + 1) = cos x 2

d. sin y = cos 10 x

e. tan (3 y + 2) = sin 5 x

f. sec 2 y = csc x

g. x cos y = sin (x + y )

h. x 2 sin 2 y = cos (3x + 5 y )

i. x tan y = cot x

a. Given sin y = cot 3x , let’s take the derivative on both sides of the equation to obtain cos y ⋅

d d y = − csc 2 3 x ⋅ 3 x dx dx

Hamilton Education Guides

; cos y ⋅ y ′ = − csc 2 3x ⋅ 3 ; y ′ = −

dy = y′ . dx

3 csc 2 3 x cos y

148

)

Calculus I

3.1 Differentiation of Trigonometric Functions

dy = y′ . dx

b. Given cos 3 y = tan 5 x , let’s take the derivative on both sides of the equation to obtain − sin 3 y ⋅

d d 3 y = sec 2 5 x ⋅ 5 x dx dx

; − sin 3 y ⋅ 3 y ′ = sec 2 5 x ⋅ 5 ; − 3 sin 3 y ⋅ y ′ = 5 sec 2 5 x ; y ′ = −

5 sec 2 5 x 3 sin 3 y

c. Given sin ( y + 1) = cos x 2 , let’s take the derivative on both sides of the equation to obtain cos ( y + 1) ⋅

d ( y + 1) = − sin x 2 ⋅ d x 2 dx dx

; cos ( y + 1) ⋅ y ′ = − sin x 2 ⋅ 2 x ; y ′ = −

2 x sin x 2

cos ( y + 1)

d. Given sin y = cos 10 x , let’s take the derivative on both sides of the equation to obtain cos y ⋅

d d y = − sin 10 x ⋅ 10 x dx dx

; cos y ⋅ y ′ = − sin 10 x ⋅10 ; y ′ = −

dy = y′ . dx

dy = y′ . dx

10 sin 10 x cos y dy = y′ . dx

e. Given tan (3 y + 2) = sin 5 x , let’s take the derivative on both sides of the equation to obtain sec 2 (3 y + 2 ) ⋅

d (3 y + 2) = cos 5 x ⋅ d 5 x dx dx

; sec 2 (3 y + 2) ⋅ 3 y ′ = cos 5 x ⋅ 5 ; y ′ =

5 cos 5 x

3 sec 2 (3 y + 2 )

f. Given sec 2 y = csc x , let’s take the derivative on both sides of the equation to obtain 2 sec y ⋅

; y′ =

d d x sec y = − csc x cot x ⋅ dx dx

− csc x cot x 2

2 sec y tan y

; y′ = −

; 2 sec y ⋅ sec y tan y ⋅

d y = − csc x cot x ⋅1 dx

dy = y′ . dx

; 2 sec 2 y tan y ⋅ y ′ = − csc x cot x

csc x cot x 2 sec 2 y tan y

g. Given x cos y = sin (x + y ) , let’s take the derivative on both sides of the equation to obtain cos y ⋅

d d d x + x ⋅ cos y = cos (x + y ) ⋅ (x + y ) dx dx dx

; cos y ⋅1 + x ⋅ − sin y ⋅

dy = y′ . dx

d  d d y = cos (x + y ) ⋅  x + y dx dx   dx

; cos y ⋅1 + x ⋅ − sin y ⋅ y ′ = cos (x + y ) ⋅ (1 + y ′) ; cos y − x sin y y ′ = cos (x + y ) + y ′ cos (x + y ) ; − x sin y y ′ − y ′ cos (x + y ) = cos (x + y ) − cos y ; x sin y y ′ + y ′ cos (x + y ) = cos y − cos (x + y )

; − y ′ [x sin y + cos (x + y ) ] = cos y − cos (x + y ) ; y ′ = −

cos y − cos ( x + y ) x sin y + cos ( x + y )

h. Given x 2 sin 2 y = cos (3x + 5 y ) , let’s take the derivative on both sides of the equation to obtain sin 2 y ⋅

d d d 2 x + x 2 ⋅ sin 2 y = − sin (3 x + 5 y ) ⋅ (3 x + 5 y ) dx dx dx

Hamilton Education Guides

; sin 2 y ⋅ 2 x + x 2 ⋅ cos 2 y ⋅

dy = y′ . dx

d 2 y = − sin (3 x + 5 y ) dx

149

Calculus I

3.1 Differentiation of Trigonometric Functions

d d  ×  3x + 5 y  dx   dx

; sin 2 y ⋅ 2 x + x 2 ⋅ cos 2 y ⋅ 2 y ′ = − sin (3x + 5 y ) ⋅ (3 + 5 y ′) ; 2 x sin 2 y + 2 y ′ x 2 cos 2 y

= −3 sin (3 x + 5 y ) − 5 y ′ sin (3 x + 5 y ) ; 2 y ′ x 2 cos 2 y + 5 y ′ sin (3 x + 5 y ) = −3 sin (3 x + 5 y ) − 2 x sin 2 y

[

; y ′ 2 x 2 cos 2 y + 5 sin (3x + 5 y )

]

= −3 sin (3 x + 5 y ) − 2 x sin 2 y

; y′ = −

3 sin (3 x + 5 y ) + 2 x sin 2 y

[2 x

2

cos 2 y + 5 sin (3 x + 5 y )

i. Given x tan y = cot x , let’s take the derivative on both sides of the equation to obtain tan y ⋅

d d d x x + x ⋅ tan y = − csc 2 x ⋅ dx dx dx

x ⋅ y ′ ⋅ sec 2 y = − csc 2 x − tan y

; y′ = −

; tan y ⋅1 + x ⋅ sec 2 y ⋅

d y = − csc 2 x ⋅1 dx

]

dy = y′ . dx

; tan y + x ⋅ sec 2 y ⋅ y ′ = − csc 2 x

csc 2 x + tan y x sec 2 y

Example 3.1-6: Find the first and second derivative of the following trigonometric functions: a. y = sin 5 x

b. y = sin 2 5 x

c. y = cos ( 10 x + 3)

d. y = tan 2 x

e. y = tan x 2

f. y = tan 2 x

g. y = sin 3x + cos 3x

h. y = x sin x

i. y = tan (5 x + 1 )

j. y = sin x − x cos x

k. y = 2 + sin 2 x

l. y = sin x 2

(

)

2

Solutions: a.

dy dx

d2y dx

b.

2

dy dx

= =

d2y dx

d (sin 5 x ) dx

=

2

d (5 cos 5 x ) dx

(

d sin 2 5 x dx

=

= cos 5 x ⋅

)

d (5 x ) dx

d 5x dx

= − 5 sin 5 x ⋅

=

d (sin 5 x )2 dx

d (10 sin 5 x cos 5 x ) dx

= 10

= cos 5 x ⋅ 5 = 5 cos 5 x

= − 5 sin 5 x ⋅ 5 = −25 sin 5 x

= 2 sin 5 x ⋅

d (sin 5 x ) dx

d (sin 5 x cos 5 x ) dx

(

= 2 sin 5 x ⋅ cos 5 x ⋅ d

d (5 x ) dx

= 10 sin 5 x cos 5 x d

  = 10 cos 5 x ⋅ (sin 5 x ) + sin 5 x ⋅ ( cos 5 x ) dx dx  

= 10 (cos 5 x ⋅ 5 cos 5 x − 5 sin 5 x ⋅ sin 5 x ) = 10 5 cos 2 5 x − 5 sin 2 5 x

)

(

= 50 cos 2 5 x − sin 2 5 x

)

Note that since sin 2α = 2 sin α cos α the first derivative can also be written as:

Hamilton Education Guides

150

Calculus I

c.

dy dx

= 5 (2 sin 5 x cos 5 x ) = 5 (sin 2 ⋅ 5 x ) = 5 sin 10 x . Thus,

dy dx

=

d2y dx

d.

3.1 Differentiation of Trigonometric Functions

=

d2y dx

d ( tan 2 x ) dx

=

2

= − sin ( 10 x + 3) ⋅

d [ 10 sin ( 10 x + 3) ] dx

= −

2

dy dx

d [ cos ( 10 x + 3) ] dx

= − 10 cos ( 10 x + 3) ⋅

= sec 2 2 x ⋅

(

d 2 sec 2 2 x dx

)

= 2⋅

d ( 10 x + 3) dx

d (2 x ) dx

d2y dx

= 5 ⋅ cos 10 x ⋅

2

d 10 x dx

= 50 cos 10 x

= − sin ( 10 x + 3) ⋅10 = − 10 sin ( 10 x + 3)

d ( 10 x + 3) dx

= − 10 cos ( 10 x + 3) ⋅10 = − 100 cos ( 10 x + 3)

= sec 2 2 x ⋅ 2 = 2 sec 2 2 x

d ( sec 2 x )2 dx

= 2 ⋅ 2 sec 2 x ⋅

d sec 2 x dx

= 4 sec 2 x ⋅ sec 2 x tan 2 x ⋅

d 2x dx

= 4 sec 2 x ⋅ sec 2 x tan 2 x ⋅ 2 = 8 sec 2 x sec 2 x tan 2 x = 8 sec 2 2 x tan 2 x e.

dy dx

(

d tan x 2 dx

=

d2y

=

dx 2

(

)

= sec 2 x 2 ⋅

(

d 2 x sec 2 x 2 dx

)

)

=

(

( )

= sec 2 x 2 ⋅ 2 x = 2 x sec 2 x 2

(

)

d x2 dx

(

2 d 2 x sec x 2 dx

= sec x 2

) dxd ( sec x ) = ( sec x )

= sec x 2 2 ⋅ 2 + 2 x ⋅ 2 sec x 2 ⋅

2 2

2

)

2

⋅

(

)

2 d d sec x 2 2x + 2x ⋅ dx dx

(

)

d 2 x ⋅ 2 + 2 x ⋅ 2 sec x 2 ⋅ sec x 2 tan x 2 ⋅ dx

(

= sec 2 x 2 ⋅ 2 + 2 x ⋅ 2 sec 2 x 2 tan x 2 ⋅ 2 x = 2 sec 2 x 2 + 8 x 2 sec 2 x 2 tan x 2 = 2 sec 2 x 2 1 + 4 x 2 tan x 2 f.

dy dx

d2y dx

(

d tan 2 x dx

=

=

2

)=

d (tan x )2 dx

(

d 2 tan x sec 2 x dx

)

= 2 tan x ⋅

 = 2  sec 2 x ⋅ 

(

= 2 sec 4 x + tan x ⋅ 2 sec x ⋅ sec x tan x g.

dy dx

=

d2y dx

2

d ( sin 3x + cos 3x dx

=

)=

d (3 cos 3x − 3 sin 3x ) dx

Hamilton Education Guides

)

d tan x dx

d d  tan x + tan x ⋅ sec 2 x  dx dx 

(

= 2 sec 4 x + 2 sec 2 x tan 2 x

d d sin 3 x + cos 3 x dx dx

=

= 2 tan x ⋅ sec 2 x ⋅

= cos 3x ⋅

d d 3 cos 3 x − 3 sin 3 x dx dx

d x dx

)

= 2 tan x ⋅ sec 2 x ⋅1 = 2 tan x sec 2 x

 = 2  sec 2 x ⋅ sec 2 x + tan x ⋅ 2 sec x ⋅ 

)

(

= 2 sec 2 x sec 2 x + 2 tan 2 x

d d 3 x − sin 3 x ⋅ 3x dx dx

= − 3 sin 3x ⋅

d  sec x  dx 

)

= 3 cos 3 x − 3 sin 3 x

d d 3 x − 3 cos 3 x ⋅ 3 x dx dx

151

Calculus I

3.1 Differentiation of Trigonometric Functions

= − 3 sin 3x ⋅ 3 − 3 cos 3x ⋅ 3 = − 9 sin 3x − 9 cos 3x = − 9(sin 3 x + cos 3 x ) h.

dy dx

=

d2y dx

d (x sin x ) dx

=

2

= sin x ⋅

d (sin x + x cos x ) dx

d d x + x ⋅ sin x dx dx

=

= sin x ⋅1 + x ⋅ cos x = sin x + x cos x

d d (x cos x ) sin x + dx dx

= cos x ⋅

d d d   x +  cos x ⋅ x + x ⋅ cos x  dx dx dx  

= cos x ⋅1 + (cos x ⋅1 − x ⋅ sin x ) = cos x + cos x − x sin x = 2 cos x − x sin x i.

dy dx

d [ tan ( 5 x + 1 ) ] dx

=

d2y dx

=

2

= sec 2 ( 5 x + 1) ⋅

[

d 5 sec 2 ( 5 x + 1) dx

dy dx

d ( sin x − x cos x dx

=

= sec 2 ( 5 x + 1) ⋅ 5 = 5 sec 2 ( 5 x + 1)

] = 5 ⋅ dxd [ sec ( 5x + 1) ] = 5 ⋅ 2 sec ( 5x + 1)⋅ dxd sec ( 5x + 1) 2

= 10 sec ( 5 x + 1) ⋅ sec ( 5 x + 1) tan ( 5 x + 1) ⋅

j.

d ( 5x + 1 ) dx

)=

d (5 x + 1) dx

= 10 sec 2 ( 5 x + 1) tan ( 5 x + 1) ⋅ 5 = 50 sec2 ( 5 x + 1) tan ( 5 x + 1)

d d sin x − (x cos x ) dx dx

= cos x ⋅

d d d   x −  cos x ⋅ x + x ⋅ cos x  dx dx dx  

= cos x ⋅1 − (cos x ⋅1 − x ⋅ sin x ) = cos x − cos x + x sin x = x sin x d2y dx

k.

2

dy dx

= =

d (x sin x ) dx

d dx

(

2 + sin 2 x

d  d ×  2 + sin 2 x  dx   dx

2

d y dx 2

=

=

= sin x ⋅

=

)

=

d d x + x ⋅ sin x dx dx

1 d (2 + sin 2 x ) 2 dx

=

= sin x ⋅1 + x ⋅ cos x = sin x + x cos x 1 1 (2 + sin 2 x ) 2 −1 ⋅ d (2 + sin 2 x ) 2 dx

1 1 (2 + sin 2 x )− 2 × ( 0 + 2 cos 2 x ) 2

d  cos 2 x dx  2 + sin 2 x 

   

=

=

1 1 ⋅ ⋅ 2/ cos 2 x 2/ (2 + sin 2 x ) 12

1 1 (2 + sin 2 x )− 2 2

=

cos 2 x 2 + sin 2 x

 (2 + sin 2 x ) 12 ⋅ d cos 2 x  −  cos 2 x ⋅ d (2 + sin 2 x ) 12      dx dx     2 + sin 2 x

 (2 + sin 2 x ) 12 ⋅ − sin 2 x ⋅ d 2 x  −  cos 2 x ⋅ 1 (2 + sin 2 x ) 12 −1 ⋅ d (2 + sin 2 x )     dx 2 dx     2 + sin 2 x

Hamilton Education Guides

=

=

 (2 + sin 2 x ) 12 ⋅ − sin 2 x ⋅ 2      2 + sin 2 x

152

Calculus I

3.1 Differentiation of Trigonometric Functions

1 −  cos 2 x ⋅ 12/ (2 + sin 2 x )− 2 ⋅ 2/ cos 2 x    2 + sin 2 x

l.

dy dx

(

d sin x 2 dx

=

)

2

= 2 sin x 2 ⋅

=

1    −1  2  − 2 sin 2 x ⋅ (2 + sin 2 x ) 2  −  cos 2 x ⋅ (2 + sin 2 x ) 2      2 + sin 2 x

d sin x 2 dx

= 2 sin x 2 cos x 2 ⋅

d 2 x dx

= 2 sin x 2 cos x 2 ⋅ 2 x = 4 x sin x 2 cos x 2

Note that since sin 2α = 2 sin α cos α the first derivative can also be written as:

(

dy dx

d2y dx

)

= 2 x 2 sin x 2 cos x 2 = 2 x sin 2 x 2 and the second derivative is equal to the following: =

2

(

d 2 x sin 2 x 2 dx

)

 = 2 sin 2 x 2 ⋅ 

(

= 2 sin 2 x 2 + x ⋅ cos 2 x 2 ⋅ 4 x

)

d d  x + x ⋅ sin 2 x 2  dx dx 

(

= 2 sin 2 x 2 + 4 x 2 cos 2 x 2

)

 = 2 sin 2 x 2 ⋅1 + x ⋅ cos 2 x 2 ⋅ 

d  2x 2  dx 

= 2 sin 2 x 2 + 8 x 2 cos 2 x 2

Example 3.1-7: Find the derivative of the following trigonometric functions: a. y = sin x 2 + 4 x 3 + 7

b. y = 10 sin (− 5 x )

c. y = cot (5 x + 3)

d. y = 5 cos (4 x − 1)

e. y = sin 2π x + 3x 2

f. y = sec 3x + 10 x 2

g. y = sin x − cos x + 3

h. y = 5 sin 8 x − 3x 2 + 2 x

i. y = 7 csc (x + 10) + 10 x

a.

dy dx

=

(

d sin x 2 + 4 x 3 + 7 dx

)=

(

d d d 7 sin x 2 + 4x 3 + dx dx dx

= 2 x cos x 2 + 12 x 2 = 2 x cos x 2 + 6 x dy dx

=

d [ 10 sin (− 5 x ) ] dx

= 10 cos (− 5 x ) ⋅

c.

dy dx

=

d [ cot (5 x + 3) ] dx

= − csc 2 (5 x + 3) ⋅

d.

dy dx

=

d [ 5 cos (4 x − 1) ] dx

e.

dy dx

=

d sin 2π x + 3 x 2 dx

f.

dy dx

=

d sec 3 x + 10 x 2 dx

(

Hamilton Education Guides

)

d (− 5 x ) dx

=

= cos x 2 ⋅ 2 x + 12 x 2

= 10 cos (− 5 x ) ⋅ −5 = − 50 cos (− 5 x )

d (5 x + 3) dx

= − 5 sin (4 x − 1) ⋅

)=

d 2 x + (4 ⋅ 3) x 2 + 0 dx

)

b.

(

= cos x 2 ⋅

= − csc 2 (5 x + 3) ⋅ 5 = − 5 csc 2 ( 5 x + 3)

d (4 x − 1) dx

d d sin 2π x + 3 x 2 dx dx

d d sec 3 x + 10 x 2 dx dx

= − 5 sin (4 x − 1) ⋅ 4 = − 20 sin (4 x − 1)

= cos 2π x ⋅

d 2π x + 6 x dx

= sec 3x tan 3x ⋅

= 2π cos 2π x + 6 x

d 3 x + 20 x dx

= 3 sec 3 x tan 3 x + 20 x

153

Calculus I

g.

dy dx

=

3.1 Differentiation of Trigonometric Functions

(

d sin dx

= cos x ⋅

1 2 x

)

=

x − cos x + 3

+ sin

x⋅

1

=

2 x

(

d sin dx

cos x 2 x

x−

+

d d cos x + 3 dx dx

sin

x

=

2 x

cos

x + sin

d d d 2x 5 sin 8 x − 3 x 2 + dx dx dx

] =

d d 7 csc (x + 10) + 10 x dx dx

dy dx

=

d 5 sin 8 x − 3 x 2 + 2 x dx

i.

dy dx

=

d [ 7 csc (x + 10) + 10 x dx

d dx

x + sin

x⋅

d dx

x +0

x

2 x

)=

h.

= cos x ⋅

= 5 cos 8 x ⋅

d 8x − 6 x + 2 dx

= − 7 csc (x + 10) cot (x + 10) ⋅

= 40 cos 8 x − 6 x + 2 d (x + 10) + 10 dx

= − 7 csc (x + 10) cot (x + 10) ⋅1 + 10 = − 7 csc ( x + 10) cot ( x + 10) + 10 Example 3.1-8: Given the following trigonometric functions find

dy dx

.

a. y = 1+ cos t and x = t + sin t

b. y = 1 + t 2 and x = t sin t

c. y = sin 3 θ and x = cos 2 θ

d. y = sin 3 α and x = cos 3 α

e. y = t e t and x = e t cos t

f. y = 5 ln 3t 2 and x = ln t 3

g. y = e t cos t and x = t ln t

h. y = sin 2 t and x = sec 4 t

i. y = cos 2 (θ + 1) and x = sin θ 2

j. y = 3 sin ω and x = cos 3ω 2

k. y = 3 t 2 and x = 1 + t

l. y = θ − θ and x = θ 2 cos θ

Solutions: a. Given y = 1+ cos t and x = t + sin t then =

d ( t + sin t ) dt

=

d d t + sin t dt dt

d t sin t dt

= sin t ⋅

d d t + t ⋅ sin t dt dt

dy dt

dx dθ

Hamilton Education Guides

=

d ( 1+ cos t ) dt

dy dt dx dt

=

(

d 1+ t 2 dt

)

=

=

=

d d 1+ cos t dt dt

dy ⋅ dt dt ⋅ dx

d d 1+ t 2 dt dt

=

dy dθ

d cos 2 θ dθ

=

=

d sin 3 θ dθ

d ( cos θ ) 2 dθ

=

d (sin θ ) 3 dθ

= 2 cos θ ⋅

dy dx

=

= − sin t and

d cos θ dθ

dy dt dx dt

dx dt

− sin t 1 + cos t dx dt

= 0 + 2t = 2t and

= sin t ⋅1 + t ⋅ cos t = sin t + t cos t So

c. Given y = sin 3 θ and x = cos 2 θ then = 3 sin 2 θ cos θ and

=

= 1+ cos t Therefore

b. Given y = 1 + t 2 and x = t sin t then =

dy dt

=

dy dx

= 3 sin 2 θ ⋅

=

2t sin t + t cos t

d sin θ dθ

= 2 cos θ ⋅ − sin θ 154

Calculus I

3.1 Differentiation of Trigonometric Functions

= − 2 sin θ cos θ

dy dθ dx dθ

Therefore

dy dx

=

dy dα

d. Given y = sin 3 α and x = cos 3 α then dx dα

= 3 sin 2 α cos α and

=

d cos 3 α dα

=

dy dx

e. Given y = t e t and x = e t cos t then

dy dt

=

= e t ( 1 + t ) and

dx dt

=

d t e cos t dt

= cos t ⋅

dy dt dx dt

= e t ( cos t − sin t ) Therefore

=

dx dt

=

d ln t 3 dt

=

1 t

3

⋅

d 3 t dt

=

1 t

3

⋅ 3t 2

=

g. Given y = e t cos t and x = t ln t then

d sin 3 α dα

d (sin α ) 3 dα

=

=

= 3 cos 2 α ⋅

− 3 sin α cos α

= et ⋅

3t 2

e

3 t

=

3

dy dt

=

= 5⋅

1 3t

2

1 t

= ln t + 1 Therefore

dy dt dx dt

h. Given y = sin 2 t and x = sec 4 t then

= 2 sin t ⋅ cos t ⋅

d dt

Hamilton Education Guides

t

=

dy dx

dy dt

= 2 sin t ⋅ cos t ⋅ −

=

=

= cos t ⋅

2

t

=

= 5⋅ dy dx

1

⋅ 6t

3t

2

=

10 t 3 t

30t

=

3t

=

2

=

10 t

and

10 3

d d t e + e t ⋅ cos t dt dt

dx dt

d t ln t dt

=

= ln t ⋅

d d t + t ⋅ ln t dt dt

e t ( cos t − sin t ) ln t + 1

d sin 2 dt

1

= − tan α

= e t ⋅1 + t ⋅ e t = e t + t e t

d 2 3t dt

⋅

dy dt dx dt

= cos t ⋅ e t + e t ⋅ − sin t = e t cos t − e t sin t = e t ( cos t − sin t ) and

= ln t ⋅1 + t ⋅

= 3 cos 2 α ⋅ − sin α

1+ t cos t − sin t

=

Therefore

d t e cos t dt

d sin α dα

= cos t ⋅ e t + e t ⋅ − sin t = e t cos t − e t sin t

( cos t − sin t )

d 5 ln 3t 2 dt

sin α cos α

d d t + t ⋅ et dt dt

et ( 1+ t )

t

= 3 sin 2 α ⋅

d cos α dα

= −

2

d d t e + e t ⋅ cos t dt dt

=

3 2

3 sin 2 α cos α

d t te dt

=

t

= − sin θ

=

dy dx

dy dt

f. Given y = 5 ln 3t 2 and x = ln t 3 then

3 sin 2 θ cos θ − 2 sin θ cos θ

d ( cos α ) 3 dα

=

dy dα dx dα

= − 3 sin α cos 2 α Therefore

=

= −

t

sin

=

(

d sin dt

t cos t

t

t

) 2 = 2 sin

and

dx dt

=

t⋅

d sin dt

d sec 4 dt

t

t

155

Calculus I

=

3.1 Differentiation of Trigonometric Functions

(

d sec dt

= −

t

2 sec 4

)4

= 4 sec 3 t ⋅

t tan

t

d sec dt

t

dy dt dx dt

Therefore

t

= 4 sec 3 t ⋅ sec t tan t ⋅

d cos (θ + 1) dθ

dy dθ dx dθ

dy dx

=

=

dx dθ

=

d 3ϕ 2 dt

=

(

d 1+ dt

t

)

=

d d 1+ dt dt

t

dy dt

= 0+

dx dθ

=

Therefore

(

d θ 2 cos θ dt

dy dθ dx dθ

=

Hamilton Education Guides

)

dy dx

= cos θ ⋅

=

=

1 2

t

d 3 2 t dt

dy dθ

2

=

2 θ −1 2 θ 2

2θ cos θ − θ sin θ

=

2 sec

t cos 4

t

t

t tan

t

d [ cos (θ + 1) ] 2 dθ

=

= −2 cos (θ + 1) ⋅ sin (θ + 1) ⋅1 = cos θ 2 ⋅ 2 θ = 2 θ cos θ 2

dy dx

=

(

dy dt dx dt

d θ− θ dθ

d d 2 θ + θ 2 ⋅ cos θ dt dt

t

sin

dx dϕ

=

d cos 3ϕ 2 dt

3 cos ϕ

=

− 6ϕ sin 3ϕ

)

=

=

dy dx

=

2 33 t 1 2

d d θ− θ dθ dθ

t

2 1 3t 3

=

= 1−

cos ϕ

= −

2

d 23 2 2 −1 2 −1 t = t3 = t 3 = dt 3 3

So

t

t

=

= 3 cos ϕ and

dy dϕ dx dϕ

=

1

=

d 2 θ dt

d 3 sin ϕ dϕ

=

t

2

θ cos θ 2

= − sin 3ϕ 2 ⋅ 6ϕ = − 6ϕ sin 3ϕ 2 So

l. Given y = θ − θ and x = θ 2 cos θ then and

dy dϕ

t

1

= 4 sec 4 t ⋅ tan t ⋅ −

sin (θ + 1) cos (θ + 1)

= −

2 θ cos θ 2

t

t tan

d (θ + 1) dθ

= cos θ 2 ⋅

−2 sin (θ + 1) cos (θ + 1)

k. Given y = 3 t 2 and x = 1 + t then

dx dt

− 2 sec

t cos 4

d cos 2 (θ + 1) dθ

=

d sin θ 2 dt

j. Given y = 3 sin ϕ and x = cos 3ϕ 2 then = − sin 3ϕ 2 ⋅

− sin

= 2 cos (θ + 1) ⋅ − sin (θ + 1) ⋅

= − 2 sin (θ + 1) cos (θ + 1) and Therefore

=

dy dθ

i. Given y = cos 2 (θ + 1) and x = sin θ 2 then = 2 cos (θ + 1) ⋅

dy dx

=

d dt

2ϕ sin 3ϕ 2

=

2⋅2

2 3

3 t

t 3

1⋅ 3 t

1 2 θ

=

=

and

4

t 3

3 t

2 θ −1 2 θ

= cos θ ⋅ 2θ + θ 2 ⋅ − sin θ = 2θ cos θ − θ 2 sin θ

(

2 θ −1

2 θ 2θ cos θ − θ 2 sin θ

)

156

Calculus I

3.1 Differentiation of Trigonometric Functions

Section 3.1 Practice Problems – Differentiation of Trigonometric Functions 1. Find the derivative of the following trigonometric functions: a. y = sin (3x + 1)

b. y = 5 cos x 3

c. y = x 3 cos x 2

d. y = sin 5 x ⋅ tan 3x

e. y = tan 2 x 3

f. y = cot ( x + 3)3

g. y = x 5 cos x 7

h. y = sec 4 x 3

i. y = sec 3x 2 + x 3

j. y = tan x 5

k. y = tan 5 x

l. y = csc x3 + 1

(

)

2. Find the derivative of the following trigonometric functions: a. y =

tan x csc x

b. y =

(

)

sin x 3 + 1 x2

c. y =

sec x csc x 3

d. y = x 5 tan x 3

e. y = x 5 sin x 2

f. y = ( x + 5) 2 cos x

g. y = x 2 tan 3 x 5

h. y = x + sin x 3

i. y = sin 1 + x 5

j. y = cot 2 x 3

k. y = sin 3 ( 1 + 5 x )

l. y = x 5 csc x 3

(

)

3. Find the derivative of the following trigonometric functions: a. y = sin 7 x

b. y = cos x 3

c. y = cot x 3

d. y = x 3 tan x 2

e. y = cot ( x + 9)

f. y = sin 2 x 3 + 5 x + 2

g. y = sin x + 3

h. y = sin x 2 + cos x 3

i. y = x 2 sin x 3

Hamilton Education Guides

(

)

157

Calculus I

3.2

3.2 Differentiation of Inverse Trigonometric Functions

Differentiation of Inverse Trigonometric Functions

The differential formulas involving inverse trigonometric functions are defined as: Table 3.2-1: Differentiation Formulas for Inverse Trigonometric Functions 1

d arc sin u dx

=

d sin −1 u dx

=

d arc cos u dx

=

d cos −1 u dx

= −

d arc tan u dx

=

d tan −1 u dx

=

du dx

⋅

1− u 2 1 1− u 1 1+ u

2

⋅

2

⋅

du dx

du dx

d arc cot u dx

=

d cot −1 u dx

= −

d arc sec u dx

=

d sec −1 u dx

=

d arc csc u dx

=

d csc −1 u dx

= −

1 1+ u

⋅

2

du dx

1

du u −1 dx ⋅

2

u

1

u

du u − 1 dx ⋅

2

Let’s differentiate some inverse trigonometric functions using the above differentiation formulas. Example 3.2-1: Find the derivative of the following inverse trigonometric functions: a. y = arc sin (3x − 4 )

b. y = arc sin x 2

d. y = x 2 arc sin 2 x

e. y =

g. y = ( 1 + x ) cot −1 x

h. y = arc tan

j. y =

arc sin x cos x

(

)

c. y = tan −1 x 2 + 1

sin −1 x 3x

f. y = cos x 2 + arc cos x 2 5 x

i. y =

cos −1 x cos x

l. y = arc cot 3 x 2

k. y = x 5 + arc tan x

Solutions: dy dx

a. Given y = arc sin (3x − 4) then

=

3

(

1 − 9 x 2 + 16 − 24 x

=

)

b. Given y = arc sin x 2 then

(

1 1 + x 4 + 2x +1

⋅ 2x

dy dx

)

=

3

=

= dy dx

(

d arc sin x 2 dx

=

[

1

=

1 − ( 3x − 4)

2

⋅

3

d (3x − 4) = dx

1 − ( 3 x − 4 )2

3 − 9 x 2 + 24 x − 15

)=

1 1− x 4

)] =

(

d tan −1 x 2 + 1 dx

⋅

d 2 x dx

(

1 2

1

=

)

1+ x +1

2

1− x 4

⋅

(

⋅ 2x

=

)

=

d x 2 +1 dx

2x 1 − x4

(

1 2

)

1+ x +1

2

⋅ 2x

2x 4

x + 2x + 2

d. Given y = x 2 arc sin 2 x then

Hamilton Education Guides

d [arc sin (3x − 4) ] dx

1 − 9 x 2 − 16 + 24 x

c. Given y = tan −1 x 2 + 1 then

=

=

dy dx

=

(

d x 2 arc sin 2 x dx

)

= arc sin 2 x ⋅

d 2 d x + x 2 ⋅ arc sin 2 x dx dx

158

Calculus I

3.2 Differentiation of Inverse Trigonometric Functions

1

= arc sin 2 x ⋅ 2 x + x 2 ⋅

1 − (2 x )2

dy d  sin −1 sin x e. Given y = then = dx  3 x 3x dx −1

3x

=

1− x 2

1− x 2

=

9 x2

9x 2

−1

d 2 x + dx

1− x 4

⋅

dy dx

h. Given y = arc tan

=

x2

⋅

2

−5

x + 25 x

i. Given y =

cos x ⋅

=

j. Given

2

= −

1− x

2

= cot −1 x −

1+ x 2

5 x

cos −1 x cos x

−1

−1

dy dx

then

(

5 x 2/

x 2/ x 2 + 25

then

dy dx

=

=

(

Hamilton Education Guides

dy dx

=

d dx

)

)(

d 3x sin −1 x − sin −1 x ⋅ dx

(3x )

(

)

2

)

1− x

2

− sin −1 x ⋅ 3

9x 2

d d cos x 2 + arc cos x 2 dx dx

=

  ⋅ 2 x = − 2 x  sin x 2 + 4  1− x  −1

=

1

3 x2 1 − x2

1

[ ( 1 + x) cot x]

3x ⋅

1 − 4x 2

x − 1 − x 2 sin −1 x

=

d cos x 2 + arc cos x 2 dx

= cot −1 x ⋅

   4  1− x  1

d (1 + x ) + (1 + x )⋅ d cot −1 x dx dx

1+ x 1+ x2

=

1

1+

()

5 2 x

⋅

d 5 dx x

=

1 1+

25 x2

⋅

−5 x

2

=

1 2

x + 25 x2

⋅

−5 x2

5 2

x + 25

d  cos −1 x  = dx  cos x  − cos x

=

cos 2 x

then

=

d  5  arc tan  dx  x

= −

)

− cos −1 x ⋅ − sin x

arc sin x y= cos x

=

d dx

9x 2 1 − x 2

d 2 x = − sin x 2 ⋅ 2 x − dx

g. Given y = ( 1 + x ) cot −1 x then = cot −1 x ⋅1 + ( 1 + x ) ⋅

dy dx

( 3x ⋅

2x 2

⋅ 2 = 2 x arc sin 2 x +

1 − 4x 2

3 x − 3 1 − x 2 sin −1 x

=

f. Given y = cos x 2 + arc cos x 2 then

= − sin x 2 ⋅

x  =  

3 x −3 1− x 2 sin −1 x

− 3 sin −1 x

1

d 2 x = 2 x arc sin 2 x + x 2 ⋅ dx

⋅

1− x

2

d  arc sin  dx  cos x

( cos x ⋅

d dx

)(

d cos x cos −1 x − cos −1 x ⋅ dx

( cos x )

+ sin x cos −1 x cos 2 x

x   

=

=

)

2

− cos x +

1 − x 2 sin x cos−1 x 1 − x 2 cos2 x

( cos x ⋅ dxd arc sin x )− ( arc sin x ⋅ dxd cos x ) ( cos x )2

159

Calculus I

3.2 Differentiation of Inverse Trigonometric Functions

1− x 2

=

cos 2 x dy dx

1

4

= 5x +

1+

( x )2

1+

4 x3

⋅

2 23 −1 x 3

=

cos 2 x

=

(

d x 5 + arc tan dx

x

)

=

1 − x 2 sin x arc sin x

cos x +

1 − x 2 cos2 x d 5 d x + arc tan dx dx

x

−1

1 x 2 d 12 1 1 1 12 −1 ⋅ x = 5x 4 + ⋅ x x = 5x + = 5x 4 + = 5 x4 + 2 (1 + x ) 1 + x dx 1+ x 2 2 x (1 + x )

d ⋅ dx

4

dy dx

l. Given y = arc cot 3 x 2 then

1

+ sin x arc sin x

1− x 2

=

k. Given y = x 5 + arc tan x then

= −

cos x

− ( arc sin x ⋅ − sin x )

1

cos x ⋅

1

= −

1+

4 x3

⋅

=

d 3 arc cot x 2 dx

2 − 13 x 3

1

= −

1+

4 x3

⋅

2 d arc cot x 3 = − dx

=

2 3

3 x

= −

1 3

1+ x

4

⋅

1 4

1+ x 3

2

⋅

d 23 x dx

= −

3

3 x

2

(

3

3 x 1+ x 3 x

)

Example 3.2-2: Find the derivative of the following inverse trigonometric functions: a. y = arc cos

(

x2 a

)

d. y = tan −1 x 2 + 3 g. y =

3x 2 arc tan x

j. cos y = sin −1 x

b. y = arc cot

c. y = cot −1 x − tan x

e.

1 x 3 5 arc sin x y= + 5x 3 x

f. y = cos x + tan −1 x

h. y = sin x + sin −1 x

i. sin 3 y = arc tan 5 x

k. sin ( y + 1) = arc cos x 2

l. cot y = tan −1 x

Solutions: a. Given y = arc cos

= −

1 1−

4

⋅

x a2

1 3

2x a

x2 a

= −

b. Given y = arc cot

x 5

Hamilton Education Guides

dy dx

then

1 2

a −x a2

4

then

⋅

2x a

dy dx

d x2 arc cos dx a

=

= −

=

= −

1 −  

1 1 a

2

a −x

1

4

⋅

2x a

x d 1  arc cot  5 dx  3

=

= −

2 x2   a 

⋅

d x2 dx a

= −

2x a/

= −

a/ 2

a −x

4

⋅

1 d  x  arc cot  5 3 dx 

1 1 −  

=

2 x2   a 

⋅

2x a

2x 2

a − x4

1 −1 3 1+ x

(5 )

2

⋅

d x dx 5

160

Calculus I

=

3.2 Differentiation of Inverse Trigonometric Functions

1 −1 1 ⋅ 3 1 + x2 5

=

25

1 15

−1 25 + x 25

1 −25 15 25 + x 2

=

2

c. Given y = cot −1 x − tan x then

(

(

1 + x 4 + 9 + 6x 2

e. Given

=

2x

  x⋅  

)

1− x 2

dy dx

=

d tan −1 x 2 + 3 dx

then

=

then

( arc tan x ⋅ 6 x ) −  3x 2 ⋅

dy dx

)

d d cot −1 x − tan x dx dx

=

=

(

1

1+ x 2 + 3

=

d  arc sin x  + 5x 3   x dx  

x

+ 15 x

f. Given y = cos x + tan −1 x then



(

)

)

2

⋅

= −

(

d x2 + 3 dx

)

1 1+ x

=

(

2

− sec 2 x

2x

)

1+ x 2 + 3

2

x + 6 x 2 + 10

x2

g. Given

(

d cot −1 x − tan x dx

2x

  − ( arc sin x ⋅1 )  

3x 2 y= arc tan x

)

4

arc sin x y= + 5x 3 x

1

(

3 25 + x 2

=

)

=

5

dy dx

d. Given y = tan −1 x 2 + 3 then

=

= −

dy dx

1 1+ x 2

  

arc tan 2 x

h. Given y = sin x + sin −1 x then

2

dy dx

=

=

dy dx

=

=

1− x 2

(

d cos x + tan −1 x dx

  x 

=

1+ x

arc tan 2 x

)

=

1− x2

x − arc sin x x

2

1− x

d d tan −1 x cos x + dx dx

=

+ 15 x 2

x2

+ 15 x 2

2 6 x arc tan x − 3 x 2

=

=

− arc sin x − x2

d  3 x 2 dx  arc tan

( dxd arc sin x ⋅ x )− ( dxd x ⋅ arc sin x )

( arc tan x ⋅

d dx

2

+ 15 x 2

1

= − sin x +

)(

d arc tan x 3 x 2 − 3 x 2 ⋅ dx

1+ x2

)

arc tan 2 x

=

(

d sin x + sin −1 x dx

)

(

)

6 x 1 + x 2 arc tan x − 3 x 2

( 1 + x ) arc tan 2

=

2

x

d d sin x + sin −1 x dx dx

= cos x +

1 1− x2

i. Given sin 3 y = arc tan 5 x let’s take the derivative of both sides of the equation to obtain: d 1 d cos 3 y ⋅ 3 y = ⋅ 5x 2 dx dx 1 + (5 x )

Hamilton Education Guides

; cos 3 y ⋅ 3 y ′ =

1 1 + 25 x 2

⋅5

; 3 cos 3 y ⋅ y ′ =

5 1 + 25 x 2

; y′ =

5 1+ 25 x 2

3 cos 3 y

161

Calculus I

;

y′ =

3.2 Differentiation of Inverse Trigonometric Functions 5 1+ 25 x 2 3 cos 3 y 1

; y′ =

5 ⋅1

; y′ =

(1 + 25x ) ⋅ 3 cos 3 y 2

5

(

3 1 + 25 x 2

) cos 3 y

j. Given cos y = sin −1 x let’s take the derivative of both sides of the equation to obtain: 1

1 d d − sin y ⋅ y= ⋅ x dx 1 − x 2 dx

; y′ =

1 ⋅1 1 − x 2 ⋅ − sin y

; − sin y ⋅ y ′ =

1 1− x 2

⋅1

; − sin y ⋅ y ′ =

1 1− x 2

; y′ =

1− x 2

− sin y

1

; y′ =

1− x 2 −sin y 1

1

; y′ = −

1 − x 2 sin y

k. Given sin ( y + 1) = arc cos x 2 let’s take the derivative of both sides of the equation to obtain: cos ( y + 1) ⋅

−1

d ( y + 1) = dx

1− x

−2 x

; y′ =

1− x 4

cos ( y + 1)

4

⋅

d 2 x ; cos ( y + 1) ⋅ y ′ = dx

−1 1− x

4

⋅ 2 x ; cos ( y + 1) ⋅ y ′ =

−2 x 1− x 4

−2 x

; y′ =

1− x 4

−2 x ⋅1

; y′ =

cos ( y +1) 1

1 − x 4 ⋅ cos ( y + 1)

; y′ = −

2x 1 − x 4 cos ( y + 1)

l. Given cot y = tan −1 x let’s take the derivative of both sides of the equation to obtain: d d 1 − csc y ⋅ ⋅ y= x 2 dx 1 + x dx 2

; y′ =

1 ⋅1

(1 + x ) ⋅ − csc 2

2

y

;

2

− csc y ⋅ y ′ =

; y′ = −

1 1+ x 2

⋅1

;

2

− csc y ⋅ y ′ =

1 1+ x 2

; y′ =

1 1+ x 2 2

− csc y

;

y′ =

1 1+ x 2

−csc 2 y 1

1

(1 + x ) csc 2

2

y

Example 3.2-3: Find the derivative of the following inverse trigonometric functions: b. y = arc sin 2 x

c. y = sin −1 x 2

d. y = arc tan 2 x 3 + 1

e. y = arc cot 3 2 x

f. y = cot x ⋅ cot −1 x

g. y = arc csc 10 x

h. y = arc csc x10

i. y = arc csc10 x

a. y = x 3 sin −1 x

(

)

Solutions: a.

dy dx

=

d 3 x sin −1 x dx

= sin −1 x ⋅

Hamilton Education Guides

d d 3 x + x 3 ⋅ sin −1 x dx dx

= sin −1 x ⋅ 3x 2 + x 3 ⋅

1 1− x 2

= 3 x 2 sin −1 x +

x3 1− x2

162

Calculus I

3.2 Differentiation of Inverse Trigonometric Functions

b.

dy dx

=

d arc sin 2 x dx

c.

dy dx

=

d sin −1 x 2 dx

d.

dy dx

=

d arc tan 2 x 3 + 1 dx

[

d ( arc sin x )2 dx

=

1

=

1− x 4

)]

(

(

e.

dy dx

= f.

=

3 (arc cot 2 x )2 ⋅

dy dx

=

−1

2

= − csc x cot

−1

x−

)

cot x 1+ x 2

g.

dy dx

=

d ( arc csc 10 x ) dx

h.

dy dx

=

d arc csc x10 dx

i.

dy dx

=

d arc csc10 x dx

)=−

(

)=

10 ( arc csc x )9 ⋅ = 1 x

−1

Hamilton Education Guides

2arc sin x

=

1− x 2

1− x2

1 − x4

(

) dxd arc tan ( x + 1)

= 2 arc tan x 3 + 1 ⋅

)

= 2 arc tan x 3 + 1 ⋅

= 3 (arc cot 2 x )2 ⋅

(

1

3

)

1+ x3 +1

d arc cot 2 x dx

2

⋅ 3x 2

=

(

6 x 2 arc tan x 3 + 1

(

1+ x3 +1

= 3 (arc cot 2 x )2 ⋅

−1

1 + (2 x )

2

)

2

⋅

)

d 2x dx

1 + 4x 2 d d cot x + cot x ⋅ cot −1 x dx dx

(1 + x ) csc 2

−

1 10 x

(10 x )2 1

x

10

=

2

= cot −1 x ⋅ − csc 2 x + cot x ⋅

−1 1+ x 2

x cot −1 x + cot x

1+ x2

x

20

d 10 x − 1 dx ⋅

= −

1 100 x 2 − 1

1/ 0/ x

d 10 x = − dx x10 −1 ⋅

d ( arc csc x )10 dx

x 2 −1

1

− 6 ( arc cot 2 x )2

=

(

2

= 2arc sin x ⋅

2x

(

)

(

= cot −1 x ⋅

= −

)]

(

d x3 +1 2 dx

=

⋅2

1 + 4x 2

(

[

d (arc cot 2 x )3 dx

=

d cot x ⋅ cot −1 x dx

1− x 4

d ( arc sin x ) dx

⋅ 2x =

d arc tan x 3 + 1 dx

⋅

3

d arc cot 3 2 x dx

1

d 2 x = dx

⋅

=

) 1 + ( x 1 + 1)

= 2 arc tan x 3 + 1 ⋅

= 2arc sin x ⋅

1 x

= 10 ( arc csc x )9 ⋅

20

−1

⋅ 1/ 0/

= −

⋅10 x 9 = −

d arc csc x dx

1 x

100 x 2 − 1

10 x

x 20 − 1

= 10 ( arc csc x )9 ⋅

−1 x

x 2 −1

− 10 ( arc csc x ) 9 x

x2 −1

163

Calculus I

3.2 Differentiation of Inverse Trigonometric Functions

Example 3.2-4: Given that the inverse sine function is given by y = sin −1 x ⇔ x = sin y where −

π 2

≤ y≤

π 2

and −1 ≤ x ≤ 1 , show that

Solution: Given sin y = x then

1 d sin −1 x = dx 1− x 2

d d sin y = x dx dx

Since y = sin −1 x we can state that

; cos y ⋅

d y =1 dx

dy 1 = dx cos y

;

. dy 1 = dx cos y

d 1 sin −1 x = dx cos y

;

To express cos y in terms of x we can use the relation sin 2 y + cos 2 y = 1 ; cos 2 y = 1 − sin 2 y ; cos 2 y = 1 − x 2 . Therefore, cos y = 1 − x 2 and we have shown that

d 1 sin −1 x = = cos y dx

1 1 − x2

Example 3.2-5: Given that the inverse sine function is given by y = cos −1 x ⇔ x = cos y where 0≤ y ≤π

and −1 ≤ x ≤ 1 , show that

Solution: Given cos y = x then

d −1 cos −1 x = dx 1− x 2

d d x cos y = dx dx

Since y = cos −1 x we can state that

.

d y =1 dx

; − sin y ⋅ dy −1 = dx sin y

;

;

dy 1 −1 = = dx − sin y sin y

d −1 cos −1 x = dx sin y

To express sin y in terms of x we can use the relation sin 2 y + cos 2 y = 1 ; sin 2 y = 1 − cos 2 y ; sin 2 y = 1 − x 2 . Therefore, sin y = 1 − x 2 and we have shown that

d −1 = cos−1 x = dx sin y

−1 1 − x2

Example 3.2-6: Given that the inverse tangent function is given by y = tan −1 x ⇔ x = tan y where −

π

y

2

π 2

for all x , show that

d 1 tan −1 x = dx 1+ x 2

.

Solution: Given tan y = x then

d d tan y = x dx dx

Since y = tan −1 x we can state that

; sec 2 y ⋅

d y =1 dx

dy 1 = dx sec 2 y

;

;

dy 1 = dx sec 2 y

1 d tan −1 x = dx sec 2 y

To express sec 2 y in terms of x we can use the relation sec 2 y − tan 2 y = 1 ; sec 2 y = 1 + tan 2 y ; sec 2 y = 1 + x 2 . Therefore, we have shown that

1 1 d tan −1 x = = 2 dx sec y 1 + x 2

Example 3.2-7: Given that the inverse tangent function is given by y = cot −1 x ⇔ x = cot y where 0

y

π for all x , show that

Hamilton Education Guides

d −1 cot −1 x = dx 1+ x 2

.

164

Calculus I

3.2 Differentiation of Inverse Trigonometric Functions

Solution: d d cot y = x dx dx

; − csc 2 y ⋅

Since y = cot −1 x we can state that

dy −1 = dx csc 2 y

Given cot y = x then

d y =1 dx

;

;

dy 1 −1 = = 2 dx − csc y csc 2 y

d −1 cot −1 x = dx csc 2 y

To express csc 2 y in terms of x we can use the relation csc 2 y − cot 2 y = 1 ; csc 2 y = 1 + cot 2 y ; csc 2 y = 1 + x 2 . Therefore, we have shown that

1 d −1 cot −1 x = =− 2 dx 1+ x2 csc y

Note: Similar steps can be taken to show that: 1.

d 1 sec −1 x = dx x x 2 −1 0≤ y

2.

π 2

where the inverse secant function is given by y = sec −1 x ⇔ x = sec y for 3π 2

or π ≤ y

d −1 csc −1 x = dx x x 2 −1 0

y

π 2

or π

y≤

and x ≥ 1 or x ≤ −1 and

where the inverse cosecant function is given by y = csc −1 x ⇔ x = csc y for 3π 2

and x ≥ 1 or x ≤ −1 .

Section 3.2 Practice Problems – Differentiation of Inverse Trigonometric Functions 1. Find the derivative of the following inverse trigonometric functions: a. y = sin −1 3x

(

d. y = arc sin x 3 + 2 g. y =

b. y = x 3 + arc cos 5 x

)

tan −1 x x

e. y = arc cot h. y =

1 x

3

x3 − arc sin x x+5

c. y = x 5 arc tan x 4 f. y =

arc sin 3 x x2

i. y = x + cos −1 x

2. Find the derivative of the following inverse trigonometric functions: a. y = x 2 + arc sin ax

b. y = cos −1 6 x

c. y = x + arc sin x 3

d. y = arc tan 2 x

e. y = x − arc tan x 2

f. y = x 2 − arc sin x

g. y = tan −1 x 3

h. y = tan −1 5 x

i. y = arc sin 5 x 2

Hamilton Education Guides

165

Calculus I

3.3

3.3 Differentiation of Logarithmic and Exponential Functions

Differentiation of Logarithmic and Exponential Functions

When differentiating logarithmic functions the following two rules should be kept in mind: 1. The derivative of a logarithmic function to the base e , i.e., log e x = ln x is equal to: dy d ( log e x ) = 1 ⋅ log e e = 1 ⋅1 = 1 = dx dx x x x

or,

dy d ( ln x ) = 1 = x dx dx

(1 )

2. The derivative of a logarithmic function to the base other than e , i.e., log a x is equal to: dy d ( log a x ) = 1 log a x = dx dx x

(2)

Let’s differentiate some logarithmic functions using the above differentiation formulas. Example 3.3-1: Find the derivative of the following logarithmic functions: b. y = ln x 3

a. y = ln 5 x

(

)

c. y = 5 ln 3x 2

)

(

d. y = ln x 2 + x + 3

e. y = ln x 2 + x + 1

g. y = x ln x − x

h. y = x 2 ln x − x 3

j. y =

ln x 2 x

f. y = x 2 ln (x + 1)

4

i. y = ln ( ln x + 1)

x +1   x  −3

k. y = ln 

3

l. y = ln (sin 3x + cos 5 x )

Solutions: a. Given y = ln 5 x then

dy dx

=

d ( ln 5 x ) dx

b. Given y = ln x 3 then

dy dx

=

d ln x 3 dx

dy dx

c. Given y = 5 ln 3x 2 then

(

)

(

)

d. Given y = ln x 2 + x + 3 e. Given y = ln x 2 + x + 1

=

(x

1 2

)

+ x +1

4

[(

4

(

=

then then

)

)

1

=

x

(

d 5 ln 3 x 2 dx dy dx

dy dx

3

)

⋅

=

]

1 ⋅5 5x

=

( )

d x3 dx

= 5⋅

1 3x

1

=

x

)]

[ (

)]

d ln x 2 + x + 1 dx

(

)

4 x 2 + x +1

(x

2

5/ 5/ x

⋅ 3x 2

( )

[ (

=

3

=

d 3x 2 dx

⋅

2

d ln x 2 + x + 3 dx

=

⋅ 4 x 2 + x + 1 3⋅ (2 x + 1)

Hamilton Education Guides

1 d ⋅ ( 5x) 5 x dx

=

3/

)

x

3x

2

3 x

=

3/ =1

5

=

⋅ 6x

30 x/

=

3x

(

2/ =1

=

)

=

d x2 + x + 3 x + x + 3 dx ⋅

2

=

(x

⋅ (2 x + 1)

+ x +1

3 x 2/

=

1

= 4

1 x

=

4/ =1

1 2

)

+ x +1

=

4

⋅

x + x +1

10 x

2x + 1 x + x+3

)

=

=

2

d x 2 + x +1 dx

4 (2 x + 1) 2

(

3/ 0/ 3/ x

4

8x + 4 x2 + x +1

166

Calculus I

3.3 Differentiation of Logarithmic and Exponential Functions

dy dx

f. Given y = x 2 ln (x + 1) then

 = [ ln (x + 1) ⋅ 2 x] +  x 2 ⋅ 

d 1  ⋅ (x + 1) x + 1 dx  dy dx

g. Given y = x ln x − x then 1 d (x ) + ln x ⋅1 − 1 x dx 

 = x ⋅ ⋅ 

( )

d x3 dx

( )

d d    x 2  +  x 2 ⋅ [ ln (x + 1) ] dx dx   

 = ln ( x + 1) ⋅ 

 = [ ln (x + 1) ⋅ 2 x] +  x 2 ⋅ 

d (x ln x − x ) dx

=

]

1  ⋅1 x +1 

= 2 x ln ( x + 1) +

d (x ln x ) − d (x ) dx dx

=

 = x ⋅ 

x2 x +1

d (ln x ) + ln x d (x ) − d (x ) dx dx  dx

 x =  / + ln x  − 1 = 1 + ln x − 1 = ln x  x/



dy dx

h. Given y = x 2 ln x − x 3 then

−

[

d x 2 ln (x + 1) dx

=

=

(

d x 2 ln x − x 3 dx

=

(

)

( )

d d x3 x 2 ln x − dx dx

 = ln x ⋅ 

( )

d d  x 2 + x 2 ⋅ (ln x ) dx dx 

x 2/ =1 



1

1 d (x ) − 3x 2 x dx 

  2 = 2 x ln x + x 2 ⋅ ⋅1 − 3x 2 = 2 x ln x +  − 3x x  x/   

 = ln x ⋅ 2 x + x 2 ⋅ ⋅ 

)

= 2 x ln x + x − 3x 2 = − 3 x 2 + x ( 2 ln x + 1) i. Given y = ln ( ln x + 1) then

=

1 1  ⋅ ⋅ 1 + 0 ln x + 1  x 

j. Given y = 2 x 4/ = 2

=

x

2/

1 x +1 x −3

x3

then

(

− 3 x 2 ln x 2

)

x6

 x +1  y = ln    x−3

k. Given

=

ln x 2

=

⋅

dy dx

1 1 ⋅ ln x + 1 x

dy dx

=

then

=

d  ln x 2 dx  x 3

   

(

x6

=

1 d  d (1) ⋅  (ln x ) + ln x + 1  dx dx 

=

1  1 d ⋅  ⋅ ( x ) + 0 ln x + 1  x dx 

x ( ln x + 1)

2 x 2 − 3 x 2 ln x 2

dy dx

=

1

=

[ (x − 3)⋅1 ] − [ (x + 1)⋅1 ] = (x − 3)2

Hamilton Education Guides

1 d ⋅ (ln x + 1) ln x + 1 dx

=

=

)

[x

⋅

) ] − [ ln x

(

d ln x 2 ⋅ dx

2

( )]

d x3 ⋅ dx

x6

=

d  x +1  ln   dx  x − 3 

1 1 x +1 x −3

3

(

x 2 2 − 3 ln x 2 x 6/ = 4

=

1 x +1 x −3

(x − 3) − (x + 1) ( x − 3)

2

)

=

d  x +1  ⋅   dx  x − 3 

=

=

=

[

 x 3 ⋅ 1 ⋅ 2 x  − ln x 2 ⋅ 3 x 2   x2

]

x6

2 − 3 ln x 2 x4

1 x +1 x −3

⋅

[ (x − 3)⋅ dxd (x + 1) ] − [ (x + 1)⋅ dxd (x − 3) ]

1 ⋅ (x − 3) (x − 3) − (x + 1) ⋅ (x + 1)⋅1 (x − 3)2

( x − 3) 2 =

x − 3 x/ − 3 − x/ − 1 ⋅ x + 1 (x − 3)2 167

Calculus I

=

3.3 Differentiation of Logarithmic and Exponential Functions

(x − 3) ⋅ x +1

−4

(x − 3)2/ =1

= −

4

( x + 1) ( x − 3)

l. Given y = ln (sin 3x + cos 5 x ) then

+

d (cos 5 x ) ] dx

=

dy dx

1 d ⋅ (sin 3 x + cos 5 x ) sin 3 x + cos 5 x dx

=

1 ⋅ (3 cos 3 x − 5 sin 5 x ) sin 3 x + cos 5 x

=

=

1 d ⋅[ sin 3 x sin 3 x + cos 5 x dx

3 cos 3 x − 5 sin 5 x sin 3 x + cos 5 x

Example 3.3-2: Find the derivative of the following logarithmic expressions:

(

)

b. y = sin ( ln 3x ) − cos ( ln 5 x )

a. y = x 3 + 1 ⋅ ln x 2

(

)

ln x 3

d. y = ln x 3 − 1 − ln (sec x )

e. y =

g. y = sin ( ln x ) + x

h. y = sin ( ln x ) + cos ( ln x )

j. y = ln x 3 + ln (csc x )

k. y = ln 

Solutions:

(

)

( )(

)

 1 d  + ⋅ x 2 ⋅ x3 +1  2 dx x 

=

(

dy dx

1 3

i. y = x 5 ( ln x + 3)

(

2

dy dx

d d ln 3 x + sin ( ln 5 x ) ⋅ ln 5 x dx dx

= cos ( ln 3x ) ⋅

1 1 ⋅ 3 + sin ( ln 5 x ) ⋅ ⋅ 5 3x 5x dy dx



Hamilton Education Guides

(



)(

)

)

= d

=

2

= 3 x ln x +

(

2 x3 +1

d [ sin ( ln 3x ) − cos ( ln 5 x ) ] dx

= cos ( ln 3x ) ⋅

)

x

=

d d sin ( ln 3 x ) − cos ( ln 5 x ) dx dx

1 d (3x ) + sin ( ln 5 x )⋅ 1 ⋅ d (5 x ) ⋅ 5 x dx 3 x dx

3/ cos ( ln 3 x ) 5/ sin ( ln 5 x ) + 3/ x 5/ x

( )

2

d

= d

cos ( ln 3 x ) sin ( ln 5 x ) + x x 1 d

 ( ln x )⋅ x 2  − (x ) = 2 x ⋅ ln x + ⋅ (x )⋅ x 2  − 1 x 2 ⋅ ln x + =  dx x dx    dx  dx

  = 2 x ⋅ ln x + ⋅1 ⋅ x 2  − 1 = 2 x ln x + 

)

  1 3 x ln x +  ⋅ 2 x/ ⋅ x 3 + 1  2/ =1 x  2

= cos ( ln 3x ) ⋅

c. Given y = x 2 ln x − x then

l. y = ln (sec x + csc x )

  d d ln x 2 ⋅ x 3 + 1  = 3 x 2 ln x 2 x 3 + 1 ⋅ ln x 2  +  =     dx  dx

b. Given y = sin ( ln 3x ) − cos ( ln 5 x ) then



f. y = ln (sec 5 x + tan 5 x )

+x

x5

 x2    x + 1  

a. Given y = x 3 + 1 ⋅ ln x 2 then

1 x

c. y = x 2 ln x − x

x 2/ =1   −1 x/ 

= 2 x ln x + x − 1

168

Calculus I

3.3 Differentiation of Logarithmic and Exponential Functions

(

)

dy dx

d. Given y = ln x 3 − 1 − ln (sec x ) then

=

⋅

e. Given y =

=

=

(

)

d d 1 x 3 −1 − ⋅ (sec x ) 3 sec x dx x − 1 dx 1

ln x 3 x5

dy dx

then

+x

=

(

)

 x 5 ⋅ 1 ⋅ d x 3  − ln x 3 ⋅ 5 x 4   x 3 dx   x10

( 3 x )− ( 5 x 4

4

ln x 3

)

x10

+1

1

=

x 3 −1

⋅ 3x 2 −

 d  ln x 3 + x  dx  x 5 

+1

)

=

1 ⋅ sec x tan x sec x

d  ln x 3 dx  x 5

 + d x  dx 

(

 x 5 ⋅ 1 ⋅ 3 x 2  − 5 x 4 ln x 3   x3  

=

x10

(

x 4/ 3 − 5 ln x 3

=

[ (

d ln x 3 − 1 − ln (sec x ) dx

=

)

x 1/ 0/ = 6

+1

=

3 − 5 ln x 3 x6

)

+1

]

(

3 x2

=

x3 − 1

(x

=

)

d d ln x 3 − 1 − ln (sec x ) dx dx

=

5

− tan x

)(

d ln x 3 − ln x 3 ⋅ d x 5 ⋅ dx dx

x10

(

 3 x 7/ = 4  − 5 x 4 ln x 3   x 3/  

=

x10

1 d ⋅ [ (sec 5 x + tan 5 x ) ] sec 5 x + tan 5 x dx

=

d [ ln (sec 5 x + tan 5 x ) ] dx

=

d 1  d ⋅  sec 5 x + tan 5 x  dx sec 5 x + tan 5 x  dx 

=

d d 1   ⋅  sec 5 x tan 5 x ⋅ 5 x + sec 2 5 x ⋅ 5 x  dx  dx sec 5 x + tan 5 x 

=

sec 5 x tan 5 x ⋅ 5 + sec 2 5 x ⋅ 5 sec 5 x + tan 5 x

=

g. Given y = sin (ln x ) + x then 1 d x +1 x dx

= cos (ln x ) ⋅ ⋅

5 sec 5 x (tan 5 x + sec 5 x )

= 1 x

= cos (ln x ) ⋅ ⋅1 + 1 =

h. Given y = sin (ln x ) + cos (ln x ) then =

d d sin (ln x ) + cos (ln x ) dx dx 1 x

1 x

i. Given y = x 5 (ln x + 3) then Hamilton Education Guides

dy dx

= cos (ln x ) ⋅

= cos (ln x ) ⋅ ⋅1 − sin (ln x ) ⋅ ⋅1 = 1 3

d [ sin (ln x ) + x] dx

dy dx

=

=

d d sin (ln x ) + x dx dx

= cos (ln x ) ⋅

d (ln x ) + 1 dx

cos ( ln x ) +1 x d [ sin (ln x ) + cos (ln x )] dx

d (ln x ) − sin (ln x )⋅ d (ln x ) dx dx

cos (ln x ) sin (ln x ) − x x

=

=

= 5 sec 5 x

(sec 5 x + tan 5 x )

dy dx

+1

+1

dy dx

f. Given y = ln (sec 5 x + tan 5 x ) then

)

) +1

=

=

d d sin (ln x ) + cos (ln x ) dx dx 1 d 1 d x x − sin (ln x ) ⋅ ⋅ x dx x dx

= cos (ln x ) ⋅ ⋅

1 [ cos ( ln x ) − sin ( ln x ) ] x

1d 5 d (ln x + 3)⋅ x 5  x ⋅ (ln x + 3) + 3  dx dx 

=

1 4 d  5 d 5 x (ln x + 3) +  ln x + 3  x  3 dx dx    169

Calculus I

3.3 Differentiation of Logarithmic and Exponential Functions

=

1 4 1 d  5 5 x ( ln x + 3) +  ⋅ (x ) + 0  ⋅ x  3  x dx  

=

1 5 x 4 ( ln x + 3) + x 4 3

[

]

=

1 4 1  5 5 x ( ln x + 3) +  ⋅1 ⋅ x  3 x  

=

=

5 4 x4 x ( ln x + 3) + 3 3

=

5 x 4 ln x 15 x 4 x 4 + + 3 3 3

Given y = x 5 ( ln x + 3) = x 5 ln x + x 5 then

dy dx

=

=

1 3

1 3

=

1 4 1 d  5 x ⋅ (ln x ) + ⋅ (x ) ⋅ x 5  + 5 x 4  x dx 3 

=

1 5 x 4 ln x + x 4 + 5 x 4 3

=

5 x 4 ln x x 4 + 15 x 4 + 3 3

[

]

5 x 4 ln x x 4 + + 5x 4 3 3

=

=

1 x

3

⋅

k. Given

=

( )

d 3 1 d x + ⋅ (csc x ) dx csc x dx

 x2   y = ln     x +1

1 ⋅ (x + 1) x 2 ⋅1

⋅

dy dx

1 4 x 5/ = 4  4 5 x ln x +  + 5x 3  x/ 

=

5 x 4 ln x x 4 5 x 4 + + 3 3 1

then

dy dx

x

=

[

d ln x 3 + ln (csc x ) dx

= 1

=

=

⋅ 3x 2 +

3

x2 x +1

]

1 ⋅ − csc x cot x csc x

d  x 2  ⋅  dx  x + 1 

1

dy dx

=

1 ⋅ (sec x tan x − csc x cot x ) sec x + csc x

Hamilton Education Guides

)(

(

)

x 4 ⋅1 + 5 x 4 ⋅ 3 5 x 4 ln x + 3 3 ⋅1

=

1 1

x2 x +1

⋅

=

=

d d ln x 3 + ln (csc x ) dx dx

3 x 2/ x

−

3/ = 2

csc x cot x csc x

[ (x + 1)⋅ (x ) ]− [x d dx

2

2

=

d [ ln (sec x + csc x ) ] dx

=

d ( x + 1) ⋅ dx

3 − cot x x

]

(x + 1)2

[ (x + 1)⋅ 2 x ] − [ x 2 ⋅1 ] (x + 1) ⋅ 2 x 2 + 2 x − x 2 (x/ + 1/ ) ⋅ x 2 + 2 x = = = x2 x 2 (x + 1)2/ =1 (x + 1)2 (x + 1)2

l. Given y = ln (sec x + csc x ) then

=

or,

5 x 4 ln x 16 x 4 + 3 3

j. Given y = ln x 3 + ln (csc x ) then

=

=

5 x 4 ln x 16 x 4 + 3 3

d 1d 5 (ln x )⋅ x 5  + d x 5 x ⋅ ( ln x ) +  dx 3  dx  dx

1 4 1  5 x ⋅ ( ln x ) + ⋅1 ⋅ x 5  + 5 x 4  x 3 

=

1 4 x 5/ = 4  5 x ( ln x + 3) +  3  x/ 

=

x/ (x + 2 )

x 2/ =1 (x + 1)

=

x+2

x ( x + 1)

1 d ⋅ (sec x + csc x ) sec x + csc x dx

sec x tan x − csc x cot x sec x + csc x

170

Calculus I

3.3 Differentiation of Logarithmic and Exponential Functions

Example 3.3-3: Find the derivative of the following logarithmic functions: a. y =

ln (sec 5 x ) x

b. y = ln csc x 2

2

(

d. y = tan ln x 2

)

c. y = ln

(

e. y = cot ln 3x 5

)

f. y = tan (ln x ) − cot (ln x )

(

g. y = sin 5 x + ln (sin 5 x )

h. y = ln x 2 + 1

j. y = ln 3 x − 1

k. y = ln 1 + x 2 + 2 arc tan x

(

Solutions: a.

dy dx

=

= b.

dy dx

=

c.

dy dx

×

d.

dy dx

=  

d ln sec 5 x dx x2

x2 sec 5 x

2

(

1 c/ s/ c/ x 2/

=

)

=

=

1 csc x

2

[ (

d tan ln x 2 dx

(

2

)x

=

[ (

2 x/ 2/ =1

⋅

d csc x 2 dx

=

)]

Hamilton Education Guides

 x 2 ⋅ 1 ⋅ d (sec 5 x )  − [ ln (sec 5 x ) ⋅ 2 x ] sec 5 x dx  

=

x4

)

⋅ tan 5 x ⋅ 5 − 2 x ln sec 5 x x

=

4

( 5x

2

)

tan 5 x − 2 x ln sec 5 x x4

1

d 2 x ⋅ − csc x 2 ⋅ cot x 2 ⋅ dx csc x

=

2

1 csc x

⋅ − csc x 2 ⋅ cot x 2 ⋅ 2 x

2

= − 2 x cot x 2

d  x +1   ⋅  x +1 dx  sin x    1

=

1 1 x +1 sin x

[⋅ sin x ⋅ dxd (x + 1) ]− [ (x + 1)⋅ dxd sin x] (sin x )

s/ i/n/ x/ sin x − (x + 1) cos x ⋅ x +1 (sin x )2/ =1

) ] = sec ( ln x )⋅ dxd ln x

d cot ln 3 x 5 dx

2

l. y = ln (arc tan 3x ) + 3x 2

x3

sin x

=

(x

)

5 x tan 5 x − 2 ln (sec 5 x )

=

[ sin x ⋅1 ] − [ (x + 1)⋅ cos x] = (sin x )2 =

=

4

⋅ − c/ s/ c/ x 2/ ⋅ cot x 2 ⋅ 2 x

d  x +1   ln  dx  sin x 

]

x4

x 4/ = 3 d ln csc x 2 dx

)

][

x/ ( 5 x tan 5 x − 2 ln sec 5 x )

=

i. y = x ⋅ ln x 2 + 3

d ln sec 5 x − ln sec 5 x ⋅ d x 2 ⋅ dx dx

d 5 x  − 2 x ln sec 5 x ⋅ sec 5 x ⋅ tan 5 x ⋅ dx 

= sec ln x ⋅ dy dx

=

x

2

e.

[x

x +1 sin x

2

2

(

)

2 sec 2 ln x 2

2

(

=

2

1 sin x − (x + 1) cos x ⋅ x +1 sin x

) x1 ⋅ dxd x

= sec 2 ln x 2 ⋅

2

2

=

1 ⋅ sin x (x + 1)⋅1

=

sin x − ( x + 1) cos x ( x + 1) sin x

(

) x1 ⋅ 2 x

= sec 2 ln x 2 ⋅

2

x

(

) dxd ( ln 3x )

= − csc 2 ln 3x 5 ⋅

5

(

) 3x1

= − csc 2 ln 3x 5 ⋅

5

⋅

( )

d 3x 5 dx

171

Calculus I

3.3 Differentiation of Logarithmic and Exponential Functions

2

(

5

) 3x

= − csc ln 3x ⋅ f.

dy dx

⋅15 x

5

4

= −

d [ tan (ln x ) − cot (ln x ) ] dx

=

(

5

1

1/ 5/ x 4/ csc 2 ln 3x 5 3/ x 5/ = 1

1 d 1 d x + csc 2 (ln x ) ⋅ ⋅ x x dx x dx

dy dx

d [ sin 5 x + ln (sin 5 x ) ] dx

=

= cos 5 x ⋅ 5 +

h.

dy dx

=

i.

(

⋅

)

x

( x + 1) 2

dy dx

(

)1



(

) + 2x ⋅ x

(

1 −1 

= 1

= ln x 2 + 3 ⋅ x 2  +  x 2 ⋅ 2 

=

j.

ln x 2 + 3

dy dx

=

= 1 3

 

1 2

x +3

(

d ln 3 x − 1 dx

⋅

1

x − 1 3(x − 1)

2 3

)

=

=

Hamilton Education Guides

)

(

)

x 2 +1



(

1   sin 5 x 

)

1 −1 d 1 2 x + 1 2 ⋅ (2 x + 1) dx x 2 −1 2

1

1

⋅

= 5 cos 5 x 1 +

( x + 1) 2

1 2

⋅

=

x

1

⋅

( x + 1) ( x + 1) 1 2

2

2

1 2

⋅

)

2 x

+

2x

=

1

=

(x − 1)

1 3

⋅

1 3(x − 1)

2 3

)

1

d

 

1

(

d

)



(

 

)1

= ln x 2 + 3 ⋅ x 2 

−1

2



  1  1 ⋅ 2x   + x 2 ⋅ 2 x +3   

x

2

x +3

d 3 x −1 dx x −1 ⋅

(



= ln x 2 + 3 ⋅ x 2  +  x 2 ⋅ ln x 2 + 3  dx dx 

 d x2 + 3  2 x + 3 dx  1

(

1

3

(

 d  12 2  x ⋅ ln x + 3  dx  

ln x 2 + 3

==

2

1 2x 2

)

x



5 cos 5 x sin 5 x

x +1

)]



(

⋅

)

(

⋅

= 5 cos 5 x +

1 d x 2 +1 2 = x 2 + 1 dx

1

=

]

2

x ⋅ ln x 2 + 3

[

d dx

=

[

1 sec 2 ( ln x ) + csc 2 ( ln x ) x

d 1 d 5x + ⋅ (sin 5 x ) dx sin 5 x dx

= cos 5 x

−1 2/ x 2 x +1 2 = x 2 + 1 2/

1

d (ln x ) + csc 2 (ln x ) d (ln x ) dx dx

1 x

1 ⋅ cos 5 x ⋅ 5 sin 5 x

x 2 +1

⋅

=

= sec 2 (ln x ) ⋅

x

=

1+ 1 2 2

x

1 x

d x 2 + 1 dx

1 −1 1 2 x + 1 2 1 ⋅ 2x x2 + 1 2

1

=

)

= sec 2 (ln x ) ⋅ ⋅1 + csc 2 (ln x ) ⋅ ⋅1 =

= 5 cos 5 x +

1

d   2  ln x +1  = dx  

=

(

5 csc 2 ln 3 x 5

d d sin 5 x + ln (sin 5 x ) dx dx

=

1 d ⋅ cos 5 x ⋅ 5 x sin 5 x dx

= −

d d tan (ln x ) − cot (ln x ) dx dx

=

= sec 2 (ln x ) ⋅ ⋅ g.

)

1

3

1 d (x − 1) 3 dx x −1

⋅

1 3 (x − 1)

1+ 2 3 3

=

=

1

3

1 1 (x − 1) 3 −1 3 x −1

1 3 (x − 1)

1+ 2 3

⋅

=

1 3 (x − 1)

3/ 3/

= =

1

3

2 1 (x − 1)− 3 3 x −1

⋅

1 3 ( x − 1)

172

Calculus I

k.

dy dx

= l.

dy dx

=

=

3.3 Differentiation of Logarithmic and Exponential Functions

[ (

1 1+ x 2

=

]=

)

d ln 1 + x 2 + 2 arc tan x dx

⋅ 2x + 2 ⋅

1 1+ x 2

⋅1

=

2x

+

1+ x 2

[

d ln (arc tan 3 x ) + 3 x 2 dx

]

)

(

d d (2 arc tan x ) ln 1 + x 2 + dx dx

2 1+ x 2

=

2x + 2 1+ x 2

=

1 1 d ⋅ ⋅ 3x + 6 x 2 arc tan 3 x 1 + (3 x ) dx

=

1 1+ x

2

⋅

(

)

d d 1 x ⋅ 1+ x 2 + 2 ⋅ 2 dx 1 + x dx

2 (1 + x ) 1+ x2

( )

=

1 1 ⋅ ⋅ 3 + 6x arc tan 3 x 1 + 9 x 2

=

d d ln (arc tan 3 x ) + 3x 2 dx dx

=

=

1 d ⋅ (arc tan 3 x ) + 6 x arc tan 3 x dx 3

( 1 + 9 x ) arc tan 3 x 2

+ 6x

When differentiating exponential functions the following two rules should be kept in mind: dy du = eu ⋅ dx dx

Given y = e u , then Given y = a u , then

dy du = a u ⋅ ln a ⋅ dx dx

Let’s differentiate some exponential functions using the above differentiation formulas. Example 3.3-4: Find the derivative of the following exponential functions: b. y = x e 2 x

c. y = x 2 e 5 x

e. y = e sin 5 x

f. y = e 5 x sin 2 x

g. y = e 2 x cos 3x

h. y = e ln x tan x

i. y = e x + arc sin x

j. y = e 5 x arc cos x

k. y = e e

a. y = e10 x d. y = (x + 8) e 3 x

2

x

l. y = e cos 5 x e −3 x

Solutions: a. Given y = e10 x then b. Given y = x e 2 x then

dy dx dy dx

= =

(

d e10 x dx

(

)

d x e 2x dx

= e10 x ⋅

)

d ( 10 x ) dx

= e 2x ⋅

= e10 x ⋅10 = 10 e 10 x

( )

d (x ) + x ⋅ d e 2 x dx dx

= e 2 x ⋅1 + x ⋅ e 2 x ⋅

d 2x dx

= e 2x + x ⋅ e 2x ⋅ 2 = e 2x + 2x e 2x = e 2 x ( 1 + 2 x) c. Given y = x 2 e 5 x then

dy dx

=

(

d x 2 e 5x dx

)

= e 5x ⋅

( )

( )

d d x2 + x2 ⋅ e 5x dx dx

= e 5x ⋅ 2x + x 2 ⋅ e 5x ⋅

d 5x dx

= e 5 x ⋅ 2 x + x 2 ⋅ e 5 x ⋅ 5 = 2 x e 5 x + 5 x 2 e 5 x = x e 5 x (2 + 5 x )

Hamilton Education Guides

173

Calculus I

3.3 Differentiation of Logarithmic and Exponential Functions

dy dx

2

d. Given y = (x + 8) e 3 x then

2 2 2 d  (x + 8) e 3x  = e 3x ⋅ d (x + 8) + (x + 8)⋅ d  e 3x  dx  dx dx     

=

  

(

2 2 2 d 2 2 2   = e 3 x ⋅1 + (x + 8) ⋅ e 3 x ⋅ 3x 2  = e 3 x + (x + 8) ⋅ e 3 x ⋅ 6 x = e 3 x [ 1 + 6 x (x + 8)] = e 3 x 6 x 2 + 48 x + 1 dx    

(

dy dx

=

d e sin 5 x dx

f. Given y = e 5 x sin 2 x then

dy dx

=

e. Given y = e sin 5 x then

 = sin 2 x ⋅ e 5 x ⋅ 





dy dx

=

d x ⋅ + dx 2

d ⋅ x 1 − x 2 dx

=

j. Given y = e 5 x arc cos x then



=  arc cos x ⋅ e 5 x ⋅ 

dy dx

x e2

dy dx

 d   5x  +  e 5x ⋅ − dx   

Hamilton Education Guides

=

=

)

 = cos 3x ⋅ 

[

)

 =  tan x ⋅ 

ln x

1 1− x 2

⋅1

(

1− x 2

( )

d d    e 2 x  + e 2 x ⋅ (cos 3 x ) dx dx   

]

=

d dx

  x  

(

) + e

ln x

d ( tan x ) dx 

⋅

]=e

ln x  tan

 

x

x

 + sec 2 x  

d 2x d d d x  = e + arc sin x arc sin x = ex + dx dx dx dx 

e2 + 2

)

(

d e ln x dx

[

x

d e 5 x arc cos x dx

⋅

]

1 ⋅  + e ln x ⋅ sec 2 x x

d  x  e + arc sin dx 

1

d (sin 2 x ) dx 

][

] = tan x ⋅ e

1 ⋅ + 2

⋅

= cos 3x ⋅ e 2 x ⋅ 2 + e 2 x ⋅ − sin 3x ⋅ 3 = e 2 x (2 cos 3 x − 3 sin 3 x )

(

[



5x

][

d e ln x tan x dx

=

( ) + e

d e 5x dx

 = sin 2 x ⋅

[

= e sin 5 x ⋅ 5 cos 5 x = 5 cos 5 x e sin 5 x

= sin 2 x ⋅ e 5 x ⋅ 5 + e 5 x ⋅ cos 2 x ⋅ 2 = e 5 x ( 5 sin 2 x + 2 cos 2 x )

(

d (ln x ) + e ln x ⋅ sec 2 x dx 

1

)

d e 2 x cos 3 x dx

=

i. Given y = e x + arc sin x then

x e2

(

d e 5 x sin 2 x dx

d d    2 x + e 2 x ⋅ − sin 3 x ⋅ 3 x  dx  dx  

h. Given y = e ln x tan x then  =  tan x ⋅ e ln x ⋅

dy dx

d (sin 5 x ) dx

= e sin 5 x ⋅

d d    5 x + e 5 x ⋅ cos 2 x ⋅ 2 x  dx   dx 

g. Given y = e 2 x cos 3x then  = cos 3x ⋅ e 2 x ⋅

)

)

1 1− x2

 = arc cos x ⋅ 

( ) + e

d e 5x dx

)

 

= arc cos x ⋅ e 5 x ⋅ 5 +  e 5 x ⋅ −  

5x

⋅

d ( arc cos x ) dx 

  ⋅1 1 − x 2  1

174

Calculus I

3.3 Differentiation of Logarithmic and Exponential Functions

= arc cos x ⋅ 5e 5 x − e 5 x ⋅

dy dx

x

k. Given y = e e then

1 1− x 2

=

l. Given y = e cos 5 x e −3 x then  = e −3 x ⋅ e cos 5 x ⋅ 

d ex e dx

dy dx

=

x

= ee ⋅

d x e dx

(

1− x 2

[

][



= ee ⋅ e x ⋅

d x dx

) = e

d e cos 5 x dx

d ( cos 5 x ) + e cos 5 x ⋅ e −3x ⋅ d (− 3x ) dx dx   

1

 = e 5 x  5arc cos x − 

x

d e cos 5 x e −3 x dx

   1 − x 2 



e 5x

= 5e 5 x arc cos x −

−3 x

⋅

x

x

= e e ⋅ e x ⋅1 = e e e x = e e

(

) + e

cos 5 x

 = e −3 x ⋅ e cos 5 x ⋅ − sin 5 x ⋅ 

⋅

(

x

+x

)

d  e −3 x  dx 

[

]

d  5 x + e cos 5 x ⋅ e −3 x ⋅ −3 dx 

]

= e −3 x ⋅ e cos 5 x ⋅ − sin 5 x ⋅ 5 + e cos 5 x ⋅ e −3 x ⋅ −3 = − 5 sin 5 x e −3 x e cos 5 x − 3e cos 5 x e −3 x = − e cos 5 x e −3 x (5 sin 5 x + 3) = − ecos 5 x − 3 x ( 5 sin 5 x + 3) Example 3.3-5: Differentiate the following exponential functions 2

e 3x cot 5 x

a. y = e ln x − e − x

b. y =

d. y = e 3 x ⋅ arc cos x 2

e. y = ln e x + arc tan x + x 3

f. y = e x − arc sin x

h. y = arc sin e x

i. y = x arc sin e 3 x

2

Solutions: 2

a. Given y = e ln x + e − x then

= e

b. Given

=

1

dy dx

d  ln x 2  − e −x  e dx  

=

1

d 2 ⋅ ⋅ x − e −x ⋅ −1 2 dx x

= e

e 3x y= cot 5 x

d  e 3 x  = = dx  cot 5 x 

[ cot 5x ⋅ e

3x

2

2

g. y = e ln x + x 3

ln x 2

c. y = x 3 e −5 x

then

dy dx

ln x 2

⋅

x 2/ =1

][

d 3 x − e 3 x ⋅ − csc 2 5 x ⋅ d 5 x ⋅ dx dx

cot 2 5 x

c. Given y = x 3 e −5 x then

Hamilton Education Guides

dy dx

=

(

⋅ 2 x/ + e

]

−x

=

d ln x 2 d − x e − e dx dx

=

2e ln x + e−x x

2

= e ln x ⋅

d d ln x 2 − e − x ⋅ (− x ) dx dx

2

[ cot 5x ⋅ ( e ) ]− [ e d dx

3x

3x

d ( cot 5 x ) ⋅ dx

]

cot 2 5 x

=

cot 5 x ⋅ e 3 x ⋅ 3 + e 3 x ⋅ csc 2 5 x ⋅ 5 cot 2 5 x

=

(

e 3 x 3 cot 5 x + 5 csc 2 5 x cot 2 5 x

)

d d d d 3 −5 x       x e =  x 3 ⋅ e −5 x  +  e −5 x ⋅ x 3  =  x 3 ⋅ e −5 x ⋅ (− 5 x ) dx dx dx  dx     

175

)

Calculus I

3.3 Differentiation of Logarithmic and Exponential Functions

[

+ e −5 x ⋅ 3x 2

] = (x

3

)(

⋅ e −5 x ⋅ −5 + e −5 x ⋅ 3 x 2 dy dx

d. Given y = e 3 x arc cos x 2 then = arc cos x 2 ⋅ e 3 x ⋅

= 3e 3 x arc cos x 2 −

(

e

x2

⋅

2

f. Given y = e x − arc sin x then 2

= e x ⋅ 2x −

1 1− x

dy dx

dy dx

2

dy dx

1

d 2 ⋅ ⋅ x + 3x 2 2 dx x

= e

h. Given y = arc sin e x then

i. Given y = x arc sin e 3 x then

= arc sin e 3 x ⋅1 + x ⋅

1 1 − e 6x

Hamilton Education Guides

⋅

⋅ 2x

2 d  d d d 3 x2 3 arc tan x + ln e x + x  ln e + arc tan x + x  = dx  dx dx dx 

d 2 1 + 3x 2 x + 2 dx 1+ x

=

ex e

2

⋅ 2x +

x2

1 1+ x

2

+ 3x 2

= 3x 2 + 2x +

1 1+ x2

1− x2 2 d d  ln x 2 d ln x 2 d 3  + x3  = x = e ln x ⋅ ln x 2 + 3 x 2 e + e dx  dx dx dx 

=

⋅

1 x 2/ =1

⋅ 2 x/ + 3 x

(

2

d arc sin e x dx

=

dy dx

1− x 4

1

ln x 2

dy dx

=

1

2 d 1 d  x2 d x2 d  x2 − e − arc sin x = e x ⋅  e − arc sin x  = dx dx dx  dx  1− x 2

=

2

2

= e

e

2

⋅ex ⋅

x2

= 2x e x −

g. Given y = e ln x + x 3 then

ln x 2

1

=

d d 3x e + e 3 x ⋅ arc cos x 2 dx dx

  4  1− x 



d x2 1 e + + 3x 2 2 dx 1+ x

= arc cos x 2 ⋅

2x

= e 3 x  3 arc cos x 2 −

1− x 4

)

d 2 x = arc cos x 2 ⋅ e 3 x ⋅ 3 + e 3 x ⋅ − dx



2 xe 3 x

2

1

⋅

1− x 4

e. Given y = ln e x + arc tan x + x 3 then

=

= − 5 x 3 e −5 x + 3x 2 e −5 x = x 2e −5 x (− 5 x + 3)

d e 3 x arc cos x 2 dx

= 1

d 3x + e 3 x ⋅ − dx

)

=

d 3x e dx

(

)

= e

2 ⋅ + 3x 2 x

1

=

d x arc sin e 3 x dx

ln x 2

1− e 2x

⋅

1 1 − e 6x

=

d x e = dx

) = arc sin e

= arc sin e 3 x + x ⋅

2

2e ln x + 3x 2 x

3x

⋅ e 3x ⋅

⋅

1 1− e 2x

⋅ex ⋅

d x = dx

ex 1 − e 2x

d d x + x ⋅ arc sin e 3 x dx dx

d 3x dx

= arc sin e 3 x +

3 xe 3 x 1 − e 6x

176

Calculus I

3.3 Differentiation of Logarithmic and Exponential Functions

In order to find the derivative of the functions of the form y = x g (x ) first multiply both sides of the equation by natural logarithm ( log e = ln ) and then apply the logarithmic rules prior to taking the derivative. The following examples show how to differentiate this class of functions:

•

Example 3.3-6: Find the derivative of the following functions: a. y = x x , x 0 1

d. y = x x , x 0

b. y = x sin x , x 0

c. y = (sin x ) x , x 0

e. y = x ln x , x 0

f. y = x e

Solutions: a. The function y = x x is equivalent to ln y = ln x x ; ln y = x ln x thus ;

1 dy d d ⋅ = ln x ⋅ x + x ⋅ ln x y dx dx dx

;

1 1 ⋅ y ′ = ln x ⋅1 + x ⋅ y x

;

1 ⋅ y ′ = ln x + 1 y

x

d d x ln x ln y = dx dx

; y ′ = y ( ln x + 1) ; y ′ = x x ( ln x + 1)

b. The function y = x sin x is equivalent to ln y = ln x sin x ; ln y = sin x ln x thus: d d ln y = sin x ln x dx dx

;

1 dy d d ⋅ = ln x ⋅ sin x + sin x ⋅ ln x y dx dx dx

;

sin x 1 ⋅ y ′ = ln x cos x + y x



; y ′ = y  ln x cos x + 

sin x   x 

;

1 1 ⋅ y ′ = ln x ⋅ cos x + sin x ⋅ y x

 ; y′ = x sin x  ln x cos x + 

sin x   x 

c. The function y = (sin x ) x is equivalent to ln y = ln ( sin x ) x ; ln y = x ln ( sin x ) thus: d d ln y = x ln ( sin x ) dx dx

;

;

1 dy d d ⋅ = ln ( sin x ) ⋅ x + x ⋅ ln (sin x ) y dx dx dx

x cos x 1 ⋅ y ′ = ln ( sin x ) + y sin x

;

1 1 ⋅ y ′ = ln ( sin x ) ⋅1 + x ⋅ ⋅ cos x y sin x

; y ′ = y [ ln ( sin x ) + x cot x ] ; y′ = (sin x )x [ ln ( sin x ) + x cot x ] 1

1

1 x

d. The function y = x x is equivalent to ln y = ln x x ; ln y = ln x thus: d d 1  ln y =  ⋅ ln x  dx  x dx 

;

 1 − ln y ′ = y   x2

x  

;

; y′ =

Hamilton Education Guides

1 dy d 1 1 d ⋅ = ln x ⋅ + ⋅ ln x dx x x dx y dx

1 xx

 1 − ln   x2

x  

;

1 1 1 1 ⋅ y ′ = ln x ⋅ − + ⋅ y x2 x x

;

ln x 1 1 ⋅ y′ = − + y x2 x2

1

= y′ =

xx

x2

( 1 − ln x )

177

Calculus I

3.3 Differentiation of Logarithmic and Exponential Functions

e. The function y = x ln x is equivalent to ln y = ln x ln x ; ln y = ln x ⋅ ln x = ln y = ln 2 x thus: d d 2 ln y = ln x dx dx

;

2 ln x 1 ⋅ y′ = y x

1 dy d ⋅ = 2 ln x ⋅ ln x y dx dx

;

;

1 d 1 x ⋅ y ′ = 2 ln x ⋅ ⋅ x dx y

1 1 ⋅ y ′ = 2 ln x ⋅ ⋅1 y x

=

1 1 ⋅ y ′ = 2 ln x ⋅ y x

=

 ln x    x 

 2 ln x    x 

; y ′ = 2 x ln x 

; y′ = y 

x

x

f. The function y = x e is equivalent to ln y = ln x e ; ln y = e x ln x thus: d d x ln y = e ln x dx dx

;

1 dy d x d ⋅ = ln x ⋅ e + e x ⋅ ln x y dx dx dx

;

1 1 ⋅ y ′ = ln x ⋅ e x ⋅1 + e x ⋅ ⋅1 y x

1 ex ⋅ y ′ = e x ln x + y x

;

d 1 d 1 x x +ex ⋅ ⋅ ⋅ y ′ = ln x ⋅ e x ⋅ x dx dx y

;

x 1  1  ; y ′ = y e x  ln x +  ; y ′ = x e e x  ln x + 

x



x



Example 3.3-7: Find the derivative of the following functions: a. y = x 3e

b. y = (sin x ) π

c. y = x

d. y = x 2π

e. y = x ln a

f. y = x sin θ + x 3

5

Solutions: a. Given y = x 3e then

dy dx

= 3e x 3e −1

Note that we could have solved this problem, and other problems in this example, by first multiplying both sides of the equation by natural logarithm ( log e = ln ) and taking the derivative as shown below. However, this is a more difficult way of solving this class of functions and is not recommended. y = x 3e

d d ln y = ln x 3e dx dx

; ln y = ln x 3e then

(

; y ′ = y 3e x 3e−1 ⋅ x −3e

)

b. Given y = (sin x )π then c. Given y = x

5

5

(

; y ′ = y 3e x 3e−1−3e dy dx

= x 2 then

Hamilton Education Guides

)

= π (sin x )π −1 ⋅

dy dx

=

5 52 −1 x 2

=

;

1 dy 1 d 3e ⋅ = ⋅ x y dx x 3e dx

(

; y ′ = y 3e x −1

)

;

1 1 ⋅ y′ = ⋅ 3e x 3e −1 3e y x

(

; y ′ = x 3e 3e x −1

d sin x dx

= π (sin x ) π − 1 cos x

5 32 x 2

5 x3 2

=

=

5 x 2

)

; y ′ = 3e x 3e −1

x

178

Calculus I

3.3 Differentiation of Logarithmic and Exponential Functions

dy dx

d. Given y = x 2π then

dy dx

e. Given y = x ln a then

d 2π x dx

=

d ln a x dx

=

dy dx

f. Given y = x sin θ + x 3 then

= 2π x 2π − 1 = ln a x ln a − 1

(

d sin θ x + x3 dx

=

)

d sin θ d 3 x x + dx dx

=

= sin θ x sin θ − 1 + 3 x 2

Note: The problems in this example can be written in the standard form of y = x a where

dy dx

is

equal to y ′ = a x a −1 . Therefore, students should not get confused if the constant a is replaced with numbers such as

1 1 , e, e2 , α , π 3

3,

remains the same. •

, η , sin θ , etc. The process of finding the derivative

The derivative of the functions of the form a u = e u ln a is equal to the following: d u a dx

=

d u ln a e dx

= e u ln a ⋅

d u ln a dx

= e u ln a ⋅ ln a ⋅

d u dx

u

= e ln a ⋅ ln a ⋅

du dx

= a u ln a

du dx

Thus,

du d u a = a u ln a dx dx

The following examples show how to differentiate this class of functions: Example 3.3-8: Find the derivative of the following functions: a. y = 2 x 1

d. y = 10 x

b. y = 5 sin x

c. y = π

e. y = 10 ln x

f. y = 5 e

x x3

Solutions: a. Given y = 2 x then

dy dx

b. Given y = 5 sin x then c. Given y = π

d. Given y

x

1 = 10 x

dy dx

then

dy dx

then

dy dx

Hamilton Education Guides

d x 2 dx

=

=

=

=

= 2 x ⋅ ln 2 ⋅

d sin x 5 dx

d x π dx 1 d 10 x dx

d x dx

= 5 sin x ⋅ ln 5 ⋅

= π x ⋅ ln π ⋅

=

= 2 x ⋅ ln 2 ⋅1 = ( ln 2) 2 x

1 10 x

d x dx

d sin x dx

= 5 sin x ⋅ ln 5 ⋅ cos x = ( ln 5) 5 sin x cos x

= π x ⋅ ln π ⋅1 = ( ln π ) π

d 1 ⋅ ln 10 ⋅ dx x

=

1 10 x

⋅ ln 10 ⋅ −

1 x2

x

1

=

( ln 10 ) 10 x − x2

179

Calculus I

3.3 Differentiation of Logarithmic and Exponential Functions

e. Given y = 10 ln x then f. Given y = 5 e

x3

then x3

= ( ln 5 ) 3 x 2 5 e e x

dy dx dy dx

= =

d 10 ln x dx

d e x3 5 dx

= 10 ln x ⋅ ln 10 ⋅ x3

= 5 e ⋅ ln 5 ⋅

d ln x dx

d x3 e dx

= 10 ln x ⋅ ln 10 ⋅ x3

3

= 5 e ⋅ ln 5 ⋅ e x ⋅

1 x

=

d 3 x dx

( ln 10 ) 10 ln x x x3

3

= 5 e ⋅ ln 5 ⋅ e x ⋅ 3x 2

3

Section 3.3 Practice Problems – Differentiation of Logarithmic and Exponential Functions 1. Find the derivative of the following logarithmic functions:

(

)

a. y = ln 10 x

b. y = 10 ln 5 x 3

c. y = ln x 2 + 3

d. y = x 3 ln x

e. y = x ln x − 5 x 2

f. y = x 3 + 2 ln x 2

g. y = sin ln x 2

h. y = ln csc x

i. y = ln x 3

j.

( ) y = cos ( ln x ) 2

k. y = ln

x +1 x −1

(

)

l. y = x 3 ln x + 5 x

2. Find the derivative of the following exponential and trigonometric functions:

(

)

b. y = x 2 + 3 e3 x

c. y = esin 3 x

e. y = e9 x sin 5 x

f. y = e2 x arc sin x

g. y = (x − 5) e− x

h. y = e ln (x +1)

i. y =

j. y = 3x ⋅ e x

k. y = x3arc sin x

l. y = e5 x cos (5 x + 1)

a. y = x3e2 x d. y = eln x

2

Hamilton Education Guides

ex tan x

180

Calculus I

3.4

3.4 Differentiation of Hyperbolic Functions

Differentiation of Hyperbolic Functions

Hyperbolic functions are defined as: Table 3.4-1: Hyperbolic Functions sinh x

=

e x − e −x 2

coth x

=

cosh x sinh x

=

cosh x

=

e x + e −x 2

sec h x

=

1 cosh x

=

tanh x

=

sinh x cosh x

csc h x

=

1 sinh x

=

=

e x − e −x x

e +e

−x

e x + e−x e x − e −x 2

e x + e −x 2 e x − e −x

where x ≠ 0

where x ≠ 0

The differential formulas involving hyperbolic functions are defined as: Table 3.4-2: Differentiation Formulas for Hyperbolic Functions d sinh u dx d cosh u dx d tanh u dx

du dx du sinh u ⋅ dx du sec h 2 u ⋅ dx

d coth u dx d sec h u dx d csc h u dx

= cosh u ⋅ = =

= − csc h 2 u ⋅

du dx

du dx du − csc h u coth u ⋅ dx

= − sec h u tanh u ⋅ =

Let’s differentiate some hyperbolic functions using the above formulas: Example 3.4-1: Find the derivative of the following hyperbolic functions: a. y = 5 sinh 6 x

b. y = sinh x 3 + cosh 3x

c. y = cosh 2 5 x 3

d. y = x 3 sinh x

e. y = sinh x 2 + cosh x 2

f. y = tanh 3x 2 − 1

h. y = ln cosh 3x

i. y = ln tanh 3x 3

1 6

g. y = sinh 3x + 10 x

(

)

Solutions: a. Given y = 5 sinh 6 x then

dy dx

=

b. Given y = sinh x 3 + cosh 3x then = cosh x 3 ⋅

d 3 d x + sinh 3 x ⋅ 3 x dx dx

c. Given y = cosh 2 5 x 3 then = 2 cosh 5 x 3 ⋅ sinh 5 x 3 ⋅ Hamilton Education Guides

dy dx

d 5x 3 dx

d (5 sinh 6 x ) dx

dy dx

=

= 5 cosh 6 x ⋅

(

d ( 6x) dx

d sinh x 3 + cosh 3 x dx

)

=

= 5 cosh 6 x ⋅ 6 = 30 cosh 6 x

(

)

d d (cosh 3x ) sinh x 3 + dx dx

= cosh x 3 ⋅ 3x 2 + sinh 3x ⋅ 3 = 3 x 2 cosh x 3 + 3 sinh 3 x =

(

d cosh 2 5 x 3 dx

) = dxd ( cosh 5x )

3 2

= 2 cosh 5 x 3 ⋅

d cosh 5 x 3 dx

= 2 cosh 5 x 3 ⋅ sinh 5 x 3 ⋅15 x 2 = 30 x 2 sinh 5 x 3 cosh 5 x 3 181

Calculus I

3.4 Differentiation of Hyperbolic Functions

dy dx

d. Given y = x 3 sinh x then

= sinh x ⋅ 3x 2 + x 3 ⋅ cosh x ⋅

d dx

d 2 d 2 x + sinh x 2 ⋅ x dx dx

(

)

f. Given y = tanh 3x 2 − 1 then 1 6

g. Given y = sinh 3x + 10 x then =

d 1 cosh 3 x ⋅ 3 x + 10 dx 6

dy dx

h. Given y = ln cosh 3x then =

1 ⋅ sinh 3 x ⋅ 3 cosh 3 x

= 3⋅

dy dx

=

sec h 2 3 x 3 tanh 3 x 3

⋅ 9x 2 =

dy dx

d 3 d x + x 3 ⋅ sinh dx dx

=

(

)=

d sinh x 2 + cosh x 2 dx

x

= 3 x 2 sinh x +

x 3 cosh

x

2 x

d d sinh x 2 + cosh x 2 dx dx

= cosh x 2 ⋅ 2 x + sinh x 2 ⋅ 2 x = 2 x cosh x 2 + 2 x sinh x 2

[

) ] = sec h (3x − 1)⋅ dxd (3x − 1)

(

dy dx

=

d tanh 3 x 2 − 1 dx

dy dx

=

d 1   sinh 3 x + 10 x  dx  6 

=

sinh 3 x cosh 3 x

i. Given y = ln tanh 3x 3 then

= sinh x ⋅

1 − 12 2

3 cosh 3 x + 10 6

=

)

x

= 3x 2 sinh x + x 3 cosh x ⋅ x

x

e. Given y = sinh x 2 + cosh x 2 then = cosh x 2 ⋅

(

d x 3 sinh dx

=

=

=

2

2

2

(

= 6 x sec h 2 3 x 2 − 1

d 1  d  sinh 3 x  + 10 x dx  6  dx

=

1 cosh 3 x + 10 2

d ( ln cosh 3x ) dx

=

d 1 ⋅ ( cosh 3 x ) cosh 3 x dx

=

1 d ⋅ sinh 3 x ⋅ 3 x cosh 3 x dx

)

=

= 3 tanh 3 x

(

d ln tanh 3 x 3 dx

)

=

1 tanh 3 x

3

⋅

(

d tanh 3 x 3 dx

sec h 2 3 x 3 tanh 3 x

3

⋅

d 3x 3 dx

9 x 2 sec h 2 3 x 3 tanh 3 x 3

Example 3.4-2: Find the derivative of the following hyperbolic functions: a. y = coth d. y =

1

b. y =

x2

sinh e 3 x cosh x 3

g. y = sinh x 3 + cosh x

sinh x 2 cosh (x + 1)

c. y =

sinh 10 x 2 cosh 3 x 5

e. y = x 2 sinh x 2

f. y = (x + 1) cosh x

h. y = x 3 sinh 2 5 x

i. y = csc h x + 3 x 2

Solutions: a. Given y = coth

1 x

2

then

Hamilton Education Guides

dy dx

=

d  1    coth dx  x2 

= − csc h 2

1 x

2

⋅

d 1 dx x 2

= − csc h 2

1 x

2

⋅

−2 x

3

=

2 x

3

csc h 2

1 x2

182

)

Calculus I

b. Given

=

=

=

sinh x 2 y= cosh (x + 1)

[cosh (x + 1) cosh x

=

then

=

d  sinh x 2  dx  cosh (x +1) 

=

][

[ cosh (x + 1)⋅

d dx

][

d cosh ( x + 1) sinh x 2 − sinh x 2 ⋅ dx

]

cosh 2 (x + 1)

d x 2 − sinh x 2 sinh (x + 1) ⋅ d (x + 1) ⋅ dx dx

]

[cosh (x + 1) cosh x

=

2

][

⋅ 2 x − sinh x 2 sinh (x + 1) ⋅1

cosh 2 (x + 1)

2 x cosh x 2 cosh ( x + 1) − sinh ( x + 1) sinh x 2 cosh 2 ( x + 1)

y=

[ cosh 3x

sinh 10 x 2

then

cosh 3 x 5 5

dy dx

⋅ cosh 10 x 2

d dx

d  sinh 10 x 2 dx  cosh 3 x 5

=

   

][

=

10 x 2 − sinh 10 x 2 ⋅ sinh 3 x 5

[ cosh 3x

d dx

[

− sinh 10 x 2 ⋅ sinh 3x 5 ⋅15 x 4

]

=

cosh 2 3 x 5

sinh e 3 x cosh x 3

then

dy dx

=

5

][

d sinh 10 x 2 − sinh 10 x 2 ⋅ d cosh 3 x 5 ⋅ dx dx

]

cosh 2 3 x 5 3x 5

]

cosh 2 3 x 5

d. Given y =

=

2

dy dx

cosh 2 (x + 1)

c. Given

=

3.4 Differentiation of Hyperbolic Functions

=

[ cosh 3x

5

⋅ cosh 10 x 2 ⋅ 20 x

]

cosh 2 3 x 5

20 x cosh 10 x 2 cosh 3 x 5 − 15 x 4 sinh 3 x 5 sinh 10 x 2 cosh 2 3 x 5

d  sinh e 3 x dx  cosh x 3

   

=

d e 3 x − sinh e 3 x ⋅ sinh x 3 ⋅ d x 3 cosh x 3 ⋅ cosh e 3 x ⋅ dx dx

cosh 2 x 3

[ cosh x

3

][

d sinh e 3 x − sinh e 3 x ⋅ d cosh x 3 ⋅ dx dx

]

cosh 2 x 3

=

cosh x 3 ⋅ cosh e 3 x ⋅ 3 e 3 x − sinh e 3 x ⋅ sinh x 3 ⋅ 3 x 2 cosh 2 x 3

3 e 3 x cosh e 3 x cosh x 3 − 3 x 2 sinh x 3 sinh e 3 x cosh 2 x 3

e. Given y = x 2 sinh x 2 then = sinh x 2 ⋅ 2 x + x 2 ⋅ cosh x 2 ⋅

dy dx

=

d 2 x dx

f. Given y = (x + 1) cosh x then

dy dx

(

d x 2 sinh x 2 dx

)

= sinh x 2 ⋅

d 2 d x + x 2 ⋅ sinh x 2 dx dx

= 2 x sinh x 2 + x 2 ⋅ cosh x 2 ⋅ 2 x = 2 x sinh x 2 + 2 x 3 cosh x 2 =

d [ (x + 1) cosh x] dx

= cosh x ⋅

d (x + 1) + (x + 1)⋅ d cosh x dx dx

= cosh x ⋅1 + (x + 1) ⋅ sinh x = cosh x + ( x + 1) sinh x Hamilton Education Guides

183

]

Calculus I

3.4 Differentiation of Hyperbolic Functions

dy dx

g. Given y = sinh x 3 + cosh x then =

=

d   sinh dx 

x 3 + cosh

d x  = sinh dx 

x3 +

d cosh dx

x

3 3 3 1 1 1 1 1 −1 3 3 −1 d d d 32 d 12 sinh x 2 + cosh x 2 = cosh x 2 ⋅ x + sinh x 2 ⋅ x = cosh x 2 ⋅ x 2 + sinh x 2 ⋅ x 2 2 2 dx dx dx dx

3

3 2

1

1

1 − 12 2

=

3 1 3 12 1 −1 x cosh x 2 + x 2 sinh x 2 2 2

dy dx

=

d x 3 sinh 2 5 x dx

= cosh x 2 ⋅ x 2 + sinh x 2 ⋅ x h. Given y = x 3 sinh 2 5 x then

= sinh 2 5 x ⋅ 3x 2 + x 3 ⋅ 2 sinh 5 x ⋅

(

d sinh 5 x dx

)

= sinh 2 5 x ⋅

=

3 x cosh 2

x3 +

1 2 x

sinh

x

d d 3 x + x 3 ⋅ sinh 2 5 x dx dx

= sinh 2 5 x ⋅ 3x 2 + x 3 ⋅ 2 sinh 5 x ⋅ cosh 5 x ⋅

d 5x dx

= sinh 2 5 x ⋅ 3x 2 + x 3 ⋅ 2 sinh 5 x ⋅ cosh 5 x ⋅ 5 = 3 x 2 sinh 2 5 x + 10 x 3 sinh 5 x cosh 5 x i. Given y = csc h x + 3 x 2 then = − csc h x coth x ⋅

dy dx

=

d  3 2   csc h x + x  dx  

=

d 3 2 d csc h x + x dx dx

=

d d 23 csc h x + x dx dx

2 2 −1 2 2 −1 d x+ x3 = − csc h x coth x ⋅1 + x 3 = − csc h x coth x + 3 3 3 dx 3 x

Example 3.4-3: Find the first and second derivative of the following hyperbolic functions: c. y = cosh (10 x + 3)

a. y = sinh 8 x

b. y = sinh 2 3x

d. y = cosh x

e. y = sinh x 2 + 1

f. y = x sinh x

g. y = cosh 3x − x 2

h. y = tanh 5 x

i. y = coth x 2

(

)

Solutions: a.

dy dx

d2y dx

b.

dy dx

d (sinh 8 x ) dx

=

2

= =

= cosh 8 x ⋅

d ( 8 cosh 8 x ) dx

(

d sinh 2 3 x dx

)

=

d ( 8x) dx

= 8 sinh 8 x ⋅

= cosh 8 x ⋅ 8 = 8 cosh 8 x

d ( 8x) dx

d ( sinh 3x )2 dx

= 8 sinh 8 x ⋅ 8 = 64 sinh 8 x

= 2 sinh 3x ⋅

d sinh 3 x dx

= 2 sinh 3x ⋅ cosh 3x ⋅

d 3x dx

= 2 sinh 3x ⋅ cosh 3x ⋅ 3 = 6 sinh 3 x cosh 3 x

Hamilton Education Guides

184

Calculus I

d2y dx

3.4 Differentiation of Hyperbolic Functions

d (6 sinh 3x cosh 3x ) dx

=

2

 = 6 cosh 3x ⋅ cosh 3x ⋅ 

(

dy dx

d2y dx

d.

=

2

d y

(

d cosh dx

1

1

− − sinh x 2 ⋅ 12 x 2

1

=

e.

1 ⋅ 2

dy dx

1 ⋅ cosh 2

=

d2y dx

2

x

)

1

dy dx

=

2 x

)

d ( 10 x + 3) dx

= 10 cosh ( 10 x + 3) ⋅10 = 100 cosh ( 10 x + 3)

d 12 x = sinh dx

1

1

1

1

1 ⋅ 4

(

)]

(

(

)

=

cosh

x−

x⋅

1 − 12 x 2

1

=

sinh 2

x x

1

1

1

1

1

= sinh x ⋅

Hamilton Education Guides

sinh x

) dxd ( x + 1)

= 2

2

[

(

1

1

)

d d x + sinh x ⋅ x dx dx

x − sinh x 4x

=

(

x

=

(

)

x cosh

= cosh x 2 + 1 ⋅ 2 x = 2 x cosh x 2 + 1

)]

d x cosh x 2 + 1 dx

) (

x cosh

x

x

= cosh x 2 + 1 ⋅

d 2 x cosh x 2 + 1 dx

d x sinh x dx

= sinh ( 10 x + 3) ⋅10 = 10 sinh ( 10 x + 3)

= sinh x ⋅

d   2 cosh x 2 + 1 ⋅1 + x ⋅ sinh x 2 + 1 ⋅ x 2 +1  dx  

f.

)

2 2 d 2 2 d 2 2 d 2 1 x ⋅ cosh x ⋅ dx x 1 x ⋅ dx sinh x − sinh x ⋅ dx x = ⋅ = ⋅ 2 x x 2

1

1

d sinh x 2 + 1 dx

(



d d  sinh 3 x + sinh 3 x ⋅ cosh 3 x  dx dx 

− − x/ ⋅ cosh x 2 ⋅ 1 − sinh x 2 ⋅ 1 2 2 1 2 2 1 2 1 x ⋅ cosh x ⋅ 2 x − sinh x ⋅ 2 x 1 2 x/ 2 x = ⋅ = ⋅ 2 x x 2

1

[

( x)

1 d sinh ⋅ 1 2 dx x2

=

x

=

d dx

1 x2

x 2 − sinh x 2 ⋅ 1

(

d (10 x + 3) dx

= 10 cosh ( 10 x + 3) ⋅

= sinh x ⋅

d  sinh x  dx  2 x   

=

2

= 18 cosh 2 3 x + sinh 2 3 x

= sinh ( 10 x + 3) ⋅

 = 6 cosh 3x ⋅

= 6(cosh 3x ⋅ cosh 3x ⋅ 3 + sinh 3x ⋅ sinh 3x ⋅ 3)

(

)

d [ 10 sinh ( 10 x + 3) ] dx

=

2

dy dx

dx

d cosh ( 10 x + 3) dx

=

d ( sinh 3x cosh 3x ) dx

d d  3 x + sinh 3 x ⋅ sinh 3 x ⋅ 3 x  dx dx 

= 6 3 cosh 2 3x + 3 sinh 2 3x c.

= 6

[

(

)

x − sinh

4x x

) (

)

d d   = 2 cosh x 2 + 1 ⋅ x + x ⋅ cosh x 2 + 1  dx dx  

(

)

(

= 2 cosh x 2 + 1 + 2 x 2 sinh x 2 + 1

= sinh x ⋅1 + cosh x ⋅

d x⋅x dx

)]

= sinh x + x cosh x

185

x

Calculus I

d2y dx

3.4 Differentiation of Hyperbolic Functions

=

2

d (sinh x + x cosh x ) dx

d d x cosh x sinh x + dx dx

=

= cosh x ⋅

d d d x+ x + cosh x ⋅ cosh x ⋅ x dx dx dx

= cosh x ⋅1 + cosh x ⋅1 + sinh x ⋅ x = cosh x + cosh x + x sinh x = 2 cosh x + x sinh x g.

dy dx

=

d2y dx

h.

=

2

dy dx

=

d2y dx

(

d cosh 3 x − x 2 dx

2

=

d (3 sinh 3x − 2 x ) dx

d ( tanh 5 x ) dx

=

)

d 2 d cosh 3 x − x dx dx

=

d d 3 sinh 3 x − 2x dx dx

= sec h 2 5 x ⋅

(

d 5 sec h 2 5 x dx

= sinh 3x ⋅

d (5 x ) dx

d 3x − 2 x dx

= 3 cosh 3x ⋅

= sinh 3x ⋅ 3 − 2 x = 3 sinh 3 x − 2 x

d 3x − 2 dx

= 3 cosh 3x ⋅ 3 − 2 = 9 cosh 3 x − 2

= sec h 2 5 x ⋅ 5 = 5 sec h 2 5 x

) = 5 ⋅ 2 sec h 5x ⋅ dxd sec h 5x

= 5 ⋅ 2 sec h 5 x ⋅ − sec h 5 x tanh 5 x ⋅

d 5x dx

= − 5 ⋅ 2 sec h 5 x ⋅ sec h 5 x tanh 5 x ⋅ 5 = − 50 sec h 2 5 x tanh 5 x i.

dy dx

d2y dx

(

d coth x 2 dx

=

2

=

)

= − csc h x 2 ⋅

(

d − 2 x csc h x 2 dx

(

d 2 x dx

= − csc h x 2 ⋅ 2 x = − 2 x csc h x 2

) = − 2 dxd ( x csc h x )

 = − 2  csc h x 2 ⋅

2

= − 2 csc h x 2 ⋅1 − x ⋅ csc h x 2 coth x 2 ⋅ 2 x

)



(

d d  x + x ⋅ csc h x 2  dx dx 

= − 2 csc h x 2 − 2 x 2 coth x 2

)

(

= − 2 csc h x 2 1 − 2 x 2 coth x 2

)

Section 3.4 Practice Problems – Differentiation of Hyperbolic Functions Find the derivative of the following hyperbolic functions:

(

a. y = cosh x 3 + 5

)

d. y = ln (sinh x ) g. y =

tanh 2 x x

j. y = sinh 3 x 2

Hamilton Education Guides

b. y = x 3 sinh x

c. y = (x + 6) sinh x 3

e. y = e sinh x

f. y = cosh 3x 5

h. y = coth

1 x

3

k. y = tanh 5 x

(

) coth ( x

i. y = x 2 + 9 tanh x l. y = x 5

3

+1

)

186

Calculus I

3.5

3.5 Differentiation of Inverse Hyperbolic Functions

Differentiation of Inverse Hyperbolic Functions

Inverse hyperbolic functions are defined as: Table 3.5-1: Inverse Hyperbolic Functions 1  x +1 ln   2  x −1 

sinh −1 x

= ln  x + x 2 + 1 

for all x

coth −1 x

=

cosh −1 x

= ln  x + x 2 − 1 

x ≥1

sec h −1 x

= ln  

tanh −1 x

=

csc h −1 x

= ln  +  x



 



1 1+ x  ln   2  1− x 

x2 1

x2 1

 1+ 1− x 2 x 

 1

   

0 x ≤1

1 + x 2   x 



x≠0

The differential formulas involving inverse hyperbolic functions are defined as: Table 3.5-2: Differentiation Formulas for Inverse Hyperbolic Functions d sinh −1 u dx

=

d cosh −1 u dx

=

d tanh −1 u dx

=

1

du u 2 +1 dx 1 du ⋅ 2 u −1 dx ⋅

1 1− u

2

⋅

u 1

du dx

u

1

d coth −1 u dx

=

d sec h −1u dx

=

d csc h −1u dx

=

1 1− u

2

⋅

du dx

−1 u 1− u

2

du dx

⋅

−1 1+ u 2

u

1

u

⋅

0 u 1

du dx

u≠0

Let’s differentiate some inverse hyperbolic functions using the above formulas: Example 3.5-1: Find the derivative of the following inverse hyperbolic functions: a. y = sinh −1 10 x

b. y = cosh −1 x 2

c. y = sinh −1 x

d. y = x tanh −1 x 3

e. y = sinh −1 e 2 x

f. y = x 3 + coth −1 x

g. y = cosh −1 x + e 2 x

h. y =

Solutions:

sinh −1 x 3x

a. Given y = sinh −1 10 x then

dy dx

=

d sinh −1 10 x dx

b. Given y = cosh −1 x 2 then

dy dx

=

d cosh −1 x 2 dx

=

dy dx

=

d sinh −1 x dx

=

c. Given y = sinh −1 x then

Hamilton Education Guides

=

i. y = sec h −1e x 1

(10 x )2

d 10 x = + 1 dx

1

(x )

2 2

1

⋅

⋅

d 2 x dx

⋅

d dx

−1

( x ) 2 +1

=

x =

1 100 x 2 + 1 1 4

x −1

⋅ 2x

⋅10

⋅

100 x 2 + 1

2x

=

d 12 x x + 1 dx 1

10

=

x4 −1

=

1 12 −1 x x +1 2 1

⋅

187

Calculus I

=

3.5 Differentiation of Inverse Hyperbolic Functions

1 − 12 x x +1 2 1

⋅

1

=

1

⋅

x +1 2 x dy dx

d. Given y = x tanh −1 x 3 then

= tanh −1 x 3 ⋅1 + x ⋅

1 1− x6

⋅

d 3 x dx

=

1 e 4x +1

⋅ e 2x ⋅ 2

=

2

x −1

+

d 2x e dx

h. Given

sinh −1 x y= 3x

3x

− 3 sinh −1 x

=

x 2 +1

9x 2

=

1 2

x −1

then

=

=

d sinh −1 e 2 x dx

)

= tanh −1 x 3 ⋅

1 1− x6

=

⋅ 3x 2

d d x + x ⋅ tanh −1 x 3 dx dx

= tanh −1 x 3 +

1

⋅

(e )

2x 2

+1

3 x3

1 − x6

1

d 2x e = dx

e

4x

+1

⋅ e 2x ⋅

d 2x dx

e4x + 1

g. Given y = cosh −1 x + e 2 x then 1

(

d x tanh −1 x 3 dx

=

2e 2 x

f. Given y = x 3 + coth −1 x then

=

x +1

2 x

= tanh −1 x 3 + x ⋅

dy dx

e. Given y = sinh −1 e 2 x then

1

=

dy dx

dy dx dy dx

= =

+ e 2x ⋅

(

d x 3 + coth −1 x dx

(

d 2x dx

d  sinh −1 = dx  3 x

9x 2 x 2 +1 dy dx

= −

=

d cosh −1 x + e 2 x dx

1

1

=

2

x −1

d 3 d x + coth −1 x dx dx

)=

= 3x 2 +

d 2x d e cosh −1 x + dx dx

+ e 2x ⋅ 2

=

1 x

2

−1

=

1− x2

2

x −1

d 2x e dx

+

+ 2e 2 x

3x ⋅

1

x 2 +1

− sinh −1 x ⋅ 3

9x 2

x − x 2 + 1 sinh −1 x 3x 2 x 2 + 1

d x 1 e = − ⋅ex = − dx 2 x x 2 e 1− e e x 1− e x

( )

1

1

=

d sinh −1 x − sinh −1 x ⋅ d 3 x 3 x ⋅ dx x  dx = = 2  x 9 

3 x − 3 x 2 + 1 sinh −1 x

i. Given y = sec h −1e x then

)

⋅

1 1 − e 2x

Example 3.5-2: Find the derivative of the following inverse hyperbolic functions: a. y = sinh −1 x 2 + ln x 3

b. y = tanh −1 sin x

c. y = coth −1 cos x

d. y = sec h −1 x ⋅ sec x

e. y = sinh −1 3e x

f. y = coth −1 ( 10 x + 3)

g. y = cosh −1 ln x

h. y = tanh −1 2 x + ln e x

i. y = tanh −1 x + x

j. y = e x sinh −1 x

k. y = x cosh −1 x

l. y = tanh −1 x 3 + 5 x

Hamilton Education Guides

188

Calculus I

3.5 Differentiation of Inverse Hyperbolic Functions

Solutions: a.

dy dx

+

= 1

x

3

(

d sinh −1 x 2 + ln x 3 dx

⋅3 x2

=

2x

3 x 2/

+

4

x +1

dy dx

=

d tanh −1 sin x dx

=

c.

dy dx

=

d coth −1 cos x dx

=

d.

dy dx

=

d sec h −1 x ⋅ sec x dx

dy dx

e.

sec x

3/ =1

x

b.

= −

1− x 2

=

d sinh −1 3e x dx

d cos x dx 1 − cos x

=

2

= sec x ⋅

( 3e )

x 2

g.

dy dx

=

d cosh −1 ln x dx

h.

dy dx

=

d tanh −1 2 x + ln e x dx

i.

dy dx

=

1

( ln x )2

(

=

2 1 − 4x 2

d tanh −1 x + x dx

)

Hamilton Education Guides

)

=

+1

=

=

2

1 − sin x

⋅ cos x

1 2

1 − cos x

2

d 3e x = dx

⋅

d ( x + 3) dx

d ln x − 1 dx ⋅

9e 2 x + 1

=

=

d d tanh −1 x + x dx dx

cos x

=

2

1 − sin x

=

⋅ 2x

cos x

= −

2

1 − cos x

−1 x

1− x 2

1 cos x

=

2

− sin x

= sec x ⋅

⋅ 3e x ⋅

d x = dx

1

(

1 − x 2 + 9 + 6x

⋅1 =

)

1

1 d ⋅ ⋅ x ln 2 x − 1 x dx

d d tanh −1 2 x + ln e x dx dx

=

cos x

=

⋅ − sin x

1

⋅

1 − 2x 2

x 4 +1

sin x 2

sin x

= −

1 sin x

+ sec h −1 x ⋅ sec x tan x

1 − x2

x

2 +1 − 4x 2

1

=

1 − x 2 sec x tan x sec h−1 x

− sec x + x

+1

1 − ( x + 3)

=

d 2 1 d 3 ⋅ x + x x 3 dx x 4 + 1 dx ⋅

d d sec h −1 x + sec h −1 x ⋅ sec x dx dx

1

(

ex

⋅

1

=

=

1

=

1

1

=

3 x

d sin x 1 − sin x dx

d coth −1 ( x + 3) dx

1 − 4x 2

x +1

+

⋅

=

ex

4

2

dy dx

+

2x

=

+ sec x tan x sec h −1 x =

x

2

d d sinh −1 x 2 + ln x 3 dx dx

1

f.

=

)=

=

1

1 − (2 x )

2

⋅

=

1 9e 2 x + 1

3e x

⋅ 3e x ⋅1 =

1

= −

2

− x − 6x − 8 1

1 ⋅ ⋅1 ln 2 x − 1 x

d 1 d x 2x + ⋅ e dx e x dx

=

=

9e 2 x + 1 1 2

x + 6x + 8

1 x

ln 2 x − 1

1 1 − 4x

2

⋅2+

1 e

x

⋅ex

− 4x 2 + 3 − 2x 2 + 1

=

1 1− x 2

⋅

d x +1 dx

=

1 1− x 2

⋅1 + 1

=

1 1− x2

+1

189

Calculus I

j.

dy dx

=

3.5 Differentiation of Inverse Hyperbolic Functions

(

d e x sinh −1 x dx

)

1

= sinh −1 x ⋅ e x ⋅1 + e x ⋅

(

= sinh −1 x ⋅

x 2 +1

)

k.

dy dx

=

d x cosh −1 x dx

l.

dy dx

=

d tanh −1 x 3 + 5 x dx

(

⋅1

d x d e + e x ⋅ sinh −1 x dx dx

= cosh −1 x ⋅

)



d d x + x ⋅ cosh −1 x dx dx

Example 3.5-3: In the following examples find

dy dx

1

= cosh −1 x ⋅1 + x ⋅

1

=

1

= e x  sinh −1 x +

x 2 +1

d d tanh −1 x 3 + 5 x dx dx

=

   2 x +1 



ex

= e x sinh −1 x +

1 d d ⋅ x +ex ⋅ x dx x 2 + 1 dx

= sinh −1 x ⋅ e x ⋅

1− x6

⋅

d 3 x +5 dx

=

x 2 −1 1

1− x6

x

= cosh −1 x +

⋅ 3x 2 + 5

=

3x 2 1− x6

x2 −1

+5

:

a. x = t + sinh −1 t and y = cos t

b. x = cosh −1 t and y = t 3 + 3t 2 + t

c. x = sinh −1 t 2 and y = sin 2 t

d. x = tanh −1 α and y = sin α

e. x = θ coth −1 θ and y = θ

f. x = t 2 sinh −1 t and y = t e t

g. x = cosh −1 e t and y = e 2t

h. x = t 3 + 3t and y = cosh −1 t

Solutions: a. Given x = t + sinh −1 t and y = cos t then

dy dt

=

d cos t dt

= − sin t So

dy dt dx dt

=

dx dt

− sin t 1+

(

t 2 +1

=

− 1+

t 2 +1

b. Given x = cosh −1 t and y = t 3 + 3t 2 + t then

2

= 3t + 6t + 1 So

dy dt dx dt

=

3t 2 + 6t + 1 1

=

Hamilton Education Guides

sin t 1

= −

t 2 +1 t 2 +1

dx dt

=

3 t 2 + 6 t +1 1 1

=

d cosh −1 t dt

(3t

2

1 t 2 +1

sin t ⋅

t 2 +1

1 ⋅1 +

t 2 +1

1

=

2

t −1

)

+ 6 t +1 ⋅

=

t 2 −1

1 ⋅1

t 2 +1

= −

and

t 2 +1

1+

dy dt

(

and

sin t

t2 +1

1+

t2 +1

=

(

d 3 t + 3t 2 + t dt

= 3t2 + 6t +1

)

)

t 2 −1

t 2 −1

t 2 −1

c. Given x = sinh −1 t 2 and y = sin 2 t then

) = 1+

d t + sinh −1 t dt

=

dx dt

=

d sinh −1 t 2 dt

=

1 4

t +1

and

dy dt

=

d sin 2 t dt

190

Calculus I

=

=

3.5 Differentiation of Inverse Hyperbolic Functions

d (sin t )2 dt

= 2 sin t ⋅

2 sin t cos t ⋅

t 4 +1

d sin t dt

t 4 +1

2 sin t cos t

=

1 ⋅1

Therefore

cos α

=

1

=

1−α 2

cos α 1 1

dx dα

=

1

−1

= coth θ ⋅1 + θ ⋅

Therefore

2

1−θ

dy dθ dx dθ

d θ ⋅ dθ

=

= coth θ +

θ

2

−1

= et ⋅

=

d d t + t ⋅ et dt dt

1 2

t +1

1−θ

2

dx dt

=

2t t 2 + 1 sinh −1 t + t 2

t 2 +1

=

t +1

=

(

−1

2

(

1⋅ 1 − θ 2

 2  1−θ 

)

θ +θ

and

)

= sinh −1 t ⋅

2t t 2 + 1 sinh −1 t + t 2 2

t +1 dy dt dx dt

=

)

d d θ +θ ⋅ coth −1 θ dθ dθ

 coth −1 θ + θ  ⋅1 

)

= cos α

= 1 − α 2 cos α

= coth −1 θ ⋅

1−θ

(

t2

)

d sin α dα

=

=

dy dθ

=

d θ dθ

=1

1 −θ 2

(1 − θ ) coth 2

−1

θ +θ

d 2 2 d t + t ⋅ sinh −1 t dt dt

and

dy dt

=

d t et dt

et ( 1 + t ) 2t t 2 + 1 sinh −1 t + t 2

t2 +1

et (1 + t) t 2 +1 2t t 2 + 1 sinh −1 t + t 2

g. Given x = cosh −1 e t and y = e 2t then

Hamilton Education Guides

)

2

(

dy dα

1

(1 − θ ) coth

= e t ⋅1 + t ⋅ e t = e t ( 1 + t ) Therefore

et (1 + t) 1

(

=

d 2 t sinh −1 t dt

=

2

=

and

2

cos α 1 − α 2

(

θ +θ 1−θ 2

= 2t sinh −1 t +

t 4 +1

t 4 +1

1−α

d θ coth −1 θ dθ

1 −1

)

1 ⋅1

=

f. Given x = t 2 sinh −1 t and y = t e t then

= sinh −1 t ⋅ 2t + t 2 ⋅

(

cos α ⋅ 1 − α 2

dx dθ

(1 − θ )coth

2 sin t cos t 1 1

=

1

1

= tanh −1 α =

1−α 2

e. Given x = θ coth −1 θ and y = θ then

2 sin t cos t

=

= 2 t 4 + 1 sin t cos t

1

d. Given x = tanh −1 α and y = sin α then

dy dα dx dα

dy dt dx dt

= 2 sin t cos t Therefore

dx dt

=

d cosh −1 e t dt

=

1 e

2t

d t e −1 dt ⋅

=

1 e

2t

−1

⋅ et

191

Calculus I

et

=

=

3.5 Differentiation of Inverse Hyperbolic Functions

and

e 2t − 1

2e 2t e 2t − 1 et

dy dt

=

=

d 2t e dt

= e 2t ⋅

2e 2 t ⋅ e − t e 2 t − 1 1

d 2t dt

= e 2t ⋅ 2 = 2e 2t Therefore

d cosh −1 dt

=

( 3t

t

1

d dt t −1

=

1

2

)

+ 3 ⋅ 2 t t −1

=

(

1

=

t

⋅

=

2e 2t et

e 2t − 1

= 2e 2t − t e 2t − 1 = 2e t e 2t − 1

h. Given x = t 3 + 3t and y = cosh −1 t then =

dy dt dx dt

⋅

(

dx dt

=

d 3 t + 3t dt

1

=

1

t −1 2 t

)

2 t t −1

=

d 3 d t + 3t dt dt

Therefore

= 3t 2 + 3 and

dy dt dx dt

=

dy dt

1 2 t t −1 3t 2 + 3

1 2

)

6 t +1

t

t −1

Section 3.5 Practice Problems – Differentiation of Inverse Hyperbolic Functions Find the derivative of the following inverse hyperbolic functions: b. y = cosh −1 ( x + 5)

a. y = x sinh −1 3x d. y =

sinh −1 x

(

x3

e. y =

)

g. y = x 2 + 3 coth −1 x j. y = sinh −1 7 x + ln e x

2

Hamilton Education Guides

sinh −1 x cosh 3 x

c. y = tanh −1 x f. y =

cosh −1 x x5

h. y = e 3 x cosh −1 x

i. y = x 3 + tanh −1 x 5

( )

l. y = coth −1(3x + 5)

k. y = tanh −1 e2 x

192

Calculus I

3.6

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

Evaluation of Indeterminate Forms Using L’Hopital’s Rule

In Chapter 1 the concept of limits and how they are used in finding a solution to an expression as the limit of a variable, within the expression, approaches a constant or infinity was addressed. In Chapter 2 and this Chapter finding the derivative of various class of functions was discussed. In this section, we will use what we have learned in the previous chapters in order to solve expressions of the form referred to as indeterminate forms. We stated earlier that the derivative of a differentiable function f (x ) is defined as lim ∆ x →0

f (x + ∆ x ) − f (x ) (x + ∆ x ) − x

( 1)

As the limit ∆ x → 0 the above expression reduces to: lim ∆ x →0

f (x + ∆ x ) − f (x ) (x + ∆ x ) − x

=

f (x + 0) − f (x ) f (x ) − f (x ) 0 = = (x + 0) − x x−x 0

Since the limit on both the numerator and the denominator is zero, equation ( 1) is referred to as indeterminate form of the type equal to ∞ , i.e.,

∞ ∞

0 0

. Similarly, when both the numerator and the denominator are

the resulting expression is also referred to as indeterminate form. Note that

indeterminate forms of the type 0 ⋅ ∞ , ∞ − ∞ , 0 0 , ∞ 0 , and 1∞ need to be transformed to one of the types

0 0

or

∞ ∞

first. These types of indeterminate forms will be discussed later in this section.

To solve expressions that result in indeterminate forms of the type is referred to as the L’Hopital’s rule.

0 0

or

∞ ∞

we apply a rule that

L’Hopital’s Rule – Given that f (x ) and g (x ) are differentiable and g (x ) ≠ 0 and assuming that lim x → a f (x ) = f (a ) = 0 and lim x → a g (x ) = g (a ) = 0 , then lim x → a

f ′ (x ) f (x ) = lim x → a g (x ) g ′ (x )

is referred to as the L’Hopital’s rule provided that f ′ (a ) and g ′ (a ) exist and g ′ (a ) ≠ 0 Note that we proceed to differentiate f (x ) and g (x ) , i.e., continue to apply the L’Hopital’s rule, as long as we still obtain the indeterminate forms of the type

0 0

or

∞ ∞

as the limit x approaches

to a . We stop differentiation as soon as either the numerator or the denominator has a finite nonzero limit. The following examples show application of the L’Hopital’s rule. Example 3.6-1: Evaluate the limit of the following functions: a. lim x → 0 d. lim x → 0

e x −1 x3

=

e x + e −x − 2 sin x

Hamilton Education Guides

b. lim x → ∞ e. lim x → 3

ln x x

c. lim x → 1

x 3 − x 2 − x − 15 3

2

x − 2x − 9

f. lim x → 0

x 4 − x 3 + x 2 −1 x 4 − 2x 2 + x

x − sin x x3 193

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

2x − π cos x

g. lim x → π

2

h. lim t → 1

t 3 −1

i. lim x → 0

3

4t − t − 3

sin 3 x x

Solutions: e x −1

a. lim x → 0

x

e 0 −1

=

3

0

e x ⋅1

ln x x

b. lim x → ∞ rule lim x → ∞

c. lim x →1

lim x →1

3x

ln x x

ln ∞ ∞

2

x − 2x + x

x 4 − x 3 + x 2 −1 x 4 − 2x 2 + x

=

=

1 − 2 ⋅1 + 1

=

d (− x ) − 0 e x + e − x ⋅ dx

Hamilton Education Guides

=

2

( e − 1) x

d dx

x3

= lim x →0

d dx

d 1 e x − dx

= lim x →0

3x 2

d x−0 e x ⋅ dx

3x 2

1 = ∞ 0

d dx

1⋅ d x dx

( x − x + x − 1) ( x − 2x + x) 4

d dx

3

4

2

2

4 ⋅1 − 4 ⋅1 + 1

e x + 1x − 2 e

sin x

e x + e −x − 2 sin x

=

=

1 x

= lim x →∞

1−1+1−1 1− 2 +1

3

=

x

1

4 ⋅13 − 3 ⋅12 + 2 ⋅1

= lim x → 0

cos x

3⋅ 0

=

2

= lim x →∞

4

the L’Hopital’s rule lim x → 0

lim x → 0

e0

d dx

which is an indeterminate function. Let’s apply the L’Hopital’s

= lim x →1

4x − 4x +1

e x + e −x − 2 sin x

2

which is an indeterminate function. Let’s apply the

14 − 13 + 12 − 1

4 x 3 − 3x 2 + 2 x − 0 3

=

d ln x dx d x dx

= lim x →∞

4

3x

∞ ∞

=

x 4 − x 3 + x 2 −1

= lim x →1

d. lim x → 0

=

ex

0 0

= lim x →0

x3

= lim x→0

2

=

e x −1

L’Hopital’s rule lim x →0

= lim x →0

1− 1 0

=

3

=

=

= lim x →1

e 0 + 10 − 2 e

sin 0

e x + e − x ⋅ −1 lim x → 0 cos x

1

2−2 2−2

4−3+ 2 4 − 4 +1

= lim x → 0

⋅1

d dx

=

=

=

(e

x

0 0 d dx

6−3 1

1 x

= lim x →∞

=

1 ∞

= 0

d x3 + d x2 − d 1 x 4 − dx dx dx d dx

d x2 + d x x 4 − 2 dx dx

=

3 1

0

+ e −x − 2 sin x

1

Apply the L’Hopital’s rule

1 + 11 − 2

d dx

1 x

= lim x →∞

=

)

= 3

1+1− 2 0

=

= lim x → 0

e x − e−x lim x → 0 cos x

=

2−2 0 d dx

lim x → 0

0 0

=

Apply

d e −x − d 2 e x + dx dx d dx

ex −

sin x

1 ex

cos x

194

Calculus I

=

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

e0 −

1 e0

=

cos 0

e. lim x → 3

1− 1 1

=

1

x 3 − x 2 − x − 15 x − 2x − 9

= lim x → 3

3x − 4 x − 0

x − sin x x

= lim x → 0

= lim x → 3

0

(x − sin x ) d dx

x

= lim x → 3

3

0−0 0

=

3

= lim x → 0

d dx

3

=

0 + sin x 6x

d sin x dx lim x → 0 d 6x dx

g. lim x → π

2

2x − π cos x d dx

= lim x → π

2

h. lim t →1

=

3

d dx

2

3x − 4 x

0 0

x

3

3

0 0

= lim x → 0

cos x 6

=

cos 0 6

=

1 6

cos π

cos x

4t − t − 3

3x 2

=

π −π

=

0

= lim x → π

2

13 − 1 3

4 ⋅1 − 1 − 3

Hamilton Education Guides

=

=

0 0

2−0 − sin x

3⋅3 − 4 ⋅3

1 − cos x

sin 0 6⋅0

=

d dx

= lim x → 3

2

= lim x → 0

=

2 ⋅ π2 − π

0 0

=

Apply the L’Hopital’s d x 2 − d x − d 15 x 3 − dx dx dx d x2 − d 9 x 3 − 2 dx dx

d dx

27 − 6 − 1 27 − 12

=

=

Apply the L’Hopital’s rule lim x → 0

d sin x x − dx d dx

27 − 27 27 − 27

3 ⋅ 32 − 2 ⋅ 3 −1

=

sin x 6x

2

t 3 −1

2

= lim x → 0

d π 2 x − dx

d dx

2

3x 2 − 2 x − 1

=

=

( x − x − x − 15) ( x − 2 x − 9)

d dx

Apply the L’Hopital’s rule again lim x → 0

= lim x → 0

27 − 9 − 3 − 15 27 − 18 − 9

=

2

3 − 2⋅3 − 9

0 − sin 0

=

3

d dx

3

x 3 − 2x 2 − 9

2

= 0

3 3 − 3 2 − 3 − 15

x 3 − x 2 − x − 15

3x 2 − 2 x − 1 − 0

0 1

=

=

2

3

rule lim x → 3

f. lim x → 0

1 − 11

1 − cos x 3x

= lim x → 0

=

2

d dx

d dx

2 − sin π

=

2 −1

=

0 0

2

=

3x 2

=

20 15

=

1 −1 4−4

2x − π cos x

=

4 3

x − sin x x3 1−1 0

= lim x → 0

=

0 0

d 1 − d cos dx dx d 3x 2 dx

Apply the L’Hopital’s rule again lim x → 0

2

1−1 4 −1− 3

3⋅ 0

2

( 1 − cos x )

Apply the rule lim x → π

=

1 − cos 0

27 − 7 15

(

sin x 6x

d 2x − π dx d cos x 2 dx

= lim x → π

x

)

= −2

Apply the L’Hopital’s rule

195

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

t 3 −1

lim t →1

4t 3 − t − 3 3t 2

= lim t →1

i.

12t − 1

= lim x → 0

2

12 ⋅1 − 1

sin 0 0

=

d dt

3 ⋅ 12

=

2

sin 3 x x

lim x → 0

d dt

= lim t →1

d 3x cos 3 x ⋅ dx

1

=

0 0

(t

3

)

−1

( 4t

3

=

3 12 − 1

−t −3

= lim t →1

)

d t3 − d 1 dt dt d 4t 3 − d t − d dt dt dt

3t 2 − 0 12t 2 − 1 − 0

3 11

=

Apply the L’Hopital’s rule

cos 3 x ⋅ 3 1

= lim x → 0

3

= lim t →1

lim x → 0

sin 3 x x

=

lim x → 0

d dx

sin 3 x d dx

x

= lim x → 0 3 cos 3x = 3 lim x → 0 cos 3x = 3 cos (3 ⋅ 0)

= 3 cos 0 = 3 ⋅ 1 = 3 Example 3.6-2: Evaluate the limit of the following functions: a. limθ → π

sin θ π −θ

d. lim x → 0

x

2

b. lim x → 0

c. lim t → 5

x+ x

e. lim x → 0 (csc x − cot x )

e x −1 tan 5 x tan 7 x

g. lim x → 0

x

h. lim x → 0

sin x 2 x

t −5 t2 −5

f. lim x → 0

e 2x −1 tan 3 x

i. lim x → 0

x(cos x − 1) sin x − x

Solutions: a. limθ → π

2

b. lim x → 0

sin π2 2 sin θ 1 = = π = π π π −θ π−2 2

x x+ x

= lim x → 0

c. lim t → 5

d.

lim x → 0

d dx

d dx

x+

t −5 2

t −5 x e x −1

0

=

x d dx

=

=

0

0+

x

0 0

Apply the L’Hopital’s rule lim x → 0

= lim x → 0

5−5 2

5 −5 0 e0 −1

Hamilton Education Guides

=

=

0 25 − 5

=

0 1−1

1 1+

=

=

0 0

1

=

2 x

1 1+

1

2⋅ 0

=

1 1+

1 0

=

1 1+

1 0

x x+ x

=

1 1+ ∞

= lim x → 0

=

d dx

d dx

x

(x+ x)

1 = 0 ∞

0 = 0 20

Apply the L’Hopital’s rule

lim x → 0

x e x −1

=

lim x → 0

d dx

d dx

x

( e −1) x

196

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

= lim x → 0

d dx x

d dx

x

e −

= lim x → 0

d 1 dx

1

= lim x → 0

x

e −0

 1 cos x   −  sin x sin x 

e. lim x → 0 (csc x − cot x ) = lim x → 0 

f. lim x → 0

e 2x −1 tan 3 x

= lim x → 0

= lim x → 0

g. lim x → 0

=

d dx

d dx

i. lim x → 0

=

tan 5 x

5 ⋅ cos 2 7 x 2

7 ⋅ cos 5 x

=

=

d x2 cos x 2 ⋅ dx d dx

x

x(cos x − 1) sin x − x

tan 0 tan 0

sin 0 0

= lim x → 0

= lim x → 0

Hamilton Education Guides

= lim x → 0

=

0 0

= lim x → 0

=

7 ⋅ cos 0

d 3x sec 2 3 x ⋅ dx

=

=

1−1 0

=

( 1 − cos x ) d dx

=

sin x

= lim x → 0

2e 2⋅0 ⋅ cos 2 3 ⋅ 0 3

=

=

sec 2 5 x ⋅ 5 sec 2 7 x ⋅ 7

5 ⋅ (cos 0 )2 7 ⋅ (cos 0 )

2

=

= lim x → 0

5 ⋅ 12 7 ⋅1

2

Apply the L’Hopital’s rule

cos x 2 ⋅ 2 x 1

x cos x − x sin x − x

d dx

2e 2 x − 0

2e 2 x cos 2 3 x 3

= lim x → 0

2

0 0

1− cos 0 sin 0

Apply the L’Hopital’s rule lim x → 0

5 ⋅ cos 2 0

=

=

=

cos 0 2 ⋅ 2 ⋅ 0 1

0 ⋅ cos 0 − 0 sin 0 − 0

=

=

1⋅ 2 ⋅ 0 1

0 ⋅1 − 0 0−0

=

=

lim x → 0

=

0−0 0−0

e 2x −1 tan 3 x

=

2e 0 ⋅ cos 2 0 3

=

2 3

tan 5 x tan 7 x

=

5 cos 2 5 x lim x → 0 7 cos 2 7 x

5 7

sin x 2 x

=

lim x → 0

d dx

sin x 2 d dx

x

0 = 0 1

=

0 0

x

sec 2 3 x ⋅ 3

7 sec 2 7 x

=

Apply

2e 2 x

5 sec 2 5 x

5 ⋅1 7 ⋅1

0 0

d 1 − d cos dx dx lim x → 0 d sin x dx

Apply the L’Hopital’s rule lim x → 0

= lim x → 0

tan 3 x

d 7x sec 2 7 x ⋅ dx

7 ⋅ cos 5 ⋅ 0

0 0

=

d 5x sec 2 5 x ⋅ dx

2

=

d dx

3 cos 2 3 x

=

1− 1 0

d 1 e 2 x − dx

d dx

2e 2 x

5 ⋅ cos 2 7 ⋅ 0

sin 0 2 0

1 − cos x sin x

= 1

1 − cos x sin x

=

= lim x → 0

tan 7 x

e

1 1

=

0

e0 −1 tan 0

= lim x → 0 tan 5 ⋅ 0 tan 7 ⋅ 0

1

= lim x → 0 tan x = tan 0 = 0

= lim x → 0

tan 3 x

3 ⋅ 12 cos 3 x

sin x 2 x

= lim x → 0

−1

2e 2 x

d lim x → 0 dx d dx

lim x → 0

)

2x

=

=

sin x cos x

= lim x → 0 e 2⋅0 − 1 tan 3 ⋅ 0

=

(e

tan 5 x tan 7 x

= lim x → 0

h.

0 + sin x cos x

e

x

= lim x → 0

the L’Hopital’s rule to the expression lim x → 0

= lim x → 0

1

Apply the L’Hopital’s 197

Calculus I

rule

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

( x cos x − x ) = ( sin x − x )

d dx lim x → 0 d dx

= lim x → 0

( cos x − x sin x ) − 1

lim x → 0

=

cos x − 1

( cos x ⋅ dxd x + x ⋅ dxd cos x )− dxd x d dx

sin x −

( cos 0 − 0 ⋅ sin 0) − 1 cos 0 − 1

L’Hopital’s rule again to the expression lim x → 0

= lim x → 0

= lim x → 0 = lim x → 0

lim x → 0

d dx

d x sin x − d 1 cos x − dx dx d dx

d 1 cos x − dx

(

2 sin x + x cos x sin x

= lim x → 0

)

2 ⋅ 0 + 0 ⋅1 0

= lim x → 0

2 cos x + cos x ⋅1 − x ⋅ sin x cos x

3 cos 0 − 0 ⋅ sin 0 cos 0

=

3 ⋅1 − 0 ⋅ 0 1

=

d dx

0 0

=

=

3 1

1−1

=

d x sin x − 0 cos x − dx d dx

cos x − 0

( cos x ⋅1 + x ⋅ − sin x ) − 1

= lim x → 0

1−1 1−1

=

cos x − 1 0 0

d dx

= lim x → 0

= lim x → 0

− sin x − ( sin x + x cos x ) − sin x

d dx

Apply the

[( cos x − x sin x ) − 1] d (cos x − 1) dx

d x sin x cos x − dx d dx

cos x

= lim x → 0

−2 sin x − x cos x − sin x

Apply the L’Hopital’s rule again to the expression

= lim x → 0

sin x

= lim x → 0

3−0 1

( 1 − 0 ⋅ 0) − 1

cos x − 1

(2 sin x + x cos x ) d dx

x

( cos x − x sin x ) − 1

= lim x → 0

− sin x

=

d dx

= lim x → 0

d x + x ⋅ d sin x − sin x − sin x ⋅ dx dx

2 sin x + x cos x sin x

=

d dx

d x + x ⋅ d cos x 2 cos x + cos x ⋅ dx dx

2 cos x + cos x − x sin x cos x

cos x

= lim x → 0

3 cos x − x sin x cos x

= 3

Example 3.6-3: Evaluate the limit of the following functions: a. lim x → 0

e x − e −x − 4x sin 2 x − 3 x

b. lim x → 0

d. lim x → 0

x + sin 3 x x − sin 3 x

e. lim x → 1

3 ln x

h. lim x → 0

g. lim x → +∞

2 x

Hamilton Education Guides

sin x x x 3 − 2 x 2 + 3x − 2 3

x − 2x +1 2 xe x 1− e x

c. lim x → 0

e x −1

f. lim x → +∞ i. lim x → ∞

x3

ln x 2x

5+ x2 x2

198

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

Solutions: a. lim x → 0

e x − e −x − 4x sin 2 x − 3 x

= lim x → 0

=

d (− x ) − 4 e x − e − x ⋅ dx d 2x − 3 cos 2 x ⋅ dx

1+1− 4 2 cos 0 − 3

b. lim x → 0

sin x

sin 0 0

cos x

=

1

2 x

0 0

=

− 4x

sin 2 x − 3 x

=

e0 −

1 e0

− 4⋅0

(e

x

sin 2 ⋅ 0 − 3 ⋅ 0 d dx

= lim x → 0

= lim x → 0

−2 −1

=

(2 ⋅1) − 3

=

x

= lim x → 0

1+1− 4

=

1 ex

e x − e −x − 4x sin 2 x − 3 x

L’Hopital’s rule lim x → 0

= lim x → 0

ex −

d dx

(

=

1 − 11 − 0 0−0

− e −x − 4x

)

)

1− 1 0

= lim x → 0

( sin 2 x − 3x )

e x − e − x ⋅ −1 − 4 cos 2 x ⋅ 2 − 3

=

ex +

= lim x → 0

1 ex

=

0 0

d dx

d e −x − d 4x e x − dx dx

−4

2 cos 2 x − 3

d dx

=

Apply the

d 3x sin 2 x − dx

e0 +

1 e0

−4

2 cos (2 ⋅ 0 ) − 3

= 2

Apply the L’Hopital’s rule lim x → 0

cos x lim x → 0 1 1 2 x

= lim x → 0

sin x x

= lim x → 0

d sin x dx d x dx

cos x ⋅ 2 x 2 x cos x = lim x → 0 = lim x → 0 2 x cos x 1 ⋅1 1

= 2 ⋅ 0 ⋅ cos 0 = 0 ⋅ cos 0 = 0 ⋅1 = 0 c. lim x → 0

=

e x −1 x3

lim x → 0

d. lim x → 0

d dx

=

e 0 −1 03

d 1 e x − dx

3x

=

0 + sin (3 ⋅ 0 ) 0 − sin (3 ⋅ 0 )

(x + sin 3x ) = (x − sin 3x )

= lim x → 0

d dx d dx

= lim x → 0

1 + 3 cos 3 x 1 − 3 cos 3 x

Hamilton Education Guides

=

0 0

=

= lim x → 0

2

x + sin 3 x x − sin 3 x

=

1− 1 0

Apply the L’Hopital’s rule lim x → 0

ex −0 3x

2

=

0+0 0−0

lim x → 0

d dx d dx

1 + 3 cos ( 3 ⋅ 0) 1 − 3 cos ( 3 ⋅ 0)

e0

=

3⋅ 0

=

=

2

0 0

x−

=

x3

= lim x → 0

sin 3 x

1 + 3 cos 0 1 − 3 cos 0

= lim x → 0

=

d dx

( e − 1) x

d dx

x3

= ∞

Apply the L’Hopital’s rule lim x → 0

d sin 3 x x + dx d dx

1 0

e x −1

d 3x 1 + cos 3 x ⋅ dx d 1 − cos 3 x ⋅ dx

1 + ( 3 ⋅1) 1 − ( 3 ⋅1)

=

1+ 3 1− 3

=

3x 4 −2

x + sin 3 x x − sin 3 x

= lim x → 0

1 + cos 3 x ⋅ 3 1 − cos 3 x ⋅ 3

= −2

199

Calculus I

e. lim x → 1

lim x → 1

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

x 3 − 2 x 2 + 3x − 2 3

x − 2x +1 x 3 − 2 x 2 + 3x − 2

f.

3 x 2 − 2 ⋅ 2 x + 3 ⋅1 − 0 3 x − 2 ⋅1 + 0

= lim x → +∞

=

1 x

2

2 x

2⋅

lim x → 0

i. lim x → ∞

3 x x

1− e0

2 xe x x

(

2 ex + x ex −e 5+ x2 x

x

)

=

= lim x → ∞

Hamilton Education Guides

(x

3

)

)

= lim x → 1

− 2x +1

3x 2 − 4 x + 3 2

3x − 2

=

1 ∞

4−4 2−2

=

3 ⋅ 12 − 4 ⋅ 1 + 3

=

2

3 ⋅1 − 2

0 0

= d dx

3 x 1 x

d dx

=

3 x

d dx

x

d x+ d 1 x 3 − 2 dx dx

3− 4+3 3− 2

ln x lim x → + ∞ 2x

=

=

2 1

= 2

d ln x dx lim x → + ∞ d 2x dx

= 0

= lim x → + ∞

d dx

Apply the L’Hopital’s rule

d x2 + 3 d x − d 2 x 3 − 2 dx dx dx

3 ln x

Apply the L’Hopital’s rule lim x → + ∞

3 x x

d 3 ln x dx d 2 x dx

3⋅ ∞ ∞ = Apply the L’Hopital’s rule ∞ ∞

=

3 2 x

= lim x → + ∞

2 x

= lim x → + ∞

= lim x → + ∞

1

3 2 x

=

3 2⋅ ∞

=

3 = 0 ∞

2 ⋅ 0 ⋅1 2 xe x 0 = Apply the L’Hopital’s rule lim x → 0 1−1 0 1− e x

=

= lim x → 0

(1 − e )

2

∞ ∞

= lim x → + ∞ 2 ⋅ 0 ⋅ e0

=

− 2 x 2 + 3x − 2

d dx

1 2⋅∞

=

2 x

1− e x

d dx

1 2x

=

3

Apply the L’Hopital’s rule

= lim x → + ∞

1

2 xe x

= lim x → 0

2⋅ ∞

3 ⋅ 1x

d dx

=

3 ln ∞

=

again lim x → + ∞

h. lim x → 0

∞ ∞

= lim x → +∞

3 ln x

= lim x → + ∞

ln ∞ 2⋅∞

(x

= lim x → 1

2

ln x lim x → + ∞ 2x

g. lim x → + ∞

1 − 2 ⋅1 + 1 d dx

1− 2 + 3 − 2 1− 2 +1

=

3

= lim x → 1

x 3 − 2x +1

= lim x → 1

13 − 2 ⋅12 + 3 ⋅1 − 2

=

(

d x + x⋅ d ex 2 e x ⋅ dx dx d 1− d dx dx

(

2 e0 + 0 ⋅ e0 −e 5 x

2

)

+1

=

ex

=

0

5 ∞

2

)

+1

= lim x → 0

2 ( 1 + 0) −1

=

5 +1 ∞

=

=

(

2 e x ⋅1 + x ⋅ e x

)

0−ex

2 = − 2 −1

0 +1

=

1

= 1

200

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

Example 3.6-4: Evaluate the limit of the following functions: a. lim x → 10 d. lim x → ∞ g. lim x → ∞ j. lim x → 0

e x − e10 x − 10

b. lim x → 5

x4 + x3 + 5

e. lim x → ∞

e x +1 ex

x 3 − 25 x x − 125

3x 2 − 1

3e x + 5 x

e 5 x − e −5 x 5 sin x

k. lim x → 0

x 2x 2

f. lim x → ∞

e 2x

7 x + 5 ln x x + 2 ln x

h. lim x → +∞

ln x

c. lim x → ∞

3

5e x + 2 x

i. lim x →+ ∞

e x − e −x x

l. lim x → 0

2 x 2 − ln x 3 x 2 + 3 ln x

cos x − 1 sin x

Solutions: a. lim x →10

e x − e10 x − 10

= lim x →10

b. lim x → 5

(e

x

d dx

(x (x

ln x

= lim x → ∞

3

)

1 x 1

5 3 − 25 ⋅ 5 5 3 − 125

) − 125) ln ∞

∞ ∞

= lim x → ∞

2 x

d. lim x → ∞

x4 + x3 + 5

= lim x → ∞

e x +1 d dx

(x d dx

4

=

x

Hamilton Education Guides

=

Apply the L’Hopital’s rule lim x →10

d dx d dx

d e10 dx d 10 dx

0 0

=

d 25 x x 3 − dx 3

x −

d 125 dx

= lim x →10

ex −0 1− 0

2 x x

= lim x → ∞

=

= lim x → ∞

∞ ∞ d dx

2 x

e x − e10 x − 10

= lim x →10 e x = e 10

Apply the L’Hopital’s rule lim x → 5

= lim x → 5

3 x 2 − 25

=

2 ∞

=

2 ∞

=

2

3x − 0

Apply the L’Hopital’s rule lim x → ∞

e∞ +1

)

d ex − dx d x− dx

125 − 125 125 − 125

∞4 + ∞3 + 5

+ x3 + 5

( e + 1)

0 0

= lim x → 5

=

∞

=

= lim x →10

− 25 x

3

=

x

− e10

=

x 3 − 125 d dx

e10 − e10 10 − 10

( x − 10)

d dx

x 3 − 25 x

= lim x → 5

c. lim x → ∞

d dx

=

ln x x

3 ⋅ 5 2 − 25 3⋅5

2

= lim x → ∞

d dx

d 1 e x + dx

x 3 − 125

75 − 25 75

=

50 75

=

2 3

d ln x dx d x dx

= 0

Apply the L’Hopital’s rule lim x → ∞ d x3 + d 5 x 4 + dx dx

=

x 3 − 25 x

= lim x → ∞

x4 + x3 + 5

4 x 3 + 3x 2 + 0 ex +0

e x +1

= lim x → ∞

4 x 3 + 3x 2 ex

201

Calculus I

=

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

4 ⋅ ∞3 + 3⋅ ∞ 2 e∞ d dx

= lim x → ∞

∞ ∞

=

( 4x

3

Apply the L’Hopital’s rule again to the expression lim x → ∞

+ 3x 2

d dx

e

)

d dx

= lim x → ∞

x

d 3x 2 4 x 3 + dx d dx

e

x

Apply the L’Hopital’s rule again to the expression d 12 x 2 + d dx dx d ex dx

= lim x → ∞

to the expression

= lim x → ∞

24 e

e. lim x → ∞

= lim x → ∞

e

=

∞

3x 2 − 1 e 2x d dx

e

24 x + 6 e

24 ∞

x

e 2⋅∞

2x

24 x + 6 e

=

2x 2

=

5e x + 2 x

= lim x → ∞

d dx

=

= lim x → ∞

d dx

( 5e

2⋅∞2 5⋅ e∞ + 2 ⋅ ∞

2x 2 x

+ 2x

)

=

=

lim x → ∞

∞ ∞

24 ⋅ ∞ + 6 e

d dx

6x − 0 2e

2x

e

=

∞ ∞

= lim x → ∞

6x

x

=

ex

=

∞ ∞

=

lim x → ∞

2e

2x

6x 2e

= lim x → ∞

12 ⋅ ∞ 2 + 6 ⋅ ∞

=

e

∞

lim x → ∞

d dx

(12 x d dx

∞ ∞

=

2

+ 6x

)

ex

Apply the L’Hopital’s rule again d dx

d 6 24 x + dx d dx

= lim x → ∞

x

e

2x

=

6⋅∞ 2⋅e

d 6x dx d 2e 2 x dx

2⋅∞

∞ ∞

=

= lim x → ∞

3x 2 − 1

24 + 0

d 2x 2 dx lim x → ∞ d 5e x + d dx dx

2x

= lim x → ∞

4x 5e x + 2

4x x

5e + 2

= lim x → ∞

=

ex

d dx

(3x − 1) 2

d dx

e 2x

Apply the L’Hopital’s

6 4e

2x

=

4⋅e

2⋅∞

=

6 ∞

= 0

5e x + 2 x

=

∞

5⋅e + 2

( 5e

6

2x 2

4⋅∞

d dx

d dx

= lim x → ∞

e 2x

Apply the L’Hopital’s rule lim x → ∞

L’Hopital’s rule again to the expression lim x → ∞

Hamilton Education Guides

∞

(24 x + 6) d dx

12 x 2 + 6 x

lim x → ∞

Apply the L’Hopital’s rule lim x → ∞

rule again to the expression lim x → ∞

f. lim x → ∞

x

e

x

ex

= 0

3⋅ ∞ 2 −1

=

d 1 3 x 2 − dx d dx

= lim x → ∞

lim x → ∞

24

=

x

6x

12 x 2 + 6 x

= lim x → ∞

4 x 3 + 3x 2

4x x

+2

)

∞ ∞

Apply the

= lim x → ∞

d 4x dx d 5e x + d dx dx

202

2

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

= lim x → ∞ g. lim x → ∞

4

= lim x → ∞

x

5e + 0 ex

d dx

= lim x → ∞

d dx

( 3e

+ 5x

= lim x → ∞

)

5⋅e

∞ ∞

=

3⋅ e∞ + 5⋅ ∞

ex

x

5e

4

=

x

e∞

=

3e x + 5 x

4

4 = 0 ∞

=

∞

Apply the L’Hopital’s rule lim x → ∞

d dx

ex

d dx x

= lim x → ∞

d 5x 3e + dx

L’Hopital’s rule again to the expression lim x → ∞

= lim x → ∞ h. lim x → +∞

ex x

3e + 0

7 x + 5 ln x x + 2 ln x

= lim x → +∞

i. lim x → + ∞

= lim x → ∞

d dx d dx

2 x 2 − ln x 3 x 2 + 3 ln x

= lim x → + ∞

d dx

=

∞ ∞

=

d lim x → + ∞ dx d dx

3e

= lim x → +∞

(7 x + 5 ln x ) = (x + 2 ln x )

d dx

ex

=

x

= lim x → ∞

7 ⋅ ∞ + 5 ⋅ ln ∞ ∞ + 2 ⋅ ln ∞

lim x → + ∞

2 ⋅ ∞ 2 − ln ∞

( 2 x − ln x) ( 3x + 3 ln x) 2

2

∞ ∞

= lim x → + ∞

3e + 5

∞ ∞

j. lim x → 0

1 x d 3 dx x

6x +

e 5 x − e −5 x 5 sin x

= lim x → + ∞

= lim x → 0

L’Hopital’s rule lim x → 0

Hamilton Education Guides

4+ 6−

e 5 x − 51x e

5 sin x

e 5 x − e −5 x 5 sin x

1 x2 3 x2

=

d dx

d dx

( 3e

=

+5

ex x

+5

7 + 5x

= lim x → +∞

d 2 x 2 − d ln x dx dx d 3 x 2 + d 3 ln x dx dx

=

6−

1 ∞2 3 ∞2

e 5⋅0 − 51⋅0 e

= lim x → 0

=

=

5 ⋅ sin 0 d dx

Apply the d ex dx lim x → ∞ d 3e x + d dx dx

=

)

1+

7 + ∞5

=

2 x

2 ∞

1+

=

Apply the L’Hopital’s rule lim x → + ∞

4+

∞ ∞

Apply the L’Hopital’s rule lim x → +∞

= lim x → + ∞

Apply the L’Hopital’s rule again to the expression lim x → + ∞

d 4 x − dx

3⋅ e

∞

1 3

=

=

x

3e x + 5 x

e∞

=

= lim x → ∞

3e x + 5

d 7 x + d 5 ln x dx dx d x + d 2 ln x dx dx

=

3 ⋅ ∞ 2 + 3 ln ∞

1 3

ex

ex

ex

(e

5x

d dx

4+0 6−0

e0 −

1 e0

5⋅0 − e −5 x

5 sin x

4 6

=

)

=

=

1 − 11 0

4 x − 1x 6x +

3 x

4 x − 1x

6 x + 3x

=

7+0 1+ 0

7 x + 5 ln x x + 2 ln x

=

7 1

= 7

2 x 2 − ln x 3 x 2 + 3 ln x

4 ⋅ ∞ − ∞1 6⋅∞ +

3 ∞

=

∞−0 ∞+0

( 4 x − 1x ) = lim x → + ∞ d ( 6x + 3 ) dx x d dx

2 3

=

= lim x → 0

1− 1 0

=

0 0

d e 5x − d dx dx d 5 sin dx

Apply the e −5 x x

203

5

Calculus I

=

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

5e 5 x + 5e −5 x lim x → 0 5 cos x

= lim x → 0

k. lim x → 0

e x − e −x x

= lim x → 0

lim x → 0

e x − e −x x

= lim x → 0

= lim x → 0 l. lim x → 0

ex + 1

1 ex

= lim x → 0

x d dx

(e

1 ex

x



=

cos 0 − 1 sin 0

sin x

=

=

0

d dx

)

=

=

=

=

1 − 11

1 e

=

0

d dx

= 1+

0

1− 1 0

5⋅ e0 + 5 ⋅1

=

0 0

d e −x e x − dx

1 1

d dx

5 e0

=

5 ⋅1 + 15

sin x

= lim x → 0

5+5 5

10 5

=

= 2

Apply the L’Hopital’s rule e x + e −x 1

= lim x → 0

x

=

5

= 1+ 1 = 2

Apply the L’Hopital’s rule lim x → 0

d 1 cos x − dx d dx

5 e5⋅0

5 ⋅ cos 0

1 e0

= e0 + 0 0

5 ⋅ e 5⋅0 +

= lim x → 0

x

1   ex 

1−1 0

= lim x → 0

e0 −

− e −x

d dx

= lim x → 0  e x +

(cos x − 1) d dx

ex −

5 e5 x

5 cos x



cos x − 1 sin x d dx

5e 5 x +

− sin x − 0 cos x

=

− sin 0 cos 0

cos x − 1 sin x

=

0 1

= 0

Example 3.6-5: Evaluate the limit of the following functions: a. lim x → 3 d. lim t → 0

x 3 − 27 x2 −9

sin t 2 t

g. lim t → 0

t − sin t

j. lim x → 0

3 − 3 cos x

t

x 4 − 16 2x − 4

c. lim x → ∞

e. lim u → 0

sin 10u u

f. lim t → 0

t − sin t

i. lim x →0

sin x

l. lim x → 0

cos x − 1 cos 3 x − 1

1+ x −1

h. lim x → +∞

3

x+x

b. lim x → 2

k. lim x → ∞

2

x

2

2 + x2 x

3

6x 2 − 5x 7x 2 +1

t2 x3

Solutions: a. lim x → 3

x 3 − 27 x2 −9

=

( (

3 3 − 27 32 − 9

) )

d x 3 − 27 dx lim = x →3 d x2 −9 dx

b. lim x → 2

x 4 − 16 2x − 4

=

=

=

27 − 27 9−9

lim x → 3

=

0 0

Apply the L’Hopital’s rule lim x → 3

d x 3 − d 27 dx dx d x2 − d 9 dx dx

= lim x → 3

3x 2 − 0 2x − 0

= lim x → 3

3x 2 2x

x 3 − 27 x2 −9

=

3 ⋅ 32 2⋅3

=

27 6

=

2 4 − 16 16 − 16 x 4 − 16 0 = = Apply the L’Hopital’s rule lim x → 2 2⋅2 − 4 4−4 2x − 4 0

Hamilton Education Guides

204

9 2

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

(x

d dx d dx

= lim x → 2

c. lim x → ∞

4

− 16

)

( 2 x − 4)

6 x 2 − 5x 7x 2 +1

=

6⋅∞2 − 5⋅∞ 7 ⋅ ∞ 2 +1

( (

) )

d 6 x 2 − 5x dx lim = x→∞ d 7x 2 +1 dx

=

∞ ∞

=

lim x → ∞

=

=

= lim x → 2

6⋅∞2 − 5⋅∞

lim x → ∞

=

7 ⋅ ∞ 2 +1

d 6 x 2 − d 5x dx dx d 7x 2 + d 1 dx dx

∞ ∞

4x 3 − 0 2−0

sin t 2 t

lim t → 0

e. lim u → 0

d 12 x − d dx dx d 14 x dx

=

d du

sin 10 ⋅ 0 0

sin 10u d du

=

u

0 0

=

12 14

=

6 x 2 − 5x

Apply the L’Hopital’s rule lim x → ∞

= lim x → ∞

12 x − 5 14 x + 0

= lim x → ∞

12 x − 5 14 x

=

12 ⋅ ∞ − 5 14 ⋅ ∞

d dx

( 12 x − 5)

12 x − 5 14 x

= lim x → ∞

7x 2 +1

d 14 x dx

6 7

Apply the L’Hopital’s rule lim t → 0

sin t 2 t

= lim t → 0

d dt

sin t 2 d dt

t

= lim t → 0 cos t 2 ⋅ 2t = lim t → 0 2t cos t 2 = 2 ⋅ 0 ⋅ cos 0 2 = 2 ⋅ 0 ⋅ 1 = 0

1

=

12 − 0 14

= lim x → ∞

sin 0 2 0

d t2 cos t 2 ⋅ dt

sin 10u u

= lim u → 0

5

4x 3 4 ⋅ 23 32 = = = 16 2 2 2

= lim x → 2

Apply the L’Hopital’s rule again to the expression lim x → ∞

d. lim t → 0

=

d x 4 − d 16 dx dx d 2x − d 4 dx dx

= lim x → 2

=

sin 0 0

= lim u → 0

0 0

=

Apply the L’Hopital’s rule lim u → 0

d 10u cos 10u ⋅ du

1

= lim u → 0

cos 10u ⋅10 1

sin 10u u

= lim u → 0 10 cos 10u = 10 cos 10 ⋅ 0

= 10 cos 0 = 10 ⋅1 = 10 f. lim t → 0

t − sin t t2

= lim t → 0

lim t → 0

=

1 − cos t 2t

1 − cos t 2t

=

0 − sin 0

=

0 0

1 − cos 0 2⋅0

=

02

=

lim t → 0

Hamilton Education Guides

d dt

Apply the L’Hopital’s rule lim t → 0 1− 1 0

=

( 1 − cos t ) d dt

2t

0 0

=

t − sin t t2

= lim t → 0

d dt

( t − sin t ) d t2 dt

Apply the L’Hopital’s rule again to the expression

lim t → 0

d 1 − d cos t dt dt d 2t dt

= lim t → 0

0 + sin t 2

=

sin 0 2

=

0 2

= 0

205

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

t − sin t

g. lim t → 0

=

t

lim t → 0

0 − sin 0

=

3

0

( t − sin t )

d dt

0−0 0

=

3

d dt

= lim t → 0

d t3 dt

=

lim t → 0

sin t 6t

3t

( =

x3

cos x 3x

2

x

sin 0

lim x → 0

k. lim x → ∞

d dx

=

2

cos 0 3⋅ 0

=

2

2

=

2 + ∞2 ∞3

Hamilton Education Guides

−0

2x

0 0

= cos t 6

=

2

3t d dt

=

d dt

3t 2

cos 0 6

=

1− 1 0

0 0

=

= lim x → 0

=

t3

1− 1 0

= lim t → 0

0 0

=

Apply the

d 1 − d cos t dt dt d 3t 2 dt

1+ x −1

Apply the L’Hopital’s rule lim x → 0

1 2 1+ x

2x

1 6

=

1 2 1+ 0

2⋅0

=

1 0

=

sin x x3

1 2

0

=

1 2 0 1

=

= lim x → 0

1 ⋅1 2⋅0

=

1 0

x2

= ∞

d sin x dx d x3 dx

= ∞

2

=

3 − 3 ⋅1 0

lim x → 0

=

3⋅ 0

=

2

t − sin t

Apply the rule again to the expression

Apply the L’Hopital’s rule lim x → 0

=

1 − cos 0

( 1 − cos t )

0 0

0+0

(x+ x )

2 + x2

=

sin 0 6⋅0

1 − cos t

= lim t → 0

3t 2

0

1 2 1+ x

3 − 3 cos 0

( 3 − 3 cos x )

d dx

x3

=

03

3 − 3 cos x x+x

2

1 − cos t

1 −1

=

02

)

=

= lim t → 0

2

= lim t → 0

1+ 0 −1

=

1+ x −1 d dx

sin x

= lim x → 0

=

d dx

sin t 6t

d sin t dt d 6t dt

= lim t → 0

x2

= lim x → 0

j. lim x → 0

= lim t → 0

1+ x −1

h. lim x → 0

i. lim x → 0

0 + sin t 6t

Apply the L’Hopital’s rule lim t → 0

d sin t t − dt

rule again to the expression lim t → 0

= lim t → 0

0 0

∞ ∞

d dx

=

3−3 0

d 3 cos x 3 − dx

d dx

x+

d dx

x

2

=

0 0

Apply the L’Hopital’s rule lim x → 0

= lim x → 0

0 + 3 sin x 1 + 2x

Apply the L’Hopital’s rule lim x → ∞

=

3 sin 0 1+ 2 ⋅ 0

2 + x2 x3

=

3⋅ 0 1

= lim x → ∞

=

d dx

3 − 3 cos x x + x2 0 1

= 0

(2 + x ) 2

d dx

x3

206

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

= lim x → ∞

d x2 2 + dx

d dx

d dx

x

3

0 + 2x

= lim x → ∞

3x

2

= lim x → ∞

2x 3x

2 3x

= lim x → ∞

2

=

2 3⋅ ∞

=

2 ∞

= 0

Note: Think of the division of 2 by ∞ as the division of 2 by a very large number. In that case, the result would then be very close to zero. l. lim x → 0

cos x − 1 cos 3 x − 1

=

cos 0 − 1 cos ( 3 ⋅ 0 ) − 1

=

d dx lim x → 0 d dx

( cos x − 1) = ( cos 3x − 1)

=

sin 0 3 sin (3 ⋅ 0 )

=

lim x → 0

sin x 3 sin 3 x

sin 0 3 sin 0

=

cos 0 − 1 cos 0 − 1

=

1−1 1−1

d cos x − d 1 dx dx lim x → 0 d cos 3 x − d 1 dx dx

0 3⋅ 0

= lim x → 0

=

=

0 0

d sin dx d 3 sin dx

x 3x

=

0 0

Apply the L’Hopital’s rule lim x → 0

= lim x → 0

− sin x − 0 − 3 sin 3 x − 0

= lim x → 0

cos x − 1 cos 3 x − 1

sin x 3 sin 3 x

Apply the L’Hopital’s rule again to the expression

= lim x → 0

cos x 9 cos 3 x

=

cos 0 9 cos (3 ⋅ 0 )

=

cos 0 9 cos 0

1 9 ⋅1

=

1 9

=

Solving indeterminate types of the form 0 ⋅ ∞ and ∞ − ∞ : Indeterminate forms of the type 0 ⋅ ∞ and ∞ − ∞ can sometimes be transformed to the indeterminate forms of the type

0 0

or

∞ ∞

by changing the given expression algebraically. For example, the

expression lim x → ∞ x 2 e −2 x result in the indeterminate form of the type ∞ ⋅ 0 as the limit x approaches infinity , i.e., lim x → ∞ x 2 e −2 x = ∞ 2 ⋅ e −2⋅∞ = ∞ 2 ⋅ rewriting the given expression in the form of lim x → ∞ type

∞ ∞

lim x → ∞

as the limit x approaches infinity. x2 e

2x

=

d x2 dx lim x → ∞ d e 2x dx

= lim x → ∞

e

e 2x

=

= lim x → ∞

1 e

2x

d ⋅ dx

d 2x ⋅ dx

= lim x → ∞

1 ∞

= ∞ ⋅ 0 . However, by

result in the indeterminate form of the

2x

= lim x → ∞

Applying the L’Hopital’s rule again to the expression lim x → ∞ d x lim x → ∞ dx d e 2x dx

e

= ∞⋅

Thus, the L’Hopital’s rule can be applied, i.e.,

2x 2x

x2

1 2⋅∞

1 e

2x

⋅2

2x e

2x

x e

= lim x → ∞

⋅2

= lim x → ∞

x e

2x

we obtain: lim x → ∞

2x

1 2e

2x

=

1 2e

2⋅∞

=

∞

=

e

2⋅∞

=

∞ ∞

x e

2x

1 1 = = 0 2⋅∞ ∞

The following are additional examples of expressions that result in the indeterminate forms of the type 0 ⋅ ∞ and ∞ − ∞ and can be changed to the indeterminate forms of the type

Hamilton Education Guides

0 0

or

∞ ∞

:

207

.

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule



4

Example 3.6-6: Evaluate the limit of the algebraic expression lim x → 2 

2

 x −4

Solution:  4 1   lim x → 2  − 2  x −4 x−2

4

=

2

2 −4

−

−

1  . x − 2 

4 1 1 4 1 − = = − = ∞ − ∞ Since the limit leads to 4−4 2−2 2−2 0 0

the indeterminate form of the type ∞ − ∞ we need to see if the given expression can be rewritten in a different algebraic form. Let’s change the algebraic expression to the form that leads to the indeterminate form of the type

0 0

by taking the common denominator of the

algebraic expression and simplifying the numerator and the denominator. = lim x → 2

− 22 + 4⋅ 2 − 4

−4 + 8 − 4 8−8−8+8

=

=

23 − 2 ⋅ 2 2 − 4 ⋅ 2 + 8

= lim x → 2

d dx d dx

(x

(− x 3

2

+ 4x − 4

)

− 2x 2 − 4x + 8

(x

−2 12 − 4

=

−2 8

= −

=

0 0

2

)

)

)

− 4 (x − 2 )

4x − 8 − x 2 + 4

= lim x → 2

x 3 − 2x 2 − 4x + 8

Apply the L’Hopital’s rule to lim x → 2

= lim x → 2

L’Hopital’s rule again to lim x → 2

=

(

4 (x − 2 ) − x 2 − 4

 4 1   lim x → 2  − 2 2  x −  x −4

−2 x + 4 2

3x − 4 x − 4

−2 x + 4 2

3x − 4 x − 4

=

= lim x → 2

−2 ⋅ 2 + 4 2

3⋅ 2 − 4 ⋅ 2 − 4 d dx

d dx

(− 2 x + 4)

( 3x

2

− 4x − 4

x 3 − 2x 2 − 4x + 8

−4 + 4 12 − 8 − 4

)

= lim x → 2

=

0 0

Apply the

−2 6x − 4

=

−2 6⋅2 − 4

1 4

Solution: 1 x

x 3 − 2x 2 − 4x + 8

− x 2 + 4x − 4

=

Example 3.6-7: Evaluate the limit of the algebraic expression lim x → ∞ x sin

lim x → ∞ x sin

− x 2 + 4x − 4

= lim x → 2

= ∞ ⋅ sin

1 ∞

1 x

.

= ∞ ⋅ sin 0 = ∞ ⋅ 0 Since the limit leads to the indeterminate form of

the type ∞ ⋅ 0 we need to see if the given expression can be rewritten in a different algebraic form. Let’s change the algebraic expression to the form that leads to the indeterminate form of the type

0 0

lim x → ∞ x sin

by letting x = 1 x

1 t

1 t

= lim t → 0 sin

Hamilton Education Guides

and by allowing t → 0 . 1 1 t

1 t

= lim t → 0 sin t = lim t → 0

sin t t

=

sin 0 0

=

0 Apply the 0

208

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

L’Hopital’s rule to lim t → 0

sin t t

d dt

= lim t → 0

sin t d dt

t

= lim t → 0

cos t 1

= lim t → 0 cos t = cos 0 = 1

Solving indeterminate types of the form 1∞ , ∞ 0 , and 0 0 : Similarly, indeterminate forms of the type 1∞ , ∞ 0 , and 0 0 may first be transformed to the indeterminate forms of the type lim x →a y = lim x →a f ( x )

0 0

or

∞ ∞

by multiplying both sides of the function 1

by logarithm ( ln ) . For example, lim x→0 (cos x ) x result in the indeterminate 1

1

form of the type 1∞ as the limit x approaches zero, i.e., lim x→0 (cos x ) x = cos 0 0 1

1

= cos 0 0 = 1∞ . However, if we let y = (cos x ) x and take the logarithm of both sides of the equation the solution has an indeterminate limit of the type 1

ln y = ln (cos x ) x

=

ln 1 0

=

0 0

1 x

1

1

, i.e., given y = (cos x ) x , then ln (cos 0 ) ln (cos x ) = 0 x 0 type the L’Hopital’s 0

and ln y = lim x→0 ln (cos x ) x = lim x→0 ln (cos x ) ; lim x→0

. Since the limit result in an indeterminate form of the ln (cos x ) lim x →0 x

can be applied to the expression = lim x→0

0 0

1 ⋅ − sin x cos x

= lim x→0 −

sin x cos x

= lim x→0

d dx

ln (cos x ) d dx

x

= lim x→0

1 ⋅ d cos x dx

rule

cos x

1

= lim x→0 − tan x = − tan 0 = 0 . Thus, ln y = 0 . To solve

for y we raise both sides of the equation by e and solve for y , i.e., e ln y = e 0 ; y = e 0 ; y = 1 . Therefore, indeterminate forms of the type 1∞ , ∞ 0 , and 0 0 may first be transformed to the indeterminate forms of the type

0 0

or

∞ ∞

by rewriting the function lim x→a y = lim x→a f (x ) in a

different form and solving the algebraic expression. The following are additional examples of expressions that result in the indeterminate forms of the type 1∞ , ∞ 0 , and 0 0 which may be rewritten in a different form and solved: Example 3.6-8: Evaluate the limit of the algebraic expression lim x → 0 x sin x . Solution: lim x → 0 x sin x y = x sin x

= 0 sin 0 = 0 0 Since the limit leads to the indeterminate form of the type 0 0 let

then by taking the logarithm of both sides of the equation we obtain lim x→0 ln y

= lim x → 0 ln x sin x = lim x → 0 sin x ln x = lim x → 0

Hamilton Education Guides

ln x csc x

=

ln 0 csc 0

=

∞ ∞

Apply the L’Hopital’s rule to

209

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

lim x → 0

ln x csc x

= lim x → 0 −

= lim x → 0

1 ⋅ sin 2 x x ⋅ cos x

= lim x → 0 −

again to the expression

= lim x → 0 −

d ln x dx d csc x dx

= lim x → 0

sin 2 x x cos x

lim x → 0 −

2 sin x ⋅ cos x cos x ⋅1 + x ⋅ − sin x

= −

= −

sin 2 x x cos x

=

1 x

− csc x cot x

sin 2 0 0 ⋅ cos 0

= −

lim x → 0 −

2 ⋅ sin 0 ⋅ cos 0 cos 0 − 0 ⋅ sin 0

= −

= lim x → 0

sin 0 0 ⋅1

cos x

− sin1 x ⋅ sin x

= lim x → 0

−

1 x cos x

sin 2 x

0 Apply the L’Hopital’s rule 0

=

d sin 2 x dx d x cos x dx

2 ⋅ 0 ⋅1 1− 0 ⋅ 0

1 x

=

= −

lim x → 0 −

0 1− 0

= −

d sin x 2 sin x ⋅ dx

d x + x ⋅ d cos x cos x ⋅ dx dx

0 1

= 0

Thus, ln y = 0 ; e ln y = e 0 ; y = e 0 ; y = 1 x

1 Example 3.6-9: Evaluate the limit of the algebraic expression lim x → ∞  cos  . x



Solution: 1  lim x → ∞  cos  x 

x

=

1   cos  ∞  

∞

= (cos 0)∞ = 1∞ Since the limit leads to the indeterminate form

of x

1 the type 1∞ let y =  cos  then by taking the logarithm of both sides of the equation we x



obtain lim x→∞ ln y = lim x → ∞

=

ln (cos 0 ) 0

= lim x → ∞

= lim x → ∞

= −

0 1

=

0 0

ln 1 0

=

1

1 ⋅ d cos dx x

cos (1 x )

− 1

x2

1 1 ⋅ − sin cos (1 x ) x 1

1  ln  cos  x 

x

= lim x → ∞

1  x ln  cos  x 

Apply the L’Hopital’s rule to lim x → ∞ 1

= lim x → ∞

= lim x → ∞

cos (1 x )

1 cos (1 x )

⋅ − sin

1 d 1 ⋅ x dx x

− 1

x2

⋅ − sin

1 x

= lim x → ∞

1  ln  cos  x  1 x

1

= lim x → ∞

= lim x→∞ −

cos (1 x )

sin ( 1 x ) cos ( 1 x )

1  ln  cos  x  1 x

= lim x → ∞ ⋅ − sin

=

1  ln  cos  ∞  1 ∞

d ln  cos 1    dx  x d 1 dx x

1 1 ⋅− x x2

− 1

x2

= −

sin ( 1 ∞ ) cos ( 1 ∞ )

= −

sin 0 cos 0

= 0 Thus, ln y = 0 ; e ln y = e 0 ; y = e 0 ; y = 1

Hamilton Education Guides

210

Calculus I

3.6 Evaluation of Indeterminate Forms Using L’Hopital’s Rule

Example 3.6-10: Evaluate the limit of the algebraic expression lim x → π ( tan x )cos x . 2

Solution: lim x → π ( tan x )cos x 2

= ( tan 2 )

π π cos 2

=

 sin π  2    cos π  2  

cos π

2

 1    0

=

0

= ∞ 0 Since the limit leads to the

indeterminate form of the type ∞ 0 let y = ( tan x )cos x then by taking the logarithm of both sides of the equation we obtain lim x→ π ln y = lim x→ π ln ( tan x )cos x = lim x→ π cos x ln ( tan x )

=

ln ( tan x ) sec x

lim x → π

2

d dx

= lim x→ π

2

ln ( tan x ) d dx

sec x

sec x

= lim x→ π

2

tan x

2

=

(

ln tan π2

)

sec π

2

= lim x→ π

2

= lim x→ π

2

1 cos x 2

sin x cos 2 x

2

2

2

=

ln ∞ ∞

1 ⋅ d tan x dx

=

ln ( tan x ) ∞ Apply the L’Hopital’s rule to lim x→ π ∞ sec x 2

tan x

= lim x→ π

sec x tan x

= lim x→ π

2

2

cos 2 x 2

cos x sin x

1 tan x

⋅ sec 2 x

sec x tan x

= lim x→ π

2

= lim x→ π

2

cos x 2

sin x

=

sec 2 x sec x tan 2 x

cos π2

sin 2 π2

=

0 1

= 0

Thus, ln y = 0 ; e ln y = e 0 ; y = e 0 ; y = 1 Section 3.6 Practice Problems - Evaluation of Indeterminate Forms Using L’Hopital’s Rule Evaluate the limit for the following functions by applying the L’Hopital’s rule, if needed: a. lim x → ∞ d. lim t → 2 g. lim x → 8 j. limt → ∞

ln x x

2

=

b. lim x → 0

t−2 2

t + 2t − 1 x −8 2

x − 64 8t + 3 4t − 2

=

=

=

Hamilton Education Guides

sin x x

e −1

=

c. lim x → 0

1 − cos x

=

f. lim t → 0

t cos t 2 sin t

=

=

i. lim x → 0

sin 8 x 3x

=

l. lim

1 − sin x 1 + cos 2 x

e. lim x → π

cos x

h. lim x →π

sin x π −x

k. lim x → 0

cos x − 1

π 2 x−

2

x

2

=

x→

π 2

x3

=

=

211

Calculus I

Quick Reference to Chapter 4 Problems

Chapter 4

Integration (Part I) Quick Reference to Chapter 4 Problems 4.1

Integration Using the Basic Integration Formulas ................................................. 213

∫(x 4.2

6

)

+ x 2 + 3 dx

∫ 3 x 2 −1

dx

=;

cos 2 x + sin 2 x dx sin 2 x

)

x + 5 x dx

=

∫

4x 3 + 6x x 4 + 3x 2 + 5

dx = ;

2x −1

∫ 4 x 2 − x − 3 dx

=

∫e

=;

sin 5 x

∫e

cos 5 x dx = ;

tan 5 x

sec 2 5 x dx =

Integration of Expressions Resulting in Inverse Trigonometric Functions ......... 259

∫ 4.5

∫ (3 x +

=;

Integration of Trigonometric Functions .................................................................. 232

∫ 4.4



1

Integration Using the Substitution Method ............................................................. 222 x

4.3

 1

∫  x 5 + x 2 + 6  dx

=;

dx 16 − 9 x 2

=;

∫

x2 9+ x

6

dx

=;

∫

dx 2

9 x + 25

=

Integration of Expressions Resulting in Exponential or Logarithmic Functions 273

∫ (e

Hamilton Education Guides

x

)

+3

2

e x dx

=;

x+6

∫ x + 5 dx

=;

1

∫ x 2 + 10 x + 24 dx

=

212

Calculus I

4.1 Integration Using the Basic Integration Formulas

Chapter 4 – Integration (Part I) The objective of this chapter is to improve the student’s ability to solve problems involving integration of various classes of functions. Integration using the basic integration formulas is addressed in Section 4.1. Integration of functions using the Substitution method is discussed in Section 4.2. How to find the integral of trigonometric functions is addressed in Section 4.3. Integration of expressions resulting in inverse trigonometric functions is addressed in Section 4.4. Finally, the steps in integrating expressions that result in exponential or logarithmic functions is discussed in Section 4.5. Each section is concluded by solving examples with practice problems to further enhance the student’s ability.

4.1

Integration Using the Basic Integration Formulas dy A function F (x ) is called a solution of the differential equation = f (x ) over a defined interval dx

I

if the function F (x ) is differentiable at every point on that interval and if

d F (x ) = f (x ) dx

at

every point on that interval as well. Note that the function F (x ) is referred to as the antiderivative of f (x ) . The set of all antiderivatives of a function f (x ) is called the indefinite

∫ f (x ) dx .

The

∫ f (x ) dx = F (x ) + c .

The

integral of f with respect to x . The indefinite integral is shown by the notation solution to the indefinite integral is given by F (x ) + c and is shown as symbol

is referred to as the integral sign. The function f (x ) is called the integrand of the

∫

integral. The dx indicates that the variable of integration is x and c is the constant of integration. In the following examples we will solve problems using the following basic indefinite integration formulas: ∫ a dx = ax + c a ≠ 0

∫ a f (x ) dx

= a ∫ f (x ) dx

a≠0

∫ [ f (x ) + g (x ) ] dx = ∫ f (x ) dx + ∫ g (x ) dx

∫

x n dx =

x n +1 +c n +1

n ≠ −1

Let’s integrate some integrals using the above basic integration formulas. Example 4.1-1: Evaluate the following integrals. a. d. g.

∫ dx = 2 ∫ 10a dx = 5 ∫ (a + b ) x dx

b. e. =

h.

∫ a dx = ∫ 3x dx = 3 ∫ ( x + x ) dx =

∫ 5 dx = f.. ∫ x 3 dx = i. ∫ ( x 6 + x 2 + 3) dx = c.

Solutions: a.

∫ dx

=

∫x

0

dx =

1 0+1 x +c 0 +1

Hamilton Education Guides

= x+c 213

Calculus I

4.1 Integration Using the Basic Integration Formulas

Check: Let y = x + c , then y ′ = x1−1 + 0 = x 0 = 1 b.

∫ a dx

a 0+1 x +c 0 +1

= a ∫ dx = a ∫ x 0 dx =

= ax + c

Check: Let y = ax + c , then y ′ = ax1−1 + 0 = ax 0 = a ⋅1 = a c.

∫ 5 dx = 5∫ dx

5 0+1 x +c 0 +1

= 5∫ x 0 dx =

= 5x + c

Check: Let y = 5 x + c , then y ′ = 5 x1−1 + 0 = 5x 0 = 5⋅1 = 5 d.

∫

∫

∫

10a 2 dx = 10a 2 dx = 10a 2 x 0 dx =

10a 2 0+1 x +c 0 +1

= 10a 2 x + c

Check: Let y = 10a 2 x + c , then y ′ = 10a 2 x1−1 + 0 = 10a 2 x 0 = 10a 2 ⋅1 = 10a 2 e.

∫ 3x dx

= 3∫ x dx = 3∫ x1dx =

3 1+1 x +c 1+1

3 2

Check: Let y = x 2 + c , then y ′ = f.

∫ x dx 3

1 3+1 x +c 3 +1

=

=

1 4

∫ (a + b ) x

5

= (a + b ) ∫ x 5 dx =

dx

3 ⋅ 2 x 2−1 + 0 2

= 3x

1 ⋅ 4 x 4−1 + 0 4

= x3

1 4 x +c 4

Check: Let y = x 4 + c , then y ′ = g.

3 2 x +c 2

=

a + b 5+1 x +c 5 +1

=

a+b 6 x +c 6

or we can find the solution to the above integral in the following way – which is rather long:

∫ (a + b ) x

5

∫ ( ax

dx =

 1 5+1  +b x + c2   5 +1 

Check: Let y = h.

∫(x

3

)

+ x dx

=

5

=

)

+ bx 5 dx =

3

5

 1 5+1  + c1  x dx + bx 5 dx = a x 5 dx + b x 5 dx = a   5 +1 

∫

b a 6 x + a c1 + x 6 + b c 2 6 6

a+b 6 x +c , 6

∫x

∫ ax

∫

∫(x

6

)

+ x 2 + 3 dx =

=

Hamilton Education Guides

6

∫

6

6

6

1 1+1 1 3+1 x +c x + 1+1 3 +1

1 2

∫x

a+b 6 a b x +c =  +  x 6 + (a c1 + b c 2 ) =

6

Check: Let y = x 4 + x 2 + c , then y ′ = i.

∫

a+b then y ′ = ⋅ 6 x 6−1 + 0 = (a + b ) x 5

dx + x dx

1 4

∫

∫

dx + x 2 dx + 3dx =

=

1 4 1 2 x + x +c 4 2

1 1 ⋅ 4 x 4−1 + ⋅ 2 x 2−1 + 0 4 2

= x3 + x

1 6+1 1 2+1 3 0+1 + x x + x +c 6 +1 2 +1 0 +1

=

1 7 1 3 x + x + 3x + c 7 3 214

Calculus I

4.1 Integration Using the Basic Integration Formulas

1 1 ⋅ 7 x 7 −1 + ⋅ 3 x 3−1 + 3 x1−1 + 0 7 3

1 3

1 7

Check: Let y = x 7 + x 3 + 3x + c , then y ′ =

= x6 + x2 + 3

Example 4.1-2: Evaluate the following integrals. dx

a.

∫ x3

d.

∫

g.

3

b.

x 2 dx 1

∫

=

t

=

e.

=

dt

h.

 1



1

∫  x 5 + x 2 + 6  dx =

c.

∫

x + 5 x dx =

f.

∫ (5 x + x

∫ (3 x + 1

∫ 5 x2

)

dx

=

i.

Solutions: a.

dx

∫ x3

=

∫x

−3

1 x −3+1 + c − 3 +1

dx =

1 = − x −2 + c = − 2

1

∫

x 5 dx

 1   + z 2  dz 3 2   z 

= 3

)

+ 10 dx =

=

+c

2x 2

1 1 1 Check: Let y = − x −2 + c , then y ′ = − ⋅ −2 x −2−1 + 0 = x −3 = 3 2

2

b.

 1



1

1

x

1

∫  x 5 + x 2 + 6  dx = ∫ x 5 dx + ∫ x 2 dx + ∫ 6 d x +

6 0+1 x +c 1+ 0

1 = − x −4 − x −1 + 6 x + c = − 4

=

∫x

−

1 + 6x + c x

1 4x

4

1 4

−5

∫

∫

dx + x −2 dx + 6 x 0 dx =

1 1 x − 2+1 x −5+1 + − 2 +1 − 5 +1

1 4

Check: Let y = − x −4 − x −1 + 6 x + c , then y ′ = − ⋅ −4 x − 4−1 + x −1−1 + 6 x1−1 + 0 = x −5 + x −2 + 6x 0 x −5 + x −2 + 6

=

1 x

5

+

1 x2

+6

1  Exception: Note that we can not apply the same integration technique  ∫ x n dx = x n +1 + c 

n +1



in order to find the integral of 1

c.

∫ x dx = ∫ x

−1

dx =

∫

∫

x 5 dx

x 5 dx

=

1 x −1+1 + c −1+1

1

5 6

=

1 1 + 15

1 +1

x5

1

∫ x dx .

+c

6

Check: Let y = x 5 + c , then y ′ = d.

∫

3

x 2 dx

=

∫

2

=

x 3 dx

3 5

5

1 1 + 23

2 +1

x3

+c

Check: Let y = x 3 + c , then y ′ = Hamilton Education Guides

This is because division by zero is undefined, i.e.,

1 0 x +c. 0

= =

1 5+1 5

x

1+ 5 5

As we will see in Section 4.5, +c =

1 6 5

6

x5 +c =

6−5 1 5 6 65 −1 ⋅ x +0 = x 5 = x5 = 6 5

=

1 3+ 2 3



x

2+3 3

+c =

1 5 3

5

x3 +c =

1

∫ x dx = ln

x +c.

5 65 x +c 6

x5 3 53 x +c 5

5−3 2 3 5 53 −1 3 ⋅ x + 0 = x 3 = x 3 = x2 5 3

215

Calculus I

e.

4.1 Integration Using the Basic Integration Formulas

∫ (3 x + +

)

5 1+1 x +c 1+1

1

=

x

3+1 3

1   1  x 3 + x 2 + 5 x  dx    

∫

x + 5 x dx =

1+ 3 3

4

3 4

1

+

2 +1 2 3

2 3

x

1+ 2 2

=

∫

1 dx

∫

1

1

f.

∫(

5

1 5+1 5

)

x + x 3 + 10 dx

3

x

1+ 5 5

+

6

=

1

∫

t

∫

1 dx

∫

=

∫

1 1

∫

dt =

t2

t

5

∫ 5 x2

dx

−1

=

2 dt

1 1 − 12

t

1− 1

2

1 2

=

∫

1 2

dx =

x5

∫

x

−2

5 dx

=

1 1 − 52

3

5 3

Check: Let y = x 5 + c , then y ′ = i.

∫ =

 1   + z 2  dz 3 2   z 

1 3− 2 3

z

3− 2 3

x2

= x

4 −3 3

+x

3− 2 2

+ 5x

=

1 +1

1 1+

1 5

x5

+

1 3+1 x + 10 x + c 1+ 3

6 −5 5 6 56 −1 1 ⋅ x + ⋅ 4 x 4−1 + 10 x1−1 + 0 = x 5 + x 3 + 10 x 0 6 5 4

x + x 3 + 10

1

1

∫

+ x 3 dx + 10 dx

+c

=

1 2 −1 2

1 −1

Check: Let y = 2t 2 + c , then y ′ = 2 ⋅ t 2 + 0 = t h.

1 + 12

3 4 43 −1 2 3 32 −1 5 ⋅ x + ⋅ x + ⋅ 2 x 2−1 + 0 4 3 3 2 2

x5

1 4

= x 5 + x 3 + 10 = dt

1 +1

1

5 6 1 1 4 x + 10 x + c = x 5 + x 4 + 10 x + c 6 4 4

5 6

1

1 + 13

+

x + x + 5x

Check: Let y = x 5 + x 4 + 10 x + c , then y ′ =

g.

x3

2

5 2

 15   x + x 3 + 10  dx    

∫

=

3

∫

1 +1

1

=

3 43 2 32 5 2 1 43 1 32 5 2 5 2 x + x + x +c x + x + x +c = x +c = 4 3 2 4 3 2 2

+

Check: Let y = x 3 + x 2 + x 2 + c , then y ′ = = x 3 + x 2 + 5x =

1

+ x 2 dx + 5 x dx

x3

+

=

∫

   1 2 + z  2  dz 3 z 

=

t

2 −1 2

1− 2 2

1

= 2t 2 + c = 2 t + c

+c

=t

−1

2

1

= t

x

1− 2 5

+c

=

1 5− 2 5

x

5− 2 5

1 2

+c =

=

1 t

55 3 5 53 x +c x +c = 3 3

3−5 −2 1 5 3 53 −1 ⋅ x +0 = x 5 = x 5 = 2 = 3 5 x5

∫

 −2   z 3 + z 2  dz    

=

∫

z

−2

3 dz

∫

+ z 2 dz

1 5

=

x2 1

1−

2 3

z

1− 2 3

+

1 2+1 z +c 2 +1

1 1 1 3 1 z + c = 3z 3 + z 3 + c = 3 3 z + z 3 + c 3 3 3 1

1 3

1 3

1 −1

1 3

Check: Let y = 3z 3 + z 3 + c , then y ′ = 3 ⋅ z 3 + ⋅ 3 z 3−1 + 0 = z

Hamilton Education Guides

1−3 3

+ z2

= z

−2 3

+ z2

=

1 3

z

2

+ z2

216

Calculus I

4.1 Integration Using the Basic Integration Formulas

Example 4.1-3: Evaluate the following integrals. a.

2 ∫ (1 + x )

d.

∫ x ( 2 x + 1 ) dx

g.

3 ∫ ( a − x ) dx =

x dx = 2

=

b.

∫(x

)

e.

∫ ( a + b ) dx =

+ 1 ( x − 1) dx =

2

3

y4 + 4y3 − 6y2

∫

h.

y2

=

dy

c.

∫ ( 5x − 3) dx =

f.

∫ ( a + x ) dx

i.

2

3

∫

=

y 3 + 3y 2 + 5 y2

dy

=

Solutions: a.

∫ (1 + x ) 2

1 2

x dx =

∫

5

∫(

=

∫x

=

2 2 x3 + x7 + c 3 7

dx + x 2 dx

2 3

)

x + x 2 x dx = 1 +1

1

=

1 + 12

= 3

x2

+

∫

∫

5 +1

1

x2

1 + 32

+c

7

Check: Let y = x 2 + x 2 + c , then y ′ = b.

∫(x =

2

)

+ 1 (x − 1) dx

=

∫ (x

2 +1 2

3

)

− x 2 + x − 1 dx

1 3

=

∫x

3

25 3 30 2 x − x + 9x + c 3 2

Check: Let y = d.

∫ x ( 2 x + 1 ) dx 2

=

=

=

∫ x ( 4x

2

2+5 2

x

5+ 2 2

+c

=

∫

1

∫

x 2 dx + x

2+ 1

2 dx

2 32 2 72 x + x +c 3 7

2 2 x x + x3 x + c 3 7

=

2 x x 2 x3 x + +c 3 7

=

∫ 25x

2

∫

∫

∫x

3

∫

∫

∫

dx − x 2 dx + x dx − x 0 dx

1 4 1 3 1 2 x − x + x − x+c 4 3 2 1 1 1 ⋅ 4 x 4−1 − ⋅ 3 x 3−1 + ⋅ 2 x 2−1 − 1 + 0 4 3 2

∫

∫

dx − 30 x dx + 9 dx

= x 3 − x 2 + x −1

25 2+1 30 1+1 9 0+1 x − x + x +c 2 +1 1+1 0 +1

=

25 then y ′ = ⋅ 3 x 3−1 − 15 ⋅ 2 x 2−1 + 9 + 0 = 25 x 2 − 30 x + 9

)

+ 4 x + 1 dx

4 3+1 4 2+1 1 1+1 x + x + x +c 3 +1 2 +1 1+1

Hamilton Education Guides

1

∫

=

25 3 x − 15 x 2 + 9 x + c 3

25 3 x − 15 x 2 + 9 x + c , 3

=

=

+

∫

1 2

2 2 ∫ ( 5x − 3) dx = ∫ ( 25x − 30 x + 9) dx

1+ 2 2

1

dx − x 2 dx + x dx − dx =

Check: Let y = x 4 − x 3 + x 2 − x + c , then y ′ = c.

x

1

x 2 dx + x 2 x 2 dx

5 3− 2 7−2 1 2 3 23 −1 2 7 72 −1 ⋅ x + ⋅ x +0 = x 2 + x 2 = x2 + x2 3 2 7 2

1 1+1 1 0+1 1 3+1 1 2+1 x x x x − +c − + 1+ 0 1+ 3 1+ 2 1+1 1 4

1

=

2 2 x2 ⋅ x + x2 ⋅ x2 ⋅ x2 ⋅ x + c 3 7

2 7

∫

x dx + x 2 x dx =

=

3

=

∫ ( 4x

3

)

+ 4 x 2 + x dx

4 4 4 3 1 2 x + x + x +c 2 4 3

=

∫ 4x

3

4 3

∫

∫

dx + 4 x 2 dx + x dx 1 2

= x4 + x3 + x2 + c

217

Calculus I

4.1 Integration Using the Basic Integration Formulas

4 3

4 3

1 2

1 2

Check: Let y = x 4 + x 3 + x 2 + c , then y ′ = 4 x 4−1 + ⋅ 3x 3−1 + ⋅ 2 x 2−1 + 0 = 4 x 3 + 4 x 2 + x e.

= ( a + b )3 ∫ x 0 dx = ( a + b )3 ⋅

∫ ( a + b ) dx = ( a + b ) ∫ dx 3

3

1 0+1 +c x 0 +1

= ( a + b )3 x + c

Check: Let y = ( a + b ) 3 x + c , then y ′ = ( a + b )3 x1−1 + 0 = ( a + b )3 x 0 = ( a + b )3 f.

∫ ( a + x ) dx 3

∫

+ a 3 dx =

=

∫ (a

3

)

+ x 3 + 3a 2 x + 3ax 2 dx

∫(x

=

3

)

+ 3ax 2 + 3a 2 x + a 3 dx

1 3+1 3a 2+1 3a 2 1+1 a 3 0+1 + x x + +c x + x 3 +1 2 +1 1+1 0 +1

1 4

Check: Let y = x 4 + ax 3 +

∫x

=

3

∫

∫

dx + 3ax 2 dx + 3a 2 x dx

1 4 3a 2 2 x + ax 3 + x + a3x + c 4 2

=

3a 2 2 1 3a 2 ⋅ 2 x 2−1 + a 3 x1−1 + 0 x + a 3 x + c , then y ′ = ⋅ 4 x 4−1 + a ⋅ 3 x 3−1 + 2 4 2

= x 3 + 3a x 2 + 3a 2 x + a 3 x 0 = x 3 + 3a x 2 + 3a 2 x + a 3 = a 3 + x 3 + 3a 2 x + 3ax 2 = (a + x )3 g.

3 3 3 2 2 ∫ ( a − x ) dx = ∫ ( a − x + 3a x − 3ax )dx

∫

+ a 3 dx = −

∫ (− x

=

3

)

1 3+1 3a 2+1 3a 2 1+1 a 3 0+1 − + x + x x x +c 1+1 3 +1 2 +1 0 +1 1 4

Check: Let y = − x 4 − ax 3 +

3a 2 2 x + a3x + c , 2

∫

∫

∫

− 3ax 2 + 3a 2 x + a 3 dx = − x 3 dx − 3ax 2 dx + 3a 2 x dx 1 4

= − x 4 − ax 3 +

3a 2 2 x + a3x + c 2

1 4

then y ′ = − ⋅ 4 x 4−1 − a ⋅ 3x 3−1 +

3a 2 ⋅ 2 x 2−1 + a 3 x1−1 + 0 2

= − x 3 − 3a x 2 + 3a 2 x + a 3 x 0 = − x 3 − 3a x 2 + 3a 2 x + a 3 = a 3 − x 3 + 3a 2 x − 3ax 2 = (a − x )3 h.

∫ =

y4 + 4y3 − 6y2 y2

dy =

∫

   

y4 y2

+

4y3 y2

−

1 4 1+1 6 y 2+1 + y − y 0+1 + c 2 +1 1+1 0 +1

6 y 2  dy y 2 

=

=

1 3

∫ =

y 3 + 3y 2 + 5 y

2

dy =

∫

   

y3 y

2

+

3y 2 y

2

+

5  dy y 2 

1 1+1 3 5 y + y 0+1 + y − 2+1 + c − 2 +1 1+1 0 +1

Check: Let w =

1 2 y + 3 y − 5 y −1 + c , 2

Hamilton Education Guides

2

=

=

)

+ 4 y − 6 dy

1 3 4 2 6 y + y − y+c 3 2 1

Check: Let w = y 3 + 2 y 2 − 6 y + c , then w′ = i.

∫(y

=

∫y

2

1 ⋅ 3 y 3−1 + 2 ⋅ 2 y 2−1 − 6 y 1−1 + 0 3

1 2 y + 3 y − 5 y −1 + c 2

∫

∫

dy + 4 y dy − 6 dy

1 3 y + 2y2 − 6y + c 3

−2 ∫ ( y + 3 + 5 y ) dy

then w′ =

=

=

=

= y2 + 4y − 6

∫ y dy + ∫ 3 dy + ∫ 5 y

−2

dy

1 2 5 y + 3y − + c 2 y

1 ⋅ 2 y 2−1 + 3 y 1−1 − 5 ⋅ − y −1−1 + 0 2

= y + 3 + 5 y −2

218

Calculus I

4.1 Integration Using the Basic Integration Formulas

Example 4.1-4: Evaluate the following integrals. a.

∫(x

3

− 6 x dx =

)

b.

∫

d.

∫(x

2

− 5 x dx =

)

e.

∫5

g.

∫ ( 2 x − 1)

dx =

h.

∫ ( 2 x + 1)

3

c.

∫ 30

=

f.

∫

dx =

i.

2 3 ∫ 6 x ( x + 1) dx =

x − 1 dx = 2

x + 1 dx 2

x + 5 dx =

3 x + 2 dx =

Solutions: a.

∫(x =

3

)

− 6 x dx

=

∫

1  3  x − 6x 2  

  dx  

∫

=

∫

3

1 4

∫

x − 1 dx =

∫ (x − 1)

1 2

2 3

dx

1

=

1 2

1 +1 1 3+1 6 x − x2 +c 1 +1 3 +1 2

∫ 30

x + 5 dx

=

1

+1

3

∫ 30 ( x + 5)

1 2

=

dx

1 3 3 −1 ⋅ 4 x 4−1 − 4 ⋅ x 2 + 0 4 2

(x − 1) 2 +1 + c =

Check: Let y = (x − 1) 2 + c then y ′ = c.

=

=

1 4 6 1+22 x x − +c 1+ 2 4 2

3 1 4 1 1 1 4 12 32 x − x + c = x 4 − 4x 2 + c = x 4 − 4 x 3 + c = x − 4 x x + c 4 4 4 3 4

Check: Let y = x 4 − 4 x 2 + c then y ′ = b.

1

x 3 dx − 6 x 2 dx

30 1 2

3 2 (x − 1) 2 + c 3

3 2 3 ⋅ ( x − 1) 2 −1 + 0 3 2

1

+1

( x + 5) 2 +1 + c =

=

1

= x 3 − 6x 2 = x 3 − 6 x 2 3

(x − 1)3 + c =

= ( x − 1)

30 1+ 2 2

( x + 5)

3− 2 2

1+ 2 2

2 ( x − 1) 3

1

= ( x − 1) 2 = +c

=

x −1 + c

x −1

3 60 ( x + 5) 2 + c 3

= 20 ( x + 5 ) 3 + c = 20( x + 5) x + 5 + c Check: Let y = d.

∫(x =

2

3 60 ( x + 5) 2 + c 3

)

∫

− 5 x dx =

1 3 10 x − x3 +c 3 3

∫

then y ′ = 1

x 2 dx − 5 x 2 dx

=

1 3

Check: Let y = x 3 −

=

3 60 3 ⋅ ( x + 5) 2 −1 + 0 3 2

1 2+1 5 − x x 1 2 +1 +1 2

1 +1 2

(

)

1 3 10 x − x x +c 3 3

=

+c

= 30 ( x + 5) =

3− 2 2

1

= 30 ( x + 5) 2 = 30 x + 5

1 3 5 1+22 x − x +c 1+ 2 3 2

=

1 3 10 32 x − x +c 3 3

1 x x 2 − 10 x + c 3

1 10 32 1 10 3 3 −1 3 30 3−2 2 = x2 −5 x 2 x + c then y ′ = ⋅ 3 x 3−1 − ⋅ x 2 + 0 = x 2 − x 3 3 3 3 2 6

= x2 −5 x e.

2

∫5

x + 1 dx

=

2 5

∫ ( x + 1)

Hamilton Education Guides

1 2

dx

=

1 2 1 ( ⋅ x + 1) 2 +1 + c 5 1 +1 2

=

1+ 2 2 1 ( ⋅ x + 1) 2 + c 5 1+ 2

2

=

3 2 2 ⋅ ( x + 1) 2 + c 5 3

219

Calculus I

=

4.1 Integration Using the Basic Integration Formulas

3 4 ( x + 1) 2 + c 15

Check: Let y = f.

∫ =

( x + 1) 3 + c =

3 4 ( x + 1) 2 + c 15

∫ ( 3x + 2)

3 x + 2 dx = 2 3

4 15

=

1 2

dx

then y ′ = 1

=

1 2

2 ( 3x + 2 ) 3 + c = ( 3 x + 2) 3

2 3

4 ( x + 1) x + 1 + c 15

∫ ( 2 x − 1)

∫

3

dx

=

+1

3

1+ 2 2

( 3x + 2)

1+ 2 2

+c

6 6

3 2 3 ⋅ ( 3 x + 2 ) 2 −1 + 0 3 2

=

]

− 13 + 3 ⋅ (2 x )2 ⋅1 − 3 ⋅ 2 x ⋅12 dx

=

∫ ( 8x

= ( 3x + 2) 3

8 3+1 12 2+1 6 1+1 1 0+1 x x − x − x +c + 3 +1 2 +1 1+1 0 +1

∫

− 6 x dx − dx =

3− 2 12 ( x + 1) 2 30

=

=

1 2 ( x + 1) 2 5

=

2 x +1 5

3 2 ( 3x + 2) 2 + c 3

3x + 2 + c

3

∫ [ (2 x )

1

1

( 3x + 2) 2 +1 + c =

Check: Let y = ( 3x + 2) 2 + c then y ′ = g.

3 4 3 ⋅ ( x + 1) 2 −1 + 0 15 2

3− 2 2

1

= ( 3x + 2) 2 = 3x + 2

)

− 1 + 12 x 2 − 6 x dx

= 8∫ x 3 dx + 12∫ x 2 dx

= 2 x4 + 4 x3 − 3 x2 − x + c

Check: Let y = 2 x 4 + 4 x 3 − 3x 2 − x + c , then y ′ = 2 ⋅ 4 x 4−1 + 4 ⋅ 3x 3−1 − 3 ⋅ 2 x 2−1 − x1−1 + 0 = 8 x 3 + 12 x 2 − 6 x − 1 = (2 x − 1)3 h.

∫ ( 2 x + 1) =

2

∫ ( 4x

dx =

4 3 4 2 x + x + x+c 3 2

2

=

)

∫

4 3

2 3 ∫ 6 x ( x + 1) dx

=

∫ ( 6x

∫

1 0+1 4 1+1 4 2+1 x + x x +c + 0 +1 1+1 2 +1

4 3 x + 2x 2 + x + c 3

Check: Let y = x 3 + 2 x 2 + x + c then y ′ = i.

∫

+ 4 x + 1 dx = 4 x 2 dx + 4 x dx + dx =

5

)

+ 6 x 2 dx

4 ⋅ 3 x 3−1 + 2 ⋅ 2 x 2−1 + 1 + 0 3

= 6∫ x 5 dx + 6∫ x 2 dx =

= 4 x 2 + 4 x + 1 = (2 x + 1)2

6 5+1 6 2+1 x + x +c 5 +1 2 +1

=

6 6 6 3 x + x +c 6 3

2 = x6 + x3 + c = x 6 + 2x 3 + c 1

Check: Let y = x 6 + 2 x 3 + c , then y ′ = 6 x 6−1 + 2 ⋅ 3x 3−1 + 0 = 6 x 5 + 6 x 2 In the next section we will discuss a more systematic method of integration, referred to as the substitution method, in evaluating more difficult integrals.

Hamilton Education Guides

220

Calculus I

4.1 Integration Using the Basic Integration Formulas

Section 4.1 Practice Problems – Integration Using the Basic Integration Formulas 1. Evaluate the following indefinite integrals: 2

a.

∫ − 3 dx =

b.

∫ 5k dx =

c.

∫ 3 x dx

=

d.

∫ x dx =

e.

∫ ax dx =

f.

∫(x

)

g.

∫ x4

h.

∫  x4 − x2  dx =

i.

∫

c.

∫ 7 x2

f.

∫ x ( 3x − 1 )

5

dx

=

7

 1

1 

3

4

+ x3 dx =

x 6 dx

=

2. Evaluate the following indefinite integrals:  3 x 2 + x  dx  

a.

∫ 

d.

∫ (1 + x )

g.

3 ∫ ( 2 + x ) dx =

x dx

1

=

b.

∫− 2

=

e.

∫ (2x

h.

3 ∫ ( 2 − x ) dx =

Hamilton Education Guides

t 2

dt

=

)

+ 1 (x − 1) dx

=

i.

1

∫

dx

y5 + 4 y 2 y2

= 2

dx

dy

= =

221

Calculus I

4.2

4.2 Integration Using the Substitution Method

Integration Using the Substitution Method

A method of solving integrals in which a change in variable can often change a difficult integral into a simpler form is referred to as the Substitution method. In this section we will learn how to use this method. Given the integral ∫ f [ g (x ) ]⋅ g ′(x ) dx , where f (x ) and g ′(x ) are continuous functions, use the following steps in order to solve the integral by the Substitution method: First - Let u = g (x ) , then

du = g ′(x ) dx

which implies du = g ′(x ) ⋅ dx and dx = du g ′(x )

Second - Replace g (x ) with u and dx with

du . g ′(x )

to obtain du

∫ f [ g (x ) ]⋅ g ′(x ) dx = ∫ f [ u ]⋅ g ′(x )⋅ g ′(x ) = ∫ f [ u ]⋅ du

Third - Solve the integral by integrating the function f (u ) with respect to u using integration formulas. Fourth - Change the solution back to its original form by replacing u with g (x ) . Fifth - Check the answer by taking the derivative of the solution. In the following examples we will solve integrals using the above substitution method. Example 4.2-1: Evaluate the following indefinite integrals: a.

4 5 2 ∫ 5x (x − 3) dx =

b.

2 2 3 ∫ x (x + 1) dx

d.

∫ 2x

e.

∫x

x 2 + 3 dx

g.

∫ 3 5x − 1 dx

h.

∫ 6x

5

1 − x 2 dx

=

=

=

=

3 x 2 + 2 dx

=

c.

3 3 4 ∫ 4 x ( x − 1) dx =

f.

∫

i.

∫

10 x + 1 dx 1 x+6

dx

=

=

Solutions:

(

)

du dx

a. Given ∫ 5 x 4 x 5 − 3 2 dx let u = x 5 − 3 , then and dx =

du 5x

4

Check: Let y =

∫ 5x

=

(

4

⋅u 2 ⋅

)

du 5x

3 1 5 x −3 +c , 3

4

=

∫u

2

du

then y ′ =

2 3 2 3 ∫ x ( x + 1) dx let u = x + 1 , then

and dx =

du 3x 2

(

) = 5x

d x5 − 3 dx

4

which implies that du = 5 x 4 dx

. Substituting the equivalent values of x 5 − 3 and dx back into the integral we obtain

2 4 5 ∫ 5x (x − 3) dx

b. Given

=

1 2+1 u +c 2 +1

=

(

)

=

1 3 u +c 3

1 ⋅ 3 x 5 − 3 3−1⋅ 5 x 4 + 0 3 du dx

=

(

(

(

1 5 x −3 3

= 5x 4 x 5 − 3

) = 3x

d x3 +1 dx

=

2

)

)

3

+c

2

which implies that du = 3x 2 dx

. Substituting the equivalent values of x 3 + 1 and dx back into the integral we obtain

Hamilton Education Guides

222

Calculus I

4.2 Integration Using the Substitution Method

2 2 2 du 2 3 ∫ x ( x + 1) dx = ∫ x ⋅ u ⋅ 3x 2

Check: Let y =

(

)

3 1 3 x +1 + c , 9

(

=

1 u 2 du 3

∫

(

du 4x 3

∫ (

)

)

3

∫

=

Check: Let y =

4x 3 ⋅ u 3 ⋅

(

)

1 4 x −1 4

4

du

=

∫

du . − 2x

=

Check: Let y = −

∫x

and dx =

4x 3

+c ,

∫

∫

x x 2 + 3 dx

dx =

∫

du 10

∫

)

) = 4x

3

)

(

)

= x2 x3 + 1

2

which implies that du = 4 x 3 dx

1 3+1 u +c 3 +1

(

)

du dx

=

d 1− x 2 dx

=

1 4 u +c 4

4 −1

⋅ 4x 3 + 0

(

=

(

1 4 x −1 4

(

)

= 4x 3 x 4 −1

)

4

+c

3

) = −2 x which implies that du = −2 x dx

∫

du − 2x

(

3 2

2 1− x 2 3

)

1

= − ∫ u 2 du = −

+c,

1 +1

1 1 2

+1

u2

let u = x 2 + 3 , then

du dx

3

3 2

+c

) = 2x which implies that du = 2x dx

= 2x 1 − x 2

⋅ −2 x + 0

)

)

(

3 −1 2

(

(

2 1− x2 3

)

d 2 x +3 dx

=

3

2 = − u2 +c = −

(

2 3 1− x 2 3 2

then y ′ = − ⋅

+c

1 2

= 2x 1 − x 2

Substituting the equivalent values of x 2 + 3 and dx back into the integral we obtain =

Check: Let y = f. Given

(

(

(

3 1 3 x +1 + c 9

1 3 x + 1 2 ⋅ 3x 2 3

=

d 4 x −1 dx

1 ⋅ 4 x 4 −1 4

then y ′ =

2x ⋅ u ⋅

x 2 + 3 dx

du . 2x

=

=

Substituting the equivalent values of 1 − x 2 and dx back into the integral we obtain

1 − x 2 dx

e. Given

du dx

u 3 du =

d. Given ∫ 2 x 1 − x 2 dx let u = 1 − x 2 , then

2x

)

1 3 u +c 9

. Substituting the equivalent values of x 4 − 1 and dx back into the integral we obtain

4 x 3 x 4 − 1 dx

and dx =

=

1 ⋅ 3 x 3 + 1 3−1⋅ 3 x 2 + 0 9

then y ′ =

c. Given ∫ 4 x 3 x 4 − 1 3 dx let u = x 4 − 1 , then and dx =

1 1 2+1 u +c 3 2 +1

=

∫

x⋅ u ⋅

du 2x

(

3 2

)

1 2 x +3 3

=

+c ,

1 1 u 2 du 2

∫

then y ′ =

10 x + 1 dx let u = 10 x + 1 , then

1 +1 1 1 ⋅ u2 +c 2 1 +1 2

=

(

)

1 3 2 ⋅ x +3 3 2

du dx

=

3 −1 2

d (10 x + 1) dx

=

(

1 2 1 2 32 1 3 x +3 ⋅ u +c = u2 +c = 3 2 3 3

⋅ 2x + 0

(

)

)

3 2

+c

1

= x x2 + 3 2 = x x2 + 3

= 10 which implies that du = 10 dx and

. Substituting the equivalent values of 10 x + 1 and dx back into the integral we obtain

10 x + 1 dx

=

∫

du u⋅ 10

Hamilton Education Guides

1

=

∫

1 +1 3 1 1 1 1 2 23 1 32 u2 ( 10 x + 1) 2 + c ⋅ u2 +c = ⋅ u +c = u +c = du = 15 10 1 + 1 10 3 15 10 2

223

Calculus I

4.2 Integration Using the Substitution Method

Check: Let y = g. Given dx =

∫

3

du 5

3 1 (10 x + 1) 2 + c , 15

∫ 3 5x − 1 dx

3 1 3 ⋅ (10 x + 1) 2 −1 ⋅10 + 0 15 2

then y ′ = du dx

let u = 5 x − 1 , then

∫

5 x − 1 dx =

3

u⋅

du 5

=

1 1 u 3 du 5

∫

4 3 ( 5 x − 1) 3 + c , 20

=

1 +1 1 1 ⋅ u3 +c 5 1 +1

du . 6x

and dx = obtain

4 3 4 ⋅ ( 5 x − 1) 3 −1 ⋅ 5 + 0 20 3

then y ′ =

du dx

(

d 3x 2 + 2 dx

=

3 43 u +c 20

=

4 3 ( 5 x − 1) 3 + c 20

=

1

= ( 5 x − 1) 3 = 3 5 x − 1

) = 6 x which implies that du = 6 x dx

Substituting the equivalent values of 3x 2 + 2 and dx back into the integral we

5

6 x 3 x 2 + 2 dx

1

6x ⋅ 5 u ⋅

(

5 3x 2 + 2 6

)

du 6x

6 5

1

+c ,

then y ′ = du dx

1 +1

1

= ∫ u 5 du =

let u = x + 6 , then

dx

x+6

∫

=

Check: Let y =

∫

1 3 34 ⋅ u +c 5 4

=

3

h. Given ∫ 6 x 5 3x 2 + 2 dx let u = 3x 2 + 2 then

i. Given

= 5 which implies that du = 5 dx and

. Substituting the equivalent values of 5 x − 1 and dx back into the integral we obtain

Check: Let y =

∫

d (5 x − 1) dx

=

1

= (10 x + 1) 2 = 10 x + 1

1 5

+1

u5

=

)

6 −1 5

(

5 6 ⋅ 3x 2 + 2 6 5 d ( x + 6) dx

=

+c

(

5 5 65 3x 2 + 2 u +c = 6 6 ⋅ 6x + 0

(

= 6 x 3x 2 + 2

)

1 5

)

6 5

+c

= 6 x 5 3x 2 + 2

= 1 which implies that du = dx . Substituting

the equivalent values of x + 6 and dx back into the integral we obtain

∫

1 x+6

dx

=

∫

1 u

du

=

∫

u

−1

2 du

=

1 1 − 12

u

1− 1

2

+c

1 2

1

=

1 2 −1 2

u

2 −1 2

1

+c

1

1

= 2u 2 + c = 2 ( x + 6) 2 + c 1

1

Check: Let y = 2 ( x + 6) 2 + c , then y ′ = 2 ⋅ ( x + 6) 2 −1 ⋅1 + 0 = ( x + 6)− 2 =

( x + 6)

1 2

=

1 x+6

Example 4.2-2: Evaluate the following indefinite integrals: a. d. g.

∫

∫ ∫

10 x−5

=

b.

dx =

e.

dx

x2 x3 +1

4x 3 + 6x x 4 + 3x 2 + 5

dx =

Hamilton Education Guides

h.

∫

∫

x x2 −3

5x 4 x5 + 3

x

∫ 3 x 2 − 1 dx =

=

c.

dx =

f.

∫ 5 x 5 + 3 dx =

i.

∫ 4 x 2 − x − 3 dx

dx

3x 2 + 4 x

∫ 3 x 3 + 2 x 2 − 1 dx

=

5x 4

2x −1

=

224

Calculus I

4.2 Integration Using the Substitution Method

Solutions: a. Given

10

∫

x−5

du dx

let u = x − 5 , then

dx

d ( x − 5) dx

=

= 1 which implies that du = dx . Substituting

the equivalent values of x − 5 and dx back into the integral we obtain

∫

10 x−5

10

∫

=

dx

= 10∫ u

du

u

−1

2 du

10

=

u

1 2

1−

1− 1

+c

2

1 2

1

10

=

2 −1 2

u

2 −1 2

1

1

= 20u 2 + c = 20( x − 5) 2 + c

+c

1

10

1

Check: Let y = 20( x − 5) 2 + c , then y ′ = 20 ⋅ ( x − 5) 2 −1 + 0 = 10( x − 5) − 2 = b. Given dx =

∫

x

∫

du . 2x

dx

2

x −3

x2 −3

=

dx

x

du u 2x

∫

(

)

1 2

+c ,

x

(

d 2 x −3 dx

) = 2 x which implies that du = 2 x dx and

2

1 1 22−1 u ⋅ +c 2 2−1

=

2

then y ′ = −1 ,

1 1 1− 12 u +c ⋅ 2 1− 1

=

then

(

=

)

)

(

(

)

1 2

+c

2

1 −1 −1 1 2 1 2 x −3 2 +0 = x − 3 2 ⋅ 2x = 2 2

du dx

(

1

= u 2 +c = x2 − 3

(x

x 2

)

−3

x

=

1 2

2

x −3

) = 2 x which implies that du = 2 x dx and

d 2 x −1 dx

Substituting the equivalent values of x 2 − 1 and dx back into the integral we obtain

x

∫ 3 x 2 −1

=

dx

∫

Check: Let y = =

∫

and dx =

∫

∫

∫ 3 x 2 − 1 dx let u = x

du . 2x

d. Given

−1 1 u 2 du 2

=

⋅

Check: Let y = x 2 − 3

dx =

=

x−5

Substituting the equivalent values of x 2 − 3 and dx back into the integral we obtain

x

c. Given

du dx

let u = x 2 − 3 , then

( x − 5)

10

=

1 2

x2 x3 +1

(

)

3 2 x −1 4

(x − 1) 2

3

x +1 du 2

=

⋅

x

x2

3x

du 3 u 2x x

∫

=

1 1 1− 13 ⋅ +c u 2 1− 1

=

3

+ c , then y ′ =

(

)

3 2 2 ⋅ x −1 4 3

(

3 2 3 2 1 1 33−1 x −1 ⋅ u +c = u3 +c = − 3 1 4 4 2

)

2 3

+c

3

2 −1 3

⋅ 2x + 0

=

)

(

1 2 x −1 2

2 −3 3

(

)

⋅ 2x = x x 2 −1

−1

3

x

=

1 3

2 3

−1 1 u 3 du 2

3

x 2 −1

dx let u = x 3 + 1 , then

du dx

=

(

) = 3x

d x3 +1 dx

2

which implies that du = 3x 2 dx

. Substituting the equivalent values of x 3 + 1 and dx back into the integral we obtain

dx =

∫

Check: Let y =

x2

⋅

du

=

u 3x 2

(

)

2 3 x +1 3

Hamilton Education Guides

1 2

−1 1 u 2 du 3

+c,

∫

=

1 1 1− 12 ⋅ u +c 3 1− 1 2

( 2

)

2 1 then y ′ = ⋅ x 3 + 1 3

1 −1 2

=

1 1 22−1 ⋅ u +c 3 2−1

⋅ 3x 2 + 0

=

2 12 u +c 3

)

⋅ 3x 2

2

=

(

1 3 x +1 3

1− 2 2

=

(

2 3 x +1 3

(

)

= x 2 x3 +1

)

−1

2

225

1 2

+c

Calculus I

4.2 Integration Using the Substitution Method

x2

= e. Given

3

5x 4

∫

5x 4

obtain 5x 4 5

x +3

x3 +1

∫

5x 4

du

⋅

u 5x 4

(

)

1 2

Check: Let y = 2 x 5 + 3

f. Given

5x 4

( x + 3) 5

5x 4

obtain 5x 4

∫ 5 x5 + 3

dx =

5x 4

∫ 5u

⋅

du 5x 4

(

)

(x + 3) 5

=

1 5

4x 3 + 6x

∫

u

4

2

x + 3x + 5

(

5

1−

1 2

u

5

4 5

∫

=

2

1

u

1 − 12

du dx

then

u

+c

1

=

2 −1 2

(

5 du

=

5x 4

which implies that du = 5 x 4 dx

1− 1

2

(

+c

1

=

u

2 −1 2

2 −1 2

)

(

1

= 2u 2 + c = 2 x 5 + 3

+c

(

1 −1 1 5 x + 3 2 ⋅ 5x 4 + 0 = x 5 + 3 2

)

1− 2 2

⋅ 5x 4

)

1 2

+c

(

)

= 5x 4 x 5 + 3

−1

2

(

) = 5x

d x5 + 3 dx

=

4

which implies that du = 5 x 4 dx

1 1 − 15

u

1− 1

5

(

1

=

5−1 5

)

4 −1 5

du dx

=

5 4 5 ⋅ x +3 4 5

dx let u = x 4 + 3 x 2 + 5 , then

u

+c

u

5 −1 5

+c =

5 54 u +c 4

)

(

⋅ 5x 4 + 0

= x5 + 3

4 −5 5

=

(

5 5 x +3 4

)

4 5

+c

(

⋅ 5x 4

)

= 5x 4 x 5 + 3

−1

5

x5 + 3

2 −1 2

du 3

4x + 6x

4

2

x + 3x + 5 1

)

1 2

+c ,

) = 4x

3

+ 6x

which implies

. Substituting the equivalent values of x 4 + 3x 2 + 5 and dx

4x 3 + 6x

∫

(

d x 4 + 3x 2 + 5 dx

dx =

(

∫

4x 3 + 6x

= 2u 2 + c = 2 x 4 + 3x 2 + 5

+c

Check: Let y = 2 x 4 + 3x 2 + 5 Hamilton Education Guides

−1

+ c , then y ′ =

)

1− 1

=

2 du

+3,

back into the integral we obtain 1

−1

then y ′ = 2 ⋅

that du = 4 x 3 + 6 x dx and dx =

=

4

x5 + 3

5 5 x +3 4

5x 4

=

) = 5x

. Substituting the equivalent values of x 5 + 3 and dx back into the integral we

Check: Let y =

g. Given

+c ,

∫ 5 x 5 + 3 dx let u = x du

∫

=

5x 4

=

1 2

5x 4

and dx =

(

d x5 + 3 dx

=

. Substituting the equivalent values of x 5 + 3 and dx back into the integral we

dx =

=

du dx

dx let u = x 5 + 3 , then

x5 + 3 du

and dx =

∫

(x + 1)

x2

=

1 2

then y ′ = 2 ⋅

(

u

)

1 2

+c

⋅

du 4x 3 + 6x

=

∫

du u

=

∫

u

−1

2 du

= 2 x 4 + 3x 2 + 5 + c

) (

)

(

1 −1 1 4 x + 3x 2 + 5 2 4 x 3 + 6 x + 0 = x 4 + 3x 2 + 5 2

)

1− 2 2

226

Calculus I

4.2 Integration Using the Substitution Method

) = (x

(

× 4x 3 + 6x

h. Given

4

3x 2 + 4 x

∫ 3 x 3 + 2 x 2 − 1 dx

(

+ 3x 2 + 5

)

1 − 13

u

1− 1

3

+c

=

Check: Let y =

1 3−1 3

du 3x + 4 x

)

) = (x

2x −1

∫ 4 x 2 − x − 3 dx

1 4 −1 4

u

4 −1 4

+c

and dx =

=

Check: Let y =

∫ 3 x 3 + 2x 2 −1

(

3

2 3

) ( 3x

+ 2x 2 −1

du . 2x −1

−1

3

(

)

−1

4

3 4

4

du dx

+ 3x 2 + 5

=

)

1 2

4x 3 + 6x

=

x 4 + 3x 2 + 5

(

) = 3x

d x 3 + 2x 2 −1 dx

2

which implies

+ 4x

∫

dx =

(

3x 2 + 4 x 3

)

+ 4x

2 3

u

=

+c

(

du dx

(x

=

du 3x 2 + 4 x

3x 2 + 4 x 3

)

+ 2x 2 −1

(

d x2 − x −3 dx

du

∫3u ∫

(

2 −1 3

=

1 3

=

=

u

−1

3 du

)

2 33 x 3 + 2x 2 − 1 + c 2

) (3x

3 2 3 ⋅ x + 2x 2 −1 2 3

)=

⋅

2

)

+ 4x + 0

(

)

2 −3 3

= x 3 + 2x 2 −1

3x 2 + 4 x 3

x 3 + 2x 2 −1

) = 2 x − 1 which implies that

Substituting the equivalent values of x 2 − x − 3 and dx back into =

)

Hamilton Education Guides

2

∫ 4 x 2 − x − 3 dx

4 2 x − x−3 3

)

(x

4x 3 + 6x

3 3 x + 2x 2 −1 2

=

+ c , then y ′ =

2x −1

(

)=

3x 2 + 4 x

4 34 4 2 u +c = x − x −3 3 3

= x2 − x −3

+ 6x

let u = x 2 − x − 3 , then

the integral we obtain

=

3

. Substituting the equivalent values of x 3 + 2 x 2 − 1 and dx

3 23 u +c 2

+c =

(

(

du = (2 x − 1) dx

3−1 3

3 3 x + 2x 2 −1 2

× 3x 2 + 4 x

i. Given

u

2

2

back into the integral we obtain

1

−1

let u = x 3 + 2 x 2 − 1 , then

that du = 3x 2 + 4 x dx and dx =

=

) ( 4x

+c ,

( 2 x − 1) =

∫ 3 4

2x −1 4

+c

u

⋅

du 2x −1

du

∫ 4 u = ∫u

=

(

44 x2 − x − 3 3

=

( 4

)

4 3 then y ′ = ⋅ x 2 − x − 3 3

(x

2x −1 2

)

− x−3

1 4

=

3 −1 4

−1

4 du

=

(2 x − 1) + 0 =

(x

)

3

1 1 − 14

u

1− 1

4

+c

+c

2

)

− x−3

3− 4 4

(2 x − 1)

2x −1 4

x2 − x −3

227

Calculus I

4.2 Integration Using the Substitution Method

Example 4.2-3: Evaluate the following indefinite integrals: x2 + 2

a.

∫ 5 x 3 + 6x + 3

d.

∫

z

g.

∫ (4 − x 2 ) 2

dz

2z 2 −1 x

∫

5x 4 − 4 x 3 + 1

dx =

b.

=

e.

∫

h.

2 3 ∫ x ( x − 1)

dx

=

5

4

x −x +x x3

dx =

dx =

x 4 −1

1 5 dx

=

x4 − 2

c.

∫

f.

∫ (1 + x ) 2

i.

∫

5

x − 10 x

dx =

dx

8x x2 −3

=

x

=

dx

Solutions: a. Given

x2 + 2

∫ 5 x 3 + 6 x + 3 dx let u = x

(

)

du = 3 x 2 + 6 dx

and dx =

du 2

3x + 6

1 1 1− 15 ⋅ u +c 3 1− 1

=

5

Check: Let y =

∫

(

) = 13 (x

4

x −x +x

(

3

)

4 5

)

+ 6x + 3

−1

5

(

⋅3 x2 + 2

∫

2 du

=

)

1 1 − 12

u

1− 1

2

(

+c

=

1 2 −1 2

Check: Let y = 2 x 5 − x 4 + x

Hamilton Education Guides

)=

dx let u = x 5 − x 4 + x , then du 4

5x − 4 x 3 + 1

and dx back into the integral we obtain −1

x2 + 2 5

⋅

) = 3x

=

1 3

+c

=

3 x2 + 2

)

(

)

4 5

(

) ⋅ (3x

u

(

du

5 4 3 ⋅ x + 6x + 3 12 5

+ c , then y ′ =

that du = 5 x 4 − 4 x 3 + 1 dx and dx =

u

∫

dx =

5 3 5 54 1 1 55−1 x + 6x + 3 u +c = ⋅ +c = u 12 12 3 5−1

5x 4 − 4 x 3 + 1 5

(

d x 3 + 6x + 3 dx

=

)

1 2

u

2

+6

du

∫5u

which implies that

2 −1 2

+c,

+c

∫

(x du dx

x2 + 2 3

=

1 5

−1 1 u 5 du 3

∫

(

)

2

)

+6 +0

=

(

)

1 3 x + 6x + 3 3

4 −5 5

x2 + 2

=

5

x 3 + 6x + 3

(

d x5 − x4 + x dx

) = 5x

4

− 4x 3 + 1

which implies

. Substituting the equivalent values of x 5 − x 4 + x

5x 4 − 4 x 3 + 1 5

)

+ 6x + 3

4 −1 5

=

4 5 5 x 3 + 6x + 3 + c 12

5

× 3x 2 + 6

b. Given

du dx

then

. Substituting the equivalent values of x 3 + 6 x + 3 and dx back

∫ 5 x 3 + 6x + 3

5 3 x + 6x + 3 12

(

+ 6x + 3 ,

x2 + 2

into the integral we obtain

=

3

4

x −x +x

dx =

∫

5x 4 − 4 x 3 + 1

(

1

= 2u 2 + c = 2 x 5 − x 4 + x

( 2

1 then y ′ = 2 ⋅ x 5 − x 4 + x

u

)

1 2

+c

) ( 5x 1 −1 2

4

⋅

du 4

3

5x − 4 x + 1

=

∫

du u

= 2 x5 − x4 + x + c

)

− 4x 3 + 1 + 0

(

= x5 − x4 + x

)

−1

2

228

Calculus I

4.2 Integration Using the Substitution Method

(

)=

5x 4 − 4 x 3 + 1

× 5x 4 − 4 x 3 + 1

c. Given

x4 − 2

∫

x 5 − 10 x

(

)

du = 5 x 4 − 10 dx

(x

and dx =

1 1 1− 12 u +c ⋅ 5 1− 1

= d. Given

(

5 x −2

and dz =

∫

∫

du 4x

3

x3 x 4 −1

)

x4 − 2

∫

(

−1

1 2

du dx

⋅5 x4 − 2

2

=

(

d 5 x − 10 x dx

∫

u

⋅

) = 5x

du

(

5 x4 − 2

(

(

2 1 5 ⋅ x − 10 x 5 2

then y ′ =

)=

x4 − 2

2 12 2 5 u +c = x − 10 x 5 5

=

(x

x4 − 2 5

− 10 x du dz

let u = 2 z 2 − 1 , then

dz

=

dx

x 5 − 10 x

+c,

x5 − x4 + x 4

which implies that

− 10

)

=

1 2

) +c

) ( 5x 1 −1 2

4

=

1 5

∫ 2 5

du

−1 1 u 2 du 5

∫

=

u

x 5 − 10 x + c

)

− 10 + 0

(

1 5 x − 10 x 5

=

) (5x −1

2

4

− 10

)

x4 − 2

=

1 2

)

=

x 5 − 10 x

(

d 2z 2 −1 dz

) = 4 z which implies that du = 4z dz

. Substituting the equivalent values of 2 z 2 − 1 and dx back into the integral we obtain z

du ⋅ = u 4z

∫

dz =

Check: Let y =

dx =

)

(

1 5 x − 10 x 5

du 4z

2z 2 −1

∫

)

(

2z −1

z

e. Given

2 5 x − 10 x 5

2

)

. Substituting the equivalent values of x 5 − 10 x and dx back

2

z

∫

du 4

1 1 22−1 ⋅ u +c 5 2−1

=

2

Check: Let y =

− x4 + x

dx let u = x 5 − 10 x , then

into the integral we obtain

=

5

5x 4 − 4 x 3 + 1

=

1 2

x3 4

x −1

(

)

1 2z 2 −1 2

1 2

∫

u

−1

2

4

du =

(

1 1 1 1 1 1− 12 1 1 22−1 2z 2 − 1 ⋅ u +c = ⋅ u +c = u2 +c = 1 − 2 1 2 2 4 4 1− 2

+ c , then y ′ =

dx let u = x 4 − 1 , then

du dx

1 2

+c

2

(

)

(

) = 4x

1 −1 1 1 ⋅ 2z 2 −1 2 ⋅ 4z + 0 = 2 2

=

)

d x 4 −1 dx

3

( 2 z − 1) 2

−1

2

⋅z

=

z 2z 2 −1

which implies that du = 4 x 3 dx and

. Substituting the equivalent values of x 4 − 1 and dx back into the integral we obtain dx =

∫

Check: Let y =

x3

⋅

du

=

u 4x 3

(

)

1 4 x −1 2

Hamilton Education Guides

1 2

∫

u

−1

2

4

du =

(

)

1 4 1 1 1 1 22−1 1 1 1− 12 x −1 u +c = u2 +c = ⋅ u +c = ⋅ 1 2 − 1 2 2 4 4 1−

+ c , then y ′ =

2

(

1 2

+c

2

)

1 −1 1 1 4 ⋅ x −1 2 ⋅ 4x 3 + 0 = 2 2

(x − 1) 4

−1

2

⋅ x3

=

x3 x 4 −1

229

Calculus I

4.2 Integration Using the Substitution Method

dx

∫ (1 + x ) 2

f. Given

du dx

let u = 1 + x , then

x

(

)=

d 1+ x dx

=

1

which implies that du =

2 x

dx 2 x

and dx = 2 x du . Substituting the equivalent values of 1 + x and dx back into the integral we obtain dx

∫ (1 + x ) 2

x

2 x du

∫

=

u

x

and dx = obtain

du . − 2x

+c,

1+ x

∫ (4 − x2 ) 2

=

x

2

Check: Let y = −

g. Given

2

du

x

Check: Let y =

3x

∫ ( x − 1) x

2

3

2

dx =

∫

∫

du . 2x

8x 2

x −3

2 2 1−2 2 +c u + c = − 2u −1 + c = − + c = − 1− 2 u 1+ x

∫

= 2 u −2 du =

then y ′ = −

1 2 x

0−

⋅2

(1 + x ) 2 du dx

1 x

=

+0

(1 + x ) 2

(

d 4 − x2 dx

=

=

1

(

x 1+ x

)2

) = −2x which implies that du = −2 xdx

(

1 5 dx

= −

+c,

)

1 1 1 1 1 1 u − 2 du = − ⋅ u − 2+1 + c = u −1 + c = +c = 2 − 2 +1 2 2 2u 2 4− x2

∫

(

then y ′ =

0 + 4x

(

4 4 − x2

let u = x 3 − 1 , then

du dx

)

+0

2

=

(

(

4 4 − x2

) = 3x

d x 3 −1 dx

=

4x

2

)

=

2

+c

)

x

(4 − x )

2 2

which implies that du = 3x 2 dx

. Substituting the equivalent values of x 3 − 1 and dx back into the integral we obtain

1 5 dx

Check: Let y = i. Given

1

2 4 − x2

2 3 ∫ x ( x − 1)

du

u

2

Substituting the equivalent values of 4 − x 2 and dx back into the integral we

x

and dx =

2 du

dx let u = 4 − x 2 , then

∫ ( 4 − x 2 )2 dx = ∫ u 2 ⋅ − 2 x

h. Given

∫

8x 2

x −3

1

∫

=

x 2u 5

(

⋅

)

5 3 x −1 18 dx

du 3x 2 6 5

1

=

+c,

∫

(

5 1 1 1+ 15 u5 1 1 55+1 5 65 x3 −1 u +c = ⋅ du = ⋅ u +c = u +c = 18 3 1+ 1 3 3 5+1 18 5 5

then y ′ =

let u = x 2 − 3 , then

du dx

(

)

5 6 3 ⋅ x −1 18 5

=

(

6 −1 5

⋅ 3x 2 + 0

=

(

)

1 3 x −1 3

6 −5 5

⋅ 3x 2

(

)

= x 2 x 3 −1

1 5

) = 2 x which implies that du = 2 xdx and

d x2 −3 dx

Substituting the equivalent values of x 2 − 3 and dx back into the integral we obtain dx

=

∫

8 x du ⋅ u 2x

(

Check: Let y = 8 x 2 − 3 Hamilton Education Guides

)

= 4∫ u 1 2

+c ,

−1

2 du

= 4⋅

1 1 − 12

then y ′ = 8 ⋅

(

u

1− 1

+c

2

1

= 4 ⋅ 2−1 u

2 −1 2

+c

(

1

= 8u 2 + c = 8 x 2 − 3

)

1 2

+c

2

)

1 2 x −3 2

1 −1 2

⋅ 2x + 0

(

= 4 x2 −3

)

−1

2

⋅ 2x

=

8x x2 −3 230

)

6 5

+c

Calculus I

4.2 Integration Using the Substitution Method

Section 4.2 Practice Problems – Integration Using the Substitution Method 1. Evaluate the following indefinite integrals: a.

2 3 ∫ t (1 + t ) dt

d.

∫ (

g.

x x2 − 2

)

1 5

= =

dx

2 3 2 ∫ x (1 − x ) dx =

b.

∫18x

e.

∫

h.

∫

2

6 x3 − 5 dx

3x 2

=

c.

∫

4

dx

=

1 − t 2 dt

=

x+5 t

dx

=

f.

∫2

dx

=

i.

∫ x (2x

=

c.

∫

2 2 ∫ x (1 − x ) dx

x +3 x3

5

x +3

8

9

)

2

+ 1 dx

=

2. Evaluate the following indefinite integrals: a.

2 3 ∫ 6 x (2 x − 1)dx =

d.

∫x

g.

∫

3

3 x 2 − 1 dx

x5 6

dx

=

=

x +3

Hamilton Education Guides

b.

∫x

1 + x 2 dx

e.

∫

x

h.

∫

2

dx

=

f.

dx

=

i.

x +1 3x 2

x −1

∫

5

7 x + 1 dx =

5x4 + 6 x 5

2

= dx

=

x + 3x + 1

231

Calculus I

4.3

4.3 Integration of Trigonometric Functions

Integration of Trigonometric Functions

In the following examples we will solve problems using the formulas below: Table 4.3-1: Integration Formulas for Trigonometric Functions 1.

∫ sin x dx = − cos x + c

2.

∫ cos x dx = sin x + c

3.

∫ tan x dx = ln

sec x + c

4.

∫ cot x dx = ln

5.

∫ sec x dx = ln

6.

∫ csc x dx = ln

csc x − cot x + c

7.

∫ tan x sec x dx = sec x + c

8.

∫ cot x csc x dx = − csc x + c

9.

∫ sin

sin x + c

x sin 2 x + +c 2 4

10.

∫ cos

2

x dx =

13.

∫ sec

2

x dx = tan x + c

sec x + tan x + c

11.

∫ tan

2

x dx = tan x − x + c

14.

∫ csc

2

x dx = − cot x + c

2

∫ cot

12.

x dx = 2

x sin 2 x +c − 4 2

x dx = − cot x − x + c

Additionally, the following formulas (identities) hold for the trigonometric functions: 1. Unit Circle Formulas sin 2 x + cos 2 x = 1

sec 2 x − tan 2 x = 1

csc 2 x − cot 2 x = 1

2. Addition Formulas sin (x + y ) = sin x cos y + cos x sin y

(1 )

sin (x − y ) = sin x cos y − cos x sin y

(2)

cos (x + y ) = cos x cos y − sin x sin y

(3 )

cos (x − y ) = cos x cos y + sin x sin y

(4)

tan (x + y ) =

tan x + tan y 1 − tan x tan y

tan (x − y ) =

tan x − tan y 1 + tan x tan y

Note that from the identities ( 1 ) and ( 2 ) above we can obtain the formulas sin x cos y =

1 [ sin (x − y ) + sin (x + y ) ] 2

cos x sin y =

1 [ sin (x + y ) − cos (x − y ) ] 2

and from the identities ( 3 ) and ( 4 ) above we can obtain the formulas sin x sin y =

1 [ cos (x − y ) − cos (x + y ) ] 2

cos x cos y =

1 [ cos (x − y ) + cos (x + y ) ] 2

3. Half Angle Formulas sin 2

1 1 x = ( 1 − cos x ) 2 2

or

1 − cos x = 2 sin 2

1 x 2

therefore

sin 2 x =

1 ( 1 − cos 2 x ) 2

cos 2

1 1 x = ( 1 + cos x ) 2 2

or

1 + cos x = 2 cos 2

1 x 2

therefore

cos 2 x =

1 ( 1 + cos 2 x ) 2

4. Double Angle Formulas sin 2 x = 2 sin x cos x

or

Hamilton Education Guides

sin x cos x =

1 sin 2 x 2

cos 2 x = cos 2 x − sin 2 x

and

232

Calculus I

4.3 Integration of Trigonometric Functions

(

)

1 1 sin 2 x cos 2 x = 12 sin 2 x 2 = sin 2 2 x = ( 1 − cos 4 x ) 4 8

Also, the tangent, cotangent, secant, and cosecant functions are defined by tan x =

sin x cos x

cot x =

cos x 1 = tan x sin x

sec x =

1 cos x

csc x =

1 sin x

We should also note that sine, tangent, cotangent, and cosecant are odd functions. This implies that sin (− x ) = − sin x

tan (− x ) = − tan x

cot (− x ) = − cot x

csc (− x ) = − csc x

on the other hand, cosine and secant are even functions. This implies that sec (− x ) = sec x

cos (− x ) = cos x

Finally, we need to know how to differentiate the trigonometric functions (addressed in Chapter 3, Section 3.1) in order to check the answer to the given integrals below. The derivatives of trigonometric functions are repeated here and are as follows: Table 4.3-2: Differentiation Formulas for Trigonometric Functions du d sin u = cos u ⋅ dx dx du d cos u = − sin u ⋅ dx dx du d tan u = sec 2 u ⋅ dx dx

d du cot u = − csc 2 u ⋅ dx dx du d sec u = sec u tan u ⋅ dx dx d du csc u = − csc u cot u ⋅ dx dx

Let’s integrate some trigonometric functions using the above integration formulas. Example 4.3-1: Evaluate the following indefinite integrals: 1

a.

∫ sin 3x dx =

b.

∫ sin 8 x dx =

c.

∫ cos 5x dx =

d.

∫ cos 4 x dx =

1

e.

∫ (sin 4 x + cos 2 x ) dx =

f.

∫ csc 5x dx

g.

∫ csc

h.

∫ csc

i.

∫x

k.

∫ x csc

l.

2 2 ∫ (2 x + 1) csc (x + x ) dx =

j.

2

5 x dx =

2 2 ∫ x sec ( x + 1) dx =

1 x dx 2

2

2

=

x 2 dx =

2

=

sec 2 x 3 dx =

Solutions: a. Given ∫ sin 3x dx let u = 3x , then

∫ sin 3x dx

=

du

∫ sin u ⋅ 3

Hamilton Education Guides

=

du d = 3x dx dx

1 sin u du 3

∫

= 3 which implies dx =

du 3

. Therefore,

1 3

1 = − cos u + c = − cos 3 x + c 3

233

Calculus I

4.3 Integration of Trigonometric Functions

1 3

1 d d cos 3 x + c 3 dx dx

1 3

Check: Let y = − cos 3x + c , then y ′ = − ⋅ 3 ⋅ sin 3 x = sin 3 x 3 x du d x x = sin dx let u = , then 8 dx dx 8 8

= − ⋅ − sin 3x ⋅

d 3x + 0 dx

1 3

= − ⋅ − sin 3x ⋅ 3

= b. Given

∫

x

∫ sin 8 dx = ∫ sin u ⋅ 8 du x 8

8 x ⋅ sin 8 8

= sin

du

x d d c cos + dx 8 dx

x +c 8 x d x +0 8 dx 8

= − 8 ⋅ − sin ⋅

x 1 8 8

= − 8 ⋅ − sin ⋅

x 8

c. Given ∫ cos 5 x dx let u = 5 x , then

∫ cos 5x dx = ∫ cos u ⋅ 5

which implies dx = 8 du . Therefore,

= 8∫ sin u du = −8 cos u + c = − 8 cos

Check: Let y = −8 cos + c , then y ′ = − 8 ⋅ =

1 8

=

du d = 5x dx dx

1 cos u du 5

∫

=

1 5

Check: Let y = sin 5 x + c , then y ′ =

=

= 5 which implies dx =

1 sin u + c 5

. Therefore,

1 sin 5 x + c 5

=

1 d d ⋅ sin 5 x + c 5 dx dx

5 ⋅ cos 5 x = cos 5 x 5 x x du d x cos dx let u = , then = 4 4 dx dx 4

du 5

=

1 d ⋅ cos 5 x ⋅ 5x + 0 5 dx

=

1 ⋅ cos 5 x ⋅ 5 5

= d. Given

∫

x

∫ cos 4 dx

=

∫ cos u ⋅ 4 du x 4

4 x ⋅ cos 4 4

= cos

1 4

which implies dx = 4 du . Therefore,

= 4∫ cos u du = 4 sin u + c = 4 sin

Check: Let y = 4 sin + c , then y ′ = 4 ⋅ =

=

d x d c sin + dx 4 dx

x +c 4 x d x +0 4 dx 4

= 4 ⋅ cos ⋅

x 1 4 4

= 4 ⋅ cos ⋅

x 4

e. Given ∫ (sin 4 x + cos 2 x ) dx = ∫ sin 4 x dx + ∫ cos 2 x dx let: a. u = 4 x , then

du du du d = 4 ; du = 4dx ; dx = = 4x ; 4 dx dx dx

b. v = 2 x , then

dv dv d dv = 2x ; = 2 ; dv = 2dx ; dx = 2 dx dx dx

Therefore, +

1 sin v + c 2 2

du

and

. dv

∫ sin 4 x dx + ∫ cos 2 x dx = ∫ sin u ⋅ 4 + ∫ cos v ⋅ 2 1 4

=

1 1 sin u du + cos v dv 4 2

∫

∫

1 4

= − cos u + c1

1 2

1 1 = − cos 4 x + sin 2 x + c1 + c 2 = − cos 4 x + sin 2 x + c 4

Hamilton Education Guides

2

234

Calculus I

4.3 Integration of Trigonometric Functions

1 4

1 2

1 d 1 d d cos 4 x + ⋅ sin 2 x + c 4 dx 2 dx dx

Check: Let y = − cos 4 x + sin 2 x + c then y ′ = − ⋅

4 2 1 d 2 x + 0 = ⋅ sin 4 x + ⋅ cos 2 x = sin 4 x + cos 2 x ⋅ cos 2 x ⋅ 4 2 2 dx du du d csc 5 x dx let u = 5 x , then = 5 x = 5 which implies du = 5dx ; dx = 5 dx dx

=

d 1 4x ⋅ sin 4 x ⋅ dx 4

+

∫

f. Given

∫ csc 5x dx

=

du

∫ csc u ⋅ 5

=

1 csc u du 5

∫

=

1 ln csc u − cot u + c 5

1 5

Check: Let y = ln csc 5 x − cot 5 x + c , then y ′ = ×

d (csc 5 x − cot 5 x ) + d c dx dx

=

1 5 csc 5 x ( csc 5 x − cot 5 x ) ⋅ 5 csc 5 x − cot 5 x

g. Given ∫ csc 2 5 x dx let u = 5 x , then

∫ csc

2

=

5 x dx

∫ csc

2

u⋅

du 5

=

d 1 d c ⋅ ln csc 5 x − cot 5 x + dx 5 dx

=

)

1 1 ⋅ 5 csc 5 x − cot 5 x

1 1 ⋅ ⋅ − csc 5 x ⋅ cot 5 x ⋅ 5 + csc 2 5 x ⋅ 5 + 0 5 csc 5 x − cot 5 x 5 csc 5 x 5

=

= csc 5 x

du d du du = 5x ; = 5 ; du = 5dx ; dx = dx dx 5 dx

1 csc 2 u du 5

∫

1 5

. Therefore,

1 5

1 = − cot u + c = − cot 5 x + c 5

1 d d c cot 5 x + 5 dx dx

Check: Let y = − cot 5 x + c , then y ′ = − ⋅ 5 ⋅ csc 2 5 x = csc 2 5 x 5 1 1 csc 2 x dx let u = x , 2 2

1 ln csc 5 x − cot 5 x + c 5

=

(

=

. Therefore,

1 5

= − ⋅ − csc 2 5 x ⋅

d 5x + 0 dx

=

1 ⋅ csc 2 5 x ⋅ 5 5

= h. Given

∫ csc

2

∫

1 x dx 2

=

∫ csc

2

u ⋅ 2du

then

du d x du 1 ; = = dx dx 2 dx 2

= 2∫ csc 2 u du = − 2 cot u + c = − 2 cot

x 2

Check: Let y = −2 cot + c , then y ′ = − 2 ⋅ = i. Given

∫x

2

∫x

2

2 x csc 2 2 2

= csc 2

sec 2 x 3 dx

sec 2 x 3 dx =

∫x

x 2

sec 2 u ⋅

du 3x

2/

=

1 3

3x 2 ⋅ sec 2 x 3 3

Hamilton Education Guides

x d x +0 2 dx 2

= − 2 ⋅ − csc 2 ⋅

1 x 2 du du d 3 du then = x ; = 3x 2 ; du = 3 x 2 dx ; dx = dx dx dx 3x 2 1 sec 2 u du 3

Check: Let y = tan x 3 + c , then y ′ = =

d x d cot + c dx 2 dx

1 x+c 2 x 1 2 2

= 2 csc 2 ⋅

= csc 2

let u = x 3 , 2

; 2du = dx ; dx = 2du . Therefore,

∫

=

1 tan u + c 3

d 1 d c ⋅ tan x 3 + dx 3 dx

=

=

. Therefore,

1 tan x 3 + c 3

1 d 3 ⋅ sec 2 x 3 ⋅ x +0 dx 3

=

1 ⋅ sec 2 x 3 ⋅ 3 x 2 3

= x 2 sec 2 x 3

235

Calculus I

4.3 Integration of Trigonometric Functions 2 2 ∫ x sec (x + 1) dx

j. Given

2 2 ∫ x sec (x + 1) dx

= 1 2

(

∫ x sec

2

(

)

u⋅

du 2x

=

1 sec 2 u du 2

∫

)

∫ x csc

∫ x csc

x 2 dx

2

x 2 dx let

=

∫ x csc

2

)

(

(

)

(

=

du 2x

u⋅

=

∫

d 1 d c cot x 2 + dx 2 dx

) (

)

Therefore,

1 2

= − ⋅ − csc 2 x 2 ⋅

d 2 x +0 dx

=

1 ⋅ csc 2 x 2 ⋅ 2 x 2

= x csc 2 x 2

(

du . 2x +1

(

1 d ⋅ sec 2 x 2 + 1 ⋅ x 2 +1 + 0 2 dx

1 2

2

)

l. Given ∫ (2 x + 1) csc 2 x 2 + x dx let u = x 2 + x , then dx =

)

1 = − cot u + c = − cot x 2 + c

1 csc 2 u du 2

1 2

2x ⋅ csc 2 x 2 2

=

)

Check: Let y = − cot x 2 + c , then y ′ = − ⋅ =

(

1 tan x 2 + 1 + c 2

=

2x ⋅ sec 2 x 2 + 1 = x sec 2 x 2 + 1 2 du d 2 du du . u = x 2 , then = x ; = 2 x ; du = 2 x dx ; dx = dx dx dx 2x

1 ⋅ sec 2 x 2 + 1 ⋅ 2 x 2

k. Given 2

(

1 tan u + c 2

=

1 d d ⋅ tan x 2 + 1 + c 2 dx dx

Check: Let y = tan x 2 + 1 + c , then y ′ = =

); dudx = 2 x ; du = 2 x dx ; dx = du2 x . Therefore,

du d 2 = x +1 dx dx

let u = x 2 + 1 , then

Therefore,

2 2 ∫ (2 x + 1) csc ( x + x ) dx

(

=

(

du d x2 + x = dx dx

∫ (2 x + 1) csc

2

); dudx = 2 x + 1 ; du = (2 x + 1) dx ;

u⋅

du = (2 x + 1)

∫ csc

2

u du

)

= − cot u + c = − cot x 2 + x + c

(

)

( ( x + x)

Check: Let y = − cot x 2 + x + c , then y ′ = −

(

)

= csc 2 x 2 + x ⋅ (2 x + 1) = (2 x + 1) csc 2

)

d d c cot x 2 + x + dx dx

(

) dxd ( x

= csc 2 x 2 + x ⋅

)

2

+ x +0

2

dx =

2

Example 4.3-2: Evaluate the following indefinite integrals: a.

∫ sec

2

3 x dx =

b.

∫ sec 3x dx

d.

∫ sin

5

(x + 1) cos (x + 1) dx =

e.

∫ cos

5

g.

∫ cos

h.

∫x

cos x 3 dx

4

x x sin dx 3 3

=

2

=

c.

∫ x csc x

x sin x dx =

f.

∫ cos

i.

∫e

=

x

5

3 x sin 3 x dx =

sec e x dx

=

Solutions: a. Given ∫ sec 2 3x dx let u = 3x , then

∫ sec

2

3 x dx =

∫ sec

2

u⋅

Hamilton Education Guides

du 3

=

du du d du = 3x ; = 3 ; du = 3 dx ; dx = dx dx dx 3

1 sec 2 u du 3

∫

=

1 tan u + c 3

=

. Therefore,

1 tan 3 x + c 3

236

Calculus I

4.3 Integration of Trigonometric Functions

1 3

Check: Let y = tan 3x + c , then y ′ =

=

du

∫ sec u ⋅ 3

=

=

1 sec 2 3 x ⋅ 3 + 0 3

1 sec u du 3

∫

1 ln sec u + tan u + c 3

=

=

3 sec 2 3 x 3

=

du d du du = 3x ; = 3 ; du = 3 dx ; dx = dx dx dx 3

b. Given ∫ sec 3x dx let u = 3x , then

∫ sec 3x dx

d d tan 3 x + c dx dx

= sec 2 3x

. Therefore,

1 ln sec 3 x + tan 3 x + c 3

1 ⋅ 3 sec 3 x tan 3 x + 3 sec 2 3 x + 0 3 sec 3 x + tan 3 x

1 1 Check: Let y = ln sec 3x + tan 3x + c , then y ′ = ⋅ 3

= c. Given

1 3 sec 3 x (sec 3 x + tan 3 x ) ⋅ 3 sec 3 x + tan 3 x

∫ x csc x

∫ x csc x

2

dx =

2

=

du

1 2

=

1 2

= sec 3x

du du d 2 du . = 2 x ; du = 2 x dx ; dx = x ; = dx 2x dx dx

dx let u = x 2 , then

∫ x csc u ⋅ 2 x

3 sec 3 x 3

∫ csc u du

1 ln csc u − cot u + c 2

=

Check: Let y = ln csc x 2 − cot x 2 + c , then y ′ = =

(

1 2 x csc x 2 csc x 2 − cot x 2 ⋅ 2 csc x 2 − cot x 2

) = 2 x csc x 2

2

∫ sin

5

du . cos ( x + 1)

(x + 1) cos (x + 1) dx =

∫u

5

= x csc x 2 du d du = sin ( x + 1) ; = cos ( x + 1) ; dx dx dx

Therefore,

cos ( x + 1) ⋅

du cos ( x + 1)

1 6

Check: Let y = sin 6 (x + 1) + c , then y ′ = e. Given ∫ cos 5 x sin x dx let u = cos x , then

∫ cos

5

x sin x dx

=

∫u 1 6

5

sin x ⋅

du − sin x

=

∫u

5

du =

1 6 u +c 6

1 ⋅ 6 sin 5 (x + 1) ⋅ cos (x + 1) + 0 6 du d du = cos x ; = − sin x dx dx dx

=

1 sin 6 ( x + 1) + c 6

= sin 5 (x + 1) cos (x + 1) ; dx =

du − sin x

. Therefore,

1 6

1 = − ∫ u 5 du = − u 6 + c = − cos 6 x + c 6

1 6

Check: Let y = − cos 6 x + c , then y ′ = − ⋅ 6 cos 5 x ⋅ − sin x + 0 = f. Given ∫ cos 5 3x sin 3x dx let u = cos 3x , then

Hamilton Education Guides

1 ln csc x 2 − cot x 2 + c 2

1 1 ⋅ ⋅ −2 x csc x 2 cot x 2 + 2 x csc 2 x 2 + 0 2 2 csc x − cot x 2

d. Given ∫ sin 5 (x + 1) cos (x + 1) dx let u = sin (x + 1) , then du = cos ( x + 1) ⋅ dx ; dx =

=

Therefore,

6 cos 5 x sin x 6

du d du = cos 3 x ; = −3 sin 3 x dx dx dx

= cos 5 x sin x

; dx =

du − 3 sin 3 x

. Thus,

237

Calculus I

∫ cos

5

4.3 Integration of Trigonometric Functions

Check: Let y = − g. Given ∫ cos 4

∫ cos

∫u

3 x sin 3 x dx =

du − 3 sin 3 x

sin 3 x ⋅

1 cos 6 3 x + c , 18

x x sin dx 3 3

x x sin dx 3 3

4

5

∫u

= 3 5

4

= −

x 3

sin

3 du x ⋅− 3 sin x

∫

3 6

du d x = cos dx dx 3

;

du 1 x = − sin dx 3 3 3 5

5

3 5

x 3

x 1 3 3

x 3

Check: Let y = − cos 5 + c , then y ′ = − ⋅ 5 cos 4 ⋅ − sin ⋅ + 0 = h. Given

∫x

2

∫x

2

cos x 3 dx let u = x 3 , then

cos x 3 dx

=

∫x

2

cos u ⋅

du 3x

2

=

∫

1 3

Check: Let y = sin x 3 + c , then y ′ =

∫e

x

sec e x dx

=

∫e

x

sec u ⋅

du ex

=

=

(

x

sec e + tan e

x

sin

=

=

)=e

x

x 3

. Therefore,

x +c 3

15 x x cos 4 sin 3 3 15

= cos 4

x x sin 3 3

. Therefore,

1 sin x 3 + c 3

3 2 x cos x 3 + 0 3

= x 2 cos x 3

du du d x du = = e x ; du = e x ⋅ dx ; dx = e ; dx dx dx ex

∫ sec u du

e x sec e x sec e x + tan e x

1 ⋅ sin u + c 3

1 ⋅ cos x 3 ⋅ 3 x 2 + c 3

. Therefore,

= ln sec u + tan u + c = ln sec e x + tan e x + c

Check: Let y = ln sec e x + tan e x + c , then y ′ = =

3du

du du d 3 du x ; = 3x 2 ; du = 3 x 2 ⋅ dx ; dx = = dx dx dx 3x 2

1 cos u du 3

i. Given ∫ e x sec e x dx let u = e x , then

; dx = −

3 = − 3∫ u 4 du = − u 5 + c = − cos 5

3

1 cos 6 3 x + c 18

1 18 ⋅ 6 cos 5 3 x ⋅ −3 sin 3 x + 0 = cos 5 3 x sin 3 x = cos 5 3 x sin 3 x 18 18

then y ′ = −

let u = cos , then

1 1 = − ⋅ u6 + c = −

1 u 5 du 3

1 sec e + tan e x x

⋅ e x sec e x tan e x + e x sec 2 e x + 0

sec e x

Example 4.3-3: Evaluate the following indefinite integrals: a.

∫ tan

7

x sec 2 x dx

d.

∫ cot

5

2 x csc 2 2 x dx =

g. j.

(x − 1) sec 2 (x − 1) dx =

b.

∫ tan

e.

∫ tan 8x dx =

∫ sec 9 x tan 9 x dx =

h.

∫ sec 2 tan 2 dx

∫x

k.

∫ x cot 3x

2

=

cot x 3 dx =

Hamilton Education Guides

4

x

x

2

dx =

=

c.

∫ cot

f.

∫ tan 4 dx =

i.

∫ 5 sec 3x tan 3x dx

l.

∫ sec

3

x csc 2 x dx

=

x

x

dx x

=

=

238

Calculus I

4.3 Integration of Trigonometric Functions

Solutions: a.

∫ tan

7

x sec 2 x dx let u = tan x , then

∫ tan

7

x sec 2 x dx

∫u

=

7

⋅ sec 2 x ⋅

du d = tan x dx dx

du 2

sec x

1 8

Check: Let y = tan 8 x + c , then y ′ = b.

∫ tan

4

; dx = =

(x − 1) sec 2 (x − 1) dx let du sec

2

(x − 1)

1 4+1 u +c 4 +1

∫ tan

1 5 u +c 5

=

4

∫u

csc x

. Therefore,

1 7 +1 u +c 7 +1

; du = sec 2 x dx ; dx =

4

1 8 u +c 8

=

sec 2 x

. Thus,

1 tan 8 x + c 8

= (tan x )7 sec 2 x = tan 7 x sec 2 x

du d = tan (x − 1) dx dx

∫u

=

du

;

du = sec 2 (x − 1) c ; du = sec 2 (x − 1) dx dx

⋅ sec 2 (x − 1) ⋅

du

=

sec (x − 1) 2

∫u

4

du

1 tan 5 ( x − 1) + c 5

c. Given ∫ cot 3 x csc 2 x dx let u = cot x , then 2

=

(x − 1) sec 2 (x − 1) dx =

1 5

du

du

then

Check: Let y = tan 5 (x − 1) + c , then y ′ =

; dx = −

7

du = sec 2 x dx

1 ⋅ 8 (tan x )8−1 ⋅ sec 2 x + 0 8

u = tan (x − 1) ,

. Therefore,

=

=

;

∫ cot

3

1 ⋅ 5 [ tan (x − 1) ]5−1 ⋅ sec 2 (x − 1) + 0 5 du d = cot x dx dx

x csc 2 x dx =

∫u

3

;

= tan 4 (x − 1) sec 2 (x − 1)

du = − csc 2 x c ; du = − csc 2 x dx dx

⋅ csc 2 x ⋅

−du 2

csc x

= − ∫ u 3 du =

−1 3+1 u +c 3 +1

1 4

1 = − u 4 + c = − cot 4 x + c 4

1 4

1 4

Check: Let y = − cot 4 x + c , then y ′ = − ⋅ 4(cot x )4−1 ⋅ − csc 2 x + 0 = cot 3 x csc 2 x du d = cot 2 x dx dx

d. Given ∫ cot 5 2 x csc 2 2 x dx let u = cot 2 x , then ; dx = −

du 2

2 csc 2 x

. Therefore,

1 1 5+1 u +c 2 5 +1

= − ⋅

Check: Let y = − e. Given

= −

∫ cot

1 6 u +c 12

1 cot 6 2 x + c , 12

2 x csc 2 2 x dx =

= −

∫u

5

du = −2 csc 2 2 x dx

⋅ csc 2 2 x ⋅

−du 2

2 csc 2 x

; du = −2 csc 2 2 x dx

= −

1 u 5 du 2

∫

1 cot 6 2 x + c 12

then y ′ = −

∫ tan 8x dx let u = 8x , then

Hamilton Education Guides

5

;

du d = 8x dx dx

1 ⋅ 6(cot 2 x )6−1 ⋅ 2 ⋅ − csc 2 2 x + 0 12

;

du =8 dx

; du = 8 dx ; dx =

du 8

= cot 5 2 x csc 2 2 x . Therefore,

239

Calculus I

4.3 Integration of Trigonometric Functions

∫ tan 8x dx

=

du

∫ tan u ⋅ 8

=

1 tan u du 8

∫

=

1 ln sec u + c 8

1 8

Check: Let y = ln sec 8 x + c , then y ′ = f. Given

x

x

∫ tan 4 dx let u = 4 , then

x

1 ln sec 8 x + c 8

=

1 1 ⋅ ⋅ sec 8 x tan 8 x ⋅ 8 + 0 8 sec 8 x

du d x = dx dx 4

;

du 1 = dx 4

= 4∫ tan u du = 4 ln sec u + c = 4 ln sec

Check: Let y = 4 ln sec

x +c, 4

g. Given ∫ sec 9 x tan 9 x dx let u = 9 x , then du

∫ sec 9 x tan 9 x dx = ∫ sec u ⋅ tan u ⋅ 9

=

1 sec

x 4

⋅ sec

du d = 9x dx dx

= tan 8 x

; 4du = dx ; dx = 4du . Therefore,

∫ tan 4 dx = ∫ tan u ⋅ 4du

then y ′ = 4 ⋅

8 tan 8 x 8

=

x x 1 tan ⋅ + 0 4 4 4 du =9 dx

;

1 sec u tan u du 9

∫

x +c 4

; du = 9dx ; dx =

1 sec u + c 9

=

4 x tan 4 4

=

=

= tan du 9

x 4

. Therefore,

1 sec 9 x + c 9

1 1 9 Check: Let y = sec 9 x + c then y ′ = ⋅ sec 9 x tan 9 x ⋅ 9 + 0 = ⋅ sec 9 x tan 9 x = sec 9 x tan 9 x 9

h. Given ∫ sec x

9

x x tan dx 2 2

x

∫ sec 2 tan 2 dx

=

x 2

let u =

, then

∫ sec u ⋅ tan u ⋅ 2du

9

du d x = dx dx 2

x x 1 tan ⋅ + 0 2 2 2

x 2

i. Given ∫ 5 sec 3x tan 3x dx let u = 3x , then = 5∫ sec u ⋅ tan u ⋅

du 3

5 3

Check: Let y = sec 3x + c , then y ′ = j. Given

∫x

2

∫x

2

du 1 = dx 2

; 2du = dx ; dx = 2du . Therefore,

= 2∫ sec u tan u du = 2 sec u + c = 2 sec

Check: Let y = 2 sec + c , then y ′ = 2 ⋅ sec

∫ 5 sec 3x tan 3x dx

;

=

du d = 3x dx dx

;

du =3 dx

5 sec u tan u du 3

∫

2 x x ⋅ sec tan 2 2 2

=

=

5 ⋅ sec 3 x tan 3 x ⋅ 3 + 0 3

=

x +c 2

= sec

; du = 3dx ; dx = 5 sec u + c 3

=

x x tan 2 2

du 3

. Therefore,

5 sec 3 x + c 3

15 ⋅ sec 3 x tan 3 x 3

= 5 sec 3x tan 3x

du du d 3 du . Therefore, cot x 3 dx let u = x 3 , then = 3x 2 ; du = 3 x 2 dx ; dx = x ; = dx dx dx 3x 2

cot x 3 dx =

∫x

2

⋅ cot u ⋅

1 3

du 3x

2

=

1 cot u ⋅ du 3

∫

Check: Let y = ln sin x 3 + c , then y ′ =

Hamilton Education Guides

=

1 ln sin u + c 3

=

1 1 ⋅ ⋅ cos x3 ⋅ 3 x 2 + 0 3 sin x3

1 ln sin x 3 + c 3

=

3 x 2 cos x3 ⋅ 3 sin x3

= x 2 cot x 3

240

Calculus I

4.3 Integration of Trigonometric Functions

∫ x cot 3x

k. Given

∫ x cot 3x

2

let u = 3x 2 , then

dx

du

∫ x ⋅ cot u ⋅ 6 x

=

dx

2

=

du d = 3x 2 dx dx

1 cot u ⋅ du 6

∫

1 6

Check: Let y = ln sin 3x 2 + c , then y ′ = dx

l. Given ∫ sec x

∫ sec

dx

x

x

x

2 x du

1 ln sin u + c 6

sec x + tan

x

⋅

=

6 x cos 3 x 2 = x cot 3x 2 ⋅ 6 sin 3 x 2

= 2∫ sec u ⋅ du = 2 ln sec u + tan u + c = 2 ln sec x + tan x + c

x

1

Therefore,

1 ln sin 3 x 2 + c 6

=

1 1 ⋅ cos 3 x 2 ⋅ 6 x + 0 ⋅ 6 sin 3 x 2

Check: Let y = 2 ln sec x + tan x + c , then y ′ = 2 ⋅ = 2⋅

du . 6x

; du = 6 x dx ; dx =

1 du d 12 du 1 − 12 ; dx = 2 x du . Therefore, = x = = x ; dx 2 dx dx 2 x

1

let u = x 2 , then

∫ sec u ⋅

=

=

du = 6x dx

;

(

sec x tan

x + sec x

2 x

1 sec x + tan

)=

2 sec x

=

2 x

x

⋅

x + sec2 x

sec x tan

2 x

+0

sec x x

Example 4.3-4: Evaluate the following indefinite integrals: 3

dx

a.

∫ sec

d.

∫ cos 3x dx =

g.

∫ 5(sin 3x csc 3x ) dx

j.

∫

x2

3

x

3 cos 1 x 2

sin

1 x dx 2

dx

∫ csc

e.

∫

=

h.

2 ∫ a ( cos 2t sec 2t + t ) dt

=

k.

∫

1

e

5

b.

=

x3

5

x

2

=

cos 2 x + sin 2 x dx sin 2 x

= =

e tan 5 x sec 2 5 x dx =

c.

∫ 5x

f.

∫ 

i.

∫e

l.

∫

2

csc x 3 dx =

 1 + cos x   dx sin x  sin 5 x

=

cos 5 x dx =

1 cot 3 x

e2

csc 2 3 x dx

=

Solutions: a. Given ∫ sec 3 x 2

∫

dx

3

sec x 2

3

x

=

dx

2

let u = x 3 , then

3

x

∫

3 3 x du sec u ⋅ ⋅ 2 3x

=

3 du d 23 du 2 − 13 2 ; dx = 3 x du . Therefore, = x ; = x = 3 2 dx dx dx 3 3 x

3 sec u ⋅ du 2

∫

3 2

=

3 ln sec u + tan u + c 2

Check: Let y = ln sec 3 x 2 + tan 3 x 2 + c , then y ′ =

=

3 1 2 ⋅ ⋅ ⋅ 2 sec 3 x 2 + tan 3 x 2 3

Hamilton Education Guides

3

3 3 3 ln sec x 2 + tan x 2 + c 2 3

3

2 sec x 2 tan x 2 + 2 sec 2 3 1 ⋅ ⋅ 2 sec 3 x 2 + tan 3 x 2 33 x

sec x 2  sec x 2 + tan  3 x 3

=

3

x 2  

=

3

3

sec x 2 6 sec x 2 = ⋅ 3 3 6 x x

241

3

x2

Calculus I

4.3 Integration of Trigonometric Functions

dx

b. Given ∫ csc 5 x 3

∫

dx

5

csc x 3

5

x2

5

=

x2

∫

3

du 3 − 52 du d 53 3 5 ; dx = 5 x 2 du . Thus, = x = = x ; 5 dx 5 dx dx 3 5 x2

let u = x 5 , then 5

5 5 x 2 du 5 5 5 3 5 3 csc u ⋅ ⋅ csc u ⋅ du = ln csc u − cot u + c = ln csc x − cot x + c = 3 3 5 x2 3 3

∫

5

1 5 3 = ⋅ ⋅ ⋅ 3 csc 5 x 3 − cot 5 x 3 5

5 5 5 csc x 3  csc x 3 − cot x 3    5

c. Given ∫ 5 x 2 csc x 3 dx let u = x 3 , then

∫ 5x

2

∫

csc x 3 dx = 5 x 2 ⋅ csc u ⋅ 5 3

du 3x

2

x2

du d 3 x = dx dx

;

5 csc u ⋅ du 3

=

∫

=

Check: Let y = ln csc x 3 − cot x 3 + c , then y ′ = = d. Given

(

5 3 x 2 csc x3 csc x3 − cot x3 ⋅ 3 csc x3 − cot x3

) = 15x 3

2

1

1

du

∫ cos 3x dx = ∫ sec 3x dx = ∫ sec u ⋅ 3

=

1 3

e. Given

∫

cos 2 x + sin 2 x dx sin 2 x

du d = 2x dx dx

∫

1 3 sec 3 x ( sec 3 x + tan 3 x ) ⋅ 3 sec 3 x + tan 3 x

;

du =2 dx

cos 2 x + sin 2 x dx sin 2 x

=

3 ⋅ sec 3 x 3

 cos 2 x  + 1 dx  sin 2 x 

= ∫ 

=

;

du =3 dx

3x 2

. Therefore,

5 ln csc x 3 − cot x 3 + c 3

; du = 3dx ; dx =

1 ln sec u + tan u + c 3

= sec 3x =

=

du 3

. Thus,

1 ln sec 3 x + tan 3 x + c 3

1 cos 3 x

= ∫ (cot 2 x + 1) dx = ∫ cot 2 x dx + ∫ dx let u = 2 x , then

. Therefore,

∫ cot 2 x dx + ∫ dx

=

1 cot u du + x 2

Hamilton Education Guides

=

du

1 − sec 3 x tan 3 x ⋅ 3 + sec2 3 x ⋅ 3 +0 ⋅ sec 3 x + tan 3 x 3

du 2

∫

; du = 3x 2 dx ; dx =

= 5 x 2 csc x 3

; du = 2dx ; dx =

=

5

5 − csc x3 cot x3 ⋅ 3 x 2 + csc2 x3 ⋅ 3 x 2 ⋅ +0 3 csc x3 − cot x3

1 sec u du 3

∫

5

5

csc x 3 15 csc x 3 = ⋅ = 5 2 15 5 x 2 x

5 ln csc u − cot u + c 3

du d = 3x dx dx

Check: Let y = ln sec 3x + tan 3x + c , then y ′ = =

du = 3x 2 dx

⋅ csc x 3

∫ cos 3x dx = ∫ sec 3x dx let u = 3x , then

5

− 3 csc x 3 cot x 3 + 3 csc 2 5 1 ⋅ ⋅ 5 3 csc 5 x 3 − cot 5 x 3 5 x2

5 3

Check: Let y = ln csc 5 x 3 − cot 5 x 3 + c , then y ′ =

=

1 ln sin u + x + c 2

=

1 ln sin 2 x + x + c 2

242

x3

Calculus I

4.3 Integration of Trigonometric Functions

1 cos 2 x ⋅ 2 ⋅ +1+ 0 2 sin 2 x

1 2

Check: Let y = ln sin 2 x + x + c , then y ′ = f.

 1 + cos x   dx sin x 

∫ 

cos x 

 1

∫  sin x + sin x  dx

=

∫ csc x dx + ∫ cot x dx

=

Check: Let y = ln csc x − cot x + ln sin x + c , then y ′ = = g.

csc x ( csc x − cot x ) cos x + csc x − cot x sin x

= csc x +

1

∫ 5 (sin 3x csc 3x ) dx = 5∫ sin 3x ⋅ sin 3x dx

= 5∫

cos x sin x

=

sin 3 x dx sin 3 x

=

2 cos 2 x ⋅ +1 2 sin 2 x

= cot 2 x + 1

= ln csc x − cot x + ln sin x + c

− csc x cot x + csc 2 x cos x + +0 sin x csc x − cot x

cos x 1 + sin x sin x

=

1+ cos x sin x

= 5∫ dx = 5 x + c

Check: Let y = 5 x + c , then y ′ = 5 ⋅ x1−1 + 0 = 5 ⋅ x 0 = 5 h.

2 ∫ a ( cos 2t sec 2t + t ) dt

= a ∫ cos 2t ⋅

1 dt + a t 2 dt cos 2t

∫

a 3

= a ∫ dt + a ∫ t 2 dt = at + t 3 + c

a 3

a 3

Check: Let y = at + t 3 + c , then y ′ = a t 1−1 + ⋅ 3t 2 + 0 = a + at 2 du d = sin 5 x dx dx

i. Given ∫ e sin 5 x cos 5 x dx let u = sin 5 x , then

∫

e sin 5 x cos 5 x dx =

∫

e u cos 5 x ⋅

du 5 cos 5 x

=

1 5

Check: Let y = e sin 5 x + c , then y ′ = j. Given ∫ e ; dx = −

∫

e

3 cos 1 x 2

2du 3 sin

3 cos 1 x 2

1 2

sin

x

sin

1 x dx 2

∫

eu du 5

=

;

1 u e du 5

∫

1 sin 5 x ⋅e ⋅ cos 5 x ⋅ 5 + 0 5

1 2

let u = 3 cos x , then

du = 5 cos 5 x dx

= =

1 u e +c 5

; dx = =

;

. Thus,

1 sin 5 x +c e 5

5 sin 5 x cos 5 x ⋅e 5

du d 1 = 3 cos x 2 dx dx

du 5 cos 5 x

= e sin 5 x cos 5 x

du 3 1 = − sin x dx 2 2

3 2

1 2

; du = − sin x dx

. Therefore,

1 x dx 2

=

∫

e u sin

1 −2du x⋅ 2 3 sin 1 x 2

= −

2 u e du 3

∫

2 3 cos 12 x +c 3

2 = − eu + c = − e 3

3 cos 1 x 1 2 3 cos 12 x 2 3 cos 12 x 1 1 2 sin x + c , then y ′ = − e ⋅ −3 sin x ⋅ + 0 = e 2 3 3 2 2

Check: Let y = − e

k. Given ∫ e tan 5 x sec 2 5 x dx let u = tan 5 x , then ; dx =

du 5 sec 2 5 x

du d = tan 5 x dx dx

;

du = 5 sec 2 5 x dx

; du = 5 sec 2 5 x dx

. Therefore,

Hamilton Education Guides

243

Calculus I

∫e

tan 5 x

4.3 Integration of Trigonometric Functions

∫e

sec 2 5 x dx =

u

sec 2 5 x ⋅

du 5 sec 2 5 x

1 5

Check: Let y = e tan 5 x + c , then y ′ =

1 u e du 5

∫

=

1 2

∫

2du 3 csc 2 3 x

1 cot 3 x

e2

du d 1 cot 3 x = dx dx 2

=

1 tan 5 x +c e 5

= e tan 5 x sec 2 5 x ;

3 2

du 3 = − csc 2 3 x dx 2

; du = − csc 2 3x dx

. Therefore,

csc 2 3 x dx

=

2 3

∫

e u csc 2 3 x ⋅

1 cot 3 x

Check: Let y = − e 2 •

1 u e +c 5

1 tan 5 x e ⋅ sec 2 5 x ⋅ 5 + 0 5

l. Given ∫ e cot x csc 2 x dx let u = cot 3x , then ; dx = −

=

−2du

= −

3 csc 2 3 x

+ c , then y ′ = −

2 u e du 3

∫

2 3

1 cot 3 x

2 = − eu + c = − e 2 3

+c

1 cot 3 x 2 12 cot 3 x 1 e ⋅ − csc 2 3 x ⋅ 3 + 0 = e 2 csc 2 3 x 3 2

To integrate even powers of sin x and cos x use the following identities: sin 2 x =

sin 2 x + cos 2 x = 1

1 ( 1 − cos 2 x ) 2

cos 2 x =

1 ( 1 + cos 2 x ) 2

To integrate odd powers of sin x and cos x use the following equalities:

∫ sin ∫ •

2 n +1

x dx =

cos 2n +1 x dx =

∫ sin ∫

2n

n n 2 2 ∫ ( sin x ) sin x dx = ∫ (1 − cos x ) sin x dx

x sin x dx =

∫ ( cos x )

cos 2n x cos x dx =

2

n

cos x dx

=

∫ (1 − sin x ) 2

n

( let u = cos x )

cos x dx

( let u = sin x )

To integrate products of sin mx , sin nx , cos mx , and cos nx (where m and n are integers) use the trigonometric identities below: 1

∫ sin mx sin nx dx = ∫ 2 [ cos (m − n)x − cos (m + n)x ] dx 1

∫ cos mx cos nx dx = ∫ 2 [ cos (m − n)x + cos (m + n)x ] dx 1

∫ sin mx cos nx dx = ∫ 2 [ sin (m − n)x + sin (m + n)x ] dx •

To integrate tan n x , set tan n x

•

To integrate cot n x , set cot n x

•

(

)

(

)

= tan n −2 x tan 2 x = tan n −2 x sec 2 x − 1 = tan n −2 x sec 2 x − tan n − 2 x = cot n −2 x cot 2 x = cot n −2 x csc 2 x − 1 = cot n −2 x csc 2 x − cot n −2 x

To integrate sec n x For even powers, set

sec n x

(

)

= sec n −2 x sec 2 x = tan 2 x + 1

n−2 2

sec 2 x

For odd powers change the integrand to a product of even and odd functions, i.e., write Hamilton Education Guides

244

Calculus I

∫ sec •

4.3 Integration of Trigonometric Functions

3

x dx as

∫ sec

2

x sec x dx (see Example 4.3-6, problem letter h).

To integrate csc n x csc n x

For even powers, set

(

)

= csc n −2 x csc 2 x = cot 2 x + 1

n−2 2

csc 2 x

For odd powers change the integrand to a product of even and odd functions, i.e., write

∫ csc

3

x dx as

∫ csc

2

x csc x dx (see Example 4.3-6, problem letter i).

In the following examples we use the above general rules in order to solve integral of products and powers of trigonometric functions: Example 4.3-5: Evaluate the following indefinite integrals: b.

d.

∫ sin 5x cos 7 x dx = ∫ sin 3x sin 5x dx =

g.

∫ sin

j.

∫ sin

a.

c.

e.

∫ sin x cos x dx = ∫ cos 3x cos 5x dx =

f.

∫ cos 3x cos 2 x dx 5 ∫ sin x dx =

3

x dx =

h.

∫ cos

5

x dx =

i.

∫ tan

4

x dx =

7

x dx

=

k.

∫ sec

4

x dx

=

l.

∫ cos

3

x dx

=

=

Solutions: a.

1

∫ sin 5x cos 7 x dx = ∫ 2 [ sin (5 − 7)x + sin (5 + 7)x ] dx =

1 1 sin 2 x dx sin 12 x dx − 2 2

∫

∫

Check: Let y = −

= b.

=

=

1

∫ 2 [ sin (− 2 x ) + sin (12 x ) ] dx

1 1 1 1 ⋅ − cos 12 x − ⋅ − cos 2 x + c 2 12 2 2

1 1 1 1 cos 12 x + cos 2 x + c , then y ′ = − ⋅ −12 sin 12 x + ⋅ −2 sin 2 x + 0 24 4 4 24

1 1 sin 12 x − sin 2 x 2 2

=

1 1 sin 12 x + sin (− 2 x ) 2 2

=

1 1 sin (5 + 7 )x + sin (5 − 7 )x 2 2

1

1

=

∫ 2 [ sin (1 − 1)x + sin (1 + 1)x ] dx = ∫ 2 [ sin (0 x ) + sin (2 x ) ] dx

1 sin 2 x dx 2

=

1 1 ⋅ − cos 2 x + c 2 2

∫

1 4

=

=

1

1

1 sin 2 x 2

=

Hamilton Education Guides

1 2

12 2 sin 12 x − sin 2 x 24 4

= sin 5 x cos 7 x

∫ (0 + sin 2 x ) dx

1 ⋅ 2 sin x cos x 2

∫ 2 [ cos (3 − 2)x + cos (3 + 2)x ] dx = ∫ 2 (cos x + cos 5x ) dx =

=

=

1 4

1 4

1 1 1 ⋅ sin x + ⋅ sin 5 x + c 2 2 5

∫ (sin 12 x − sin 2 x ) dx

= − cos 2 x + c

Check: Let y = − cos 2 x + c , then y ′ = − ⋅ −2 sin 2 x + 0 =

∫ cos 3x cos 2 x dx

1 2

1 1 cos 12 x + cos 2 x + c 24 4

∫ sin x cos x dx =

c.

= −

=

=

= sin x cos x

1 1 cos x dx + cos 5 x dx 2 2

∫

∫

1 1 sin x + sin 5 x + c 2 10 245

Calculus I

4.3 Integration of Trigonometric Functions

1 2

Check: Let y = sin x +

= d.

1 1 1 sin 5 x + c , then y ′ = ⋅ cos x + ⋅ 5 cos 5 x + 0 10 2 10

1 1 cos x + cos 5 x 2 2 1

= cos 3x cos 2 x

1

1 1 cos 2 x dx − cos 8 x dx 2 2

∫

∫

1 4

Check: Let y = sin 2 x −

=

=

1 1 1 1 ⋅ sin 2 x − ⋅ sin 8 x + c 2 2 2 8

=

=

1 [ cos (− 2 x ) − cos (8 x ) ] 2

1

∫ (cos 2 x − cos 8x ) dx

8 2 cos 2 x − cos 8 x 16 4

=

1 [ cos (3 − 5)x − cos (3 + 5)x ] 2

=

1 2

1 1 sin 2 x − sin 8 x + c 4 16

1 1 1 sin 8 x + c , then y ′ = ⋅ cos 2 x ⋅ 2 − ⋅ cos 8 x ⋅ 8 + 0 4 16 16

1 1 cos 2 x − cos 8 x 2 2

=

1

= sin 3x sin 5 x 1

∫ cos 3x cos 5x dx = ∫ 2 [ cos (3 − 5)x + cos (3 + 5)x ] dx = ∫ 2 [ cos (− 2 x ) + cos (8x ) ] dx = ∫ 2 (cos 2 x + cos 8x ) dx =

1 1 cos 2 x dx + cos 8 x dx 2 2

∫

∫

1 4

Check: Let y = sin 2 x +

= f.

1 5 cos x + cos 5 x 10 2

∫ sin 3x sin 5x dx = ∫ 2 [ cos (3 − 5)x − cos (3 + 5)x ] dx = ∫ 2 [ cos (− 2 x ) − cos (8x ) ] dx =

e.

1 1 cos (3 − 2 )x + cos (3 + 2 )x 2 2

=

=

∫ sin

5

∫ sin

4

1 1 1 1 ⋅ sin 2 x + ⋅ sin 8 x + c 2 8 2 2

=

1 1 sin 2 x + sin 8 x + c 16 4

1 1 1 sin 8 x + c , then y ′ = ⋅ cos 2 x ⋅ 2 + ⋅ cos 8 x ⋅ 8 + 0 4 16 16

1 1 cos 2 x + cos 8 x 2 2

x dx =

=

=

x sin x dx =

du = − sin x dx

1 [ cos (− 2 x ) + cos (8 x ) ] 2

=

1 [ cos (3 − 5)x + cos (3 + 5)x ] 2

2 2 2 2 ∫ ( sin x ) sin x dx = ∫ (1 − cos x ) sin x dx .

du sin x

8 2 cos 2 x + cos 8 x 16 4

=

= cos 3x cos 5 x

Let u = cos x , then

du d = cos x dx dx

;

∫ sin

2 du 4 2 2 2 2 ∫ (1 − cos x ) sin x dx = ∫ (1 − u ) sin x dx = ∫ ( u − 2u + 1)sin x ⋅ − sin x

5

x dx =

(

; du = − sin x dx ; dx = −

)

. Therefore,

1 5

2 3

1 2 = − ∫ u 4 − 2u 2 + 1 du = − ∫ u 4 du + 2∫ u 2 du − ∫ du = − u 5 + u 3 − u + c = − cos 5 x + cos 3 x − cos x + c 5

2 3

1 5

3

1 5

2 3

Check: Let y = − cos 5 x + cos 3 x − cos x + c , then y ′ = − ⋅ 5 cos 4 x ⋅ − sin x + ⋅ 3 cos 2 x ⋅ − sin x + sin x + 0

(

)

(

= sin x cos 4 x − 2 sin x cos 2 x + sin x = sin x cos 4 x − 2 cos 2 x + 1 = sin x 1 − cos 2 x

(

= sin x sin 2 x g.

∫ sin

3

x dx =

∫ sin

2

)

2

2

= sin x sin 4 x = sin 5 x

x sin x dx =

Hamilton Education Guides

)

2 ∫ (1 − cos x ) sin x dx .

Let u = cos x , then

du d = cos x dx dx

;

du = − sin x dx

246

Calculus I

4.3 Integration of Trigonometric Functions

; du = − sin x dx ; dx = −

∫ sin =

3

x dx

=

1 3 u −u +c 3

du sin x

. Therefore,

2 ∫ (1 − cos x ) sin x dx

du 2 ∫ (1 − u ) sin x ⋅ − sin x

=

)

(

= − ∫ 1 − u 2 du =

∫u

2

∫

du − du

1 cos 3 x − cos x + c 3

=

1 3

1 ⋅ 3 cos 2 x ⋅ − sin x + sin x + 0 3

Check: Let y = cos 3 x − cos x + c , then y ′ =

(

)

= − sin x cos 2 x + sin x

= sin x 1 − cos 2 x = sin x sin 2 x = sin 3 x h.

∫ cos

5

x dx =

∫ cos

4

2 2 2 2 ∫ ( cos x ) cos x dx = ∫ (1 − sin x ) cos x dx .

x cos x dx =

du cos x

du d = sin x dx dx

;

∫ cos

2 du 2 2 2 ∫ (1 − sin x ) cos x dx = ∫ (1 − u ) cos x ⋅ cos x

=

5

∫u

x dx

4

=

du = cos x dx

∫

; du = cos x dx ; dx =

1 5 2 3 u − u +u +c 3 5

∫

du − 2 u 2 du + du =

=

. Therefore, =

2 2 ∫ (1 − u ) du

∫ (u

=

4

Check: Let y = sin 5 x − sin 3 x + sin x + c , then y ′ =

(

1 2 ⋅ 5 sin 4 x ⋅ cos x − ⋅ 3 sin 2 x ⋅ cos x + cos x + 0 5 3

)

(

= cos x sin 4 x − 2 cos x sin 2 x + cos x = cos x sin 4 x − 2 sin 2 x + 1 = cos x 1 − sin 2 x

(

= cos x cos 2 x i.

∫ tan =

4

x dx =

∫ tan

2

∫ tan

2

)

2

∫ ( sec

2

2

∫ tan

4

x sec 2 x dx = x dx =

∫ tan

2

∫u

2

)

=

∫ tan

du d = tan x dx dx

sec 2 x ⋅

du 2

sec x

∫

;

2

∫

du = sec 2 x dx

= ∫ u 2 du =

∫

Check: Let y = tan 3 x − tan x + x + c , then y ′ =

(

)

(

∫

x sec 2 x dx − sec 2 x dx + dx .

x sec 2 x dx − sec 2 x dx + dx =

1 3

2

2 2 2 2 2 ∫ tan x ( sec x − 1) dx = ∫ tan x sec x dx − ∫ tan x dx

x − 1 dx

integral let u = tan x , then

∫ tan

)

= cos x cos 4 x = cos 5 x

x tan 2 x dx =

x sec 2 x dx −

)

− 2u 2 + 1 du

2 1 sin 5 x − sin 3 x + sin x + c 3 5

2 3

1 5

Let u = sin x , then

)(

1 3 u 3

To solve the first

; du = sec 2 x dx ; dx = =

1 tan 3 x . 3

sec 2 x

. Therefore,

Grouping the terms we find

1 tan 3 x − sec 2 x dx + dx 3

∫

du

∫

3 tan 2 x ⋅ sec 2 x − sec 2 x + 1 + 0 3

=

1 tan 3 x − tan x + x + c 3

= tan 2 x ⋅ sec 2 x − sec 2 x + 1

)

= sec 2 x tan 2 x − 1 + 1 = tan 2 x + 1 tan 2 x − 1 + 1 = tan 4 x − tan 2 x + tan 2 x − 1 + 1 = tan 4 x

Hamilton Education Guides

247

Calculus I

j.

∫ sin

7

4.3 Integration of Trigonometric Functions

x dx =

∫ sin

6

x sin x dx =

du d cos x = dx dx

;

∫ sin

3 2 ∫ (1 − cos x ) sin x dx

7

x dx =

du = − sin x dx

3 3 2 2 ∫ ( sin x ) sin x dx = ∫ (1 − cos x ) sin x dx .

(

; du = − sin x dx ; dx = −

)

= − ∫ 1 − 3u 2 + 3u 4 − u 6 du = =

∫u

6

=

du sin x

2 3 ∫ (1 − u ) sin x dx

∫

∫

. Therefore, =

∫ (1 − 3u

)

+ 3u 4 − u 6 sin x ⋅

du − sin x

1 3 cos 7 x − cos 5 x + cos 3 x − cos x + c 7 5 1 7

3 5

+ 3 cos 2 x ⋅ − sin x + sin x + 0

(

∫ sec

4

x dx

1 3 ⋅ 7 cos 6 x ⋅ − sin x − ⋅ 5 cos 4 x ⋅ − sin x 7 5

= − sin x cos 6 x + 3 sin x cos 4 x − 3 sin x cos 2 x + sin x

)

(

= sin x − cos 6 x + 3 cos 4 x − 3 cos 2 x + 1 = sin x 1 − cos 2 x

=

2

1 1 1 7 u − 3⋅ u 5 + 3⋅ u 3 − u + c 3 5 7

∫

du − 3 u 4 du + 3 u 2 du − du =

Check: Let y = cos 7 x − cos 5 x + cos 3 x − cos x + c , then y ′ =

k.

Let u = cos x , then

(

)

3

(

= sin x sin 2 x

)

3

= sin x sin 6 x = sin 7 x

)

= ∫ sec 2 x sec 2 x dx = ∫ sec 2 x 1 + tan 2 x dx = ∫ sec 2 x dx + ∫ tan 2 x sec 2 x dx

2 2 2 2 2 2 ∫ (1 + tan x ) dx + ∫ tan x sec x dx = ∫ dx + ∫ tan x dx + ∫ tan x sec x dx .

du = sec 2 x dx

let u = tan x , then

du d tan x = dx dx

;

∫ tan

2

∫u

du

∫ sec

4

x sec 2 x dx = x dx =

2

∫ dx + ∫ tan

sec 2 x ⋅ 2

1 tan 3 x + tan x + (x − x ) + c 3

2

sec x

; du = sec 2 x dx ; dx =

= ∫ u 2 du =

∫

x dx + tan 2 x sec 2 x dx =

1 3 u 3

=

1 tan 3 x . 3

∫ dx + ∫ tan

2

x dx +

To solve the third integral

du sec 2 x

. Therefore,

Grouping the terms we find 1 tan 3 x 3

1 3

= x + tan x − x + tan 3 x + c

1 tan 3 x + tan x + c 3

=

3 1 Check: Let y = tan 3 x + tan x + c , then y ′ = tan 2 x ⋅ sec 2 x + sec 2 x + 0 = tan 2 x ⋅ sec 2 x + sec 2 x 3

(

3

)

= sec 2 x tan 2 x + 1 = sec 2 x sec 2 x = sec 4 x l.

∫ cos

3

x dx =

2 2 ∫ ( cos x ) cos x dx = ∫ (1 − sin x ) cos x dx .

; du = cos x dx ; dx =

∫ cos

3

x dx =

du cos x

du d = sin x dx dx

;

du = cos x dx

. Therefore,

2 ∫ (1 − sin x ) cos x dx

Hamilton Education Guides

Let u = sin x , then

=

du 2 ∫ (1 − u )cos x ⋅ cos x

=

2 ∫ (1 − u ) du

= − ∫ u 2 du + ∫ du

248

Calculus I

4.3 Integration of Trigonometric Functions

1 3

1 = − u 3 + u + c = − sin 3 x + sin x + c 3

1 1 Check: Let y = − sin 3 x + sin x + c , then y ′ = − ⋅ 3 sin 2 x ⋅ cos x + cos x + 0 = − cos x sin 2 x + cos x

(

3

)

2

(

3

)

2

3

= cos x 1 − sin x = cos x cos x = cos x Example 4.3-6: Evaluate the following indefinite integrals: b.

d.

∫ cos x dx = 3 ∫ tan x dx =

g.

∫ cot

j.

∫ tan

a.

4

c.

e.

∫ cos 5x dx = 4 ∫ cot x dx =

f.

∫ sin x dx = 6 ∫ tan x dx =

2

4

3

x dx =

h.

∫ sec

3

x dx =

i.

∫ csc

3

x dx =

5

x dx

=

k.

∫ cot

5

x dx

=

l.

∫ cot

6

x dx

=

Solutions: a.

∫

cos 4 x dx =

∫ ( cos x ) 2

2

=

dx

∫

2

1   2 ( 1 + cos 2 x ) dx  

=

1 1 2 dx + cos 2 2 x dx + cos 2 x dx 4 4 4

+

1 1 cos 4 x dx + sin 2 x 8 4

∫

∫

∫

∫

Check: Let y =

=

x 1 + 4 4

=

1 4

∫ ( 1 + cos 2 x ) dx = 4 ∫ (1 + cos 1 1

1

∫ 2 ( 1 + cos 4 x ) dx + 2 ⋅ 2 sin 2 x

1 x x 1 1 + + ⋅ sin 4 x + sin 2 x + c 4 8 8 4 4

=

1

2

=

=

2

)

2 x + 2 cos 2 x dx

x 1 1 + ⋅ dx 4 4 2

∫

3 1 1 sin 4 x + sin 2 x + c x+ 8 32 4

1 3 4 cos 4 x 2 cos 2 x 3x 1 + sin 4 x + sin 2 x + c , then y ′ = + + +0 4 4 32 8 8 32

=

3 1 2 + cos 4 x + cos 2 x 8 8 4

1 1 1 2 1 1 1 2 2 1 1 =  +  + cos 4 x + cos 2 x = + ( 1 + cos 4 x ) + cos 2 x = + ⋅ ( 1 + cos 4 x ) + cos 2 x 4

=

8 8

(

∫ cos

2

4

1 1 2 + ⋅ cos 2 2 x + cos 2 x 4 4 4

=

x 2

2

(

1 1 sin 10 x + c 2 10

Check: Let y =

) = 14 ( 1 + cos 2 x)

2

5 x dx

=

2

1

∫ dx −∫ 2 ( 1 − cos 10 x ) dx

4

1 =  ( 1 + cos 2 x ) 2

2



= x−

1 1 dx + cos 10 x dx 2 2

∫

∫

x sin 10 x  1 1 +c = x1 −  + sin 10 x + c = +



2

20

2

20

x sin 10 x 1 1 + + c , then y ′ = + ⋅ cos 10 x ⋅10 + 0 2 20 2 20

Hamilton Education Guides

4 2

4

2

= x− + ⋅

4

4

8

1 1 + cos 2 2 x + 2 cos 2 x 4

) = cos x 5 x dx = ∫ (1 − sin 5 x )dx = ∫ dx − ∫ sin = cos 2 x

b.

4

=

1 10 ⋅ cos 10 x + 2 20

=

1 1 + cos 10 x 2 2

249

Calculus I

4.3 Integration of Trigonometric Functions

1 1 1 1 1 = 1 −  + cos 10 x = 1 −  − cos 10 x  = 1 − ( 1 − cos 10 x ) = 1 − sin 2 5 x = cos 2 5 x

c.

∫

sin 4 x dx =

2

2

2



∫ ( sin x ) 2

2

∫

=

dx

2

1 1 2 cos 2 2 x dx − dx + cos 2 x dx 4 4 4

+

1 1 cos 4 x dx − sin 2 x 8 4

∫

∫

∫

Check: Let y =

=

2



1   2 ( 1 − cos 2 x ) dx  

=

∫

2

x 1 + 4 4

=

1 4

=

∫ ( 1 − cos 2 x ) dx 2

1

=

1 4

∫ (1 + cos

1 1

=

)

2 x − 2 cos 2 x dx

x 1 1 + ⋅ dx 4 4 2

∫ 2 ( 1 + cos 4 x ) dx − 2 ⋅ 2 sin 2 x =

x x 1 1 1 + + ⋅ sin 4 x − sin 2 x + c 4 8 8 4 4

2

∫

1 1 3 x+ sin 4 x − sin 2 x + c 4 32 8

3 4 cos 4 x 2 cos 2 x 1 3x 1 − +0 + sin 4 x − sin 2 x + c , then y ′ = + 8 32 4 8 32 4

=

2 3 1 + cos 4 x − cos 2 x 4 8 8

2 1 1 1 1 1 1 2 2 1 1 =  +  + cos 4 x − cos 2 x = + ( 1 + cos 4 x ) − cos 2 x = + ⋅ ( 1 + cos 4 x ) − cos 2 x 4

8 8

(

= sin 2 x

∫ tan

3

x dx

=

)

2

∫ tan

2

2

∫ tan

3

x tan x dx =

x dx =

∫

− tan x dx =

∫ tan

x tan x dx

∫ sec 2

=

4

4

8

(

1 1 + cos 2 2 x − 2 cos 2 x 4

4

4 2

) = 14 ( 1 − cos 2 x)

2

1 =  ( 1 − cos 2 x ) 2

2



= sin 4 x

integral let u = tan x , then

∫ sec

4

1 1 2 + ⋅ cos 2 2 x − cos 2 x 4 4 4

=

d.

4

2

∫ ( sec

=

2

du d = tan x dx dx

x ⋅u ⋅

du 2

sec x

x tan x dx =

∫

=

;

= ∫ sec 2 x tan x dx − ∫ tan x dx . To solve the first

du = sec 2 x dx

= ∫ u du =

∫ ( sec

1 tan 2 x − tan x dx 2

)

x − 1 tan x dx

2

1 2 u 2

; du = sec 2 dx ; dx = =

)

x − 1 tan x dx =

1 tan 2 x . 2

∫ ( sec

2

du sec 2 x

. Thus,

Combining the term

)

x tan x − tan x dx =

∫ sec

2

x tan x dx

1 tan 2 x − ln sec x + c 2

1 1 1 Check: Let y = tan 2 x − ln sec x + c , then y ′ = − ⋅ 2 tan x ⋅ sec 2 x − ⋅ sec x tan x + 0 2

2

= tan x sec 2 x − e.

∫ cot

4

x dx =

∫ cot

2

sec x tan x sec x

x cot 2 x dx =

(

)

sec x

(

)

= tan x sec 2 x − tan x = tan x sec 2 x − 1 = tan x tan 2 x = tan 3 x 2 2 2 2 2 ∫ cot x ( csc x − 1) dx = ∫ cot x csc x dx − ∫ cot x dx

= ∫ cot 2 x csc 2 x dx − ∫ csc 2 x − 1 dx = ∫ cot 2 x csc 2 x dx − ∫ csc 2 x dx + ∫ dx . To solve the first integral let u = cot x , then

Hamilton Education Guides

du d = cot x dx dx

;

du = − csc 2 x dx

; du = − csc 2 x dx ; dx = −

du csc 2 x

. Therefore,

250

Calculus I

∫ cot

2

∫ cot

4

4.3 Integration of Trigonometric Functions

x csc 2 x dx = x dx =

∫ cot

2

∫u

2

1 3

1 3

du

csc 2 x ⋅ −

= − ∫ u 2 du = − u 3 = − cot 3 x . Grouping the terms we find

2

csc x

1 1 x csc 2 x dx − csc 2 x dx + dx = − cot 3 x − csc 2 x dx + dx = − cot 3 x + cot x + x + c 3 3

∫

∫

∫

∫

1 3 Check: Let y = − cot 3 x + cot x + x + c , then y ′ = − cot 2 x ⋅ − csc 2 x − csc 2 x + 1 = cot 2 x csc 2 x − csc 2 x + 1 3

(

)

(

)(

3

)

= csc 2 x cot 2 x − 1 + 1 = cot 2 x + 1 cot 2 x − 1 + 1 = cot 4 x − cot 2 x + cot 2 x − 1 + 1 = cot 4 x f.

∫ tan

6

x dx =

∫ tan

4

4 2 4 2 4 4 2 ∫ tan x ( sec x − 1) dx = ∫ ( tan x sec x − tan x ) dx = ∫ tan x sec x dx

x tan 2 x dx =

∫

− tan 4 x dx . In example 4.3-5, problem letter i, we found that

∫ tan =

6

x dx =

∫ tan

4

∫ tan

4

x tan 2 x dx =

4 2 ∫ tan x ( sec x − 1) dx

 1 x sec 2 x dx −  tan 3 x − tan x + x + c   3

=

=

∫ tan

4

∫ tan

4

x dx =

1 tan 3 x − tan x + x + c . 3

Thus,

∫

x sec 2 x dx − tan 4 x dx

1 1 tan5 x − tan 3 x + tan x − x + c 5 3

3 1 1 1 Check: Let y = tan 5 x − tan 3 x + tan x − x + c , then y ′ = ⋅ 5 tan 4 x ⋅ sec 2 x − tan 2 x ⋅ sec 2 x + sec 2 x − 1 + 0 5

3

5

(

)

3

= tan 4 x ⋅ sec 2 x − tan 2 x ⋅ sec 2 x + sec 2 x − 1 = sec 2 x tan 4 x − tan 2 x + 1 − 1

(

)(

)

= 1 + tan 2 x tan 4 x − tan 2 x + 1 − 1 = tan 4 x − tan 2 x + 1 + tan 6 x − tan 4 x + tan 2 x − 1 = tan 6 x g.

∫ cot

3

x dx =

∫ cot

2

x cot x dx =

integral let u = cot x , then

∫ csc

2

∫ cot

3

∫

x cot x dx =

x dx =

∫ cot

− cot x dx = −

∫ csc 2

2

∫ ( csc

2

du d = cot x dx dx

x ⋅u ⋅ −

du 2

csc x

x cot x dx =

∫ ( csc

1 cot 2 x − cot x dx 2

∫

1 2

)

x − 1 cot x dx =

;

du = − csc 2 x dx

∫ csc

2

∫

x cot x dx − cot x dx . To solve the first

; du = − csc 2 dx ; dx = −

Hamilton Education Guides

. Thus,

1 2

1 2

2

)

x − 1 cot x dx =

∫ ( csc

2

)

x ⋅ cot x − cot x dx =

∫ csc

2

x cot x dx

1 2

= − cot 2 x − ln sin x + c 1 2

cos x sin x

csc 2 x

= − ∫ u du = − u 2 = − cot 2 x . Combining the term

Check: Let y = − cot 2 x − ln sin x + c , then y ′ = − ⋅ 2 cot x ⋅ − csc 2 x − = cot x csc 2 x −

du

(

1 ⋅ cos x + 0 sin x

)

= cot x csc 2 x − cot x = cot x csc 2 x − 1 = cot x cot 2 x = cot 3 x

251

Calculus I

h.

∫ sec

3

4.3 Integration of Trigonometric Functions

x dx =

∫ sec

2

x sec x dx =

∫ ( tan

2

)

x + 1 sec x dx =

∫ tan

2

∫

x sec x dx + sec x dx =

∫ tan x ⋅ tan x sec x dx

∫

+ sec x dx . To solve the first integral let u = tan x and dv = tan x sec x dx , then du = sec 2 x dx and

∫ dv = ∫ tan x sec x dx we obtain

∫ tan

2

which implies v = sec x . Using the integration by parts formula ∫ u dv = u v − ∫ v du

x sec x dx =

∫ tan x ⋅ tan x sec x dx = tan x ⋅ sec x − ∫ sec x ⋅ sec

Combining the terms we have ∫ sec 3 x dx =

∫

+ sec x dx . Moving the

∫ sec

3

2

∫

x dx = tan x sec x − sec 3 x dx

∫ tan x ⋅ tan x sec x dx + ∫ sec x dx = tan x sec x − ∫ sec

3

x dx .

x dx term from the right hand side of the equation to the left hand

side we obtain ∫ sec 3 x dx + ∫ sec 3 x dx = 2∫ sec 3 x dx = tan x sec x + ∫ sec x dx . Therefore,

∫ sec

3

x dx =

(

1 tan x sec x + sec x dx 2

∫

1 2

)=

1 1 tan x sec x + ln sec x + tan x + c 2 2 sec 2 x ⋅ sec x + sec x tan x ⋅ tan x 2

1 2

Check: Let y = tan x sec x + ln sec x + tan x + c , then y ′ =

i.

∫ csc

3

sec 3 x + sec x tan 2 x sec x ( sec x + tan x ) + 2 2(sec x + tan x )

+

sec x tan x + sec 2 x +0 2(sec x + tan x )

=

sec 3 x + sec x tan 2 x + sec x 2

=

2 sec 3 x = sec 3 x 2

x dx =

∫

+ csc x dx .

∫ csc

2

x csc x dx =

=

=

(

) = sec

sec 3 x + sec x tan 2 x + 1 2

3

=

sec 3 x + sec x tan 2 x sec x + 2 2

x + sec x sec 2 x 2

=

sec 3 x + sec 3 x 2

2 2 ∫ (1 + cot x ) csc x dx = ∫ cot x csc x dx + ∫ csc x dx = ∫ cot x ⋅ cot x csc x dx

To solve the first integral let u = cot x and dv = cot x csc x dx , then du = − csc 2 x dx and

∫ dv = ∫ cot x csc x dx

which implies v = − csc x . Using the integration by parts formula ∫ u dv = u v − ∫ v du

we obtain ∫ cot 2 x csc x dx = ∫ cot x ⋅ cot x csc x dx = cot x ⋅ − csc x − ∫ csc x ⋅ csc 2 x dx = − cot x csc x − ∫ csc 3 x dx Combining the terms we have ∫ csc 3 x dx = ∫ cot x ⋅ cot x csc x dx + ∫ csc x dx = − cot x csc x − ∫ csc 3 x dx .

∫

+ csc x dx . Moving the

∫ csc

3

x dx term from the right hand side of the equation to the left hand

side we obtain ∫ csc 3 x dx + ∫ csc 3 x dx = 2∫ csc 3 x dx = − cot x csc x + ∫ csc x dx . Therefore,

∫ csc

3

x dx

=

(

1 − cot x csc x + csc x dx 2

Hamilton Education Guides

∫

)

1 2

1 2

= − cot x csc x + ln csc x − cot x + c

252

Calculus I

4.3 Integration of Trigonometric Functions

(

− − csc 2 x ⋅ csc x − csc x cot x ⋅ cot x 2

1 1 Check: Let y = − cot x csc x + ln csc x − cot x + c , then y ′ = 2

j.

∫ tan

5

2

+

− csc x cot x + csc 2 x +0 2(csc x − cot x )

=

csc 3 x + csc x cot 2 x + csc x 2

=

2 csc 3 x = csc 3 x 2

x dx =

∫ tan

3

=

csc 3 x + csc x cot 2 x csc x ( csc x − cot x ) + 2(csc x − cot x ) 2

=

(

du d = tan x dx dx

∫ tan

du

x sec 2 x dx

= ∫ u 3 sec 2 x ⋅

2

sec x

problem letter d, we found that

∫ tan =

5

x dx =

∫ tan

3

x tan 2 x dx =

1 tan 4 x − tan 3 x dx 4

∫

=

∫ tan

;

du = sec 2 x dx

= ∫ u 3 du = 3

x dx

=

x + csc x csc 2 x 2

1 4 u +c 4

; du = sec 2 x dx ; dx = =

1 tan 4 x + c . 4

1 tan 2 x − ln sec x + c . 2

=

sec 3 x + sec 3 x 2

To solve the

du sec 2 x

. Therefore,

Also, in Example 4.3-6,

Grouping the terms together

3 2 3 3 2 ∫ tan x ( sec x − 1) dx = ∫ tan x sec x dx − ∫ tan x dx

1  1 tan 4 x −  tan 2 x − ln sec x + c  4  2

1 2

1 4

3

csc 3 x + csc x cot 2 x csc x + 2 2

3 2 3 3 2 ∫ tan x ( sec x − 1) dx = ∫ tan x sec x dx − ∫ tan x dx .

x tan 2 x dx =

first integral let u = tan x , then 3

) = csc

csc 3 x + csc x cot 2 x + 1 2

=

)

Check: Let y = tan 4 x − tan 2 x + ln sec x + c , then y ′ =

(

=

1 1 tan 4 x − tan 2 x + ln sec x + c 4 2

sec x tan x 1 2 ⋅ 4 tan 3 x ⋅ sec 2 x − tan x ⋅ sec 2 x + +0 4 2 sec x

)

(

)(

)

= tan 3 x sec 2 x − tan x sec 2 x + tan x = sec 2 x tan 3 x − tan x + tan x = 1 + tan 2 x tan 3 x − tan x + tan x = tan 3 x − tan x + tan 5 x − tan 3 + tan x = tan 5 x k.

∫ cot

5

x dx =

∫ cot

3

first integral let u = cot x , then

∫ cot

3

x csc 2 x dx =

3 2 3 2 3 ∫ cot x ( csc x − 1) dx = ∫ cot x csc x dx − ∫ cot x dx .

x cot 2 x dx =

∫u

3

du d cot x = dx dx

csc 2 x ⋅ −

du 2

csc x

;

du = − csc 2 x dx

; du = − csc 2 x dx ; dx = −

To solve the du csc 2 x

. Thus,

1 4

1 4

= − ∫ u 3 du = − u 4 + c = − cot 4 x + c . Also, in example 4.3-6, 1 2

problem letter g, we found that ∫ cot 3 x dx = − cot 2 x − ln sin x + c . Grouping the terms together

∫ cot

5

x dx

=

∫ cot

3

x cot 2 x dx

=

3 2 3 2 3 ∫ cot x ( csc x − 1) dx = ∫ cot x csc x dx − ∫ cot x dx

1 4

1 2

1 1   1 = − cot 4 x − ∫ cot 3 x dx = − cot 4 x −  − cot 2 x − ln sin x + c  = − cot 4 x + cot 2 x + ln sin x + c 4

Hamilton Education Guides

4

 2



253

Calculus I

4.3 Integration of Trigonometric Functions

1 1 − 4 ⋅ cot 3 x ⋅ − csc 2 x 2 ⋅ cot x ⋅ − csc 2 x cos x Check: Let y = − cot 4 x + cot 2 x + ln sin x + c , then y ′ = + + +0 4

2

(

4

)

(

2

)(

)

sin x

= cot 3 x csc 2 x − cot x csc 2 x + cot x = csc 2 x cot 3 x − cot x + cot x = 1 + cot 2 x cot 3 x − cot x + cot x = cot 3 x − cot x + cot 5 x − cot 3 + cot x = cot 5 x l.

∫ cot

6

x dx

(

)

= ∫ cot 4 x cot 2 x dx = ∫ cot 4 x csc 2 x − 1 dx =

∫ ( cot

4

∫

)

x csc 2 x − cot 4 x dx

− cot 4 x dx . In example 4.3-6, problem letter e, we found that

∫ cot

Thus, =

∫ cot

4

6

∫ cot

x dx =

4

x cot 2 x dx =

4 2 ∫ cot x ( csc x − 1) dx

1 5

∫ cot

4

4

1 x dx = − cot 3 x + cot x + x + c . 3

∫

x csc 2 x dx − cot 4 x dx

1 3

1 5

 1  x csc 2 x dx −  − cot 3 x + cot x + x + c   3 

=

∫ cot

= ∫ cot 4 x csc 2 x dx

= − cot 5 x + cot 3 x − cot x − x + c

1 3

Check: Let y = − cot 5 x + cot 3 x − cot x − x + c , then y ′ = −

(

5 ⋅ cot 4 x ⋅ − csc 2 x 3 ⋅ cot 2 x ⋅ − csc 2 x + + csc 2 x − 1 5 3

)

= cot 4 x ⋅ csc 2 x − cot 2 x ⋅ csc 2 x + csc 2 x − 1 = csc 2 x cot 4 x − cot 2 x + 1 − 1

(

)(

)

= 1 + cot 2 x cot 4 x − cot 2 x + 1 − 1 = cot 4 x − cot 2 x + 1 + cot 6 x − cot 4 x + cot 2 x − 1 = cot 6 x Example 4.3-7: Evaluate the following indefinite integrals: c.

e.

∫ sin x cos x dx = 2 ∫ tan x sec x dx =

f.

∫ sin x cos x dx = 5 3 ∫ tan x sec x dx =

x sec 4 x dx =

h.

∫ tan

3

x sec 4 x dx =

i.

∫ tan

3

x sec 3 x dx =

x csc 2 x dx

k.

∫ cot

3

x csc 3 x dx

l.

∫ cot

3

x csc 4 x dx

=

1 1 dx − cos 4 x dx 8 8

b.

d.

∫ sin x cos x dx = 3 2 ∫ sin x cos x dx =

g.

∫ tan

j.

∫ cot

a.

2

5

2

2

=

2

5

=

4

2

=

Solutions: a.

∫ sin =

2

1 sin 2 2 x dx 4

x 1 1 − ⋅ sin 4 x + c 8 8 4 1 8

= b.

∫ sin

2

∫ ( 1 − cos 4 x ) dx

1 4 cos 4 x 1 +0 sin 4 x + c , then y ′ = − 32 8 32

=

1 1 − cos 4 x 8 8

=

∫ sin

Hamilton Education Guides

=

∫ 2 ( 1 − cos 4 x ) dx

∫

∫

1 1 x− sin 4 x + c 32 8

1 1 ⋅ ( 1 − cos 4 x ) 4 2

x cos 5 x dx

1

1 8

=

Check: Let y = x −

1 4

=

∫

x cos 2 x dx =

2

=

1 ⋅ sin 2 2 x 4

=

1 ( 1 − cos 4 x ) 8

= sin 2 x cos 2 x

x cos 4 x cos x dx

=

2 2 2 2 2 2 ∫ sin x (cos x ) cos x dx = ∫ sin x (1 − sin x ) cos x dx

254

Calculus I

4.3 Integration of Trigonometric Functions

=

2 4 2 ∫ sin x (1 + sin x − 2 sin x ) cos x dx

=

∫ sin

2

=

∫ ( sin

)

2

x cos x + sin 6 x cos x − 2 sin 4 x cos x dx

∫

∫

x cos x dx + sin 6 x cos x dx − 2 sin 4 x cos x dx = 1 7

1 3

2 5

Check: Let y = sin 3 x + sin 7 x − sin 5 x + c , then y ′ =

2 1 1 sin 3 x + sin 7 x − sin 5 x + c 5 7 3

3 sin 2 x cos x 7 sin 6 x cos x 10 sin 4 x cos x − +0 + 5 7 3

(

= sin 2 x cos x + sin 6 x cos x − 2 sin 4 x cos x = sin 2 x cos x 1 + sin 4 x − 2 sin 2 x

(

= sin 2 x cos x 1 + sin 2 x c.

∫ sin −

4

x cos 2 x dx

)

∫

=

1 8

2

sin 2 x dx

1

1

=

)

2

= sin 2 x cos x cos 4 x = sin 2 x cos 5 x

∫ ( 12 sin 2 x )

=

∫ 2 ( 1 − cos 4 x ) dx − 8 ∫ sin

1 1 1 1 x − ⋅ sin 4 x − sin 3 2 x + c 16 16 4 48

=

= sin 2 x cos x cos 2 x

∫ ( sin x cos x )

=

1 sin 2 2 x cos 2 x dx 8

Check: Let y =

(

2

2

)

2

⋅

1 ( 1 − cos 2 x ) dx 2

2 x cos 2 x dx

=

=

1 sin 2 2 x dx 8

∫

1 1 1 1 dx − cos 4 x dx − ⋅ sin 3 2 x 8 6 16 16

∫

∫

1 1 1 x− sin 3 2 x + c sin 4 x − 48 16 64

1 4 cos 4 x 6 sin 2 2 x cos 2 x 1 1 1 − − +0 x − sin 4 x − sin 3 2 x + c , then y ′ = 16 64 48 48 16 64

1 (1 − cos 4 x ) − 1 sin 2 2 x cos 2 x 8 16

=

1 1 1 ⋅ (1 − cos 4 x ) − sin 2 2 x cos 2 x 8 8 2

=

1 1 ⋅ sin 2 2 x − sin 2 2 x cos 2 x 8 8

2

1 1 =  sin 2 x  ⋅ ( 1 − cos 2 x ) = ( sin x cos x )2 sin 2 x = sin 2 x cos 2 x ⋅ sin 2 x = sin 4 x cos 2 x 2

d.

∫ sin

3



x cos 2 x dx =

2

3 5 3 2 3 5 ∫ sin x (1 − sin x )dx = ∫ ( sin x − sin x ) dx = ∫ sin x dx − ∫ sin x dx .

1 5

In Example,

2 3

4.3-5, problem letters f and g, we found that ∫ sin 5 x dx = − cos 5 x + cos 3 x − cos x + c and ∫ sin 3 x dx =

∫ sin =

3

1 cos 3 x − cos x + c . 3

x cos 2 x dx =

∫ sin

3

Therefore,

∫

x dx − sin 5 x dx =

1 1 2 cos 5 x + cos 3 x − cos 3 x − cos x + cos x + c 5 3 3 1 5

1 3

2 1   1 cos 3 x − cos x + c −  − cos 5 x + cos 3 x − cos x + c  3 3  5 

=

Check: Let y = cos 5 x − cos 3 x + c , then y ′ =

1 1 cos 5 x − cos 3 x + c 5 3 1 3 ⋅ 5 cos 4 x ⋅ − sin x − ⋅ cos 2 x ⋅ − sin x + 0 5 3

(

)

= − sin x cos 4 x + sin x cos 2 x = sin x cos 2 x 1 − cos 2 x = sin x cos 2 x sin 2 x = sin 3 x cos 2 x

Hamilton Education Guides

255

Calculus I

e.

∫ tan

2

4.3 Integration of Trigonometric Functions

∫ ( sec

x sec x dx =

2

)

x − 1 sec x dx =

h, we found that ∫ sec 3 x dx =

∫ tan

2

∫ sec

x sec x dx =

∫ sec

∫

x dx − sec x dx . In Example 4.3-6, problem letter

1 1 tan x sec x + ln sec x + tan x + c . Therefore, 2 2

∫

3

3

x dx − sec x dx =

1 1 tan x sec x + ln sec x + tan x − ln sec x + tan x + c 2 2

1 1 tan x sec x − ln sec x + tan x + c 2 2 1 2

(

f.

∫ tan

5

=

sec x ( sec x + tan x ) 1 sec 3 x + sec x tan 2 x − 2 (sec x + tan x ) 2

=

1 sec x sec 2 x + tan 2 x − 1 2

(

)

(

x sec 3 x dx =

) = 12 sec x ( tan

∫ tan

4

x sec 2 x sec x tan x dx =

=

∫ ( sec

=

1 1 2 sec 7 x + sec 3 x − sec 5 x + c 7 3 5

4

)

x + 1 − 2 sec 2 x sec 2 x sec x tan x dx

sec x 1 1 sec 3 x + sec x tan 2 x − 2 2 2

x + tan 2 x

∫ ( sec

∫ ( sec

6

2

( = ( sec

)

x −1

2

2

x

= tan 2 x sec x

sec 2 x sec x tan x dx

)

x + sec 2 x − 2 sec 4 x sec x tan x dx

1 3

Check: Let y = sec 7 x − sec 5 x + sec 3 x + c , then y ′ = 3 + sec 2 x ⋅ sec x tan x + 0 3

) = 12 sec x ⋅ 2 tan

1 2 1 sec 7 x − sec 5 x + sec 3 x + c 3 5 7

=

2 5

1 7

=

2

=

)

sec x tan x + sec 2 x 1 sec 3 x + sec x tan 2 x − 2 (sec x + tan x ) 2

1 2

Check: Let y = tan x sec x − ln sec x + tan x + c , then y ′ =

7 10 sec 6 x ⋅ sec x tan x − sec 4 x ⋅ sec x tan x 7 5

= sec 6 x ⋅ sec x tan x − 2 sec 4 x ⋅ sec x tan x + sec 2 x ⋅ sec x tan x

)

(

)

= sec 6 x + sec 2 x − 2 sec 4 x sec x tan x = sec 4 x + 1 − 2 sec 2 x sec 2 x sec x tan x

g.

∫ tan

5

2

)

x −1

x sec 4 x dx =

2

sec 2 x sec x tan x

∫ tan

∫

+ tan 5 x sec 2 x dx =

5

= tan 4 x sec 2 x sec x tan x = tan 5 x sec 3 x

x sec 2 x sec 2 x dx =

5 2 2 ∫ tan x ( tan x + 1) sec x dx

=

∫ tan

7

x sec 2 x dx

1 1 tan 8 x + tan 6 x + c 8 6 1 6

1 8

Check: Let y = tan 8 x + tan 6 x + c , then y ′ =

1 1 ⋅ 8 tan 7 x ⋅ sec 2 x + ⋅ 6 tan 5 x ⋅ sec 2 x + 0 6 8

(

)

= tan 7 x sec 2 x + tan 5 x sec 2 x = tan 5 x sec 2 x tan 2 x + 1 = tan 5 x sec 2 x sec 2 x = tan 5 x sec 4 x h.

∫ tan

3

x sec 4 x dx =

∫ tan

Hamilton Education Guides

3

x sec 2 x sec 2 x dx =

3 2 2 ∫ tan x ( tan x + 1) sec x dx

=

∫ tan

5

x sec 2 x dx

256

Calculus I

4.3 Integration of Trigonometric Functions

∫

+ tan 3 x sec 2 x dx =

1 1 tan 6 x + tan 4 x + c 6 4

1 1 1 1 Check: Let y = tan 6 x + tan 4 x + c , then y ′ = ⋅ 6 tan 5 x ⋅ sec 2 x + ⋅ 4 tan 3 x ⋅ sec 2 x + 0 6

4

5

6

2

3

2

(

∫ tan

∫ sec

2

)

∫ tan =

3

∫u

4

2

x sec 2 x ⋅ tan x sec x dx =

∫ ( sec

2

)

2

x − 1 sec 2 x ⋅ tan x sec x dx =

∫ sec

x ⋅ tan x sec x dx . To solve the first and the second integral let u = sec x , then

du = sec x tan x dx

;

∫ tan

x sec 3 x dx =

3

2

4

= tan x sec x + tan x sec x = tan x sec x tan x + 1 = tan 3 x sec 2 x sec 2 x = tan 3 x sec 4 x i.

3

; du = sec x tan x dx ; dx =

x sec 3 x dx

=

⋅ tan x sec x ⋅

∫ tan

2

du sec x tan x

x sec 2 x ⋅ tan x sec x dx

du d = sec x dx dx

∫

4

x ⋅ tan x sec x dx − sec 2 x ⋅ tan x sec x dx

du du − u 2 ⋅ tan x sec x ⋅ tan x sec x tan x sec x

∫

1 1 sec 5 x − sec 3 x + c 5 3 1 1 Check: Let y = sec 5 x − sec 3 x + c , 5 3

x ⋅ tan x sec x dx

. Therefore,

∫ sec

=

4

=

∫u

4

∫

du − u 2 du =

1 5 1 3 u − u +c 3 5

=

then y ′ =

5 3 sec 4 x ⋅ sec x tan x − sec 2 x ⋅ sec x tan x + 0 5 3

(

)

(

)

= sec 4 x ⋅ sec x tan x − sec 2 x ⋅ sec x tan x = sec 4 x − sec 2 x sec x tan x = sec 2 x − 1 sec 2 x sec x tan x = tan 2 x sec 2 x sec x tan x = tan 3 x sec 3 x j. Given ∫ cot 2 x csc 2 x dx let u = cot x , then Therefore,

∫ cot

2

x csc 2 x dx

=

∫u

2

csc 2 x ⋅

du d = cot x dx dx

du − csc 2 x

;

du = − csc 2 x dx

∫ cot ∫

3

x csc 3 x dx =

− csc 2 x cot x dx

∫ cot x cot 1 5

2

x csc 3 x dx =

csc 2 x

1 3

3

Check: Let y = − cot 3 x + c , then y ′ = − ⋅ 3 cot 2 x ⋅ − csc 2 x + 0 =

k.

−du

1 = − ∫ u 2 du = − u 3 + c = − cot 3 x + c

1 3

1 3

; du = − csc 2 x dx ; dx =

∫ cot x ( csc

2

)

3 ⋅ cot 2 x csc 2 x 3

x − 1 csc 2 x dx

=

∫ csc

4

= cot 2 x csc 2 x x cot x dx

1 3

= − csc 5 x + csc 3 x + c 1 5

1 3

1 5

1 3

Check: Let y = − csc 5 x + csc 3 x + c , then y ′ = − ⋅ 5 csc 4 x ⋅ − csc x cot x + ⋅ 3 csc 2 x ⋅ − csc x cot x + 0

(

)

= csc 5 x ⋅ cot x − csc 3 x ⋅ cot x = csc 3 x cot x csc 2 x − 1 = csc 3 x cot x cot 2 x = cot 3 x csc 3 x

Hamilton Education Guides

257

.

Calculus I

l.

∫ cot

3

4.3 Integration of Trigonometric Functions

x csc 4 x dx

=

∫ cot

∫

+ cot 3 x csc 2 x dx = − 1 6

3

x csc 2 x csc 2 x dx

=

3 2 2 ∫ cot x ( cot x + 1) csc x dx

=

∫ cot

5

x csc 2 x dx

1 1 cot 6 x − cot 4 x + c 4 6 1 4

1 6

1 4

Check: Let y = − cot 6 x − cot 4 x + c , then y ′ = − ⋅ 6 cot 5 x ⋅ − csc 2 x − ⋅ 4 cot 3 x ⋅ − csc 2 x + 0

(

)

= cot 5 x csc 2 x + cot 3 x csc 2 x = cot 3 x csc 2 x cot 2 x + 1 = cot 3 x csc 2 x csc 2 x = cot 3 x csc 4 x Section 4.3 Practice Problems – Integration of Trigonometric Functions 1. Evaluate the following integrals: 3x dx 5

b.

∫ 3cos 2 x dx =

c.

∫ (sin 5x − cos 7 x ) dx

3 x dx =

e.

∫ x sec

2 2

f.

∫ 8sec 5x dx =

x cos x dx =

h.

∫ cos

3

x sin x dx =

i.

∫ 2 cos x dx

b.

∫ tan

9

x sec2 x dx =

c.

∫ cot

∫ x cot x dx

f.

∫3x

i.

∫e

a.

∫ sin

d.

∫ 2 csc

g.

∫ sin

3

2

=

x dx =

x

2

=

=

2. Evaluate the following integrals: a.

∫e

d.

∫ sec 2 x tan 2 x dx =

e.

g.

∫ (sin 5x csc 5x ) dx =

h.

∫ (cos 5t sec 5t + 3t

3x

sec e3 x dx =

2

= 2

)

+ t dt =

5

x csc2 x dx =

1 2 csc x3dx cot x

=

csc 2 x dx =

3. Evaluate the following integrals: a.

∫ sin 3x cos 5x dx =

b.

∫ cos 6 x cos 4 x dx =

c.

∫ sin

d.

∫ tan 2 x dx = 4 ∫ cot 3x dx =

e.

∫ cos ax dx = 3 ∫ cot 2 x dx =

f.

∫ tan 5x dx = 6 ∫ cot 3x dx =

g.

4

Hamilton Education Guides

h.

2

i.

5

3 x dx =

3

258

Calculus I

4.4

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

Integration of Expressions Resulting in Inverse Trigonometric Functions

In the following examples we will solve problems using the following formulas: 1

∫

2

a −x 1

x 1 arc tan + c a a

∫ a 2 + x 2 dx = ∫x

1 2

x −a

dx

2

x 1 arc sin + c a a

=

dx

2

=

x 1 sin −1 + c a a

=

x 1 tan −1 + c a a

=

1 x arc sec + c a a

=

1 x sec −1 + c a a

Let’s integrate some algebraic expressions using the above integration formulas. Example 4.4-1: Evaluate the following indefinite integrals: a. d. g.

dx

∫

16 − 9 x 2

x 2 dx

∫ ∫

= =

e.

=

h.

16 − x 6 dx 9 x 2 + 25

dx

∫

b.

∫ ∫

36 − x 2 x2 9+ x

dx

6

x 3 dx 49 + 4 x 8

=

c.

∫

=

f.

∫

=

i.

∫

x 3 dx 1− x8 x 4

x +5 x 2 dx 5 + 9x 6

=

dx =

=

Solutions:

∫

a. First - Write the given integral in its standard form =

∫

dx a2 − x2

= arc sin

x +c, a

i.e.,

∫

dx 16 − 9 x 2

dx 4 − (3 x ) 2 2

du d = 3 x = 3 x which implies dx dx 1 du 1 du 1 u = = arc sin ⋅ 3 3 4 42 − u 2 3 42 − u 2

Second - Use substitution method by letting u = 3x , then du = 3 x dx

; dx =

du 3

. Therefore,

∫

dx 4 2 − (3 x ) 2

=

∫

∫

1 3x u 1 +c arc sin + c = arc sin 3 4 3 4 3x 1 let y = arc sin + c then 4 3 4 1

Third - Write the answer in terms of the original variable, i.e., x . Fourth - Check the answer by differentiating the solution, i.e., y′

=

1 3

1 1−

( )

3x 2 4

⋅

d 3x +0 dx 4

=

1 3

1 2

x 1 − 916

⋅

3 4

=

3 12

1 16 −9 x 2 16

b. First - Write the given integral in its standard form =

∫

∫

=

1 4

dx a2 − x2

16 − 9 x 2 = arc sin

=

16 − 9 x 2

x +c, a

i.e.,

∫

dx 36 − x 2

dx 62 − x2

Hamilton Education Guides

259

Calculus I

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

Second - Use substitution method by letting u = x , then Therefore,

∫

dx 36 − x 2

=

∫

du

du d = x =1 dx dx

which implies du = dx .

u 6

= arc sin + c

62 − u 2

x +c 6

u 6

Third - Write the answer in terms of the original variable, i.e., x . arc sin + c = arc sin x 6

Fourth - Check the answer by differentiating the solution, i.e., let y = arc sin + c then y′

1

=

(6x )2

1−

⋅

d x +0 dx 6

=

1 1−

x2 36

⋅

1 6

=

1 6

1

1 6

=

36 − x 2 36

∫

c. First - Write the given integral in its standard form =

∫

6 36 − x 2 dx a2 − x2

1

=

36 − x 2

= arc sin

( )

12 − x 4

du = 4 x 3 dx

; dx =

du 3

4x

. Therefore,

x 3 dx

∫

( )

12 − x 4

2

=

x3

∫

du d 4 x = 4x 3 = dx dx

⋅

du

12 − u 2 4 x

3

Fourth - Check the answer by differentiating the solution, i.e., let 1 4

1

( )

1− x 4

⋅ 2

d 4 1 x +0 = dx 4

1

⋅ 4x 3 =

1− x8

x3

4 4

1− x8

d. First - Write the given integral in its standard form =

∫

∫

x3

=

=

1 4

which implies du

∫

12 − u 2

1 u arc sin + c 4 1

1 1 u arc sin + c = arc sin x 4 + c 4 1 4 1 y = arc sin x 4 + c then 4

a2 − x2

= arc sin

x +c, a

i.e.,

x 2 dx

∫

16 − x 6

x 2 dx

( )

42 − x3

du = 3 x 2 dx

2

; dx =

du 3x

2

. Therefore,

∫

x 2 dx

( )

42 − x3

=

2

∫

du d 3 x = 3x 2 = dx dx

x2

⋅

du

4 2 − u 2 3x

Third - Write the answer in terms of the original variable, i.e., x .

2

=

1 3

which implies

∫

du

u 1 arc sin + c 4 3

1 3

1 3

1 3 1 −  x4   

2

⋅

d x3 +0 dx 4

Hamilton Education Guides

=

1 3

1 6

x 1 − 16

⋅

3x 2 4

=

1 4

x2 16 − x 6 16

=

1 4

4x 2 16 − x 6

=

=

42 − u 2

Fourth - Check the answer by differentiating the solution, i.e., let y = arc sin =

=

1− x8

dx

Second - Use substitution method by letting u = x 3 , then

y′

1− x8

2

Third - Write the answer in terms of the original variable, i.e., x .

=

x 3 dx

∫

i.e.,

x 3 dx

Second - Use substitution method by letting u = x 4 , then

y′

x +c, a

=

u 1 arc sin + c 3 4

x3 1 arc sin +c 3 4

x3 +c 4

then

x2

16 − x 6

260

Calculus I

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

e. First - Write the given integral in its standard form =

1

dx

x2

x

∫ a 2 + x 2 = a arc tan a + c , i.e., ∫

9 + x6

dx

dx

∫ 3 2 + (x 3 ) 2 du d 3 x = 3x 2 which implies = dx dx 1 du 1 du u 1 ⋅ = = arc tan + c 2 2 2 2 2 3 3 9 3 + u 3x 3 +u

Second - Use substitution method by letting u = x 3 , then du = 3 x 2 dx

; dx =

du 3x

2

dx

∫ 3 2 + (x 3 ) 2

. Therefore,

=

∫

∫

Third - Write the answer in terms of the original variable, i.e., x .

1 u arc tan + c 9 3

1 9

Fourth - Check the answer by differentiating the solution, i.e., let y = arc tan y′

1 9

=

d x3 +0 3  2 dx 3  x 1+  3    1

=

⋅

1 1 3x 2 1 x2 1 9x 2 x2 ⋅ = = = 9 1 + x6 3 9 9+ x 6 9 9 + x6 9 + x6 9

∫

x 5+ x

4

dx

=

x3 +c 3

then

9

f. First - Write the given integral in its standard form =

1 x3 arc tan +c 9 3

=

x dx

1

dx

x

x

∫ a 2 + x 2 = a arc tan a + c , i.e., ∫

4

x +5

dx

∫ ( 5 ) 2 + (x 2 ) 2 du d 2 x = 2 x which implies du = 2 x dx = dx dx u 1 x du 1 du = = arc tan +c ⋅ 2 2 2 2x 2 2 5 2 5 +u 5 5 +u

Second - Use substitution method by letting u = x 2 , then ; dx =

du . Therefore, 2x

x dx

∫ ( 5 ) 2 + (x 2 ) 2

=

∫( )

∫( )

Third - Write the answer in terms of the original variable, i.e., x .

1 2 5

Fourth - Check the answer by differentiating the solution, i.e., let y =

1 2 5

u 5

arc tan

+c = x2

1 2x 1 5x x 1 2x 1 d x2 = = = ⋅ +0 = 4 4 4 2 10 5+ x 5 5+ x dx 5 2 2 5 1+ x 5 2 5 5+ x4 1 +  x  5 5  5 x dx 1 First - Write the given integral in its standard form = arc tan + c , 2 2 a a a +x y′ =

g.

arc tan

1

1

∫

dx 25 + 9 x

2

=

2 5

arc tan

x2 5

+c

+ c then

⋅

∫

=

5

1

dx

∫ 9 (25 + x 2 ) 9

=

1 9

∫

dx 25 9

+x

2

=

1 9

i.e.,

∫

dx 2

9 x + 25

dx

∫ ( 5) 2 + x2 3

du d = x = 1 which implies du = dx . dx dx 1 3u 1 3 3u ⋅ arc tan +c = arc tan +c 15 5 9 5 5

Second - Use substitution method by letting u = x , then Therefore,

1 9

dx

∫ ( 5) 2 +u2 3

=

u 1 1 ⋅ arc tan + c 5 9 5 3

3

=

Third - Write the answer in terms of the original variable, i.e., x .

Hamilton Education Guides

1 3u arc tan +c 15 5

=

3x 1 arc tan +c 5 15

261

Calculus I

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

3x 1 arc tan + c then 5 15 1 75 1 = 2 75 25 + 9 x 25 + 9 x 2

Fourth - Check the answer by differentiating the solution, i.e., let y = y′

d 3x 1 1 ⋅ +0 2 dx 5 15 1 + 3 x

=

(5 )

=

1 1 3 ⋅ 2 15 1 + 9 x 5

3 75

=

25

1 25+9 x 25

3 25 75 25 + 9 x 2

=

2

=

Note that another way of solving this class of problems is by rewriting the integral in the following way:

∫

dx 9 x + 25

dx

∫

25 + 9 x du 3

; dx =

du = 3dx

=

=

2

u 1 arc tan + c 5 15

=

2

∫

dx

5 + (3 x ) 2

dx

∫

. Therefore,

5 + (3 x ) 2

=

2

∫

x 3 dx

∫ 4 (49 + x 8 ) 4

1 4

1 3

∫

∫ a2 + x2

=

x 1 arc tan + c , a a

5 +u

2

⋅

du 3

dx

x 3 dx

∫

+ x8

49 4

=

x 3 dx

1 4

=

; dx =

du 4x

u 1 1 ⋅ arc tan + c 7 7 16 2

5 +u

3

. Therefore,

=

2

1 4

x 3 dx

∫ (7 )2 + (x 4 )2 2

2u 1 2 ⋅ arc tan +c 7 16 7

=

1 4

=

x3

du

∫ (7 )2 + u 2 ⋅ 4 x 3 2

1 56

d 2x 4 +0 4  2 dx 7  2 x 1+  7    1

1 1 u ⋅ arc tan + c 3 5 5

⋅

=

i.e.,

∫

x 3 dx 49 + 4 x 8

=

1 16

which implies du

∫ (7 )2 + u 2 2

2u 1 arc tan +c 7 56 1 2u 1 2 x4 arc tan +c arc tan +c = 56 7 56 7

Fourth - Check the answer by differentiating the solution, i.e., let y = =

=

du d 4 = x = 4x 3 dx dx

Third - Write the answer in terms of the original variable, i.e., x .

y′

2

∫ (7 )2 + (x 4 )2 2

Second - Use substitution method by letting u = x 4 , then du = 4 x 3 dx

du 2

1 3x arc tan +c 15 5

=

=

which implies

=

1 2

h. First - Write the given integral in its standard form =

du d = 3x = 3 dx dx

. Now, let u = 3x , then

2

1 2x 4 arc tan +c 7 56

then

1 1 8x 3 x3 1 1 49 x 3 x3 = = = ⋅ 56 1 + 4 x8 7 49 49+ 4 x8 49 49 + 4 x 8 49 + 4 x 8 49

49

or, the alternative approach would be to rearrange the integral in the following way:

∫

x 3 dx 49 + 4 x

; dx = =

8

du 8x

3

=

∫

x 3 dx 2

( )

7 + 2x

. Therefore,

1 u arc tan + c 56 7

=

4 2

∫

. Now, let u = 2x 4 , then x 3 dx

( )

7 2 + 2x

4 2

=

∫

x3 2

7 +u

2

⋅

du d = 2 x 4 = 8x 3 dx dx

du 8x

3

=

1 8

∫

du 2

7 +u

2

which implies du = 8 x 3 dx =

1 1 u ⋅ arc tan + c 8 7 7

2 x4 1 arc tan +c 56 7

i. First - Write the given integral in its standard form

Hamilton Education Guides

dx

∫ a2 + x2

=

x 1 arc tan + c , a a

i.e.,

∫

x 2 dx 5 + 9x 6

262

Calculus I

=

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

x 2 dx

x 2 dx

1

∫ 9 (5 + x 6 ) = 9 ∫ 9

+ x6

5 9

=

1 9

x 2 dx

∫

( )

2 3 2  5  3  + x  

du d 3 x = 3x 2 = dx dx

Second - Use substitution method by letting u = x 3 , then du = 3 x 2 dx

; dx =

du 3x

2

u 1 1 arc tan +c ⋅ 5 27 5

. Therefore,

x 2 dx

∫

  

5 3

1 3 3u ⋅ +c arc tan 27 5 5

=

3

3

1 9

( )

2

 + x3   1

=

9 5

1 9

=

2

arc tan

x2

∫

  

3u

 + u 2 3x  

1 9 5

arc tan

3u 5

Fourth - Check the answer by differentiating the solution, i.e., let y = y′

=

1

1

9 5

3 1 +  3 x  5  

d 3x 3 +0 dx 5

⋅

2

=

1

1

9 5 1 + 9 x6 5

⋅

9x 2 5

=

du 2

=

1 27

=

1

du

∫

  

5 3

2

 +u2  

=

+c

5

Third - Write the answer in terms of the x variable, i.e.,

5 3

⋅

2

which implies

+c 1

9 5

9 5

arc tan

3 x3

arc tan

3x 3 5

5

+c

then

+c

1 x2 x2 1 5x 2 = = 5 5+ 9 x 6 5 5 + 9x 6 5 + 9x 6 5

or, the alternative approach would be to rearrange the integral in the following way:

∫

x 2 dx 5 + 9x

du

; dx = =

6

9x

1 9 5

2

x 2 dx

∫ ( 5 )2 + (3x 3 )2 .

=

. Therefore, u

arc tan

5

+c

=

Now, let u = 3x 3 , then x 2 dx

∫ ( 5 )2 + (3x 3 )2 1 9 5

arc tan

3 x3 5

=

du d 3x 3 = 9 x 2 = dx dx

x2

du

∫ ( 5 )2 + u 2 ⋅ 9 x 2

=

1 9

which implies du = 9 x 2 dx du

∫ ( 5 )2 + u 2

=

1 1 u ⋅ arc tan +c 9 5 5

+c

Example 4.4-2: Evaluate the following indefinite integrals: a. d. g.

∫

∫ ∫

dx

=

4

x x −4

dx x 6x 2 − 9 ex 9e

2x

+16

b.

=

dx

=

∫

e.

∫

h.

∫

dx 2

x 25 x − 9

dx x 16 x 4 − 25 dx 25 − (x + 4 )2

∫

=

c.

=

f.

∫

=

i.

∫

dx

=

x x 2 − 25 ex e 2x + 4

dx

=

dx 3 − ( x − 2 )2

=

Solutions: a. First - Write the given integral in its standard form

∫

dx x x4 − 4

=

∫

∫x

dx 2

x −a

2

=

1 x arc sec + c , a a

i.e.,

dx

x

(x )

Hamilton Education Guides

2 2

− 22

263

Calculus I

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

du d 2 = x = 2 x which dx dx du 1 1 du = ⋅ 2 x2 u 2 − 22 x u 2 − 2 2 2x

Second - Use substitution method by letting u = x 2 , then du = 2 x dx

=

1 2

∫

du . 2x

; dx = du 2

u u −2

=

2

dx

∫

Therefore,

(x )

2 2

x

1 1 u ⋅ arc sec + c 2 2 2

=

−2

=

2

∫

implies

∫

1 u arc sec + c 4 2

1 1 x2 u +c arc sec + c = arc sec 4 4 2 2 x2 1 Fourth - Check the answer by differentiating the solution, i.e., let y = arc sec + c then 2 4 1 4x 1 1 d x2 x 1 1 1 2x 1 = = = ⋅ +0 = y′ = ⋅ 4 x2 x4 − 4 4 2 dx 2 4 x 2 x 4 −4 4 x2 x4 2 2 x x4 − 4  x 2  −1 −1 x 2 4 2 4   2  2 

Third - Write the answer in terms of the original variable, i.e., x .

b. First - Write the given integral in its standard form

∫

dx x 25 x 2 − 9

dx

∫

=

=

9 25 x x 2 − 25

∫

dx 9 5 x x 2 − 25

=

∫x 1 5

∫

dx x2 − a2

dx x x2 −

=

x 1 arc sec + c , a a

i.e.,

(53 )2

du d = x = 1 which implies du = dx . dx dx 1 5u u 1 1 +c ⋅ arc sec + c = arc sec 3 3 3 5 3

Second - Use substitution method by letting u = x , then Therefore,

1 5

dx

∫

2

x x −

()

3 2 5

=

1 5

∫

du 2

u u −

()

3 2 5

=

5

5

1 5u 5x 1 arc sec + c = arc sec +c 3 3 3 3 5x 1 Fourth - Check the answer by differentiating the solution, i.e., let y = arc sec + c then 3 3 d 5x 1 1 1 1 5 1 5 1 9 5 = = y′ = ⋅ +0 = ⋅ = dx 3 3 5x 5x 2 3 5 x 25 x 2 3 9 5 x 25 x 2 −9 9 5 x 25 x 2 − 9 x 25 x 2 − 9 −1 −1 3 9 3 9

Third - Write the answer in terms of the original variable, i.e., x .

3

(3 )

or, the alternative approach would be to rearrange the integral in the following way:

∫

dx x 25 x 2 − 9

; dx = =

=

du 5

∫

dx x

. Therefore,

u 1 arc sec + c 3 3

=

(5x )2 − 3 2

∫

. Now, let u = 5 x , then

dx x

(5x )2 − 3 2

=

∫

1

u 5

du u 2 − 32 5

dx x x 2 − 25

=

∫

=

1 5

∫

which implies du = 5dx

5 du u u 2 − 32

=

∫

du u u 2 − 32

1 5x arc sec +c 3 3

c. First - Write the given integral in its standard form

∫

⋅

du d = 5x = 5 dx dx

∫x

dx x2 − a2

=

1 x arc sec + c , a a

i.e.,

dx x x 2 − 52

Hamilton Education Guides

264

Calculus I

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

du d = x =1 dx dx

Second - Use substitution method by letting u = x , then Therefore,

dx

∫

=

x x 2 − 52

du

∫

u u 2 − 52

which implies du = dx .

1 u arc sec + c 5 5

=

1 u 1 x arc sec + c = arc sec + c 5 5 5 5 x 1 Fourth - Check the answer by differentiating the solution, i.e., let y = arc sec + c then 5 5 1 5 1 d x 1 1 1 1 1 1 1 = = y′ = ⋅ +0 = ⋅ = 2 5 dx 5 5 x x 2 5 x x2 5 5 2 − 25 x x − 25 x x 2 − 25 −1 x x 25 −1 5 25

Third - Write the answer in terms of the original variable, i.e., x .

(5 )

5

∫

dx x 6x 2 − 9

=

dx

∫

6 x x 2 − 96

dx

∫

=

6 x x 2 −  

3 6

  

dx

∫x

d. First - Write the given integral in its standard form

x −a

1

dx

6∫

x x 2 −  

3 6

  

1

6∫

=

2

du u u 2 −  

3 6

  

2

=

1

=

6

1 x arc sec + c , a a

x x 2 −  

3 6

  

2

du d = x =1 dx dx

1

⋅

3 6

u

arc sec

which implies du = dx . +c

3 6

=

1 3

Fourth - Check the answer by differentiating the solution, i.e., let y = arc sec 1 3

1   

6x 3

6x  3 

2

⋅ −1

d 6x +0 dx 3

1 3

=

1 6x 3

6x 9

2

⋅ −1

6 3

=

6 9

1 6u arc sec +c 3 3

1 6u 1 arc sec + c = arc sec 3 3 3

Third - Write the answer in terms of the original variable, i.e., x .

=

i.e.,

dx

6∫

Second - Use substitution method by letting u = x , then Therefore,

2

1

=

2

2

1

=

2

6x 3

6 x −9 9

6 9

6x +c 3

6x +c 3

then y ′

9 6 x 6x 2 − 9

=

1 x 6x 2 − 9

or, the alternative approach would be to rearrange the integral in the following way:

∫

dx x 6x 2 − 9

; dx =

du 6

=

∫

dx

x

( 6x)2 − 3 2

. Therefore,

∫

. Now, let u = 6 x , then

dx

( 6x)

2

=

∫

1

⋅

2

2

e. First - Write the given integral in its standard form

∫x

=

∫

1 u arc sec + c 3 3

dx 4

x 16 x − 25

=

=

x

1 arc sec 3

∫

Hamilton Education Guides

u −3

du 6

=

6 du

1

6∫

which implies du = 6 dx

2

u u −3

2

=

∫

du u u 2 − 32

6x +c 3

dx 16 x x

− 32

u 6

du d = 6x = 6 dx dx

4

25 − 16

=

∫

dx

4x

(x ) − ( ) 2 2

5 2 4

=

dx x2 − a2

1 4

∫

=

1 x arc sec + c , a a

i.e.,

dx

x

(x ) − ( ) 2 2

5 2 4

265

Calculus I

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

du d 2 = x = 2 x which implies dx dx 1 du 1 du = ⋅ 8 2 2x 2 x u 2 − 54 x 2 u 2 − 54

Second - Use substitution method by letting u = x 2 , then du = 2 x dx

=

1 8

∫

; dx = du

du . 2x

dx

∫

x

2 2

4

=

4

1 4

=

(x ) − ( )

1 1 u ⋅ arc sec + c 5 5 8

=

(54 )2

u u2 −

1 4

Thus,

5 2 4

∫

∫

()

1 4u arc sec +c 10 5 4x 2 1 4u 1 arc sec +c arc sec +c = 5 10 5 10

Third - Write the answer in terms of the original variable, i.e., x .

1 4x 2 +c arc sec 10 5

Fourth - Check the answer by differentiating the solution, i.e., let y = y′

1 10

=

1 2

4 x2 5

d 4x 2 +0 dx 5

⋅

 4 x 2  −1  5   

=

1 10

()

1 4 x2 5

⋅

16 x 4 25

−1

8x 5

8 50

=

x 4 x2 5

1

=

16 x 4 − 25 25

then

x 16 x 4 − 25

or, the alternative approach would be to rearrange the integral in the following way:

∫

dx x 16 x 4 − 25

du = 8 x dx

=

1 8

∫

du 8x

; dx = du

u 4

=

u 2 − 52

dx

∫

=

(4x )

2 2

x

−5

. Therefore, 1 8

∫

2

∫

dx

x

(4x )

=

1 2

4du u u 2 − 52

2 2

∫

∫

=

− 52 du

u u 2 − 52

f. First - Write the given integral in its standard form =

∫

ex 4 + e 2x

dx

∫

=

ex

( )

2

2 + e

x 2

=

=

1 8

u 1 1 ⋅ arc sec + c 5 2 5

=

1 2

x u −5

du e

x

. Therefore,

∫

⋅

2

du 8x

1

dx

1 2

1 x 1 +  e2   

2

⋅

ex

( )

22 + e

d ex +0 dx 2

=

x 2

dx

=

∫

ex 2

2 +u

2

⋅

du e

u 2 − 52

1 4x 2 arc sec +c 10 5

ex e 2x + 4

dx

x

du d x e = ex = dx dx

∫

=

du 2

2 +u

2

=

which implies du = e x dx

1 u arc tan + c 2 2

1 ex +c arc tan 2 2 1 ex y = arc tan + c then 2 2 ex ex

1 u arc tan + c 2 2

=

1 1 ex 1 ex 1 4e x = = = = 2x ⋅ 2 1 + e2 x 2 4 4+ e 2 x 4 4 + e 2x 4 + e 2x e +4 4

4

g. First - Write the given integral in its standard form

Hamilton Education Guides

x

x

Fourth - Check the answer by differentiating the solution, i.e., let =

du 2

dx

Third - Write the answer in terms of the x variable, i.e.,

y′

∫

which implies

∫ a 2 + x 2 = a arc tan a + c , i.e., ∫

Second - Use substitution method by letting u = e x , then ; dx =

du d 4 x 2 = 8x = dx dx

. Now, let u = 4x 2 , then

dx

∫ a2 + x2

=

x 1 arc tan + c , a a

i.e.,

∫

ex 9 e 2 x +16

dx

266

Calculus I

=

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

ex

e x dx

∫ 16 + 9 e 2 x dx = ∫ 9 ( 16 + e 2 x ) 9

=

1 9

e x dx

∫

16 9

+ e 2x

e x dx

1 9

=

∫ ( 4 )2 + ( e x )2 3 du d x e = ex = dx dx

Second - Use substitution method by letting u = e x , then ; dx = =

du e

x

u 1 1 ⋅ arc tan + c 4 9 4

=

3

3

e x dx

1 9

ex

∫ ( 4 )2 + ( e x )2 3

=

1 9

1 3 3u +c ⋅ arc tan 9 4 4

=

1 3u +c arc tan 12 4

. Therefore,

du

∫ ( 4 )2 + u 2 ⋅ e x 3

=

1 9

which implies du = e x dx

du

∫ ( 4 )2 + u 2 3

1 3u 3e x 1 arc tan +c = arc tan +c 12 4 4 12 1 3e x i.e., let y = arc tan + c then 12 4 ex 1 16e x ex = = = 16 16 + 9e 2 x 16 + 9e 2 x 9e 2 x + 16

Third - Write the answer in terms of the original variable, i.e., x . Fourth - Check the answer by differentiating the solution, y′

=

1 12

1 x 1 +  3e4   

2

⋅

d 3e x +0 dx 4

1 1 1 3e x = ⋅ 2 x 16 12 1 + 9e 4

=

16

ex 16 +9e 2 x 16

or, the alternative approach would be to rearrange the integral in the following way: ex

∫ 16 + 9 e 2 x ; dx = =

du 3e

x

dx

=

∫

e x dx 4 + 3e

. Therefore,

1 u arc tan + c 12 4

=

∫

u = x+4.

Therefore,

1 5

5 25 − (x + 4 )2

( )

4 2 + 3e

x 2

∫

=

x+4 +c 5

2

4 +u

2

⋅

du 3e

=

x

1 3

∫

in its standard form

2

∫

=

dx 25 − u 2 1

then y ′ = 1−

∫

= ⋅

(x+5 4 )2

du 2

4 +u

∫

dx 2

a −x

dx 52 − u 2

d x+4 +0 dx 5

u 1 1 ⋅ arc tan + c 4 3 4

=

2

= arc sin

2

x +c a

by letting

u 5

= arc sin + c = arc sin 1

= 1−

⋅

( x + 4 )2

1 1 = 5 5

x+4 +c 5

1 25−( x + 4 )2 25

25

25 − (x + 4 )2

∫

Check: Let y = arc sin

Hamilton Education Guides

25 − (x + 4 )

ex

which implies du = 3e x dx

1

=

Therefore,

∫

25 − (x + 4 )2

i. Write the given integral u = x−2.

e x dx

dx

∫

Check: Let y = arc sin =

x 2

1 3e x arc tan +c 4 12 dx

h. Write the given integral

du d 3e x = 3e x = dx dx

. Now, let u = 3e x , then

( )

2

∫

dx 3 − ( x − 2 )2

dx 3 − ( x − 2 )2 x−2 3

+c

=

in its standard form

∫

then y ′ =

dx

= arc sin

( 3 )2 − u 2 1 1 −  x − 2   3 

∫

2

⋅

dx a2 − x2 u 3

d x−2 +0 dx 3

= arc sin

+c

= arc sin

=

1 1−

(x−2)

2

3

x +c a

x−2 3

⋅

1 3

by letting

+c

=

1 3

1 3−( x − 2 )2 3

267

Calculus I

=

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

3

1 3

1

=

3 − ( x − 2 )2

3 − ( x − 2 )2

Example 4.4-3: Evaluate the following indefinite integrals: dx

∫

9 + (x + 3)2

d.

∫

(x + 1)3 4 + (x + 1)8

g.

∫

a.

=

b.

∫

dx =

e.

∫

=

h.

∫

dx 2

x − 8 x + 17

(

x dx

49 + x 2 + 9 dx x x 2 − 36

)

=

2

=

dy 2

y + 20 y + 120

=

c.

∫

f.

∫

i.

∫

( x − 2 )2 9 + ( x − 2 )6

dx =

dx

=

(x − 3) (x − 3)2 − 49 dt 2

t + 6t + 13

=

Solutions: a. First - Write the given integral in its standard form

1

dx

du d ( x + 3) = 1 = dx dx

Second - Use substitution method by letting u = x + 3 , then du = dx .

Therefore,

∫

dx

9 + ( x + 3)

2

=

∫

du 2

3 +u

2

=

x

∫ a 2 + x 2 = a arc tan a + c , i.e., ∫

1 u arc tan + c 3 3

= 1 3

1 1 3 1 + x +3

⋅

(3)

2

1 d  x+3 +0 =  3 dx  3 

1 1+

⋅

1

( x +3) 2 3

=

9

9 1 9 9 + (x + 3)2

b. First - Write the given integral in its standard form =

∫

x dx

(

72 + x2 + 9

=

x+3 +c 3

then

1

9 + ( x + 3) 2

1

dx

x dx

x

∫ a 2 + x 2 = a arc tan a + c , i.e., ∫

)

(

49 + x 2 + 9

)

2

2

Second - Use substitution method by letting u = x 2 + 9 , then du = 2 x dx

which implies

1 x+3 arc tan +c 3 3

Fourth - Check the answer by differentiating the solution, i.e., let y = arc tan =

9 + (x + 3)2

u 1 arc tan + c 3 3

Third - Write the answer in terms of the x variable, i.e.,

y′

dx

; dx =

du . 2x

Therefore,

∫

x dx 2

(

2

7 + x +9

)

2

=

∫

x 2

7 +u

Third - Write the answer in terms of the x variable, i.e.,

2

⋅

du 2x

=

(

∫

1 u arc tan + c 14 7

Fourth - Check the answer by differentiating the solution, i.e., let y =

Hamilton Education Guides

)

du d 2 = x + 9 = 2 x which implies dx dx du 1 1 1 u = ⋅ arc tan + c 2 72 + u2 2 7 7

=

1 x2 + 9 arc tan +c 14 7

1 x2 + 9 arc tan +c 14 7

then

268

Calculus I

y′

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

1 14

=

1 1 d  x 2 + 9  +0 = 2   dx 14 7 2 x 2 +9   1 +  x 7+9  1 + 49   1

⋅

(

)

⋅

2

2x 1 = 7 49

c. First - Write the given integral in its standard form =

(x − 2 ) 2 9 + [ (x − 2 )3 ] 2

∫

dx

=

∫

( x − 2 )2 2 3 2 + [ (x − 2 )3 ]

49 x

(

49 + x 2 + 9

dx

=

1 3

∫

du 2

3 +u

2

=

du

1

3 (x − 2 )

u 1 1 ⋅ arc tan + c 3 3 3

=

(

49 + x 2 + 9

( x − 2 )2 2 3 2 + [ (x − 2 )3 ]

∫

x

dx

=

( x − 2 )2 ⋅

∫

2

3 +u

2

1 u arc tan + c 9 3

=

1  ( x − 2 )3  1+  3   

2

⋅

d  (x − 2 )3 3 dx 

 +0  

1 9

=

1 1+

⋅

(x + 1)3 2 4 + [(x + 1 )4 ]

∫

=

dx

∫

(x + 1)3 2 2 2 + [(x + 1)4 ]

( x − 2 )6 9

=

1 4

∫

du 2

2 +u

2

=

du

4 (x + 1)

3

1 1 u ⋅ arc tan + c 4 2 2

=

dx

1

1  ( x +1)4  1+  2   

2

⋅

d  (x + 1)4 dx  2

Hamilton Education Guides

 +0  

3

(x + 1)3 4 + (x + 1)8

. Thus,

∫

(x + 1)3 2 2 2 + [(x + 1)4 ]

dx

du d (x + 1)4 = 4 (x + 1)3 = dx dx

=

∫

(x + 1)3 2

2 +u

dx

2

⋅

which

du

4(x + 1)3

1 u arc tan + c 8 2

1 u arc tan + c 8 2

= 1 8

1 8

(x − 2)3 + c then

x

( x + 1) 4 + c 1 arc tan 8 2

Fourth - Check the answer by differentiating the solution, i.e., let y = arc tan =

3(x − 2 )2

dx

Third - Write the answer in terms of the x variable, i.e.,

y′

du

∫ a 2 + x 2 = a arc tan a + c , i.e., ∫

Second - Use substitution method by letting u = (x + 1)4 , then implies du = 4 (x + 1)3 dx ; dx =

which

2 ( x − 2 )2 3(x − 2 )2 1 27 (x − 2) = = 3 27 9 + (x − 2)6 9 + ( x − 2 )6

d. First - Write the given integral in its standard form =

dx

( x − 2) 3 + c 1 arc tan 9 3

1 9

1 9

( x − 2 )2 9 + ( x − 2 )6

du d (x − 2)3 = 3 (x − 2)2 = dx dx

Fourth - Check the answer by differentiating the solution, i.e., let y = arc tan =

2

1 u arc tan + c 9 3

Third - Write the answer in terms of the x variable, i.e.,

y′

)

dx

. Thus,

2

)

x

∫ a 2 + x 2 = a arc tan a + c , i.e., ∫

Second - Use substitution method by letting u = (x − 2)3 , then implies du = 3 (x − 2)2 dx ; dx =

=

2

=

1 8

1 1+

( x +1) 8 4

⋅

(x + 1)4 2

+c

then

3 (x + 1)3 4(x + 1)3 1 16 (x + 1) = = 2 16 4 + (x + 1)8 4 + (x + 1)8

269

Calculus I

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

=

∫

1

∫x

e. First - Write the integral in its standard form

dx =

x2 − a2

1 x arc sec + c , a a

∫

i.e.,

dx x x 2 − 36

dx x x2 − 62

du d = x =1 dx dx

Second - Use substitution method by letting u = x , then dx

∫

Therefore,

∫

=

x x2 − 62

du u u 2 − 62

which implies du = dx .

u 1 arc sec + c 6 6

=

1 u arc sec + c 6 6

Third - Write the answer in terms of the original variable, i.e., x .

1 6

=

x 1 arc sec + c 6 6

x 6

Fourth - Check the answer by differentiating the solution, i.e., let y = arc sec + c then y′

=

1 6 x

1

( 6x ) 2 − 1

6

⋅

d x 1 +0 = dx 6 6

1 2

x 36

x 6

⋅ −1

1 6

1 36

=

6

∫

(x − 3) (x − 3)2 − 49

=

x −36 36

x

f. First - Write the given integral in its standard form dx

2

∫x

∫

1 36 x

1 2

x −a

36 2

x − 36

dx =

2

=

∫

x 1 arc sec + c , a a

dx

(x − 3) (x − 3)

2

− 72

∫

=

du 2

u u −7

2

=

du d = x −3 =1 dx dx

u 1 arc sec + c 7 7

1 u arc sec + c 7 7 1 7

=

1

x −3 7

( )

x −3 2 7

⋅ −1

d x−3 +0 dx 7

=

1 7

1 x −3 7

( x −3 ) − 1 49 2

⋅

1 1 = 49 7

7 x−3

( x −3 )

2

49

− 49

=

=

x−3 1 arc sec +c 7 7

x−3 +c 7

1 49 x − 3

then y ′

49

(x − 3)2 − 49

1 x−3

(x − 3)2 − 49

g. First - Write the integral in its standard form =

i.e.,

which implies du = dx .

Fourth - Check the answer by differentiating the solution, i.e., let y = arc sec 1 7

x 2 − 36

x

(x − 3) (x − 3)2 − 7 2

Third - Write the answer in terms of the original variable, i.e., x .

=

1

dx

Second - Use substitution method by letting u = x − 3 , then Therefore,

=

dx

∫ (x 2 − 8x + 16)+ 1 = ∫

dx

( x − 4 )2 + 1

=

dx

1

Therefore,

Hamilton Education Guides

dx

∫ 1 + ( x − 4 )2

=

dx 2

x − 8 x + 17

dx

∫ 1 + ( x − 4 )2

Second - Use substitution method by letting u = x − 4 , then du = dx .

x

∫ a 2 + x 2 = a arc tan a + c , i.e., ∫

du

∫ 1+ u 2

du d (x − 4 ) = 1 = dx dx

which implies

= arc tan u + c

270

Calculus I

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

Third - Write the answer in terms of the x variable, i.e., arc tan u + c = arc tan ( x − 4) + c Fourth - Check the answer by differentiating the solution, i.e., let y = arc tan (x − 4) + c then y′

=

1

1 + (x − 4 )

2

⋅

d (x − 4 ) + 0 dx

1

=

1 + (x − 4 )

2

1

=

⋅1

1 + ( x − 4 )2

=

dy

dy

∫ ( y 2 + 20 y + 100)+ 20 = ∫

( y + 10)

2

20 + ( y + 10 )

dy

∫ ( 20 )2 + ( y + 10)2

Therefore,

=

du

∫ ( 20 )2 + u 2 1

Third - Write the answer in terms y , i.e.,

20

20

=

1

1

20

y +10  1 +    20 

2

⋅

d  y + 10   +0 dy  20 

1

=

1

20 1 + ( y +10 ) 20

dy

1 20

=

+c

u

arc tan 1 20

20

dt

∫ ( t 2 + 6t + 9)+ 4 = ∫

dt

( t + 3)

2

+4

=

20

4 + ( t + 3)

2

=

∫

Therefore,

∫

dt

2 + ( t + 3) 2

2

=

∫

du 2

2 +u

arc tan

2

=

=

(2 )

Hamilton Education Guides

1 1+

( t +3) 4

2

⋅

+c

then

20 + ( y + 10 )2

dt 2

t + 6t + 13

dt

du d = ( t + 3) = 1 dt dt

which implies

1 u arc tan + c 2 2 1 u arc tan + c 2 2

= 1 2

1 d t +3 1 1 ⋅  +0 = 2 2 dy  2  2 1 + t +3

20 1

x

1 t+3 arc tan +c 2 2

Fourth - Check the answer by differentiating the solution, i.e., let w = arc tan =

y + 10

2 + ( t + 3)2

Third - Write the answer in terms of the variable t , i.e.,

w′

+c

2

Second - Use substitution method by letting u = t + 3 , then du = dt .

20

20

1 20 20 20 + ( y + 10 )2

1

dx

dt

∫

=

1

⋅

y + 10

arc tan 1

which implies

+c

∫ a 2 + x 2 = a arc tan a + c , i.e., ∫

i. First - Write the integral in its standard form =

2

y + 20 y + 120

∫ ( 20 )2 + ( y + 10)2

Fourth - Check the answer by differentiating the solution, i.e., let w = w′

2

du d ( y + 10) = 1 = dy dy

=

u

arc tan

=

2

Second - Use substitution method by letting u = y + 10 , then du = dy .

dy

x

dy

∫

=

+ 20

1

dx

∫ a 2 + x 2 = a arc tan a + c , i.e., ∫

h. First - Write the integral in its standard form

t +3 +c 2

then

1 4 1 1 = = 2 2 4 4 + ( t + 3) 4 + ( t + 3)2

271

Calculus I

4.4 Integration of Expressions Resulting in Inverse Trigonometric Functions

Section 4.4 Practice Problems – Integration of Expressions Resulting in Inverse Trigonometric Functions

1. Evaluate the following indefinite integrals: dx

a.

∫

d.

∫ 9 x 2 +16

=

25 − 9 x 2 dx

=

dx

b.

∫

e.

∫ 7 + 9x 6

4 − x2 x 2 dx

= =

c.

∫

f.

∫x

x 2 dx 25 − x 6 dx x 4 − 25

= =

2. Evaluate the following indefinite integrals: a.

∫x

d.

∫

dx x 2 −16

=

dx 25 − (x − 3)2

Hamilton Education Guides

dx

b.

∫x

e.

∫ 25 + (x + 4)2

7x 2 − 4 dx

ex

=

c.

∫ 4 e 2 x + 9 dx =

=

f.

∫ x 2 − 10 x + 26

dx

=

272

Calculus I

4.5

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions

Integration of Expressions Resulting in Exponential or Logarithmic Functions

In the following examples we will solve problems using the following formulas: 1

∫ x dx = ln

∫ ln xdx ∫a

x

dx =

x +c

= x ln x − x + c

ax +c ln a

∫e

x

a 0 and a ≠ 1

dx = e x + c

Let’s integrate some exponential and algebraic expressions using the above integration formulas. Example 4.5-1: Evaluate the following indefinite integrals: a.

dx

∫ x+5

=

x2

d.

∫ 2x 3 + 1

g.

∫  x



3

−

b. =

dx

2x   dx x 2 +1

=

dx

∫ 5x +1 = 4x 3

e.

∫ x 4 − 3 dx =

h.

∫ xe

x2

dx

=

x

c.

∫ x 2 − 3 dx =

f.

∫  x + 1 + x − 1  dx

i.

∫ 2x

2 

 5

2 x3

e

dx

=

=

Solutions: a. Given dx

∫ x+5

dx

∫ x+5 =

let u = x + 5 , then

1

∫ u ⋅ du

du d ( x + 5) = dx dx

dx

∫ 5x +1 let u = 5x + 1 , then 1 du

dx

∫ 5x +1 = ∫ u ⋅ 5

=

1 5

1

∫ u du

=

du d (5 x + 1) = dx dx

1 ln u + c 5

1 5

x

x

∫ x 2 − 3 dx let u = x

∫ x 2 − 3 dx

=

; dx = du . Therefore,

1 1 +0 = x+5 x+5

Check: Let y = ln 5 x + 1 + c , then y ′ =

c. Given

du =1 dx

= ln u + c = ln x + 5 + c

Check: Let y = ln x + 5 + c , then y ′ =

b. Given

;

x du

∫ u ⋅ 2x

Hamilton Education Guides

=

1 2

2

−3,

1

∫ u du

then =

=

;

du =5 dx

; du = 5 dx ; dx =

du 5

. Therefore,

1 ln 5 x + 1 + c 5

1 1 5 1 1 = ⋅ ⋅5 + 0 = ⋅ 5x + 1 5 5x + 1 5 5x + 1

(

) ; dudx = 2x ; du = 2x dx ; dx = du2 x . Thus,

du d = x2 −3 dx dx

1 ln u + c 2

=

1 ln x 2 − 3 + c 2

273

Calculus I

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions

x 1 1 2 x = 2 ⋅ ⋅ 2x + 0 = ⋅ 2 2 2 x −3 2 x −3 x −3

1 2

Check: Let y = ln x 2 − 3 + c , then y ′ = x2

∫ 2 x 3 + 1 dx let u = 2 x

d. Given x2

∫ 2x 3 + 1

dx

∫

=

x 2 du ⋅ u 6x 2

=

3

+1 ,

1 6

∫ u du

then

1

=

(

1 ln u + c 6

4x 3

∫ x 4 − 3 dx let u = x

4x 3

∫ x4 −3

dx

∫

=

4 x 3 du ⋅ u 4x 3

4

−3,

1

∫ u du

=

(

2  5 +  dx −1  + x x 1 

5

1 x4 −3

5

) ; dudx = x

⋅ 4x 3 + 0 =

1



∫x

3

dx −

2x   dx x +1

2x

∫ x 2 + 1 dx

Check: Let y =

h. Given

∫

2

2

∫ xe

xe x dx

=

x2

∫

dx

=

=

∫x

3

dx −

. Thus,

−3

; du = 4 x 3 dx ; dx =

du 4x 3

. Thus,

4x 3 x4 −3

2 5 5 2 +0 = + + x +1 x −1 x +1 x −1

2x

∫ x 2 + 1 dx let u = x

1 4 2 x du x − ⋅ 4 u 2x

∫

=

Therefore,

= 5 ln u1 + 2 ln u 2 + c = 5 ln x + 1 + 2 ln x − 1 + c

Check: Let y = 5 ln x + 1 + 2 ln x − 1 + c , then y ′ = 

4

2

∫ x + 1 dx + ∫ x − 1 dx = 5∫ u1 du1 + 2∫ u 2 du 2

g. Given ∫  x 3 −

6x 2

∫ x + 1 dx + ∫ x − 1 dx let u1 = x + 1 and u 2 = x − 1 respectively.

= 1

2

du

= ln u + c = ln x 4 − 3 + c

Check: Let y = ln x 4 − 3 + c , then y ′ =

f. Given ∫ 

; du = 6 x 2 dx ; dx =

1 ln 2 x 3 + 1 + c 6

=

du d x4 −3 = dx dx

then

2

1 1 6 x2 x2 = ⋅ ⋅ 6x 2 + 0 = ⋅ 6 2x 3 +1 6 2x 3 +1 2x 3 + 1

1 6

Check: Let y = ln 2 x 3 + 1 + c , then y ′ =

e. Given

) ; dudx = 6x

du d = 2x 3 + 1 dx dx

1 4 1 x − du 4 u

∫

2

=

+1,

then

du = 2x dx

1 4 x − ln u + c 4

=

; dx =

du . 2x

Therefore,

1 4 x − ln x 2 + 1 + c 4

1 4 2x 1 1 x − ln x 2 + 1 + c , then y ′ = ⋅ 4 x 3 − ⋅ 2x + 0 = x 3 − 2 2 4 4 x +1 x +1

let u = x 2 , then

xe u ⋅ 1 2

du 2x 2

=

1 u e du 2

∫

du d 2 = x dx dx

=

Check: Let y = e x + c , then y ′ =

Hamilton Education Guides

1 u e +c 2

; =

du = 2x dx

; du = 2 x dx ; dx =

du . 2x

Therefore,

1 x2 e +c 2

2 1 x2 2 ⋅ e ⋅ 2x + 0 = ⋅ x e x 2 2

= xex

2

274

Calculus I

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions

du d 3 = x dx dx

3

i. Given ∫ 2 x 2 e x dx let u = x 3 , then

∫ 2x

2 x3

e

dx

=

∫ 2x

2 u

2 3

3

e ⋅

du 3x

=

2

2 u e du 3

∫

Check: Let y = e x + c , then y ′ =

=

du = 3x 2 dx

;

2 u e +c 3

; du = 3x 2 dx ; dx =

du 3x 2

. Therefore,

2 x3 e +c 3

=

3 2 x3 6 ⋅ e ⋅ 3x 2 + 0 = ⋅ x 2 e x 3 3

= 2x 2 e x

3

Example 4.5-2: Evaluate the following indefinite integrals: a.

∫

e y 5 + 3e y dy

d.

∫

e − x dx

g.

∫(x

2

=

=

)

+ e x − e −3 x dx =

ex

b.

∫ 1− e x

e.

∫

h.

dx

e −3 x dx

=

=

2e 3 x

c.

∫ 1 + e 3x dx =

f.

e −5 x + 2 dx = 2

∫

1

1 ex

∫ 1 + x 2 dx =

∫

i.

e

x2

=

dx

x3

Solutions:

(

∫

e y 5 + 3e y dy

Check: Let y =

=

∫

ey ⋅ u ⋅

(

2 5 + 3e y 9

)

3 2

ex

du 3e y

ex

∫ 1− e x

=

dx

ex

du

∫ 1− e x ⋅ − e x

1 1 u 2 du 3

∫

, then y ′ =

∫ 1 − e x dx let u = 1 − e

b. Given

=

x

, then

2e 3 x

2e 3 x

∫ 1 + e 3x dx

∫ 1 + e 3x

dx

=

∫

2 3

(

(

)

3 −1 2

; dy = =

(

3 y e 5 + 3e y 3

⋅ 3e y =

) ; dudx = −e

y

x

2 32 u 9

)

1 2

du 3e y

=

. Therefore,

(

2 5 + 3e y 9

)

3 2

= e y 5 + 3e y

; du = −e x dx ; dx = −

du ex

. Thus,

1 = − ∫ du = − ln u + c = − ln 1 − e x + c u

=

2 3

1

∫ u du

1 1− e x

⋅ −e x + 0 =

(

du d = 1+ e 3 x dx dx

=

Check: Let y = ln 1 + e 3 x + c , then y ′ =

Hamilton Education Guides

1 2 32 ⋅ u 3 3

=

2

du d = 1− e x dx dx

let u = 1+ e 3 x , then

2e 3 x du ⋅ u 3e 3 x

1 1 1+ 12 ⋅ u 3 1+ 1

=

2 3 ⋅ 5 + 3e y 9 2

Check: Let y = − ln 1 − e x + c , then y ′ = −

c. Given

) ; dudy = 3e

du d = 5 + 3e y dy dy

a. Given ∫ e y 5 + 3e y dy let u = 5 + 3e y , then

2 ln u + c 3

ex 1− e x

) ; dudx = 3e =

3x

; du = 3e 3 x dx ; dx =

du 3e 3 x

. Thus,

2 ln 1 + e 3 x + c 3

2 1 6 e 3x 2e 3 x ⋅ 3e 3 x + 0 = ⋅ = 3 1 + e 3x 3 1 + e 3x 1 + e 3x

275

Calculus I

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions

d. Given ∫ e − x dx let u = − x , then

∫e

−x

dx =

∫e

du d (− x ) = dx dx

∫

u

du = −1 dx

;

; du = −dx ; dx = −du . Therefore,

= − e−x + c

⋅ −du = − e u du = − e u + c

Check: Let y = −e − x + c , then y ′ = − e −x ⋅ −1+ 0 = e − x du d (− 3x ) = dx dx

e. Given ∫ e −3 x dx let u = −3x , then

∫e

−3 x

dx =

∫e

u

⋅−

du 3

= −

du = −3 dx

;

∫

∫

1 3

e −5 x + 2 dx = 2

∫(x

2

eu du ⋅− 5 2

∫

= −

1 e u du 10

∫

= −

1 u e +c 10

)

+ e x − e −3 x dx

∫x

=

2

∫

∫

dx + e x dx − e −3 x dx =

1

1

ex

1

∫

x2

dx = x +

Check: Let y =

i.

1

ex

∫ 1 + x 2 dx = ∫ dx + ∫ x 2 dx let u = x , then 1+

1 x2

∫

e

∫

ex

x3 1 2

x3

dx

=

∫

eu

∫ x2

1 x−e x

dx let u =

1 − 5 x+2 e +c 10

1 x2

⋅ − x 2 du

x3 ⋅− du 2 x3

eu

1 2

Check: Let y = − e

1 x2

Hamilton Education Guides

1 du =− dx x2

;

; x 2 du = −dx ; dx = − x 2 du . Thus,

1

1 1− e x

du d 1 = dx dx x 2

=

du d 1 = dx dx x

= x − ∫ e u du = x − e u + c = x − e x + c

+ c , then y ′ =

, then

1 3 1 x + e x + e −3 x + c 3 3

1 1 ⋅ 3 x 2 + e x + ⋅ e −3 x ⋅ −3 + 0 = x 2 + e x − e −3 x 3 3

1 3

1 3

ex

= −

1 1 5 −5 x + 2 1 −5 x + 2 = e −5 x + 2 e + c , then y ′ = − e −5 x + 2 ⋅ −5 + 0 = e 2 10 10 10

Check: Let y = x 3 + e x + e −3 x + c then y ′ =

h.

3 −3 x = e −3 x e 3

e −5 x + 2 du d dx let u = −5 x + 2 , then (− 5 x + 2) ; du = −5 ; du = −5dx ; dx = − du . Thus, = 2 5 dx dx dx

Check: Let y = −

g.

. Therefore,

1 3

3

1 3

∫

du 3

1 = − e u + c = − e −3 x + c

1 u e du 3

Check: Let y = − e −3 x + c , then y ′ = − e −3 x ⋅ −3 + 0 =

f. Given

; du = −3dx ; dx = −

1 u − e du 2

∫

;

⋅−

1 x2

1

+ 0 = 1+

du 2 =− dx x3

=

1 + c , then y ′ = − ⋅ e 2

⋅−

2 x

x2

; x 3 du = −2dx ; dx = −

1 − eu + c 2 1 x2

ex

3

=

x3 du . Therefore, 2

1

1 2 − ex +c 2

+0 =

2 ⋅e 2

1 x2

1

⋅

1 x

3

=

e

x2

x3 276

Calculus I

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions

Example 4.5-3: Evaluate the following indefinite integrals: 2

a. d. g.

e

x3

∫ 3x 4

∫ (e

x

dx =

b.

∫

)

e.

∫ (e

− 1 e x dx =

x3

∫ 1 + 4x 4

=

dx

Solutions: a. Given

2 3

ex

∫ 3x 4

x3

∫ 3x 4 dx

∫

=

h.

2

dx let u =

2

e

2 +5

ex

, then

x3

x4 ⋅− du 6 3x 4 eu

=

x2

)

+ 3 2 e x dx =

x

x2

∫ 1− x3

du d 2 = dx dx x 3

1 − e u du 18

∫

dx =

=

dx

;

=

1 − eu + c 18

=

b. Given

∫

2 +5

ex

x2

∫

2 +5

x2

eu

∫ x2

dx =

2 +5, x

dx let u =

⋅−

x2 du 2

then

= −

1 u e du 2

∫

;

du 2 =− dx x2 2 +5

1 2

1 = − eu + c = − e x 2

c. Given

∫

e

∫

− 1

x2

dx let u = −

x3

− 1

x2

x3

dx =

∫

eu x3 ⋅ du x3 2

1

, then

du d −2 =− x dx dx

1 u e du 2

1 u e +c 2

x2

=

∫

=

=

;

du = 2x − 3 dx

∫ (e

x

∫ (e

)

x

)

− 1 e x dx

− 1 e x dx =

let u = e x − 1 , then

∫ u ⋅e

Hamilton Education Guides

x

⋅

du e

x

=

∫ u du

=

x4 du . 6

=

Therefore,

2 3

ex

=

3x 4

x2 du . Thus, 2

=

2 +5

ex

x2

; du = 2 x − 3 dx ; dx =

1

(

=

∫ 1 − 3x 5 dx

x3 du . Thus, 2

1

) ; dudx = e

du d = e x −1 dx dx

1 2 u +c 2

10 x 4

+c

1 − 2 1 − 2 2 − 2 1 Check: Let y = e x + c , then y ′ = e x ⋅ 2 x −3 + 0 = ⋅ e x ⋅ 2 2 2 x3

d. Given

)

− 1 3 e − x dx =

1 − x2 e +c 2

1

1

−x

; x 2 du = −2dx ; dx = −

1 2 +5 2 2 +5 1 1 2 +5 2 Check: Let y = − e x + c , then y ′ = − ⋅ e x ⋅ − 2 + 0 = ⋅ e x ⋅ 2 2 2 2 x x

e

∫ (e

dx =

1 x3 +c e 18 2

du d  2  =  + 5 dx dx  x 

f.

x3

2

−

6 x3 1 1 6 1 3 3 Check: Let y = − e x + c , then y ′ = − ⋅ e x ⋅ − +0 = ⋅e ⋅ 4 18 18 18 x x4

ex

x2

∫

; x 4 du = −6dx ; dx = −

2

2

− 1

c.

i.

du 6 =− dx x4

e

(

1 x e −1 2

)

2

x

=

e

− 1

x2

x3

; du = e x dx ; dx =

du ex

. Thus,

+c

277

Calculus I

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions

Check: Let y =

e. Given

∫ (e

∫ (e

)

x

)

2

∫u

+ 3 e x dx =

∫ (e

f. Given

)

−x

)

+ 3 2 e x dx

x

Check: Let y =

∫ (e

(

−x

2

let u = e x + 3 , then du

⋅ex ⋅

(

e

∫

=

Check: Let y =

=

∫u

3

(

⋅ e −x ⋅ −

)

1 −x e −1 4

4

∫ 1− x3

=

1 3 u +c 3

(

x 3 du ⋅ u 16x 3

∫

du e

−x

x2 du ⋅− u 3x 2

∫ 1 − 3x 5 dx

∫ 1 − 3x 5

=

ex

. Thus,

)

)

(

(

)

)

= e x + 3 2e x

)

4

(

(

)

1 −x e −1 4

)

=

1 16

3

= −

4

(

du d = 1 + 4x 4 dx dx

, then 1

∫ u du

=

1 ln u + c 16

=

) ; dudx = 16x

3

∫

, then 1 3

(

du d = 1− x3 dx dx

1

∫ u du

let u = 1 − 3x 5 , then

2 3

du 16x 3

= −

10 15

(

)

3

= e −x −1 e −x

. Therefore,

1 ln 1 + 4 x 4 + c 16

) ; dudx = −3x

2

; dx = −

=

du

x3 1 + 4x 4

. Therefore,

3x 2

1 3

1 = − ln u + c = − ln 1 − x 3 + c 3

(

du d 1 − 3x 5 = dx dx

1

∫ u du

) ; dudx = −15x

4

=

x2 1− x3

; dx = −

du 15x 4

. Therefore,

2 3

2 = − ln u + c = − ln 1 − 3 x 5 + c 3

2 1 30 x4 ⋅ −15 x 4 + 0 = ⋅ 3 1 − 3x 5 3 1 − 3x 5

Check: Let y = − ln 1 − 3x 5 + c , then y ′ = − ⋅

Hamilton Education Guides

+c

)

; dx =

1 1 3 x2 ⋅ 3x 2 + 0 = ⋅ 3 1− x3 3 1− x3

10 x 4 du ⋅− u 15 x 4

4

(

1 1 16 x3 1 ⋅ ⋅16 x 3 + 0 = ⋅ ln 1 + 4 x 4 + c , then y ′ = 16 1 + 4 x 4 16 1 + 4 x 4 16

10 x 4

dx

du

3 1 x e +3 +c 3

=

1 = − ∫ u 3 du = − u 4 + c = −

1 3

10 x 4

)

; du = e x dx ; dx =

x

Check: Let y = − ln 1 − x 3 + c , then y ′ = − ⋅

i. Given

(

= e x −1 e x

3 3 1 4 −x + c , then y ′ = − ⋅ 4 e − x − 1 ⋅ −e − x + 0 = e −1 e −x 4 4

∫ 1 − x 3 dx let u = 1 − x dx

) ; dudx = e

(

x2

x2

du =

(

∫ 1 + 4 x 4 dx let u = 1 + 4 x

h. Given

2

)

x3

∫ 1 + 4x 4

∫u

=

x

(

du d = ex +3 dx dx

)

3

dx

)

du du d du . Therefore, − 1 3 e − x dx let u = e −x − 1 , then = e −x − 1 ; = −e − x ; dx = − dx dx dx e −x

Check: Let y = −

x3

(

2 1 x 1 3 x e + 3 3 + c , then y ′ = ⋅ 3 e x + 3 2 ⋅ e x + 0 = e +3 ex 3 3 3

− 1 e − x dx

g. Given

)

(

1 x 1 2 x e −1 e x e − 1 2 + c , then y ′ = ⋅ 2 e x − 1 ⋅ e x + 0 = 2 2 2

=

10 x 4 1 − 3x 5

278

Calculus I

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions

Example 4.5-4: Evaluate the following indefinite integrals: x+3

a.

∫ x + 1 dx

d.

∫a

g.

∫

3x

dx

=

=

3

a x x 2 dx

=

x+6

b.

∫ x + 5 dx =

e.

∫a

h.

∫

x2

c.

x+8

∫ x + 4 dx =

=

x dx

4

a x x 3 dx 2

=

f.

∫a

i.

∫

x 2 +5

3a x

3

x dx

+1 2

=

x dx

=

Solutions: a.

x+3

∫ x + 1 dx

=

∫

(x + 1) + 2 dx x +1

x +1

2

2

∫ x + 1 dx + ∫ x + 1 dx = ∫ dx + ∫ x + 1 dx

=

2 ⋅1 + 0 x +1

Check: Let y = x + 2 ln x + 1 + c , then y ′ = 1 +

b.

x+6

∫ x + 5 dx = ∫

(x + 5) + 1 dx x+5

x+5

1

∫ x + 5 dx + ∫ x + 5 dx

=

c.

x+8

∫ x + 4 dx = ∫

(x + 4) + 4 dx

=

x+4

x+4

4

∫ x + 4 dx + ∫ x + 4 dx

Check: Let y = x + 4 ln x + 4 + c , then y ′ = 1 +

d. Given ∫ a 3 x dx let u = 3x , then

∫

a 3 x dx =

∫

au ⋅

Check: Let y =

du 3

=

1 a u du 3

∫

du d = 3x dx dx

;

1 au +c 3 ln a

=

1

=

= x + ln x + 5 + c x+6 x+5

=

4

∫ dx + ∫ x + 4 dx

4 ⋅1 + 0 x+4

du =3 dx

x + 5 +1 x+5

=

x+3 x +1

=

∫ dx + ∫ x + 5 dx

=

1 ⋅1 + 0 x+5

Check: Let y = x + ln x + 5 + c , then y ′ = 1 +

x +1+ 2 x +1

=

= x + 2 ln x + 1 + c

=

x+4+4 x+4

= x + 4 ln x + 4 + c =

; du = 3dx ; dx =

x+8 x+4

du 3

. Therefore,

1 a 3x +c 3 ln a

=

3 ln a 3 x 1 1 a 3x ⋅ a 3 x ln a ⋅ 3 + 0 = ⋅a + c , then y ′ = 3 ln a 3 ln a 3 ln a

= a 3x

Reminder: In Section 3.3 “Differentiation of Exponential and Logarithmic Functions” we learned that the derivative of a u = e u ln a is equal to the following: d u a dx

=

d u ln a e dx

= e u ln a ⋅

d u ln a dx

2

e. Given ∫ a x x dx let u = x 2 , then

∫

2

a x x dx

=

∫

au ⋅ x ⋅

du 2x

Hamilton Education Guides

=

du d 2 = x dx dx

1 a u du 2

∫

u

= e ln a ⋅ ln a ⋅

=

;

du dx

= a u ln a

du = 2x dx

1 au +c 2 ln a

du dx

. Therefore,

; du = 2 x dx ; dx =

du . 2x

d u du a = a u ln a dx dx

Therefore,

2

=

1 ax +c 2 ln a

279

Calculus I

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions 2

2 2 2 ln a x 2 1 ax 1 Check: Let y = + c , then y ′ = ⋅ a x ln a ⋅ 2 x + 0 = ⋅a ⋅ x = ax x 2 ln a 2 ln a 2 ln a 2

f. Given ∫ a x

∫a

x 2 +5

+5

=

x dx

∫

(

du d 2 x +5 = dx dx

let u = x 2 + 5 , then

x dx

du a ⋅x⋅ 2x u

=

1 a u du 2

∫

=

1 au +c 2 ln a

) ; dudx = 2x ; du = 2 x dx ; dx = du2 x . Therefore, 2

1 a x +5 +c = 2 ln a

2

Check: Let y =

2 2 2 ln a x 2 +5 1 a x +5 1 + c , then y ′ = ⋅ a x +5 ln a ⋅ 2 x + 0 = ⋅a ⋅ x = a x +5 x 2 ln a 2 ln a 2 ln a

du d 3 = x dx dx

3

g. Given ∫ a x x 2 dx let u = x 3 , then

∫a

x3 2

x dx

∫a

=

u

2

⋅x ⋅

du 3x 2

=

1 a u du 3

∫

=

du = 3x 2 dx

;

1 au +c 3 ln a

; du = 3x 2 dx ; dx = 3

3

h. Given

∫

∫

4

4

a x x 3 dx 2

∫

=

let u = x 4 , then

au ⋅ x3 ⋅

du 8x 3

=

du d 4 = x dx dx

1 a u du 8

∫

=

du = 4x 3 dx

;

1 au +c 8 ln a

. Therefore,

3x 2

1 ax +c = 3 ln a

3 3 ln a x3 2 1 ax 1 + c , then y ′ = ⋅ a x ln a ⋅ 3 x 2 + 0 = Check: Let y = ⋅a ⋅ x 3 ln a 3 ln a 3 ln a

a x x 3 dx 2

du

3

= a x x2

; du = 4 x 3 dx ; dx =

du 4x 3

. Therefore,

4

1 ax +c 8 ln a

=

4

4 4 ln a x 4 3 1 ax 1 4 1 Check: Let y = + c , then y ′ = ⋅ a x ln a ⋅ 4 x 3 + 0 = ⋅ a ⋅ x = a x x3 8 ln a 8 ln a 8 ln a 2

i. Given ∫ 3a x

∫ 3a

3

+1 2

x 3 +1 2

x dx

x dx

let u = x 3 + 1 , then

= 3∫ a ⋅ x ⋅ u

2

du 3x 2

=

(

) ; dudx = 3x

du d x3 +1 = dx dx

3 a u du 3

∫

=

2

; du = 3x 2 dx ; dx =

du 3x 2

. Thus,

3

a x +1 +c = ln a

au +c ln a

3

Check: Let y =

3 3 ln a x3 +1 a x +1 1 + c , then y ′ = ⋅ a x +1 ln a ⋅ 3 x 2 + 0 = ⋅a ⋅ 3 x 2 = 3a x +1 x 2 ln a ln a ln a

Example 4.5-5: Evaluate the following indefinite integrals: x2

a.

∫a

d.

∫ 5a

2 x dx

5 x +3

dx

x +1

+ e x +1 dx =

c.

∫ (a

x

)

f.

∫ x + 1 dx =

=

b.

∫ (a

=

e.

∫ (e

Hamilton Education Guides

)

5

+ 1 e x dx

=

2 x +3

)

+ 3 dx =

x+2

280

Calculus I

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions

Solutions: 2

a. Given ∫ a x 2 x dx let u = x 2 , then

∫a

x2

=

2 x dx

du a ⋅ 2x ⋅ 2x

∫

u

∫a

=

u

du d 2 = x dx dx au +c ln a

du =

du = 2x dx

;

; du = 2 x dx ; dx =

du . 2x

Thus,

2

ax +c ln a

=

2

2 2 ln a x 2 ax 1 Check: Let y = + c , then y ′ = ⋅ a ⋅ 2x = a x 2x ⋅ a x ln a ⋅ 2 x + 0 = ln a ln a ln a

b. Given

∫

∫ (a

x +1

)

+ e x +1 dx =

∫

a x +1 dx + e x +1 dx

∫

∫ (a

x +1

∫

a u du + e u du

)

2 x +3

+ 3 dx =

∫

a 2 x +3 dx + 3 dx =

∫

au ⋅

∫a

2 x +3

du + 3x 2

du + 3x 5

∫

5a 5 x +3 dx = 5 a u ⋅

1 a u du + 3 x 2

∫

∫ (e

x

)

5

x

)

5

+ 1 e x dx

+ 1 e x dx

=

1 au + 3x + c 2 ln a

=

;

du =2 dx

=

du d (5 x + 3) = dx dx

5 a u du 5

∫

=

∫

a u du =

du =5 dx

;

au +c ln a

; du = 5 dx ; dx =

=

=

∫u

5

Hamilton Education Guides

let u = e x + 1 , then ⋅ex ⋅

du e

x

=

∫u

5

(

du =

1 6 u +c 6

=

(

; dx =

du 2

. Thus,

du 5

= a 2 x +3 + 3 . Thus,

a 5 x+3 +c ln a

) ; dudx = e

du d = e x +1 dx dx

= a x +1 + e x +1

a 2 x+3 + 3x + c 2 ln a

5 ln a 5 x +3 1 a 5 x +3 ⋅ a 5 x +3 ln a ⋅ 5 + 0 = ⋅a + c , then y ′ = ln a ln a ln a

Check: Let y =

∫ (e

du d (2 x + 3) = dx dx

∫

=

; dx = du . Thus,

a x +1 + e x +1 + c ln a

=

dx + 3 dx let u = 2 x + 3 , then

d. Given ∫ 5a 5 x +3 dx let u = 5 x + 3 , then

e. Given

au + eu + c ln a

=

du =1 dx

;

2 ln a 2 x +3 1 a 2 x +3 ⋅ a 2 x + 3 ln a ⋅ 2 + 3 + 0 = ⋅a +3 + 3 x + c , then y ′ = 2 ln a 2 ln a 2 ln a

Check: Let y =

∫

du d (x + 1) = dx dx

∫

dx + e x +1 dx let u = x + 1 , then

ln a x +1 1 a x +1 ⋅ a x +1 ln a ⋅1 + e x +1 ⋅1 + 0 = ⋅a + e x +1 + e x +1 + c , then y ′ = ln a ln a ln a

Check: Let y =

c. Given

∫

=

∫a

x

= 5a 5 x + 3

; du = e x dx ; dx =

du ex

. Therefore,

)

6 1 x e +1 + c 6

281

Calculus I

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions

Check: Let y =

f.

x+2

∫ x + 1 dx = ∫

(

(

)

)

(

)

5 5 6 1 6 x 1 x e + 1 + c , then y ′ = ⋅ 6 e x + 1 ⋅ e x + 0 = e +1 e x 6 6 6

(x + 1) + 1 dx x +1

x +1

1

1

∫ x + 1 dx + ∫ x + 1 dx = ∫ dx + ∫ x + 1 dx

=

Check: Let y = x + ln x + 1 + c , then y ′ = 1 +

1 ⋅1 + 0 x +1

=

x +1+1 x +1

(

)

5

= e x +1 e x

= x + ln x + 1 + c x+2 x +1

=

In the following examples we will solve problems using the following two formulas: 1

1 x−a ln +c 2a x+a

1

a+x 1 +c ln a−x 2a

∫ x 2 − a 2 dx = ∫ a 2 − x 2 dx =

Example 4.5-6: Evaluate the following indefinite integrals: 1

a.

∫ x 2 − 4 dx =

d.

∫ x 2 + 10 x + 24 dx =

g.

∫ x 2 + 18x + 75 dx =

j.

∫ 16 − ( t + 3) 2 dt

1

c.

∫ (x − 1) 2 − 16 dx

1

f.

∫ x 2 − 4 x + 1 dx =

i.

∫ 4 − (x − 1) 2 dx

l.

∫ x2 −5 + ∫ 4 − x2

b.

∫ 36 x 2 − 25 dx =

1

e.

∫ x 2 − 4 x + 3 dx =

1

h.

∫ 16 − 9 x 2 dx =

k.

∫ x 2 + 3x dx + ∫ 36 − x 2 dx

1

=

1

2x + 3

1

=

1

=

1

1

dx

=

3 dx

=

Solutions: a.

1

1

∫ x 2 − 4 dx = ∫ x 2 − 2 2 dx 1 4

Check: Let y = ln =

b.

1

∫ 36 x 2 − 25

=

Check: Let y =

1 x−2 ln +c 2⋅2 x+2

x−2 +c x+2

1 1 4 ⋅ ⋅ 4 x−2 x+2 dx

=

=

then y ′ =

(x − 2)(x + 2)

∫ 36 (x 2 − 25 )

dx

36

6x − 5 1 +c ln 60 6x + 5

Hamilton Education Guides

x−2 1 +c ln x+2 4

1 1 1 ⋅ (x + 2 ) − 1 ⋅ (x − 2 ) ⋅ ⋅ +0 4 x−2 ( x + 2 )2 x+2

1

1

=

=

1 36

=

1 2

x + 2x − 2x − 4 1

∫ x 2 − ( 5 )2

then y ′ =

dx

6

=

=

=

1 x+2 x+2− x+2 ⋅ ⋅ 4 x−2 (x + 2)2

1 2

x −4

x − 56 1 6x − 5 1 1 ln +c ln +c = 5 5 60 6x + 5 36 2 ⋅ x + 6 6

1 1 6 ⋅ (6 x + 5) − 6 ⋅ (6 x − 5) ⋅ ⋅ +0 60 6 x −5 (6 x + 5) 2 6 x +5

=

1 6x + 5 ⋅ 60 6 x − 5

282

Calculus I

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions

×

c.

36 x + 30 − 36 x + 30

(6 x + 5)

1

1

∫ (x − 1)2 − 16 dx = ∫ (x − 1)2 − 4 2 dx 1 8

8 1 1 ⋅ ⋅ 8 x−5 x+3

=

1

(x + 4)(x + 6)

1 1 2 ⋅ ⋅ 2 x − 3 x −1

= 1

∫ x 2 − 4 x + 1 dx

=

Check: Let y = ×

=

2 3

1

(x − 3)(x − 1) dx

x−2− 3

+c

x−2+ 3

x − 0.27 x − 0.27 − x + 3.73 ⋅ x − 3.73 (x − 0.27 )2

=

1 2

x − 0.27 x − 3.73 x + 1

1

1

=

=

1 ln 2 ⋅ 2.45

(x + 9) − 2.45 (x + 9) + 2.45

Hamilton Education Guides

+c

=

(x − 1)2 − 16 1 x+4 ln +c 2 x+6

+c =

x + 6 x + 4 x + 24

=

=

1 2

x + 10 x + 24

(x − 2) − 1 (x − 2) + 1

1 ln 2 ⋅1

1 1 1 ⋅ (x − 1) − 1 ⋅ (x − 3) +0 ⋅ ⋅ 2 x −3 (x − 1)2

=

1 2

x − x − 3x + 3

=

=

=

1 x−3 ln +c 2 x −1

1 x −1 x −1− x + 3 ⋅ ⋅ 2 x − 3 (x − 1)2

1 x − 4x + 3

1

1 ⋅ 3.46

+c

2

∫ (x − 2)2 − ( 3 )2 dx

then y ′ =

(x + 6 )

=

1 x −3.73 x −0.27

1 1 3.46 ⋅ ⋅ 3.46 x − 3.73 x − 0.27

⋅

=

1 2 3

ln

( x − 2) − ( x − 2) +

3

1 ⋅ (x − 0.27 ) − 1 ⋅ (x − 3.73)

(x − 0.27 )

+c

3

2

+0

=

1 3.46

3.46 3.46(x − 3.73)(x − 0.27 )

1 2

x − 4x + 1

∫ x 2 + 18x + 75 dx = ∫ (x 2 + 18x + 81)− 6 =

36 x − 25

1

2 x+4

x −1

1

ln

=

( x + 5) − 1 ( x + 5) + 1

1 ln 2 ⋅1

=

2

then y ′ =

∫ (x 2 − 4 x + 4)− 3 1

x − 2 x − 15

=

1

=

1

x−3 +c x −1

1 2

(x + 6 )

x+6

1

1 2

g.

2

1

=

2

1 x+6 x+6− x−4 1 1 1 ⋅ (x + 6 ) − 1 ⋅ (x + 4 ) then y ′ = ⋅ x + 4 ⋅ ⋅ +0 = ⋅ 2 2

∫ x 2 − 4 x + 3 dx = ∫ (x 2 − 4 x + 4)− 1 dx = ∫ (x − 2)2 − 1 dx Check: Let y = ln

f.

x + 3 x − 5 x − 15 1

x+4 +c x+6

2 1 1 ⋅ ⋅ 2 x+4 x+6

=

1

=

(x + 3)

8 x −5

=

2

1

1 2

36 x + 30 x − 30 x − 25

(x + 3)

x +3

1

=

(x − 5)(x + 3)

1

1 2

x−5 1 +c ln x+3 8

=

+c

∫ x 2 + 10 x + 24 dx = ∫ (x 2 + 10 x + 25)− 1 dx = ∫ (x + 5)2 − 1 dx Check: Let y = ln

e.

8

1

=

=

(6 x − 5)(6 x + 5)

(x − 1) − 4 (x − 1) + 4

1 ln 2⋅4

=

1

=

1 x +3 x +3− x +5 1 1 1 ⋅ (x + 3) − 1 ⋅ (x − 5) then y ′ = ⋅ x −5 ⋅ ⋅ +0 = ⋅ 2 2

x−5 +c x+3

Check: Let y = ln

d.

1 1 60 ⋅ ⋅ 60 6 x − 5 6 x + 5

=

2

dx

=

1

∫ (x + 9)2 − ( 6 )2 dx

=

1 2⋅ 6

ln

(x + 9 ) − (x + 9 ) +

6 6

+c

x + 6.55 1 +c ln x + 11.45 4.9 283

Calculus I

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions

Check: Let y =

1 x + 6.55 1 ⋅ (x + 11.45) − 1 ⋅ (x + 6.55) 1 x + 11.45 1 1 ln + c then y ′ = ⋅ +0 = ⋅ ⋅ 2 + x 6 . 55 4.9 x + 11.45 4.9 x + 6.55 4.9 (x + 11.45) x +11.45

×

h.

1

∫ 16 − 9 x 2 =

x + 11.45 − x − 6.55

(x + 11.45)

dx

=

Check: Let y = ×

i.

1

∫ 9 (16 − x 2 ) +c

1 4 + 3x +c ln 24 4 − 3x

12 − 9 x + 12 + 9 x

(4 − 3x )

1 9

=

2

1+ x +c 3− x

1 1 4 ⋅ ⋅ 4 1+ x 3 − x

1

1 8

= 2x + 3

(1 + x )(3 − x )

=

1 6+ x ln +c 12 6− x

(7 + t )( 1 − t ) =

∫

=

1

=

(4 + 3x )(4 − 3x )

−x

+c

1 4 − 3x ⋅ 24 4 + 3 x

=

1

=

16 − 12 x + 12 x − 9 x

1 2 + x −1 ln +c 4 2 − x +1

=

1 1 1 ⋅ (3 − x ) + 1 ⋅ (1 + x ) ⋅ ⋅ +0 4 1+ x (3 − x )2

1 3 − x + 3x − x

=

2

1 3 + 2x − x

=

2

= =

1 7 − 7t + t − t

∫

1

=

2

=

=

=

7 − 6t − t du

2

= dx

2

=

1 16 − 9 x 2

1 1+ x ln +c 4 3− x

1 3 − x 3 − x +1+ x ⋅ ⋅ 4 1 + x (3 − x )2

1

4 − (x − 1)2

4+t +3 1 +c ln 4−t −3 8

1 1 1 ⋅ ( 1 − t ) + 1 ⋅ (7 + t ) +0 ⋅ ⋅ 8 7 +t ( 1 − t )2

=

1 7+t ln +c 8 1− t

1 1− t 1− t + 7 + t ⋅ ⋅ 8 7 + t ( 1 − t )2 1

16 − ( t + 3)2

∫ u + ∫ 62 − x2

=

du

∫u

+

1 6+ x ln 2⋅6 6− x

6+ x 1 +c ln 12 6− x

Check: Let y1 = ln x 2 + 3x + c then y1' =

Hamilton Education Guides

1 1 3 ⋅ (4 − 3 x ) + 3 ⋅ (4 + 3 x ) ⋅ ⋅ +0 24 4+3 x (4 − 3x )2

2 x + 3 du dx + 2 u 2x + 3 6 − x2

= ln x 2 + 3 x +

+x

1 4 + 3x ln +c 24 4 − 3x

=

1−t

1

4 3 4 3

1 1 ⋅ ln 9 2⋅ 4 3

3

4 + ( t + 3) 1 ln +c 2⋅4 4 − ( t + 3)

then y ′ =

1

∫ ( 4 ) 2− x2

=

dx

3− x

7+t +c 1− t

=

1

2 + (x − 1) 1 +c ln 2 − (x − 1) 2⋅2

=

∫ x 2 + 3x dx + ∫ 36 − x 2 dx = ln u +

=

1

8 1 1 ⋅ ⋅ 8 7 + t 1− t

1 9

4 −3 x

1

=

=

9

then y ′ =

∫ 16 − ( t + 3)2 dt = ∫ 4 2 − ( t + 3)2 dx Check: Let y = ln

∫ 16 − x 2

dx

24 1 1 ⋅ ⋅ 24 4 + 3 x 4 − 3 x

1

1

=

1

then y ′ =

∫ 4 − (x − 1)2 dx = ∫ 2 2 − (x − 1)2 dx 1 4

k.

=

3(4 + 3 x ) 3 ln +c 72 3(4 − 3 x )

=

Check: Let y = ln

j.

dx

9

4+3 x 3 4 −3 x 3

1 3 ⋅ ln 9 8

1 1 4.9 1 1 = = ⋅ ⋅ (x + 6.55)(x + 11.45) x 2 + 18 x + 75 4.9 x + 6.55 x + 11.45

=

2

2x + 3 2

x + 3x

+0

=

2x + 3 2

x + 3x

. Let y 2 =

1 6+ x ln +c 12 6− x

284

Calculus I

4.5 Integration of Expressions Resulting in Exponential or Logarithmic Functions

1 1 1 ⋅ (6 + x ) + 1 ⋅ (6 − x ) ⋅ ⋅ +0 12 6+ x (6 − x )2

then y 2' =

1 6− x 6+ x+6− x ⋅ ⋅ 12 6 + x (6 − x )2

=

6− x

1

=

l.

3 dx

dx

∫ x2 −5 + ∫ 4 − x2 Check: Let y1 = ×

=

(6 + x )(6 − x )

36 − 6 x + 6 x − x

3 dx

dx

1 2⋅ 5

ln

(x + 5 )

2

3 4

Let y 2 = ln

x− 5

=

1

=

1

=

2 5

x− 5

ln

1

+

x+ 5 1

⋅

⋅

2 5 x− 5

(

⋅

1

2 5

⋅

3 2+ x +c ln 2− x 4

) ( (x + 5 )2

1

=

(x − 5 )(x + 5 )

=

x + 5x − 5x − 5

=

(2 + x )(2 − x )

3

=

4 − 2x + 2x − x

2

=

⋅

x+ 5

2 5 x− 5

1

2− x

3

1

2

3 1 1 ⋅ (2 + x ) + 1 ⋅ (2 − x ) +0 ⋅ ⋅ 4 2+ x (2 − x )2

then y 2' =

)+ 0 =

1⋅ x + 5 − 1⋅ x − 5

x+ 5

2 5 x− 5 x+ 5

2+ x +c 2− x

4 3 1 ⋅ ⋅ 4 2+ x 2− x

36 − x 2

+ c then y1' =

x+ 5

1 1 12 ⋅ ⋅ 12 6 + x 6 − x

1

=

2

∫ x 2 − ( 5 )2 + ∫ 2 2 − x 2

=

x+ 5 −x+ 5

=

1

=

=

1 2

x −5

.

3 2− x 2+ x+2− x ⋅ ⋅ 4 2+ x (2 − x )2

3 4 − x2

Section 4.5 Practice Problems – Integration of Expressions Resulting in Exponential or Logarithmic Functions

1. Evaluate the following indefinite integrals. a. d. g.

∫

dx 2x +1

∫

3   1 +  dx  x + 3 x −5 

∫ 3e

− ax

=

dx =

=

x

b.

∫ x2 − a

e.

∫

h.

∫(x

2

dx

xe3 x dx

=

c.

=

f.

x3

∫ x4 − 1 dx = e5 x

∫ 1 − e5x dx = 1 3

)

i.

b.

∫ ( e − 1) e dx =

c.

∫ 1 + x3 dx =

e.

∫ x + 6 dx =

f.

∫ x + 5 dx =

h.

∫3a

i.

∫ (e

k.

∫ x2 + 6 x + 8 dx

l.

∫ 9 − (x − 1) 2 dx =

3

+ e2 x − e−5 x dx =

ex

∫ 5x 4 dx =

2. Evaluate the following indefinite integrals. a.

∫ (3e

5x

)

+ 5 e5 x dx =

x4

d.

∫ 1 + x5 dx =

g.

∫a

j.

∫ (x + 1) 2− 25 dx =

x2 +k

x dx

=

1

Hamilton Education Guides

5 x

x

x+7

2 x3 +5 2 x dx

1

= =

x2

x+9

2x

)

3

+ 3 e 2 x dx

=

1

285

Calculus I

Quick Reference to Chapter 5 Problems

Chapter 5

Integration (Part II) Quick Reference to Chapter 5 Problems 5.1

Integration by Parts ................................................................................................... 287

∫e 5.2

2x

=;

∫e

x

sin x dx

∫ x cos 3x dx

=;

=

Integration Using Trigonometric Substitution ........................................................ 308

∫ 5.3

cos 2 x dx

x 2 dx

=;

36 − x 2

dx

∫ ( 9 + x 2 )2

dx

∫ x4

=;

x 2 −1

=

Integration by Partial Fractions ............................................................................... 320 Case I - The Denominator Has Distinct Linear Factors 320

∫

x +1 dx x (x − 2 )(x + 3)

=;

1

∫ (x + 1)(x + 2)

∫

=;

dx

x 2 +1 dx x(x − 1)(x + 1)

=

Case II - The Denominator Has Repeated Linear Factors 327 1

x+3

∫ x (x − 1)2 dx = ; ∫ x 2 (x − 1) dx

=;

5

∫ x (x − 1)2 dx =

Case III - The Denominator Has Distinct Quadratic Factors 334 x2 − x + 3

∫ x (x 2 + 1)

dx

=;

1

∫ x (x 2 + 25)

dx = ;

1

∫ x 2 (x 2 + 16)

dx =

Case IV - The Denominator Has Repeated Quadratic Factors 344 x2

∫ (x 2 + 1)2 5.4

dx

=;

x 2 +1

∫ (x 2 + 4)2

dx

=;

x3

∫ (x 2 + 2)2 dx

=

Integration of Hyperbolic Functions ........................................................................ 350 1

∫ cosh 5 x dx

Hamilton Education Guides

=;

∫ (sinh 4 x + cosh 2 x ) dx

=;

∫x

2

csc h 2 x 3 dx =

286

Calculus I

5.1 Integration by Parts

Chapter 5 – Integration (Part II) The objective of this chapter is to improve the student’s ability to solve additional problems involving integration. A method used to integrate functions, known as Integration by Parts, is addressed in Section 5.1. Integration of functions using the Trigonometric Substitution method is discussed in Section 5.2. Integration of functions using the Partial Fractions technique is addressed in Section 5.3. Four different cases, depending on the denominator having distinct linear factors, repeated linear factors, distinct quadratic factors, or repeated quadratic factors, are addressed in this section. Finally, integration of hyperbolic functions is discussed in Section 5.4. Each section is concluded by solving examples with practice problems to further enhance the student’s ability.

5.1

Integration by Parts

Integration by parts is a technique for replacing hard to integrate integrals by ones that are easier to integrate. This technique applies mainly to integrals that are in the form of ∫ f (x ) g (x ) dx where in most cases f (x ) can be differentiated several times to become zero and g (x ) can be integrated several times without difficulty. For example, given the integral

∫x

e dx the function f (x ) = x 2

2 3x

can be differentiated three times to become zero and the function g (x ) = e 3 x can be integrated several times easily. On the other hand, integrals such as

∫e

2x

cos 2 x dx and

∫e

−3 x

sin 3 x dx do not

fall under the category described above. In this section we will learn how to apply Integration by Parts method in solving various integrals. The formula for integration by parts comes from the product rule, i.e., where u and v are differentiable obtain

d ( uv ) = u dv + v du dx dx dx functions of x , multiplying

both sides of the equation by dx we

d ( uv ) = u dv + v du

rearranging the terms we then have

u dv = d ( uv ) − v du

integrating both sides of the equation we obtain

∫ u dv = uv − ∫ v du

The above formula is referred to as the Integration by Parts Formula. Note that in using the above equality we must first select dv such that it is easily integrable and second ensure that ∫ u dv is easier to evaluate than ∫ u dv . In the following examples we will solve problems using the Integration by Parts method.

Example 5.1-1: Evaluate the following indefinite integrals: a.

∫xe

d.

∫ ln x dx

x

dx =

=

Hamilton Education Guides

b.

∫x

e dx =

c.

∫xe

e.

∫ x ln x dx =

f.

∫x

2 3x

3

−2 x

dx =

ln x dx =

287

Calculus I

g.

∫e

2x

5.1 Integration by Parts

cos 2 x dx =

∫e

h.

x

sin x dx =

i.

∫ x cos 3x dx =

Solutions: a. Given

∫ xe

x

dx let u = x and dv = e x dx then du = dx and

∫ dv = ∫ e

x

dx which implies v = e x .

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ xe

x

∫

dx = xe x − e x dx = xe x − e x + c

= e x ( x − 1) + c

Check: Let y = e x (x − 1) + c , then y ′ = e x ⋅ (x − 1) + 1 ⋅ e x + 0 = xe x − e x + e x = xe x − e x + e x = xe x b. Given

∫x

1 3

let u = x 2 and dv = e 3 x dx then du = 2 x dx and ∫ dv = ∫ e 3 x dx which implies v = e 3 x .

2 3x

e dx

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫

x 2 e 3x 2 1 1 3x x 2 e 3 x dx = x 2 ⋅ e 3 x − e ⋅ 2 x dx = − 3 3 3 3

∫

To integrate v=

1 3x e . 3

∫xe

3x

3x

(1 )

dx

let u = x and dv = e 3 x dx then du = dx and ∫ dv = ∫ e 3 x dx which implies

dx

Therefore,

∫xe

∫

xe 3 x e 3 x 1 3x 1 x e 3 x dx = x ⋅ e 3 x − − +c e dx = 3 9 3 3

(2)

∫

Combining equations ( 1 ) and ( 2 ) together we obtain:

∫

x 2 e 3 x dx =

x 2 e 3x 2 − 3 3

∫

x e 3 x dx

1 3

=

 1 2 3x 2 3x 2 3x x 2 e 3 x 2  xe 3 x e 3 x e +c x e − xe + − − + c =   27 9 3 3 3 3 9 

+

c. Given

2 ⋅ 3e 3 x + 0 27

∫xe

−2 x

v=−

1 −2 x . e 2

∫xe

dx

−2 x

=

) (

(

1 2 2 3x 2 2 x ⋅ e 3 x + 3e 3 x ⋅ x 2 − 1 ⋅ e 3 x + 3e 3 x ⋅ x e + c , then y ′ = 3 9 27 9 2 2 2 2 3x xe + x 2 e 3 x − e 3 x − xe 3 x + e 3 x = x 2 e 3 x 3 9 3 9

Check: Let y = x 2 e 3 x − xe 3 x +

dx let u = x and dv = e −2 x dx then du = dx and

∫ dv = ∫ e

−2 x

)

dx which implies

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1 2

= x ⋅ − e −2 x + 1 2

1 e − 2 x dx 2

∫

1 4

1 2

= − x e −2 x +

1 e − 2 x dx 2

Check: Let y = − x e −2 x − e −2 x + c , then y ′ = − xe − 2 x +

1 −2 x e 2

Hamilton Education Guides

∫

(

1 2

1 4

= − x e−2 x − e−2 x + c

)

1 1 1 ⋅ e − 2 x − 2e − 2 x ⋅ x − ⋅ −2e − 2 x + 0 2 4

1 2

= − e −2 x

= xe −2 x

288

Calculus I

5.1 Integration by Parts

1 x

d. Given ∫ ln x dx let u = ln x and dv = dx then du = dx and ∫ dv = ∫ dx which implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ ln x dx

1 = ln x ⋅ x − ∫ x ⋅ dx = x ln x − ∫ dx = x ln x − x + c x

1 Check: Let y = x ln x − x + c , then y ′ =  1 ⋅ ln x + ⋅ x  − 1 + 0 = ln x + 1 − 1 = ln x 

e. Given

x



1

1

∫ x ln x dx let u = ln x and dv = x dx then du = x dx and ∫ dv = ∫ x dx which implies v = 2 x

2

.

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1 2

1 2 1 ⋅ dx x

∫ x ln x dx = ln x ⋅ 2 x − ∫ 2 x 1 2

=

1 2 1 x ln x − 2 2

1 4

Check: Let y = x 2 ln x − x 2 + c , then y ′ = f. Given

∫x

3

∫ x dx

1 2 1 x ln x − x 2 + c 2 4

=

1 2 1 1  2 x ⋅ ln x + ⋅ x  − ⋅ 2 x + 0 2 x  4

1 dx x

ln x dx let u = ln x and dv = x 3 dx then du =

1 2

1 2

= x ln x + x − x = x ln x 1 4

and ∫ dv = ∫ x 3 dx which implies v = x 4 .

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1 4

1 4 1 ⋅ dx x

∫ x ln x dx = ln x ⋅ 4 x − ∫ 4 x 1 4

Check: Let y = x 4 ln x − −

1 3 x 4

1 4 x +c, 16

=

1 4 1 x ln x − 4 4

then y ′ =

∫x

3

dx

=

1 4 1 4 x ln x − x +c 16 4

1 3 1 4 1 3  4 x ⋅ ln x + ⋅ x  − ⋅ 4 x + 0 4 16 x 

1 4

= x 3 ln x + x 3

= x 3 ln x

g. Given ∫ e 2 x cos 2 x dx let u = e 2 x and dv = cos 2 x dx then du = 2e 2 x dx and ∫ dv = ∫ cos 2 x dx which 1 2

implies v = sin 2 x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we 1 2

obtain ∫ e 2 x cos 2 x dx = e 2 x ⋅ sin 2 x −

1 sin 2 x ⋅ 2e 2 x dx 2

∫

=

e 2 x sin 2 x − e 2 x sin 2 x dx 2

(1 )

∫

To integrate ∫ e 2 x sin 2 x dx let u = e 2 x and dv = sin 2 x dx then du = 2e 2 x dx and ∫ dv = ∫ sin 2 x dx which 1 2

1 2

implies v = − cos 2 x . Thus, ∫ e 2 x sin 2 x dx = e 2 x ⋅ − cos 2 x +

∫

+ e 2 x cos 2 x dx

1 cos 2 x ⋅ 2e 2 x dx 2

∫

= −

e 2 x cos 2 x 2

( 2 ) . Combining equations ( 1 ) and ( 2 ) together we obtain:

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289

Calculus I

5.1 Integration by Parts

e 2 x sin 2 x − e 2 x sin 2 x dx 2

∫

∫

e 2 x cos 2 x dx =

+

e 2 x cos 2 x − e 2 x cos 2 x dx . 2

∫

 e 2 x sin 2 x  e 2 x cos 2 x + e 2 x cos 2 x dx  − −   2 2  

∫

=

∫

Check: Let y = −

e 2 x sin 2 x e 2 x cos 2 x + 2 2

(

and

)

1 2x e sin 2 x + e 2 x cos 2 x + c , 4

2 sin 2 x ⋅ e 2 x + 0 4

=

e 2 x sin 2 x 2

Taking the ∫ e 2 x cos 2 x dx from the right hand side of the equation

to the left hand side we obtain ∫ e 2 x cos 2 x dx + ∫ e 2 x cos 2 x dx = 2 e 2 x cos 2 x dx =

=

∫e

2x

cos 2 x dx =

then y ′ =

e 2 x sin 2 x e 2 x cos 2 x + 2 2

which implies

(

)

1 2x e sin 2 x + e 2 x cos 2 x + c 4

2 2 2 2x e ⋅ sin 2 x + cos 2 x ⋅ e 2 x + e 2 x ⋅ cos 2 x 4 4 4

2 2x 4 2 e sin 2 x + e 2 x cos 2 x − e 2 x sin 2 x 4 4 4

=

4 2x e cos 2 x 4

= e 2 x cos 2 x

h. Given ∫ e x sin x dx let u = e x and dv = sin x dx then du = e x dx and ∫ dv = ∫ sin x dx which implies v = − cos x .

∫e

x

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫

(1 )

∫

sin x dx = e x ⋅ − cos x − − cos x ⋅ e x dx = − e x cos x + e x cos x dx

To integrate ∫ e x cos x dx let u = e x and dv = cos x dx then du = e x dx and ∫ dv = ∫ cos x dx which

(2)

implies v = sin x . Thus, ∫ e x cos x dx = e x ⋅ sin x − ∫ sin x ⋅ e x dx = e x sin x − ∫ e x sin x dx Combining equations ( 1 ) and ( 2 ) together we obtain:

∫e

x

∫

∫

sin x dx = − e x cos x + e x cos x dx = − e x cos x + e x sin x − e x sin x dx

Taking the ∫ e x sin x dx from the right hand side of the equation to the left hand side we obtain

∫e

x

and

∫

∫

sin x dx + e x sin x dx = − e x cos x + e x sin x which implies 2 e x sin x dx = − e x cos x + e x sin x

∫e

x

sin x dx = −

1 x 1 e cos x + e x sin x + c 2 2

1 2

1 2

1 2

1 2

1 2

1 2

Check: Let y = − e x cos x + e x sin x + c , then y ′ = − e x ⋅ cos x + sin x ⋅ e x + e x ⋅ sin x + cos x ⋅ e x 1 2

1 2

1 2

1 2

= − e x cos x + e x sin x + e x sin x + e x cos x = i. Given

∫ x cos 3x dx let u = x

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1 x 1 e sin x + e x sin x 2 2

= e x sin x

and dv = cos 3x dx then du = dx and ∫ dv = ∫ cos 3x dx which implies

290

Calculus I

v=

5.1 Integration by Parts

1 sin 3 x . 3

∫ x cos 3x dx

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1 3

= x ⋅ sin 3x −

1 sin 3 x dx 3

1 3

∫

=

1 1 x sin 3 x + cos 3 x + c 3 9

1 9

Check: Let y = x sin 3x + cos 3x + c , then y ′ = +

1 3 x cos 3 x − sin 3 x 3 3

1 ( 1⋅ sin 3x + cos 3x ⋅ 3 ⋅ x ) − 1 ⋅ sin 3x ⋅ 3 + 0 3 9

=

1 sin 3 x 3

= x cos 3x

Example 5.1-2: Evaluate the following indefinite integrals: a.

∫ x sec

d.

∫ arc sin 6 x dx

g.

∫

x dx 2

2

b.

∫ x sec

=

e.

=

h.

5 x dx =

arc cos

(x +1) dx =

x

c.

∫ 3 sin 2 x dx =

∫ 5 arc sin y dy =

1

f.

∫ arc cos x dx =

∫

arc tan 10 x dx =

i.

2

x ex

∫ (1 + x )2 dx =

Solutions:

∫ x sec

a. Given v=

1 tan 5 x . 5

∫ x sec

2

2

5 x dx let u = x and dv = sec 2 5 x dx then du = dx and

1 5

= x ⋅ tan 5 x − 1 5

Check: Let y = x tan 5 x −

b. Given

2

5 x dx which implies

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

5 x dx

=

∫ dv = ∫ sec

1 tan 5 x dx 5

∫

=

1 1 x tan 5 x − ln sec 5 x + c 5 25

1 ln sec 5 x + c , 25

tan 5 x 5 x sec 2 5 x 5 sec 5 x tan 5 x + − 5 5 25 sec 5 x

∫ x sec

2

(x +1) dx let

u=x

then y ′ = =

(

)

sec 5 x tan 5 x 1 1 ⋅ tan 5 x + sec 2 5 x ⋅ 5 ⋅ x − ⋅5 + 0 5 25 sec 5 x

tan 5 x tan 5 x + x sec 2 5 x − 5 5

= x sec 2 5 x

and dv = sec 2 (x + 1) dx then du = dx and ∫ dv = ∫ sec 2 (x + 1) dx

which implies v = tan (x + 1) . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ x sec

2

(x +1) dx =

x ⋅ tan (x + 1) − tan (x + 1) dx = x tan ( x + 1) − ln sec ( x + 1) + c

∫

Check: Let y = x tan (x + 1) − ln sec (x + 1) + c , then y ′ = 1 ⋅ tan (x + 1) + sec 2 (x + 1) ⋅ x −

sec (x + 1) tan (x + 1) +0 sec (x + 1)

= tan (x + 1) + x sec 2 (x + 1) − tan (x + 1) = x sec 2 (x + 1) c. Given

x

x

∫ 3 sin 2 x dx let u = 3

Hamilton Education Guides

and dv = sin 2 x dx then du =

dx 3

and ∫ dv = ∫ sin 2 x dx which implies 291

Calculus I

v=−

5.1 Integration by Parts

1 cos 2 x . 2

x

∫ 3 sin 2 x dx

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

=

1 6

1 dx x 1 ⋅ − cos 2 x + cos 2 x 3 3 2 2

∫

= − x cos 2 x +

1 sin 2 x + c 12

1 6

1 1 1 sin 2 x + c , then y ′ = − ( 1 ⋅ cos 2 x − sin 2 x ⋅ 2 ⋅ x ) + cos 2 x ⋅ 2 + 0 12 12 6 x 1 1 1 − cos 2 x + x sin 2 x + cos 2 x = sin 2 x 3 3 6 6

Check: Let y = − x cos 2 x + =

6 dx

d. Given ∫ arc sin 6 x dx let u = arc sin 6 x and dv = dx then du =

and ∫ dv = ∫ dx which

1 − (6 x )2

implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ arc sin 6 x dx

= arc sin 6 x ⋅ x − ∫ x ⋅

To integrate

∫

dx = −

= −

dw 72 x

x dx 1 − 36 x

Therefore,

2 −1 1 1 ⋅ w 2 − 2 1 72

2

= −

2

∫

6 dx

x dx

= x arc sin 6 x − 6∫

1 − 36 x 2

(1 )

1 − 36 x 2 dw = −72 x dx

use the substitution method by letting w = 1 − 36 x 2 then x dx 1 − 36 x 2

=

x

dw − w 72 x

∫

= −

1 72

(

1 12 1 2 12 1 ⋅ w = − w = − 1 − 36 x 2 72 1 36 36

)

∫

dw

−1 1 w 2 dw 72

∫

= −

w

= −

and

1− 1 1 1 ⋅ w 2 72 1 − 1 2

1 2

(2)

Combining equations ( 1 ) and ( 2 ) together we obtain

∫

arc sin 6 x dx = x arc sin 6 x − 6

Check: Let y = x arc sin 6 x + = arc sin 6 x + e. Given v= y. 1

x dx

∫

1 − 36 x 2

(

1 1 − 36 x 2 6

6x 1 − 36 x 2

= x arc sin 6 x +

−

)

1 2

+c ,

6x

1

)

1 2

+ c = x arc sin 6 x +

6x

then y ′ = arc sin 6 x +

1 − 36 x 2

−

1 12

(

1 1 − 36 x 2 6

72 x 1 − 36 x 2

+0

= arc sin 6 x

1 − 36 x 2

∫ 5 arc sin y dy let u = arc sin y

(

6 1 − 36 x 2 36

and dv = dy then du =

dy 1− y 2

and ∫ dv = ∫ dy which implies

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ 5 arc sin y dy =

1 1 arc sin y ⋅ y − 5 5

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∫ y⋅

dy 1− y

2

=

1 1 y arc sin y − 5 5

∫

y dy 1− y

2

(1 )

292

)

1 2

+c

Calculus I

5.1 Integration by Parts

To integrate dy = − 1 2

dw 2y

∫

y dy

use the substitution method by letting w = 1 − y 2 then

1− y 2

Therefore,

1

= − ⋅ 2−1 w

2 −1 2

∫

1 2 2 1

y dy 1− y 2 1

y

dw − w 2y

∫

=

= −

(

1

= − ⋅ w 2 = − w 2 = − 1− y 2

2

)

1 2

∫

dw

= −

w

dw = −2 y dy

and

1− 1 1 1 w 2 2 1− 1 2

−1 1 w 2 dw 2

∫

= − ⋅

1 2

(2)

Combining equations ( 1 ) and ( 2 ) together we obtain:

∫

1 arc sin y dy 5

1 1 y arc sin y − 5 5

=

1 5

Check: Let w = y arc sin y + =

1 1 arc sin y + 5 5

y dy

∫

1− y 2

(

1 1− y 2 5

y 1− y 2

1 2

+c ,

y

1 5

−

)

(

1 1 y arc sin y + 1 − y 2 5 5

=

1− y 2

then w′ =

1 1 arc sin y + 5 5

To integrate

1 2

+c

y 1 − y2

−

1 1 ⋅ ⋅ 5 2

2y 1 − y2

+0

− dx 1− x 2

and ∫ dv = ∫ dx which implies

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ arc cos x dx = arc cos x ⋅ x − ∫ x ⋅

dx = −

1 2

1 arc sin y 5

=

f. Given ∫ arc cos x dx let u = arc cos x and dv = dx then du = v=x.

)

dw 2x

∫

x dx

1

2 −1 2

1− x

1− x 2

∫

1 2 2 1

x dx 1− x 2

=

∫

x

dw w − 2x

(

1

1

= − ⋅ w 2 = − w 2 = − 1− x 2

2

x dx

= x arc cos x + ∫

2

1− x

(1 )

2

dw = −2 x dx

use the substitution method by letting w = 1 − x 2 then

Therefore,

= − ⋅ 2−1 w

−dx

= −

)

1 2

∫

dw w

= −

−1 1 w 2 dw 2

∫

and

1− 1 1 1 w 2 2 1− 1 2

= − ⋅

1 2

(2)

Combining equations ( 1 ) and ( 2 ) together we obtain:

∫

arc cos x dx = x arc cos x +

∫

(

x dx 1− x 2

Check: Let y = x arc cos x − 1 − x 2 −

x 1− x 2

+

Hamilton Education Guides

x 1− x 2

)

1 2

(

= x arc cos x − 1 − x 2 +c,

)

1 2

then y ′ = arc cos x −

+c

x 1− x

2

−

1 2

−2 x 1− x 2

+ 0 = arc cos x

= arc cos x

293

Calculus I

5.1 Integration by Parts

g. Given ∫ arc cos implies v = x .

x dx 2

x

x

To integrate

∫

x dx 1−

2

x 2

−dx

= x arc cos +

()

x 2 2

2 1−

1 2

we obtain

x dx

∫

1−

(1 )

()

x 2 2

2x 1 x dw ⋅ =− =− 2 2 2 dx

2

(2x )2

2 x

2 −1 2

and ∫ dv = ∫ dx which

use the substitution method by letting w = 1 − (2x ) then

and dx = − dw Therefore,

= − 2 ⋅ 2−1 w

− dx

and dv = dx then du =

2 2 1 − (2x ) Using the integration by parts formula ∫ u dv = u v − ∫ v du

∫ arc cos 2 dx = arc cos 2 ⋅ x − ∫ x ⋅

1

x 2

let u = arc cos

x dx

∫

=

()

x 2 2

1−

x −2dw w x

∫

= − 2∫

−1

dw

= − 2∫ w 2 dw = − 2 ⋅

w

1 1 − 12

1− 1

w

1

1 2 2 2 1 = − 2 ⋅ w 2 = − 4w 2 = − 41 − (2x )  1  

2

(2)

Combining equations ( 1 ) and ( 2 ) together we obtain:

∫

arc cos

x x 1 dx = x arc cos + 2 2 2

Check: Let

x dx

∫

1−

x y = x arc cos − 2 1 − 2  x 2

x

= arc cos −

2 1−

(2x )2

()

()

x 2 2 1

x 22 2 

x

+

2 1−

4

x

= x arc cos − 1 − 2 2 +c,

(2x )2  x 2

1 2

+ c = x arc cos

x

then y ′ = arc cos − = arc cos

(2x )2

2 1−

(2x )2

−

x  − 2 1 − 2 

x 1−

(2x )2

⋅−

( 2x ) 2 

1 2

1 +0 2

x 2

h. Given ∫ arc tan 10 x dx let u = arc tan 10 x and dv = dx then du =

10 dx 1 + (10 x )2

and ∫ dv = ∫ dx which

implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ arc tan 10 x dx To integrate And dx =

= arc tan 10 x ⋅ x − ∫ x ⋅ x dx

∫ 1 + (10 x )2

dw 200 x

Thus,

10 dx 1 + (10 x )

2

= x arc tan 10 x − 10∫

x dx

(1 )

1 + (10 x )2

use the substitution method by letting w = 1 + (10 x )2 then x dx

∫ 1 + (10 x )2

=

x dw

∫ w 200 x

=

1 200

∫

dw w

=

1 ln w 200

=

dw = 200 x dx

1 ln 1 + (10 x )2 200

(2)

Combining equations ( 1 ) and ( 2 ) together we obtain:

Hamilton Education Guides

294

+c

Calculus I

5.1 Integration by Parts

∫ arc tan 10 x dx

= x arc tan 10 x − 10∫

Check: Let y = x arc tan 10 x − x

arc tan 10 x +

x ex

∫ (1 + x )2 dx let u = xe

i. Given

which implies v = − x ex

∫ (1 + x )2 =

1 + ( 10 x )

dx = xe x ⋅

1 . 1+ x

−1 − 1+ x

− xe x + e x + xe x +c 1+ x

Check: Let y =

=

2

x

x dx 1 + (10 x )2

= x arc tan 10 x −

1 ln 1 + (10 x )2 + c , 20

−

x

1 ln 1 + (10 x ) 2 + c 20

then y ′ = arc tan 10 x +

x

1 + ( 10 x )

−

2

20 x 1 +0 20 1 + ( 10 x )2

= arc tan 10 x

1 + ( 10 x )2

and dv = ( 1 + x )−2 dx then du = e x ( 1 + x ) dx and ∫ dv = ∫ ( 1 + x )−2 dx

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫

−1 ⋅ e x ( 1 + x ) dx ( 1 + x)

=

− xe x + e x dx 1+ x

∫

=

− xe x +ex +c 1+ x

=

− xe x + e x ( 1 + x ) +c 1+ x

ex +c 1+ x

ex +c , 1+ x

then y ′ =

e x ( 1 + x) − e x

( 1 + x )2

+0 =

e x + xe x − e x

=

( 1 + x )2

xe x

( 1 + x )2

Example 5.1-3: Evaluate the following indefinite integrals: b.

∫ sin

2

=

c.

∫ arctan x dx =

∫ sin ( ln x ) dx =

e.

∫x

2

e x dx =

f.

∫x

∫x

h.

∫e

−x

i.

∫e

a.

∫ sin

d. g.

3

2

x dx

=

cos 3 x dx =

x dx

cos x dx =

3

sin x dx =

−3 x

sin 3 x dx =

Solutions: a. Given ∫ sin 3 x dx = ∫ sin 2 x ⋅ sin x dx let u = sin 2 x and dv = sin x dx then du = 2 sin x cos x dx and

∫ dv = ∫ sin x dx which implies v = − cos x .

Using the integration by parts formula ∫ u dv = u v − ∫ v du

we obtain ∫ sin 3 x dx = sin 2 x ⋅ − cos x + ∫ cos x ⋅ 2 sin x cos x dx = − sin 2 x cos x + 2∫ cos 2 x sin x dx

(1 )

To integrate ∫ cos 2 x sin x dx use the integration by parts method again, i.e., let u = cos 2 x and dv = sin x dx

∫ cos

2

then du = −2 sin x cos x dx and ∫ dv = ∫ sin x dx which implies v = − cos x . Therefore,

∫

∫

x sin x dx = cos 2 x ⋅ − cos x − cos x ⋅ 2 sin x cos x dx = − cos 3 x − 2 cos 2 x sin x dx . Taking the

integral − 2∫ cos 2 x sin x dx from the right hand side of the equation to the left side we obtain Hamilton Education Guides

295

Calculus I

∫ cos

5.1 Integration by Parts

1 3

∫

2

(2)

= − cos 3 x . Therefore, ∫ cos 2 x sin x dx = − cos 3 x

x sin x dx + 2 cos 2 x sin x dx

Combining equations ( 1 ) and ( 2 ) together we have

∫ sin

3

2 1 x dx = − sin 2 x cos x + 2 cos 2 x sin x dx = − sin 2 x cos x + 2 ⋅ − cos 3 x + c = − sin 2 x cos x − cos 3 x + c 3 3

∫

(

)

− 1 − cos 2 x cos x −

1 2 2   cos 3 x + c = − cos x +  cos 3 x − cos 3 x  + c = cos 3 x − cos x + c 3 3 3  

Note that another method of solving the above problem (as was shown in Section 4.3) is in the following way:

∫ sin

3

∫ sin

x dx =

Therefore,

(

∫ sin

3

2

x dx =

)

= − ∫ 1 − u 2 du =

2 ∫ (1 − cos x )⋅ sin x dx

x ⋅ sin x dx =

∫u

2

∫ sin

2

x ⋅ sin x dx =

−1 du =

1 3 u −u +c 3

let u = cos x , then

2 ∫ (1 − cos x )⋅ sin x dx

=

du = − sin x dx

and dx = −

du sin x

.

du 2 ∫ (1 − u )⋅ sin x ⋅ − sin x

1 cos 3 x − cos x + c 3

=

1 1 Check: Let y = cos 3 x − cos x + c , then y ′ = ⋅ 3 cos 2 x ⋅ − sin x + sin x + 0 = − cos 2 x sin x + sin x

(

3

3

)

2

3

2

= sin x 1 − cos x = sin x sin x = sin x b. Given ∫ sin 2 x dx = ∫ sin x ⋅ sin x dx let u = sin x and dv = sin x dx then du = cos x dx and ∫ dv = ∫ sin x dx which implies v = − cos x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ sin

2

∫

∫

x dx = sin x ⋅ − cos x + cos x ⋅ cos x dx = − sin x cos x + cos 2 x dx = − sin x cos x +

2 ∫ (1 − sin x ) dx

= − sin x cos x + x − ∫ sin 2 x dx . Taking the integral ∫ sin 2 x dx from the right hand side of the equation to the left side we have ∫ sin 2 x dx + ∫ sin 2 x dx = − sin x cos x + x . Therefore,

∫

2 sin 2 x dx

= − sin x cos x + x and

∫ sin

2

x dx

1 4

x 2

1 x = − sin x cos x + + c = − sin 2 x + + c 2

2

or, we can solve the given integral in the following way:

∫ sin

2

x dx

=

1

∫ 2 ( 1 − cos 2 x ) dx 1 4

x 2

=

1 1 dx − cos 2 x dx 2 2

∫

∫

1 4

= 1 2

x 1 sin 2 x − ⋅ +c 2 2 2 1 2

Check: Let y = − sin 2 x + + c , then y ′ = − cos 2 x ⋅ 2 + + 0 = − cos 2 x +

Hamilton Education Guides

1 4

x 2

= − sin 2 x + + c 1 1 = ( 1 − cos 2 x ) = sin 2 x or, 2 2

296

Calculus I

5.1 Integration by Parts

1 2

1 2

x 2

1 2

1 2

1 2

1 2

Let y = − sin x cos x + + c , then y ′ = − cos x cos x + sin x sin x + + 0 = − cos 2 x + sin 2 x +

= −

(

)

1 1 1 1 1 1 1 1 − sin 2 x + sin 2 x + = − + sin 2 x + sin 2 x + 2 2 2 2 2 2 2

c. Given ∫ arctan x dx let u = arc tan x and dv = dx then du = v=x.

dx 1+ x 2

=

1 1 sin 2 x + sin 2 x 2 2

1 2

= sin 2 x

and ∫ dv = ∫ dx which implies

Substituting the integral with its equivalent value ∫ u dv = u v − ∫ v du we obtain

∫ x arctan x dx

= arc tan x ⋅ x − ∫ x ⋅

To integrate

∫ 1+ x 2

Therefore,

x dx

x dx

∫ 1+ x 2

dx 1+ x

2

= x arc tan x − ∫

x dx

(1 )

1+ x 2

use the substitution method by letting w = 1 + x 2 then

=

x dw

∫ w 2x

=

1 2

∫

dw w

=

1 ln w 2

=

dw = 2x dx

and dx =

1 ln 1 + x 2 2

dw 2x

(2)

Combining equations ( 1 ) and ( 2 ) together we obtain:

∫ x arctan x dx

= x arc tan x −

x dx

∫ 1+ x 2

1 2

= x arc tan x − ln 1 + x 2 + c

1 2

Check: Let y = x arc tan x − ln 1 + x 2 + c , then y ′ = arc tan x + +

x 1+ x

2

−

x 1+ x 2

x 1+ x

2

−

1 2x +0 2 1+ x 2

= arc tan x

= arc tan x

d. Given ∫ sin ( ln x ) dx let u = sin ( ln x ) and dv = dx then du =

cos ( ln x ) dx x

and ∫ dv = ∫ dx which

implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ sin ( ln x ) dx = sin ( ln x )⋅ x − ∫ x ⋅

cos ( ln x ) dx x

= x sin ( ln x ) − ∫ cos ( ln x ) dx

(1 )

To integrate ∫ cos ( ln x ) dx use the integration by parts formula again, i.e., let u = cos ( ln x ) and dv = dx

then du =

− sin ( ln x ) dx x

= cos ( ln x ) ⋅ x + ∫ x ⋅

sin ( ln x ) dx x

and ∫ dv = ∫ dx which implies v = x . Therefore, ∫ cos ( ln x ) dx = x cos ( ln x ) + ∫ sin ( ln x ) dx

(2)

Combining equations ( 1 ) and ( 2 ) together we have

∫ sin ( ln x ) dx = x sin ( ln x ) − ∫ cos ( ln x ) dx

= x sin ( ln x ) − x cos ( ln x ) − ∫ sin ( ln x ) dx

Taking the integral − ∫ sin ( ln x ) dx from the right hand side of the equation to the left hand side

Hamilton Education Guides

297

Calculus I

5.1 Integration by Parts

we obtain ∫ sin ( ln x ) dx + ∫ sin ( ln x ) dx = x sin ( ln x ) − x cos ( ln x ) + c Therefore, 2 sin ( ln x ) dx = x sin ( ln x ) − x cos ( ln x ) + c

∫

Check: Let y =

∫x

∫ sin ( ln x ) dx

=

x x sin ( ln x ) − cos ( ln x ) + c 2 2

sin ( ln x ) x cos ( ln x ) cos ( ln x ) x sin ( ln x ) x x sin ( ln x ) − cos ( ln x ) + c , then y ′ = − + +0 + 2 2 2x 2 2x 2

sin ( ln x ) cos ( ln x ) cos ( ln x ) sin ( ln x ) sin ( ln x ) sin ( ln x ) = = sin ( ln x ) + − + + 2 2 2 2 2 2

= e. Given

and thus

2

e x dx let u = x 2 and dv = e x dx then du = 2 x dx and

∫ dv = ∫ e

x

dx which implies v = e x .

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫x

2

∫

(1 )

∫

e x dx = x 2 ⋅ e x − e x ⋅ 2 x dx = x 2 e x − 2 x e x dx

To integrate

∫ xe

x

dx use the integration by parts formula again, i.e., let u = x and dv = e x dx

Then du = dx and ∫ dv = ∫ dx which implies v = e x . Therefore,

∫ xe

x

∫

(2)

∫

dx = x ⋅ e x − e x ⋅ dx = xe x − e x dx = xe x − e x

Combining equations ( 1 ) and ( 2 ) together we have

∫x

2

(

∫

)

e x dx = x 2 e x − 2 x e x dx = x 2 e x − 2 xe x − e x + c

= x 2 e x − 2 xe x + 2e x + c

Check: Let y = x 2 e x − 2 xe x + 2e x + c , then y ′ = 2 xe x + x 2 e x − 2e x − 2 xe x + 2e x + 0 = x 2 e x

f. Given

∫x

3

sin x dx let u = x 3 and dv = sin x dx then du = 3 x 2 dx and

∫ dv = ∫ sin x dx which implies

v = − cos x .

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫x

= x 3 ⋅ − cos x + ∫ cos x ⋅ 3x 2 dx = − x 3 cos x + 3∫ x 2 cos x dx

3

sin x dx

To integrate

∫x

2

(1 )

cos x dx use the integration by parts formula again, i.e., let u = x 2 and dv = cos x dx

then du = 2 x dx and ∫ dv = ∫ cos x dx which implies v = sin x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫x

2

∫

∫

cos x dx = x 2 ⋅ sin x − sin x ⋅ 2 x dx = x 2 sin x − 2 x sin x dx

To integrate

∫ x sin x dx

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(2)

use the integration by parts formula again, i.e., let u = x and dv = sin x dx

298

Calculus I

5.1 Integration by Parts

then du = dx and ∫ dv = ∫ sin x dx which implies v = − cos x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

(3 )

∫ x sin x dx = x ⋅ − cos x + ∫ cos x ⋅ dx = − x cos x + sin x Combining equations ( 1 ) , ( 2 ) and ( 3 ) together we have

∫x

3

sin x dx

= − x 3 cos x + 3∫ x 2 cos x dx = − x 3 cos x + 3x 2 sin x − 6∫ x sin x dx = − x 3 cos x + 3 x 2 sin x

+ 6 x cos x − 6 sin x + c

(

Check: Let y = − x 3 cos x + 3x 2 sin x + 6 x cos x − 6 sin x + c , then y ′ = − 3x 2 cos x + x 3 sin x

(

)

+ 6 x sin x + 3 x 2 cos x + (6 cos x − 6 x sin x ) − 6 cos x + 0

g. Given

∫x

2

implies v =

∫x

2

= x 3 sin x

cos 3 x dx let u = x 2 and dv = cos 3 x dx then du = 2 x dx and sin 3 x 3

cos 3 x dx

To integrate

)

∫ dv = ∫ cos 3x dx which

. Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

= x2 ⋅

sin 3 x sin 3 x ⋅ 2 x dx − 3 3

∫

=

2 1 2 x sin 3 x − 3 3

(1 )

∫ x sin 3x dx

∫ x sin 3x dx use the integration by parts formula again, i.e., let u = x

then du = dx and ∫ dv = ∫ sin 3x dx which implies v =

− cos 3 x . 3

and dv = sin 3x dx

Using the integration by parts

formula ∫ u dv = u v − ∫ v du we obtain

∫ x sin 3x dx = x ⋅

− cos 3 x cos 3 x + ⋅ dx 3 3

∫

1 3

= − x cos 3x +

1 cos 3 x dx 3

(2)

∫

Combining equations ( 1 ) and ( 2 ) together we have

∫x =

2

cos 3 x dx =

2 1 2 x sin 3 x − 3 3

∫ x sin 3x dx

=

1 2 2 1 2 1 x sin 3 x − ⋅ − x cos 3 x − ⋅ cos 3 x dx 3 3 3 3 3

∫

2 1 2 2 x sin 3 x + x cos 3 x − sin 3 x + c 27 3 9

(

)

1 2 sin 3 x + c , then y ′ = 2 x sin 3 x + 3 x 2 cos 3 x 3 27 2 6 2 3 2 2 2 + ( cos 3 x − 3 x sin 3 x ) − cos 3 x + c = x sin 3 x + x 2 cos 3 x + cos 3 x − x sin 3 x − cos 3 x 9 27 3 3 9 3 9 1 3

2 9

Check: Let y = x 2 sin 3x + x cos 3x −

=

3 2 x cos 3 x 3

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= x 2 cos 3x

299

Calculus I

5.1 Integration by Parts

h. Given ∫ e − x cos x dx let u = cos x and dv = e − x dx then du = − sin x dx and ∫ dv = ∫ e − x dx which implies v = −e − x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫e

−x

cos x dx

(1 )

= cos x ⋅ −e − x − ∫ e − x ⋅ sin x dx = − e − x cos x − ∫ e − x sin x dx

To integrate ∫ e − x sin x dx use the integration by parts formula again, i.e., let u = sin x and then du = cos x dx and ∫ dv = ∫ e − x dx which implies v = −e − x . Therefore,

dv = e − x dx

∫e

−x

(2)

∫

∫

sin x dx = sin x ⋅ −e − x + e − x ⋅ cos x dx = − e − x sin x + e − x cos x dx

Combining equations ( 1 ) and ( 2 ) together we have

∫e

−x

∫

∫

cos x dx = − e − x cos x − e − x sin x dx = − e − x cos x + e − x sin x − e − x cos x dx

Taking the integral

∫e

−x

cos x dx from the right hand side of the equation to the left hand side

we obtain ∫ e − x cos x dx + ∫ e − x cos x dx = − e − x cos x + e − x sin x therefore

∫

2 e − x cos x dx = − e − x cos x + sin x e − x and thus

Check: Let y =

=

1 −x e 2

∫e

−x

cos x dx = −

1 −x 1 e cos x + e − x sin x + c 2 2

(

) (

)

− e− x cos x e− x sin x 1 1 + + c , then y ′ = − − e − x cos x − e − x sin x + − e − x sin x + e − x cos x + 0 2 2 2 2 1 −x 1 −x 1 −x 1 −x 1 −x cos x + e sin x − e sin x + e cos x = e cos x + e cos x = e − x cos x 2 2 2 2 2

i. Given ∫ e −3 x sin 3x dx let u = sin 3x and dv = e −3 x dx then du = 3 cos 3x dx and ∫ dv = ∫ e −3 x dx which 1 3

implies v = − e −3 x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫e

−3 x

sin 3 x dx

To integrate dv = e −3 x dx

∫e

−3 x

1 3

= sin 3x ⋅ − e −3 x +

∫e

−3 x

1 −3 x e ⋅ 3 cos 3 x dx 3

∫

1 3

= − e −3 x sin 3x + ∫ e −3 x cos 3x dx

(1 )

cos 3 x dx use the integration by parts formula again, i.e., let u = cos 3 x and 1 3

then du = − 3 sin 3x and ∫ dv = ∫ e −3 x dx which implies v = − e −3 x . Therefore,

1 1 1 −3 x cos 3 x dx = cos 3 x ⋅ − e −3 x − ⋅ 3 sin 3 x dx = − e −3 x cos 3 x − e −3 x sin 3 x dx e 3 3 3

∫

∫

(2)

Combining equations ( 1 ) and ( 2 ) together we have

∫e

−3 x

1 1 1 sin 3 x dx = − e −3 x sin 3 x + e −3 x cos 3 x dx = − e −3 x sin 3 x − e −3 x cos 3 x − e −3 x sin 3 x dx 3 3 3

Taking the integral

∫

∫e

Hamilton Education Guides

−3 x

∫

sin 3 x dx from the right hand side of the equation to the left hand side

300

Calculus I

5.1 Integration by Parts

1 3

1 3

we obtain ∫ e −3 x sin 3x dx + ∫ e −3 x sin 3x dx = − e −3 x sin 3x − e −3 x cos 3x therefore 1 1 2 e −3 x sin 3 x dx = − e −3 x sin 3 x − e −3 x cos 3 x and thus 3 3

∫

1 6

1 6

=

(

−3 x

sin 3 x dx = −

1 −3 x 1 sin 3 x − e − 3 x cos 3 x e 6 6

(

)

1 − 3e −3 x sin 3 x + 3e −3 x cos 3 x 6 1 −3 x 1 1 1 sin 3 x − e −3 x cos 3 x + e −3 x cos 3 x + e −3 x sin 3 x e 2 2 2 2

Check: Let y = − e − 3 x sin 3x − e − 3 x cos 3x , then y ′ = − −

∫e

)

1 − 3e −3 x cos 3 x − 3e −3 x sin 3 x + 0 = 6 1 −3 x 1 sin 3 x + e −3 x sin 3 x = e −3 x sin 3 x e 2 2

Example 5.1-4: Evaluate the following indefinite integrals: x

=

b.

∫ (x − 3)(3x − 1)

x dx =

e.

∫x

h.

∫ cos

a.

∫ 3 (x + 1)

d.

∫ x sec

g.

∫5e

2

4

dx

1 x sin 3 x dx

=

c.

∫ (x + 1) csc

x − 5 dx =

f.

∫ ln (x

−1

i.

∫ tan

3

dx =

3 x dx =

2

−1

2

x dx =

)

+1 dx =

5 x dx =

Solutions: a. Given

x

∫ 3 (x + 1)

4

dx =

1 x (x + 1)4 dx 3

∫

let u = x and dv = (x + 1)4 dx then du = dx and ∫ dv = ∫ (x + 1)4 dx

1 5

which implies v = (x + 1)5 . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1 x (x + 1)4 dx 3

∫

=

Check: Let y = +

x (x + 1)5 1 ⋅ − 3 5 3⋅5

x (x + 1)5 5 ( ) x + ⋅ 1 dx = ∫ 15

−

1 1 ⋅ (x + 1)5+1 + c 15 6

[

=

x ( x + 1) 5 1 ( x + 1) 6 + c − 90 15

]

5 x (x + 1)5 1 1 (x + 1)5 + 5 x(x + 1)4 − 6 (x + 1)5 + 0 = (x + 1) − (x + 1)6 + c , then y ′ = 15 90 15 90 15

5 5x (x + 1)4 − (x + 1) = 5 x (x + 1)4 = x (x + 1)4 3 15 15 15

b. Given ∫ (x − 3)(3x − 1)3 dx let u = x − 3 and dv = (3x − 1)3 dx then du = dx and ∫ dv = ∫ (3x − 1)3 dx which implies v =

1 (3x − 1)4 . 12

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

(3x − 1) 3 ∫ (x − 3)(3x − 1) dx = (x − 3)⋅

4

12

=

( x − 3)(3 x − 1) 4 12

−

−

1 12

(x − 3) (3x − 1)4 4 ( ) 3 x − 1 dx = ∫ 12

−

1 1 ⋅ (3 x − 1)4+1 + c 12 15

1 (3 x − 1) 5 + c 180

Hamilton Education Guides

301

Calculus I

5.1 Integration by Parts

Check: Let y = =

(x − 3)(3x − 1)4 12

−

[

]

1 (3x − 1)5 + c , then y ′ = 1 (3x − 1)4 + 12(x − 3)(x + 1)3 − 15 (3x − 1)4 + 0 180 180 12

3 1 (3x − 1)4 + 12(x − 3)(x + 1) − 1 (3x − 1)4 = (x − 3)(x + 1)3 12 12 12

c. Given ∫ (x + 1) csc 2 x dx let u = x + 1 and dv = csc 2 x dx then du = dx and ∫ dv = ∫ csc 2 x dx which implies v = − cot x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ (x + 1) csc

2

x dx = (x + 1) ⋅ − cot x + cot x dx = − ( x + 1) cot x + ln sin x + c

∫

[

]

Check: Let y = −(x + 1) cot x + ln sin x + c , then y ′ = − cot x − (x + 1) csc 2 x +

cos x + 0 = − cot x sin x

+ (x + 1) csc 2 x + cot x = (x + 1) csc 2 x

∫ x sec

d. Given

v = tan x .

∫ x sec

2

2

x dx

let u = x and dv = sec 2 x dx then du = dx and ∫ dv = ∫ sec 2 x dx which implies

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫

x dx = x ⋅ tan x − tan x dx = x tan x − ln sec x + c

Check: Let y = x tan x − ln sec x + c , then y ′ = tan x + x sec 2 x −

sec x tan x + 0 = tan x + x sec 2 x − tan x sec x

= x sec 2 x

e. Given v=

∫x

∫x

x − 5 dx let u = x and dv = x − 5 dx then du = dx and

3 2 ( x − 5) 2 3

∫ dv = ∫

x − 5 dx which implies

. Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

x − 5 dx = x ⋅

5 2 2 − ⋅ ( x − 5) 2 + c 3 5

3 3 2 (x − 5) 2 − 2 (x − 5) 2 dx 3 3

∫

=

=

3 3 2 2 1 (x − 5) 2 +1 + c x ( x − 5) 2 − ⋅ 3 3 3 1+

2

=

3 2 x ( x − 5) 2 3

3 5 4 2 x ( x − 5) 2 − ( x − 5) 2 + c 15 3

3 5 3 3 1 2 2 3 4 5 4 2 x (x − 5) 2 − (x − 5) 2 + c , then y ′ = (x − 5) 2 + ⋅ x(x − 5) 2 − ⋅ (x − 5) 2 + 0 3 3 2 15 2 15 3 3 3 1 1 2 2 = ( x − 5) 2 + x ( x − 5) 2 − ( x − 5) 2 = x ( x − 5) 2 = x x − 5 3 3

Check: Let y =

(

)

(

)

f. Given ∫ ln x 2 +1 dx let u = ln x 2 + 1 and dv = dx then du = v=x.

2x 2

x +1

dx

and ∫ dv = ∫ dx which implies

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

Hamilton Education Guides

302

Calculus I

5.1 Integration by Parts

∫ (

)

(

) ∫

2x

ln x 2 +1 dx = ln x 2 + 1 ⋅ x − x ⋅

(

)

= x ln x 2 + 1 − 2∫ dx + 2∫

1 2

x +1

(

dx

2

x +1

dx

(

)

x2

= x ln x 2 + 1 − 2∫

(

2

x +1

dx

(

)

(

g. Given

−2+

2

x +1

1 e x sin 3 x dx 5

∫

 

  dx x +1 1

2

)

)

2x 2

)

= x ln x 2 + 1 − 2 x + 2 tan −1 x + c

)

Check: Let y = x ln x 2 + 1 − 2 x + 2 tan −1 x + c , then y ′ = 1 ⋅ ln x 2 + 1 + = ln x 2 + 1 +

(

= x ln x 2 + 1 − 2∫ 1 −

(

2

)

= ln x 2 + 1 +

2

x +1

2x 2 − 2x 2 − 2 + 2 2

x +1

2x 2

x +1

⋅x−2+

(

)

2 1+ x 2

= ln x 2 + 1 +

+0

(

0

)

= ln x 2 + 1

2

x +1

let u = e x and dv = sin 3x dx then du = e x dx and ∫ dv = ∫ sin 3x dx which

1 3

implies v = − cos 3x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1 e x sin 3 x dx 5

∫

=

1 1 x 1 e ⋅ − cos 3 x + cos 3 x ⋅ e x dx 15 3 5

∫

= −

1 1 x e x cos 3 x dx e cos 3 x + 15 15

(1 )

∫

To integrate ∫ e x cos 3x dx use the integration by parts method again, i.e., let u = e x and dv = cos 3 x dx

∫e

x

1 3

then du = e x dx and ∫ dv = ∫ cos 3x dx which implies v = sin 3x . Therefore,

1 1 1 1 cos 3 x dx = e x ⋅ sin 3 x − sin 3 x ⋅ e x dx = e x sin 3 x − e x sin 3 x dx 3 3 3 3

∫

(2)

∫

Combining equations ( 1 ) and ( 2 ) together we have 1 e x sin 3 x dx 5

∫

= −

1 1 x e x cos 3 x dx e cos 3 x + 15 15

Taking the integral − obtain = −

∫

1 e x sin 3 x dx 45

∫

1 1 e x sin 3 x dx + e x sin 3 x dx 5 45

∫

∫

1 x 1 x e sin 3 x + c e cos 3 x + 15 45

= −

1 x 1 x 1 e cos 3 x + e sin 3 x − e x sin 3 x dx 15 45 45

∫

from the right hand side of the equation to the left side we = −

1 x 1 x e cos 3 x + e sin 3 x + c . 15 45

which implies

∫e

x

sin 3 x dx

= −

Therefore,

2 e x sin 3 x dx 9

∫

3 x 1 x e cos 3 x + e sin 3 x + c 10 10

3 x 1 1 3 3 e cos 3 x + e x sin 3 x + c , then y ′ = − e x ⋅ cos 3 x + sin 3 x ⋅ 3 ⋅ e x + e x ⋅ sin 3 x 10 10 10 10 10 9 1 1 3 3 9 + cos 3 x ⋅ 3 ⋅ e x + 0 = − e x cos 3 x + e x sin 3 x + e x sin 3 x + e x cos 3 x = e x sin 3 x 10 10 10 10 10 10 1 10 x 9 +1 x + e x sin 3 x = e sin 3 x = e x sin 3 x e sin 3 x = 10 10 10

Check: Let y = −

h. Given ∫ cos −1 3x dx let u = cos −1 3x and dv = dx then du = −

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3 1 − 9x 2

dx and

∫ dv = ∫ dx

which

303

Calculus I

5.1 Integration by Parts

implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ cos

−1

∫

To integrate

implies dx = − 1

1 6

3x 1 − 9x 2

dw . 18 x

= − ⋅ 2w 2 = −

dx

Thus,

1 − 9x 2 3

∫ tan

−1

1 − 9x

3x

= x cos −1 3x + ∫

2

dx

1 − 9x 2

dw = −18 x dx

use the substitution method by letting w = 1 − 9 x 2 then 3x

∫

dx

1 − 9x 2

and

∫

=

∫

3x w

⋅−

dw 18 x

cos −1 3 x dx = x cos −1 3 x +

1 − 9x 2

Check: Let y = x cos −1 3x −

i. Given

3 dx

∫

3 x dx = cos −1 3 x ⋅ x + x ⋅

3

= −

1 6

= −

w

3x

∫

3x 1 − 9x 5

2

dx

1 + 25 x 2

1 6

∫

1

dw

1

= −

w2

+

18 x 6 1 − 9x

2

−1 1 w 2 dw 6

∫

1 − 9 x2 +c 3

dx = x cos−1 3 x −

1 − 9x 2

+ c , then y ′ = cos −1 3 x −

5 x dx let u = tan −1 5 x and dv = dx then du =

∫

dw

which

+ 0 = cos −1 3 x

and ∫ dv = ∫ dx which

implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ tan

−1

∫

5 x dx = tan −1 5 x ⋅ x − x ⋅ 5x

5 dx 1 + 25 x

2

= x tan −1 5 x − ∫

5x 1 + 25 x 2

dx

To integrate

∫ 1 + 25x 2 dx

use the substitution method by letting w = 1 + 25 x 2 then

implies dx =

dw 50 x

∫ 1 + 25x 2 dx = ∫ w ⋅ 50 x

∫ tan

= x tan −1 5 x − ∫

−1

5 x dx

. Thus,

Check: Let y = x tan −1 5 x −

5x

5 x dw

5x 1 + 25 x 2

dx

=

= x tan−1 5 x −

1 10

∫

dw w

=

1 ln w 10

=

dw = 50 x dx

1 ln 1 + 25 x 2 10

which

and

1 ln 1 + 25 x 2 + c 10

5x 1 50 x 1 − + 0 = tan −1 5 x ln 1 + 25 x 2 + c , then y ′ = tan −1 5 x + 2 10 2 10 1 + 25 x 1 + 25 x

Example 5.1-5: Evaluate the following indefinite integrals: a.

∫ sinh

d.

∫ cos 5x cos 7 x dx

−1

5 x dx =

=

b.

∫ x tan

e.

∫5e

−1

x dx =

x −x dx

=

c.

∫ sin x sin 7 x dx =

f.

∫ x sinh 3x dx =

Solutions: a. Given ∫ sinh −1 5 x dx let u = sinh −1 5 x and dv = dx then du =

5 1 + 25 x 2

dx and

∫ dv = ∫ x dx which

implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain Hamilton Education Guides

304

Calculus I

∫ sinh

5.1 Integration by Parts

−1

dw = 50 x dx

1 10

1

∫

1 w2

dw

=

(1 )

1 + 25 x 2

use the substitution method by letting w = 1 + 25 x 2 then

1 + 25 x 2

which implies dx =

5 x dx

= x sinh −1 5 x − ∫

1 + 25 x 2

5 x dx

∫

To get the integral of

=

5 dx

∫

5 x dx = sinh −1 5 x ⋅ x − x ⋅

dw 50 x

. Therefore,

∫

5 x dx 1 + 25 x

=

2

5x

dw 50 x w

∫

⋅

=

5 50

∫

1 w

dw

1− 1 −1 1 1 1 2 12 1 1 1 ⋅ w 2 = w = w2 = w 2 dw = 1 + 25 x 2 1 10 1 − 10 10 5 5

∫

(2)

2

Combining equations ( 1 ) and ( 2 ) together we have

∫

sinh −1 5 x dx = x sinh −1 5 x −

Check: Let y = x sinh −1 5 x − 5x

sinh −1 5 x +

b. Given

∫ x tan

−1

1 + 25 x

x dx

2

∫

5 x dx 1 + 25 x 2

(

)

1 1 + 25 x 2 5 −

= x sinh −1 5 x −

1 2

5x 1 + 25 x

(

1 1 + 25 x 2 5

+ c , then y ′ = sinh −1 5 x +

)

1 2

+c

5x 1 + 25 x

2

−

50 x 10 1 + 25 x

+0

2

= sinh −1 5 x

2

let u = tan −1 x and dv = x dx then du =

1 1+ x 2

dx

and ∫ dv = ∫ x dx which

1 2

implies v = x 2 . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ x tan =

−1

x dx = tan −1 x ⋅

x2 1 − 2 2

∫x

2

⋅

1 2 1 1 1 x tan −1 x − dx + dx 2 2 2 1+ x 2

∫

Check: Let y =

∫

dx 1+ x 2

=

=

1 2 1 x2 x tan −1 x − dx 2 2 1+ x 2

∫

=

1 2 1 x tan −1 x − 2 2



1

1 2 1 1 x tan −1 x − x + tan −1 x + c 2 2 2

1 2 1 1 1 1 1 1 x2 1 x tan −1 x − x + tan −1 x + c , then y ′ = ⋅ 2 x ⋅ tan −1 x + +0 − + ⋅ 2 2 2 2 1+ x 2 2 2 1+ x 2 2

x tan −1 x +

1 1 1 x2 1 1 1 1 x 2 +1 1 ⋅ + ⋅ − = x tan −1 x + ⋅ − = x tan −1 x + − = x tan −1 x 2 2 2 1+ x 2 2 1+ x 2 2 2 1+ x 2 2

c. Given ∫ sin x sin 7 x dx let u = sin x and dv = sin 7 x dx then du = cos x dx and ∫ dv = ∫ sin 7 x dx which 1 7

implies v = − cos 7 x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain 1

1

1

1

∫ sin x sin 7 x dx = sin x ⋅ − 7 cos 7 x + 7 ∫ cos 7 x ⋅ cos x dx = − 7 sin x cos 7 x + 7 ∫ cos x ⋅ cos 7 x dx

(1 )

To integrate ∫ cos x ⋅ cos 7 x dx use the integration by parts method again, i.e., let u = cos x and Hamilton Education Guides



∫ 1 − 1 + x 2 dx

305

Calculus I

5.1 Integration by Parts

dv = cos 7 x dx

1 7

then du = − sin x dx and ∫ dv = ∫ cos 7 x dx which implies v = sin 7 x . Therefore, 1

1

∫ cos x ⋅ cos 7 x dx = cos x ⋅ 7 sin 7 x + 7 ∫ sin 7 x ⋅ sin x dx

=

1 1 cos x sin 7 x + sin x sin 7 x dx 7 7

(2)

∫

Combining equations ( 1 ) and ( 2 ) together we have 1

1

1

1

1

∫ sin x sin 7 x dx = − 7 sin x cos 7 x + 7 ∫ cos x ⋅ cos 7 x dx = − 7 sin x cos 7 x + 49 cos x sin 7 x + 49 ∫ sin x sin 7 x dx 1 sin x sin 7 x dx 49

∫

Taking the integral

∫ sin x sin 7 x dx

to the left hand side and simplifying we have

1 49 1 49 sin x cos 7 x + ⋅ cos x sin 7 x + c 7 48 49 48

= − ⋅

Check: Let y = −

= −

7 1 sin x cos 7 x + cos x sin 7 x + c 48 48

7 49 1 7 cos x ⋅ cos 7 x + sin 7 x ⋅ sin x cos x sin 7 x + c , then y ′ = − sin x cos 7 x + 48 48 48 48

−

1 7 49 − 1 1 49 sin x ⋅ sin 7 x + cos 7 x ⋅ cos x + 0 = sin 7 x ⋅ sin x sin 7 x ⋅ sin x − sin x ⋅ sin 7 x = 48 48 48 48 48

=

48 sin 7 x ⋅ sin x = sin 7 x sin x 48

d. Given ∫ cos 5 x cos 7 x dx let u = cos 5 x and dv = cos 7 x dx then du = −5 sin 5 x dx and ∫ dv = ∫ cos 7 x dx 1 7

which implies v = sin 7 x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ cos 5x cos 7 x dx

1 7

= cos 5 x ⋅ sin 7 x +

5 sin 7 x ⋅ sin 5 x dx 7

∫

=

1 5 cos 5 x sin 7 x + sin 5 x ⋅ sin 7 x dx 7 7

(1 )

∫

To integrate ∫ sin 5 x ⋅ sin 7 x dx use the integration by parts method again, i.e., let u = sin 5 x and dv = sin 7 x dx

1 7

then du = 5 cos 5 x dx and ∫ dv = ∫ sin 7 x dx which implies v = − cos 7 x . Therefore, 1

5

1

5

∫ sin 5x ⋅ sin 7 x dx = sin 5x ⋅ − 7 cos 7 x + 7 ∫ cos 7 x ⋅ cos 5x dx = − 7 sin 5x cos 7 x + 7 ∫ cos 5x cos 7 x dx

(2)

Combining equations ( 1 ) and ( 2 ) together we have

∫ cos 5x cos 7 x dx =

1 5 5 25 1 cos 5 x sin 7 x + sin 5 x ⋅ sin 7 x dx = cos 5 x sin 7 x − cos 5 x cos 7 x dx sin 5 x cos 7 x + 7 7 49 49 7

Taking the integral

∫ cos 5x cos 7 x dx Check: Let y =

=

∫

25 cos 5 x cos 7 x dx 49

∫

∫

to the left hand side and simplifying we have

1 49 5 49 ⋅ cos 5 x sin 7 x − ⋅ sin 5 x cos 7 x + c 7 24 49 24

=

7 5 cos 5 x sin 7 x − sin 5 x cos 7 x + c 24 24

7 35 5 49 cos 5 x sin 7 x − sin 5 x ⋅ sin 7 x + sin 5 x cos 7 x + c , then y ′ = − cos 7 x ⋅ cos 5 x 24 24 24 24

Hamilton Education Guides

306

Calculus I

e. Given

5.1 Integration by Parts

−

49 25 49 − 25 35 25 cos 5 x cos 7 x − cos 5 x cos 7 x sin 5 x ⋅ sin 7 x + 0 = cos 5 x cos 7 x = cos 5 x ⋅ cos 7 x + 24 24 24 24 24

=

24 cos 5 x cos 7 x = cos 5 x cos 7 x 24 x −x dx

∫5e

x −x e dx 5

=

and dv = e − x dx then du = dx and ∫ dv = ∫ e − x dx which implies

x dx ⋅ −e − x + e − x ⋅ 5 5

∫

Check: Let y = −

f. Given

1 5

x 5

Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

v = −e − x .

∫

let u =

= −

xe − x e − x − +c 5 5

= −

e− x ( x + 1) + c 5

xe − x e − x xe − x e − x xe − x e − x − + c , then y ′ = − + +0 = + 5 5 5 5 5 5

∫ x sinh 3x dx let u = x

and dv = sinh 3x dx then du = dx and ∫ dv = ∫ sinh 3x dx which

1 3

implies v = cosh 3x dx . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ x sinh 3x dx

1 3

1 3

= x ⋅ cosh 3x − ∫ cosh 3x ⋅ dx = 1 3

1 1 x cosh 3 x − cosh 3 x ⋅ dx 3 3

∫

1 9

Check: Let y = x cosh 3x − sinh 3x + c , then y ′ = 1 + x sinh 3 x − cosh 3 x 3

=

1 1 x cosh 3 x − sinh 3 x + c 9 3

1 1 1 1 cosh 3 x + ⋅ 3 x sinh 3 x − ⋅ 3 cosh 3 x + 0 = cosh 3 x 3 3 9 3

= x sinh 3x

Section 5.1 Practice Problems – Integration by Parts 1. Evaluate the following integrals using the integration by parts method.

d.

∫ xe dx = ∫ x sin 5x dx =

g.

∫ cos ( ln x ) dx

j.

∫xe

a.

4x

− ax

=

dx =

x

f.

∫ ( 5 − x ) e dx = 3 3x ∫ x e dx =

=

i.

∫ ln x dx =

sin 3 x dx =

l.

∫e

c.

∫ arc tan x dx = −2 x ∫ e cos 3x dx =

i.

∫ 3 cos

l.

∫ 5 sinh 7 x dx

b.

∫ 2 cos x dx

e.

∫x

h.

∫ 3 tan

k.

∫e

=

c.

3 − x dx =

x

x

−1

x dx

5x

5

x

cos 5 x dx =

2. Evaluate the following integrals using the integration by parts method. b.

d.

∫ x sec x dx = 3 ∫ sin 5x dx =

e.

∫ arc sin 3 y dy = 2 ∫ x cos x dx =

g.

∫ x ( 5x − 1)

h.

∫ x csc

2

x dx =

j.

∫ sinh

k.

∫ x sec

2

10 x dx

a.

2

−1

3

x dx

dx =

=

Hamilton Education Guides

=

f.

2 x

−1

5 x dx

= =

307

Calculus I

5.2

5.2 Integration Using Trigonometric Substitution

Integration Using Trigonometric Substitution

Many integration problems involve radical expressions of the form a2 − b2 x2

a2 + b2 x2

b2 x2 − a2

In such instances we can use a trigonometric substitution by letting x=

a sin t b

x=

a sec t b

respectively to obtain

= a 2 − a 2 sin 2 t = a

(1− sin t )

= a cos 2 t = a cos t

tan 2 t

= a 2 + a 2 tan 2 t = a

(1+ tan t )

= a sec 2 t = a sec t

sec 2 t − a 2

= a 2 sec 2 t − a 2 = a

( sec

= a tan 2 t = a tan t

a2 − b2 ⋅

a2 + b2 x2 =

a2 + b2 ⋅

b2 x2 − a2 =

b2 ⋅

b2

x=

sin 2 t

a2 − b2 x2 =

a2

a tan t b

a2 b2 a2 b

2

2

2

2

)

t −1

Notice that using trigonometric substitution result in elimination of the radical expression. This in effect reduces the difficulty of solving integrals with radical expressions. Reminder 1:

bx a a b b x a x = tan t , then t = tan −1 a b b x a x = sec t , then t = sec −1 a b

Given x = sin t , then t = sin −1

for −1 ≤ x ≤ 1 and −

Given

for all x and −

Given

π 2

π 2 t

≤t ≤

π 2

π 2

for x ≥ 1 or x ≤ −1 and 0 ≤ t

π 2

or π ≤ t

3π 2

Reminder 2: In solving this class of integrals the integrand in the original variable may be obtained by the use of a right triangle. For example, in a right triangle •

sin t =

opposite x = . hypotenuse a

Therefore, using the Pythagorean theorem, the adjacent side (w) is equal to

a = x 2 + w2 ; a 2 = x 2 + w2 ; w2 = a 2 − x 2 ; w = a 2 − x 2

•

cos t =

adjacent x = hypotenuse a

. Therefore, the opposite side (w) is equal to

a = x 2 + w2 ; a 2 = x 2 + w2 ; w2 = a 2 − x 2 ; w = a 2 − x 2

•

tan t =

opposite x = adjacent a

. Therefore, the hypotenuse (w) is equal to w = a 2 + x 2

•

cot t =

adjacent x = opposite a

. Therefore, the hypotenuse (w) is equal to w = a 2 + x 2

Hamilton Education Guides

308

Calculus I

•

5.2 Integration Using Trigonometric Substitution

hypotenuse x = adjacent a

sec t =

. Therefore, the opposite side (w) is equal to

x = a 2 + w2 ; x 2 = a 2 + w2 ; w2 = x 2 − a 2 ; w = x 2 − a 2

•

hypotenuse x = opposite a

csc t =

. Therefore, the adjacent side (w) is equal to

x = a 2 + w2 ; x 2 = a 2 + w2 ; w2 = x 2 − a 2 ; w = x 2 − a 2

Let’s integrate some integrals using the above trigonometric substitution method: Example 5.2-1: Use trigonometric substitution to evaluate the following indefinite integrals: a. d.

∫ x2

dx 4 − x2

dx

∫ ( 9 + x 2 )2

=

b.

=

e.

x 2 dx

∫

25 − x 2 x2

∫

2

x −1

=

dx

=

1+ x 2

c.

∫

f.

∫ x4

x2

dx =

dx 2

x −1

=

Solutions: a. Given

∫ x2

dx 4− x

2

(

let x = 2 sin t , then dx = 2 cos t dt and 4 − x 2 = 4 − 4 sin 2 t = 4 1 − sin 2 t

)

= 4 cos 2 t = 2 cos t . Substituting these values back into the original integral we obtain:

∫ x2 =

dx 4 − x2

=

1 cos t − +c 4 sin t

2 cos t dt

∫ (2 sin t )2 ⋅ 2 cos t

=

1 − 4

4− x 2 2 x 2

+c

=

2 cos t

∫ 4 sin 2 t ⋅ 2 cos t dt

=

1

∫ 4 sin 2 t dt

 1 4− x2 1  4 − x 2 ⋅ 2/  −  + c =  x 4 2/ ⋅ x   

= −  4

=

1 4

1 csc 2 t dt 4

∫

= − cot t + c

 +  c 

Check: To check the answer we start with the solution and find its derivative. The derivative should match with the integrand, i.e., the algebraic expression inside the integral. Note that not all the steps in finding the derivative is given. At this level, it is expected that students are able to work through the details that are not shown (review differentiation techniques described in Chapters 2 and 3). Let y = −

1 4

4− x x

2

+c ,

then y ′ = −

=

1 − 4

Hamilton Education Guides

4− x 2

x

2

(

1 4

− x 2 − 4− x 2

− x 2 − 4− x 2 ⋅ 4− x 2

=

1 − 4

4− x 2 2

x

−2 x 2 4− x

2

⋅ x − 1⋅ 4 − x 2 x

) = −

2

+0

= −

1 4

−2 x 2 2 4− x

2

−

4− x 2 1

x2

1 − x2 − 4 + x2 1 1 −4 = − = 4 x2 4 − x2 4 x2 4 − x2 x2 4 − x2

309

Calculus I

b. Given

5.2 Integration Using Trigonometric Substitution

(

x 2 dx

∫

25 − x

let x = 5 sin t , then dx = 5 cos t dt and 25 − x 2 = 25 − 25 sin 2 t = 25 1 − sin 2 t

2

)

= 25 cos 2 t = 5 cos t . Substituting these values back into the original integral we obtain: x 2 dx

∫

=

25 − x 2

25 sin 2 t ⋅ 5 cos t dt 5 cos t

∫

=

25 2

=

25  −1 x  25  x 25 − x 2 −   sin 2  5  2  25

=

25 2

1+ x 2

∫

x

2

∫

 +c 

=

=

25 25 t − sin t cos t + c 2 2

25 then y ′ =

1  25 − x 2 x2 − ⋅ −  2 2 25 − x 2 25 − x 2 5 

dx

25

=

−

2 25 − x 2

1

2

5

   

∫

=

25  −1 x  25  x 25 − x 2 ⋅ −  sin 2  5  2  5 5 

25 − 2 x 2 2 25 − x 2

1−

x2 25

1 1 − 2x x ⋅ − 25 − x 2 + ⋅   5 2 2 25 − x 2 2  

∫

(

=

25 − 25 + 2 x 2 2 25 − x 2

=

2x 2 2 25 − x 2

)

tan t

= ln sec t + tan t + ∫

   

=

=

25 2 25 − x 2

x2 25 − x 2

let x = tan t , then dx = sec 2 t dt and 1 + x 2 = 1+ tan 2 t = sec 2 t = sec t

sec t ⋅ 1 + tan 2 t dt 2

)

(

  2 25 − x 2 − 2 x 2 25  =  −   2 2 2 25 − x  4 25 − x 

Substituting these values back into the original integral we obtain =

 +c  

25  −1 x  x 25 − x 2 + c  sin − 2  5 2

25 x x sin −1 − 25 − x 2 + c , 2 5 2

 50 − 4 x 2 −  2  4 25 − x

c. Given

∫

1 − cos 2t 25 sin 2 t dt = 25 sin 2 t dt = 25 dt 1 2

25  1   t − sin 2t  + c 2  2 

∫ ( 1 − cos 2t ) dt =

Check: Let y =

=

=

1 cos t 2

sin t cos 2 t

∫

sec t + sec t tan 2 t

dt

2

tan t

dt

=

= ln sec t + tan t + ∫

∫

sec t tan 2 t 2

tan t cos 2 t 2

cos t sin t

dt

dt +

∫

sec t

1+ x 2 x2

∫ tan 2 t dt

=

dx =

Hamilton Education Guides

1 + x2 + x −

tan 2 t sec t

∫ sec t dt + ∫ tan 2 t dt

= ln sec t + tan t + ∫

= ln sec t + tan t + ∫ sin −2 t cos t dt = ln sec t + tan t + sin −1 t + c = ln sec t + tan t + = ln sec t + tan t − csc t + c = ln

∫

sec t ⋅ sec 2 t dt

cos t sin 2 t

dt

1 +c sin t

1 + x2 +c x

310

Calculus I

5.2 Integration Using Trigonometric Substitution

=

x

 2x + 2 1 + x 2  ⋅ 2 1 + x + x  2 1 + x 2

1 1+ x 2 dx

∫ ( 9 + x 2 )2

(

+

1 x 2 1+ x 2

=

(

) =

(

x 2 1+ x 2

)

x2 2 1+ x

2

⋅ x − 1+ x 2 x2

2 x + 1 + x 2   2 2    x −1− x ⋅ −  x 2 1+ x 2 1+ x 2 + x 2 1+ x 2  1

 

x 2 1+ x 2 + 1+ x 2

=

  1 2x + 1 − ⋅   1+ x 2 + x  2 1+ x 2 

+ c , then y ′

  2 2   x − 1+ x −   2   x 1+ x 2  

1

=

d. Given

1+ x

1+ x 2 + x −

Check: Let y = ln

2

(

1+ x 2 1+ x 2

=

(

x 2 1+ x 2

)

)=

    

1 + x2 x2

let x = 3 tan t , then dx = 3 sec 2 t dt and 9 + x 2 = 9 + (3 tan t )2 = 9 + 9 tan 2 t

)

= 9 1 + tan 2 t = 9 sec 2 t Substituting these values back into the original integral we obtain dx

∫ ( 9 + x 2 )2 =

3 sec 2 t dt

∫ ( 9 sec 2 t )2

=

=

3 sec 2 t dt

∫ 81sec 2 t sec 2 t

=

1 27

dt

∫ sec 2 t

=

1 cos 2 t dt 27

∫

 1  1 1  (t + sin t cos t ) + c = 1  tan −1 x + x ⋅ 3  t + sin 2t  + c = 54  2 54 3 54  9 + x2 9 + x2 

Check: Let

+

= e. Given

∫

1  −1 x 3x  tan + 54  3 9 + x2

1 27 + 3 x 2 − 6 x 2 ⋅ 2 54 9 + x2

(

(

18

18 9 + x

x2 x 2 −1

)

)

2 2

=

=

  + c  1 ⋅ 54

1 then y ′ = ⋅ 54

3

(9 + x ) 2

+

1 31 + 

2

x 9

1 27 − 3 x 2 ⋅ 2 54 9 + x2

)

(

  

+

=

(

=

1 1 ⋅ 27 2

∫ (1 + cos 2t ) dt

  + c = 1  tan −1 x + 3 x  54  3 9+ x2 

)

1 9 1 3 9 + x 2 − 2 x ⋅ 3x = ⋅ ⋅ 2 54 54 9 + x2 3 9 + x2

(

1

(

18 9 + x 2

)

+

(

)

9 − x2

(

18 9 + x 2

)

2

=

)

9 + x2 + 9 − x2

(

18 9 + x 2

)

2

1

(9 + x )

2 2

x 2 −1 =

dx let x = sec t , then dx = sec t tan t and

Substituting these values back into the original integral we obtain

sec 2 t − 1 =

∫

x2 x 2 −1

tan 2 t = tan t

dx =

∫

sec 2 t sec t tan t dt tan t

= ∫ sec 3 t dt = ∫ sec 2 t sec t dt . Let u = sec t and dv = sec 2 t , then du = sec t tan t and v = tan t . Using the substitution formula uv − ∫ v du the integral ∫ sec 2 t sec t dt can be rewritten as

∫ sec

3

∫

∫

(

)

t dt = sec t tan t − sec t tan 2 t dt = sec t tan t − sec t sec 2 t − 1 dt = sec t tan t −

Hamilton Education Guides

  + c 

∫ ( sec

3

)

t − sec t dt

311

Calculus I

5.2 Integration Using Trigonometric Substitution

= sec t tan t − ∫ sec 3 t dt + ∫ sec t dt Note that ∫ sec 3 t dt = sec t tan t − ∫ sec 3 t dt + ∫ sec t dt therefore by moving − ∫ sec 3 t dt to the left hand side of the equality we obtain

∫ sec

∫

∫

∫

∫

t dt + sec 3 t dt = sec t tan t + sec t dt thus 2 sec 3 t dt = sec t tan t + sec t dt and

3

(

1 sec t tan t + sec t dt 2

∫

) = 12 ( sec t tan t + ln

1 2

)+ c

sec t + tan t

=



= f. Given

=

)

1 4x 2 − 2 + 2 4x 2 = = 2 2 x 2 −1 4 x 2 −1

∫ x4

dx x −1

=

   2 x −1 

1  4 x 2 − 2 + 2  2 x 2 −1 

1

x2 x 2 −1 x 2 −1

let x = sec t , then dx = sec t tan t dt and

2

t dt

 x 2 − 1 + ln x + x 2 − 1  + c 

2 x + x 2 − 1  1 1 2 x 2 −1 + 2x 2 1   + 2 2 2 2 2 2 x −1 2 x −1 x + x −1

(

3

1  2 2 x 2  1 1 x −1 + +   2 2 x 2 −1  2 x + x 2 −1 



Check: Let y =  x x 2 − 1 + ln x + x 2 − 1  + c , then y ′ =

  2x  × 1 +   2  2 x −1 

1  x 2

∫ sec

= sec 2 t − 1 = tan 2 t = tan t

Substituting these values back into the original integral we obtain

∫ x4 =

dx 2

x −1

sec t tan t dt

∫ sec 4 t tan t

=

∫ (cos t − sin

2

Check: Let y = x 2 − x 2 +1

= =

)

t cos t dt =

x 2 −1

x2

=

dt

∫ sec 3 t

=

∫ cos t dt − ∫ sin

2

∫ cos

3

t dt =

t cos t dt =

3

x − 1 1  x 2 − 1  −  + c , then y ′ = 3 x x    2

 x 2 −1  ⋅ −  x2   

x 2 − x 2 +1 x 4 x 2 −1

Hamilton Education Guides

=

x 2 − x 2 +1 x 2 −1

x2

=

2

t cos t dt =

1 sin t − sin 3 t + c 3 2 x2 2

2 x −1

− x 2 −1 x

2

2 ∫ (1 − sin t ) cos t dt 3

=

  x2 −1 1  x2 −1  −   + c 3 x x   2

3  x 2 − 1  ⋅ −  3 x   

 x 2 −1  1 ⋅ −  x2  2 2 2  x x −1 x −1 

1 x2

∫ cos

=

1 x 2 x 2 −1

2 x2 2

2 x −1

− x 2 −1 x2

−

x 2 −1 x 4 x 2 −1

1 x4 x2 −1

312

Calculus I

5.2 Integration Using Trigonometric Substitution

Example 5.2-2: Use trigonometric substitution to evaluate the following indefinite integrals: a. d. g.

∫

dx

=

b.

∫x

2

dx =

e.

∫

a 2 − x 2 dx

=

h.

∫

9 − x 2 dx

(1 − )

3 x2 2

x2

∫

4+ x

2

x2

∫

9− x

dx

2

4 − x 2 dx

=

=

=

dx

c.

∫

f.

∫

i.

∫ x2

(

)

=

3

4 + 9x 2 2

x 2 − a 2 dx

1 4 + x2

=

dx

=

Solutions: a. Given

∫

∫

(

dx

)

let x = sin t , then dx = cos t dt and 1 − x 2 = 1− sin 2 t = cos 2 t . Therefore,

3

1− x 2 2

dx

∫

=

(1 − )

3 x2 2

cos t dt

(cos t )

Check: Let y = b. Given

∫x

2

(

4 1 − sin 2 t

∫x

2

2

cos t

∫ cos 3 t dt

=

1

∫ cos 2 t dt = ∫ sec

1⋅ 1 − x 2 −

x

+ c , then y ′ =

1− x 2

4 − x 2 dx

)

=

3 2

1− x

−2 x 2 1− x

2

2

x

=

1− x2

+c

2− 2 x 2 + 2 x 2

⋅x

2 1− x 2

=

2

sin t +c cos t

t dt = tan t + c =

1− x

=

2

(

2

2 1−

)

3 x2 2

=

1

(1 − x )

3

2 2

let x = 2 sin t , then dx = 2 cos t dt and 4 − x 2 = 4 − (2 sin t )2 = 4 − 4 sin 2 t

= 4 cos 2 t = 2 cos t . Therefore,

4 − x 2 dx

(

=

∫ 4 sin

)

= 4∫ 1 − cos 2 2t dt =

2

∫

t ⋅ 2 cos t ⋅2 cos t dt = 16 sin 2 t cos 2 t dt = 16

∫ 4 dt − 4∫ cos

2

2t dt =

1

∫ 4 dt − 4∫ 2 ( 1 + cos 4t ) dt

=

∫ 4 dt − ∫ 2 dt − 2∫ cos 4t dt

(

4 2

1 2

1

∫ 4 (1 − cos 2t ) (1 + cos 2t ) dt

)

= 4t − 2t − sin 4t + c = 4t − 2t − sin 2t cos 2t + c = 2t − 2 (sin t cos t ) cos 2 t − sin 2 t + c x

x

= 2 sin −1 − 2  ⋅ 2 2 

4 − x2 2

Check: Let y = 2 sin −1

=

2 4 − x2

−

 2 2 x   4 − x − x  + c 2 sin −1 − 4 − x 2 =   4 2 4  

x − 4 − x2 2

 2x − x 3   4 

(

)(

 2  + c , then y ′ =   2 1−

− 2 x 2 + x 4 + 2 − 3x 2 4 − x 2

Hamilton Education Guides

4 4 − x2

)=

2 4 − x2

−

 2 x − x3   +c   4  

x2 4

−

(

− 2x 2x − x 3 8 4 − x2

) + (2 − 3x )(4 − x ) 2

1

2 2

− 2 x 2 + x 4 + 8 − 14 x 2 + 3 x 4 4 4 − x2

313

Calculus I

5.2 Integration Using Trigonometric Substitution

8 − 4 x 4 + 16 x 2 − 8

=

c. Given

∫

4 4− x dx

( 4 + 9x )

∫

=

3

2

2 2

(

dx 4 9

− 4 x 4 + 16 x 2

=

+ x2

4 4− x

)

3 2

2

− 4 x 4 + 16 x 2 4 4− x

2

2 3

4 − x2

⋅

4− x

2

(

)

4 4 − x2 x2 4 − x2

=

4 (4 − x ) 2

= x2 4 − x2

2 3 

2 3

let x = tan t , then dx = sec 2 t dt and 4 + 9 x 2 = 4 + 9 ⋅  tan t 

2



)

(

4 9

=

= 4 + 9 ⋅ tan 2 t = 4 + 4 tan 2 t = 4 1 + tan 2 t = 4 sec 2 t . Therefore,

∫ =

(

dx

)

3 4 + 9x 2 2

1 12

dt

∫ sec t

∫

=

=

(

2 3

sec 2 t

)

3 4 sec 2 2

1 cos t dt 12

∫

dt

=

=

∫

2 3

3 42

sec 2 t 2× 3 sec 2

1 sin t + c 12

=

dt

=

t

1 12

2 3

sec 2 t

∫ 2 4 3 sec 3 t dt

3x 4 + 9x

2

+c

=

2

Check: Let y = d. Given

∫

(

= 4 1 + tan 2 t

∫

x 2 dx 4+ x

2

x 4 4 + 9x

x 2 dx 4+ x

=

∫

1⋅ 4 4 + 9 x − 4 ⋅

2

)

2

+c,

then y ′ =

(

16 4 + 9 x

=

dt

∫ 2 64 sec t

x 4 4 + 9x 18 x

2 4+9 x 2 2

2 3

2 3

dt

∫ 8 sec t

+c

2

)

(

4 4 +9 x 2 −36 x 2

⋅x

)

=

=

(

4+9 x 2

16 4 + 9 x

2

)

=

1

( 4 + 9x )

3

2 2

let x = 2 tan t , then dx = 2 sec 2 t dt and 4 + x 2 = 4 + (2 tan t )2 = 4 + 4 tan 2 t = 4 sec 2 t = 2 sec t . Therefore,

4 tan 2 t ⋅ 2 sec 2 t dt 2 sec t

(

(

)

)

= 4∫ tan 2 t sec t dt = 4∫ sec 2 t − 1 sec t dt = 4∫ sec 3 t − sec t dt

= 4∫ sec 3 t dt − 4∫ sec t dt . To solve ∫ sec 3 t dt = ∫ sec 2 t ⋅ sec t dt use substitution method by letting u = sec t and dv = sec 2 t then du = sec t tan t dt and v = tan t . Using the substitution formula uv − ∫ v du we obtain ∫ sec 2 t ⋅ sec t dt = sec t tan t − ∫ tan t ⋅ sec t tan t dt

(

)

(

)

= sec t tan t − ∫ sec t tan 2 t dt = sec t tan t − ∫ sec t sec 2 t − 1 dt = sec t tan t − ∫ sec 3 t − sec t dt = sec t tan t − ∫ sec 3 t dt + ∫ sec t dt . Again at this point we know that ∫ sec 3 t dt = sec t tan t − ∫ sec 3 t dt + ∫ sec t dt bringing − ∫ sec 3 t dt into the left hand side of the equation we obtain ∫ sec 3 t dt + ∫ sec 3 t dt = sec t tan t + ∫ sec t dt . Therefore 2∫ sec 3 t dt = sec t tan t + ∫ sec t dt thus ∫ sec 3 t dt =

[

1 sec t tan t + sec t dt 2

∫

] = 12 [ sec t tan t + ln sec t + tan t

].

Now substituting this

value into 4∫ sec 3 t dt − 4∫ sec t dt we have

Hamilton Education Guides

314

Calculus I

5.2 Integration Using Trigonometric Substitution

∫

∫

1 sec t tan t + ln sec t + tan t 2

[

]− 4∫ sec t dt

= 2 [ sec t tan t + ln sec t + tan t

4 sec 3 t dt − 4 sec t dt

= 4⋅

− 4 ln sec t + tan t + c

= 2 sec t tan t + 2 ln sec t + tan t − 4 ln sec t + tan t + c = 2 sec t tan t

Check: Let y =

x 4 + x2 x + 4 + x2 1 2x 2 − 2 ln + c , then y ′ =  4 + x 2 + 2 2 2 2 4 + x2 

 2 1  2x −2 ⋅ ⋅ 1 +  x + 4 + x2 2  2 4 + x2  

=

∫

e. Given

) x +

4 + x 2  − 2 4 + x 2 − 2 x x − 4 + x2  ⋅  x + 4 + x2  4 + x2 x − 4 + x2    

(

x 4 − 4x 2 − x 4

=

− 4 4 + x2

(

= a 2 1− sin 2 t

)

∫

=

a2 ( t + sin t cos t ) + c 2

Check: Let y =

− 4 4 + x2

a cos t ⋅ a cos t dt =

4 + x2

=

a2 2

∫

a 2 cos 2 t dt =

 2 2  sin −1 x + x ⋅ a − x  a a a 

a2 x x a2 − x2 sin −1 + +c, 2 a 2

a2 a a2 − x2 − x2 ⋅ + 2 a a2 − x2 2 a2 − x2

a2 − x2 2

x2

=

= a 2 cos 2 t = a cos t . Therefore,

∫

=

− 4x 2

)

let x = a sin t , then dx = a cos t dt and a 2 − x 2 = a 2 − a 2 sin 2 t

a 2 − x 2 dx

=

 2 4 + x 2 + 2x −      x + 4 + x2  4 + x2  

4 + x 2  − 2 4 + x 2 − 2 x x4 − x3 4 + x2 − x3 4 + x2 − x2 4 + x2 x − 4 + x2  = ⋅  x + 4 + x2  4 + x2 − 4 4 + x2 x − 4 + x2    

a 2 − x 2 dx

=

(

4 + x 2  − 2 4 + x 2 − 2 x 2 + x2  =  x + 4 + x2  4 + x2    

   

(2 + x ) x + 2

=

2     + 0 = 1  4 + 2x  2  4 + x2  

(2 + x ) x + 2

=

x 4+ x2 x + 4+ x2 4 + x2 x +c − 2 ln + +c = 2 2 2 2

4 + x2 x ⋅ − 2 ln 2 2

− 2 ln sec t + tan t + c = 2 ⋅

]

a −x

2

⋅

Hamilton Education Guides

a2 − x2 2

a −x

2

=

(a

2

=

− x2 2

)

a2 2

a2 2 a2 − x2

a2 − x2

a −x

a2  1   t + sin 2t  + c 2  2 

 2  x a2 − x2 a −1 x +c + sin =  2  2 a  

then y ′ =

2

∫ ( 1 + cos 2t ) dt =

a2 2 +

1 1−

( ax )2

a 2 − 2x 2 2 a2 − x2

⋅

   + c 

1 1 − 2x 2 + a2 − x2 + a 2 2 a2 − x2

=

2a 2 − 2 x 2 2 a2 − x2

=

a2 − x2 a2 − x2

= a2 − x2 315

Calculus I

f. Given

5.2 Integration Using Trigonometric Substitution

∫

x 2 − a 2 dx

=

a 2 tan 2 t

∫

x 2 − a 2 dx

∫ a tan t ⋅ a sec t tan t dt

=

=

∫a

a2 (tan t sec t + ln sec t + tan t 2

∫

  +c  

) − a 2 ln

=

a2 a2 x x2 − a2 − ln a + c ln x + x 2 − a 2 + 2 2 2

Check: Let y =

g. Given

∫

x2 − a2 + x2 2

2 x −a

9− x

2

dx

2

−

 ⋅ 1 +  2

2x

x 2 −a 2

a2 2

2 x −a

=

2

=

sec t + tan t + c

∫

a2 (tan t sec t − ln sec t + tan t 2

)+ c

x a2 x + x2 − a2 x2 − a2 − +c ln a 2 2

=

x 2

x2 −a2 −

x a2 x2 − a2 − ln x + x 2 − a 2 + c , 2 2

a2 1 ⋅ 2 x + x2 − a2

x2

=

)

sec t tan 2 t dt = a 2 sec t sec 2 t − 1 dt = a 2 sec 3 t dt

 x a2  x x2 − a2 − ln + ⋅  a a 2 a 

x2 − a2 a

(

∫

2

=

=

a 2 sec 2 t − a 2

= a tan t . Therefore,

− a 2 sec t dt =

−

x2 − a2 =

let x = a sec t , then dx = a sec t tan tdt and

a2 ln x + x 2 − a 2 + c 2

  2 2  2 x −a 

1 then y ′ =  x 2 − a 2 + 2

2x 2



 x2 − a2 + x2 a2 1 x + x2 − a2 +0 = − ⋅ ⋅  2 x + x2 − a2  2 x2 − a2 x2 − a2 2 x 2 − 2a 2 2

2 x −a

2

x2 − a2

=

2

x −a

=

2

x2 − a2 2

x −a

2

⋅

x2 − a2 2

x −a

2

let x = 3 sin t , then dx = 3 cos t dt and 9 − x 2 = 9 − 9 sin 2 t =

x2 − a2

=

(

9 1 − sin 2 t

)

= 9 cos 2 t = 3 cos t . Substituting these values back into the original integral we obtain:

∫

x 2 dx 9 − x2

=

∫

9 sin 2 t ⋅ 3 cos t dt 3 cos t

=

9 2

=

9  −1 x  9  x  9 − x2  + c  sin −  2 3 29 

∫ ( 1 − cos 2t ) dt =

9 2

∫

9 sin 2 t dt 1

9 1   t − sin 2t  + c 2 2 

x 3

Check: Let y = sin −1 −

Hamilton Education Guides

=

=

=

= 9∫ sin 2 t dt = 9∫

9 9 t − sin t cos t + c 2 2

=

1 − cos 2t dt 2

9  −1 x  9  x 9 − x 2 ⋅ −  sin 2 3  2  3 3 

 +c  

9  −1 x  x 9− x2 + c  sin − 2 3 2

x 9 − x2 + c , 2

then y ′ =

9 2

1 1−

x2 9

1 1 ⋅ − 3 2 

9 − x2 +

x ⋅   2 9 − x2 2  − 2x

316

Calculus I

5.2 Integration Using Trigonometric Substitution

 1  ⋅ − 9 − x 2 3 

=

    

  18 − 4 x 2 −  4 9 − x2 

h. Given

∫

9 − x 2 dx

=

9 cos 2 t

∫

9 − x 2 dx

=

2 9

=

−

2 9 − x2

9 − 2x 2

=

=

2 9 − x2

(

)

  2 9 − x 2 − 2x 2 −  4 9 − x2 

9 2 9 − x2 9 − 9 + 2x 2 2 9 − x2

=

2x 2

9 − x2

=

9 2 9 − x2

x2

=

2 9 − x2

    

9 − x2

9 − 9 sin 2 t

=

=

(

9 1 − sin 2 t

)

= 9 cos t . Therefore,

Check: Let

∫ 3 cos t ⋅ 3 cos t dt = ∫ 9 cos

=

3

1 4+ x

dx

2

9 − x2 2

9 − x2 − x2 9 − x2

2

9 − x2

=

9 − x2

2

+

9 − x2

9 − x2

∫ x2

=

t dt =

9 2

∫ ( 1 + cos 2t ) dt =

1

2

9 9 − x2

2

+

( 3x )2

9 − x2

(9 − x ) =

9 − x2

1−

9 − 2x 2 2

9 1   t + sin 2t  + c 2 2 

=

 9− x2  +c 2  

 x 9 −1 x  2 sin 3 +  

9 then y ′ =

9 − x2

⋅

9 − x2

+c,

=

2

9 − x 2  +c 3  

 9  −1 x x + ⋅ sin 2  3 3 

x x 9 y = sin −1 + 2 3

9 ⋅ 2 3

= i. Given

x2

let x = 3 sin t , then dx = 3 cos t dt and

9 ( t + sin t cos t ) + c 2

=

  −  2 9 − x 2 

9 − x2

3

9 2

2

9 − x2

9− x

2

=

1 1 ⋅ + 3 2

9 − x2 +

9 + 9 − 2x 2 9 − x2

2

=

− 2x 2 2

9 − x2

18 − 2 x 2 9 − x2

2

(

2 9 − x2

=

2

)

9 − x2

9 − x2

=

let x = 2 tan t , then dx = 2 sec 2 t dt and 4 + x 2 = 4 + 4 tan 2 t = 2 1 + tan 2 t

= 2 sec 2 t = 2 sec t . Therefore, dx

∫ x2 =

1 4

4 + x2

∫

cos t u2

⋅

=

dx

du cos t

Check: Let

2 sec 2 t

∫ 4 tan 2 t ⋅ ( 2 sec t )

=

1 4

1 y=− 4

1

∫ u2

du

4+ x x

2

dt

1 4

Hamilton Education Guides

4+ x 2

x

2

1 4

sec dt

∫ tan 2 t

1 4

= − u −1 + c = −

dt

1 +c 4u 2x

+c ,

then y ′ =

   x 2 − 4 + x 2 ⋅ 4 + x 2    

= −

=

1 − 4

2 4+ x

2

=

1 4

= −

∫

1 cos 2 t ⋅ dt cos t sin 2 t

1 +c 4 sin t

= −

1 4

=

let u = sin t

4+ x2 +c x

1 4

1 − 4

cos t

∫ sin 2 t dt

2 x2

⋅ x − 1⋅ 4 + x 2 x2

=

2 4+ x 2

− 4 + x2 x2

x 2 − 4− x 2

= −

4+ x 2

4x

2

= −

−4 4x 2 4 + x 2

=

1 x2 4 + x2 317

Calculus I

5.2 Integration Using Trigonometric Substitution

The following are additional standard forms of integration that have already been derived. Trigonometric substitution can be used in most of these cases in order to confirm the result. Table 5.2-1: Integration Formulas dx

1.

∫ a2 + x2

3.

∫ a2 − x2

5.

∫

7.

∫x

9.

∫

=

x 1 tan −1 + c a a

2.

∫ ( a 2 + x 2 )2

=

a+x 1 x+a 1 +c = ln +c ln x−a 2a a−x 2a

4.

∫ ( a 2 − x 2 )2

6.

∫

a 2 + x 2 dx =

8.

∫

a2 + x2 dx = x

dx

dx a2 + x2 2

a 2 + x 2 dx =

dx

∫x

a2 + x2

13.

∫

2

15.

∫x

∫

dx

)

a2 + x2

−

a4 x sinh −1 + c a 8

a −x 2

x2

∫x

a2 − x2

21.

∫

2

23.

∫x

25.

∫

27.

∫x ∫

x −a 2

(

dx = − sin −1

dx

dx 2

x2

dx 2

x −a

2

dx 2ax − x 2

x a2 − x2 +c − a x

x + c = ln x + x 2 − a 2 + c a

= cosh −1

(

x 2x 2 − a 2 8

dx = cosh −1

=

)

a + a2 − x2 +c x

1 a

= − ln

x 2 − a 2 dx =

x2 − a2

x +c a

a4 x 1 sin −1 − x a 2 − x 2 a 2 − 2 x 2 + c a 8 8

a 2 − x 2 dx =

a2 − x2

a + a2 + x2 +c x

1 a

= − ln = sin −1

2

19.

29.

(

x a 2 + 2x 2 8

a2 + x2 x a2 + x2 dx = sinh −1 − +c a x x

11.

17.

x + c = ln x + a 2 + x 2 + c a

= sinh −1

x − a

)

x2 − a2 −

a4 x cosh −1 + c a 8

x2 − a2 +c x

x 1 a 1 + c = cos −1 +c sec−1 a x a a  x−a +c  a 

= sin −1 

Hamilton Education Guides

dx

dx

x2

= =

2a

(a

2

2a

2

(a

2

12.

∫ x2

14.

∫

a 2 − x 2 dx =

16.

∫

a2 − x2 dx = x

18.

∫

dx = −

a2 + x2

x2 a2 − x2

= −

20.

∫ x2

22.

∫

x 2 − a 2 dx =

24.

∫

x2 − a2 dx = x

26.

∫

28.

∫ x2

30.

∫

a2 − x2

x2 x2 − a2

=

x2 − a2

a 2 + x 2 dx =

)

+

1 2a

tan −1

x +c a

dx

2

∫ a2 − x2

a +c x

a2 x x a2 + x2 +c sinh −1 + 2 2 a a2 + x2 a2x

+c

a 2 − x 2 − a ln

a + a2 − x2 +c x

a2 x 1 sin −1 − x a 2 − x 2 + c a 2 2

a2 − x2 a2x

+c

x a2 x x2 − a2 − cosh −1 + c 2 2 a

dx =

dx

−x

2

1 2a 3

a2 x x a2 − x2 + sin −1 + c a 2 2

dx =

dx

x

)

+

a 2 + x 2 − a sinh −1

∫

dx

+x

2

x a2 x a2 + x2 + sinh −1 + c 2 2 a

10.

a2 + x2

x

2

=

x 2 − a 2 − a sec −1

x +c a

a2 x x x2 − a2 + c cosh −1 + 2 a 2

x2 − a2 a2x

+c

x 2 a2   a + x2 + ln  x + x 2 + a 2  + c 2 2  

318

Calculus I

5.2 Integration Using Trigonometric Substitution

Section 5.2 Practice Problems – Integration Using Trigonometric Substitution Evaluate the following indefinite integrals: a.

∫

dx x 2 16 − x 2

=

1

b.

d.

∫ (49 + x2 )2 dx =

e.

g.

∫

h.

36 − x 2 dx

=

Hamilton Education Guides

x2

∫ ∫ ∫

9 − x2

dx =

    x2 + 5 x  dx   2   x −1 

(

dx

)

3

9 + 36 x 2 2

=

=

dx

c.

∫

f.

∫

x 2 − 25 dx

=

∫

9 − 4x 2 dx x

=

i.

x 9 + 4x 2

=

319

Calculus I

5.3

5.3 Integration by Partial Fractions

Integration by Partial Fractions

The primary objective of this section is to show that rational functions can be integrated by breaking them into simpler parts. A function F (x ) =

f (x ) , g (x )

where f (x ) and g (x ) are polynomials,

is referred to as a rational function. Depending on the degree of the f (x ) , the function F (x ) is either a proper or an improper rational fraction. •

A proper rational fraction is a fraction where the degree of the numerator f (x ) is less than the degree of the denominator g (x ) . For example, proper rational fractions.

•

x +1 2

x − x−6

,

1 3

x −1

,

1 2

x + 2x − 3

, and

x 2

x +1

are

An improper rational fraction is a fraction where the degree of the numerator f (x ) is at least as large as the degree of the denominator g (x ) . In such cases, long division is used in order to reduce the fraction to the sum of a polynomial and a proper rational fraction. For example, x3 + 2

x2 − x − 6

,

x3

x 2 +1

, and

x 4 − x 3 − x −1 x3 − x2

are improper rational fractions. (See Section 6.3 of

Mastering Algebra – An Introduction for solved problems on long division.) A fraction, depending on its classification of the denominator, can be represented in four different cases. These cases are as follows: CASE I - The Denominator Has Distinct Linear Factors In this case the linear factors of the form ax + b appear only once in the denominator. To solve this class of rational fractions we equate each proper rational fraction with a single fraction of the form

A B , ax + b cx + d

,

C ex + f

, etc. The following examples show the steps as to how this class of

integrals are solved. Example 5.3-1: Evaluate the integral

x +1

∫ x 3 + x 2 − 6 x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 3 + x 2 − 6 x into x(x − 2)(x + 3) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: x +1 3

2

x + x − 6x

=

x +1 x(x − 2 )(x + 3)

=

C A B + + x x−2 x+3

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x +1 3

2

x + x − 6x

Hamilton Education Guides

=

A (x − 2 )(x + 3) + Bx (x + 3) + Cx (x − 2 ) x(x − 2 )(x + 3)

320

Calculus I

5.3 Integration by Partial Fractions

x +1

(

) (

) (

)

= A x 2 + 3x − 2 x − 6 + B x 2 + 3x + C x 2 − 2 x = Ax 2 + Ax − 6 A + Bx 2 + 3Bx + Cx 2 − 2Cx x +1

= ( A + B + C )x 2 + ( A + 3B − 2C )x − 6 A therefore,

A+ B +C = 0

A + 3B − 2C = 1 1 6

which result in having A = − , B =

3 10

, and C = −

−6 A = 1

2 15

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x +1

A

B

1

C

1

3

1

2

1

∫ x 3 + x 2 − 6 x dx = ∫ x dx + ∫ x − 2 dx + ∫ x + 3 dx = − 6 ∫ x dx + 10 ∫ x − 2 dx − 15 ∫ x + 3 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. −

1 6

1

3

1

2

1

1

∫ x dx + 10 ∫ x − 2 dx − 15 ∫ x + 3 dx = − 6 ln

x +

2 3 ln x − 2 − ln x + 3 + c 15 10

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 1 6

Let y = − ln x + = = =

2 3 ln x − 2 − ln x + 3 + c 15 10

− 150(x − 2 )(x + 3) + 270 x (x + 3) − 120 x (x − 2 ) 900 x (x − 2 )(x + 3)

(

900 x + 900

(

900 x 3 + x 2 − 6 x

)

=

(

900 x 3 + x 2 − 6 x

Example 5.3-2: Evaluate the integral

=

)

(

(

900 x x 3 + x 2

)

900(x + 1)

)

(

1 2 1 3 ⋅1 + 0 ⋅1 − ⋅ ⋅ 15 x + 3 10 x − 2

) − 6x)

(

− 150 x 2 + x − 6 + 270 x 2 + 3 x − 120 x 2 − 2 x

=

− 150 x 2 − 150 x + 900 + 270 x 2 + 810 x − 120 x 2 + 240 x 900 x 3 + x 2 − 6 x

1 1 6 x

, then y ′ = − ⋅ ⋅1 +

=

)

(− 150 + 270 − 120)x 2 + (− 150 + 810 + 240)x + 900

(

900 x 3 + x 2 − 6 x

)

x +1 3

x + x 2 − 6x dx

∫ x 2 + 3x + 2 .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 2 + 3x + 2 into (x + 1)(x + 2) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: 1 2

x + 3x + 2

=

1

(x + 1)(x + 2)

=

A B + x +1 x + 2

Fourth - Solve for the constants A and B by equating coefficients of the like powers. 1 2

x + 3x + 2 Hamilton Education Guides

=

A (x + 2 ) + B (x + 1) (x + 1)(x + 2) 321

Calculus I

5.3 Integration by Partial Fractions

1

= A (x + 2) + B (x + 1) = Ax + 2 A + Bx + B 1 = ( A + B )x + (2 A + B ) therefore, 2A + B = 1

A+ B = 0

which result in having A = 1 and B = −1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

A

1

B

1

∫ x 2 + 3x + 2 dx = ∫ x + 1 dx + ∫ x + 2 dx = ∫ x + 1 dx − ∫ x + 2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1

1

∫ x + 1 dx − ∫ x + 2 dx

= ln x + 1 − ln x + 2 + c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = ln x + 1 − ln x + 2 + c , then y ′ = Example 5.3-3: Evaluate the integral

1 1 ⋅1 − ⋅1 + 0 x +1 x+2

=

(x + 2) − (x + 1) = (x + 1)(x + 2)

1

2

x + 3x + 2

x dx

∫ x 2 − 5x + 6 .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 2 − 5 x + 6 into (x − 2)(x − 3) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: x 2

x − 5x + 6

=

x

(x − 2)(x − 3)

=

B A + x−2 x−3

Fourth - Solve for the constants A and B by equating coefficients of the like powers. x 2

x − 5x + 6 x

=

A (x − 3) + B (x − 2 ) (x − 2)(x − 3)

= A (x − 3) + B (x − 2) = Ax − 3 A + Bx − 2 B x

= ( A + B )x − (3 A + 2 B ) therefore,

A+ B =1

3 A + 2B = 0

which result in having A = −2 and B = 3 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x dx

A

B

2

3

∫ x 2 − 5x + 6 = ∫ x − 2 dx + ∫ x − 3 dx = − ∫ x − 2 dx + ∫ x − 3 dx Hamilton Education Guides

322

Calculus I

5.3 Integration by Partial Fractions

Sixth - Integrate each integral individually using integration methods learned in previous sections. −

3

2

∫ x − 2 dx + ∫ x − 3 dx = − 2 ln

x − 2 + 3 ln x − 3 + c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = −2 ln x − 2 + 3 ln x − 3 + c , then y ′ = − 2 ⋅ =

−2 x + 6 + 3 x − 6

=

2

x − 5x + 6

1 1 ⋅1 + 0 ⋅1 + 3 ⋅ x−3 x−2

=

− 2(x − 3) + 3(x − 2 ) (x − 2)(x − 3)

x 2

x − 5x + 6

Example 5.3-4: Evaluate the integral

x 2 +1

∫ x 3 − x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 − x into x x 2 − 1 = x(x − 1)(x + 1) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: x 2 +1 3

x −x

=

x 2 +1 x(x − 1)(x + 1)

=

C B A + + x x −1 x +1

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x 2 +1 3

x 2 +1

x −x

(

=

A (x − 1)(x + 1) + Bx (x + 1) + Cx (x − 1) x(x − 1)(x + 1)

) (

) (

)

= A x 2 + x − x − 1 + B x 2 + x + C x 2 − x = Ax 2 − A + Bx 2 + Bx + Cx 2 − Cx x 2 +1

= ( A + B + C )x 2 + (B − C )x − A therefore, B −C = 0

A+ B +C =1

−A = 1

which result in having A = −1 , B = 1 , and C = 1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x 2 +1

A

B

C

1

1

1

∫ x 3 − x dx = ∫ x dx + ∫ x − 1 dx + ∫ x + 1 dx = − ∫ x dx + ∫ x − 1 dx + ∫ x + 1 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. −

1

1

1

∫ x dx + ∫ x − 1 dx + ∫ x + 1 dx

= − ln x + ln x − 1 + ln x + 1 + c

Seventh - Check the answer by differentiating the solution. The result should match the integrand.

Hamilton Education Guides

323

Calculus I

5.3 Integration by Partial Fractions

1 x

Let y = − ln x + ln x − 1 + ln x + 1 + c , then y ′ = − + =

− x 2 +1+ x 2 + x + x 2 − x

(

)

x x 2 −1

=

1 1 +0 + x −1 x +1

=

− (x − 1)(x + 1) + x (x + 1) + x (x − 1) x(x − 1)(x + 1)

x 2 +1 x3 − x

x−3

∫ x (x 2 + x − 2)

Example 5.3-5: Evaluate the integral

dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 2 + x − 2 into (x + 2)(x − 1) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: x−3 x(x + 2 )(x − 1)

=

C B A + + x x + 2 x −1

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x−3 x(x + 2 )(x − 1) x−3

(

=

) (

A (x + 2 )(x − 1) + Bx (x − 1) + Cx (x + 2 ) x(x + 2 )(x − 1)

) (

)

= A x 2 + 2 x − x − 2 + B x 2 − x + C x 2 + 2 x = Ax 2 + Ax − 2 A + Bx 2 − Bx + Cx 2 + 2Cx x−3

= ( A + B + C )x 2 + ( A − B + 2C )x − 2 A therefore,

A+ B +C = 0

A − B + 2C = 1 3 2

5 6

which result in having A = , B = − , and C = −

−2 A = −3

2 3

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x−3

∫ x (x 2 + x − 2)

dx =

A

B

C

3

1

5

1

2

1

∫ x dx + ∫ x + 2 dx + ∫ x − 1 dx = 2 ∫ x dx − 6 ∫ x + 2 dx − 3 ∫ x − 1 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 3 2

1

5

1

2

1

∫ x dx − 6 ∫ x + 2 dx − 3 ∫ x − 1 dx

=

5 2 3 ln x − ln x + 2 − ln x − 1 + c 3 2 6

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 3 2

5 6

2 3

Let y = ln x − ln x + 2 − ln x − 1 + c , then y ′ = =

9(x + 2 )(x − 1) − 5 x (x − 1) − 4 x (x + 2 ) 6 x (x + 2 )(x − 1)

Hamilton Education Guides

=

(

5 1 2 1 3 1 ⋅1 − ⋅ ⋅1 + 0 ⋅ ⋅1 − ⋅ 3 x −1 2 x 6 x+2

) ( ) ( 6 x (x + x − 2 )

9 x 2 + x − 2 − 5 x 2 − x − 4 x 2 + 2x 2

) = (9 − 5 − 4)x + (9 + 5 − 8)x − 18 6 x (x + x − 2 ) 2

2

324

Calculus I

=

(

5.3 Integration by Partial Fractions

6 x − 18

6x x 2 + x − 2

)

=

(

6(x − 3)

6x x 2 + x − 2

)

=

x−3

(

x x2 + x − 2

)

x3 + 2

∫ x 2 − x − 6 dx .

Example 5.3-6: Evaluate the integral

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. In this case the integrand is an improper rational fraction, i.e., the degree of the numerator is greater than the degree of the denominator. Applying the long division method we obtain x3 + 2 2

x − x−6

= (x + 1) +

7x + 8 2

x − x−6

To integrate the second term we proceed with the following steps: Second - Factor the denominator x 2 − x − 6 into (x − 3)(x + 2) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: 7x + 8 2

x − x−6

=

7x + 8 (x − 3)(x + 2)

=

B A + x−3 x+ 2

Fourth - Solve for the constants A and B by equating coefficients of the like powers. 7x + 8 2

x − x−6 7x + 8 7x + 8

29 5

A ( x + 2 ) + B ( x − 3) (x − 3)(x + 2)

= Ax + 2 A + Bx − 3B

= ( A + B )x + (2 A − 3B ) therefore, 2 A − 3B = 8

A+ B = 7

which result in having A =

=

and B =

6 5

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x3 + 2

7x + 8

7x + 8

∫ x 2 − x − 6 dx = ∫ (x + 1)dx + ∫ x 2 − x − 6 dx = ∫ (x + 1)dx + ∫ (x − 3)(x + 2) dx = =

1 (x + 1)2 + 29 2 5

1

6

1 (x + 1)2 + 2

A

B

∫ x − 3 dx + ∫ x + 2 dx

1

∫ x − 3 dx + 5 ∫ x + 2 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 1 (x + 1)2 + 29 2 5

1

6

1

∫ x − 3 dx + 5 ∫ x + 2 dx =

1 ( x + 1)2 + 29 ln x − 3 + 6 ln x + 2 + c 2 5 5

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Hamilton Education Guides

325

Calculus I

5.3 Integration by Partial Fractions

1 2

Let y = (x + 1)2 + = (x + 1) +

, then y ′ = (x + 1) +

29 1 6 1 ⋅ ⋅1 + ⋅ ⋅1 + 0 5 x−3 5 x+2

5(x + 1)(x − 3)(x + 2 ) + 29(x + 2 ) + 6(x − 3) 29 1 6 1 = ⋅ + ⋅ 5 (x − 3)(x + 2 ) 5 x−3 5 x+ 2

5 x 3 + 10 5 (x − 3)(x + 2 )

=

29 6 ln x − 3 + ln x + 2 + c 5 5

=

(

(

5 x3 + 2

)

5 x2 − x − 6

5 x 3 − 35 x − 30 + 29 x + 58 + 6 x − 18 5 (x − 3)(x + 2 )

x3 + 2

=

)

=

x2 − x − 6

Example 5.3-7: Evaluate the integral

1

∫ 49 − x 2 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator 49 − x 2 into (7 − x )(7 + x ) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: 1 49 − x

2

=

1

(7 − x )(7 + x )

=

A B + 7−x 7+ x

Fourth - Solve for the constants A and B by equating coefficients of the like powers. 1 49 − x

2

=

A (7 + x ) + B (7 − x ) (7 − x )(7 + x )

= A (7 + x ) + B (7 − x ) = 7 A + Ax + 7 B − Bx

1

1 = ( A − B )x + (7 A + 7 B ) therefore, 7 A + 7B = 1

which result in having A =

1 14

A− B = 0

, and B =

1 14

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

A

B

∫ 49 − x 2 dx = ∫ 7 − x dx + ∫ 7 + x dx

=

1 14

1

1

1

∫ 7 − x dx + 14 ∫ 7 + x dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 1 14

1

1

1

1

∫ 7 − x dx + 14 ∫ 7 + x dx = 14 ln

7− x +

1 ln 7 + x + c 14

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = =

(

1 1 ln 7 − x + ln 7 + x + c 14 14

7+7

14 49 + 7 x − 7 x − x 2

)

Hamilton Education Guides

=

(

14

14 49 − x 2

)

=

, then y ′ =

1 1 1 1 ⋅ + ⋅ +0 14 7 − x 14 7 + x

=

7+ x+7−x 14(7 − x )(7 + x )

1 49 − x 2 326

Calculus I

5.3 Integration by Partial Fractions

CASE II - The Denominator Has Repeated Linear Factors In this case each linear factor of the form ax + b appears n times in the denominator. To solve this class of rational fractions we equate each proper rational fraction, that appears n times in the Mn M1 M2 + + ... + 2 ax + b (ax + b ) (ax + b )n

denominator, with a sum of n partial fractions of the form

. The

following examples show the steps as to how this class of integrals are solved. Example 5.3-8: Evaluate the integral

x+3

∫ x 3 − 2 x 2 + x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 − 2 x 2 + x into x x 2 − 2 x + 1 = x(x − 1)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: x+3 3

=

2

x − 2x + x

(

x+3

=

)

x x 2 − 2x +1

x+3

x(x − 1)

2

=

C A B + + x x − 1 (x − 1)2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x+3 x 3 − 2x 2 + x x+3

(

) (

=

A (x − 1)2 + Bx (x − 1) + Cx x(x − 1)(x − 1)2

)

= A x 2 − 2 x + 1 + B x 2 − x + Cx = Ax 2 − 2 Ax + A + Bx 2 − Bx + Cx x+3

= ( A + B )x 2 + (− 2 A − B + C )x + A therefore,

A+ B = 0

A=3

−2 A − B + C = 1

which result in having A = 3 , B = −3 , and C = 4 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x+3

A

B

1

C

1

1

∫ x 3 − 2 x 2 + x dx = ∫ x dx + ∫ x − 1 dx + ∫ (x − 1)2 dx = 3∫ x dx − 3∫ x − 1 dx + 4∫ (x − 1)2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 3

1

1

1

∫ x dx − 3∫ x − 1 dx + 4∫ (x − 1)2 dx

= 3 ln x − 3 ln x − 1 −

4 +c x −1

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = 3 ln x − 3 ln x − 1 −

Hamilton Education Guides

4 +c x −1

1 x

, then y ′ = 3 ⋅ − 3 ⋅

1 4 + +0 x − 1 (x − 1)2

=

3(x − 1)2 − 3x (x − 1) + 4 x x (x − 1)2

327

Calculus I

=

5.3 Integration by Partial Fractions

3x 2 − 6 x + 3 − 3x 2 + 3x + 4 x 3

x+3

=

2

x − 2x + x

3

x − 2x 2 + x

Example 5.3-9: Evaluate the integral

dx

∫ x3 − x2 .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 3 − x 2 into x 2 (x − 1) . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: 1 3

x −x

2

=

1

x

2

(x − 1)

A B C + + 2 x x x −1

=

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 1 x3 − x2

(

=

Ax (x − 1) + B (x − 1) + Cx 2 x 2 (x − 1)

)

1 = A x 2 − x + B (x − 1) + Cx 2 = Ax 2 − Ax + Bx − B + Cx 2 1 = ( A + C )x 2 + (− A + B )x − B therefore, A+C = 0

−A + B = 0

−B = 1

which result in having A = −1 , B = −1 , and C = 1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. dx

∫ x3 − x2

=

A

B

1

C

1

1

∫ x dx + ∫ x 2 dx + ∫ x − 1 dx = − ∫ x dx − ∫ x 2 dx + ∫ x − 1 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. −

1

1

1

∫ x dx − ∫ x 2 dx + ∫ x − 1 dx = − ln

x +

1 + ln x − 1 + c x

=

1 + ln x − 1 − ln x + c x

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 1 x

Let y = + ln x − 1 − ln x + c , then y ′ = − =

− x +1+ x 2 − x 2 + x 3

x −x

2

=

x2

+

1 1 − +0 x −1 x

=

− (x − 1) + x 2 − x (x − 1) x 2 (x − 1)

1 3

x − x2

Example 5.3-10: Evaluate the integral

Hamilton Education Guides

1

5dx

∫ x 3 − 2x 2 + x . 328

Calculus I

5.3 Integration by Partial Fractions

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second – Factor the denominator x 3 − 2 x 2 + x into x(x − 1)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: 5

x(x − 1)

A B C + + x x − 1 (x − 1)2

=

2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 5 x 3 − 2x 2 + x 5

(

A (x − 1)2 + Bx (x − 1) + Cx

=

) (

x(x − 1)2

)

= A x 2 − 2 x + 1 + B x 2 − x + Cx = Ax 2 − 2 Ax + A + Bx 2 − Bx + Cx 5

= ( A + B )x 2 + (− 2 A − B + C )x + A therefore, −2 A − B + C = 0

A+ B = 0

A=5

which result in having A = 5 , B = −5 , and C = 5 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 5dx

∫ x 3 − 2x 2 + x

=

A

B

1

C

1

1

∫ x dx + ∫ x − 1 dx + ∫ (x − 1)2 dx = 5∫ x dx − 5∫ x − 1 dx + 5∫ (x − 1)2 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 5

1

1

1

∫ x dx − 5∫ x − 1 dx + 5∫ (x − 1)2 dx = 5 ln x

− 5 ln x − 1 −

5 +c x −1

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = 5 ln x − 5 ln x − 1 − =

5(x − 1)2 − 5 x (x − 1) + 5 x x(x − 1)

2

=

5 +c x −1

(

1 x

, then y ′ = 5 ⋅ ⋅1 − 5 ⋅

)

5 x 2 − 2 x + 1 − 5x 2 + 5x + 5x 3

2

x − 2x + x

Example 5.3-11: Evaluate the integral

=

1 1 ⋅1 + 5 ⋅ ⋅1 + 0 x −1 (x − 1)2

=

5 5 5 − + x x − 1 (x − 1)2

5 x 2 − 10 x + 5 − 5 x 2 + 5 x + 5 x 3

2

x − 2x + x

=

5 3

x − 2x 2 + x

x+6

∫ (x + 2)(x − 3)2 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

Hamilton Education Guides

329

Calculus I

5.3 Integration by Partial Fractions

Second – Factor the denominator. However, the denominator is already in its reduced form of (x + 2)(x − 3)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: x+6

(x + 2)(x − 3)

=

2

A B C + + x + 2 x − 3 (x − 3)2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x+6

(x + 2)(x − 3)2 x+6

(

A (x − 3)2 + B(x + 2 )(x − 3) + C (x + 2 )

=

) (

(x + 2)(x − 3)2

)

= A x 2 − 6 x + 9 + B x 2 − x − 6 + C (x + 2) = Ax 2 − 6 Ax + 9 A + Bx 2 − Bx − 6 B + Cx + 2C = ( A + B )x 2 + (− 6 A − B + C )x + (9 A − 6 B + 2C ) therefore,

x+6 A+ B = 0

9 A − 6 B + 2C = 6

−6 A − B + C = 1

which result in having A =

4 4 , B=− , 25 25

and C =

9 5

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x+6

A

B

C

∫ (x + 2)(x − 3)2 dx = ∫ x + 2 dx + ∫ x − 3 dx + ∫ (x − 3)2 dx =

4 25

1

4

1

9

1

∫ x + 2 dx − 25 ∫ x − 3 dx + 5 ∫ (x − 3)2 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 4 25

1

4

1

9

1

∫ x + 2 dx − 25 ∫ x − 3 dx + 5 ∫ (x − 3)2 dx =

4 4 9 1 ln x + 2 − ln x − 3 − +c 25 25 5 ( x − 3)

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = = +

4 9 1 4 1 4 1 9 1 4 + c , then y ′ = ln x + 2 − ln x − 3 − + +0 ⋅ − ⋅ 25 25 5 ( x − 3) 25 x + 2 25 x − 3 5 (x − 3)2

4(x − 3)2 − 4(x + 2 )(x − 3) + 45(x + 2 ) 25(x + 2 )(x − 3)2 4 x + 24 + 45 x + 90 25(x + 2 )(x − 3)

2

=

=

25 x + 150

25(x + 2 )(x − 3)

2

Example 5.3-12: Evaluate the integral

(

) (

)

4 x 2 − 6 x + 9 − 4 x 2 − x − 6 + 45(x + 2 )

=

25(x + 2 )(x − 3)2

25(x + 6 )

25(x + 2 )(x − 3)

2

=

=

4 x 2 − 24 x + 36 − 4 x 2 25(x + 2 )(x − 3)2

x+6

(x + 2)(x − 3)2

x+5

∫ x 3 + 4 x 2 + 4 x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

Hamilton Education Guides

330

Calculus I

5.3 Integration by Partial Fractions

)

(

Second - Factor the denominator x 3 + 4 x 2 + 4 x into x x 2 + 4 x + 4 = x(x + 2)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: x+5 3

2

x + 4x + 4x

=

x+5

(

x x 2 + 4x + 4

=

)

x+5

x(x + 2 )

=

2

A B C + + x x + 2 ( x + 2 )2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x+5 x 3 + 4x 2 + 4x x+5

A (x + 2 )2 + Bx (x + 2 ) + Cx

=

x(x + 2 )(x + 2 )2

) (

(

)

= A x 2 + 4 x + 4 + B x 2 + 2 x + Cx = Ax 2 + 4 Ax + 4 A + Bx 2 + 2 Bx + Cx x+5

= ( A + B )x 2 + (4 A + 2 B + C )x + 4 A therefore,

A+ B = 0

4A = 5

4 A + 2B + C = 1 5 4

5 4

which result in having A = , B = − , and C = −

3 2

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x+5

A

B

5

C

1

5

1

3

1

∫ x 3 + 4 x 2 + 4 x dx = ∫ x dx + ∫ x + 2 dx + ∫ (x + 2)2 dx = 4 ∫ x dx − 4 ∫ x + 2 dx − 2 ∫ (x + 2)2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 5 4

1

5

1

3

1

∫ x dx − 4 ∫ x + 2 dx − 2 ∫ (x + 2)2 dx =

3 1 5 5 +c ln x − ln x + 2 + ⋅ 2 x+2 4 4

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 5 4

5 4

3 1 +c 2 x+2

Let y = ln x − ln x + 2 + ⋅ =

5 x 2 + 20 x + 20 − 5 x 2 − 10 x − 6 x

(

4 x 3 + 4x 2 + 4x

)

=

(

5 1 5 1 , then y ′ = ⋅ − ⋅ − 4 x

4 x + 20

4 x 3 + 4x 2 + 4x

Example 5.3-13: Evaluate the integral

)

=

4 x+2

2(x + 2 )2

4(x + 5)

(

3

4 x 3 + 4x 2 + 4x

)

=

+0

=

5(x + 2 )2 − 5 x (x + 2 ) − 6 x 4 x ( x + 2 )2

x+5 3

x + 4x 2 + 4x

1

∫ x 5 + 2 x 4 + x 3 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 5 + 2 x 4 + x 3 into x 3 x 2 + 2 x + 1 = x 3 (x + 1)2 .

Hamilton Education Guides

331

Calculus I

5.3 Integration by Partial Fractions

Third - Write the linear factors in partial fraction form. Since both factors in the denominator are repeated, the integrand can be represented in the following way: 1 5

4

x + 2x + x

3

=

1

(

=

)

x 3 x 2 + 2x +1

1

x (x + 1) 3

A B C D E + + + + x x 2 x 3 x + 1 (x + 1)2

=

2

Fourth - Solve for the constants A , B , C , D , and E by equating coefficients of the like powers. 1 x 5 + 2x 4 + x 3

=

Ax 2 (x + 1)2 + Bx (x + 1)2 + C (x + 1)2 + Dx 3 (x + 1) + Ex 3 x 3 (x + 1)2

) (

(

)

(

)

1 = Ax 2 x 2 + 2 x + 1 + Bx x 2 + 2 x + 1 + C x 2 + 2 x + 1 + Dx 3 (x + 1) + Ex 3 1 = Ax 4 + 2 Ax 3 + Ax 2 + Bx 3 + 2 Bx 2 + Bx + Cx 2 + 2Cx + C + Dx 4 + Dx 3 + Ex 3 1 = ( A + D )x 4 + (2 A + B + D + E )x 3 + ( A + 2 B + C )x 2 + (B + 2C )x + C therefore, A+ D = 0

2A + B + D + E = 0

B + 2C = 0

A + 2B + C = 0

C =1

which result in having A = 3 , B = −2 , C = 1 , D = −3 , and E = −1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

∫ x 5 + 2 x 4 + x 3 dx

1 x

= 3∫ dx − 2∫

1 x

dx +

2

1

1

1

∫ x 3 dx − 3∫ x + 1 dx − ∫ (x + 1)2 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 1

1

1

1

1

∫ x dx − 2∫ x 2 dx + ∫ x 3 dx − 3∫ x + 1 dx − ∫ (x + 1)2 dx = 3 ln

3

x +

2 1 1 − − 3 ln x + 1 + +c x 2 x2 x +1

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 2 x

Let y = 3 ln x + − =

+

1 2x

2

− 3 ln x + 1 +

1 +c x +1

3x 2 (x + 1)2 − 2 x(x + 1)2 + (x + 1)2 − 3 x 3 (x + 1) − x 3 x 3 (x + 1)2

− 3x 3 − x 3 3

(

2

)

x x + 2x + 1

=

1 x

, then y ′ = 3 ⋅ − =

2 x

2

+

1 x

3

− 3⋅

1 1 +0 − x + 1 (x + 1)2

3x 4 + 3x 2 + 6 x 3 − 2 x 3 − 2 x − 4 x 2 + x 2 + 1 + 2 x − 3x 4

(

)

x 3 x 2 + 2x + 1

1 5

x + 2x 4 + x 3

Example 5.3-14: Evaluate the integral

1

∫ x 4 − 6 x 3 + 9 x 2 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 4 − 6 x 3 + 9 x 2 into x 2 x 2 − 6 x + 9 = x 2 (x − 3)2 . Hamilton Education Guides

332

Calculus I

5.3 Integration by Partial Fractions

Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: 1 4

3

x − 6x + 9x

2

=

1

(

x 2 x 2 − 6x + 9

=

)

1

x

2

(x − 3)

2

=

D C A B + + + x x 2 x − 3 ( x − 3) 2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 1 x 4 − 6x 3 + 9x 2

=

Ax (x − 3)2 + B (x − 3)2 + Cx 2 (x − 3) + Dx 2 x 2 (x − 3)2

(

) (

)

1 = Ax (x − 3)2 + B (x − 3)2 + Cx 2 (x − 3) + Dx 2 = Ax x 2 + 9 − 6 x + B x 2 + 9 − 6 x + Cx 2 (x − 3) + Dx 2 1

= Ax 3 + 9 Ax − 6 Ax 2 + Bx 2 + 9 B − 6 Bx + Cx 3 − 3Cx 2 + Dx 2

1 = ( A + C )x 3 + (− 6 A + B − 3C + D )x 2 + (9 A − 6 B )x + 9 B therefore, A+C = 0

−6 A + B − 3C + D = 0

which result in having A =

6 1 , B= 81 9

, C=−

6 , 81

9B = 1

9 A − 6B = 0

and D =

1 9

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. dx

∫ x 4 − 6x 3 + 9x 2

A

B

D

C

∫ x dx + ∫ x 2 dx + ∫ x − 3 dx + ∫ (x − 3)2 dx

=

=

6 dx 1 + 81 x 9

∫

6

dx

dx

1

dx

∫ x 2 − 81 ∫ x − 3 + 9 ∫ (x − 3)2

Sixth - Integrate each integral individually using integration methods learned in previous sections. 6 dx 1 + 81 x 9

∫

dx

6

dx

1

dx

∫ x 2 − 81 ∫ x − 3 + 9 ∫ (x − 3)2

=

6 1 6 1 ln x − ln x − 3 − − +c 81 9 x 81 9( x − 3 )

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = =

=

6 1 6 1 6 1 1 6 1 1 ln x − − ln x − 3 − + c , then y ′ = ⋅ + − ⋅ + +0 2 81 9 x 81 9(x − 3) 81 x 9 x 81 x − 3 9(x − 3)2

6 x(x − 3)2 + 9(x − 3)2 − 6 x 2 (x − 3) + 9 x 2 81x

2

(x − 3)

2

=

(

6 x 3 − 36 x 2 + 54 x + 9 x 2 − 54 x + 81 − 6 x 3 + 18 x 2 + 9 x 2

(

81x 2 x 2 − 6 x + 9

Hamilton Education Guides

)

) (

)

6 x x 2 − 6 x + 9 + 9 x 2 − 6 x + 9 − 6 x 3 + 18 x 2 + 9 x 2

(

81x 2 x 2 − 6 x + 9

=

(

81

81 x 4 − 6 x 3 + 9 x 2

)

)

=

1 4

x − 6x 3 + 9x 2

333

Calculus I

5.3 Integration by Partial Fractions

CASE III - The Denominator Has Distinct Quadratic Factors In this case the quadratic factors of the form ax 2 + bx + c appear only once in the denominator and are irreducible. To solve this class of rational fractions we equate each proper rational fraction with a single fraction of the form

Ax + B

2

ax + bx + c

Cx + D

,

2

cx + dx + e

,

Ex + F

2

ex + fx + g

, etc. The following

examples show the steps as to how this class of integrals are solved. Example 5.3-15: Evaluate the integral

∫

x2 − x + 3

dx .

x3 + x

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 + x into x x 2 + 1 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: x2 − x + 3 3

x +x

=

x2 − x + 3

(

)

2

x x +1

=

A Bx + C + x x 2 +1

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. x2 − x + 3 3

x +x x2 − x + 3

(

=

(

)

A x 2 + 1 + (Bx + C ) x

(

)

x x 2 +1

)

= A x 2 +1 + (Bx + C ) x = Ax 2 + A + Bx 2 + Cx

x2 − x + 3

= ( A + B )x 2 + Cx + A therefore,

A+ B =1

A=3

C = −1

which result in having A = 3 , B = −2 , and C = −1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

∫

x2 − x + 3 3

x +x

dx

=

A

1

Bx + C

1

−2 x − 1

2x

1

∫ x dx + ∫ x 2 + 1 dx = 3∫ x dx + ∫ x 2 + 1 dx = 3∫ x dx − ∫ x 2 + 1 dx − ∫ x 2 + 1 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. To solve the second integral let u = x 2 + 1 . 1

2x

1

1

2 x du

1

∫ x dx − ∫ x 2 + 1 dx − ∫ x 2 + 1 dx = 3∫ x dx − ∫ u ⋅ 2 x − ∫ x 2 + 1 dx

3

1 x

1 u

= 3∫ dx − ∫ du − ∫

1 2

x +1

dx

= 3 ln x − ln u − tan −1 x + c = 3 ln x − ln x 2 + 1 − tan −1 x + c

Hamilton Education Guides

334

Calculus I

5.3 Integration by Partial Fractions

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 1 x

Let y = 3 ln x − ln x 2 + 1 − tan −1 x + c , then y ′ = 3 ⋅ + =

(

)

3 x 2 +1 − 2x 2 − x

(

)

x x 2 +1

=

3x 2 + 3 − 2 x 2 − x x3 + x

Example 5.3-16: Evaluate the integral

=

−1

2

x +1

⋅ 2x −

1

1+ x

2

+0

=

3 2x 1 − − x x 2 +1 x 2 +1

x2 − x + 3 x3 + x

1

∫ x 3 + 25x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 + 25 x into x x 2 + 25 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: 1 3

x + 25 x

=

1

(

2

x x + 25

=

)

A Bx + C + x x 2 + 25

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 1

=

3

x + 25 x 1

(

(

)

A x 2 + 25 + (Bx + C ) x

(

2

x x + 25

)

)

= A x 2 + 25 + (Bx + C ) x = Ax 2 + 25 A + Bx 2 + Cx 1 = ( A + B )x 2 + Cx + 25 A therefore, C=0

25 A = 1

which result in having A =

1 25

, B=−

1 , 25

A+ B = 0

and C = 0

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

A

Bx + C

∫ x 3 + 25x dx = ∫ x dx + ∫ x 2 + 25 dx

=

1 25

1

1

x

∫ x dx − 25 ∫ x 2 + 25 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. To solve the second integral let u = x 2 + 25 . 1 25

=

1

1

x

∫ x dx − 25 ∫ x 2 + 25 dx =

1 25

1

1

x du

∫ x dx − 25 ∫ u ⋅ 2 x

=

1 25

1

1

1

∫ x dx − 50 ∫ u du =

1 1 ln x − ln u + c 25 50

1 1 ln x − ln x 2 + 25 + c 25 50

Hamilton Education Guides

335

Calculus I

5.3 Integration by Partial Fractions

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y =

(x

=

2

1 1 ln x − ln x 2 + 25 + c 25 50

)

+ 25 − x 2

(

25 x x 2 + 25

)

=

x 2 + 25 − x 2

(

25 x 3 + 25 x

, then y ′ =

=

)

(

25

25 x 3 + 25 x

Example 5.3-17: Evaluate the integral

1 1 1 1 ⋅ − ⋅ ⋅ 2x + 0 2 25 x 50 x + 25

1 − 25 x

(

x

25 x 2 + 25

)

1

=

)

=

3

x + 25 x

1

∫ x 4 + 16 x 2 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 4 + 16x 2 into x 2 x 2 + 16 . Third - Write the factors in partial fraction form. Since the factors in the denominator are in quadratic form, the integrand can be represented in the following way: 1 4

x + 16 x

=

2

1

(

x 2 x 2 + 16

=

)

Ax + B x

2

+

Cx + D x 2 + 16

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. 1 x 4 + 16 x 2

(

=

( Ax + B ) (x 2 + 16)+ x 2 (Cx + D )

(

x 2 x 2 + 16

)

)

1 = ( Ax + B ) x 2 + 16 + x 2 (Cx + D ) = Ax 3 + 16 Ax + Bx 2 + 16 B + Cx 3 + Dx 2 1

= ( A + C )x 3 + (B + D )x 2 + 16 Ax + 16 B therefore, B+D =0

A+C = 0

which result in having A = 0 , B =

1 16

16 A = 0

, C = 0 , and D = −

16 B = 1

1 16

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

∫ x 4 + 16 x 2 dx = ∫

Ax + B x

2

dx +

B

Cx + D

D

∫ x 2 + 16 dx = ∫ x 2 dx + ∫ x 2 + 16 dx

=

1 16

1

1

1

∫ x 2 dx − 16 ∫ x 2 + 16 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. 1 16

=

1

1

1

∫ x 2 dx − 16 ∫ x 2 + 16 dx

=

x 1 1 1 ⋅ 4 tan −1 + c ⋅− − 4 16 x 256

Hamilton Education Guides

1 16

= −

1

1

1

∫ x 2 dx − 16 ∫ 16(x 2 + 1 ) 16

dx

=

1 16

1

1

1

∫ x 2 dx − 256 ∫ (x 2 + 1 )

dx

16

x 1 1 − tan −1 + c 16 x 64 4

336

Calculus I

5.3 Integration by Partial Fractions

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = − =

1 16 x

2

x 1 1 − tan −1 + c 4 16 x 64

−

(

1

16 16 + x

2

=

)

1

, then y ′ =

(16 + x )− x 16 x (16 + x ) 2

2

2

16 x

Example 5.3-18: Evaluate the integral

1 1 1 ⋅ ⋅ +0 2 64 1 + x 4

=

16

16 + x 2 − x 2

=

2

−

2

2

16 x

=

(16 + x ) 2

1 16 x

2

−

16 1 ⋅ 256 16 + x 2

16

16 x

2

=

(16 + x ) 2

1 4

x + 16 x 2

1

∫ x 3 − 8 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 − 8 into (x − 2) x 2 + 2 x + 4 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: 1 3

x −8

=

(x − 2) (x

1 2

+ 2x + 4

)

=

A Bx + C + x − 2 x 2 + 2x + 4

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 1 3

x −8 1

(

=

(

)

A x 2 + 2 x + 4 + (Bx + C )(x − 2 )

(x − 2) (x 2 + 2 x + 4)

)

= A x 2 + 2 x + 4 + (Bx + C )(x − 2) = Ax 2 + 2 Ax + 4 A + Bx 2 − 2 Bx + Cx − 2C 1 = ( A + B )x 2 + (2 A − 2 B + C )x + (4 A − 2C ) therefore, 2 A − 2B + C = 0

A+ B = 0

which result in having A =

1 12

, B=−

1 , 12

and C = −

4 A − 2C = 1

1 3

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

∫ x3 − 8

dx

=

∫

A dx + x−2

Bx + C

∫ x 2 + 2x + 4

dx

=

1 12

∫

1 dx + x−2

1 x−1 − 12 3

1

1

1

x+4

∫ x 2 + 2 x + 4 dx = 12 ∫ x − 2 dx − 12 ∫ x 2 + 2 x + 4 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. To solve the second integral let u = x 2 + 2 x + 4 , then x+4

can be rewritten as x + 4 = (x + 1) + 3 =

Hamilton Education Guides

1 (2 x + 2) + 3 . 2

du = 2x + 2 dx

and dx =

du . 2x + 2

Also,

Therefore,

337

Calculus I

1 12

∫

5.3 Integration by Partial Fractions

1 1 dx − 12 x−2

x+4

∫ x 2 + 2x + 4

1

1

1

1

1 12

dx =

2x + 2

∫

(x + 1) + 3

1 1 dx − 12 x−2

∫ x 2 + 2x + 4

3

1

1 12

∫

=

1 12

∫ x − 2 dx − 24 ∫ x 2 + 2 x + 4 dx − 12 ∫ x 2 + 2 x + 4 dx = 12 ∫ x − 2 dx − 24 ∫

=

1 12

∫ x − 2 dx − 24 ∫ u du − 4 ∫ (x + 1)2 + 3 dx = 12 ln

=

1 1 3 x +1 ln x 2 + 2 x + 4 − ln x − 2 − tan −1 +c 24 12 4⋅3 3

1

1

dx =

1

1

1

1

x−2 −

1

1 1 dx − 12 x−2

∫

1 2

(2 x + 2) + 3

x 2 + 2x + 4

3 2 x + 2 du ⋅ − u 2 x + 2 12

dx

1

∫ (x + 1)2 + 3 dx

1 1 x +1 ln u − tan −1 +c 24 4 3 3

1 1 3 x +1 ln x − 2 − ln x 2 + 2 x + 4 − tan −1 +c 12 24 12 3

=

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = y′

=

1 1 3 x +1 ln x − 2 − ln x 2 + 2 x + 4 − tan −1 +c 12 24 12 3

3 1 1 1 1 ⋅ ⋅ (2 x + 2 ) − − ⋅ ⋅ 12 12 x − 2 24 x 2 + 2 x + 4

−

3 3 3 ⋅ ⋅ 2 12 3 + (x + 1) 3

=

1 + 12(x − 2 )

=

(

1 − 12(x − 2 )

−x − 4

12 x 2 + 2 x + 4

4+8

(

=

12(x − 2 ) x 2 + 2 x + 4

)

=

)

=

(x

1 1 +  x +1   3

x +1

(

12 x 2 + 2 x + 4

) 12(x − 2 ) (x

)

−

⋅

2

, then

(1⋅ 3 )− 0 ⋅ (x + 1) + 0 = 1 − x + 1 12(x − 2 ) ( 3 )2 12(x 2 + 2 x + 4 ) 3

(

12 x 2 + 2 x + 4

+ 2 x + 4 + (− x − 4 )(x − 2 )

2

(

2

+ 2x + 4

12

12(x − 2 ) x 2 + 2 x + 4

Example 5.3-19: Evaluate the integral

)

=

)

3

=

=

)

1 + 12(x − 2 )

(

−x −1− 3

12 x 2 + 2 x + 4

)

x 2 + 2x + 4 − x 2 + 2x − 4x + 8 12(x − 2 ) x 2 + 2 x + 4

(

)

1

1

2

2

x + 2x + 4x − 2x − 4x − 8

=

3

x −8

1

∫ x 3 + 8 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 + 8 into (x + 2) x 2 − 2 x + 4 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: 1 3

x +8

=

(x + 2) (x

1 2

− 2x + 4

)

=

A Bx + C + x + 2 x 2 − 2x + 4

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 1 3

x +8

Hamilton Education Guides

=

(

)

A x 2 − 2 x + 4 + (Bx + C )(x + 2)

(x + 2) (x 2 − 2 x + 4)

338

Calculus I

5.3 Integration by Partial Fractions

)

(

1 = A x 2 − 2 x + 4 + (Bx + C )(x + 2 ) = Ax 2 − 2 Ax + 4 A + Bx 2 + 2 Bx + Cx + 2C 1 = ( A + B )x 2 + (− 2 A + 2 B + C )x + (4 A + 2C ) therefore, A+ B = 0

4 A + 2C = 1

−2 A + 2 B + C = 0 1 12

which result in having A =

, B=−

1 , 12

and C =

1 3

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

∫ x 3 + 8 dx = ∫

A dx + x+2

1 12

Bx + C

1 dx + x+2

∫ x 2 − 2 x + 4 dx = ∫

1 x+ 1 − 12 3

1

1

1

x−4

∫ x 2 − 2 x + 4 dx = 12 ∫ x + 2 dx − 12 ∫ x 2 − 2 x + 4 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. To solve the second integral let u = x 2 − 2 x + 4 , then x−4 1 12

∫

can be rewritten as x − 4 = (x − 1) − 3 = 1 1 dx − x+2 12

x−4

∫ x 2 − 2x + 4

1 12

dx =

∫

1 (2 x − 2) − 3 . 2

du = 2x − 2 dx

(x − 1) − 3

1 1 dx − 12 x+2

∫ x 2 − 2x + 4

3

1

dx =

1 12

∫

1 1 dx − x+2 12

1 12

∫ x + 2 dx − 24 ∫ x 2 − 2 x + 4 dx + 12 ∫ x 2 − 2 x + 4 dx = 12 ∫ x + 2 dx − 24 ∫

=

1 12

∫ x + 2 dx − 24 ∫ u du + 4 ∫ (x − 1)2 + 3 dx = 12 ln

=

x −1 3 1 1 +c tan −1 ln x + 2 − ln x 2 − 2 x + 4 + 4⋅3 12 24 3

1

1

1

2x − 2

1

1

1

1

1

1

x+2 −

=

du . 2x − 2

Also,

Therefore,

=

1

and dx =

1

∫

1 2

(2 x − 2) − 3

x 2 − 2x + 4

2 x − 2 du 3 ⋅ + 2 x + 2 12 u

dx

1

∫ (x − 1)2 + 3 dx

x −1 1 1 +c tan −1 ln u + 24 4 3 3

1 1 3 x −1 ln x + 2 − ln x 2 − 2 x + 4 + tan −1 +c 12 24 12 3

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = y′

=

x −1 1 3 1 +c tan −1 ln x + 2 − ln x 2 − 2 x + 4 + 12 24 12 3

1 1 1 1 3 ⋅ − ⋅ ⋅ (2 x − 2 ) + ⋅ 12 12 x + 2 24 x 2 − 2 x + 4

+

3 3 3 ⋅ ⋅ 2 12 3 + (x − 1) 3

=

1 + 12(x + 2 )

(

=

1 − 12(x + 2 )

−x + 4

12 x 2 − 2 x + 4

Hamilton Education Guides

)

=

(x

(

1 1 +  x −1   3

x −1

12 x 2 − 2 x + 4

) 12(x + 2 ) (x

2

)

+

2

⋅

, then

(1⋅ 3 )− 0 ⋅ (x − 1) + 0 = 1 − x − 1 12(x + 2 ) ( 3 )2 12(x 2 − 2 x + 4 )

(

12 x 2 − 2 x + 4

− 2 x + 4 + (− x + 4 )(x + 2 ) 2

3

− 2x + 4

)

=

)

=

1 + 12(x + 2 )

(

−x +1+ 3

12 x 2 − 2 x + 4

)

x 2 − 2x + 4 − x 2 − 2x + 4x + 8

(

12(x + 2 ) x 2 − 2 x + 4

)

339

Calculus I

=

5.3 Integration by Partial Fractions

4+8

(

12(x + 2 ) x 2 − 2 x + 4

=

)

(

12

12(x + 2 ) x 2 − 2 x + 4

Example 5.3-20: Evaluate the integral

1

=

)

3

2

2

x − 2x + 4x + 2x − 4x + 8

=

1 3

x +8

x2

∫ 16 − x 4 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)(

)

(

)

Second - Factor the denominator 16 − x 4 into 4 − x 2 4 + x 2 = (2 − x )(2 + x ) 4 + x 2 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: x2 16 − x

=

4

x2

A B Cx + D + + 2 − x 2 + x 4 + x2

=

(2 − x )(2 + x ) (4 + x 2 )

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. x2 16 − x x2

x2 x2

4

(

(

)

)

A(2 + x ) 4 − x 2 + B(2 − x ) 4 − x 2 + (2 − x )(2 + x )(Cx + D )

=

(

(2 − x )(2 + x ) (4 + x 2 )

)

(

)

= A(2 + x ) 4 + x 2 + B(2 − x ) 4 + x 2 + (2 − x )(2 + x )(Cx + D )

= 8 A + 2 Ax 2 + 4 Ax + Ax 3 + 8B + 2 Bx 2 − 4 Bx − Bx 3 + 4Cx + 4 D − Cx 3 − Dx 2

= ( A − B − C )x 3 + (2 A + 2 B − D )x 2 + (4 A − 4 B + 4C )x + (8 A + 8B + 4 D ) therefore, 2 A + 2B − D = 1

A− B −C = 0

1 8

4 A − 4 B + 4C = 0

1 8

which result in having A = , B = , C = 0 , and D = −

8 A + 8B + 4 D = 0

1 2

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x2

A

B

1

Cx + D

1

1

1

1

1

∫ 16 − x 4 dx = ∫ 2 − x dx + ∫ 2 + x dx + ∫ 4 + x 2 dx = 8 ∫ 2 − x dx + 8 ∫ 2 + x dx − 2 ∫ 4 + x 2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1 8

∫ 2 − x dx + 8 ∫ 2 + x dx − 2 ∫ 4 + x 2 dx = 8 ∫ 2 − x dx + 8 ∫ 2 + x dx − 2 ∫ 2 2 + x 2 dx

1

=

x 1 1 1 1 ln 2 − x + ln 2 + x − ⋅ tan −1 + c 2 8 8 2 2

1

1

1

1

1

=

1

1

1

1

1

1 1 1 x ln 2 − x + ln 2 + x − tan −1 + c 8 8 4 2

Seventh - Check the answer by differentiating the solution. The result should match the integrand.

Hamilton Education Guides

340

Calculus I

5.3 Integration by Partial Fractions

1 8

1 8

1 4

1 1 1 1 1 1 1 ⋅ + ⋅ − ⋅ ⋅ +0 2 8 2 − x 8 2 + x 4 1+ x 2

x 2

Let y = ln 2 − x + ln 2 + x − tan −1 + c , then y ′ = =

+

1 1 + − 8(2 − x ) 8(2 + x )

4

(

(

)(

) (2 x =

8 4 + x2

2 x 2 − 4 x − x 3 − 16 + 4 x 2 8 4 − x2 4 + x2

=

)

2

(2 + x ) (4 + x 2 )+ (2 − x ) (4 + x 2 )− 4(2 − x )(2 + x ) = 8(2 − x )(2 + x ) (4 + x 2 )

)

+ 2 x 2 + 4 x 2 + (8 + 8 − 16 )

(

8 16 − x 4

Example 5.3-21: Evaluate the integral

=

)

8x 2

(

8 16 − x 4

)

4

8 + 2x 2 + 4x + x 3 + 8

(

)(

8 4 − x2 4 + x2

)

x2

=

16 − x 4

5

∫ x 4 − 1 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)(

)

(

)

Second - Factor the denominator x 4 − 1 into x 2 − 1 x 2 + 1 = (x − 1)(x + 1) x 2 + 1 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: 5 4

x −1

=

5

(x − 1)(x + 1) (x

2

=

)

+1

A B Cx + D + + x −1 x +1 x 2 +1

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. 5

=

4

x −1 5 5 5

(

)

(

)

A(x + 1) x 2 + 1 + B(x − 1) x 2 + 1 + (x − 1)(x + 1)(Cx + D )

(

(x − 1)(x + 1) (x

(

)

2

)

+1

)

= A(x + 1) x 2 + 1 + B(x − 1) x 2 + 1 + (x − 1)(x + 1)(Cx + D )

= Ax 3 + Ax 2 + Ax + A + Bx 3 − Bx 2 + Bx − B + Cx 3 + Dx 2 − Cx − D

= ( A + B + C )x 3 + ( A − B + D )x 2 + ( A + B − C )x + ( A − B − D ) therefore,

A+ B +C = 0

A− B + D = 0

5 4

A+ B −C = 0

5 4

which result in having A = , B = − , C = 0 , and D = −

A− B − D = 5

5 2

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 5

A

B

∫ x 4 − 1 dx = ∫ x − 1 dx + ∫ x + 1 dx + ∫

Cx + D 2

x +1

dx

=

5 4

1

5

1

5

1

∫ x − 1 dx − 4 ∫ x + 1 dx − 2 ∫ x 2 + 1 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections.

Hamilton Education Guides

341

Calculus I

5 4

5.3 Integration by Partial Fractions

1

5

1

5

1

∫ x − 1 dx − 4 ∫ x + 1 dx − 2 ∫ x 2 + 1 dx

5 5 5 ln x − 1 − ln x + 1 − tan −1 x + c 4 4 2

=

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 5 4

5 4

5 2

5 1 5 1 5 1 ⋅ − ⋅ − ⋅ ⋅1 + 0 2 4 x −1 4 x +1 2 x +1

Let y = ln x − 1 − ln x + 1 − tan −1 x + c , then y ′ = =

5 5 − − 4(x − 1) 4(x + 1)

5

(

(

)(

4 x 2 −1 4 + x 2

)

2 x 2 +1

+ 5 x 2 − 5 x + 5 − 10 x 2 + 10

)

=

(

(

(

)

)

5(x + 1) x 2 + 1 − 5(x − 1) x 2 + 1 − 10(x − 1)(x + 1)

=

(

)

4(x − 1)(x + 1) x 2 + 1

20

)

4 x 4 −1

5x 3 + 5x 2 + 5x + 5 − 5x 3

(

)(

4 x 2 −1 4 + x 2

)

5

=

Example 5.3-22: Evaluate the integral

=

4

x −1 1

∫ x 3 − 64 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x 3 − 64 into (x − 4) x 2 + 4 x + 16 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way: 1

=

3

x − 64

1

(x − 4) (x 2 + 4 x + 16)

=

Bx + C A + x − 4 x 2 + 4 x + 16

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 1 3

x − 64

(

=

(

)

A x 2 + 4 x + 16 + (Bx + C )(x − 4 )

)

(x − 4) (x 2 + 4 x + 16)

1 = A x 2 + 4 x + 16 + (Bx + C )(x − 4 ) = Ax 2 + 4 Ax + 16 A + Bx 2 − 4 Bx + Cx − 4C

1 = ( A + B )x 2 + (4 A − 4 B + C )x + (16 A − 4C ) therefore, 4 A − 4B + C = 0

A+ B = 0

which result in having A =

1 48

, B=−

1 , 48

and C = −

16 A − 4C = 1

1 6

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. 1

∫ x 3 − 64 dx = ∫

A dx + x−4

∫

1 dx = 2 48 x + 4 x + 16

Bx + C

∫

1 dx + x−4

1 x− 1 − 48 6

∫ x 2 + 4 x + 16 dx =

1 48

1

1

x +8

∫ x − 4 dx − 48 ∫ x 2 + 4 x + 16 dx

Sixth - Integrate each integral individually using integration methods learned in previous Hamilton Education Guides

342

Calculus I

5.3 Integration by Partial Fractions

sections. To solve the second integral let u = x 2 + 4 x + 16 , then 1 (2 x + 4) + 6 . 2

Also, x + 8 can be rewritten as x + 8 = (x + 2) + 6 = 1 48

∫

1 1 dx − 48 x−4

x +8

∫ x 2 + 4 x + 16

1

1

1

1

1 48

dx =

∫

1 1 dx − 48 x−4

(x + 2 ) + 6

∫ x 2 + 4 x + 16

=

1 48

∫ x − 4 dx − 96 ∫ x 2 + 4 x + 16 dx − 48 ∫ x 2 + 4 x + 16 dx =

=

1 48

∫ x − 4 dx − 96 ∫ u du − 8 ∫ (x + 2)2 + 12 dx

=

1 1 12 x+2 ln x − 4 − ln x 2 + 4 x + 16 − tan −1 +c 48 96 8 ⋅12 12

2x + 4

1

1

1

6

1

1 48

du = 2x + 4 dx

and dx =

du . 2x + 4

Therefore,

dx =

1 48

1

∫

1

∫ x − 4 dx − 96 ∫

1 1 dx − 48 x−4

1 2

(2 x + 4) + 6

∫ x 2 + 4 x + 16 dx

2 x + 4 du 6 ⋅ − 2 x + 4 48 u

1

∫ (x + 2)2 + 12 dx

x+2 1 1 1 +c tan −1 ln x − 4 − ln u − 96 48 12 8 12

=

1 1 12 x+2 ln x − 4 − ln x 2 + 4 x + 16 − tan −1 +c 48 96 96 12

=

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y =

x+2 12 1 1 tan −1 +c ln x − 4 − ln x 2 + 4 x + 16 − 48 96 96 12

12 1 1 1 1 ⋅ ⋅ (2 x + 4 ) − − ⋅ ⋅ 96 48 x − 4 96 x 2 + 4 x + 16

y′

=

−

12 12 12 ⋅ ⋅ 2 96 12 + (x + 2 ) 12

=

1 + 48(x − 4 )

=

(

=

−x − 8

48 x 2 + 4 x + 16

16 + 32

(

48(x − 4 ) x 2 + 4 x + 16

Hamilton Education Guides

)

)

=

1 − 48(x − 4 )

=

(x

(

1 +  x + 2   12 

x+2

48 x 2 + 4 x + 16

) 48(x − 4 ) (x

2

1

)

2

−

⋅

(1⋅ 12 )− 0 ⋅ (x + 2) + 0 = ( 12 )2

(

+ 4 x + 16

48

48(x − 4 ) x 2 + 4 x + 16

)

=

6

(

48 x 2 + 4 x + 16

+ 4 x + 16 + (− x − 8)(x − 4 ) 2

, then

)

=

=

)

1 − 48(x − 4 )

1 + 48(x − 4 )

2

48 x + 4 x + 16

−x − 2 − 6

48 x 2 + 4 x + 16

)

x 2 + 4 x + 16 − x 2 + 4 x − 8 x + 32

(

48(x − 4 ) x 2 + 4 x + 16

1 3

(

(

x+2 2

2

x + 4 x + 16 x − 4 x − 16 x − 64

=

)

1 3

x − 64

343

)

Calculus I

5.3 Integration by Partial Fractions

CASE IV - The Denominator Has Repeated Quadratic Factors In this case each irreducible quadratic factor of the form ax 2 + bx + c appears n times in the denominator. To solve this class of rational fractions we equate each proper rational fraction, that appears n times in the denominator, with a sum of n partial fractions of the form M 1 x + N1 2

ax + bx + c

+

M 2x + N2

( ax

2

+ bx + c

)

2

+ ... +

M n x + Nn

( ax

2

+ bx + c

this class of integrals are solved.

Example 5.3-23: Evaluate the integral

)

n

. The following examples show the steps as to how

x2

∫ x 4 + 2 x 2 + 1 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

2

Second - Factor the denominator x 4 + 2 x 2 + 1 into x 2 + 1 . Third - Write the factors in partial fraction form. Since the quadratic form in the denominator is repeated, the integrand can be represented in the following way: x2 4

2

x + 2x + 1

=

x2

(x + 1) 2

2

=

Ax + B

+

2

x +1

Cx + D

(x + 1) 2

2

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. x2 x 4 + 2x 2 + 1 x2

(

)

=

( Ax + B ) (x 2 + 1)+ Cx + D

(x + 1) 2

2

= ( Ax + B ) x 2 + 1 + Cx + D = Ax 3 + Ax + Bx 2 + B + Cx + D x2

= Ax 3 + Bx 2 + ( A + C )x + (B + D ) therefore, B =1

A=0

A+C = 0

B+D =0

which result in having A = 0 , B = 1 , C = 0 , and D = −1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x2

Ax + B

Cx + D

0 +1

0 −1

1

1

∫ x 4 + 2 x 2 + 1 dx = ∫ x 2 + 1 dx + ∫ (x 2 + 1)2 dx = ∫ x 2 + 1 dx + ∫ (x 2 + 1)2 dx = ∫ x 2 + 1 dx − ∫ (x 2 + 1)2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. To solve the second integral let x = tan t , then

Hamilton Education Guides

dx = sec 2 t dt

which implies dx = sec 2 t dt .

344

Calculus I

5.3 Integration by Partial Fractions

sec 2 t

1

1

∫ x 2 + 1 dx − ∫ (x 2 + 1)2 dx = arc tan x − ∫ (tan 2 t + 1)2 dt = arc tan x − ∫

1 2

sec t

dt

= arc tan x − ∫ cos 2 t dt = tan −1 x − 1 2

1 2

sec 2 t

= arc tan x − ∫

1 2

1 2

∫ ( 1 + cos 2t ) dt

x

= tan −1 x − ( t + sin t cos t ) + c = tan −1 x − tan −1 x − ⋅

(sec t )

1

⋅

1+ x 2

2

2

= arc tan x − ∫

sec 2 t sec 4 t

dt

1 1 = tan −1 x −  t + sin 2t  + c

+c

1+ x 2

dt

2

2

=



1 x tan −1 x − +c 2 2 x2 + 1

(

)

Or, we could use the already derived integration formulas by using Tables 5.2-1 and 5.4-3. Note – Since the objective of this section is to teach the process for solving integrals using the Partial Fractions method, in the remaining example problems, we will use the already derived integration formulas summarized primarily in Tables 5.2-1 and 5.4-3 in order to solve this class of problems. Seventh - Check the answer by differentiating the solution. The result should match the integrand. 1 2

Let y = tan −1 x − =

(

1

)

2 x 2 +1

−

x

(

− x 2 +1

(

)

2 x 2 +1

)

2 2 x 2 +1

=

(

x 2 +1+ x 2 −1

(

=

)

2 2 x 2 +1

Example 5.3-24: Evaluate the integral

)

1 1 1 1⋅ x 2 + 1 − 2 x ⋅ x 1 ⋅ − ⋅ +0 = 2 2 2 x +1 2 x 2 +1 2 x 2 +1

+ c , then y ′ =

(

2x 2

=

)

2 2 x 2 +1

(

x2

=

(x + 1)

2

2

)

(

)

−

x 2 − 2x 2 + 1

(

)

2 2 x 2 +1

x2 x 4 + 2x 2 + 1

x 2 +1

∫ x 4 + 8x 2 + 16 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

2

Second - Factor the denominator x 4 + 8 x 2 + 16 into x 2 + 4 . Third - Write the factors in partial fraction form. Since the quadratic form in the denominator is repeated, the integrand can be represented in the following way: x 2 +1 4

2

x + 8 x + 16

x 2 +1

=

(x

2

+4

)

2

=

Ax + B

+

2

x +4

Cx + D

(x

2

+4

)

2

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. x 2 +1 x 4 + 8 x 2 + 16

x 2 +1

(

Hamilton Education Guides

( Ax + B ) (x 2 + 4)+ Cx + D

(x

2

+4

)

2

= ( Ax + B ) x 2 + 4 + Cx + D = Ax 3 + 4 Ax + Bx 2 + 4 B + Cx + D x 2 +1

A=0

)

=

= Ax 3 + Bx 2 + (4 A + C )x + (4 B + D ) therefore, B =1

4A + C = 0

4B + D = 1

345

Calculus I

5.3 Integration by Partial Fractions

which result in having A = 0 , B = 1 , C = 0 , and D = −3 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x 2 +1

Ax + B

0 +1

Cx + D

0−3

3

1

∫ x 4 + 8x 2 + 16 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx = ∫ x 2 + 4 dx − ∫ (x 2 + 4)2 dx Sixth - Integrate each integral individually by using Tables 5.2-1 and 5.4-3. 1

3

∫ x 2 + 4 dx − ∫ (x 2 + 4)2 dx

=

1 x 3 x tan −1 − tan −1 − 2 2 16 2

(

3x

8 x2 + 4

)

+c =

x 5 tan −1 − 16 2

(

3x

8 x2 + 4

)

+c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = =

(

5 x tan −1 − 16 2

5

8 x2 + 4

)

(

3x

8 x2 + 4

)

3 x 2 − 2x 2 + 4 − ⋅ 2 8 x2 + 4

(

)

+ c , then y ′ =

=

(

5

8 x2 + 4

Example 5.3-25: Evaluate the integral

)

5 ⋅ 16

1

()

x 2 2

(

3 − x2 + 4 − ⋅ 2 8 x2 + 4

(

)

1 3 1⋅ x 2 + 4 − 2 x ⋅ x +0 ⋅ − ⋅ 2 +1 2 8 x2 + 4

=

)

(

)

5 x 2 + 20 + 3 x 2 − 12

(

8 x2 + 4

)

2

=

x 2 +1

(x

2

+4

)

2

=

x 2 +1 x 4 + 8 x 2 + 16

1

∫ x4 + 10 x2 + 25 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is a rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

)

(

2

Second - Factor the denominator x 4 + 10 x 2 + 25 into x 2 + 5 . Third - Write the factors in partial fraction form. Since the quadratic form in the denominator is repeated, the integrand can be represented in the following way: 1 4

=

2

x + 10 x + 25

(x

1 2

+5

)

2

Ax + B

=

+

2

x +5

Cx + D

(x

2

+5

)

2

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. 1 x 4 + 10 x 2 + 25

(

)

=

( Ax + B ) (x 2 + 5)+ Cx + D

(x

2

+5

)

2

1 = ( Ax + B ) x 2 + 5 + Cx + D = Ax 3 + 5 Ax + Bx 2 + 5 B + Cx + D 1 = Ax 3 + Bx 2 + (5 A + C )x + (5B + D ) therefore, A=0

B=0

5A + C = 0

5B + D = 1

which result in having A = 0 , B = 0 , C = 0 , and D = 1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants Hamilton Education Guides

346

Calculus I

5.3 Integration by Partial Fractions

with their specific values. 1

Ax + B

0+0

Cx + D

0 +1

1

∫ x4 + 10 x2 + 25 dx = ∫ x 2 + 5 dx + ∫ (x 2 + 5)2 dx = ∫ x 2 + 5 dx + ∫ (x 2 + 5)2 dx = ∫ (x 2 + 5)2 dx Sixth - Integrate each integral individually by using Tables 5.2-1 and 5.4-3. 1

1

∫ (x 2 + 5)2 dx = 10

5

x

tan −1

5

+

10 (x

x 2

+ 5)

+c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y =

=

(

1 10 5

5 2

50 x + 5

)

tan −1

+

x 5

+

(

x 2

10 x + 5

1 x 2 − 2x 2 + 5 ⋅ 2 10 x2 + 5

(

)

+ c , then y ′ =

)

=

(

1 2

10 x + 5

Example 5.3-26: Evaluate the integral

)

1

(

2

10 5  x    +1  5

− x2 + 5

+

1

⋅

2

10 x + 5

)

2

⋅

1

+

5

(

(

2

10 x + 5

(

2

10 x + 5

x2 + 5− x2 + 5

=

)

1⋅ x 2 + 5 − 2 x ⋅ x

)

2

=

(x

)

2

1 2

+5

)

2

+0

=

1 4

x + 10 x 2 + 25

x3

∫ x 4 + 4 x 2 + 4 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is a rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

2

Second - Factor the denominator x 4 + 4 x 2 + 4 into x 2 + 2 . Third - Write the factors in partial fraction form. Since the quadratic form in the denominator is repeated, the integrand can be represented in the following way: x3 4

2

x + 4x + 4

=

(x

x3 2

+2

)

2

=

Ax + B

+

2

x +2

Cx + D

(x

2

+2

)

2

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. x3 x 4 + 4x 2 + 4 x3

(

)

=

( Ax + B ) (x 2 + 2)+ Cx + D

(x

2

+2

)

2

= ( Ax + B ) x 2 + 2 + Cx + D = Ax 3 + 2 Ax + Bx 2 + 2 B + Cx + D x3

A =1

= Ax 3 + Bx 2 + (2 A + C )x + (2 B + D ) therefore, B=0

2A + C = 0

2B + D = 0

which result in having A = 1 , B = 0 , C = −2 , and D = 0 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

Hamilton Education Guides

347

Calculus I

5.3 Integration by Partial Fractions

x3

Ax + B

x+0

Cx + D

−2 x + 0

2x

x

∫ x 4 + 4 x 2 + 4 dx = ∫ x 2 + 2 dx + ∫ (x 2 + 2)2 dx = ∫ x 2 + 2 dx + ∫ (x 2 + 2)2 dx = ∫ x 2 + 2 dx − ∫ (x 2 + 2)2 dx Sixth - Integrate each integral individually by using Tables 5.2-1 and 5.4-3. x 1

2x

x

1

x

∫ x 2 + 2 dx − ∫ (x 2 + 2)2 dx = ∫ u ⋅ 2 x du − 2∫ u 2 ⋅ 2 x du =

1 1 ln u − u − 2+1 + c 2 − 2 +1

=

1 ln u + u −1 + c 2

=

1 2

=

1

2

1

1

1

∫ u du − 2 ∫ u 2 du = 2 ∫ u du − ∫ u

1 1 ln u + + c 2 u

−2

du

1 1 ln x 2 + 2 + +c 2 2 x +2

=

Seventh - Check the answer by differentiating the solution. The result should match the integrand. 1 2

1

Let y = ln x 2 + 2 + =

x 3 + 2x − 2x

(x

2

+2

)

2

=

(x

2

x +2

x3 2

+2

+c ,

=

)

2

then y ′ =

1 2x ⋅ − 2 x2 + 2

2x

(x

2

+2

+0 =

)

2

x 2

x +2

−

(x

2x 2

+2

)

2

=

(

)

x x 2 + 2 − 2x

(x

2

+2

)

2

x3 x 4 + 4x 2 + 4

Example 5.3-27: Evaluate the integral

2x 2 + x + 7

∫ x 4 + 8x 2 + 16 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is a rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

2

Second - Factor the denominator x 4 + 8 x 2 + 16 into x 2 + 4 . Third - Write the factors in partial fraction form. Since the quadratic form in the denominator is repeated, the integrand can be represented in the following way: 2x 2 + x + 7 4

=

2

x + 8 x + 16

2x 2 + x + 7

(x

2

+4

)

2

=

Ax + B 2

x +4

+

Cx + D

(x

2

+4

)

2

Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers. 2x 2 + x + 7 x 4 + 8 x 2 + 16

2x 2 + x + 7

(

)

( Ax + B ) (x 2 + 4)+ Cx + D

(x

2

+4

)

2

= ( Ax + B ) x 2 + 4 + Cx + D = Ax 3 + 4 Ax + Bx 2 + 4 B + Cx + D

2x 2 + x + 7

A=0

=

= Ax 3 + Bx 2 + (4 A + C )x + (4 B + D ) therefore,

B=2

4A + C = 1

4B + D = 7

which result in having A = 0 , B = 2 , C = 1 , and D = −1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

Hamilton Education Guides

348

Calculus I

5.3 Integration by Partial Fractions

2x 2 + x + 7

Ax + B

0+2

Cx + D

x −1

2

x −1

∫ x 4 + 8x 2 + 16 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx =

2

1

x

∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx − ∫ (x 2 + 4)2 dx

Sixth - Integrate each integral individually by using Tables 5.2-1 and 5.4-3. 2

1

x

∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx − ∫ (x 2 + 4)2 dx 1 2

1 2

1 1 1 x u − 2+1 − tan −1 − 2 − 2 +1 16 2

x 2

= 2 ⋅ tan −1 + ⋅ x 2

= tan −1 −

1

(

2 x2 + 4

)

−

x 1 tan −1 − 16 2

x

(

x 2

= 2 ⋅ tan −1 + ∫

8 x2 + 4

(

x 2

8 x +4

=

+c

)

)

x x 1 1 x +c ⋅ du − tan −1 − 16 2 8 x2 + 4 u2 2x

(

+ c = tan −1

15 x tan −1 + 16 2

(

)

x 1 1 x − − tan −1 − 2 2u 16 2 −4 − x

8 x2 + 4

)

=

+c

(

x 2

8 x +4

x 15 tan −1 − 16 2

)

(

+c x+4

8 x2 + 4

)

+c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = =

15 x tan −1 − 16 2

15

(

8 x2 + 4

2

2

)

+ c , then y ′ =

1 x 2 − 2 x 2 − 8x + 4 − ⋅ 2 8 x2 + 4

( ) ( 8 (2 x + x + 7 ) 2x = = (x 8(x + 4 ) 8 x2 + 4

x+4

)

2

+ x+7

2

+4

2

)

2

=

=

(

15

8 x2 + 4

)

15 ⋅ 16 +

1

()

x 2 2

(

)

1 1 ⋅ x 2 + 4 − 2 x ⋅ (x + 4 ) ⋅ − +0 2 +1 2 8 x2 + 4

x 2 + 8x − 4

(

8 x2 + 4

)

2

=

(

)

15 x 2 + 60 + x 2 + 8 x − 4

(

8 x2 + 4

=

)

2

16 x 2 + 8 x + 56

(

8 x2 + 4

)

2

2x 2 + x + 7 x 4 + 8 x + 16

Section 5.3 Practice Problems – Integration by Partial Fractions Evaluate the following integrals: a.

dx

∫ x 2 + 5x + 6 x+5

=

d.

∫ x 3 + 2x 2 + x

g.

∫ x 3 − 1 dx =

dx

=

1

Hamilton Education Guides

x 2 +1

b.

∫ x 3 − 4 x dx =

e.

1

∫ x 3 − 2x 2 + x

h.

∫ x4 − 1 dx =

1

dx

c. =

1

∫ 36 − x 2 dx = x2 + 3

f.

∫ x 2 − 1 dx =

i.

∫ x 3 + 64 dx =

1

349

Calculus I

5.4

5.4 Integration of Hyperbolic Functions

Integration of Hyperbolic Functions

In the following examples we will solve problems using the formulas below: Table 5.4-1: Integration Formulas for Hyperbolic Functions 1.

∫ sinh x dx = cosh x + c

2.

∫ cosh x dx = sinh x + c

4.

∫ coth x dx = ln

5.

∫ sec h x dx = sin

7.

∫ tanh x sec h x dx = − sec h x + c

8.

∫ coth x csc h x dx = − csc h x + c

sinh x + c

sinh 2 x x + +c 4 2

10.

∫ cosh

2

x dx =

13.

∫ sec h

2

x dx = tanh x + c

11.

∫ tanh

14.

∫ csc h

2

2

−1

(tanh x ) + c

x dx = x − tanh x + c

3.

∫ tanh x dx = ln cosh x + c

6.

∫ csc h x dx = ln

9.

∫ sinh

12.

2

∫ coth

x dx = 2

tanh

x +c 2

sinh 2 x x − +c 4 2

x dx = x − coth x + c

x dx = − coth x + c

Additionally, the following formulas, similar to the trigonometric functions, hold for the hyperbolic functions: 1. Unit Formulas tanh h 2 x + sec h 2 x = 1

cosh 2 x − sinh 2 x = 1

coth h 2 x − csc h 2 x = 1

2. Addition Formulas sinh (x ± y ) = sinh x cosh y ± cosh x sinh y tanh (x ± y ) =

cosh (x ± y ) = cosh x cosh y ± sinh h x sinh y

tanh x ± tanh y 1 ± tanh x tanh y

coth (x ± y ) =

coth x coth y ± 1 coth y ± coth x

3. Half Angle Formulas sinh

1 x= 2

cosh x − 1 2

cosh

1 x= 2

cosh x + 1 2

tanh

1 x= 2

cosh x − 1 cosh x + 1

tanh

1 x 2

sinh x cosh x + 1

=

=

x 0

sinh

cosh x − 1 1 x=− 2 2

x 0

x 0

sinh

cosh x − 1 1 x=− 2 cosh x + 1

x 0

or,

cosh x − 1 sinh x

4. Double Angle Formulas sinh 2 x = 2 sinh x cosh x tanh 2 x =

cosh 2 x

= cosh 2 x + sinh 2 x = 2 cosh 2 x − 1 = 1+ 2 sinh 2 x

2 tanh x 1 + tanh 2 x

Hamilton Education Guides

350

Calculus I

5.4 Integration of Hyperbolic Functions

Also, the hyperbolic functions are defined by sinh x =

(

2

) (e (e

1 2

(e

1 x e − e −x 2 1

sinh x 2 = tanh x = cosh x 1

sec h x =

1 = cosh x

cosh x = x x

− e −x + e −x

1 x

+e

−x

) )

)

=

=

1 2

(e

x

cosh x 2 = coth x = sinh x 1

e x + e −x

2 e +e

x

2

) (e (e

1

e x − e −x

x

(

1 x e + e −x 2

csc h x =

−x

1 = sinh x

x

+ e −x − e −x

1 −e

−x

) )

)

=

=

e x + e −x e x − e −x

2 x

e − e −x

Also note that the negative argument of the hyperbolic functions is equal to the following: sinh (− x ) = − sinh x

tanh (− x ) = − tanh x

coth (− x ) = − coth x

csc h (− x ) = − csc h x

cosh (− x ) = cosh x

sec h (− x ) = sec h x

Finally, we need to know how to differentiate the hyperbolic functions (addressed in Chapter 3, Section 3.4) in order to check the answer to the given integrals below. The derivatives of hyperbolic functions are repeated here and are as follows: Table 5.4-2: Differentiation Formulas for Hyperbolic Functions d sinh u dx d cosh u dx d tanh u dx

du dx du sinh u ⋅ dx du sec h 2 u ⋅ dx

d coth u dx d sec h u dx d csc h u dx

= cosh u ⋅ = =

= − csc h 2 u ⋅

du dx

du dx du − csc h u coth u ⋅ dx

= − sec h u tanh u ⋅ =

Let’s integrate some hyperbolic functions using the above integration formulas. Example 5.4-1: Evaluate the following integrals: 1

a.

∫ sinh 5x dx =

b.

∫ sinh 6 x dx

d.

∫ cosh 5 x dx =

1

e.

∫ (sinh 4 x + cosh 2 x ) dx

g.

∫ csc h

h.

∫ csc h

k.

∫x

j.

2

5 x dx

=

2 2 ∫ x sec h (x + 5) dx =

2

2

=

1 x dx 4

=

=

csc h 2 x 3 dx

=

c.

∫ cosh 7 x dx =

f.

∫ csc h 8x dx =

i.

∫x

l.

∫ 2 x csc h

2

sec h 2 x 3 dx 2 2

x dx

= =

Solutions: a. Given ∫ sinh 5 x dx let u = 5 x , then

Hamilton Education Guides

du d = 5x dx dx

= 5 which implies dx =

du 5

. Therefore,

351

Calculus I

5.4 Integration of Hyperbolic Functions

du

∫ sinh 5x dx = ∫ sinh u ⋅ 5

1 sinh u du 5

∫

=

=

1 cosh u + c 5

1 cosh 5 x + c 5

=

1 d d 1 1 d 1 Check: Let y = cosh 5 x + c , then y ′ = ⋅ sinh 5 x + c = ⋅ sinh 5 x ⋅ 5 x + 0 = ⋅ sinh 5 x ⋅ 5 5 dx 5 5 = ⋅ sinh 5 x = sinh 5 x 5 1 x x du d x = sinh dx let u = , then = 6 6 6 dx dx 6

∫

b. Given x

∫ sinh 6 dx = ∫ sinh u ⋅ 6 du

d x d cosh + c 6 dx dx

x 6

6 x ⋅ sinh 6 6

c. Given ∫ cosh 7 x dx let u = 7 x , then du

x +c 6 x d x +0 6 dx 6

= 6 ⋅ sinh ⋅

x 1 6 6

= 6 ⋅ sinh ⋅

x 6

= sinh

∫ cosh 7 x dx = ∫ cosh u ⋅ 7

5

dx

which implies dx = 6 du . Therefore,

= 6∫ sinh u du = 6 cosh u + c = 6 cosh

Check: Let y = 6 cosh + c , then y ′ = 6 ⋅ =

5

dx

=

1 7

du d = 7x dx dx

1 cosh u du 7

∫

Check: Let y = sinh 7 x + c , then y ′ =

=

= 7 which implies dx =

1 sinh u + c 7

=

1 d d ⋅ sinh 7 x + c 7 dx dx

7 ⋅ cosh 7 x = cosh 7 x 7 x du d x x = cosh dx let u = , then 5 dx dx 5 5

du 7

. Therefore,

1 sinh 7 x + c 7

=

1 d 7x + 0 ⋅ cosh 7 x ⋅ 7 dx

=

1 ⋅ cosh 7 x ⋅ 7 7

= d. Given

∫ x

∫ cosh 5 dx = ∫ cosh u ⋅ 5 du x 5

5 x ⋅ cosh 5 5

= cosh

1 5

which implies dx = 5 du . Therefore,

= 5∫ cosh u du = 5 sinh u + c = 5 sinh

Check: Let y = 5 sinh + c , then y ′ = 5 ⋅ =

=

d x d sinh + c 5 dx dx

x +c 5 x d x +0 5 dx 5

= 5 ⋅ cosh ⋅

x 1 5 5

= 5 ⋅ cosh ⋅

x 5

e. Given ∫ (sinh 4 x + cosh 2 x ) dx = ∫ sinh 4 x dx + ∫ cosh 2 x dx let: a. u = 4 x , then

du du du d = 4x ; = 4 ; du = 4dx ; dx = 4 dx dx dx

b. v = 2 x , then

dv dv dv d = 2 ; dv = 2dx ; dx = = 2x ; 2 dx dx dx

Therefore,

du

and

. dv

∫ sinh 4 x dx + ∫ cosh 2 x dx = ∫ sinh u ⋅ 4 + ∫ cosh v ⋅ 2

Hamilton Education Guides

=

1 1 sinh u du + cosh v dv 4 2

∫

∫

352

Calculus I

=

5.4 Integration of Hyperbolic Functions

1 1 cosh u + c1 + sinh v + c 2 4 2

1 1 cosh 4 x + sinh 2 x + c1 + c 2 4 2

= 1 2

1 4

Check: Let y = cosh 4 x + sinh 2 x + c then y ′ =

=

1 1 cosh 4 x + sinh 2 x + c 2 4

d 1 d 1 d c ⋅ cosh 4 x + ⋅ sinh 2 x + dx 4 dx 2 dx

4 2 d 1 ⋅ cosh 2 x ⋅ 2 x + 0 = ⋅ sinh 4 x + ⋅ cosh 2 x = sinh 4 x + cosh 2 x 4 2 dx 2 du du d csc h 8 x dx let u = 8 x , then = 8 x = 8 which implies du = 8dx ; dx = 8 dx dx

d 1 4x ⋅ sinh 4 x ⋅ dx 4

=

+

∫

f. Given

du

∫ csc h 8x dx = ∫ csc h u ⋅ 8

1 csc h u du 8

∫

=

1 8

(

)

=

1 ⋅ 2

1 cosh 2 4 x sinh 4 x cosh 4 x

1 1 ⋅ ⋅ sec h 2 4 x ⋅ 4 + 0 8 tanh 4 x cosh 2 4 x − sinh 2 4 x

=

1 ⋅ 2

=

1 sinh 8 x

cosh 2 4 x sinh 4 x cosh 4 x

=

2

5 x dx

=

8x 1 +c ln tanh 2 8

1 4 sec h 2 4 x ⋅ 8 tanh 4 x

=

=

=

1 sec h 2 4 x ⋅ 2 tanh 4 x

cosh 4 x 1 ⋅ 2 cosh 2 4 x ⋅ sinh 4 x

=

1 ln tanh 4 x + c 8

=

1 1 d ⋅ ⋅ (tanh 4 x ) + 0 8 tanh 4 x dx

=

1 1 − tanh 2 4 x ⋅ 2 tanh 4 x

1 2 cosh 4 x ⋅ sinh 4 x

=

1 ⋅ 2

2 1 − sinh 2 4 x

cosh 4 x sinh 4 x cosh 4 x

1 sinh 2 ⋅ 4 x

=

= csc h 8 x

g. Given ∫ csc h 2 5 x dx let u = 5 x , then

∫ csc h

=

1 d d ⋅ ln tanh 4 x + c 8 dx dx

Check: Let y = ln tanh 4 x + c , then y ′ = =

1 u ln tanh +c 8 2

=

. Therefore,

∫ csc h

2

u⋅

du 5

du d du du = 5x ; = 5 ; du = 5dx ; dx = dx dx 5 dx

∫

1 5

1 5

1 = − coth u + c = − coth 5 x + c

1 csc h 2 u du 5

=

5

1 d (coth 5 x )⋅ d 5 x + d c dx dx 5 dx

Check: Let y = − coth 5 x + c , then y ′ = − ⋅ 5 ⋅ csc h 2 5 x = csc h 2 5 x 5 1 1 csc h 2 x dx let u = x , 4 4

. Therefore,

1 5

(

)

= − ⋅ − csc h 2 5 x ⋅ 5 + 0

=

∫

h. Given

∫ csc h

2

1 x dx 4

=

∫ csc h

2

u ⋅ 4du

then

du d x du 1 ; = = dx dx 4 dx 4

1 4

= 4∫ csc h 2 u du = − 4 coth u + c = − 4 coth x + c

x 4

Check: Let y = −4 coth + c , then y ′ = − 4 ⋅ = i. Given

∫x

2

4 x csc h 2 4 4

= csc h 2

x 4

= csc h 2

sec h 2 x 3 dx let u = x 3 , then

Hamilton Education Guides

; 4du = dx ; dx = 4du . Therefore,

d x d c coth + dx 4 dx

x d x +0 4 dx 2

= − 4 ⋅ − csc h 2 ⋅

x 1 4 4

= 4 csc h 2 ⋅

1 x 4

du du d 3 du = x ; = 3x 2 ; du = 3 x 2 dx ; dx = dx dx dx 3x 2

. Therefore,

353

Calculus I

∫x

2

5.4 Integration of Hyperbolic Functions

∫x

sec h 2 x 3 dx =

2

sec h 2 u ⋅

du 3x

=

2/

1 sec h 2 u du 3

∫

1 3

Check: Let y = tanh x 3 + c , then y ′ = 3x 2 ⋅ sec h 2 x 3 3

=

1 d d ⋅ tanh x 3 + c 3 dx dx

2 2 ∫ x sec h (x + 5) dx

=

∫ x sec h

2

u⋅

du 2x

=

(

du d 2 = x +5 dx dx

1 sec h 2 u du 2

∫

)

(

1 2

∫x

k. Given

∫x

2

(

)

1 ⋅ sec h 2 x 2 + 5 ⋅ 2 x 2 2

csc h 2 x 3 dx let

∫x

csc h 2 x 3 dx =

2

1 3

=

1 tanh u + c 2

(

)

(

)

(

du 3x

2

=

1 csc h 2 u du 3

∫

∫ 2 x csc h

2 2

x dx =

)

1 tanh x 2 + 5 + c 2

=

1 d ⋅ sec h 2 x 2 + 5 ⋅ x2 + 5 + 0 2 dx

)

(

) (

)

. Therefore,

1 3

1 = − coth u + c = − coth x 3 + c 3

d 1 d c coth x 3 + dx 3 dx

1 ⋅ csc h 2 x 3 ⋅ 3 x 2 3

(

=

2x ⋅ sec h 2 x 2 + 5 = x sec h 2 x 2 + 5 2 du du d 3 du u = x 3 , then = = 3x 2 ; du = 3 x 2 dx ; dx = x ; dx dx dx 3x 2

=

csc h 2 u ⋅

l. Given ∫ 2 x csc h 2 x 2 dx let

1 ⋅ sec h 2 x 3 ⋅ 3 x 2 3

=

) ; dudx = 2 x ; du = 2 x dx ; dx = du2 x . Therefore,

Check: Let y = − coth x 3 + c , then y ′ = − ⋅ =

1 tanh x 3 + c 3

1 d 3 ⋅ sec h 2 x 3 ⋅ x +0 3 dx

d 1 d c ⋅ tanh x 2 + 5 + dx 2 dx

Check: Let y = tanh x 2 + 5 + c , then y ′ = =

=

=

= x 2 sec h 2 x 3

2 2 2 ∫ x sec h (x + 5) dx let u = x + 5 , then

j. Given

1 tanh u + c 3

=

1 3

= − ⋅ − csc h 2 x 3 ⋅

3x 2 ⋅ csc h 2 x 3 = x 2 csc h 2 x 3 3 du d 2 du du . u = x 2 , then = = 2 x ; du = 2 x dx ; dx = x ; dx dx dx 2x

d 3 x +0 dx

=

∫ 2 x csc h

2

u⋅

du 2x

=

∫ csc h

Check: Let y = − coth x 2 + c , then y ′ = −

2

u du

Therefore,

= − coth u + c = − coth x 2 + c

d d c coth x 2 + dx dx

= csc h 2 x 2 ⋅

d 2 x +0 dx

= csc h 2 x 2 ⋅ 2 x

= 2 x csc h2 x 2 Example 5.4-2: Evaluate the following indefinite integrals: a.

∫ sec h

d.

∫ sinh

g.

∫ cosh

j.

∫ sec h

m.

5 x dx =

b.

∫x

(x + 1) cosh (x + 1) dx =

e.

∫ cosh

h.

∫x

k.

∫ csc h 7 x coth 7 x dx

n.

∫ sinh

2

5

4

2

∫ cosh

x x sinh dx 2 2

=

(5 x − 1) dx = 3

x dx 5

=

Hamilton Education Guides

2

2

sinh x 3 dx = 5

x sinh x dx =

cosh x 3 dx =

3

x dx

=

=

c.

∫ x csc h x

f.

∫ cosh

i.

∫

l.

∫ tanh

o.

∫ 2 sinh x dx =

5

2

dx =

5 x sinh 5 x dx =

e 2x sec h e 2 x dx 3 2

=

10 x dx =

x

354

Calculus I

5.4 Integration of Hyperbolic Functions

Solutions: du d du du = 5x ; = 5 ; du = 5 dx ; dx = dx dx dx 5

a. Given ∫ sec h 2 5 x dx let u = 5 x , then

∫ sec h

2

5 x dx =

∫ sec h

2

u⋅

du 5

=

1 5

∫x

2

Thus,

∫x

2

=

1 d d ⋅ tanh 5 x + c 5 dx dx

=

∫

Check: Let y = tanh 5 x + c , then y ′ = b. Given

1 tanh u + c 5

1 sec h 2 u du 5

=

. Therefore,

1 tanh 5 x + c 5 1 sec h 2 5 x ⋅ 5 + 0 5

5 sec h 2 5 x 5

=

= sec h 2 5 x

du du d 3 . sinh x 3 dx let u = x 3 , then = x = 3x 2 which implies du = 3 x 2 dx ; dx = dx dx 3x 2

sinh x 3 dx =

∫x

2

sinh u ⋅

du 3x

2

1 sinh u du 3

∫

=

=

1 cosh u + c 3

=

1 cosh x 3 + c 3

1 d 1 d 1 d 1 Check: Let y = cosh x 3 + c , then y ′ = ⋅ cosh x 3 + c = ⋅ sinh x 3 ⋅ x 3 + 0 = ⋅ sinh x3 ⋅ 3x 2 3 dx

3

= c. Given

∫

2

3x ⋅ sinh x3 3

∫ x csc h x

x csc h x 2 dx

=

3

dx

= x 2 sinh x3

2

dx

∫

x csc h u ⋅

let u = x 2 , then

1 2

3

dx

Check: Let y = ln tanh

du 2x

=

1 2

∫

du d 2 du du . = x ; = 2 x ; du = 2 x dx ; dx = dx dx dx 2x

csc h u du

=

1 u ln tanh +c 2 2

=

Therefore,

x2 1 +c ln tanh 2 2

1 1 x2 d x2 1 d + c , then y ′ = ⋅ ln tanh c = ⋅ + 2 tanh 2 dx 2 2 dx 2

2

x2 2

⋅

d x2 tanh +0 2 dx

2 x 2 x 2 d x2 1 1 x x 1 − tanh 2 1 sec h 2 2 x 2 x = ⋅ = ⋅ = ⋅ ⋅ sec h ⋅ +0 = ⋅ ⋅ 2 2 2 2 tanh x 2 dx 2 2 2 2 tanh x 2 tanh x

2

1−

2

2 2 cosh 2 x 2 x2

sinh

2

2

sinh 2 x

2

2 cosh x 2

=

x ⋅ 2

2 2 cosh 2 x − sinh 2 x 2 2 2 cosh 2 x 2 x2

sinh

cosh

=

x sinh x

2

1

=

2 x2

x ⋅ 2

2 cosh 2 x 2 x2

sinh

cosh

2

2 x2

=

x ⋅ 2

2

cosh cosh 2

2

x2 2

2

x x ⋅ sinh 2 2

x

= 2 cosh

2

2

x x ⋅ sinh 2 2

=

x 2

sinh 2 ⋅ x2

= x csc h x 2

d. Given ∫ sinh 5 (x + 1) cosh (x + 1) dx let u = sinh (x + 1) , then du = cosh (x + 1) ⋅ dx ; dx =

Hamilton Education Guides

du . cosh (x + 1)

du du d = cosh (x + 1) ; = sinh (x + 1) ; dx dx dx

Therefore,

355

Calculus I

∫ sinh

5.4 Integration of Hyperbolic Functions

5

(x + 1) cosh (x + 1) dx =

∫u

5

cosh (x + 1) ⋅

du cosh (x + 1)

∫u

=

x sinh x dx

5

sinh x ⋅

du sinh x

∫u

=

1 6

5

du

Check: Let y = g. Given ∫ cosh 4

∫ cosh

4

∫u

5 x sin 5 5 x dx =

5

sinh 5 x ⋅

1 cosh 6 5 x + c , 30

x x sinh dx 2 2

x x sinh dx 2 2

=

∫u

2 5

4

du 5 sinh 5 x

then y ′ =

let u = cosh sinh

x 2

∫x

2

∫x

2

2

∫x

2

cosh u ⋅

du 3x 2

=

1 u 5 du 5

∫

du . sinh x

; dx =

=

Therefore,

6 cosh 5 x sinh x 6

=

1 1 6 ⋅ u +c 5 6

=

= cosh 5 x sinh x

; dx =

du 5 sinh 5 x

. Thus,

1 cosh 6 5 x + c 30

6 cosh 5 5 x ⋅ 5 sinh 5 x 30 +0 = cosh 5 5 x sinh 5 x = cosh 5 hx sinh 5 x 30 30 du d x du 1 x = cosh ; = sinh dx dx 2 dx 2 2

1 cosh u du 3

∫

1 3

∫

= sinh 5 (x + 1) cosh (x + 1)

1 cosh 6 x + c 6

=

2 5 u +c 5

=

x x 1 2 ⋅ 5 cosh 4 ⋅ sinh ⋅ + 0 5 2 2 2

x 2

Check: Let y = sinh x 3 + c , then y ′ =

∫

1 6 u +c 6

1 sinh 6 ( x + 1) + c 6

=

; dx =

2du sin

x 2

. Therefore,

2 x cosh 5 + c 5 2

=

10 x x cosh 4 sin 2 2 10

= cosh 4

x x sinh 2 2

du du d 3 du . Therefore, cosh x 3 dx let u = x 3 , then = x ; = 3x 2 ; du = 3 x 2 ⋅ dx ; dx = dx dx dx 3x 2

cosh x 3 dx =

i. Given

=

= 2∫ u 4 du =

Check: Let y = cosh 5 + c , then y ′ = h. Given

=

, then

x 2 du ⋅ 2 sinh x

1 6 u +c 6

du du d = 5 sinh 5 x cosh 5 x ; = dx dx dx

f. Given ∫ cosh 5 5 x sinh 5 x dx let u = cosh 5 x , then 5

du =

1 ⋅ 6 cosh 5 x ⋅ sinh x + 0 6

Check: Let y = cosh 6 x + c , then y ′ =

∫ cosh

5

du d du = cosh x ; = sinh x dx dx dx

e. Given ∫ cosh 5 x sinh x dx let u = cosh x , then 5

∫u

1 ⋅ 6 sinh 5 (x + 1) ⋅ cosh (x + 1) + 0 6

1 6

Check: Let y = sinh 6 (x + 1) + c , then y ′ =

∫ cosh

=

e 2x sec h e 2 x dx 3

e 2x sec h e 2 x dx 3

= 1 6

∫

(

)

=

=

=

1 sinh x 3 + c 3

3 2 x cosh x 3 + 0 3

= x 2 cosh x 3

du du d 2 x du e ; = = 2e 2 x ; du = 2e 2 x ⋅ dx ; dx = dx dx dx 2e 2 x 1 sec h u du 6

∫

Check: Let y = sin −1 tanh e 2 x + c , then y ′ =

Hamilton Education Guides

1 ⋅ sinh u + c 3

1 ⋅ cosh x 3 ⋅ 3 x 2 + c 3

let u = e 2 x , then

1 2x du e sec h u ⋅ 3 2e 2 x

=

=

1 sin −1 (tanh u ) + c 6

1 2

6 1 − tanh e

2x

⋅

=

d tanh e 2 x + 0 dx

. Thus,

(

)

1 sin −1 tanh e 2 x + c 6

=

sec h 2 e 2 x 2 2x

6 sec h e

⋅

d 2x e dx

356

Calculus I

5.4 Integration of Hyperbolic Functions

=

sec h 2 e 2 x

⋅ 2e 2 x

6 sec h e 2 x

2e 2 x sec h 2 e 2 x ⋅ 6 sec h e 2 x

=

du d (5 x − 1) ; du = 5 ; du = 5dx ; dx = du = 5 dx dx dx

j. Given ∫ sec h 2 (5 x − 1) dx let u = 5 x − 1 , then

∫ sec h

2

(5 x − 1) dx =

∫ sec h

2

u⋅

du 5

1 sec h 2 u du 5

∫

=

1 5

Check: Let y = tanh (5 x − 1) + c , then y ′ = 5 ⋅ sec h 2 (5 x − 1) 5

∫ csc h u coth u

=

=

1 tanh u + c 5

1 d ⋅ sec h 2 (5 x − 1) ⋅ (5 x − 1) + 0 5 dx

du 7

=

1 csc h u coth u du 7

∫

∫ tanh

2

10 x dx

10 x dx =

Check: Let y =

2

(

∫ cosh

3

x dx 5

= ∫ cosh 2

x 5

= ∫ cosh dx + ∫ sinh 2 x

du 10

=

∫ cosh 5 dx + ∫ sinh

2

∫

2

x x cosh dx 5 5

x x cosh dx 5 5

=

=

then y ′ =

) = 1 − sec h 10 x = tanh

x x cosh dx 5 5

Hamilton Education Guides

d 7x + 0 dx

=

1 ⋅ csc h 7 x coth 7 x ⋅ 7 7

2

dw cosh u

1 (u − tanh u ) + c 10

=

. Therefore,

1 (10 x − tanh 10 x ) + c 10

d 1   2 10 − sec h 10 x ⋅ 10 x  + 0 dx 10  

=

(

1 10 − sec h 2 10 x ⋅10 10

)

10 x

x x x x x = ∫ 1 + sinh 2  cosh dx = ∫  cosh + sinh 2 cosh  dx

let u =

5

5



x 5

, then



5

5

5

1 du d x du 1 ; = = ; du = dx ; dx = 5 du . 5 dx dx 5 dx 5

∫ cosh u 5du + ∫ sinh

To solve the second integral let w = sinh u , then = 5 sinh u + 5∫ w 2 cosh u ⋅

1 7

7

du du d du = 10 ; du = 10 dx ; dx = = 10 x ; dx dx 10 dx

1 tanh 2 u du 10

1 (10 x − tanh 10 x ) + c , 10

10 1 − sec h 2 10 x 10

m.

u⋅

. Therefore,

= csc h 7 x coth 7 x

let u = 10 x , then

∫ tanh

1 ⋅ sec h 2 (5 x − 1) ⋅ 5 5

1 = − csc h u + c = − csc h 7 x + c

1 7

7 ⋅ csc h 7 x coth 7 x 7 2

=

du du d du = 7x ; = 7 ; du = 7 dx ; dx = dx dx dx 7

1 7

∫ tanh

1 tanh (5 x − 1) + c 5

=

Check: Let y = − csc h 7 x + c , then y ′ = − ⋅ − csc h 7 x coth 7 x ⋅

l. Given

. Therefore,

= sec h 2 (5 x − 1)

k. Given ∫ csc h 7 x coth 7 x dx let u = 7 x , then

∫ csc h 7 x coth 7 x dx

e 2x ⋅ sec h e 2 x 3

=

2

Therefore,

∫

u cosh u 5du = 5 sinh u + 5 sinh 2 u cosh u du

dw dw d dw = sinh u ; = cosh u ; dx = cosh u dx dx dx 5 3

= 5 sinh u + 5∫ w 2 dw = 5 sinh u + w 3 + c =

thus,

x x 5 sinh 3 + 5 sinh + c 5 5 3

357

Calculus I

5.4 Integration of Hyperbolic Functions

x 5

5 3

5 3

x 1 5 5

x 5

x 1 5 5

x 5

Check: Let y = 5 sinh + sinh 3 + c , then y ′ = 5 ⋅ cosh ⋅ + ⋅ 3 sinh 2 ⋅ cosh ⋅ + 0 = n.

∫ sinh

3

5 x 15 x x ⋅ cosh + ⋅ sinh 2 ⋅ cosh 5 5 5 15 5

x dx =

∫ sinh

2

x sinh x dx =

x x x x x = cosh 1 + sinh 2  = cosh ⋅ cosh 2 = cosh 3 5

5

∫ ( cosh

)

2

x − 1 sinh x dx =

∫ ( cosh

2

5

5

5

)

x sinh x − sinh x dx

= ∫ cosh 2 x sinh x dx − ∫ sinh x dx = ∫ cosh 2 x sinh x dx − cosh x . To solve the first integral let u = cosh x , then

du du du d . = sinh x ; dx = cosh x ; = sinh x dx dx dx

∫ cosh

2

x sinh x dx − cosh x

∫u

=

2

sinh x ⋅

Therefore,

du − cosh x sinh x

=

∫u

2

du − cosh x

=

1 3 u − cosh x + c 3

cosh 3 x − cosh x + c 3

=

1 1 Check: Let y = cosh 3 x − cosh + c , then y ′ = ⋅ 3 cosh 2 x ⋅ sinh x − sinh x + 0 = cosh 2 x ⋅ sinh x − sinh x 3

(

3

)

= sinh x cosh 2 x − 1 = sinh x ⋅ sinh 2 x = sinh 3 x o. Given

1

x

∫ 2 sinh x dx = 2 ∫ x sinh x dx let u = x

and dv = sinh x dx then du = dx and ∫ dv = ∫ sinh x dx

which implies v = cosh x . Using the substitution by parts formula ∫ u dv = u v − ∫ v du we obtain 1 2

∫ x sinh x dx =

1 1 x cosh x − cosh x dx 2 2

∫

1 2

=

1 1 x cosh x − sinh x + c 2 2

1 2

Check: Let y = x cosh x − sinh x + c , then y ′ = −

cosh x 2

=

x sinh x 2

=

1 (cosh x + x sinh x ) − 1 cosh x + 0 2 2

cosh x x sinh x + 2 2

=

1 x sinh x 2

Example 5.4-3: Evaluate the following indefinite integrals: a.

∫ tanh

8

x sec h 2 x dx =

d.

∫ coth

5

3 x csc h 2 3 x dx

g. j.

∫ coth

∫ tanh 5x dx =

f.

∫ 2 tanh 3 dx =

∫ sec h 2 tanh 2 dx =

x

i.

∫ csc h

∫x

l.

∫

∫ tanh

e.

∫ sec h 5x tanh 5x dx =

h.

∫x

k.

2

coth x 3 dx

=

Hamilton Education Guides

=

(x + 3) sec h 2 (x + 3) dx =

c.

b.

2

5

x

sec h 5 x 3 dx

=

3

x csc h 2 x dx = x

2

( 1 − 2 x ) dx =

sec h x x

dx

=

358

Calculus I

5.4 Integration of Hyperbolic Functions

Solutions: a.

∫ tanh

8

x sec h 2 x dx let u = tanh x , then

∫ tanh

8

x sec h 2 x dx

=

∫u

8

⋅ sec h 2 x ⋅

du d = tanh x dx dx

du 2

sec h x

1 9

Check: Let y = tanh 9 x + c then y ′ = b.

∫ tanh

5

(x + 3) sec h 2 (x + 3) dx let

; du = sec h 2 (x + 3) dx ; dx = =

∫u

5

du =

1 5+1 u +c 5 +1

=

∫u

=

sec h

2

(x + 3)

1 6 u +c 6

=

1 6

2

csc h x

. Therefore,

∫ coth

3

=

1 8+1 u +c 8 +1

=

∫ tanh

5

1 9 u +c 9

=

du

. Thus,

sec h 2 x

1 tanh 9 x + c 9

= (tanh x )8 sec h 2 x = tanh 8 x sec h 2 x ;

du = sec h 2 (x + 3) c ; dx

(x + 3) sec h 2 (x + 3) dx =

∫u

5

⋅ sec 2 (x + 3) ⋅

du

sec (x + 3) 2

1 tanh 6 ( x + 3 ) + c 6 1 ⋅ 6 [ tanh (x + 3) ]6−1 ⋅ sec h 2 (x + 3) + 0 6

c. Given ∫ coth 3 x csc h 2 x dx let u = coth x , then du

; du = sec h 2 x dx ; dx =

du d = tanh (x + 3) dx dx

then

. Thus,

Check: Let y = tanh 6 (x + 3) + c then y ′ =

; dx = −

du

du = sec h 2 x dx

1 ⋅ 9 (tanh x )9−1 ⋅ sec h 2 x + 0 9

u = tanh (x + 3) , du

8

;

du d = coth x dx dx

x csc h 2 x dx =

∫u

3

;

= tanh 5 (x + 3) sec h 2 (x + 3)

du = − csc h 2 x c ; du = − csc h 2 x dx dx

⋅ csc h 2 x ⋅

−du 2

csc h x

= − ∫ u 3 du =

−1 3+1 u +c 3 +1

1 4

1 = − u 4 + c = − coth 4 x + c 4

1 4

1 4

Check: Let y = − coth 4 x + c then y ′ = − ⋅ 4(coth x )4−1 ⋅ − csc h 2 x + 0 = coth 3 x csc h 2 x d. Given ∫ coth 5 3x csc h 2 3x dx let u = coth 3x , then ; dx = −

du 2

3 csc h 3 x

. Therefore,

1 1 5+1 u +c 3 5 +1

= − ⋅

Check: Let y = − e. Given

= −

1 6 u +c 18

1 coth 6 3 x + c 18

∫ coth = −

3 x csc h 2 3 x dx =

∫u

5

;

du = −3 csc h 2 3 x c ; du = −3 csc h 2 3 x dx dx

⋅ csc h 2 3 x ⋅

−du 2

3 csc h 3 x

= −

1 u 5 du 3

∫

1 coth 6 3 x + c 18

then y ′ = −

∫ tanh 5x dx let u = 5x , then

Hamilton Education Guides

5

du d = coth 3 x dx dx

1 ⋅ 6 (coth 3 x )6−1 ⋅ 3 ⋅ − csc 2 3 x + 0 18

du d = 5x dx dx

;

du =5 dx

; du = 5 dx ; dx =

du 5

= coth 5 3x csc h 2 3x . Therefore,

359

Calculus I

5.4 Integration of Hyperbolic Functions

du

∫ tanh 5x dx = ∫ tanh u ⋅ 5

=

1 tanh u du 5

∫

x dx 3

x

∫ 2 tanh 3 dx

let u =

x 3

du d x = dx dx 3

, then

=

1 ln cosh 5 x + c 5

1 1 ⋅ ⋅ sinh 5 x ⋅ 5 + 0 5 cosh 5 x

1 5

Check: Let y = ln cosh 5 x + c then y ′ = f. Given ∫ 2 tanh

1 ln cosh u + c 5

=

;

du 1 = dx 3

=

5 sinh 5 x ⋅ 5 cosh 5 x

; 3du = dx ; dx = 3du . Therefore,

= 2∫ tanh u ⋅ 3du = 6∫ tanh u du = 6 ln cosh u + c = 6 ln cosh x 3

Check: Let y = 6 ln cosh + c , then y ′ = 6 ⋅

=

du

∫ sec h u ⋅ tanh u ⋅ 5

⋅ sinh

x 3

du d 5x = dx dx

g. Given ∫ sec h 5 x tanh 5 x dx let u = 5 x , then

∫ sec h 5x tanh 5x dx

1 cosh

=

= tanh 5 x

;

x 1 ⋅ +0 3 3 du =5 dx

1 sec h u tanh u du 5

∫

=

x +c 3

6 x tanh 3 3

= 2 tanh

; du = 5dx ; dx =

du 5

x 3

. Therefore, 1 5

1 = − sec h u + c = − sec h 5 x + c 5

5 sec h 5 x tanh 5 x 1 1 Check: Let y = − sec h 5 x + c , then y ′ = ⋅ sec h 5 x tanh 5 x ⋅ 5 + 0 = = sec h 5 x tanh 5 x 5

h. Given ∫ sec h x

5

5

x x tanh dx 2 2

let u =

x 2

, then

x

∫ sec h 2 tanh 2 dx = ∫ sec h u ⋅ tanh u ⋅ 2du

du d x = dx dx 2

x 2

i. Given ∫ csc h 2 ( 1 − 2 x ) dx let u = 1− 2 x , then 2

(1 − 2 x ) dx =

∫ csc h

2

u⋅−

du 2

= −

∫x

2

2

x x 1 tanh ⋅ + 0 2 2 2

du d = 1− 2 x dx dx

;

=

du = −2 dx

2 x x ⋅ sec h tanh 2 2 2

x +c 2

= sec h

; du = −2dx ; dx =

du . −2

x x tanh 2 2

Thus,

∫

1 2

∫x

; 2du = dx ; dx = 2du . Therefore,

1 1 1 csc h 2 u du = coth u + c = coth ( 1 − 2 x ) + c 2 2 2

Check: Let y = coth ( 1 − 2 x ) + c then y ′ = j. Given

du 1 = dx 2

= 2∫ sec h u tanh u du = − 2 sec h u + c = − 2 sec h

Check: Let y = −2 sec h + c then y ′ = − 2 ⋅ − sec h

∫ csc h

;

1 ⋅ − csc h 2 ( 1 − 2 x ) ⋅ −2 + 0 2

= csc h 2 ( 1 − 2 x )

du du d 3 du . Therefore, coth x 3 dx let u = x 3 , then = x ; = 3x 2 ; du = 3 x 2 dx ; dx = dx dx dx 3x 2

coth x 3 dx =

∫x

2

1 3

⋅ coth u ⋅

du 3x

2

=

1 coth u ⋅ du 3

∫

Check: Let y = ln sinh x 3 + c then y ′ =

Hamilton Education Guides

=

1 ln sinh u + c 3

=

1 1 ⋅ ⋅ cosh x3 ⋅ 3 x 2 + 0 3 sinh x3

1 ln sinh x 3 + c 3

=

3 x 2 cosh x 3 = x 2 coth x 3 ⋅ 3 sinh x 3 360

Calculus I

5.4 Integration of Hyperbolic Functions

∫x

k. Given

∫x

2

2

Check: Let y =

∫

l. Given

∫

∫x

sec h 5 x 3 dx =

=

15 sec h 5 x dx

x

x

=

⋅ sec h u ⋅

du

)

∫

⋅15 x 2

3

= 1

x

Check: Let y = 2 sin

−1

=

15 1 − tanh 2 5 x 3

=

sec h

1 x2

⋅

1 1 2x 2

. Thus,

(

)

1 sin −1 tanh 5 x 3 + c 15

sec h 2 5 x 3 d d ⋅ 5x 3 tanh 5 x 3 + 0 = dx 2 3 dx 15 sec h 5 x

1 du d 12 du 1 − 12 ; dx = 2 x du . Therefore, = = x = x ; dx 2 dx dx 2 x

)

(

2

=

15x 2

= 2∫ sec h u ⋅ du = 2 sin −1 (tanh u ) + c = 2 sin −1 tanh x + c

⋅ 2 x du

( tanh x )+ c

1

⋅

=

du

= x 2 sec h 5 x 3

1

2

then y ′ =

1

1 − tanh 2 x 2 2 sec h 2 x 2

; du = 15 x 2 dx ; dx =

1 sin −1 (tanh u ) + c 15

1

then y ′ =

15 x 2 sec h 2 5 x 3 ⋅ 15 sec h 5 x 3

let u = x 2 , then

sec h u

∫

du = 15x 2 dx

;

1 sec h u ⋅ du 15

=

2

15 x

(

sec h 2 5 x 3

dx

2

1 sin −1 tanh 5 x 3 + c 15

sec h x

sec h x

du d 5x 3 = dx dx

sec h 5 x 3 dx let u = 5x 3 , then

1 2x 2

1

1

⋅

sec h 2 x 2 1 x2

sec h

=

sec h x 2 1 x2

=

1 2 sec h 2 x 2 d 12 d ⋅ x ⋅ tanh x 2 + 0 = 1 dx dx sec h 2 x 2

sec h x x

Example 5.4-4: Evaluate the following indefinite integrals: 5

dx

a.

∫ sec h

d.

∫ cosh 7 x dx =

g.

∫ 2 sinh

j.

∫

x2

5

x

3

=

1

e

cosh x

3

3x 3x dx csc h 2 2

sinh

=

x dx = 3

x dx

b.

∫ sinh x 2

e.

∫

h.

3x ∫ ( cosh x sec h x + e ) dx

k.

∫

=

cosh 5 x + sinh 5 x dx sinh 5 x

e tanh 5 x sec h 2 5 x dx

= =

=

c.

∫x

f.

∫ 

i.

∫e

l.

∫

2

sinh x dx

=

 1 + cosh x   dx sinh x  sinh 8 x

cosh 8 x dx =

1 coth 7 x

e3

=

csc h 2 7 x dx

=

Solutions: a. Given ∫ sec h 5 x 2

∫

dx

5

sec h x 2

5

x3

=

dx 5

x3

∫

Hamilton Education Guides

2

let u = x 5 , then

sec h u ⋅

5

5 x 3 du 5

2 x3

=

5 2 du d 52 du 2 − 53 ; dx = 5 x 3 du . Therefore, = x = = x ; 5 2 dx 5 dx dx 5 x3 5 sec h u ⋅ du 2

∫

=

5 sin −1 (tanh u ) + c 2

=

5 5   sin −1  tanh x 2  + c 2  

361

Calculus I

5.4 Integration of Hyperbolic Functions

5 Check: Let y = sin −1  tanh 5 x 2  + c , then y ′ = 2   2

5 sec h 2 x 5

=

⋅

2

2 sec h x 5

b. Given

x dx

∫ sinh x 2

x dx

∫ sinh x 2

∫

=

=

2

=

3

3

∫ x csc h x 1 2

=

2

2

=

sec h x 5

∫

du sinh u

=

1 csc h u du 2

∫

x = ⋅ 2

1−

cosh

2 sinh x 2 x2

cosh

c. Given

∫x

2

sinh 2 ⋅ x2 2

x = ⋅ 2

sinh

cosh

2

x

=

2

2 2 cosh 2 x − sinh 2 x 2 2 2 cosh 2 x 2 x2 2 x2 2

5

du d 2 = x dx dx

1 1 1 x2 Check: Let y = ln tanh + c , then y ′ = ⋅ 2 tanh 2 2 2 sinh 2 x 2 2 x2

5

x5

dx let u = x 2 , then

x2 2

=

;

x3

du = 2x dx

x 2 2x ⋅ sec h ⋅ 2 2 2

2

cosh 2 x sinh cosh

; du = 2 x dx ; dx =

1 u ln tanh +c 2 2

=

1

x ⋅ 2

2 d 5 sec h 2 x 5 d 52 ⋅ tanh x 5 + 0 = ⋅ x 2 dx dx 2 5 2 sec h x

sec h x 2

=

3

sec h x 5

10 x 5

5x 5

x du ⋅ sinh u 2 x

⋅

sec h 2 x 5

2

2 1 − tanh 2 x 5 2

2

10

2

5

2 x2 2 x2

=

=

du . 2x

Thus,

1 x2 ln tanh +c 2 2

=

2 x2 2

2 x sec h ⋅ 4 tanh x2

cosh 2 x ⋅ 2 cosh 2 x 2 ⋅ sinh 2

=

x2 2

x2 2

2 x2 2

x 1 − tanh ⋅ 2 2 tanh x 2

=

x 2 cosh

x2 2

⋅ sinh

x2 2

2

x

=

sinh x 2

sinh x dx let u = x 2 and dv = sinh x dx then du = 2 x dx and

∫ dv = ∫ sinh x dx which implies

implies v = cosh x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫x

2

∫

(1 )

∫

sinh x dx = x 2 ⋅ cosh x − cosh x ⋅ 2 x dx = x 2 cosh x − 2 x cosh x dx

To integrate

∫ x cosh x dx

let u = x and dv = cosh x dx then du = dx and ∫ dv = ∫ cosh x dx which

implies v = sinh x . Using the integration by parts formula again we have

∫ x cosh x dx

(2)

= x ⋅ sinh x − ∫ sinh x ⋅ dx = x sinh x − cosh x + c

Combining equations ( 1 ) and ( 2 ) together we obtain

∫x

2

(

)

sinh x dx = x 2 cosh x − 2 x cosh x dx = x 2 cosh x − 2 (x sinh x − cosh x + c ) = x 2 + 2 cosh x − 2 x sinh x + c

∫

(

)

(

)

Check: Let y = x 2 + 2 cosh x − 2 x sinh x + c , then y ′ = 2 x cosh x + x 2 + 2 sinh h x − 2 sinh x − 2 x cosh x + 0

(

)

= x 2 + 2 sinh h x − 2 sinh x = x 2 sinh h x + 2 sinh x − 2 sinh x = x 2 sinh h x

Hamilton Education Guides

362

Calculus I

d. Given

5.4 Integration of Hyperbolic Functions

1

∫ cosh 7 x dx = ∫ sec h 7 x dx let u = 7 x , then

1

∫ cosh 7 x dx = ∫ sec h 7 x dx

=

du

∫ sec h u ⋅ 7

1 sec h u du 7

∫

=

e. Given

∫

=

7 sec h 2 7 x ⋅ 7 sec h 7 x

cosh 5 x + sinh 5 x dx sinh 5 x

du d 5x = dx dx

;

du =5 dx

∫

cosh 5 x + sinh 5 x dx sinh 5 x

=

1 ln sinh u + x + c 5

 cosh hx  + 1 dx  sinh 5 x 

 cosh hx

du 5

 1 + cosh x   dx sinh x 

=

=

tanh



cosh x 

1

=

x + ln sinh x + c , 2

g.

∫

cosh 2 x

1 ⋅ 2

1 cosh 5 x ⋅ 5 +1+ 0 ⋅ 5 sinh 5 x

then y ′ =

x ⋅ sinh x 2 2

3x 3x 2 sinh csc h dx 2 2

Hamilton Education Guides

=

cosh x + sinh x

+

cosh x sinh x

=

1 sinh 2 ⋅

x 2

3x 1 2 sinh ⋅ dx 2 sinh 3 x

∫

2

1 tanh

sec h 2 2x 2 tanh 2x

2

2 sinh x 2 cosh x 2

1 2 cosh

1 cosh 7 x

∫ csc h x dx + ∫ coth x dx

1 + ⋅ cosh x = sinh x

2

=

1 sin −1 (tanh 7 x ) + c 7

= ∫ (coth 5 x + 1) dx = ∫ coth 5 x dx + ∫ dx let u = 5 x , then

cosh 2 x − sinh 2 x

=

. Thus,

⋅

1

du

∫  sinh x + sinh x  dx

1 ⋅ sec h 2 2x ⋅ x 2 2

cosh x + sinh x

=

du 7

1 ln sinh 5 x + x + c 5

Check: Let y = ln tanh

1

1 sin −1 (tanh u ) + c 7



1 5

∫ 

; du = 7dx ; dx =

. Therefore,

Check: Let y = ln sinh 5 x + x + c , then y ′ = f.

du =7 dx

∫  sinh 5x + 1 dx = ∫ coth 5x dx + ∫ dx = ∫ coth u ⋅ 5x + x = 5 ∫ coth u du + x

= =

= sec h 7 x =

= ∫ 

; du = 5dx ; dx =

;

sec h 2 7 x d d ⋅ 7x tanh 7 x + 0 = dx 7 sec h 2 7 x 7 1 − tanh 2 7 x dx

7

sec h 2 7 x ⋅7 7 sec h 7 x

=

1

1 Check: Let y = sin −1 (tanh 7 x ) + c , then y ′ =

=

du d = 7x dx dx

=

+

cosh x sinh x

= 2∫

⋅

2 sinh x 2 cosh x 2

+

=

sinh 32x sinh 32x

cosh x sinh x

5 cosh 5 x ⋅ +1 5 sinh 5 x

= ln tanh

d tanh dx

cosh x = + sinh x

1 cosh 2 x

1 ⋅ 2

x 2

=

x 2

+

1 d ⋅ sinh x + 0 sinh x dx

cosh x 1 = ⋅ + sinh x 2

1−

sinh 2 x

2

cosh 2 x

sinh x 2 cosh x 2

cosh 2x cosh x 1 ⋅ + 2 2 cosh x ⋅ sinh x sinh x 2 2

cosh x 1 + sinh x sinh x

dx

x + ln sinh x + c 2

2 x 1 1 − tanh 2 ⋅ 2 tanh 2x

=

= coth 5 x + 1

=

1+ cosh x sinh x

= 2∫ dx = 2 x + c

363

2

Calculus I

5.4 Integration of Hyperbolic Functions

Check: Let y = 2 x + c , then y ′ = 2 ⋅ x1−1 + 0 = 2 ⋅ x 0 = 2 h.

3x ∫ ( cosh x sec h x + e ) dx



1

∫  cosh x ⋅ cosh x + e

=

1 3

  dx 

3x

=

3x 3x ∫ (1 + e ) dx = ∫ dx + ∫ e dx =

1 3

x+

1 3x e +c 3

3 3

Check: Let y = x + e 3 x + c , then y ′ = x1−1 + ⋅ e 3 x ⋅ 3 + 0 = x 0 + ⋅ e 3 x = 1+ e 3 x du d = sinh 8 x dx dx

i. Given ∫ e sinh 8 x cosh 8 x dx let u = sinh 8 x , then

∫

∫

e sinh 8 x cosh 8 x dx =

e u cosh 8 x ⋅

du 8 cosh 8 x

1 8

Check: Let y = e sinh 8 x + c , then y ′ = j. Given ∫ e

∫

e

cosh x

3

cosh x

3

sinh

sinh

∫e

du 5 sec h 2 5 x

tanh 5 x

∫

e u sinh

cosh x

x 3du ⋅ 3 sinh x 3

∫e

u

sec h 2 5 x ⋅

cosh x

3

du 2

5 sec h 5 x

1 5

1 coth 7 x

⋅ sinh

7 3

∫

1 u e +c 8

=

. Thus,

1 sinh 8 x e +c 8

8 sinh 8 x cosh 8 x ⋅e 8

= e sinh 8 x cosh 8 x

cosh x 3

+c

cosh x x 3 cosh 3x x x 1 3 sinh ⋅ +0 = e ⋅ sinh = e 3 3 3 3 3

du d = tanh 5 x dx dx

csc h 2 7 x dx

3 7

=

1 u e du 5

∫

3du 7 csc h 2 7 x

=

;

du = 5 sec h 2 5 x dx

1 u e +c 5

1 tanh 5 x ⋅ sec h 2 5 x ⋅ 5 + 0 e 5

let u = coth 7 x , then

e u csc h 2 7 x ⋅

1 coth 7 x

Check: Let y = − e 3

Hamilton Education Guides

∫

=

1 3

csc h 2 7 x dx

; du = − csc h 2 7 x dx ; dx = − 1 coth 7 x

=

=

du 8 cosh 8 x

; du = 5 sec h 2 5 x dx

. Therefore,

Check: Let y = e tanh 5 x + c , then y ′ =

e3

∫

= 3∫ e u du = 3e u + c = 3e

+ c , then y ′ = 3e

3

sec h 2 5 x dx =

l. Given ∫ e 3

1 u e du 8

1 sinh 8 x ⋅e ⋅ cosh 8 x ⋅ 8 + 0 8

k. Given ∫ e tanh 5 x sec h 2 5 x dx let u = tanh 5 x , then ; dx =

=

; dx =

x du d x x du 1 x 3du . Therefore, dx let u = cosh , then = cosh ; = sinh ; dx = 3 dx dx 3 3 3 dx 3 sin 3x

x dx = 3

Check: Let y = 3e

eu du 8

∫

=

du = 8 cosh 8 x dx

;

=

=

1 tanh 5 x e +c 5

5 tanh 5 x e sec h 2 5 x 5

du d 1 = coth 7 x dx dx 3

;

= e tanh 5 x sec h 2 5 x

du 7 = − csc h 2 7 x dx 3

. Therefore, −3du 7 csc h 2 7 x

+ c , then y ′ = −

= −

3 u e du 7

∫

3 7

1 coth 7 x

3 = − eu + c = − e 3 7

+c

1 coth 7 x 3 13 coth 7 x 1 e ⋅ − csc h 2 7 x ⋅ 7 + 0 = e 3 csc h 2 7 x 7 3

364

Calculus I

•

5.4 Integration of Hyperbolic Functions

To integrate even powers of sinh x and cosh x use the following identities: sinh 2 x =

cosh 2 x − sinh 2 x = 1

1 ( cosh 2 x − 1) 2

cosh 2 x =

1 ( cosh 2 x + 1) 2

To integrate odd powers of sin x and cos x use the following equalities:

∫ sinh

∫ •

•

2 n +1

∫

n n 2 2 ∫ ( sinh x ) sinh x dx = ∫ ( cosh x − 1) sinh x dx

cosh 2n x cosh x dx =

∫ ( cosh x ) 2

n

cosh x dx

=

∫ (1 + sinh x ) 2

n

( let u = cosh x )

cosh x dx

( let u = sinh x )

To integrate products of sinh x , sinh y , cosh x , and cosh y use the identities below: sinh x sinh y

=

1 [ cosh (x + y ) − cosh (x − y ) ] 2

cosh x cosh y

=

1 [ cosh (x + y ) + cosh (x − y ) ] 2

sinh x cosh y

=

1 [ sinh (x + y ) + sinh (x − y ) ] 2

To integrate tanh n x , set

(

)

(

)

= tanh n −2 x tanh 2 x = tanh n −2 x 1 − sec h 2 x = tanh n −2 x − tanh n −2 x sec h 2 x

To integrate coth n x , set coth n x

•

= ∫ sinh 2n x sinh x dx =

cosh 2n +1 x dx =

tanh n x

•

x dx

= coth n −2 x coth 2 x = coth n −2 x 1 + csc h 2 x = coth n −2 x + coth n −2 x csc h 2 x

To integrate sec h n x sec h n x

For even powers, set

(

= sec h n −2 x sec h 2 x = 1 − tanh 2 x

)

n−2 2

sec h 2 x

For odd powers change the integrand to a product of even and odd functions, i.e., write

∫ sec h •

3

x dx as

∫ sec h

2

x sec h x dx (see Example 5.4-6, problem letter h).

To integrate csc h n x csc h n x

For even powers, set

(

)

= csc h n −2 x csc h 2 x = coth 2 x − 1

n−2 2

csc h 2 x

For odd powers change the integrand to a product of even and odd functions, i.e., write

∫ csc h

3

x dx as

∫ csc h

2

x csc h x dx (see Example 5.4-6, problem letter i).

In the following examples we use the above general rules in order to solve integral of products and powers of hyperbolic functions: Example 5.4-5: Evaluate the following indefinite integrals: a. d.

∫ sinh 5x cosh 7 x dx = ∫ sinh 3x sinh 5x dx =

Hamilton Education Guides

b. e.

∫ sinh x cosh x dx = ∫ cosh 3x cosh 5x dx =

c. f.

∫ cosh 3x cosh 2 x dx = 5 ∫ sinh x dx = 365

Calculus I

g.

∫ sinh

j.

∫ sinh

5.4 Integration of Hyperbolic Functions 3

7

x dx =

h.

∫ cosh

5

x dx =

i.

∫ tanh

4

x dx =

x dx =

k.

∫ sec h

4

x dx =

l.

∫ cosh

3

x dx =

Solutions: a.

∫ sinh 5x cosh 7 x dx

1

∫ 2 [ sinh (5 + 7)x + sinh (5 − 7)x ] dx 1

=

1 2

=

1 1 cosh 12 x − cosh 2 x + c 4 24

=

1

1 1 sinh 12 x − sinh 2 x 2 2

1 sinh 2 x dx 2

∫

=

=

1 1 sinh 12 x + sinh (− 2 x ) 2 2

=

1

∫ 2 [ sinh ( 1 + 1)x + sinh ( 1 − 1)x ] dx

1 1 ⋅ cosh 2 x 2 2

=

1 4

1 ⋅ 2 sinh 2 x + 0 4

1

1 cosh x dx 2

∫

=

Check: Let y =

= d.

1

1 1 1 ⋅ sinh 5 x + sinh x + c 2 5 2

=

1 sinh 2 x 2

=

∫ cosh 3x cosh 2 x dx = ∫ 2 [ cosh (3 + 2)x + cosh (3 − 2)x ] dx +

=

=

1

∫ 2 (cosh 5x + cosh x ) dx

1

∫ sinh 3x sinh 5x dx = ∫ 2 [ cosh (3 + 5)x − cosh (3 − 5)x ] dx =

1 2

=

1 1 sinh 8 x − sinh 2 x + c 16 4

∫ (cosh 8x − cosh 2 x ) dx

Hamilton Education Guides

=

=

1 1 cosh 8 x dx − cosh 2 x dx 2 2

∫

1 2

= sinh 5 x cosh 7 x

∫ (sinh 2 x + 0) dx

=

= sinh x cosh x

1 cosh 5 x dx 2

∫

1 1 sinh 5 x + sinh x + c 2 10

1 1 cosh ( 3 + 2 )x + cosh ( 3 − 2 )x 2 2

=

=

1 ⋅ 2 sinh x cosh x 2

1 1 1 1 sinh 5 x + sinh x + c , then y ′ = ⋅ 5 cosh 5 x + ⋅ cosh x + 0 2 2 10 10

1 1 cosh 5 x + cosh x 2 2

12 2 sinh 12 x − sinh 2 x 24 4

1 1 sinh (5 + 7 )x + sinh (5 − 7 )x 2 2

=

∫ 2 [ sinh (2 x ) + sinh (0 x ) ] dx

=

=

1 cosh 2 x + c 4

Check: Let y = cosh 2 x + c , then y ′ =

c.

1 1 1 1 ⋅ cosh 12 x − ⋅ cosh 2 x + c 2 12 2 2

=

1 1 1 1 ⋅12 sinh 12 x − ⋅ 2 sinh 2 x + 0 cosh 12 x − cosh 2 x + c , then y ′ = 24 4 24 4

∫ sinh x cosh x dx =

1

∫ 2 [ sinh ( 12 x ) + sinh (− 2 x ) ] dx

=

∫ (sinh 12 x − sinh 2 x ) dx = 2 ∫ sinh 12 x dx − 2 ∫ sinh 2 x dx

Check: Let y =

b.

=

∫

=

5 1 cosh 5 x + cosh x 10 2

= cosh 3x cosh 2 x

1

∫ 2 [ cosh (8x ) − cosh (− 2 x ) ] dx =

1 1 1 1 ⋅ sinh 8 x − ⋅ sinh 2 x + c 2 8 2 2

366

Calculus I

5.4 Integration of Hyperbolic Functions

Check: Let y =

= e.

1 1 cosh 8 x − cosh 2 x 2 2

∫ cosh 3x cosh 5x dx 1

=

1 1 sinh 8 x + sinh 2 x + c 16 4

du d = cosh x dx dx

;

∫ sinh

∫ ( cosh

=

∫ (u

4

1 1 cosh 8 x dx + cosh 2 x dx 2 2

∫

=

1 1 cosh 8 x + cosh 2 x 2 2

∫

5

x dx =

=

∫

=

1 [ cosh (8 x ) + cosh (− 2 x ) ] 2

sinh 4 x sinh x dx = du = sinh x dx 2

)

x −1

∫u

− 2u 2 + 1 du =

4

∫ ( sinh x ) 2

2

sinh x dx

=

; du = sinh x dx ; dx =

du . sinh x

)

)

2

sinh x dx

∫

∫ (u

=

∫

2

du − 2 u 2 du + du =

−1

2

= sinh 3x sinh 5 x

1

1 1 1 1 ⋅ sinh 8 x + ⋅ sinh 2 x + c 2 8 2 2

=

=

2 8 cosh 8 x + cosh 2 x 4 16

1 [ cosh (3 + 5)x + cosh (3 − 5)x ] 2

=

∫ ( cosh

2

x −1

)

2

sinh x dx .

= cosh 3x cosh 5 x

Let u = cosh x , then

Therefore, =

∫ (u

1 5 2 3 u − u +u +c 5 3

Check: Let y = cosh 5 x − cosh 3 x + cosh x + c , then y ′ =

(

2 8 cosh 8 x − cosh 2 x 4 16

∫ 2 [ cosh (8x ) + cosh (− 2 x ) ] dx

sinh x dx

2 3

1 5

=

1 [ cosh (3 + 5)x − cosh (3 − 5)x ] 2

=

1 1 1 1 ⋅ cosh 8 x ⋅ 8 + ⋅ cosh 2 x ⋅ 2 + 0 sinh 8 x + sinh 2 x + c , then y ′ = 16 4 4 16

sinh 5 x dx =

∫

1 [ cosh (8 x ) − cosh (− 2 x ) ] 2

1

∫ 2 (cosh 8x + cosh 2 x ) dx

=

=

∫ 2 [ cosh (3 + 5)x + cosh (3 − 5)x ] dx

=

=

Check: Let y =

f.

1 1 1 1 ⋅ cosh 8 x ⋅ 8 − ⋅ cosh 2 x ⋅ 2 + 0 sinh 8 x − sinh 2 x + c , then y ′ = 16 4 16 4

4

=

)

− 2u 2 + 1 sinh x ⋅

du sinh x

1 2 cosh 5 x − cosh 3 x + cosh x + c 5 3

2 1 ⋅ 5 cosh 4 x ⋅ sinh x − ⋅ 3 cosh 2 x ⋅ sinh x + sinh x 3 5

)

(

= sinh x cosh 4 x − 2 sinh x cosh 2 x + sinh x = sinh x cosh 4 x − 2 cosh 2 x + 1 = sinh x cosh 2 x − 1

(

= sinh x sinh 2 x

∫ sinh

g. ;

3

; du = sinh x dx ; dx =

du . sinh x

∫ sinh

2

∫ ( cosh =

2

2

= sinh x sinh 4 x = sinh 5 x

∫ ( cosh

x dx =

1 3 u −u +c 3

2

x sinh x dx =

x dx =

du = sinh x dx

∫ sinh =

3

)

)

)

x − 1 sinh x dx =

2

)

x − 1 sinh x dx . Let u = cosh x , then

du d = cosh x dx dx

Therefore,

∫ (u

2

)

− 1 sinh x ⋅

du sinh x

=

∫ (u

2

)

− 1 du =

∫u

2

∫

du − du

1 cosh 3 x − cosh x + c 3

Hamilton Education Guides

367

Calculus I

5.4 Integration of Hyperbolic Functions

1 3

1 ⋅ 3 cosh 2 x ⋅ sinh x − sinh x + 0 3

Check: Let y = cosh 3 x − cosh x + c , then y ′ =

(

)

= sinh x cosh 2 x − sinh x

= sinh x cosh 2 x − 1 = sinh x sinh 2 x = sinh 3 x h.

∫ cosh then

x dx =

∫ cosh

du d sinh x = dx dx

∫ cosh =

5

∫u

5

4

=

x dx

4

x cosh x dx =

du = cosh x dx

;

2 2 2 2 ∫ ( cosh x ) cosh x dx = ∫ (1 + sinh x ) cos x dx .

; du = cosh x dx ; dx =

du cosh x

. Therefore,

2 du 2 2 2 ∫ (1 + sinh x ) cosh x dx = ∫ (1 + u ) cosh x ⋅ cosh x

∫

∫

du + 2 u 2 du + du = 1 5

1 5 2 3 u + u +u +c 5 3

=

Let u = sinh x ,

=

2 2 ∫ (1 + u ) du

∫ (u

=

4

)

+ 2u 2 + 1 du

1 2 sinh 5 x + sinh 3 x + sinh x + c 5 3

2 3

Check: Let y = sinh 5 x + sinh 3 x + sinh x + c , then y ′ =

(

6 5 sinh 4 x ⋅ cosh x + sinh 2 x ⋅ cosh x + cosh x + 0 3 5

)

(

= cosh x sinh 4 x + 2 cosh x sinh 2 x + cosh x = cosh x sinh 4 x + 2 sinh 2 x + 1 = cosh x 1 + sinh 2 x

(

= cosh x cosh 2 x i.

∫ tanh =

4

x dx =

∫ tanh

2

)

2

2 2 2 2 2 ∫ tanh x (1 − sec h x ) dx = ∫ tanh x dx − ∫ tanh x sec h x dx

2 2 2 ∫ (1 − sec h x ) dx − ∫ tanh x sec h x dx

= − ∫ tanh 2 x sec h 2 x dx − ∫ sec h 2 x dx + ∫ dx . To solve the first

du d = tanh x dx dx

integral let u = tanh x , then

4

x dx =

∫ tanh

2

;

du

∫

− tanh 2 x sec h 2 x dx = − u 2 sec h 2 x ⋅

∫ tanh

2

= cosh x cosh 4 x = cosh 5 x

x tanh 2 x dx =

∫

)

2

sec h x

du = sec h 2 x dx

; du = sec h 2 x dx ; dx = 1 3

sec h 2 x

. Thus,

1 3

= − ∫ u 2 du = − u 3 = − tanh 3 x . Therefore,

∫

∫

∫

x tanh 2 x dx = − tanh 2 x sec h 2 x dx − sec h 2 x dx + dx = −

1 3

du

1 tanh 3 x − tanh x + x + c 3

Check: Let y = − tanh 3 x − tanh x + x + c , then y ′ = − tanh 2 x sec h 2 x − sec h 2 x + 1

(

)

(

)(

)

= sec h 2 x − tanh 2 x − 1 + 1 = 1 − tanh 2 x − tanh 2 x − 1 + 1 = − tanh 2 x − 1 + tanh 4 x + tanh 2 x + 1

(

)

= − tanh 2 x + tanh 2 x + (− 1 + 1) + tanh 4 x = tanh 4 x j.

∫ sinh

7

x dx =

∫ sinh

6

du d = cosh x dx dx

;

∫ sinh

∫ ( cosh

7

x dx

=

x sinh x dx =

du = sinh x dx

Hamilton Education Guides

2

x −1

3 3 2 2 ∫ ( sinh x ) sinh x dx = ∫ ( cosh x − 1 ) sinh x dx .

; du = sinh x dx ; dx =

)

3

sinh x dx

=

∫ (u

2

du . sinh x

−1

)

3

Let u = cosh x , then

Therefore,

sinh x dx

=

∫ (u

6

)

− 3u 4 + 3u 2 − 1 sinh x ⋅

du sinh x

368

Calculus I

5.4 Integration of Hyperbolic Functions

=

∫ (u

)

=

3 1 cosh 7 x − cosh 5 x + cosh 3 x − cosh x + c 5 7

6

∫u

− 3u 4 + 3u 2 − 1 du =

6

∫

∫

∫

du − 3 u 4 du + 3 u 2 du − du =

1 1 7 1 u − 3⋅ u 5 + 3⋅ u 3 − u + c 5 3 7

3 5

1 7

Check: Let y = cosh 7 x − cosh 5 x + cosh 3 x − cosh x + c , then y ′ = + 3 cosh 2 x ⋅ sinh x − sinh x + 0

15 7 cosh 6 x ⋅ sinh x − cosh 4 x ⋅ sinh x 5 7

= sinh x cosh 6 x − 3 sinh x cosh 4 x + 3 sinh x cosh 2 x − sinh x

(

(

)

= sinh x cosh 6 x − 3 cosh 4 x + 3 cosh 2 x − 1 = sinh x cosh 2 x − 1

)

3

(

= sinh x sinh 2 x

)

3

= sinh x sinh 6 x = sinh 7 x

∫ sec h

k.

=

4

∫ sec h

x dx =

2

x sec h 2 x dx =

2 2 2 2 2 ∫ sec h x (1 − tanh x ) dx = ∫ sec h x dx − ∫ tanh x sec h x dx

2 2 2 2 2 2 ∫ (1 − tanh x ) dx − ∫ tanh x sec h x dx = ∫ dx − ∫ tanh x dx − ∫ tanh x sec h x dx .

du d = tanh x dx dx

integral let u = tanh x , then − tanh 2 x sec h 2 x dx

= − ∫ u 2 sec h 2 x ⋅

∫ sec h

2

∫

4

∫ sec h

x dx =

x sec h 2 x dx =

;

du 2

sec h x

du = sec h 2 x dx

To solve the third

; du = sec h 2 x dx ; dx = 1 3

du sec h 2 x

. Therefore,

1 3

= − ∫ u 2 du = − u 3 = − tanh 3 x . Thus,

∫ dx − ∫ tanh

2

∫

x dx − tanh 2 x sec h 2 x dx =

∫ dx − ∫ tanh

2

1 x dx − tanh 3 x 3

1 3

1 1 = x − (x − tanh x ) − tanh 3 x + c = − tanh 3 x + tanh x + (x − x ) + c = − tanh 3 x + tanh x + c 3

3

1 3

3 3

Check: Let y = − tanh 3 x + tanh x + c , then y ′ = − tanh 2 x ⋅ sec h 2 x + sec h 2 x = − tanh 2 x sec h 2 x + sec h 2 x

(

)

= sec h 2 x 1 − tanh 2 x = sec h 2 x sec h 2 x = sec h 4 x l.

∫ cosh ;

3

du = cosh x dx

∫ cosh =

3

2 2 ∫ ( cosh x ) cosh x dx = ∫ (1 + sinh x ) cosh x dx .

x dx =

; du = cosh x dx ; dx =

x dx =

1 3 u +u +c 3

du cosh x

Let u = sinh x , then

. Therefore,

du 2 2 ∫ (1 + sinh x ) cosh x dx = ∫ (1 + u )cosh x ⋅ cosh x

=

du d = sinh x dx dx

=

2 ∫ (1 + u ) du

=

∫u

2

∫

du + du

1 sinh 3 x + sinh x + c 3 1 3

Check: Let y = sinh 3 x + sinh x + c , then y ′ =

(

)

(

)

1 ⋅ 3 sinh 2 x ⋅ cosh x + cosh x + 0 3

= cosh x sinh 2 x + cosh x

= cosh x 1 + sinh 2 x = cosh x cosh 2 x = cosh 3 x Hamilton Education Guides

369

Calculus I

5.4 Integration of Hyperbolic Functions

Example 5.4-6: Evaluate the following indefinite integrals: b.

d.

∫ cosh x dx = 3 ∫ tanh x dx =

g.

∫ coth

a.

4

3

x dx =

c.

e.

∫ cosh 5x dx = 4 ∫ coth x dx =

f.

∫ sinh x dx = 6 ∫ tanh x dx =

h.

∫ sec h

i.

∫ csc h

2

3

x dx =

4

3

x dx =

Solutions: a.

∫

cosh 4 xdx =

∫ (cosh x ) 2

2

dx =

∫

2

1 1   2 (cosh 2 x + 1 ) dx = 4  

=

2 1 1 dx + cosh 2 x dx cosh 2 2 x dx + 4 4 4

+

1 1 cosh 4 x dx + sinh 2 x 4 8

∫

∫

∫

∫

Check: Let y =

=

x 1 + 4 4

∫ (cosh 2 x + 1 ) dx 2

∫ (1 + cosh

1 4

1 1

1

∫ 2 ( 1 + cosh 4 x ) dx + 2 ⋅ 2 sinh 2 x

1 x x 1 1 + + ⋅ sinh 4 x + sinh 2 x + c 4 8 8 4 4

=

=

=

=

2

)

2 x + 2 cosh 2 x dx

x 1 1 + ⋅ dx 4 4 2

∫

1 1 3 x+ sinh 4 x + sinh 2 x + c 4 32 8

1 3x 1 3 4 cosh 4 x 2 cosh 2 x + sinh 4 x + sinh 2 x + c , then y ′ = + + 4 4 8 32 32 8

3 1 2 + cosh 4 x + cosh 2 x 8 8 4

=

1 1 1 2 1 1 1 2 1 1 2 =  +  + cosh 4 x + cosh 2 x = + ( 1 + cosh 4 x ) + cosh 2 x = + ⋅ ( 1 + cosh 4 x ) + cosh 2 x 4

=

8 8

(

∫ cosh −

2

4

1 1 2 + ⋅ cosh 2 2 x + cosh 2 x 4 4 4

8

2

(

4 2

) = 14 ( cosh 2 x + 1 )

4

2

1 =  ( cosh 2 x + 1 ) 2

2



4

2

1 1 1 x dx = x + ⋅ sinh 10 x − + c 2 2 10 2

∫

Check: Let y =

4

4

1 1 + cosh 2 2 x + 2 cosh 2 x 4

=

) = cosh x 5 x dx = ∫ (1 + sinh 5 x )dx = ∫ dx + ∫ sinh

= cosh 2 x b.

4

2

1

∫ dx + ∫ 2 ( cosh 10 x − 1 ) dx =

5 x dx =

x+

1 cosh 10 x dx 2

∫

x sinh 10 x  1 1 +c = x1 −  + sinh 10 x + c = +

2



20

2

20

1 1 x sinh 10 x ⋅ cosh 10 x ⋅10 + 0 + + c , then y ′ = + 2 20 20 2

=

1 10 + ⋅ cosh 10 x 2 20

=

1 1 + cosh 10 x 2 2

1 1 1 1 1 = 1 −  + cosh 10 x = 1 +  cosh 10 x −  = 1 + ( cosh 10 x − 1 ) = 1 + sinh 2 5 x = cosh 2 5 x 

c.

∫ =

sinh 4 x dx =

2

2

2

∫ (sinh x ) 2

2

dx

=

∫

2

1   2 (cosh 2 x − 1 ) dx  

1 2 1 cosh 2 2 x dx − cosh 2 x dx + dx 4 4 4

∫

∫

Hamilton Education Guides

∫

2

2

=

1 4

1

=

1 4

∫ (cosh 2 x − 1 ) dx 2

1 1

=

1 4 x

∫ 2 ( 1 + cosh 4 x ) dx − 2 ⋅ 2 sinh 2 x + 4

∫ (cosh =

2

)

2 x − 2 cosh 2 x + 1 dx

1 1 ⋅ dx 4 2

∫

370

Calculus I

+

5.4 Integration of Hyperbolic Functions

1 x 1 cosh 4 x dx − sinh 2 x + 8 4 4

∫

Check: Let y =

3 1 1 1 x x 1 1 =  +  + ⋅ sinh 4 x − sinh 2 x + c = x + sinh 4 x − sinh 2 x + c 8

4

4 8 4

8

32

3 4 cosh 4 x 2 cosh 2 x 1 3x 1 − + sinh 4 x − sinh 2 x + c , then y ′ = + 8 32 4 8 32 4

4

=

2 3 1 + cosh 4 x − cosh 2 x 4 8 8

1 1 1 2 1 1 1 2 1 1 2 =  +  + cosh 4 x − cosh 2 x = + ( 1 + cosh 4 x ) − cosh 2 x = + ⋅ ( 1 + cosh 4 x ) − cosh 2 x 4

=

8 8

∫ tanh

3

4

1 1 2 + ⋅ cosh 2 2 x − cosh 2 x 4 4 4

(

)

x dx =

∫ tanh

= sinh 2 x d.

4

2

=

∫

∫ tanh

3

x dx =

4

4

(

1 1 + cosh 2 2 x − 2 cosh 2 x 4

4 2

) = 14 ( cosh 2 x − 1 )

4

2

1 =  ( cosh 2 x − 1 ) 2

2



= sinh 4 x 2

2 ∫ (1 − sec h x ) tanh x dx

x tanh x dx =

the first integral let u = tanh x , then − sec h 2 x tanh x dx

8

= − ∫ sec h 2 x ⋅ u ⋅

∫ tanh

2

du d = tanh x dx dx

du 2

sec h x

x tanh x dx =

;

= − ∫ sec h 2 x tanh x dx + ∫ tanh x dx . To solve

du = sec h 2 x dx

; du = sec h 2 dx ; dx =

du

. Thus,

sec h 2 x

1 2

1 2

= − ∫ u du = − u 2 = − tan 2 x . Combining the term

2 ∫ (1 − sec h x ) tanh x dx

=

∫ ( − sec h

2

)

x tanh x + tanh x dx

1 2

1 = − ∫ sec h 2 x tanh x dx + ∫ tanh x dx = − tanh 2 x + ∫ tanh x dx = − tanh 2 x + ln cosh x + c 2

1 1 Check: Let y = − tanh 2 x + ln cosh x + c , then y ′ = − ⋅ 2 tanh x ⋅ sec h 2 x + 2

2

= − tanh x sec h 2 x + e.

∫ coth

4

x dx =

∫ coth

2

sinh x cosh x

(

)

= − tanh x sec h 2 x + tanh x = tanh x 1 − sec h 2 x = tanh x tanh 2 x = tanh 3 x

x coth 2 x dx =

(

1 ⋅ sinh x + 0 cosh x

)

2 2 2 2 2 ∫ coth x ( csc h x + 1) dx = ∫ coth x csc h x dx + ∫ coth x dx

= ∫ coth 2 x csc h 2 x dx + ∫ csc h 2 x + 1 dx = ∫ coth 2 x csc h 2 x dx + ∫ csc h 2 x dx + ∫ dx . To solve the first integral let u = coth x , then ; dx = −

du 2

csc h x

du d = coth x dx dx

;

du = − csc h 2 x dx

; du = − csc h 2 x dx

. Grouping the terms together we find ∫ coth 2 x csc h 2 x dx = ∫ u 2 csc h 2 x ⋅ − 1 3

du csc h 2 x

1 3

= − ∫ u 2 du = − u 3 = − coth 3 x . Therefore,

∫ coth

4

x dx

=

∫ coth

2

Hamilton Education Guides

x coth 2 x dx

=

2 2 2 2 2 ∫ coth x ( csc h x + 1) dx = ∫ coth x csc h x dx + ∫ csc h x dx + ∫ dx

371

Calculus I

5.4 Integration of Hyperbolic Functions

1 3

1 = − coth 3 x + ∫ csc h 2 x dx + ∫ dx = − coth3 x − coth x + x + c 3

1 3

3 3

Check: Let y = − coth 3 x − coth x + x + c , then y ′ = − coth 2 x ⋅ − csc h 2 x + csc h 2 x + 1 + 0

(

)

(

)(

)

= coth 2 x csc h 2 x + csc h 2 x + 1 = csc h 2 x coth 2 x + 1 + 1 = coth 2 x − 1 coth 2 x + 1 + 1 = coth 4 x + coth 2 x − coth 2 x − 1 + 1 = coth 4 x f.

4 4 2 4 2 ∫ tanh x (1 − sec h x ) dx = ∫ ( tanh x − tanh x sec h x ) dx =

= ∫ tanh 4 x tanh 2 x dx =

∫ tanh

6

x dx

∫ tanh

4

x dx − tanh 4 x sec h 2 x dx . In example 5.4-5, problem letter i, we found that

∫ tanh

4

1 x dx = − tanh 3 x − tanh x + x + c . Therefore, 3

∫ tanh

6

∫

x dx =

∫ tanh

4

x tanh 2 x dx =

4 2 4 4 2 ∫ tanh x (1 − sec h x ) dx = ∫ tanh x dx − ∫ tanh x sec h x dx

1 3

1 5

 1  =  − tanh 3 x − tanh x + x  − ∫ tanh 4 x sec h 2 x dx = − tanh5 x − tanh 3 x − tanh x + x + c  3



5 5

1 3

1 5

3 3

Check: Let y = − tanh 5 x − tanh 3 x − tanh x + x + c , then y ′ = − tanh 4 x ⋅ sec h 2 x − tanh 2 x ⋅ sec h 2 x − sec h 2 x + 1 + 0

(

(

)

= − tanh 4 x sec h 2 x − tanh 2 x sec h 2 x − sec h 2 x + 1 = − sec h 2 x tanh 4 x + tanh 2 x + 1 + 1

)(

(

)

)

= − 1 − tanh 2 x tanh 4 x + tanh 2 x + 1 + 1 = − tanh 4 x + tanh 2 x + 1 − tanh 6 x − tanh 4 x − tanh 2 x + 1 = − tanh 4 x − tanh 2 x − 1 + tanh 6 x + tanh 4 x + tanh 2 x + 1 = tanh 6 x g.

∫ coth

3

x dx

first integral let u = coth x , then

∫ csc h ∫ coth

2

3

∫ csc h

2

x coth x dx = x dx

∫ ( csc h

= ∫ coth 2 x coth x dx =

=

∫ coth

∫ csc h 2

2

∫

)

x + 1 coth x dx

du d = coth x dx dx

x ⋅u ⋅ −

x coth x dx

2

du csc h x

=

x coth x dx + coth x dx = −

2

;

= ∫ csc h 2 x coth x dx + ∫ coth x dx . To solve the

du = − csc h 2 x dx

; du = − csc h 2 dx ; dx = −

1 2

Hamilton Education Guides

cosh x sinh x

. Thus,

1 2

∫ ( csc h

2

)

x + 1 coth x dx

1 coth 2 x + coth x dx 2

∫

=

∫ ( csc h

2

)

x ⋅ coth x + coth x dx

1 2

= − coth 2 x + ln sinh x + c

Check: Let y = − coth 2 x + ln sinh x + c , then y ′ = − ⋅ 2 coth x ⋅ − csc h 2 x + = coth x csc h 2 x +

csc h 2 x

= − ∫ u du = − u 2 = − coth 2 x . Combining the term

1 2

1 2

du

(

1 ⋅ cosh x + 0 sinh x

)

= coth x csc h 2 x + coth x = coth x csc h 2 x + 1 = coth x coth 2 x = coth 3 x

372

Calculus I

h.

5.4 Integration of Hyperbolic Functions

∫ sec h

3

∫ sec h

x dx =

2

x sec h x dx =

2 2 ∫ (1 − tanh x ) sec h x dx = − ∫ tanh x sec h x dx + ∫ sec h x dx

= − ∫ tanh x ⋅ tanh x sec h x dx + ∫ sec h x dx . To solve the first integral let u = tanh x and dv = tanh x sec h x dx , then du = sec h 2 x dx and ∫ dv = ∫ tanh x sec h x dx which implies v = − sec h x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫

∫

∫

− tanh 2 x sec h x dx = − tanh x ⋅ tanh x sec h x dx = tanh x sec h x − sec h x sec h 2 x dx = tanh x sec h x

∫

− sec h 3 x dx . Combining the terms we have

∫ tanh x ⋅ tanh x sec h x dx − ∫ sec h x dx

∫ sec h

3

x dx =

= tanh x sec h x − ∫ sec h 3 x dx + ∫ sec h x dx . Moving the

∫ sec h

3

x dx

∫ sec h

3

x dx + sec h 3 x dx = 2 sec h 3 x dx = tanh x sec h x + sec h x dx . Therefore,

∫ sec h

3

x dx =

term from the right hand side of the equation to the left hand side we obtain

∫

∫

(

∫

1 tanh x sec h x + sec h x dx 2

∫

)

1 1 tanh x sec h x + sin −1 ( tanh x ) + c 2 2

=

sec h 2 x ⋅ sec h x − sec h x tanh x ⋅ tanh x 2

1 1 Check: Let y = tanh x sec h x + sin −1 (tanh x ) + c , then y ′ = 2

+

i.

∫ csc h

3

2

sec h 2 x 2 1 − tanh 2 x

+0

=

sec h 3 x − sec h x tanh 2 x sec h 2 x + 2 2 sec h 2 x

(

sec h 3 x − sec h x tanh 2 x + sec h x 2

=

sec h 3 x + sec h 3 x 2 sec h 3 x = = sec h 3 x 2 2

x dx =

∫ csc h

2

x csc h x dx =

∫ ( coth

2

sec h 3 x − sec h x tanh 2 x sec h x + 2 2

sec h 3 x + sec h x 1 − tanh 2 x 2

=

=

=

)

x − 1 csc h x dx =

∫ coth

2

) = sec h x + sec h x sec h 3

2

x

2

∫

x csc h x dx − csc h x dx

= ∫ coth x ⋅ coth x csc h x dx − ∫ csc h x dx . To solve the first integral let u = coth x and dv = coth x csc h x dx , then du = − csc h 2 x dx and ∫ dv = ∫ coth x csc h x dx which implies v = − csc h x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ coth

2

x csc h x dx =

∫ coth x ⋅ coth x csc h x dx

= coth x ⋅ − csc h x − ∫ csc h x ⋅ csc h 2 x dx = − coth x csc h x

∫

− csc h 3 x dx . Combining the terms we have

∫ csc h

3

x dx =

∫ coth x ⋅ coth x csc h x dx − ∫ csc h x dx

Hamilton Education Guides

= − coth x csc h x − ∫ csc h 3 x dx − ∫ csc h x dx . Moving 373

Calculus I

5.4 Integration of Hyperbolic Functions

the ∫ csc h 3 x dx term from the right hand side of the equation to the left hand side we obtain

∫ csc h

3

x dx + csc h 3 x dx = 2 csc h 3 x dx = − coth x csc h x − csc h x dx . Therefore,

∫

∫

∫ csc h

3

x dx =

(

∫

1 − coth x csc h x − csc h x dx 2

∫

1 2

1 2

Check: Let y = − coth x csc h x − ln tanh −

sec h 2

x 2

4 tanh

x 2

(

sec h 2

x 2

4 tanh

x 2

(

=

1 2

1 2

= − coth x csc h x − ln tanh

x +c, 2

2 x 1 1 − tanh 2 1 = ⋅ = ⋅ x 4 4 tanh

sinh 2 x

2 cosh 2 x 2 sinh x 2 cosh x 2

1−

2

(

2 x 2

(

x +c 2

− − csc h 2 x ⋅ csc h x − csc h x coth x ⋅ coth x 2

then y ′ =

sec h csc h 3 x + csc h x coth 2 x +0 = − 2 4 tanh

)

)

)

. The 2nd term is simplified as follows:

)

x 2

cosh 2 x − sinh 2 x 2

=

cosh 2 x

1 ⋅ 4

)

1 cosh 2 x

2

1 1 2 = ⋅ = ⋅ 4 sinh 2x 4

2 sinh x 2 cosh x 2

cosh x 2

cosh cosh 2

x 2

x x ⋅ sinh 2 2

csc h x 1 1 1 1 = = = . Therefore, x x x 2 sinh x 2 2 ⋅ sinh 2 ⋅ 2 2 ⋅ 2 cosh ⋅ sinh 2 2 2 x 2

sec h csc h 3 x + csc h x coth 2 x − 2 4 tanh

(

x 2

=

csc h 3 x + csc h x coth 2 x − csc h x 2

=

sec h 3 x + sec h 3 x 2

=

2 csc h 3 x 2

=

) =

csc h 3 x + csc h x coth 2 x csc h x − 2 2

(

) = csc h x + csc h x csc h

csc h 3 x + csc h x coth 2 x − 1 2

3

2

x

2

= csc h 3 x

Example 5.4-7: Evaluate the following indefinite integrals: b.

d.

∫ sinh x cosh x dx = 3 2 ∫ sinh x cosh x dx =

g.

∫ coth

a.

2

2

2

x csc h 2 x dx =

c.

e.

∫ sinh x cosh x dx = 5 4 ∫ tanh x sec h x dx =

f.

∫ sinh x cosh x dx = 3 3 ∫ tanh x sec h x dx =

h.

∫ coth

i.

∫ coth

2

3

5

x csc h 3 x dx =

4

2

3

x csc h 4 x dx =

Solutions: a.

∫ sinh =

2

x cosh 2 x dx =

1 1 x ⋅ sinh 4 x − + c 8 4 8

Check: Let y =

=

1 sinh 2 2 x dx 4

∫

=

1 4

1

∫ 2 ( cosh 4 x − 1 ) dx

1 8

∫ ( cosh 4 x − 1) dx

=

1 1 cosh 4 x dx − dx 8 8

∫

∫

1 1 sinh 4 x − x + c 32 8

4 cosh 4 x 1 1 1 sinh 4 x − x + c , then y ′ = − +0 32 8 32 8

Hamilton Education Guides

=

=

1 1 cosh 4 x − 8 8

=

1 ( cosh 4 x − 1) 8

374

Calculus I

5.4 Integration of Hyperbolic Functions

= b.

∫ sinh

2

1 1 ⋅ ( cosh 4 x − 1) 4 2

x cosh 5 x dx =

1 ⋅ sinh 2 2 x 4

=

∫ sinh

2

= sinh 2 x cosh 2 x

x cosh 4 x cosh x dx =

=

2 4 2 ∫ sinh x (1 + sinh x + 2 sinh x ) cosh x dx

=

=

∫ sinh

∫

2

∫

2 2 2 2 2 2 ∫ sinh x (cosh x ) cosh x dx = ∫ sinh x (1 + sinh x ) cos x dx

∫ ( sinh

2

)

x cosh x + sinh 6 x cosh x + 2 sinh 4 x cosh x dx

x cosh x dx + sinh 6 x cosh x dx + 2 sinh 4 x cosh x dx 1 3

1 7

=

3 sinh 2 x cosh x 7 sinh 6 x cosh x + 3 7

2 5

Check: Let y = sinh 3 x + sinh 7 x + sinh 5 x + c , then y ′ = +

(

10 sinh 4 x cosh x = sinh 2 x cosh x + sinh 6 x cosh x + 2 sinh 4 x cosh x = sinh 2 x cosh x 1 + sinh 4 x + 2 sinh 2 x 5

(

= sinh 2 x cosh x 1 + sinh 2 x

c.

∫ sinh +

4

2 1 1 sinh 3 x + sinh 7 x + sinh 5 x + c 7 5 3

1 sinh 2 2 x cosh 2 x dx 8 1 1 8 6

= + ⋅ sinh 3 2 x = Check: Let y =

= −

2

= sinh 2 x cosh x cosh 2 x

∫ ( sinh x cosh x )

x cosh 2 x dx =

∫

(

)

= −

1 8

2

sinh 2 x dx =

1

)

2

= sinh 2 x cosh x cosh 4 x = sinh 2 x cosh 5 x

∫ ( 12 sinh 2 x )

1

∫ 2 ( cosh 4 x − 1 ) dx + 8 ∫ sinh

1 1 1 1 sinh 3 2 x + c x − ⋅ sinh 4 x + 16 16 4 48

=

2

2

⋅

1 ( cosh 2 x − 1 ) dx 2

2 x cosh 2 x dx

=

= −

1 sinh 2 2 x dx 8

∫

1 1 dx − cosh 4 x dx 16 16

∫

∫

1 1 1 sinh 4 x + sinh 3 2 x + c x− 16 64 48

1 4 cosh 4 x 6 sinh 2 2 x cosh 2 x 1 1 1 − + +0 x − sinh 4 x + sinh 3 2 x + c , then y ′ = 16 64 48 48 64 16

1 ( cosh 4 x − 1) + 1 sinh 2 2 x cos 2 x 16 8

1 1 8 2

1 8

= − ⋅ ( cosh 4 x − 1) + sinh 2 2 x cosh 2 x 2

1 1 1 1 = − ⋅ sinh 2 2 x + sinh 2 2 x cosh 2 x =  sinh 2 x  ⋅ ( cosh 2 x − 1) = ( sinh x cosh x )2 sinh 2 x 2

8

8



2

= sinh 2 x cosh 2 x ⋅ sinh 2 x = sinh 4 x cosh 2 x d.

∫ sinh

3

x cosh 2 x dx =

3 5 2 3 5 3 ∫ sinh x (1 + sinh x )dx = ∫ ( sinh x + sinh x ) dx = ∫ sinh x dx + ∫ sinh x dx .

Example, 5.4-5, problem letters f and g, we found ∫ sinh 5 x dx = 1 cosh 3 x − cosh x + c . 3

∫ sinh

3

x dx =

∫ sinh

3

x cosh 2 x dx =

Hamilton Education Guides

∫ sinh

3

In

1 2 cosh 5 x − cosh 3 x + cosh x + c and 5 3

Therefore,

∫

x dx + sinh 5 x dx =

1 2 1  cosh 3 x − cosh x + c +  cosh 5 x − cosh 3 x + cosh x + c  3 3 5 

375

)

Calculus I

=

5.4 Integration of Hyperbolic Functions

1 1 2 cosh 5 x + cosh 3 x − cosh 3 x − cosh x + cosh x + c 5 3 3 1 3

1 5

Check: Let y = cosh 5 x − cosh 3 x + c , then y ′ =

1 1 cosh 5 x − cosh 3 x + c 5 3

=

3 1 ⋅ 5 cosh 4 x ⋅ sinh x − ⋅ cosh 2 x ⋅ sinh x + 0 3 5

(

)

= sinh x cosh 4 x − sinh x cosh 2 x = sinh x cosh 2 x cosh 2 x − 1 = sinh x cosh 2 x sinh 2 x = sinh 3 x cosh 2 x e.

∫ tanh

5

∫ tanh

x sec h 4 x dx =

∫

+ tanh 5 x sec h 2 x dx = −

5

x sec h 2 x sec h 2 x dx =

5 2 2 ∫ tanh x (1 − tanh x ) sec h x dx

= − ∫ tanh 7 x sec h 2 x dx

1 1 tanh8 x + tanh6 x + c 8 6

1 8

1 6

1 8

1 6

Check: Let y = − tanh 8 x + tanh 6 x + c , then y ′ = − ⋅ 8 tanh 7 x ⋅ sec h 2 x + ⋅ 6 tanh 5 x ⋅ sec h 2 x + 0

(

)

= − tanh 7 x sec h 2 x + tanh 5 x sec h 2 x = tanh 5 x sec h 2 x 1 − tanh 2 x = tanh 5 x sec h 2 x sec h 2 x = tanh 5 x sec h 4 x f.

∫ tanh

3

x sec h 3 x dx =

∫ tanh

2

2 2 ∫ (1 − sec h x )sec h x ⋅ tanh x sec h x dx

x sec h 2 x ⋅ tanh x sec h x dx =

= − ∫ sec h 4 x ⋅ tanh x sec h x dx + ∫ sec h 2 x ⋅ tanh x sec h x dx . To solve the first and the second integral let u = sec h x , then

∫ tanh

3

du d = sec h x dx dx

x sec h 3 x dx =

∫ tanh

2

;

1 5 1 3 u − u +c 5 3

=

du sec h x tanh x

. Therefore,

∫

+ sec h 2 x ⋅ tanh x sec h x dx = − u 4 ⋅ −

=

; dx = −

x sec h 2 x ⋅ tanh x sec h x dx = − sec h 4 x ⋅ tanh x sec h x dx

∫

∫

du = − sec h x tanh x dx

tanh x sec h x tanh x sec h x du du + u 2 ⋅ − tanh x sec h x tanh x sec h x

∫

=

∫u

4

∫

du − u 2 du

1 1 sec h 5 x − sec h 3 x + c 5 3

1 1 3 5 Check: Let y = sec h 5 x − sec h 3 x + c , then y ′ = sec h 4 x ⋅ − sec h x tanh x − sec h 2 x ⋅ − sec h x tanh x 3

5

5

4

3

(

2

4

2

)

= − sec h x ⋅ sec h x tanh x + sec h x ⋅ sec h x tanh x = − sec h x + sec h x sec h x tanh x

(

)

= 1 − sec h 2 x sec h 2 x sec h x tanh x = tanh 2 x sec h 2 x sec h x tanh x = tanh 3 x sec h 3 x g. Given ∫ coth 2 x csc h 2 x dx let u = coth x , then

∫

coth 2 x csc h 2 x dx =

Hamilton Education Guides

∫

u 2 csc h 2 x 2

− csc h x

du

du d = coth x dx dx

;

du = − csc h 2 x dx

; dx =

−du csc h 2 x

. Thus,

1 3

1 = − ∫ u 2 du = − u 3 + c = − coth 3 x + c 3

376

Calculus I

5.4 Integration of Hyperbolic Functions

1 3

1 3

Check: Let y = − coth 3 x + c , then y ′ = − ⋅ 3 coth 2 x ⋅ − csc h 2 x + 0 =

h.

∫ coth

3

x csc h 3 x dx =

∫ coth x coth

∫

+ csc h 2 x coth x dx = −

2

x csc h 3 x dx =

3 ⋅ coth 2 x csc h 2 x 3

= coth 2 x csc h 2 x

2 2 4 ∫ coth x (1 + csc h x ) csc h x dx = ∫ csc h x coth x dx

1 1 csc h 5 x − csc h 3 x + c 5 3

1 5

1 3

5 5

3 3

Check: Let y = − csc h 5 x − csc h 3 x + c , then y ′ = − csc h 4 x ⋅ − csc h x coth x − csc h 2 x ⋅ − csc h x coth x

(

)

= csc h 5 x coth x + csc h 3 x coth x = csc h 3 x coth x csc h 2 x + 1 = csc h 3 x coth x coth 2 x = coth 3 x csc h 3 x i.

∫ coth

3

x csc h 4 x dx =

∫ coth

∫

− coth 3 x csc h 2 x dx = − 1 6

3

x csc h 2 x csc h 2 x dx =

5 2 3 2 2 ∫ coth x ( coth x − 1) csc h x dx = ∫ coth x csc h x dx

1 1 coth 6 x + coth 4 x + c 4 6 1 4

1 6

1 4

Check: Let y = − coth 6 x + coth 4 x + c , then y ′ = − ⋅ 6 coth 5 x ⋅ − csc h 2 x + ⋅ 4 coth 3 x ⋅ − csc h 2 x + 0

(

)

= coth 5 x csc h 2 x − coth 3 x csc h 2 x = coth 3 x csc h 2 x coth 2 x − 1 = coth 3 x csc h 2 x csc h 2 x = coth 3 x csc h 4 x Table 5.4-3 provides a summary of the basic integration formulas covered in this book. Section 5.4 Practice Problems – Integration of Hyperbolic Functions 1. Evaluate the following integrals: a.

∫ cosh 3x dx =

b.

3x ∫ (sinh 2 x − e ) dx

d.

∫x

e.

∫3x

g.

∫ cosh

2

sec h 2 x3dx = 7

( x + 1) sinh ( x + 1) dx = h.

c.

∫ csc h 5x dx =

f.

∫x

3

∫ csc h (5x + 3) coth (5x + 3) dx =

i.

∫e

x +1

∫ coth ( x + 1) csc h ( x + 1) dx = ∫ sec h ( 3x + 2) dx = 5 ∫ coth x dx =

c.

∫ e tanh e dx = cosh (3 x + 5 ) sinh (3 x + 5) dx ∫e 6 ∫ coth x dx =

(

=

)

2 3 csc h 2 x 4 + 1 dx

=

(

)

csc h 2 x 4 + 5 dx = sec h e x +1 dx =

2. Evaluate the following integrals: a. d. g.

∫ tanh x sec h x dx = 3 4 ∫ x sec h ( x + 1) dx = 5 ∫ tanh x dx = 5

2

Hamilton Education Guides

b. e. h.

6

2

f. i.

3x

3x

377

=

Calculus I

5.4 Integration of Hyperbolic Functions

Table 5.4-3: Basic Integration Formulas 1.

∫ a dx

3.

∫ a f (x ) dx = a ∫ f (x ) dx

5.

= ax + c

x n +1 +c n +1

a≠0

2.

∫x

a≠0

4.

∫ [ f (x ) + g (x ) ] dx = ∫ f (x ) dx + ∫ g (x ) dx

∫ sin x dx = − cos x + c

6.

∫ cos x dx = sin x + c

7.

∫ tan x dx = ln

sec x + c

8.

∫ cot x dx = ln

9.

∫ sec x dx = ln

sec x + tan x + c

10.

∫ csc x dx = ln

12.

∫ cot x csc x dx = − csc x + c

14.

∫ cos

2

x dx =

n

dx =

n ≠ −1

sin x + c csc x − cot x + c

11.

∫ tan x sec x dx = sec x + c

13.

∫ sin

2

x dx =

15.

∫ tan

2

x dx = tan x − x + c

16.

∫ tan

2

x dx = tan x − x + c

17.

∫ cot

2

x dx = − cot x − x + c

18.

∫ sec

2

x dx = tan x + c

19.

∫ csc

2

x dx = − cot x + c

20.

∫ x dx

21.

∫ ln xdx

22.

∫

23.

∫e

24.

∫

x

x sin 2 x − +c 4 2

= x ln x − x + c

dx = e x + c

1

x sin 2 x + +c 4 2

= ln x + c

a x dx =

ax +c ln a

1 2

a −x

2

dx =

a 0 and a ≠ 1

1 x 1 x arc sin + c = sin −1 + c a a a a

=

1 x 1 x arc tan + c = tan −1 + c a a a a

26.

∫ a 2 − x 2 dx

=

1 x−a ln +c 2a x+a

28.

∫x

∫ sinh x dx = cosh x + c

30.

∫ cosh x dx = sinh x + c

31.

∫ tanh x dx = ln cosh x + c

32.

∫ coth x dx = ln

sinh x + c

33.

∫ sec h x dx = sin

34.

∫ csc h x dx = ln

tanh

35.

∫ tanh x sec h x dx = − sec h x + c

36.

∫ coth x csc h x dx = − csc h x + c

37.

∫ sinh

2

x dx =

38.

∫ cosh

2

x dx =

39.

∫ tanh

2

x dx = x − tanh x + c

40.

∫ coth

2

x dx = x − coth x + c

41.

∫ sec h

x dx = tanh x + c

42.

∫ csc h

1

25.

∫ a 2 + x 2 dx

27.

∫ x 2 − a 2 dx

29.

1

2

−1

(tanh x ) + c

sinh 2 x x − +c 4 2

Hamilton Education Guides

1

1 x2 − a2

2

=

a+x 1 +c ln a−x 2a

dx =

1 x 1 x arc sec + c = sec−1 + c a a a a

x +c 2

sinh 2 x x + +c 4 2

x dx = − coth x + c

378

Appendix - Exercise Solutions Chapter 1 Solutions: Section 1.1 Solutions - Sequences 1. List the first four and tenth terms of the given sequence. 2n + 1 , then − 2n

a1 =

2 ⋅1 + 1 3 = − = −1.5 − 2 ⋅1 2

a2 =

7 2⋅3 +1 = − = −1.17 6 − 2⋅3

a4 =

2⋅ 4 +1 9 = − = −1.13 − 2⋅4 8

a10 =

a. Given a n = a3 =

b. Given bk = b3 =

k (k + 1) k

3 ⋅ (3 + 1) 32

2

=

, then

b1 =

12 3⋅ 4 = = 1.33 9 9

b4 =

1 ⋅ (1 + 1) 12

4 ⋅ (4 + 1) 42

c. Given d n = 3 − (− 2 )n , then

= =

1⋅ 2 2 = = 2 1 1

2 ⋅ 10 + 1 21 = − = −1.05 − 2 ⋅ 10 20

2 ⋅ (2 + 1)

b2 =

4⋅5 20 = = 1.25 16 16

5 2⋅ 2 +1 = − = −1.25 4 − 2⋅2

22

b10 =

=

10 ⋅ (10 + 1) 102

2⋅3 6 = = 1.5 4 4 =

110 = 1.1 100

d1 = 3 − (− 2 )1 = 3 − (− 2 ) = 3 + 2 = 5

d 2 = 3 − (− 2 )2 = 3 − (+ 4 ) = 3 − 4 = −1

d3 = 3 − (− 2 )3 = 3 − (−8) = 3 + 8 = 11

d 4 = 3 − (− 2 )4 = 3 − (+16 ) = 3 − 16 = −13

d10 = 3 − (− 2 )10 = 3 − (+1024 ) = 3 − 1024 = −1021

 1 d. Given k n =  −   2  1 k2 =  −   2

2

 1 k4 =  −   2

4

n

1+1 1 (− 1)2 1 1 1  1  (− 1) = − ⋅ = − ⋅ = − k1 =  −  6 2 3 2 3  2  1+ 2 1

(− 1)n+1 , then n+2

(− 1)2+1 2+2

(− 1)4+1 4+2

=

1 1 1 (− 1)3 1 = ⋅− = − ⋅ 4 4 4 4 16

3+1 1 1 1 (− 1)4 1  1  (− 1) = − ⋅ = − ⋅ = − k3 =  −  8 5 8 5 40  2 3+ 2

=

1 1 (− 1)5 1 1 = ⋅ ⋅− = − 16 6 16 6 96

 1 k10 =  −   2

3

10

(− 1)10+1

1 (− 1)11 −1 = ⋅ 1024 12 12,288

=

10 + 2

2. Write s3 , s 4 , s 5 , s 8 , and s10 for the following sequences. a. Given s n = s4 = s8 =

n(n + 1) 2n

4 ⋅ (4 + 1) 2⋅4

−1

8 ⋅ (8 + 1) 2 ⋅ 8−1

−1

= =

, then

4⋅5 2⋅4

−1

8⋅9

s3 =

80 4⋅5⋅4 = = = 40 2 2

=

2 ⋅ 8−1

8⋅9⋅8 576 = = 288 2 2

b. Given s n = (− 1)n +1 2 n − 2 , then

s5 = s10 =

3 ⋅ (3 + 1)

=

−1

2⋅3

5 ⋅ (5 + 1) 2⋅5

=

−1

10 ⋅ (10 + 1) 2 ⋅ 10−1

3⋅ 4

−1

2⋅3

5⋅6 2⋅5 =

−1

=

36 3⋅ 4⋅3 = = 18 2 2

=

150 5⋅6⋅5 = = 75 2 2

10 ⋅ 11

2 ⋅ 10−1

=

1100 10 ⋅ 11 ⋅ 10 = = 550 2 2

s3 = (− 1)3+1 ⋅ 23− 2 = (− 1)4 ⋅ 21 = 1⋅ 2 = 2

s4 = (− 1)4 +1 ⋅ 24 − 2 = (− 1)5 ⋅ 22 = −1⋅ 4 = −4

s5 = (− 1)5+1 ⋅ 25− 2 = (− 1)6 ⋅ 23 = 1⋅ 8 = 8

s8 = (− 1)8+1 ⋅ 28− 2 = (− 1)9 ⋅ 26 = −1⋅ 64 = −64

s10 = (− 1)10 +1 ⋅ 210 − 2 = (− 1)11 ⋅ 28 = −1⋅ 256 = −256

c. Given s n = s4 =

s8 =

(− 2)n +1 (n − 2) , then 2n

(− 2) (4 − 2) 4 +1

2⋅4

(− 2) (8 − 2) 8 +1

2⋅8

s3 =

= =

(− 2)

5

⋅2

8

(− 2)

Hamilton Education Guides

9

16

⋅6

= =

−32 ⋅ 2 = −8 8 −512 ⋅ 6 = −192 16

s5 =

s10 =

(− 2)3+1(3 − 2) 2⋅3

(− 2) (5 − 2) 5 +1

2⋅5

(− 2)

10 +1

= =

(10 − 2)

2 ⋅ 10

(− 2)4 ⋅ 1 6

(− 2)6 ⋅ 3 10

=

= =

(− 2)11 ⋅ 8 20

16 = 2.67 6 64 ⋅ 3 = 19.2 10

=

−2048 ⋅ 8 = −819.2 20

379

Calculus I

Chapter 1 Solutions

3. Write the first five terms of the following sequences. a. Given a n = (− 1)n +1 (n + 2 ) , then

a1 = (− 1)1+1 (1 + 2 ) = (− 1)2 ⋅ 3 = 1⋅ 3 = 3

a2 = (− 1)2 +1 (2 + 2 ) = (− 1)3 ⋅ 4 = −1⋅ 4 = −4

a3 = (− 1)3+1 (3 + 2 ) = (− 1)4 ⋅ 5 = 1⋅ 5 = 5

a4 = (− 1)4 +1 (4 + 2 ) = (− 1)5 ⋅ 6 = −1⋅ 6 = −6

a5 = (− 1)5+1 (5 + 2 ) = (− 1)6 ⋅ 7 = 1⋅ 7 = 7

 1  b. Given a i = 3   100 

i −2

−1

= 3⋅

100 1 = 3⋅ = 300 1 1 100

 1  a3 = 3   100 

3− 2

3 1  1  = 3 =  = 3⋅ 1 100 100 100  

2

 1  a5 = 3   100 

5− 2

1 3  1  = 3 =  = 3⋅ 1,000,000 1003  100 

2− 2

 1  = 3  = 3 ⋅1 = 3  100 

 1  a4 = 3   100 

4− 2

3 1  1  = 3 =  = 3⋅ 10,000 1002  100  i −1

 1  = 3   100 

0

 1  a2 = 3   100 

 1 c. Given c i = 3 −   5

1− 2

 1  a1 = 3   100 

, then

 1 c2 = 3 −   5

2 −1

1 3  1 = 3 −  = 3 ⋅ − = − = −0.6 5 5  5

 1 c4 = 3 −   5

4 −1

3 1  1 = 3 −  = 3 ⋅ − 3 = − = −0.024 125 5 5  

3

1−1

 1 = 3 −  = 3 ⋅ 1 = 3  5

0

 1 c1 = 3 −   5

, then

1

1

 1 c3 = 3 −   5

3−1

3 1  1 = 3 −  = 3 ⋅ 2 = = 0.12 25 5  5

3

 1 c5 = 3 −   5

5 −1

3 1  1 = 3 −  = 3 ⋅ 4 = = 0.0048 625 5 5  

d. Given a n = (3n − 5)2 , then

2

4

a1 = (3 ⋅ 1 − 5)2 = (3 − 5)2 = (− 2 )2 = 4

a2 = (3 ⋅ 2 − 5)2 = (6 − 5)2 = 12 = 1

a3 = (3 ⋅ 3 − 5)2 = (9 − 5)2 = 42 = 16

a4 = (3 ⋅ 4 − 5)2 = (12 − 5)2 = 7 2 = 49

a5 = (3 ⋅ 5 − 5)2 = (15 − 5)2 = 102 = 100

a +2 r

u1 = ar1− 2 + 2 = ar −1 + 2 =

e. Given u k = ar k − 2 + 2 , then u2 = ar 2 − 2 + 2 = ar 0 + 2 = a + 2

u3 = ar 3− 2 + 2 = ar1 + 2 = ar + 2

u4 = ar 4 − 2 + 2 = ar 2 + 2

u5 = ar 5− 2 + 2 = ar 3 + 2

2 f. Given bk = −3  3

k −2

1− 2

2 b1 = − 3 ⋅   3

, then

2 b2 = − 3 ⋅   3

2− 2

2 = − 3 ⋅   = −3 ⋅ 1 = −3 3

2 b4 = − 3 ⋅   3

4− 2

4 4 2 = − 3⋅  = − 3⋅ = − 9 3 3

g. Given c j =

0

2 b3 = − 3 ⋅   3

3− 2

2

2 b5 = − 3 ⋅   3

5− 2

j + j , then j +1

2 = − 3⋅  3

= − 3⋅

−1

= − 3⋅

2 3/ ⋅ 2 = − = −2 3 3/ 3

8 8 2 = − 3⋅  = − 3⋅ = − 27 9 3

c1 =

1+ 2 3 1 1 = +1 = +1 = 1+1 2 2 2

c2 =

2+6 2 2 8 = +2 = +2= 2 +1 3 3 3

c3 =

3 3 3 + 12 15 = +3 = +3= 3 +1 4 4 4

c4 =

24 4 + 20 4 4 = +4 = +4= 5 5 4 +1 5

c5 =

5 5 5 + 30 35 = +5 = +5= 6 6 5 +1 6

1   h. Given y n = 1 −   n+2

n +1

Hamilton Education Guides

, then

1+1

1   y1 = 1 −   1+ 2 

9 3 1 = − 3⋅ = − 2 2 2 3

2

2

2

 1  3 −1 2 = 1 −  =   =   = 0.67 2  3  3  3

380

Calculus I

Chapter 1 Solutions

1   y2 = 1 −   2+2

2 +1

3  1  4 −1 = 1 −  =   =   = 0.753 4  4  4 

3

3

3

1   y3 = 1 −   3+ 2

1   y4 = 1 −   4+2

4 +1

5  1  6 −1 = 1 −  =   =   = 0.835 6  6  6 

5

5

5

1   y5 = 1 −   5+2

i. Given u k = 1 − (− 1)k +1 , then

3+1

 1  5 −1 4 = 1 −  =   =   = 0.84  5  5  5

4

4

4

5 +1

 1  7 −1 6 = 1 −  =   =   = 0.866  7  7  7

6

6

6

u1 = 1 − (− 1)1+1 = 1 − (− 1)2 = 1 − 1 = 0

u2 = 1 − (− 1)2 +1 = 1 − (− 1)3 = 1 − (−1) = 1 + 1 = 2

u3 = 1 − (− 1)3+1 = 1 − (− 1)4 = 1 − 1 = 0

u4 = 1 − (− 1)4 +1 = 1 − (− 1)5 = 1 − (−1) = 1 + 1 = 2

u5 = 1 − (− 1)5+1 = 1 − (− 1)6 = 1 − 1 = 0

j. Given y k = y2 = y4 =

2

2

2 −1

4

2

4 −1

k. Given yn =

k

, then

2 k −1 2

=

2

4

=

y3 =

4 = = 0.5 8

3

2

1 9k

2 = 1 2

=

1

y1 =

y5 =

(k − 2) , then

y1 =

1

21−1 3

3−1

2

5

5 −1

2

1 91

1

=

20 3

=

2

2

5

=

2

(1 − 2)

4

1 = 1 1

= =

3 = 0.75 4

=

5 = 0.313 16

= 9 ⋅ −1 = −9

1

1

1

1

1

1

y3 = 9 3 (3 − 2 ) = 9 3 ⋅ 1 = 9 3 =

1

1

y5 = 9 5 (5 − 2 ) = 9 5 ⋅ 3 = 3 5 9

y2 = 9 2 (2 − 2 ) = 9 2 ⋅ 0 = 0 y4 = 9 4 (4 − 2 ) = 9 4 ⋅ 2 = 2 4 9

l. Given c n =

n2 − 2 , then n +1

1

c1 =

12 − 2 1− 2 1 = = − 1+1 2 2

c2 =

22 − 2 4−2 2 = = 2 +1 3 3

c3 =

32 − 2 7 9−2 = = 3 +1 4 4

c4 =

42 − 2 16 − 2 14 = = 5 4 +1 5

c5 =

52 − 2 25 − 2 23 = = 5 +1 6 6

3

9

4. Given n! read as “n factorial” which is defined as n != n (n − 1)(n − 2 )(n − 3) 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 , find a. 8 ! = 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 40,320 b. Given an =

2n + 1 , then n!

a2 =

4 +1 2⋅ 2 +1 = = 5 = 5 = 2.5 2 2 ⋅1 2! 2!

a4 =

3 9 8 +1 2⋅ 4 +1 = = = = 0.375 8 4 ⋅ 3 ⋅ 2 ⋅1 4! 4!

c. Given cn =

1 + 3n −1

(n !)

2

, then

c10 =

1 + 310 −1

(10 !)

2

=

a1 =

3 2 ⋅1 + 1 2 +1 = = = 3 1 1! 1!

a3 =

7 2⋅3 +1 6 +1 7 = = = = 1.17 3 ⋅ 2 ⋅1 6 3! 3!

1 + 39 19,684 = and 10 ! 10 ! 10 ! 10 !

d. The first, fifth, tenth, and fifteenth terms of yn =

c12 =

1 + 312 −1

(12 !)

2

=

1 + 311 177,148 = 12 ! 12 ! 12 ! 12 !

n ! (n − 1) . 2+n!

y1 =

0 1 ! (1 − 1) 1! ⋅ 0 = = = 0 3 2 +1! 2 +1!

y5 =

(5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1) ⋅ 4 = 120 ⋅ 4 = 480 = 3.934 5 ! (5 − 1) 5! ⋅4 = = 2 + 120 122 2+5! 2+5! 2 + (5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1)

y10 =

(10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1) ⋅ 9 = 3,628,800 ⋅ 9 = 32,659,200 = 8.9999 ≈ 9 10 ! ⋅ 9 10 ! (10 − 1) = = 2 + 10 ! 2 + 10 ! 2 + (10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1) 2 + 3,628,800 3,628,802

Hamilton Education Guides

381

Calculus I

Chapter 1 Solutions

or, a quicker way of solving this problem is as follows:

y10 =

10 ! (10 − 1) 10 ! ⋅ 9 10 ! ⋅ 9 1/ 0/ ! ⋅ 9 = = = 9 ≈ 2 + 10 ! 2 + 10 ! 10 ! 1/ 0/ !

y15 =

(15 ⋅ 14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1) ⋅ 14 ≈ 14 or 15 ! (15 − 1) 15 ! ⋅ 14 = = 2 + 15 ! 2 + 15 ! 2 + (15 ⋅ 14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1)

y15 =

1/ 5/ ! ⋅ 14 15 ! (15 − 1) 15 ! ⋅ 14 15 ! ⋅ 14 = = = 14 ≈ 2 + 15 ! 2 + 15 ! 15 ! 1/ 5/ !

5. Write the first three terms of the following sequences.

(2n − 3)(n + 1) , then (n − 4)n [ (2 ⋅ 2) − 3] (2 + 1) = (4 − 3) ⋅3 − 2⋅2 (2 − 4) ⋅ 2

a. Given cn = c2 =

=

1 ⋅3 3 = − 4 −4

 1  n−2 b. Given, an =    , then  n −1  2 + n 

−1 ⋅ 2 2 = −3 3

c3

=

12 3 ⋅4 = − = −3 −3 4

1 11  1  3−2 a3 =    =    = 10  25  3 −1  2 + 3 

c. Given sn = (− 1) n +1 2 n +1 , then

s1 = (− 1)1+1 ⋅ 21+1 = (− 1)2 ⋅ 22 = 1⋅ 4 = 4

s2 = (− 1)2 +1 ⋅ 22 +1 = (− 1)3 ⋅ 23 = −1⋅ 8 = −8

y2 = (− 1)2 +1 ⋅

=

 1   −1   1  1− 2  a1 =   =     which is undefined  0 3   1−1  2 +1

 1   2 − 2  1  0  a2 =    =    = 0  2 −1  2 + 2   1  4 

s3 = (− 1)3+1 ⋅ 23+1 = (− 1)4 ⋅ 24 = 1⋅ 16 = 16

k (k − 1) , then 2

d. Given yk = (− 1)k +1

[ (2 ⋅ 1) − 3] (1 + 1) = (2 − 3) ⋅ 2 − 3 ⋅1 (1 − 4) ⋅ 1 [(2 ⋅ 3) − 3] (3 + 1) = (6 − 3) ⋅ 4 = − 1⋅ 3 (3 − 4) ⋅ 3

c1 =

y1 = (− 1)1+1 ⋅

2 2 ⋅ (2 − 1) 2 ⋅1 = (− 1)3 ⋅ = − = 1 2 2 2

1 ⋅ (1 − 1) 1⋅ 0 0 = (− 1)2 ⋅ = = 0 2 2 2

y3 = (− 1)3+1 ⋅

3⋅ 2 3 ⋅ (3 − 1) 6 = (− 1)4 ⋅ = = 3 2 2 2

 1−1  0 b1 = 12 ⋅   = 1⋅   = 0 2 + 1   3

 n −1  e. Given bn = n 2   , then 2+n

 2  18  3 −1  = 3.6 b3 = 32 ⋅   = 9⋅  = 5 5  2+3

4 1  2 −1  = 1 b2 = 22 ⋅   = 4⋅  = 4 4 2+2

f. Given xa = (5 − a )a +1 2a , then

x1 = (5 − 1)1+1 ⋅ 21 = 42 ⋅ 2 = 16 ⋅ 2 = 32

x2 = (5 − 2 )2 +1 ⋅ 22 = 33 ⋅ 4 = 27 ⋅ 4 = 108

x3 = (5 − 3)3+1 ⋅ 23 = 24 ⋅ 8 = 16 ⋅ 8 = 128

Section 1.2 Solutions - Series 1. Given

n

n

∑ a = 10 and ∑ b = 25 , find i

i

i =1

a.

i =1

n

∑ (2ai + 4bi ) = i =1

b.

n

∑ (− ai + bi ) = i =1

c.

n

∑ (3a i =1

d.

n

∑ i =1

i

+ 5bi ) =

n

∑

2ai +

i =1

∑ n

− ai +

i =1

∑

bi = −

n

i

1  1  a i + bi  = 2 5  

i

i =1

i =1

n

∑ i =1

Hamilton Education Guides

1 ai + 2

∑b

n

∑ i =1

i

n

n

∑ ∑b ai +

= −10 + 25 = 15

i

i =1

n

n

∑ a + 5∑ b

= 3

= (2 ⋅ 10 ) + (4 ⋅ 25) = 20 + 100 = 120

i =1

i =1

∑ 3a + ∑ 5b

n

ai + 4

i =1

i =1

n

n

∑

4bi = 2

i =1

n

∑

n

i

i

i =1

1 1 bi = 5 2

= (3 ⋅ 10 ) + (5 ⋅ 25) = 30 + 125 = 155

i =1

n

∑ i =1

ai +

1 5

n

∑b

i

i =1

 1  1 =  ⋅ 10  +  ⋅ 25  = 5 + 5 = 10 2 5    

382

Calculus I

Chapter 1 Solutions

2. Evaluate each of the following series. a. b.

5

∑ 2 + k = (2 + 1) + (2 + 2) + (2 + 3) + (2 + 4) + (2 + 5) = 3 + 4 + 5 + 6 + 7 = 25 k =1 6

1

∑ (− 2) n =0

+

1

1

=

n +1

1

+

1

+

1

+

1

+

1

+

1

+

+

1

(− 2)0+1 (− 2)1+1 (− 2)2+1 (− 2)3+1 (− 2)4+1 (− 2)5+1 (− 2)6+1 1

+

(− 2)5 (− 2)6 (− 2)7

= −

=

1

1

+

1

+

+

1

(− 2)1 (− 2)2 (− 2)3 (− 2)4

1 1 1 1 1 1 1 = −0.5 + 0.25 − 0.125 + 0.0625 − 0.031 + 0.016 − 0.008 + − + − + − 2 4 8 16 32 64 128

= (−0.5 − 0.125 − 0.031 − 0.008) + (0.25 + 0.0625 + 0.016 ) = −0.664 + 0.328 = −0.336 c. d.

4

∑ (− 1)

= (− 1)0 +1 + (− 1)1+1 + (− 1)2 +1 + (− 1)3+1 + (− 1)4 +1 = (− 1)1 + (− 1)2 + (− 1)3 + (− 1)4 + (− 1)5 = −1/ + 1/ − 1/ + 1/ − 1 = −1

n +1

n =0 3

∑ j −3j

[

][

][

][

][

][

][

= − 3 − 3 ⋅ (− 3)2 + − 2 − 3 ⋅ (− 2 )2 + − 1 − 3 ⋅ (− 1)2 + 0 − 3 ⋅ 02 + 1 − 3 ⋅ 12 + 2 − 3 ⋅ 22 + 3 − 3 ⋅ 32

2

]

j = −3

= [− 3 − (3 ⋅ 9 ) ] + [− 2 − (3 ⋅ 4 ) ] + [− 1 − (3 ⋅ 1) ] + 0 + [1 − (3 ⋅ 1) ] + [2 − (3 ⋅ 4 ) ] + [3 − (3 ⋅ 9 ) ] = [− 3 − 27] + [− 2 − 12] + [− 1 − 3] + [1 − 3] +[2 − 12] + [3 − 27] = −30 − 14 − 4 − 2 − 10 − 24 = −84

e.

5

∑ (a + 2)

= (3 + 2 )3 + (4 + 2 )4 + (5 + 2 )5 = 53 + 64 + 75 = 125 + 1296 + 16807 = 18,228

a

a =3

f.

5

∑

(− 1)i +1 2i

i =0

= −1+ g.

(− 1)0+1 + (− 1)1+1 + (− 1)2+1 + (− 1)3+1 + (− 1)4+1 + (− 1)5+1

=

20

21

22

23

24

25

=

(− 1)1 + (− 1)2 + (− 1)3 + (− 1)4 + (− 1)5 + (− 1)6 1

2

4

8

16

32

1 1 1 1 1 = −1 + 0.5 − 0.25+ 0.125− 0.063 + 0.031 = −1.313+ 0.656 = −0.657 − + − + 2 4 8 16 32

3

∑ (2k − 3)

= [(2 ⋅ −2 ) − 3]−2 + 2 + [(2 ⋅ −1) − 3]−1+ 2 + [(2 ⋅ 0 ) − 3]0 + 2 + [(2 ⋅ 1) − 3]1+ 2 + [(2 ⋅ 2 ) − 3]2 + 2 + [(2 ⋅ 3) − 3]3+ 2

k +2

k = −2

= [− 4 − 3]0 + [− 2 − 3]1 + [0 − 3]2 + [2 − 3]3 + [4 − 3]4 + [6 − 3]5 = (− 7 )0 + (− 5)1 + (− 3)2 + (− 1)3 + 14 + 35 = 248 h.

5

1   − 1 m  m =1

∑

2

2

2

2

2

1   1   1   1   1  =  − 1 +  − 1 +  − 1 +  − 1 +  − 1 1   2   3   4   5 

2

= (1 − 1)2 + (0.5 − 1)2 + (0.333 − 1)2 + (0.25 − 1)2 + (0.2 − 1)2

= 02 + (− 0.5)2 + (0.667 )2 + (− 0.75)2 + (− 0.8)2 = 0.25+ 0.445 + 0.563+ 0.64 = 1.898 3. Find the sum of the following series within the specified range. a.

3

∑10

i

= 10−3 + 10−2 + 10−1 + 100 + 101 + 102 + 103 = 0.001 + 0.01 + 0.1 + 1 + 10 + 100 + 1000 = 1111.111

i = −3

b.

6

n −1

∑2 n =0

n

=

1 0 1 2 3 4 5 0 −1 1−1 2 −1 3 −1 4 −1 5 −1 6 −1 = −1 + 0.25 + 0.25 + 0.1875 + + + 1 + 2 + 3 + 4 + 5 + 6 = − + + + + 1 2 4 8 16 32 64 20 2 2 2 2 2 2

+0.125 + 0.08 = −0.108

c.

4

1

∑ 10 a =0

d.

5

∑ (n

a

2

=

1

100

+

) (

1

101

+

1

102

) (

+

1

103

) (

+

1

104

=

) (

1 1 1 1 1 = 1 + 0.1 + 0.01 + 0.001 + 0.0001 = 1.1111 + + + + 1 10 100 1000 10000

) (

)

− n = 12 − 1 + 22 − 2 + 32 − 3 + 42 − 4 + 52 − 5 = 0 + (4 − 2 ) + (9 − 3) + (16 − 4 ) + (25 − 5) = 2 + 6 + 12 + 20 = 40

n =1

Hamilton Education Guides

383

Calculus I

e.

Chapter 1 Solutions

6

∑ (− 1)

= (− 1)0 +1 + (− 1)1+1 + (− 1)2 +1 + (− 1)3+1 + (− 1)4 +1 + (− 1)5+1 + (− 1)6 +1 = (− 1)1 + (− 1)2 + (− 1)3 + (− 1)4 + (− 1)5

m +1

m =0

+ (− 1)6 + (− 1)7 = −1/ + 1/ − 1/ + 1/ − 1/ + 1/ − 1 = −1

f.

5

∑

1 + (− 1)k 2

k =0

= g.

=

k

1 + (− 1)0 2

0

+

1 + (− 1)1 1

2

+

1 + (− 1)2 2

2

+

1 + (− 1)3 2

3

+

1 + (− 1)4 2

+

4

1 + (− 1)5 25

=

1+1 1−1 1+1 1−1 1+1 1−1 + + + + + 1 2 4 8 16 32

2 0 2 0 2 0 = 2 + 0.5 + 0.125 = 2.625 + + + + + 1 2 4 8 16 32

6

∑ [5(a − 1) + 3] = [5(1 − 1) + 3] + [5(2 − 1) + 3] + [5(3 − 1) + 3] + [5(4 − 1) + 3] + [5(5 − 1) + 3] + [5(6 − 1) + 3] = [5 ⋅ 0 + 3] + [5 ⋅1 + 3] a =1

+[5 ⋅ 2 + 3] + [5 ⋅ 3 + 3] + [5 ⋅ 4 + 3] + [5 ⋅ 5 + 3] = 3 + 8 + 13 + 18 + 23 + 28 = 93

h.

5

 1 −   3 k =0

∑

k −1

 1 = −   3

3

0 −1

1−1

 1 + −   3

 1 + −   3

2 −1

 1 + −   3

3−1

 1 + −   3

4 −1

 1 + −   3

5 −1

 1 = −   3

−1

1

0

 1  1  1 + −  + −  + −   3  3  3

2

4

1 1 1 1  1  1 = −3 + 1 − 0.33 + 0.11 − 0.04 + 0.01 = −2.25 + + −  + −  = − 3 +1− + − 3 9 27 81 3 3    

i.

5

∑ ( j − 3 j ) = (1 − 3 ⋅1 )+ (2 − 3 ⋅ 2 )+ (3 − 3 ⋅ 3 )+ (4 − 3 ⋅ 4 )+ (5 − 3 ⋅ 5 ) = (1 − 3) + (2 − 12) + (3 − 27) + (4 − 48) + (5 − 75) 2

2

2

2

2

2

j =1

= −2 − 10 − 24 − 44 − 72 = −152 j.

4

∑ n =1

n +1 − n

4

22 32 42   2 3 4 5   1 4 9 16  n2  1 + 1 2 + 1 3 + 1 4 + 1   12 =  =  + + + − + + +  + + + + + + − n +1 2 3 4   1 + 1 2 + 1 3 + 1 4 + 1   1 2 3 4   2 3 4 5   1 n =1

∑

= (2 + 1.5 + 1.33 + 1.25) − (0.5 + 1.33 + 2.25 + 3.2 ) = 6.08 − 7.28 = −1.2 k. l.

5

∑ 5k

= 5 ⋅ 1−1 + 5 ⋅ 2−1 + 5 ⋅ 3−1 + 5 ⋅ 4−1 + 5 ⋅ 5−1 =

−1

k =1 4

∑ (− 0.1)

2i − 5

5 5 5 5 5 + + + + = 5 + 2.5 + 1.67 + 1.25 + 1 = 11.42 1 2 3 4 5

= (− 0.1)2⋅1−5 + (− 0.1)2⋅2 −5 + (− 0.1)2⋅3−5 + (− 0.1)2⋅4 −5 = (− 0.1)−3 + (− 0.1)−1 + (− 0.1)1 + (− 0.1)3

i =1

=

1

+

1

(− 0.1)3 (− 0.1)1

− 0.1 − 0.001 =

1 1 + − 0.1 − 0.001 = −1000 − 10 − 0.1 − 0.001 = −1010.101 − 0.001 − 0.1

4. Rewrite the following terms using the sigma notation. a.

1 1 1 1 1 1 + + + + + = 2 3 4 5 6 7

c. 2 + 4 + 8 + 16 + 32 + 64 = e. 0 +

1 2 3 4 5 + + + + = 2 3 4 5 6

5

1

1 2 3 4 5 6 + + + + + = 2 3 4 5 6 7

∑n+2

b.

∑2

d. 1 +

n=0 5

k +1

k =0 6

∑ i =1

n−1 n

f. 1 −

6

n

∑ n+1

n =1 6

1

1 1 1 1 1 + + + + = 2 3 4 5 6

∑k

1 1 1 1 1 + − + − = 2 3 4 5 6

∑

k =1 6

n =1

(− 1)n+1 n

Section 1.3 Solutions - Arithmetic Sequences and Arithmetic Series 1. Find the next seven terms of the following arithmetic sequences. a. Substituting s1 = 3 and d = 2 into the general arithmetic sequence s n = s1 + (n − 1)d for n = 2, 3, 4, 5, 6, and 7 we obtain s 2 = s1 + (2 − 1)d = s1 + d = 3 + 2 = 5

s3 = s1 + (3 − 1)d = s1 + 2d = 3 + (2 ⋅ 2 ) = 3 + 4 = 7

Hamilton Education Guides

384

Calculus I

Chapter 1 Solutions

s4 = s1 + (4 − 1)d = s1 + 3d = 3 + (3 ⋅ 2 ) = 3 + 6 = 9

s5 = s1 + (5 − 1)d = s1 + 4d = 3 + (4 ⋅ 2 ) = 3 + 8 = 11

s6 = s1 + (6 − 1)d = s1 + 5d = 3 + (5 ⋅ 2 ) = 3 + 10 = 13 s7 = s1 + (7 − 1)d = s1 + 6d = 3 + (6 ⋅ 2 ) = 3 + 12 = 15

Thus, the first seven terms of the arithmetic sequence are (3, 5, 7, 9, 11, 13, 15 ) b. Substituting s1 = −3 and d = 2 into the general arithmetic sequence s n = s1 + (n − 1)d for n = 2, 3, 4, 5, 6, and 7 we obtain s 2 = s1 + (2 − 1)d = s1 + d = −3 + 2 = −1

s3 = s1 + (3 − 1)d = s1 + 2d = −3 + (2 ⋅ 2 ) = −3 + 4 = 1

s4 = s1 + (4 − 1)d = s1 + 3d = −3 + (3 ⋅ 2 ) = −3 + 6 = 3

s5 = s1 + (5 − 1)d = s1 + 4d = −3 + (4 ⋅ 2 ) = −3 + 8 = 5

s6 = s1 + (6 − 1)d = s1 + 5d = −3 + (5 ⋅ 2 ) = −3 + 10 = 7 s7 = s1 + (7 − 1)d = s1 + 6d = −3 + (6 ⋅ 2 ) = −3 + 12 = 9

Thus, the first seven terms of the arithmetic sequence are (− 3, − 1, 1, 3, 5, 7, 9 ) c. Substituting s1 = 10 and d = 0.8 into the general arithmetic sequence s n = s1 + (n − 1)d for n = 2, 3, 4, 5, 6, and 7 we obtain s 2 = s1 + (2 − 1)d = s1 + d = 10 + 0.8 = 10.8

s3 = s1 + (3 − 1)d = s1 + 2d = 10 + (2 ⋅ 0.8) = 10 + 1.6 = 11.6

s4 = s1 + (4 − 1)d = s1 + 3d = 10 + (3 ⋅ 0.8) = 10 + 2.4 = 12.4 s5 = s1 + (5 − 1)d = s1 + 4d = 10 + (4 ⋅ 0.8) = 10 + 3.2 = 13.2 s6 = s1 + (6 − 1)d = s1 + 5d = 10 + (5 ⋅ 0.8) = 10 + 4 = 14

s7 = s1 + (7 − 1)d = s1 + 6d = 10 + (6 ⋅ 0.8) = 10 + 4.8 = 14.8

Thus, the first seven terms of the arithmetic sequence are (10, 10.8, 11.6, 12.4, 13.2, 14, 14.8 ) 2. Find the general term and the eighth term of the following arithmetic sequences. a. Given s1 = 3 and d = 4 , the nth term of the arithmetic sequence is equal to s n = s1 + (n − 1)d = 3 + (n − 1) ⋅ 4 = 3 + 4n − 4 = 4n − 1 . Substituting n = 8 into the general equation s n = 4n − 1 we have s8 = 4 ⋅ 8 − 1 = 32 − 1 = 31 b. Given s1 = −3 and d = 5 , the nth term of the arithmetic sequence is equal to s n = s1 + (n − 1)d = −3 + (n − 1) ⋅ 5 = −3 + 5n − 5 = 5n − 8 . Substituting n = 8 into the general equation s n = 5n − 8 we have s8 = 5 ⋅ 8 − 8 = 40 − 8 = 32 c. Given s1 = 8 and d = −1.2 , the nth term of the arithmetic sequence is equal to s n = s1 + (n − 1)d = 8 + (n − 1) ⋅ −1.2 = 8 − 1.2n + 1.2 = −1.2n + 9.2 . Substituting n = 8 into the general equation s n = −1.2n + 9.2 we have s8 = −1.2 ⋅ 8 + 9.2 = −9.6 + 9.2 = −0.4 3. Find the next six terms in each of the following arithmetic sequences. a. Given the arithmetic sequence 5, 8,  , s1 = 5 and d = 8 − 5 = 3 . Therefore, using the general arithmetic equation sn = s1 + (n − 1)d or sn +1 = sn + d the next six terms are as follows:

s3 = s2 + d = 8 + 3 = 11

s4 = s3 + d = 11 + 3 = 14

s5 = s4 + d = 14 + 3 = 17

s6 = s5 + d = 17 + 3 = 20

s7 = s6 + d = 20 + 3 = 23

s8 = s7 + d = 23 + 3 = 26

Thus, the first eight terms of the arithmetic sequence are (5, 8, 11, 14, 17, 20, 23, 26 ) b. Given the arithmetic sequence x, x + 4,  , s1 = x and d = (x + 4 ) − x = 4 . Therefore, using the general arithmetic equation sn = s1 + (n − 1)d or sn +1 = sn + d the next six terms are as follows: s3 = s2 + d = (x + 4 ) + 4 = x + 8

Hamilton Education Guides

s4 = s3 + d = (x + 8) + 4 = x + 12

385

Calculus I

Chapter 1 Solutions

s5 = s4 + d = (x + 12 ) + 4 = x + 16

s6 = s5 + d = (x + 16 ) + 4 = x + 20

s7 = s6 + d = (x + 20 ) + 4 = x + 24

s8 = s7 + d = (x + 24 ) + 4 = x + 28

Thus, the first eight terms of the arithmetic sequence are ( x , x + 4, x + 8, x + 12, x + 16, x + 20, x + 24, x + 28 ) c. Given the arithmetic sequence 3 x + 1, 3 x + 4,  , s1 = 3 x + 1 and d = (3 x + 4 ) − (3 x + 1) = 3 . Therefore, using the general arithmetic equation sn = s1 + (n − 1)d or sn +1 = sn + d the next six terms are as follows: s3 = s2 + d = (3 x + 4 ) + 3 = 3 x + 7

s4 = s3 + d = (3 x + 7 ) + 3 = 3 x + 10

s5 = s4 + d = (3 x + 10 ) + 3 = 3 x + 13

s6 = s5 + d = (3 x + 13) + 3 = 3 x + 16

s7 = s6 + d = (3 x + 16 ) + 3 = 3 x + 19

s8 = s7 + d = (3 x + 19 ) + 3 = 3 x + 22

Thus, the first eight terms of the arithmetic sequence are (3 x + 1, 3 x + 4, 3 x + 7, 3 x + 10, 3 x + 13, 3 x + 16, 3 x + 19, 3 x + 22 ) d. Given the arithmetic sequence w, w − 10,  , s1 = w and d = (w − 10 ) − w = −10 . Using the general arithmetic equation sn = s1 + (n − 1)d or sn +1 = sn + d the next six terms are as follows: s3 = s2 + d = (w − 10 ) − 10 = w − 20

s4 = s3 + d = (w − 20 ) − 10 = w − 30

s5 = s4 + d = (w − 30 ) − 10 = w − 40

s6 = s5 + d = (w − 40 ) − 10 = w − 50

s7 = s6 + d = (w − 50 ) − 10 = w − 60

s8 = s7 + d = (w − 60 ) − 10 = w − 70

Thus, the first eight terms of the arithmetic sequence are (w , w − 10, w − 20, w − 30, w − 40, w − 50, w − 60, w − 70 ) 4. Find the sum of the following arithmetic series. a. The first three terms of the given series are

20

∑ (2i − 4) = (2 ⋅10 − 4) + (2 ⋅11 − 4) + (2 ⋅12 − 4) + 

= 16 + 18 + 20 +  .

i =10

Substituting s1 , d , and n into the arithmetic series formula

Therefore, s1 = 16 , d = 18 − 16 = 2 , and n = 11 .

n 11 [2 ⋅ 16 + (11 − 1) ⋅ 2] = 5.5 ⋅ (32 + 10 ⋅ 2) = 5.5 ⋅ (32 + 20) = 5.5 ⋅ 52 = 286 S n = [2 s1 + (n − 1)d ] we can obtain S11 = 2 2 b. The first three terms of the given series are

1000

∑k

= 1 + 2 + 3 +  . Therefore, s1 = 1 , d = 2 − 1 = 1 , and n = 1000 .

k =1

Substituting s1 , d , and n into the arithmetic series formula S n = S1000 =

n [2s1 + (n − 1)d ] we obtain 2

1000 [2 ⋅ 1 + (1000 − 1) ⋅ 1] = 500 ⋅ [2 + 999] = 500 ⋅ 1001 = 500500 2

c. The first three terms of the given series are

100

∑ (2k − 3) = (2 ⋅1 − 3) + (2 ⋅ 2 − 3) + (2 ⋅ 3 − 3) + 

= −1 + 1 + 3 +  .

k =1

Therefore, s1 = −1 , d = 1 − (− 1) = 2 , and n = 100 . Sn =

Substituting s1 , d , and n into the arithmetic series formula

n [2s1 + (n − 1)d ] we obtain S100 = 100 [ (2 ⋅ −1) + (100 − 1) ⋅ 2] = 50 ⋅ (−2 + 99 ⋅ 2) = 50 ⋅ (−2 + 198) = 50 ⋅ 196 = 9800 2 2

d. The first three terms of the given series are

15

∑ 3i = (3 ⋅1) + (3 ⋅ 2) + (3 ⋅ 3) + 

= 3+6+9+  .

i =1

Therefore, s1 = 3 , d = 6 − 3 = 3 , and n = 15 .

Substituting s1 , d , and n into the arithmetic series formula

15 n [ (2 ⋅ 3) + (15 − 1) ⋅ 3] = 12.5 ⋅ (6 + 14 ⋅ 3) = 12.5 ⋅ 48 = 600 S n = [2 s1 + (n − 1)d ] we obtain S15 = 2 2

e. The first three terms of the given series are

10

∑ (i + 1) = (1 + 1) + (2 + 1) + (3 + 1) + 

= 2+3+ 4+  .

i =1

Hamilton Education Guides

386

Calculus I

Chapter 1 Solutions Substituting s1 , d , and n into the arithmetic series formula

Therefore, s1 = 2 , d = 3 − 2 = 1 , and n = 10 . Sn =

n [2s1 + (n − 1)d ] we obtain S10 = 10 [ (2 ⋅ 2) + (10 − 1) ⋅ 1] = 5 ⋅ (4 + 9) = 5 ⋅ 13 = 65 2 2

f. The first three terms of the given series are

15

∑ (2k − 1) = (2 ⋅ 5 − 1) + (2 ⋅ 6 − 1) + (2 ⋅ 7 − 1) + 

= 9 + 11 + 13 +  .

k =5

Substituting s1 , d , and n into the arithmetic series formula

Therefore, s1 = 9 , d = 11 − 9 = 2 , and n = 11 . Sn =

n [2s1 + (n − 1)d ] we obtain S11 = 11 [ (2 ⋅ 9) + (11 − 1) ⋅ 2] = 5.5 ⋅ (18 + 20) = 5.5 ⋅ 38 = 209 2 2

g. The first three terms of the given series are

10

∑ (3i + 4) = (3 ⋅ 4 + 4) + (3 ⋅ 5 + 4) + (3 ⋅ 6 + 4) + 

= 16 + 19 + 22 +  .

i =4

Substituting s1 , d , and n into the arithmetic series formula

Therefore, s1 = 16 , d = 19 − 16 = 3 , and n = 7 . Sn =

n [2s1 + (n − 1)d ] we obtain S7 = 7 [ (2 ⋅ 16) + (7 − 1) ⋅ 3] = 3.5 ⋅ (32 + 18) = 3.5 ⋅ 50 = 175 2 2

h. The first three terms of the given series are

13

∑ (3 j + 1) = (3 ⋅ 5 + 1) + (3 ⋅ 6 + 1) + (3 ⋅ 7 + 1) + 

= 16 + 19 + 22 +  .

j =5

Substituting s1 , d , and n into the arithmetic series formula

Therefore, s1 = 16 , d = 19 − 16 = 3 , and n = 9 .

9 n S n = [2 s1 + (n − 1)d ] we obtain S9 = [ (2 ⋅ 16 ) + (9 − 1) ⋅ 3] = 4.5 ⋅ (32 + 24 ) = 4.5 ⋅ 56 = 252 2 2

i. The first three terms of the given series are

18

∑ (4k − 3) = (4 ⋅ 7 − 3) + (4 ⋅ 8 − 3) + (4 ⋅ 9 − 3) + 

= 25 + 29 + 33 +  .

k =7

Therefore, s1 = 25 , d = 29 − 25 = 4 , and n = 12 . Sn =

Substituting s1 , d , and n into the arithmetic series formula

n [2s1 + (n − 1)d ] we obtain S12 = 12 [ (2 ⋅ 25) + (12 − 1) ⋅ 4] = 6 ⋅ (50 + 44) = 6 ⋅ 94 = 564 2 2

5. The first term of an arithmetic sequence is 6 and the third term is 24 . Find the tenth term. Since s1 = 6 and s3 = 24 we use the general formula s n = s1 + (n − 1)d in order to solve for d . Therefore, s3 = s1 + (3 − 1)d ; 24 = 6 + 2d ; 24 − 6 = 2d ; d =

18 ; d = 9 . Then, s10 = s1 + (10 − 1)d = s1 + 9 ⋅ d = 6 + 9 ⋅ 9 = 87 2

6. Given the first term s1 and d , find S 50 for each of the following arithmetic sequences. a. Given s1 = 2 and d = 5 , use the n th term for an arithmetic series S n = S50 =

n [2s1 + (n − 1)d ] to find S 50 . 2

50 [ (2 ⋅ 2) + (50 − 1) ⋅ 5] = 50 (4 + 245) = 25 ⋅ 249 = 6225 2 2

b. Given s1 = −5 and d = 6 , use the n th term for an arithmetic series S n = S50 =

50 [ (2 ⋅ −5) + (50 − 1) ⋅ 6] = 50 (− 10 + 294) = 25 ⋅ 284 = 7100 2 2

c. Given s1 = 30 and d = 10 , use the n th term for an arithmetic series S n = S50 =

n [2s1 + (n − 1)d ] to find S 50 . 2

n [2s1 + (n − 1)d ] to find S 50 . 2

50 [ (2 ⋅ 30) + (50 − 1) ⋅ 10] = 50 (60 + 490) = 25 ⋅ 550 = 13750 2 2

7. Find the sum of the following sequences for the indicated values. a. Given the sequence −8, 6,  the first term s1 and the common difference d are equal to s1 = −8 and d = 6 − (−8) = 14 . Thus, using the general arithmetic series S n =

Hamilton Education Guides

n [2s1 + (n − 1)d ] , S15 is equal to: 2

387

Calculus I

S15 =

Chapter 1 Solutions 15 [ (2 ⋅ −8) + (15 − 1) ⋅ 14] = 15 (− 16 + 196) = 7.5 ⋅ 180 = 1350 2 2

b. Given the sequence −20, 20,  the first term s1 and the common difference d are equal to s1 = −20 and d = 20 − (−20 ) = 40 . Thus, using the general arithmetic series S n = S100 =

n [2s1 + (n − 1)d ] , S100 is equal to: 2

100 [ (2 ⋅ −20) + (100 − 1) ⋅ 40] = 50(−40 + 3960) = 50 ⋅ 3920 = 196,000 2

Section 1.4 Solutions - Geometric Sequences and Geometric Series 1. Find the next four terms of the following geometric sequences. a. Substituting s 1 = 3 , r = 0.5 into sn = s1r n −1 we obtain s2 = 3 ⋅ r 2 −1 = 3r = 3 ⋅ 0.5 = 1.5

s3 = 3 ⋅ r 3−1 = 3r 2 = 3 ⋅ 0.52 = 3 ⋅ 0.25 = 0.75

s4 = 3 ⋅ r 4 −1 = 3r 3 = 3 ⋅ 0.53 = 3 ⋅ 0.125 = 0.375

s5 = 3 ⋅ r 5−1 = 3r 4 = 3 ⋅ 0.54 = 3 ⋅ 0.0625 = 0.1875

Thus, the first five terms of the geometric sequence are (3, 1.5, 0.75, 0.375, 0.1875 ) b. Substituting s1 = −5 , r = 2 into sn = s1r n −1 we obtain s2 = − 5 ⋅ r 2 −1 = −5r = −5 ⋅ 2 = −10

s3 = − 5 ⋅ r 3−1 = − 5r 2 = − 5 ⋅ 22 = −5 ⋅ 4 = −20

s4 = − 5 ⋅ r 4 −1 = − 5r 3 = − 5 ⋅ 23 = −5 ⋅ 8 = −40

s5 = − 5 ⋅ r 5−1 = − 5r 4 = − 5 ⋅ 24 = −5 ⋅ 16 = −80

Thus, the first five terms of the geometric sequence are (− 5, − 10, − 20, − 40, − 80 ) c. Substituting s1 = 5 , r = 0.75 into sn = s1r n −1 we obtain s2 = 5 ⋅ r 2 −1 = 5r = 5 ⋅ 0.75 = 3.75

s3 = 5 ⋅ r 3−1 = 5r 2 = 5 ⋅ 0.752 = 5 ⋅ 0.563 = 2.81

s4 = 5 ⋅ r 4 −1 = 5r 3 = 5 ⋅ 0.753 = 5 ⋅ 0.42 = 2.11

s5 = 5 ⋅ r 5−1 = 5r 4 = 5 ⋅ 0.754 = 5 ⋅ 0.316 = 1.58

Thus, the first five terms of the geometric sequence are (5, 3.75, 2.81, 2.11, 1.58 ) 2. Find the eighth and the general term of the following geometric sequences. a. Substituting s1 = 2 , r = 3 into sn = s1r n −1 the eighth and the n th term are equal to: s8 = 2r 8−1 = 2r 7 = 2 ⋅

( 3)

7

7

= 2 ⋅ 3 2 = 2 ⋅ 46.76 = 93.53 and sn = 2 ⋅

( 3)

n −1

= 2⋅3

n −1 2

b. Substituting s1 = −4 , r = 1.2 into sn = s1r n −1 the eighth and the n th term are equal to: n −1 s8 = − 4r 8−1 = − 4r 7 = − 4 ⋅ (1.2 )7 = − 4 ⋅ 1.27 = −4 ⋅ 3.583 = −14 ⋅ 33 and sn = − 4 ⋅ (1.2 )

c. Substituting s1 = 4 , r = −2.5 into sn = s1r n −1 the eighth and the n th term are equal to: s8 = 4r 8−1 = 4r 7 = 4 ⋅ (− 2.5)7 = − 4 ⋅ 2.57 = −4 ⋅ 610.35 = −2441.4 and sn = 4 ⋅ (− 2.5 )n −1

3. Find the next six terms and the n th term in each of the following geometric sequences. 1 a. Given 1, ,  , then s1 = 1 and r = 4

=

1 . Using the general geometric equation sn = s1r n −1 the next six terms are: 4 3

2

1 1 s4 = 1 ⋅ r 4 −1 = r 3 =   = 3 4 4  

4

1 1 s6 = 1 ⋅ r 6 −1 = r 5 =   = 5 4 4  

6

1 1 s8 = 1 ⋅ r 8−1 = r 7 =   = 7 4 4

1 1 s3 = 1 ⋅ r 3−1 = r 2 =   = 2 4 4   1 1 s5 = 1 ⋅ r 5−1 = r 4 =   = 4 4 4   1 1 s7 = 1 ⋅ r 7 −1 = r 6 =   = 6 4 4

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1 4 1

5

7

388

Calculus I

Chapter 1 Solutions

 1 1 1 1 1 1 1  Thus, the first eight terms of the geometric sequence are  1, , 2 , 3 , 4 , 5 , 6 , 7  and the n th term is equal to  4 4 4 4 4 4 4 

1 sn = 1 ⋅   4

n −1

1n −1

=

4n −1

1

=

4n −1

1 1 1 b. Given − , ,  , then s1 = − and r = 2 4 2

1 4 −1

=−

2

the next six terms are:

1⋅ 2 2 1 = − = − . Using the general geometric equation sn = s1r n −1 4 ⋅1 4 2

2

s3 = −

1 3−1 1 1 1  1 = − r2 = − ⋅  −  = − 3 ⋅r 2 2 2  2 2

s5 = −

1 1 5−1 1 1  1 = − r4 = − ⋅  −  = − 5 ⋅r 2 2 2  2 2

s7 = −

1 1 7 −1 1 1  1 = − r6 = − ⋅  −  = − 7 ⋅r 2 2 2  2 2

4

6

3

s4 = −

1 4 −1 1 1 1  1 = − r3 = − ⋅  −  = 4 ⋅r 2 2 2  2 2

s6 = −

1 6 −1 1 1  1 = r5 = − ⋅  −  = 6 ⋅r 2 2  2 2

s8 = −

1 8−1 1 1 1 1 = − r7 = −  −  = 8 ⋅r 2 2 2 2 2

5

7

1 1  1 1 1 1  1 1 Thus, the first eight terms of the geometric sequence are  − , 2 , − 3 , 4 , − 5 , 6 , − 7 , 8  and the n th term is 2 2 2  2 2 2 2 2  equal to sn = − c. Given

1  1 ⋅−  2  2

n −1

= −

(− 1)n −1 = − (− 1)n −1 = − (− 1)n −1 1 (− 1)n −1 ⋅ n −1 = − 2 2 2 ⋅ 2n −1 2n −1+1 2n

p 1 and r = p, − 3 p,  , then s1 = 3 3

−3 p p 3

=

−3 p 1

p 3

=

− 3 p ⋅ 3 − 9 p/ = − 9 . Using the general geometric equation p/ 1⋅ p

sn = s1r n −1 the next six terms are: p p p p 3−1 = ⋅ r 2 = ⋅ (− 9 )2 = ⋅ 92 = 3 3 p ⋅r 3 3 3 3 p p p p s5 = ⋅ r 5−1 = ⋅ r 4 = ⋅ (− 9 )4 = ⋅ 94 = 37 p 3 3 3 3 p p p p s7 = ⋅ r 7 −1 = ⋅ r 6 = ⋅ (− 9 )6 = ⋅ 96 = 311 p 3 3 3 3

p p p p 4 −1 = ⋅ r 3 = ⋅ (− 9 )3 = − ⋅ 93 = − 35 p ⋅r 3 3 3 3 p p p p s6 = ⋅ r 6 −1 = ⋅ r 5 = ⋅ (− 9 )5 = − ⋅ 95 = − 39 p 3 3 3 3 p p p s8 = ⋅ r 8−1 = r 7 = ⋅ (− 9 )7 = − ⋅ 97 = − 313 p 3 3 3

s3 =

s4 =

 1  Thus, the first eight terms of the geometric sequence are  , − 3 p, 33 p, − 35 p, 37 p, − 39 p, 311 p, − 313 p  and the n th 3 p   p p p n −1 2 n − 3 n −1 n −1 2 n −1 n −1 2 n − 2 n −1 2 n − 2 −1 = ⋅ (− 1) 3 = ⋅ (− 1) ⋅ 3 = p (− 1) 3 = p (− 1) 3 term is equal to sn = ⋅ (− 9 ) 3 3 3 4. Given the following terms of a geometric sequence, find the common ratio r .

( )

a. Substitute s1 = 25 and s 4 = ;

1

( )

 1 3 = r 3 ;  3  = r 3 3 5 5 

1

1 3

;

b. Substitute s1 = 4 and s5 =

1

1 1 1 into sn = s1r n −1 and solve for r , i.e., s4 = s1r 4 −1 ; = 25r 3 ; 5 = r 3 ; = r3 5 25 125 5

1 3× 1 5 3

=r

3× 1

3

;

1 1 =r ; r= 5 5 1

1 1 into sn = s1r n −1 and solve for r , i.e., s5 = s1r 5−1 ; = 4r 5−1 ; 64 = r 4 ; 4 64 64 1

;

( )

1 1 1  1 4 = r4 ; = r4 ; 4 = r4 ;  4  = r4 64 × 4 256 4 4 

1 4

;

1 4× 1 4 4

=r

4× 1

4

;

1

1 1 37

=r

7× 1 7

;

1 1 37

=r ; r=

Hamilton Education Guides

= r4

1 1 =r ; r= 4 4

c. Substitute s1 = 3 and s8 = 1 into sn = s1r n −1 and solve for r , i.e., s8 = s1r 8−1 ; 1 = 3r 7 ; ;

1 64 4 1

( )

1  1 7 = r7 ;   = r7 3 3

1 7

1

7

3

389

Calculus I

Chapter 1 Solutions

5. Write the first five terms of the following geometric sequences.  1 a. Given sn =  −   3

2⋅1−1

 1 s1 =  −   3

, then

 1 = −   3

2 −1

4 −1

1  1 = −  = − 3 3 3  

3

 1 s3 =  −   3

2⋅3−1

 1 = −   3

8 −1

1  1 −  = − 7  3 3

7

 1 s5 =  −   3

2⋅5 −1

 1 = −   3

2⋅2 −1

 1 = −   3

2⋅4 −1

 1 = −   3

 1 s2 =  −   3  1 s4 =  −   3

2 n −1

=

1 3

= −

5

6 −1

1  1 = −  = − 5 3  3 

10 −1

=

9

1  1 −  = − 9  3 3

 1 1 1 1 1  Thus, the first five terms of the geometric sequence are  − , − 3 , − 5 , − 7 , − 9  3 3 3 3 3  

 1  3

b. Given s n =   2⋅2 + 2

1 =   3

2⋅4 + 2

1 =   3

1 s2 =   3 1 s4 =   3

2n+2

1 s1 =   3

, then 6

2⋅1+ 2

1 =   3

2+ 2

4+ 2

1 1 =   = 6 3 3

1 s3 =   3

2⋅3+ 2

1 =   3

8+ 2

1 =   3

10

1 s5 =   3

2⋅5 + 2

1 =   3

=

1

310

=

4

1 1   = 4 3 3   8

6+ 2

1 1 =   = 8 3 3

10 + 2

=

12

1   3

1

=

312

 1 1 1 1  1 Thus, the first five terms of the geometric sequence are  4 , 6 , 8 , 10 , 12  3 3 3 3 3  

 1 5

c. Given s n =  −   

 1 = −   5

2⋅4 −3

 1 = −   5

4 −3

8−3

= −

1 5 5

1  1 = −  = − 5 5  5

2 −3

2⋅1−3

 1 = −   5

 1 s3 =  −   5

2⋅3−3

 1 = −   5

 1 s5 =  −   5

2⋅5 −3

 1 = −   5

 1 s1 =  −   5

, then

2⋅2 −3

 1 s2 =  −   5  1 s4 =  −   5

2 n −3

=

 1 −   5

−1

= −5

3

6 −3

1  1 = −  = −  5 53

10 −3

1  1 = −  = −  5 57

7

1  1 1 1  Thus, the first five terms of the geometric sequence are  − 5, − , − 3 , − 5 , − 7  5 5  5 5 

 1  2

n

1

1  1 s1 =  −  = − = −0.5 2  2

d. Given s n =  −  , then 2

1 1  1 s3 =  −  = − 3 = − = −0.125 8 2 2  

3

4

1 1  1 = −0.031 s5 =  −  = − 5 = − 32 2  2

1 1  1 = 0.25 s2 =  −  = 2 = 4 2 2  

5

1 1  1 = 0.063 s4 =  −  = 4 = 16 2  2

Thus, the first five terms of the geometric sequence are (− 0.5, 0.25, − 0.125, 0.063, − 0.031) 6. Evaluate the sum of the following geometric series. a.

6

∑3

= 31−1 + 32 −1 + 33−1 + 34 −1 + 35−1 + 36 −1 = 30 + 31 + 32 + 33 + 34 + 35 = 1 + 3 + 9 + 27 + 81 + 243 = 364

k −1

k =1

or we can use the geometric series formula S n = S6 =

b.

10

(

)

(

)

s1 1 − r n 3 where s1 = 1 , r = = 3 , and n = 6 . Therefore, 1− r 1

1 ⋅ 1 − 36 1 − 729 728 = = = 364 2 1− 3 −2

∑ (− 2)

k −3

= (− 2 )3−3 + (− 2 )4 −3 + (− 2 )5−3 + (− 2 )6 −3 + (− 2 )7 −3 + (− 2 )8−3 + (− 2 )9 −3 + (− 2 )10 −3 = (− 2 )0 + (− 2 )1 + (− 2 )2

k =3

+ (− 2 )3 + (− 2 )4 + (− 2 )5 + (− 2 )6 + (− 2 )7 = 1 − 2 + 4 − 8 + 16 − 32 + 64 − 128 = −85

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Calculus I

Chapter 1 Solutions

or we can use the geometric series formula S n = S8 =

c.

[

1 ⋅ 1 − (− 2 )8 1 − (− 2 )

8

j +1

 1 4 −   2 j =4

∑

]

)

255 1 − 256 = − = −85 1+ 2 3

= 8

 1 −   2 j =4

∑

= 4

(

s1 1 − r n −2 where s1 = 1 , r = = −2 , and n = 8 . Therefore, 1− r 1

j +1

 1  4 +1  1 5+1  1 6 +1  1 7 +1  1 8+1   1 5  1 6  = 4  −  +  −  +  −  +  −  +  −   = 4  −  +  −    2   2   2    2  2  2  2  

 1 7  1 8  1 9  + 4  −  +  −  +  −   = 4 (−0.03 + 0.012 − 0.008 + 0.004 − 0.002 ) = 4(−0.024 ) = −0.096  2   2   2   or we can use the geometric series formula S n = S5 =

d.

[

− 0.12 ⋅ 1 − (− 0.4 )5 1 − (− 0.4 )

4

∑ (− 2)

m −3

]

=

(

)

s1 1 − r n 0.012 where s1 = −0.12 , r = = −0.4 , and n = 5 . Therefore, 1− r − 0.03

−0.12 ⋅ (1 + 0.0102 ) 0.1212 = − = −0.09 1 + 0.4 1.4

= (− 2 )1−3 + (− 2 )2 −3 + (− 2 )3−3 + (− 2 )4 −3 = (− 2 )−2 + (− 2 )−1 + (− 2 )0 + (− 2 )1 =

m =1

=

1 1 − + 1 − 2 = 0.25 − 0.5 − 1 = −1.25 4 2

or we can use the geometric series formula S n = S4 =

e.

[

0.25 ⋅ 1 − (− 2 )4 1 − (− 2 )

10

∑ (− 3)

n−4

]

=

(

1

(− 2)

2

+

1 +1− 2 −2

)

s1 1 − r n −0.5 where s1 = 0.25 , r = = −2 , and n = 4 . Therefore, 1− r 0.25

3.75 0.25 ⋅ (1 − 16 ) = − = −1.25 3 1+ 2

= (− 3)5− 4 + (− 3)6 − 4 + (− 3)7 − 4 + (− 3)8− 4 + (− 3)9 − 4 + (− 3)10 − 4 = (− 3)1 + (− 3)2 + (− 3)3 + (− 3)4 + (− 3)5 + (− 3)6

n =5

= −3 + 9 − 27 + 81 − 243 + 729 = 546 or we can use the geometric series formula S n = S6 =

f.

[

− 3 ⋅ 1 − (− 3)6 1 − (− 3)

5

∑ (− 3)

k −1

]

= (− 3)1−1 + (− 3)2 −1 + (− 3)3−1 + (− 3)4 −1 + (− 3)5−1 = 1 + (− 3)1 + (− 3)2 + (− 3)3 + (− 3)4 = 1 − 3 + 9 − 27 + 81 = 61

or we can use the geometric series formula S n =

g.

5

∑4

[

1 ⋅ 1 − (− 3)5 1 − (− 3)

m

]

=

4

∑ j =1

)

= 41 + 42 + 43 + 44 + 45 = 4 + 16 + 64 + 256 + 1024 = 1364

or we can use the geometric series formula S n =

h.

(

s1 1 − r n −3 where s1 = 1 , r = = −3 , and n = 5 . Therefore, 1− r 1

244 1 + 243 = = 61 1+ 3 4

m =1

S5 =

)

−3 ⋅ (1 − 729 ) 2184 = = 546 1+ 3 4

=

k =1

S5 =

(

s1 1 − r n 9 where s1 = −3 , r = = −3 , and n = 6 . Therefore, 1− r −3

(

)

4 ⋅ 1 − 45 4 ⋅ (1 − 1024 ) 4092 = = = 1364 1− 4 −3 3

1 3j = 27 27

4

∑3 j =1

j

=

(

(

)

s1 1 − r n 16 where s1 = 4 , r = = 4 , and n = 5 . Therefore, 1− r 4

)

1 1 2 3 4 1 (3 + 9 + 27 + 81) = 120 = 4.44 3 +3 +3 +3 = 27 27 27

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391

Calculus I

Chapter 1 Solutions

or we can use the geometric series formula S n = S4 =

i.

(

)

(

)

s1 1 − r n 9 3 where, s1 = = 0.111 , r = = 3 , and n = 4 . Therefore, 1− r 27 3

8.88 0.111 ⋅ 1 − 34 0.111 ⋅ (1 − 81) = = = 4.44 1− 3 −2 2

6

1 6  2 k =3

∑

k +1

= 6

6

∑ 0.5

k +1

[

]

]

[

= 6 0.53+1 + 0.54 +1 + 0.55+1 + 0.56 +1 = 6 0.54 + 0.55 + 0.56 + 0.57 = 6 [0.063 + 0.031]

k =3

+6 [0.016 + 0.008] = 6 (0.118) = 0.708 ≈ 0.7

or we can use the geometric series formula S n =

(

)

(

)

s1 1 − r n 0.031 where, s1 = 6 ⋅ 0.54 = 0.375 , r = = 0.5 , and n = 4 . Thus, 1− r 0.063

0.375 ⋅ (1 − 0.063) 0.375 ⋅ 1 − 0.54 0.351 = = = 0.702 ≈ 0.7 1 − 0.5 0.5 0.5 7. Given the first term s1 and r , find S 8 for each of the following geometric sequences. S4 =

a. Given s1 = 3 and r = 3 use the geometric series formula S n = S8 =

(

)

3 ⋅ 1 − 38 19680 3 ⋅ (1 − 6561) = = = 9840 1− 3 2 −2

(

b. Given s1 = −8 and r = 0.5 use the geometric series formula S n = S8 =

(

)

− 8 ⋅ 1 − 0.58 7.968 −8 ⋅ (1 − 0.0039 ) = = − = −15.94 1 − 0.5 0.5 0. 5

c. Given s1 = 2 and r = −2.5 use the geometric series formula S n = S8 =

[

2 ⋅ 1 − (− 2.5)8

1 − (− 2.5) 8. Solve for x and y . a. Given

]

)

s1 1 − r n to find S8 , i.e., 1− r

(

)

(

)

s1 1 − r n to find S8 , i.e., 1− r

s1 1 − r n to find S8 , i.e., 1− r

2 ⋅ [1 − 1525.88] 2 ⋅ 1524.88 3049.76 = − = − = −871.36 1 + 2.5 3.5 3.5

=

7

∑ (ix + 2) = 30 , then (3x + 2) + (4 x + 2) + (5x + 2) + (6 x + 2) + (7 x + 2) = 30 ; (3x + 4 x + 5x + 6 x + 7 x) + 10 = 30 i =3

; 25 x + 10 = 30 ; 25 x = 30 − 10 ; x = b. Expanding Expanding

20 ; x = 0.8 25

4

∑ (ix + y ) = 20 we obtain (x + y ) + (2 x + y ) + (3x + y ) + (4 x + y ) = 20 ; 10 x + 4 y = 20 . i =1 6

∑ (ix + y )

= 10 we obtain (2 x + y ) + (3 x + y ) + (4 x + y ) + (5 x + y ) + (6 x + y ) = 10 ; 20 x + 5 y = 10 . Using

i =2

substitution method we obtain x = −2 and y = 10

Section 1.5 Solutions - Limits of Sequences and Series 1. State which of the following sequences are convergent. To see if a sequence is convergent or divergent consider the nth term of the sequence and let it approach infinity. a. lim n →∞

∞ n +1 n = ∞ ≈ lim n →∞ = 2 2 2

The sequence diverges

b. lim n →∞

n2 n2 − 1 n 2/ =1 = lim n →∞ = lim n →∞ n = ∞ ≈ lim n →∞ n n n/

The sequence diverges

c. lim n →∞ 2n +1 ≈ lim n →∞ 2n = 2∞ = ∞

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The sequence diverges

392

Calculus I

d. lim n →∞ e. lim n →∞

Chapter 1 Solutions 1

≈ lim n →∞

4n +1 n −1

≈ lim n →∞

n2

1 f. lim n →∞   5

n +1

1

1

=

4n n

4∞

=

= lim n →∞

n2

1 = 0 ∞

The sequence converges

1 1 n/ = lim n →∞ = = 0 ∞ n n 2/ =1

The sequence converges

n

1 ≈ lim n →∞   = lim n →∞ 0.2n = 0.2∞ = 0 5

The sequence converges

2. State which of the following geometric sequences are convergent.

1 1 1 1 1 , , , ,  , n ,  converges to 0 since, for large values of n , the absolute value of the difference 4 16 64 256 4

a. The sequence between

1

and 0 is very small.

4n

b. The sequence − 5, 25, − 125, 625, − 3125, , (− 5)n ,  diverges since, as n increases, the nth term increases without bound. c. The sequence 2, − 2, 2, − 2, , 2(− 1)n +1,  diverges since, as n increases, the nth term oscillates back and forth between

+2 and −2 . 1 1 1 1 , , , ,   2 4 8 2

d. The sequence 1, 1 between   2

n −1

n −1

,  converges to 0 since, for large values of n , the absolute value of the difference

and 0 is very small.

e. The sequence − 9, 27, − 81, 243, , (− 1)n 3n +1,  diverges since, as n increases, the nth term oscillates back and forth from a large positive number to a large negative number. 1 1 1 1 1 , , , , ,   3 9 27 81 3

f. The sequence 1,

n −1

,  converges to 0 since, for large values of n , the absolute value of the

n −1

1 difference between   and 0 is very small. 3 Again note that an easier way of knowing if a sequence is convergent or divergent is by considering the nth term and letting it approach to infinity as shown in practice problems 1.5-1 and 1.5-3.

3. State whether or not the following sequences converges or diverges as n → ∞ . If it does converge, find the limit. a. lim n →∞ b. lim n →∞ c. lim n →∞ d. lim n →∞

e. lim n →∞ f. lim n →∞

n2

≈ lim n →∞

3

n −4

5n + 1 2

n +1 25n

5n + 25 125

n

(n + 2)2 n2

2n 2n + 1

n

= lim n →∞

3

5n

≈ lim n →∞

≈ lim n →∞

5n +1

n2

52 n

≈ lim n →∞ ≈ lim n →∞

5n

3n

5

n2 n2

2n 2n

n

3/ =1

1 1 = = 0 n ∞

converges to 0 converges to 5

52 n ⋅ 5− n = lim n →∞ 52 n − n = lim n →∞ 5n = 5∞ = ∞ 1

= lim n →∞

= lim n →∞

= lim n →∞

= lim n →∞

5n 5n/ = lim n →∞ = lim n →∞ 5 = 5 n n/

= lim n →∞

= lim n →∞

5n

≈ lim n →∞

Hamilton Education Guides

n

2

n 2/

3n

5

n 2/ n 2/

2n/ 2n/

1 ⋅5

−n

= lim n →∞

= lim n →∞ 1 = 1

= lim n →∞ 1 = 1

1

3n − n

5

= lim n →∞

1 5

2n

=

1

5

2⋅∞

diverges =

1

1 = 0 ∞ 5 converges to 0 ∞

=

converges to 1 converges to 1

393

Calculus I

Chapter 1 Solutions

n 2 + 2n

g. lim n →∞

4

n +1

5

h. lim n →∞

≈ lim n →∞

2

n +1

n2

≈ lim n →∞

n

5 n

5

=

2

= lim n →∞

4

∞

2

n n

2

= lim n →∞

1 1 n/ = lim n →∞ = = 0 2/ n ∞ n

converges to 0

5 = 0 ∞

=

converges to 0

n n +1 n/ ≈ lim n →∞ = lim n →∞ = lim n →∞ 1 = 1 n −1 n n/

i. lim n →∞

n

j. lim n →∞

≈ lim n →∞

n3 − 1

k. lim n →∞

1 10 n

1 ∞ 10

=

m. lim n →∞ 100

n. lim n →∞ o. lim n →∞

p. lim n →∞

3 n 3

−

=

1 n

n + 100

n2 n2 n 2/ = lim n →∞ − 2 = lim n →∞ − 2/ = lim n →∞ − 1 = −1 −n⋅n n n

1

1

=

1 100 n

100

0

=

1 = 1 1

converges to 1

converges to 1

n

= lim n →∞

n3

1

1

1

100 n

100 n

100 ∞

≈ lim n →∞

n2 + 3

1

=

1 100 ∞

converges to −1

= 30 = 1

≈ lim n →∞

n3 − 10

converges to 0 converges to 1

≈ lim n →∞

= lim n →∞

3 ∞ 3

1 n/ 1 1 = lim n →∞ 2 = 2 = = 0 ∞ n3/ = 2 ∞ n

= lim n →∞

= 100 = 1

(n − 1)2 (1 − n )(1 + n )

l. lim n →∞

n

n3

converges to 1

=

n2

∞2

1 n/ 1 1 = lim n →∞ 2 = 2 = = 0 ∞ n3/ = 2 ∞ n =

converges to 0

1 1000 = = 0 ∞ ∞

converges to 0

1 1 1 − 1 = 0 − 1 = −1 − 1 ≈ lim n →∞ − 1 = n +1 n ∞ 1 1 1 r. lim n →∞ (0.25)− n = lim n →∞ = = = ∞ 0 0.25n 0.25∞

converges to −1

q. lim n →∞

n +1

s. lim n →∞

≈ lim n →∞

n +1 n

t. lim n →∞

n

= lim n →∞

n

n

+ 2 ≈ lim n →∞

∞

a. Given

1

∑ 3 8 

j

, then s1 = 3 and r =

j =0

∞

1 to obtain the sum, i.e., 3  8 j =0

∑

∞

b. Given

∑

1 . Since r 8

∞

j =0

∞

3 3  2 k =1

∑

k −1

1

∞

1 we can use the equation S∞ =

∞

∑s r = ∑s r 1

n

n =0

1

n −1

=

n =1

s1 1− r

3 3×8 24 3 3 1 = = = = = = 7 1 8 −1 7 7 1× 7 1− 8 8 8 8

j

∞

1 we can use the equation S∞ =

∑ n =0

s1r n =

∞

∑s r 1

n −1

n =1

=

s1 1− r

3 3× 4 3 3 3 3 12 = = = = = 1 = = 5 1 4 +1 5 5 1 × 5  1 1+ 1− −  4 4 4 4 4  

, then s1 = 3 and r =

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converges to 3

3

j

∑ 3 − 4 

converges to 1

n/ + 2 = lim n →∞ (1 + 2 ) = 1 + 2 = 3 n/

1  1 3 −  , then s1 = 3 and r = − . Since r 4 4  j =0

to obtain the sum, i.e.,

c. Given

j

n/ = lim n →∞ 1 = 1 n/

+ 2 = lim n →∞

n +1 n 4. Find the sum of the following geometric series.

diverges

3 . Since r 〉 1 the geometric series 2

∞

3 3  2 k =1

∑

k −1

has no finite sum.

394

Calculus I

Chapter 1 Solutions ∞

d. Given

∑ 100 k =1

Since r

∞

5 k +1

=

∑ 100 k =1

2

5 ⋅ 100

∞

k −1

=

∑ k =1 ∞

1 we can use the equation S∞ =

5  1    = 10000  100k −1 

∑

s1r n =

n =0

∞

∑ n =1

s1r n −1 =

∞

5

1

∑ 10000  100 

k −1

, then s1 =

k =1

5 1 and r = . 10000 100

∞

s1 to obtain the sum, i.e., 1− r

∑ k =1

5  1    10000  100 

k −1

5 5 5 5 × 100 5 1 500 10000 10000 10000 = = = = = = = 1 99 100 − 1 1980 10000 × 99 990000 9900 1− 100 100 100 ∞

e. Give

j

1 1   , then s1 = 1 and r = . Since r 3 3  j =0

∑

∞

to obtain the sum, i.e.,

j =0

∞

f. Given

1

∑  − 5 

j

1

∑  3 

j

∑

s1r n =

n =0

∞

∑s r 1

n −1

=

n =1

s1 1− r

1 1× 3 1 1 1 3 = = = = 1 = = 2 1 3 −1 2 1 × 2 2 1− 3 3 3 3

, then s1 = 1 and r = −

j =0

∞

1 we can use the equation S∞ =

1 . Since r 5

∞

1 we can use the equation S∞ =

∞

∑s r = ∑s r 1

n

n =0

1

n −1

n =1

=

s1 1− r

1 ∞ j 1 1 1 1× 5 1 5  1 = = = = 1 = = to obtain the sum, i.e., −  = 5 +1 6 1 6 5 × 1 6 6  1  1+ j =0 1− −  5 5 5 5  5 5. Find the sum of the following infinite geometric series.

∑

a. Given the series 5 − 1 +

1 1 1 1 1 − +  , s1 = 5 and r = − . Since r = − = = 0.2 1 we can use the equation 5 5 25 5 5

5 5×5 25 5 5 5 5 1 = = = = = = 6 6 5 +1 1 1 × 6 6  1 1+ 1− −  5 5 5 5  5 1 1 2 b. Given the series − + 2 − 8 + 32 +  s1 = − and r = 1 = −4 . Since r = − 4 = 4 is greater than one the geometric 2 2 −2 s 1 1 S∞ = 1 to obtain the sum, i.e., 5 − 1 + − + = 5 25 1− r

series −

1 + 2 − 8 + 32 +  has no finite solution. 2

1 1 1 1 1 1 1 c. Given the series 1 + + + +  s1 = 1 and r = 6 = . Since r = = = 0.17 1 we can use the equation 1 6 6 6 6 36 216

1 1× 6 s1 1 1 1 1 1 1 6 to obtain the sum, i.e., 1 + + = = = 1 = = S∞ = + + = 5 1 6 −1 5 1− r 1 × 5 5 6 36 216 1− 6 6 6 6 1 1 1 1 1 1 1 10 = . Since r = d. Given the series 1 + + +  s1 = 1 and r = + = = 0.1 1 we can use the equation 1 10 10 100 1000 10 10 1 s1 1 × 10 1 1 1 1 1 1 10 to obtain the sum, i.e., 1 + = = = 1 = = S∞ = + + + = 9 1 10 − 1 9 9 1 × 9 1− r 10 100 1000 1− 10 10 10 10 6. Write the following repeating decimals as the quotient of two positive integers. a. Given 0.666666  = 0.66 + 0.0066 + 0.000066 , which is a gemetric series, then s1 = 0.66 and r =

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0.0066 = 0.01 . Since 0.66

395

Calculus I

Chapter 1 Solutions

th ratio r is less than one, we can use the infinite geometric series equation s∞ = series 0.66 + 0.0066 + 0.000066 , i.e., s∞ =

s1 to obtain the sum of the infinite 1− r

s1 0.66 66 22 22 0.66 = = = = . Thus, 0.666666 = 99 33 33 1 − r 1 − 0.01 0.99

b. Given 3.027027027  , consider the decimal portion of the number 3.027027027  and write it in its equivalent form of

0.027027027  = 0.027 + 0.000027 + 0.000000027 . Since this is a geometric series, then s1 = 0.027 and r = = 0.001 . Since the ratio r is less than one, we can use the infinite geometric series equation s∞ = of the infinite series 0.027 + 0.000027 + 0.000000027 , i.e., s∞ = 3.027027027 = 3

3 111

s1 to obtain the sum 1− r

s1 0.027 3 27 0.027 = = = = . Thus, 1 − 0.001 999 111 0.999 1− r

c. 0.111111  = 0.11 + 0.0011 + 0.000011 , which is a gemetric series, then s1 = 0.11 and r = ratio r is less than one, we can use the infinite geometric series equation s∞ = 0.11 + 0.0011 + 0.000011 , i.e., s∞ =

0.000027 0.027

0.0011 = 0.01 . Since the 0.11

s1 to obtain the sum of the infinite series 1− r

s1 0.11 1 11 0.11 1 = = = = . Thus, 0.111111 = 9 9 99 1 − r 1 − 0.01 0.99

Section 1.6 Solutions - The Factorial Notation 1. Expand and simplify the following factorial expressions. a. 11 ! = 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 39,916,800 b. (10 − 3) ! = 7 ! = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 5040 c.

12 ! 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ! 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5/ ! 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 = = = = 3,991,680 1 5! 5! 5/ !

d.

14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ! 14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 1/ 0/ ! 14 ⋅ 13 ⋅ 12 ⋅ 11 14 ! = = = = 24,024 1 10 ! 1/ 0/ ! 10 !

e.

15 ⋅ 14 ⋅ 13 ⋅ 3 ⋅ 11 ⋅ 5 ⋅ 3 15 ! 15 ⋅ 14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ! 15 ⋅ 14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8/ ! 15 ⋅ 14 ⋅ 13 ⋅ 1/ 2/ ⋅ 11 ⋅ 1/ 0/ ⋅ 9/ = = = = = 1,351,350 8!4! 8!4! 8/ ! ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 1 4/ ⋅ 3/ ⋅ 2/ ⋅ 1

3

5 3

5 3

1/ 0/ ⋅ 9/ 10 ⋅ 9 5⋅3 10 ⋅ 9 ⋅ 8 ! 10 ⋅ 9 ⋅ 8/ ! 10 ! 10 ! 15 f. = = = = = = = / 4 ⋅ 3/ ⋅ 2 ⋅ 1 4 ⋅ 3 ⋅ 2 ⋅1 4 4 4 !8 ! 4 !8 ! 4 ! 8/ ! 4 ! (10 − 2 ) !

g.

6! 360 6⋅5⋅4⋅3 12 ! 6 ! 12 ! 6 ! 1/ 2/ ! 6 ! 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2/ ⋅ 1 = = = = = = 91 7 ⋅ 13 14 ! 14 ⋅ 13 ⋅ 12 ! 14 ⋅ 13 ⋅ 1/ 2/ ! 14 ⋅ 13 1/ 4/ ⋅ 13

h.

(7 − 3) ! 9 ! 12 ! (7 − 2 ) !

7

4!9! 4/ ! 9/ ! 1 4!9! 1 = = = = = 12 ! 5 ! 12 ⋅ 11 ⋅ 10 ⋅ 9 ! ⋅ 5 ⋅ 4 ! 12 ⋅ 11 ⋅ 10 ⋅ 9/ ! ⋅ 5 ⋅ 4/ ! 12 ⋅ 11 ⋅ 10 ⋅ 5 6600

2. Write the following products in factorial form. a. 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 7 ! 8! 3! 3. Expand the following factorial expressions. d. 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 =

b. 10 ⋅ 11 ⋅ 12 ⋅ 13 ⋅ 14 ⋅ 15 = e. 4 ⋅ 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 =

9! 3!

15 ! 9!

c. 22 ⋅ 23 ⋅ 24 ⋅ 25 = f. 35 =

25 ! 21 !

35 ! 34 !

a. 5(n !) = 5[n (n − 1) (n − 2 ) (n − 3 ) (n − 4 ) (n − 5 ) (n − 6 )  4 ⋅ 3 ⋅ 2 ⋅ 1]

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Chapter 1 Solutions

b.

(n − 7 ) ! = (n − 7 ) (n − 8) (n − 9) (n − 10) (n − 11) (n − 12)  4 ⋅ 3 ⋅ 2 ⋅ 1

c.

(n + 10) ! = (n + 10) (n + 9) (n + 8) (n + 7 ) (n + 6) (n + 5) (n + 4) (n + 3) (n + 2) (n + 1) n (n − 1) (n − 2)  4 ⋅ 3 ⋅ 2 ⋅ 1

d. (5n − 5) ! = (5n − 5 ) (5n − 6 ) (5n − 7 ) (5n − 8 ) (5n − 9 ) (5n − 10 )  4 ⋅ 3 ⋅ 2 ⋅ 1 e.

(2n − 8) ! = (2n − 8) (2n − 9) (2n − 10) (2n − 11) (2n − 12) (2n − 13)  4 ⋅ 3 ⋅ 2 ⋅ 1

f.

(2n + 6) ! = (2n + 6) (2n + 5) (2n + 4) (2n + 3) (2n + 2) (2n + 1) 2n (2n − 1) (2n − 2) (2n − 3)  4 ⋅ 3 ⋅ 2 ⋅ 1

g.

(2n − 5) !

= (2n − 5 ) (2n − 6 ) (2n − 7 ) (2n − 8 ) (2n − 9 ) (2n − 10 )  4 ⋅ 3 ⋅ 2 ⋅ 1

h. (3n + 3) ! = (3n + 3 ) (3n + 2 ) (3n + 1) 3n (3n − 1) (3n − 2 ) (3n − 3 )  4 ⋅ 3 ⋅ 2 ⋅ 1 4. Expand and simplify the following factorial expressions. a. b. c.

(n − 2) ! (n − 4) !

=

(n + 4) ! n!

(n + 5) ! (n − 2) !

= =

(n − 2) ! (n − 4) (n − 3) (n − 2) !

(n/ − 2/ ) ! (n − 4) (n − 3) (n/ − 2/ ) !

=

(n + 4) (n + 3) (n + 2) (n + 1) n ! n!

1

=

(n − 4) (n − 3)

(n + 4) (n + 3) (n + 2) (n + 1) n/ !

=

n/ !

(n + 5) (n + 4) (n + 3) (n + 2) (n + 1) (n ) (n − 1) (n − 2) ! (n − 2) !

=

= (n + 4 ) (n + 3 ) (n + 2 ) (n + 1)

(n + 5) (n + 4) (n + 3) (n + 2) (n + 1) (n ) (n − 1) (n/ − 2/ ) ! (n/ − 2/ ) !

= (n + 5 ) (n + 4 ) (n + 3 ) (n + 2 ) (n + 1) (n ) (n − 1) d.

(n − 1) (n + 1) ! (n + 2) !

e.

(3n ) ! (3n − 2) ! (3n + 1) ! (3n − 4) !

f.

(n − 1) ! (n + 2) ! (n !)2

g.

(2n − 3) ! 2(n !) (2n ) ! (n − 2) ! =

(n − 1) (n + 1) ! (n + 2) (n + 1) !

=

=

=

=

=

n −1 n+2

(3n ) ! (3n − 2) ! (3n + 1) (3n ) ! (3n − 4) (3n − 3) (3n − 2) !

(n − 1) ! (n + 2) ! (n !) (n !) =

(n − 1) (n/ + 1/ ) ! (n + 2) (n/ + 1/ ) !

=

(n − 1) ! (n + 2) ! (n !) (n ) (n − 1) !

(2n − 3) ! 2 [ (n ) (n − 1) (n − 2) ! ] (2n ) (2n − 1) (2n − 2) (2n − 3) ! (n − 2) !

(2/ n/ ) (n − 1) (2/ n/ ) (2n − 1) (2n − 2)

=

=

=

=

(3/ n/ ) ! (3/ n/ − 2/ ) ! (3n + 1) (3/ n/ ) ! (3n − 4) (3n − 3) (3/ n/ − 2/ ) ! (n/ − 1/ ) ! (n + 2) ! (n !) (n ) (n/ − 1/ ) !

=

=

1

(3n + 1) (3n − 4) (3n − 3)

1

(n + 2) ! (n !) (n)

(2/ n/ − 3/ ) ! 2 [ (n ) (n − 1) (n/ − 2/ ) ! ] [ (2n ) (2n − 1) (2n − 2) (2/ n/ − 3/ ) ! ] (n/ − 2/ ) !

=

2(n ) (n − 1) (2n ) (2n − 1) (2n − 2)

n −1

(2n − 1) (2n − 2)

5. Write the following expressions in factorial notation form. Simplify the answer. 2

 5 5! 10 5⋅2 5 ⋅ 4/ 5 ⋅ 4 ⋅ 3/ ! 5! a.   = = = = = = = 10 3 1 1 1 ⋅ 2/ 3!2! 3 ! (5 − 3) ! 3/ ! 2 !   5 3 2

10  5⋅3⋅ 2⋅7 210 1/ 0/ ⋅ 9/ ⋅ 8/ ⋅ 7 10 ! 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6/ ! 10 ! = = = = = = 210 b.   = 1 1 4/ ⋅ 3/ ⋅ 2/ ⋅ 1 6 ! (10 − 6 ) ! 6/ ! 4 ! 6!4! 6  c.   =

 8  0

1 8! 8/ ! = = = 1 0 ! (8 − 0 ) ! 0 ! 8/ ! 1 ⋅1

 8  8

1 8! 8/ ! = = = 1 1 ⋅1 8 ! (8 − 8) ! 8/ ! 0 !

d.   =

2

2

 6 2⋅5⋅2 20 6/ ⋅ 5 ⋅ 4/ 6 ⋅ 5 ⋅ 4 ⋅ 3/ ! 6! 6! e.   = = = = = = = 20 / 1 1 1 ⋅ 2 ⋅ 3 / ( ) / 3 ! 3 ! 3 ! 6 − 3 ! 3 ! 3 !  3

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 5 1 

f.   =

Chapter 1 Solutions

5 ⋅ 4/ ! 5! 5! 5 = = = = 5 1 ! (5 − 1) ! 1! 4 ! 1 ! 4/ ! 1

n   = g.   n − 5 =

n!

(n − 5) ! [n − (n − 5) ] !

=

n ⋅ (n − 1) ⋅ (n − 2 ) ⋅ (n − 3) ⋅ (n − 4 ) ⋅ (n/ − 5/ ) ! n! n! = = (n/ − 5/ ) ! 5 ! (n − 5) ! (n/ − n/ + 5) ! (n − 5) ! 5 !

n ⋅ (n − 1) ⋅ (n − 2 ) ⋅ (n − 3) ⋅ (n − 4 ) n ⋅ (n − 1) ⋅ (n − 2 ) ⋅ (n − 3) ⋅ (n − 4 ) n ⋅ (n − 1) ⋅ (n − 2 ) ⋅ (n − 3 ) ⋅ (n − 4 ) = = 5! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 120

 2n   =  2n − 1

(2n − 1) ! [2n − (2n − 1) ] !

 3n   =  3n − 3

(3n − 3) ! [3n − (3n − 3) ] !

h. 

i. 

=

2n !

3n !

=

=

2n ! 2n ! 2n ⋅ (2/ n/ − 1/ ) ! = = = 2n (2n − 1) ! (2/ n/ − 2/ n/ + 1) ! (2n − 1) ! 1 ! (2/ n/ − 1/ ) ! 3n !

(3n − 3) ! (3/ n/ − 3/ n/ + 3) !

=

3n !

(3n − 3) ! 3 !

n ⋅ (3n − 1) ⋅ (3n − 2 ) 3/ n ⋅ (3n − 1) ⋅ (3n − 2 ) = 1 ⋅ 2 ⋅ 3/ 2

n   = j.   n − 6

n!

(n − 6) ! [n − (n − 6) ] !

=

n!

(n − 6) ! (n/ − n/ + 6) !

=

n!

(n − 6) ! 6 !

=

=

3n ⋅ (3n − 1) ⋅ (3n − 2 ) ⋅ (3/ n/ − 3/ ) ! (3/ n/ − 3/ ) ! ⋅ 1 ⋅ 2 ⋅ 3

n ⋅ (n − 1) ⋅ (n − 2 ) ⋅ (n − 3) ⋅ (n − 4 ) ⋅ (n − 5) ⋅ (n/ − 6/ ) ! (n/ − 6/ ) ! 6 !

n ⋅ (n − 1) ⋅ (n − 2 ) ⋅ (n − 3 ) ⋅ (n − 4 ) ⋅ (n − 5 ) n ⋅ (n − 1) ⋅ (n − 2 ) ⋅ (n − 3) ⋅ (n − 4 ) ⋅ (n − 5) = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 720 6. Expand the following binomial expressions. =

a.

( x − 2 )4

 4  4  4  4  4  4  4  4  4 =   x 4 +   x 4 −1 ⋅ (− 2 ) +   x 4 − 2 (− 2 )2 +   x 4 −3 (− 2 )3 +   x 4 − 4 (− 2 )4 =   x 4 − 2  x3 + 4  x 2 − 8  x1 0 1 2 2 0 1 3 4                 3

 4 4! 4 2⋅4! 3 4⋅4! 2 4! 2⋅4! 3 4⋅4! 8⋅4 ! 16 ⋅ 4 ! = + 16  x 0 = x − x + x x4 − x + x2 − x+ 4 ( ) ( ) ( ) ( ) ( ) − − − − − 0 !4! 3! 2!2! 0 ! 4 0 ! 1 ! 4 1 ! 2 ! 4 2 ! 3 ! 4 3 ! 4 ! 4 4 !   −

8⋅4 ! 16 ⋅ 4 ! 4/ ! 4 2 ⋅ 4 ⋅ 3/ ! 3 4 ⋅ 4 ⋅ 3 ⋅ 2/ ! 2 8 ⋅ 4 ⋅ 3/ ! 16 ⋅ 4/ ! = = x 4 − 8 x 3 + 24 x 2 − 32 x + 16 x+ x − x + x − x+ 3 !1 ! 4!0! 0 ! 4/ ! 3/ ! 1 ⋅ 2 ⋅ 2/ ! 3/ ! 1 ! 4/ ! 0 !

7 7 7 7 7 7 7 7 b. (u + 2 )7 =  u 7 +  u 7 −1 ⋅ 2 +  u 7 − 2 ⋅ 22 +  u 7 −3 ⋅ 23 +  u 7 − 4 ⋅ 24 +  u 7 −5 ⋅ 25 +  u 7 − 6 ⋅ 26 +  u 7 − 7 ⋅ 27 6 5 2 3 4 0 1 7               7 7 7 7 7 7 7 7 2⋅7 ! 6 4⋅7 ! 7! =  u 7 + 2 u 6 + 4 u 5 + 8 u 4 + 16 u 3 + 32 u 2 + 64 u + 128 u 0 = u7 + u + u5 0 1 2 3 4 5 6 7 ( ) ( ) ( ) − 0 ! 1 ! 7 1 ! 0 ! 7 − 2 ! 7 − 2 !                 +

8⋅7 ! 16 ⋅ 7 ! 3 32 ⋅ 7 ! 2 64 ⋅ 7 ! 128 ⋅ 7 ! 7 ! 7 2⋅7 ! 6 4⋅7 ! 5 8⋅7 ! 4 = u4 + u + u + u+ u + u + u + u 3 ! (7 − 3) ! 4 ! (7 − 4 ) ! 5 ! (7 − 5) ! 6 ! (7 − 6 ) ! 7 ! (7 − 7 ) ! 0! 7! 1! 6 ! 2! 5! 3! 4!

+

128 ⋅ 7 ! 16 ⋅ 7 ! 3 32 ⋅ 7 ! 2 64 ⋅ 7 ! = u 7 + (2 ⋅ 7 ) u 6 + (7 ⋅ 12 ) u 5 + (8 ⋅ 35) u 4 + (16 ⋅ 35) u 3 + (32 ⋅ 21) u 2 + (64 ⋅ 7 )u + 128 u + u + u+ 5! 2! 6 ! 1! 7! 0! 4! 3!

= u7 + 14u6 + 84u5 + 280u4 + 560u 3 + 672u 2 + 448u + 128 c.

( y − 3)5

5  5 5  5 5  5 5  5 =   y 5 +   y 5−1 ⋅ (− 3) +   y 5− 2 (− 3)2 +   y 5−3 (− 3)3 +   y 5− 4 (− 3)4 +   y 5−5 (− 3)5 =   y 5 − 3  y 4 0 1 2 3 4 5 0 1               

5  5 5  5 5! 3⋅5 ! 9⋅5 ! 27 ⋅ 5 ! 2 81 ⋅ 5 ! + 9  y 3 − 27  y 2 + 81  y1 − 243  y 0 = y5 − y4 + y3 − y + y 0 ! (5 − 0 ) ! 1 ! (5 − 1) ! 2 ! (5 − 2 ) ! 3 ! (5 − 3) ! 4 ! (5 − 4 ) !  2  3  4  5 −

5 ! 5 3 ⋅ 5 ! 4 9 ⋅ 5 ! 3 27 ⋅ 5 ! 2 81 ⋅ 5 ! 243 ⋅ 5 ! 243 ⋅ 5 ! = = y 5 − (3 ⋅ 5) y 4 + (9 ⋅ 10 ) y 3 − [27 ⋅ 10]y 2 y − y + y − y + y− 0! 5! 1! 4 ! 2!3! 3! 2! 4 ! 1! 5! 0! 5 ! (5 − 5) !

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Chapter 1 Solutions

+ (81 ⋅ 5) y − 243 = y 5 − 15 y 4 + 90 y 3 − 270 y 2 + 405 y − 243

7. Use the general equation for binomial expansion to solve the following exponential numbers to the nearest hundredth. a. (0.95)5 = (1 − 0.05)5 therefore, a = 1 , b = −0.05 , n = 5 . Using the general binomial expansion formula

(a + b )n

n n n  n  n − r +1 r −1 n a =  a n +  a n −1b +  a n − 2b 2 +  +  b +  +  b n we obtain the following: 0 1  2  r − 1 n

 5 5 5  5  5 5 =   ⋅ 15 +   ⋅ 14 ⋅ (− 0.05) +   ⋅ 13 ⋅ (− 0.05)2 +   ⋅12 ⋅ (− 0.05)3 +   ⋅12 ⋅ (− 0.05)4 +   ⋅ (− 0.05)5  0 1   2  3  5  4 5  5 5 5! 5! 5! 5! 5! 5! =   − 0.05  + 0.0025  −  = − 0.05 + 0.0025 − = − − 0.05 + 0.0025 1 0 2 ( ) ( ) ) ( 0 ! 5 0 ! 1 ! 5 1 ! 2 ! 5 2 ! − − − 2 ! 3! 4 ! 5 !      

(1 − 0.05)5

2

5/ ! 5 ⋅ 4/ ! 5 ⋅ 4/ ⋅ 3/ ! = − 0.05 + 0.0025 −  = 1− 0.25+ 0.025 −  ≈ 0.775 5/ ! 4/ ! 2/ ⋅ 1 ⋅ 3/ !

Therefore, (0.95)5 to the nearest hundredth is equal to 0.78 . (Note that this is an estimate.) b.

(2.25)7

= (2 + 0.25)7 therefore, a = 2 , b = 0.25 , n = 7 . Using the general binomial expansion formula

(a + b )n

n n n  n  n − r +1 r −1 n a =  a n +  a n −1b +  a n − 2b 2 +  +  b +  +  b n we obtain the following: 0 1  2  r − 1 n

(2 + 0.25)7

7 7 7 7 7 7 7 7 =   27 +  26 ⋅ 0.25 +  25 ⋅ 0.252 +  24 ⋅ 0.253 +  23 ⋅ 0.254 +   22 ⋅ 0.255 +  2 ⋅ 0.256 +  ⋅ 0.257 0 1 2 3 4 6 5               7

7 7 7 7 7 0.25 ⋅ 7! 0.0313 ⋅ 7! 128 ⋅ 7! 16 ⋅ 7! 2 ⋅ 7! = 128 + 16  + 2 + 0.25 + 0.0313  +  = + + + + + 0 1 2 3 4 ( ) ( ) ( ) − − − 7 2 ! 3 ! (7 − 3) ! 4! (7 − 4 ) ! 0 ! 7 0 ! 1 ! 7 1 ! 2 !           =

128 ⋅ 7! 16 ⋅ 7! 2 ⋅ 7! 0.25 ⋅ 7! 0.0313 ⋅ 7! 128 ⋅ 7/ ! 16 ⋅ 7 ⋅ 6/ ! 2/ ⋅ 7 ⋅ 6 ⋅ 5/ ! 0.25 ⋅ 7 ⋅ 6/ ⋅ 5 ⋅ 4/ ! 0.0313 ⋅ 7 ⋅ 6/ ⋅ 5 ⋅ 4/ ! + + + = + + + + + + + 4! 3! 3! 4! 2! 5! 6! 7! 7/ ! 6/ ! 2/ ⋅ 1 ⋅ 5/ ! 3/ ⋅ 2/ ⋅ 1 ⋅ 4/ ! 4/ !⋅ 3/ ⋅ 2/ ⋅ 1

= 128+112+ 42 + 8.75 + 1.095 +  ≈ 291.845 Therefore, (2.25)7 to the nearest hundredth is equal to 291.85 . (Note that this is an estimate.) c. (1.05)4 = (1 + 0.05)4 therefore, a = 1 , b = 0.05 , n = 4 . Using the general binomial expansion formula

(a + b )n

n n n  n  n − r +1 r −1 n  a =   a n +   a n −1b +   a n − 2b 2 +  +  b +  +   b n we obtain the following: 0 1  2  r − 1 n

(1 + 0.05)4 =

 4  4  4  4  4  4  4  4  4 =   ⋅ 14 +   ⋅13 ⋅ 0.05 +   ⋅12 ⋅ 0.052 +   ⋅11 ⋅ 0.053 +   ⋅ 0.054 =  + 0.05  + 0.0025 + 0.000125 +  0 1 2 0 1 2 3 4                 3

4! 0.05 ⋅ 4! 0.0025 ⋅ 4! 0.000125 ⋅ 4! 4! 0.05 ⋅ 4! 0.0025 ⋅ 4! 0.000125 ⋅ 4! + + + + = + + + + 3 ! 1! 2 ! 2! 3! 4! 0 ! (4 − 0) ! 1! (4 − 1) ! 2 ! (4 − 2 ) ! 3 ! (4 − 3) ! 2

4/ ! 0.05 ⋅ 4 ⋅ 3/ ! 0.0025 ⋅ 4/ ⋅ 3 ⋅ 2/ ! 0.000125 ⋅ 4 ⋅ 3/ ! = + + + +  = 1+ 0.2+ 0.015 + 0.0005 +  = 1.2155 4/ ! 3/ ! 2/ ⋅ 1 ⋅ 2/ ! 3/ ! Therefore, (1.05)4 to the nearest hundredth is equal to 1.22 . 8. Find the stated term of the following binomial expressions. a. To find the eighth term of (x + 3)12 first identify the a, b, r , and n terms, i.e., a = x , b = 3 , r = 8 , and n = 12 .

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Chapter 1 Solutions

 n  n − r +1 r −1 n! a Then, use the equation  = b a n − r +1b r −1 (r − 1) ! (n − r + 1) !  r − 1 12  5 7 1/ 2/ ⋅11 ⋅ 1/ 0/ ⋅ 9 ⋅ 8 ⋅ 7/ ! 12 ! 5 12!   x ⋅ 3 = = (2187 ⋅ 792 )x5 = 1,732,104 x 5 x5 ⋅ 37 = x ⋅ 2187 = 2187 x 5 ⋅ 7 ( ) 7 ! 5 ! 7/ ! ⋅ 5/ ⋅ 4/ ⋅ 3/ ⋅ 2/ ⋅1 − 7 ! 12 7 !   b. To find the ninth term of (x − y )10 first identify the a, b, r , and n terms, i.e., a = x , b = − y , r = 9 , and n = 10 .

 n  n − r +1 r −1 n! a Then, use the equation  = b a n − r +1b r −1 (r − 1) ! (n − r + 1) !  r − 1 10  2 10 ! 2 8 10 ⋅ 9 ⋅ 8/ ! 2 8 90 2 8 10!   x ⋅ (− y )8 = x y = 45 x 2 y 8 x 2 ⋅ y8 = x y = x y = 8! 2 ! 2 8/ ! ⋅ 2 ⋅1 8 ! (10 − 8) ! 8  c. To find the seventh term of (u − 2a )11 first identify the a, b, r , and n terms, i.e., a = u , b = −2a , r = 7 , and n = 11 .

 n  n − r +1 r −1 n! a Then, use the equation  = b a n − r +1b r −1 (r − 1) ! (n − r + 1) !  r − 1 3 2

11 11 ⋅ 1/ 0/ ⋅ 9/ ⋅ 8/ ⋅ 7 ⋅ 6/ ! 11! 11!  u 5 ⋅ (− 2a )6 = = (64 ⋅ 462 )a 6u 5 = 29,568a 6 u5 64a 6 u 5 = 64a 6 u 5 64a 6u 5 = 6! 5! 6/ ! 5/ ⋅ 4/ ⋅ 3/ ⋅ 2/ ⋅1 6 ! (11 − 6 ) ! 6  d. To find the twelfth term of (x − 1)18 first identify the a, b, r , and n terms, i.e., a = x , b = −1 , r = 12 , and n = 18 .

 n  n − r +1 r −1 n! Then, use the equation  = b a n − r +1b r −1 a (r − 1) ! (n − r + 1) !  r − 1 2

3

2

18  7 / 8/ ⋅ 17 ⋅ 1/ 6/ ⋅ 1/ 5/ ⋅ 1/ 4/ ⋅ 13 ⋅ 1/ 2/ ⋅ 1/ 1/ ! / ⋅ 2/ ⋅ 13 18! 11 7 17 ⋅ 2 ⋅ 3 7 18 ! 7 71 = − (17 ⋅ 13 ⋅ 2 )x 7 11  x ⋅ (− 1) = − x 11 ! (18 − 11) ! = − x 11!9 ! = − x 1/ 1/ !⋅ 9/ ⋅ 8/ ⋅ 7/ ⋅ 6 ⋅ 5/ ⋅ 4/ ⋅ 3/ ⋅ 2/ ⋅ 1 = − x / 6   = − 442x 7

Hamilton Education Guides

400

Chapter 2 Solutions: Section 2.1 Practice Problems – The Difference Quotient Method 1. Find the derivative of the following functions by using the difference quotient method.

[(x + h) − 1]− (x − 1) = x/ 2

f (x + h ) − f (x ) = h

a. Given f (x ) = x 2 − 1 , then

= h + 2 x . Therefore, f ′(x ) = lim h→0

h/ (h + 2 x ) + h 2 + 2hx − 1/ − x/ 2 + 1/ h 2 + 2hx = = h/ h h

2

h

f (x + h ) − f (x ) h

f (x + h ) − f (x ) = h

b. Given f (x ) = x 3 + 2 x − 1 , then

2

= lim h→0 (h + 2 x ) = 0 + 2 x = 2 x

[(x + h) + 2(x + h) − 1]− [x 3

3

]

+ 2x − 1

h

(

)

h3 + 3 x 2 h + 3 xh 2 + 2h h/ h 2 + 3 x 2 + 3 xh + 2 x/ 3 + h3 + 3 x 2 h + 3 xh 2 + 2/ x/ + 2h − 1/ − x/ 3 − 2/ x/ + 1/ = = = h 2 + 3 x 2 + 3 xh + 2 . h h h/ f (x + h ) − f (x ) Therefore, f ′(x ) = lim h→0 = lim h →0 h 2 + 3 x 2 + 3 xh + 2 = 02 + 3 ⋅ x 2 + 3 ⋅ 0 ⋅ h + 2 = 3 x 2 + 2 h

=

(

)

(x + h )(x − 1) − x(x + h − 1) (x + h − 1)(x − 1)

x+h x − f (x + h ) − f (x ) x x + h − 1 x −1 = , then = c. Given f (x ) = h x −1 h x/ 2 − x/ + h/ x/ − h − x/ 2 − h/ x/ + x/

=

(

h x − x + hx − h − x + 1

d. Given f (x ) = −

=

1

(

2

x

2

2

hx x + h + 2hx

f ′(x ) = lim h →0

)

=

2

(

h/ (h + 2 x ) 2

2

h/ x x + h + 2hx

)

1

x 2 − 2 x + hx − h + 1

(x + h )(x − 1) − x(x + h − 1) h [ (x + h − 1)(x − 1)]

. Therefore,

−

1

(x + h )

2

+

− x 2 + (x + h )2

1 x

2

h =

)

4

x 2 (x + h )2 h

=

h + 2x

x + h 2 x 2 + 2hx3

− x 2 + (x + h )2

=

=

hx (x + h ) 2

2

(

− x 2 + x 2 + h 2 + 2hx 2

(

2

2

hx x + h + 2hx

)

)

. Therefore,

f (x + h ) − f (x ) 2 2 x/ 0 + 2x h + 2x   = lim h →0  4 = 4/ =3 =  = 4 2 2 3 2 2 3 h x + 0 ⋅ x + 2⋅0⋅ x x x3  x + h x + 2hx 

e. Given f (x ) = 20 x 2 − 3 , then

[

][

] = [20(x

20(x + h )2 − 3 − 20 x 2 − 3 f (x + h ) − f (x ) = h h

2

) ][

]

+ 2hx + h 2 − 3 − 20 x 2 − 3 h

40hx + 20h 2 2/ 0/ x + 40hx + 20h 2 − 3/ − 2/ 0/ x + 3/ h/ (40 x + 20h ) = = = 40 x + 20h h h h/ 2/

2/

Therefore, f ′(x ) = lim h→0 f. Given f (x ) =

=

(

= −

h/ x − 2 x + hx − h + 1

f (x + h ) − f (x ) = h

, then

− x/ 2 + x/ 2 + h 2 + 2hx 2

h/ 2

=

f (x + h ) − f (x ) −1 −1 1 −1   = lim h →0  2 = 2 = −  = 2 h x − 2x + 1 x − 2x + 0 ⋅ x − 0 + 1 ( x − 1)2  x − 2 x + hx − h + 1 

f ′(x ) = lim h→0

=

= −

)

2

h

x3 , then

f (x + h ) − f (x ) = lim h →0 (40 x + 20h ) = 40 x + (20 ⋅ 0 ) = 40 x + 0 = 40 x h

f (x + h ) − f (x ) = h

x3 + h3 + 3 x 2 h + 3 xh 2 − x3 h x3 + h3 + 3 x 2 h + 3 xh 2 + x3

Hamilton Education Guides

=

(x + h )3 − h

(

x3

=

h/ h 2 + 3 x 2 + 3 xh

(x + h )3 −

)

h/ x3 + h3 + 3 x 2 h + 3 xh 2 + x3

h

=

x3

⋅

(x + h )3 + (x + h )3 +

x3 x3

=

h

h 2 + 3 x 2 + 3 xh x3 + h3 + 3 x 2 h + 3 xh 2 + x3

(x + h )3 − x3 (x + h )3 + x3 . Therefore,

401

Calculus I

Chapter 2 Solutions

 f (x + h ) − f (x ) = lim h →0  h  

f ′(x ) = lim h→0

3x 2

=

x3 + x3

= 10 ⋅

2 x3

10

g. Given f (x ) =

= 10 ⋅

3x 2

=

(

3

x−5 − x+h−5

f (x + h ) − f (x ) h

10 x+h−5 h

=

x−5 + x+h−5

⋅

h⋅ x + h −5 ⋅ x −5

(h/ ⋅

3

3 2 − 32 3 2− 2 3 x ⋅x = = x x 2 2 2

=

2x 2

, then

x−5

= 10 ⋅

x−5 + x+h−5

− h/

)(

x+h−5 ⋅ x−5 ⋅

x−5 + x+h−5

  f (x + h ) − f (x ) = lim h →0  h  

f ′(x ) = lim h→0

=

3x 2

=

  =  x3 + h3 + 3 x 2 h + 3 xh 2 + x3  h 2 + 3 x 2 + 3 xh

−10

)(

x−5 ⋅ x−5 ⋅

x−5 + x−5

)

=

(

=

)

(h ⋅

4 −3 2

x 3 + 03 + 3 x 2 ⋅ 0 + 3 x ⋅ 0 2 + x 3

1

3 3 2 x x = 2 2

=

10

−

02 + 3x 2 + 3x ⋅ 0

10 x − 5 − 10 x + h − 5

x−5 =

x+h−5 ⋅ x−5 h

x/ − 5/ − x/ − h + 5/

)(

x+h−5 ⋅ x−5 ⋅

x−5 + x+h−5

−10

(

)(

x+h−5 ⋅ x−5 ⋅

x−5 + x+h−5

   = x−5 + x+h−5  

− 10

)(

x+h−5 ⋅ x−5 ⋅

5

−10

(x − 5) ⋅ (2

)

x−5

)

=

− 1/ 0/ 2/ (x − 5)

(

x−5

)

=

h⋅ x + h −5 ⋅ x −5

)

. Therefore,

)

−10

)(

(

x+0−5 ⋅ x−5 ⋅

−5

=

1+ 1

(x − 5)

10 x − 5 − 10 x + h − 5

=

2

−5

( x − 5)

3 2

x−5 + x+0−5 5

= −

( x − 5)3

 a (x + h ) + b   ax + b  cx(ax + ah + b ) − (cx + ch ) (ax + b )  −  ( ) + c x h cx f (x + h ) − f (x ) ax + b cx(cx + ch )    , then =  = h. Given f (x ) = cx h h h

=

a/ c/ x/ 2 + a/ c/ h/ x/ + b/ c/ x/ − a/ c/ x/ 2 − b/ c/ x/ − a/ c/ h/ x/ − bch −bc/ h/ −b acx 2 + achx + bcx − acx 2 − bcx − achx − bch = = = 2 . c/ h/ x(cx + ch ) chx(cx + ch ) chx(cx + ch ) cx + chx

Therefore, f ′(x ) = lim h→0

f (x + h ) − f (x ) b −b b   = lim h →0  2 = − 2  = − 2 h cx + c ⋅ 0 ⋅ x cx  cx + chx 

2. Compute f ′(x ) for the specified values by using the difference quotient equation as the lim h→0 . a. Given f (x ) = x3 , then using f ′(x ) = lim h →0 f ′(1) = lim h →0

(1 + h )3 − 13 = h

lim h →0

(x + h )3 − x3 at x = 1 f (x + h ) − f (x ) we obtain f ′(x ) = lim h →0 h h

(

b. Given f (x ) = 1+ 2 x , then using f ′(x ) = lim h →0 f ′(0 ) = lim h →0

f ′(−1) = lim h →0

(

f (x + h ) − f (x ) 1 + 2(x + h ) − (1 + 2 x ) we obtain f ′(x ) = lim h →0 at x = 0 h h

1 + 2(0 + h ) − (1 + 2 ⋅ 0 ) 1/ + 2h − 1/ 2h/ = lim h →0 = lim h →0 = lim h →0 2 = 2 h h/ h

c. Given f (x ) = x 3 + 1 , then using f ′(x ) = lim h →0

= lim h →0

)

1/ + h3 + 3h + 3h 2 − 1/ h/ h 2 + 3 + 3h = lim h →0 = lim h →0 h 2 + 3 + 3h = 02 + 3 + 3 ⋅ 0 = 3 h h/

[(− 1 + h) + 1]− [(− 1) + 1] 3

)

3

h

(

[

](

)

(x + h )3 + 1 − x3 + 1 f (x + h ) − f (x ) we obtain f ′(x ) = lim h →0 h h

= lim h →0

(− 1/ + h

3

)

(

at x = −1

)

+ 3h + 3h 2 + 1/ − (− 1/ + 1/ ) h3 + 3h + 3h 2 − 0 = lim h →0 h h

)

h/ h 2 + 3 + 3h = lim h →0 h 2 + 3 + 3h = 02 + 3 + 3 ⋅ 0 = 3 h/

Hamilton Education Guides

402

)

Calculus I

Chapter 2 Solutions

d. Given f (x ) = x 2 (x + 2 ) = x 3 + 2 x 2 , then using f ′(x ) = lim h →0

[(x + h) + 2(x + h) ]− (x 3

f ′(x ) = lim h →0

3

+ 2x2

)

at x = 2 f ′(2) = lim h →0

h

[8 + h

= limh →0

2

3

]

+ 12h + 6h 2 + 8 + 2h 2 + 8h − 16

(

= lim h →0

h

)

f (x + h ) − f (x ) we obtain h

[(2 + h) + 2(2 + h) ]− (2 3

2

3

+ 2 ⋅ 22

)

h

(

1/ 6/ + h3 + 20h + 8h 2 − 1/ 6/ h/ h 2 + 8h + 20 = lim h →0 h h/

)

= lim h →0 h 2 + 8h + 20 = 02 + 8 ⋅ 0 + 20 = 20 1

e. Given f (x ) = x −2 + x −1 + 1 =

x

2

1 + 1 + h + (1 + h )2

(1 + h )

2

f ′(1) = lim h →0

=

2

3/ + h + 3h − 3/ − 3h − 6h

(

2

h 1 + h + 2h

(−2 ⋅ 0) − 3 =

( 10 + h )−

1 10 + 10

1 + 1 + 12 2

1

10

h

=

1 + x + h + (x + h )2

= lim h →0

2

− 2h − 3h

(

2

h 1 + h + 2h

)

1 + x + x2 x2

at x = 1

h

1 + h 2 + 2h h

= lim h →0

−

(x + h )2

2 + h + 1 + h 2 + 2h

x + 2 , then using f ′(x ) = lim h →0

at x = 10 f ′(10) = lim h →0

=

)

)

= lim h →0

−3

= lim h →0

h/ (−2h − 3)

(

2

h/ 1 + h + 2h

)

1 2 10

=

h

( 10 + h )+ 10 ( 10 + h )+ 10

⋅

= lim h →0

(

(

2

h 1 + h + 2h

= lim h →0

f (x + h ) − f (x ) we obtain f ′(x ) = lim h →0 h

( 10 + h + 2)− ( 10 + 2) = lim

(

3 + h 2 + 3h − 3 1 + h 2 + 2h

−3 = −3 1

1 + 02 + 2 ⋅ 0

= lim h →0

−

h

2

f. Given f (x ) =

(

1 x + x2 x + x 2 + x3 1 + x + x2 x/ 1 + x + x 2 = = = , then using +1 = + 1 x x3 x3 x3/ = 2 x2

f (x + h ) − f (x ) we obtain f ′(x ) = lim h →0 h

f ′(x ) = lim h →0

= lim h →0

+

(

)

)

−2h − 3

1 + h 2 + 2h

)(

x+h +2 − h

x +2

)

10 + h + 2/ − 10 + 2/ 10 + h − 10 = lim h →0 h h

h →0

1/ 0/ + h − 1/ 0/

h ⋅ 10 + h + 10

= lim h →0

)

(

h/

h/ ⋅ 10 + h + 10

)

=

1 10 + 0 + 10

1 1 = = 0.158 6.32 2 ⋅ 3.16

Section 2.2 Solutions - Differentiation Rules Using the Prime Notation 1. Find the derivative of the following functions. Compare your answers with the practice problem number one in Section 2.1. a. Given f (x ) = x 2 − 1 , then f ′(x ) = 2 x 2 −1 − 0 = 2 x b. Given f (x ) = x 3 + 2 x − 1 , then f ′(x ) = 3x 3−1 + 2 ⋅1x1−1 − 0 = 3x 2 + 2 x 0 = 3 x 2 + 2 c. Given f (x ) =

[1 ⋅ (x − 1)] − [1 ⋅ x] = x/ − 1 − x/ = − 1 x , then f ′(x ) = x −1 (x − 1)2 (x − 1)2 ( x − 1)2

d. Given f (x ) = −

1 x

2

, then f ′(x ) = −

(0 ⋅ x )− (2 x ⋅1) = − 0 − 2 x = 2

x

4

x

4

2 x/ x

4/ =3

=

2

x3

e. Given f (x ) = 20 x 2 − 3 , then f ′(x ) = (20 ⋅ 2)x 2 −1 − 0 = 40 x 3

f. Given f (x ) = x 3 = x 2 , then f ′(x ) =

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3 3− 2 3 32 −1 3 3 1 = x 2 = x2 = x x 2 2 2 2

403

Calculus I

Chapter 2 Solutions

10

g. Given f (x ) = =

−5

(x − 5)

x−5

=

1 +1 2

=

1

(x − 5) 2

−5

(x − 5)

0 ⋅ (x − 5) 12  − 10 ⋅ 1 (x − 5)− 12  −1     −5 2  = 0 − 5(x − 5) 2 = , then f ′(x ) =  1 x−5 x−5 (x − 5) 2 ⋅ (x − 5)

10

−5

=

3 2

(x − 5)

3

= −

5

( x − 5)

x−5

ax + b a ⋅ cx − c ⋅ (ax + b ) acx − acx − bc b bc , then f ′(x ) = = = − 2/ =1/ 2 = − 2 2 2 cx (cx ) (cx ) cx c x

h. Given f (x ) =

2. Differentiate the following functions: a. Given f (x ) = x 2 + 10 x + 1 , then f ′(x ) = 2 x 2−1 + 10 x 1−1 + 0 = 2 x + 10 x 0 = 2 x + 10 b. Given f (x ) = x 8 + 3 x 2 − 1 , then f ′(x ) = 8 x 8−1 + (3 ⋅ 2 )x 2−1 − 0 = 8 x 7 + 6 x c. Given f (x ) = 3 x 4 − 2 x 2 + 5 , then f ′(x ) = (3 ⋅ 4 )x 4−1 − (2 ⋅ 2 )x 2−1 + 0 = 12 x 3 − 4 x

(

)

d. Given f (x ) = 2 x 5 + 10 x 4 + 5 x = 2 x 5 + 20 x 4 + 10 x , then f ′(x ) = (2 ⋅ 5)x 5−1 + (20 ⋅ 4 )x 4−1 + (10 ⋅1)x 1−1 = 10 x 4 + 80 x 3 + 10 x 0 4

3

= 10 x + 80 x + 10

( )

( )

e. Given f (x ) = a 2 x 3 + b 2 x + c 2 , then f ′(x ) = 3 ⋅ a 2 x3−1 + 1 ⋅ b 2 x1−1 + 0 = 3a 2 x 2 + b 2 x 0 = 3a 2 x 2 + b 2 f. Given f (x ) = x 2 (x − 1) + 3 x = x3 − x 2 + 3 x , then f ′(x ) = 3 x 3−1 − 2 x 2−1 + 3 x1−1 = 3 x 2 − 2 x + 3 x 0 = 3 x 2 − 2 x + 3

(

)(

[ (

)

)]

)] [ (

g. Given f (x ) = x3 + 1 x 2 − 5 , then f ′(x ) = 3 x 2 x 2 − 5 + 2 x x3 + 1

(

)

h. Given f (x ) = 3 x 2 + x − 1 (x − 1) , then f ′(x ) =

= 3 x 4 − 15 x 2 + 2 x 4 + 2 x = 5 x 4 − 15 x 2 + 2 x

[ (6 x + 1) ⋅ (x − 1) ] + [1 ⋅ (3x 2 + x − 1) ]

= 6 x 2 − 6 x + x − 1 + 3x 2 + x − 1

= (6 + 3)x 2 + (− 6 + 1 + 1)x + (− 1 − 1) = 9 x 2 − 4 x − 2

(

)

i. Given f (x ) = x x 3 + 5 x 2 − 4 x = x 4 + 5 x 3 − 4 x , then f ′(x ) = 4 x 4−1 + (5 ⋅ 3)x 3−1 − (4 ⋅1)x 1−1 = 4 x 3 + 15 x 2 − 4 x 0 = 4 x 3 + 15 x 2 − 4 j. Given f (x ) = k. Given f (x ) = =

9x 6 + 6x 9x

4

=

l. Given f (x ) =

[(3x + 0)⋅ x]− [1 ⋅ (x + 1)] 2

x3 +1 , then f ′(x ) = x

x 5 + 2x 2 −1 3x 2

(

3/ x/ 3 x5 + 2 3 = 9/ x

4/ =3

3

, then f ′(x ) =

) = 3x

5

[(5x

x 4

=

2

3x 3 − x 3 − 1 x

) ][ ( (3x )

=

2

)]

+ 4 x ⋅ 3x 2 − 6 x ⋅ x5 + 2 x 2 − 1

2x3 − 1 x2

=

2 2

15 x 6 + 1/ 2/ x/ 3/ − 6 x 6 − 1/ 2/ x/ 3/ + 6 x 9x4

+2

3x3

[ ]

[ 2 x ⋅ (4 x − 1) ] − 4 ⋅ x 2 x2 x2 , then f ′(x ) = = (x − 1) + 3x 4 x − 1 (4 x − 1)2

=

8x 2 − 2x − 4x 2

(4 x − 1)

2

=

4x 2 − 2x

(4 x − 1)

2

=

2 x (2 x − 1)

(4 x − 1)2

1 x2  m. Given f (x ) = x 2  2 +  = 2 x 2 + = 2 x 2 + x , then f ′(x ) = (2 ⋅ 2 )x 2−1 + x 1−1 = 4 x + x 0 = 4 x + 1 x x  n. Given f (x ) = (x + 1) ⋅ =

(4 − 2)x 2 − 4 x − 2 (x − 1)2

o. Given f (x ) =

2x 2x 2 + 2x , then f ′(x ) = = x −1 x −1

=

[ (4 x + 2) ⋅ (x − 1) ] − [1 ⋅ (2 x 2 + 2 x ) ] (x − 1)2

=

4 x 2 − 4 x + 2/ x/ − 2 − 2 x 2 − 2/ x/

(x − 1)2

2x2 − 4x − 2

x 3 + 3x − 1 x4

Hamilton Education Guides

( x − 1)2

[ (3x + 3)⋅ x ]−  4 x ⋅ (x 2

, then f ′(x ) =

4

3

x8

3

)

 + 3x − 1  

=

3 x 6 + 3 x 4 − 4 x 6 − 12 x 4 + 4 x 3 x8

404

Calculus I

=

Chapter 2 Solutions

− x 6 − 9x 4 + 4x 3 x

=

8

(

(

x 3/ − x 3 − 9 x + 4 x

8/ =5

)= − x

3

+ 9x − 4 x5

(

)

)(

2 3  3    2x 3 + 5   , then f ′(x ) = 2 x ⋅  2 x + 5  +  6 x ⋅ x − 2 x + 5 p. Given f (x ) = x 2 − 1  2   x   x  x  

(

)

(

) ⋅ (x − 1) = 4 x  2



3

+ 10

)

5 3 2 5 3 2  3   6x 3 − 2x 3 − 5  2  ⋅ x − 1 = 4 x 3 + 10 +  4 x − 5  x 2 − 1 = 4 x 3 + 10 + 4 x − 4 x − 5 x + 5 = 8 x − 4 x + 5 x + 5 + 2 2 2 2  x    x x x    

q. Given f (x ) = =

[(

][(

)

12 x3 + 2 x ⋅ (x − 1) − 1 ⋅ 3 x 4 + x 2 + 2 3x 4 + x 2 + 2 , then f ′(x ) = x −1 (x − 1)2

)]

12 x 4 − 12 x 3 + 2 x 2 − 2 x − 3 x 4 − x 2 − 2

=

(x − 1)2

9 x 4 − 12 x 3 + x 2 − 2 x − 2

( x − 1)2

r. Given f (x ) = x −1 +

1 x −2

= x −1 + x 2 , then f ′(x ) = − x −1−1 + 2 x 2−1 = − x −2 + 2 x = −

1

+ 2x

x2

3. Compute f ′(x ) at the specified value of x . Compare your answers with the practice problem number two in Section 2.1. a. Given f (x ) = x3 , then f ′(x ) = 3x3−1 = 3x 2

at x = 1

f ′(x ) = 3 ⋅ 12 = 3

b. Given f (x ) = 1+ 2 x , then f ′(x ) = 0 + (2 ⋅1)x1−1 = 2x 0 = 2

at x = 0

f ′(x ) = 2

c. Given f (x ) = x 3 + 1 , then f ′(x ) = 3x 3−1 + 0 = 3x 2

at x = −1

f ′(x ) = 3 ⋅ (− 1)2 = 3

d. Given f (x ) = x 2 (x + 2) = x 3 + 2 x 2 , then f ′(x ) = 3x 3−1 + (2 ⋅ 2)x 2−1 = 3x 2 + 4 x at x = 2

f ′(x ) = 3 ⋅ 2 2 + 4 ⋅ 2 = 3 ⋅ 4 + 8 = 12 + 8 = 20

e. Given f (x ) = x −2 + x −1 + 1 , then f ′(x ) = − 2 x −2−1 − x −1−1 + 0 = − 2 x −3 − x −2 = − at x = 1

f ′(x ) = −

2 13

−

1

1 12

f ′(x ) =

1 2 10

=

x

3

−

1 x2

= −2 − 1 = −3

f. Given f (x ) = x + 2 = x 2 + 2 , then f ′(x ) = at x = 10

2

1 12 −1 1 1− 2 1 −1 x +0 = x 2 = x 2 = 2 2 2

1 1 2x 2

=

1 2 x

1 1 = = 0.158 2 ⋅ 3.16 6.32

4. Find f ′(0 ) and f ′(2 ) for the following functions: a. Given f (x ) = x 3 − 3 x 2 + 5 , then f ′(x ) = 3 x 3−1 − (3 ⋅ 2 )x 2−1 + 0 = 3 x 2 − 6 x

(

)

( ) b. Given f (x ) = (x + 1)(x − 1) , then f ′(x ) = [3 x ⋅ (x − 1) ]+ [1 ⋅ (x + 1) ] = 3 x − 3 x + x + 1 = 4 x − 3 x + 1 Therefore, f ′(0 ) = (4 ⋅ 0 ) − (3 ⋅ 0 ) + 1 = 0 − 0 + 1 = 1 and f ′(2 ) = (4 ⋅ 2 )− (3 ⋅ 2 )+ 1 = 32 − 12 + 1 = 21 c. Given f (x ) = x (x + 1) = x + x , then f ′(x ) = 3 x = 3x + x = 3x + 1 +x Therefore, f ′(0 ) = (3 ⋅ 0 )+ 1 = 0 + 1 = 1 and f ′(2 ) = (3 ⋅ 2 )+ 1 = 12 + 1 = 13 Therefore, f ′(0 ) = 3 ⋅ 0 2 − (6 ⋅ 0 ) = 0 − 0 = 0 and f ′(2 ) = 3 ⋅ 2 2 − (6 ⋅ 2 ) = 12 − 12 = 0 2

3

3

2

3

3

2

3

2

3

2 −1

2

1−1

0

3

3

2

2

2

2

d. Given f (x ) = 2 x 5 + 10 x 4 − 4 x , then f ′(x ) = (2 ⋅ 5)x 5−1 + (10 ⋅ 4 )x 4−1 − 4 x 1−1 = 10 x 4 + 40 x 3 − 4 x 0 = 10 x 4 + 40 x3 − 4

(

)(

)

(

)(

)

Therefore, f ′(0 ) = 10 ⋅ 0 4 + 40 ⋅ 0 3 − 4 = −4 and f ′(2 ) = 10 ⋅ 2 4 + 40 ⋅ 2 3 − 4 = 160 + 320 − 4 = 476

Hamilton Education Guides

405

Calculus I

Chapter 2 Solutions

e. Given f (x ) = 2 x −2 − 3 x −1 + 5 x , then f ′(x ) = (2 ⋅ −2 )x −2 −1 + (− 3 ⋅ −1)x −1−1 + 5 x1−1 = − 4 x −3 + 3 x −2 + 5 x 0

4

= − 4 x −3 + 3 x −2 + 5 = − f ′(2 ) = −

4

+

23

3 22

x

3

+5 = −

(

)

+

3 x

+ 5 . Therefore, f ′(0 ) = −

2

4 0

+

3

4 3 + + 5 = −0.5 + 0.75 + 5 = 5.25 8 4

3 02

+ 5 which is undefined due to division by zero and

f. Given f (x ) = x −2 x 5 − x 3 + x = x 3 − x + x = x 3 , then f ′(x ) = 3 x 3−1 = 3x 2 Therefore, f ′(0 ) = 3⋅ 0 2 = 0 and f ′(2 ) = 3⋅ 2 2 = 12

x

g. Given f (x ) =

1+ x

Therefore, f ′(0 ) =

2

, then f ′(x ) = 1− 02

(1 + 0 )

2 2

[1 ⋅ (1 + x )]− [2 x ⋅ x] (1 + x ) 2

=

2 2

= 1 and f ′(2 ) =

1− 22

(1 + 2 )

(1 + x )

(1 + 4)

2

=

−3 5

1 − x2

=

2 2

1− 4

=

2 2

1 + x 2 − 2x 2

(1 + x )

2 2

= −

2

3 25

(0 ⋅ x ) − (1 ⋅1) + 3x 3−1 = − 1 + 3x 2 1 + x 3 , then f ′(x ) = x x2 x2 1 1 1 Therefore, f ′(0 ) = − 2 + 3 ⋅ 02 which is undefined due to division by zero and f ′(2 ) = − 2 + 3 ⋅ 22 = − + 12 = 11.75 4 0 2

h. Given f (x ) =

( )

ax 2 + bx , then f ′(x ) = cx − d

i. Given f (x ) =

=

acx 2 − 2adx − bd

(cx − d )

f ′(2 ) =

( )

2

[ (2ax + b ) ⋅ (cx − d ) ] − [c ⋅ (ax 2 + bx ) ] (cx − d )2

. Therefore, f ′(0 ) =

(ac ⋅ 2 )− (2ad ⋅ 2) − bd 2

(c ⋅ 2 − d )2

=

(ac ⋅ 0 )− (2ad ⋅ 0) − bd 2

(c ⋅ 0 − d )

=

2

2acx 2 − 2adx + b/ c/ x/ − bd − acx 2 − b/ c/ x/

=

(cx − d )2

0 − 0 − bd

=

(0 − d )

2

−bd/ d

/ 1 2=

= −

b and d

4ac − 4ad − bd

(2c − d )2

5. Given f (x ) = x 2 + 1 and g (x ) = 2 x − 5 find h(x ) and h ′(x ) .

(

)

a. Given h(x ) = x 3 f (x ) where f (x ) = x 2 + 1 , then h(x ) = x 3 x 2 + 1 = x 5 + x 3 and h ′(x ) = 5 x 4 + 3 x 2

(

)

b. Given f (x ) = 3 + h(x ) where f (x ) = x 2 + 1 , then h(x ) = f (x ) − 3 = x 2 + 1 − 3 = x 2 − 2 and h ′(x ) = 2 x c. Given 2 g (x ) = h(x ) − 1 where g (x ) = 2 x − 5 , then h(x ) = 2 g (x ) + 1 = 2(2 x − 5) + 1 = 4 x − 10 + 1 = 4 x − 9 and h ′(x ) = 4 d. Given 3h(x ) = 2 x g (x ) − 1 where g (x ) = 2 x − 5 , then h(x ) = h ′(x ) =

[ (8 x − 10) ⋅ 3] − [0 ⋅ (4 x 2 − 10 x − 1) ] 32

=

2 x g (x ) − 1 2 x (2 x − 5) − 1 4 x 2 − 10 x − 1 = = and 3 3 3

8 x − 10 3(8 x − 10 ) = 9 3

e. Given 3[ f (x ) ]2 − 2h(x ) = 1 where f (x ) = x 2 + 1 , then h(x ) =

(

)

(

)

)

(

2 1 3 1 3 1 − 1 + 3[ f (x ) ]2 = − + [ f (x ) ]2 = − + x 2 + 1 and 2 2 2 2 2

h ′(x ) = 3 x 2 + 1 ⋅ 2 x = 6 x 3 + 6 x

(

)

(

f. Given h(x ) = g (x ) ⋅ 3 f (x ) where f (x ) = x 2 + 1 and g (x ) = 2 x − 5 , then h(x ) = (2 x − 5) ⋅ 3 x 2 + 1 = (2 x − 5) 3 x 2 + 3

)

= 6 x 3 − 15 x 2 + 6 x − 15 and h′(x ) = 18 x 2 − 30 x + 6 g. Given 3h(x ) − f (x ) = 0 where f (x ) = x 2 + 1 , then h(x ) =

Hamilton Education Guides

f (x ) x2 + 1 2 = and h ′(x ) = x 3 3 3

406

Calculus I

Chapter 2 Solutions

(

)

h. Given 2 g (x ) + h(x ) = f (x ) where f (x ) = x 2 + 1 and g (x ) = 2 x − 5 , then h(x ) = f (x ) − 2 g (x ) = x 2 + 1 − 2(2 x − 5) = x 2 + 1 − 4 x + 10 = x 2 − 4 x + 11 and h ′(x ) = 2 x − 4

(

)

i. Given f (x ) = x 3 + 5 x 2 + h(x ) where f (x ) = x 2 + 1 , then h(x ) = f (x ) − x 3 − 5x 2 = x 2 + 1 − x 3 − 5 x 2 = − x3 + (− 5 + 1)x 2 + 1 = − x 3 − 4 x 2 + 1 and h ′(x ) = − 3 x 2 − 8 x

(

3x3 − x3 − 1

=

[

][ (

)]

3x 2 ⋅ x − 1 ⋅ x3 + 1 x3 + 1 x3 +1 − x 2 + 1 and h ′(x ) = − f (x ) where f (x ) = x 2 + 1 , then h(x ) = x x x2

j. Given h(x ) =

x2

− 2x =

2x3 − 1 x2

)

− 2x

− 2x

(

)

k. Given h(x ) = 2 f (x ) + g (x ) where f (x ) = x 2 + 1 and g (x ) = 2 x − 5 , then h(x ) = 2 x 2 + 1 + (2 x − 5) = 2 x 2 + 2 + 2 x − 5 = 2 x 2 + 2 x − 3 and h ′(x ) = 4 x + 2

[ (

l. Given [h(x )] − f (x ) = 10 where f (x ) = x 2 + 1 , then [ h(x ) ]2 = 10 + f (x ) ; h(x ) =

(

)

= x 2 + 11

1 2

and h ′(x ) =

(

(

)

)

1 −1 −1 1 2 x + 11 2 ⋅ 2/ x = x x 2 + 11 2 2/

[4 ⋅ (x + 1)]− [2 x ⋅ (4 x − 10) ] = 4 x + 4 − 8x + 20 x = − 4 x + 20 x + 4 (x + 1) (x + 1) (x + 1) 3 f (x ) 1 n. Given = where f (x ) = x + 1 , then h(x ) = 3 x f (x ) = 3 x (x + 1) = 3 x h(x ) x 2

2

2

2

2

2

2

2

2

2

2

o. Given f (x ) =

2

[0 ⋅ (x + 1)]− [2 x ⋅1] − 0 = 0 − 2 x (x + 1) (x + 1) 2

2

2

3

+ 3 x and h ′(x ) = 9 x 2 + 3

f (x ) 1 1 , which is equivalent to ; f ( x ) ⋅ [ h ( x ) + 4] = 1 ⋅ 1 ; f ( x ) h ( x ) + 4 f ( x ) = 1 = h(x ) + 4 1 h(x ) + 4 1 1 1 − 4 f (x ) 1 4 f (x ) = = −4 = 2 − − 4 and h ′(x ) f (x ) f (x ) f (x ) f (x ) x +1

; f (x )h(x ) = 1 − 4 f (x ) , and f (x ) = x 2 + 1 , then h(x ) =

=

1 2

2 g (x ) 4 x − 10 2 g (x ) 2(2 x − 5) where f (x ) = x 2 + 1 and g (x ) = 2 x − 5 , then h(x ) = = = and h ′(x ) f (x ) h(x ) x2 + 1 x2 + 1

m. Given f (x ) =

=

)]

10 + f (x ) = 10 + x 2 + 1

2

2

2

= −

2x

(x + 1)

2

2

Section 2.3 Solutions - Differentiation Rules Using the

d dx

Notation

dy for the following functions: dx dy d 5 d d d 5 a. Given y = x 5 + 3 x 2 + 1 , then = x + 3 x 2 + 1 = 5 x 4 + (3 ⋅ 2 )x + 0 = 5 x 4 + 6 x x + 3x 2 + 1 = dx dx dx dx dx dy d d d 2 2 2 b. Given y = 3 x + 5 , then = 3x + 5 = 3 x + 5 = (3 ⋅ 2 )x + 0 = 6 x dx dx dx dx dy d 3 d 3 d −1 1 d  3 1 3 1 c. Given y = x − , then = x − x −1 = x = 3 x 2 + x −1−1 = 3 x 2 + x −2 = 3 x 2 + x − x −  = x dx dx dx dx x dx  x2

1. Find

(

)

(

)

(

(

)

)

(

)

 3 d 2  2 d 3   1 − x dx x  −  x dx 1 − x  dy d  x 2     = d. Given y = , then = =  2 dx dx  1 − x3  1 − x3 1 − x3

x2

=

2 x − 2 x 4 + 3x 4

(1 − x )

3 2

=

x4 − 2x

(1 − x )

Hamilton Education Guides

3 2

=

(

x x3 − 2

(1 − x )

3 2

)

(

)

[ (1 − x )⋅ 2 x]−  x ⋅ (− 3x ) (1 − x ) 3

2

2

 

3 2

407

Calculus I

Chapter 2 Solutions

d   d    (x − 1) ⋅ dx 1 − 1 ⋅ dx (x − 1)  1 dy d 1 1 d d         e. Given y = 4 x 2 + , then = 4x2 +   4x2 +  =  = 8x + x −1 dx dx  x −1 dx dx  x − 1  (x − 1)2

( )

= 8x +

0 −1

= 8x −

(x − 1)

2

1

( x − 1)2

(

) (

) ( ( )

) (

) [(x + 1)⋅ (2 x + 2)] (x + 1)

d 2 d 3  3   2   x + 1 dx x + 2 x  −  x + 2 x dx x + 1  dy d  x 2 + 2 x   =    , then = = f. Given y = 3 2 3 dx dx  x3 + 1  x +1 x +1

x2 + 2x

[(

) ] = 2 x + 2 x + 2 x + 2 − 3x − 6 x = − x − 4 x + 2 x + 2 (x + 1) (x + 1) (x + 1) d dy (x + 5x − 2 x ) = dxd x g. Given y = x (x + 5 x − 2 ) = x + 5 x − 2 x , then = dx dx − x 2 + 2 x ⋅ 3x 2

4

3

2

3

4

2

5

4

5

3

)

(

=

[(x − 1)⋅ (3x

2

)] [(

4

[(

3

5

1 x

3

]

)

(

) ]

, then

(x − 1)(x + 3) x2

(

=

− 2x + 6x x

4

d d 5 x 4 − 2 x3 = 5 x 4 + 20 x 3 − 6 x 2 dx dx

dy d d 3  = x3 + 3 x 2 (x − 1) =  (x − 1) x + 3x 2 dx dx dx 

) + (x

(

3

(

=

x 2 + 3x − x − 3 x2

) (

=

−2 x/ (x − 3) x

4/ =3

+ 3x 2

) dxd (x − 1) 

)

)

dy d d −3 d 3 1  d  = 5 x − x −3 = 5x − x = 5 + 3 x −3−1 = 5 + 3 x −4 = 5 + 4  5 x − 3  = dx dx dx dx dx  x  x

=

)

x 2 + (3 − 1)x − 3 x2

=

x2 + 2x − 3 x2

, then

dy d  x2 + 2x − 3  =   dx dx  x2 

d 2  2 d 2   2 2 2  x dx x + 2 x − 3  −  x + 2 x − 3 dx x      = x ⋅ (2 x + 2 ) − x + 2 x − 3 ⋅ 2 x = x4 x4 2

+

+ 6 x + x3 + 3 x 2 ⋅ 1 = 3 x3 + 6 x 2 − 3 x 2 − 6 x + x3 + 3 x 2 = 4 x3 + 6 x 2 − 6 x = 2 x 2 x 2 + 3 x − 3

i. Given y = 5 x −

j. Given y =

2

3

h. Given y = x 2 (x + 3)(x − 1) = x3 + 3 x 2 (x − 1) , then

2

3

3

2

3

3

4

3

3

= −

[

] [(

) ] = 2x

3

+ 2 x 2 − 2 x3 − 4 x 2 + 6 x x4

2 ( x − 3) x3

(

) (

)

d   d 2   2 2 3 ⋅ dx x − x  −  x − x ⋅ dx 3 dy d  x 2 − x   x − 1  x2 − x    = [3 ⋅ (2 x − 1) ] − x − x ⋅ 0 k. Given y = x , then = =  = 9 dx dx  3  3 32  3 

[(

2x − 1 3/ (2 x − 1) 3(2 x − 1) − 0 = = 9/ = 3 9 3 dy d 2 d    d  −1 2 l. Given y = x (x + 3) , then = x (x + 3)−1 = (x + 3)−1 x 2  +  x 2 (x + 3)−1  = dx dx dx dx    

) ]

=

[

= 2 x (x + 3)−1 − x 2 (x + 3)−2 =

]

[(x + 3)

−1

][ (

⋅ 2 x + x 2 ⋅ − (x + 3)−2

2x x2 − x + 3 ( x + 3 )2

x−3 dy  x −3 d  x   x  d  x −3  x  x −3 ⋅ =  m. Given y =   , then   +    =  5 dx  5  dx  1 + x   1 + x  dy  5  1+ x   5 

(1 + x ) d

d x − x (1 + x ) dx dx (1 + x )2

d d 5 ( x − 3) − ( x − 3) 5 x − 3 [(1 + x ) ⋅ 1] − [x ⋅ 1] x [5 ⋅ 1] − [(x − 3) ⋅ 0] x x − 3 x/ + 1 − x/ x 5/ dx dx = = ⋅ ⋅ + + ⋅ ⋅ + ⋅ 25 1+ x 5 1+ x 5 52 (1 + x )2 (1 + x )2 1 + x 2/ 5/ 5

=

x−3 x 1 x−3 x 1 ⋅ = + ⋅ + 5 (1 + x )2 1 + x 5 5(1 + x )2 5(1 + x )

Hamilton Education Guides

408

)]

Calculus I

Chapter 2 Solutions

d 3  3 d   3 3  (x − 1) dx x  −  x dx (x − 1)  dy x 1 d d x       = 3x 2 +  n. Given y = x3 1 + , then = x3 +   = x3 + dx x −1 x −1 dx dx  x − 1  (x − 1)2  = 3x 2 +

[(x − 1)⋅ 3x ]− [x ⋅1] 2

3

(x − 1)2

= 3x 2 +

3x3 − 3x 2 − x3

= 3x 2 +

(x − 1)2

2 x3 − 3x 2

= 3x2 +

(x − 1)2

x 2 (2 x − 3 )

( x − 1)2

(

)

(

)

d d  2    2  3 x + x dx (2 x − 1) − (2 x − 1) dx 3 x + x  dy 2x − 1 d  2x − 1  1  2x − 1     o. Given y =  , then =   =   = 2 dx dx  3 x 2 + x  x  3x + 1  3x 2 + x 3x 2 + x

[(3x + x)⋅ 2]− [(2 x − 1)⋅ (6 x + 1) ] = 6 x (3x + x) 2

=

2

2

2

(

(

)

(3x

(bx )

2

2abx 2 + b/ 2/ x/ − abx 2 − b/ 2/ x/ − bc

=

) = 6x

+ 2 x − 12 x 2 + 2 x − 6 x − 1 2

+x

)

2

dy ax 2 + bx + c d  ax 2 + bx + c  p. Given y = , then = =  bx dx dx  bx 

− ax 2 + bx + c ⋅ b

)

(

2 2

b x

2

2

2

=

3x 6 − 9 x 2 − 4 x 6 + 8 x3

(x − 3)

2

4

r. Given y =

=

=

(

5x

(1 + x )2

b x

) (

=

(

b/ ax 2 − c

)

(1 + x )4

2

2

=

b

2/ =1 2

x

) ( ( )

) = ax

2

(bx )2

−c

bx 2

) (

) [(x − 3)⋅ 3x ]− [(x − 2)⋅ 4 x ] (x − 3) 4

3

3

2

4

2

− x6 + 8 x3 − 9 x2

(x

4

−3

)

2

 (1 + x )2 d 5 x  − 5 x d (1 + x )2     dy d  5x  dx   dx  = , then = =  dx dx  (1 + x )2  (1 + x )4

5 x 2 + 2 x + 1 − 10 x(1 + x )

− 6x2 + 6x + 1

2

d 3 d 4  4   3   x − 3 dx x − 2  −  x − 2 dx x − 3  dy d  x3 − 2   =   , then = =  q. Given y = 4 2 dx dx  x 4 − 3  x −3 x4 − 3

x3 − 2

=

(bx )2

2 2

(

+ 2/ x/ − 12 x 2 − 2/ x/ + 6 x + 1

(3x + x) (3 x + x ) d   d   bx dx (ax + bx + c ) − (ax + bx + c ) dx bx  bx ⋅ (2ax + b )    

abx 2 − bc

=

2

=

5 x 2 + 1/ 0/ x/ + 5 − 1/ 0/ x/ − 10 x 2

(1 + x )4

=

[(1 + x) ⋅ 5]− [5x ⋅ 2(1 + x) ] 2

(1 + x )4

− 5x2 + 5

(1 + x )4

2. Find the derivative of the following functions:

(

)

( )

a.

d 2 d d 3t + 5t = 3t 2 + (5t ) = (3 ⋅ 2 ) t 2 −1 + (5 ⋅ 1) t1−1 = 6t + 5t 0 = 6t + 5 dt dt dt

b.

d d d (5 x ) + d (− 2) = (6 ⋅ 3)x 3−1 + (5 ⋅1)x1−1 + 0 = 18 x 2 + 5 x 0 = 18 x 2 + 5 6x 3 + 6 x3 + 5 x − 2 = dx dx dx dx

c.

d d d 3 d 3 u + 2u 2 + 5 = 2u 2 + 5 = 3u 3−1 + (2 ⋅ 2 )u 2−1 + 0 = 3u 2 + 4u u + du du du du

(

)

( )

(

)

( )

(

 d 2 5 ⋅ dt t + 2t d  t 2 + 2t  =  d. dt  5 

(

( )

) −  (t 52

2

) dtd 5

+ 2t ⋅

=

[5 ⋅ (2t + 2)] − [(t 2 + 2t )⋅ 0] 25

=

5 (2t + 2 ) − 0 2t + 2 5/ (2t + 2 ) = = or, 25 5 2/ 5/ 5

)

1 2t + 2 1 d 2 1 d d  d  t 2 + 2t  = t + 2t =  t 2 + 2t  = (2t + 2 ) = 5 5 5 dt 5  dt dt  dt  5 

(

) (

)

d 2  2 d 3   3 2 2 3  s ⋅ dt s + 3s − 1  −  s + 3s − 1 ⋅ dt s  d  s 3 + 3s − 1   = s ⋅ 3s + 3 − s + 3s − 1 ⋅ 2 s    e. =  ds  s4 s2 s4 

Hamilton Education Guides

[ (

)] [(

) ]

=

3 s 4 + 3s 2 − 2 s 4 − 6 s 2 + 2 s s4

409

Calculus I

Chapter 2 Solutions

s 4 − 3s 2 + 2 s s

(

)

s/ s 3 − 3s + 2 s3 − 3s + 2 = 4/ =3 s s3

=

4

 (1 + w) ⋅ d w2  − w2 ⋅ d (1 + w) 2 2  2     d  3 w  d 3 d  w  dw   dt  = 3w2 + [(1 + w) ⋅ 2 w] − w ⋅ 1 f. = = 3w 2 +  w + w + dw  1 + w  dw dw  1 + w  (1 + w)2 (1 + w)2

[ ]

= 3w2 +

2 w + 2 w2 − w2

= 3w 2 +

(1 + w)

2

[

[(

)]

(

w 2 + 2w

(1 + w )2

)] (

)(

) (

)(

) (

) (

)(

)(

)

d d 3 2 d 2 d 2 t3 + t 2 t 2 − 3 = t 2 − 3 t + t + t3 + t 2 t − 3 = t 2 − 3 ⋅ 3t 2 + 2t + t 3 + t 2 ⋅ 2t t (t + 1) t 2 − 3 = dt dt dt dt

g.

= 3t 4 + 2t 3 − 9t 2 − 6t + 2t 4 + 2t 3 = 5t 4 + 4t 3 − 9t 2 − 6t , or

[

[(

)]

(

)]

(

)] = (x + 5) dxd (x + 1) + (x + 1) dxd (x + 5) = [(x + 5)⋅1]+ [(x + 1) ⋅ 2 x] = x

)(

(

)

d 2 d d 5 4 d d 5 d 4 d t3 + t 2 t 2 − 3 = t (t + 1) t 2 − 3 = t + t − 3t 3 − 3t 2 = t + t − 3 t 3 − 3 t 2 = 5t 4 + 4t 3 − 9t 2 − 6t dt dt dt dt dt dt dt

[

d (x + 1) x 2 + 5 dx

h.

or,

[

(

d (x + 1) x 2 + 5 dx

2

2

2

)] = dxd (x

3

)

2

+ 5 + 2x2 + 2x = 3 x 2 + 2 x + 5

d d 3 d 2 d x + 5x + 5 = 3 x 2 + 2 x + 5 x + dx dx dx dx

+ x2 + 5x + 5 =

d   d d 2  2 d     (1 − u ) ⋅ du u  − u ⋅ du (1 − u ) (1 + u ) ⋅ du u  − u ⋅ du (1 + u ) d  u2 u  d  u 2  d  u     −    i. − −  =    = du 1 − u 1 + u  du  1 − u  du  1 + u  (1 − u )2 (1 + u )2

[ (1 − u ) ⋅ 2u ] − [u 2 ⋅ −1] [ (1 + u ) ⋅ 1 ] − [u ⋅ 1] − (1 − u )2 (1 + u )2

=

(

=

2u − 2u 2 + u 2

(1 − u )

2

) (

−

1+ u − u

(1 + u )

2

)

2u − u 2

=

1

−

(1 − u )2 (1 + u )2

=

u(2 − u )

1

−

(1 − u)2 (1 + u)2

d   d   3 3 2 2 2 3 2 r ⋅ dr 3r − 2r + 1  −  3r − 2r + 1 ⋅ dr r  d  3r 3 − 2r 2 + 1    = r ⋅ 9r − 4r − 3r − 2r + 1 ⋅ 1   j. =  dr  r r2 r2 

9r 3 − 4r 2 − 3r 3 + 2r 2 − 1

=

r2

=

[ (

)] [(

6r 3 − 2r 2 − 1 r2

(

(

)

)

d 2  2 d 3  3   s + 1 ⋅ ds 3s  − 3s ⋅ ds s + 1  d  3s 2 1 d  3s 2  d  1     − k. −   = −  =  2 3 ds  s 3 + 1  ds  s 2  ds  s 3 + 1 s 2  s +1

[(s + 1)⋅ 6s]− [3s (s + 1) 3

=

2

⋅ 3s 2

2

3

) ]

] − (s ⋅ 0)− (1 ⋅ 2s ) = 6s 2

s

4

(

4

+ 6s − 9s 4

(s + 1)

2

3

)

−

0 − 2s s

4

=

− 3s 4 + 6 s

(s + 1) 3

2

 2 d   d 2  s ⋅ ds 1 − 1 ⋅ ds s      4 s +

2 s/ s

4/ =3

= −

(

3s s 3 − 2

(s + 1) 3

2

)+ 2

s3

d 3  3 d d 2    2 d   (1 − u ) ⋅ du u  − u ⋅ du (1 − u ) u ⋅ du (u + 1) − (u + 1) ⋅ du u  d  u 3  d  u + 1  d  u3 u + 1 −      l. − − 2  =   =   du  1 − u  du  u 2  du 1 − u u4 u  (1 − u )2

= =

[(1 − u )⋅ 3u ]− [u ⋅ −1] − [u ⋅1]− [(u + 1)⋅ 2u] = 3u 2

3

2

(1 − u )2 u 2 (− 2u + 3) u/ (u + 2 ) + 4/ =3 u (1 − u )2

u4

= −

u 2 (2u − 3 )

(1 − u)

+

− 3u 3 + u 3

(1 − u )2

−

u 2 − 2u 2 − 2u u4

=

− 2u 3 + 3u 2

(1 − u )2

−

− u 2 − 2u u4

u+2

u3 3. Find the derivative of the following functions at the specified value.

a.

(

)

( )

2

2

( )

d 3 d 3 d d x + 3x 2 + 1 = x + 3 x 2 + (1) = 3 x3−1 + (3 ⋅ 2 )x 2 −1 + 0 = 3 x 2 + 6 x dx dx dx dx

Hamilton Education Guides

410

Calculus I

Chapter 2 Solutions

(

[

)

d 3 x + 3 x 2 + 1 = 3 ⋅ 22 + 6 ⋅ 2 = 12 + 12 = 24 dx

at x = 2

)] (

(

)

)

) (

(

d (x + 1) x 2 − 1 = x 2 − 1 d (x + 1)+ (x + 1) d x 2 − 1 = x 2 − 1 ⋅ 1+ (x + 1) ⋅ 2 x = x 2 − 1+ 2 x 2 + 2 x = 3x 2 + 2 x − 1 dx dx dx

b.

[

)] ( )

(

d (x + 1) x 2 − 1 = 3 ⋅ 12 + (2 ⋅ 1) − 1 = 3 + 2 − 1 = 4 dx

at x = 1

[

]

(

)

( )

(

)

d d d d 3s 3 − 3s 2 = 3s 3 + − 3s 2 = 9 s 2 − 6 s 3s 2 (s − 1) = ds ds ds ds

c.

[

] ( )

d 3s 2 (s − 1) = 9 ⋅ 02 − (6 ⋅ 0 ) = 0 ds

at s = 0

)

) (

(

d 2 d    2  (t − 1) dt t + 1  −  t + 1 dt (t − 1) d t2 +1      = d.   = dt  t − 1  (t − 1)2

=

=

2t 2 − 2t − t 2 − 1

(t − 1)2

=

t 2 − 2t − 1

(t − 1)2

(− 1)2 + (− 2 ⋅ −1) − 1 = 1 + 2 − 1 = 2 = 1 d t2 +1    = 4 4 2 dt  t − 1  (− 1 − 1)2

at t = −1

d e. du

[ (t − 1) ⋅ 2t ] − [(t 2 + 1)⋅ 1] (t − 1)2

 2 d 3  3 d 2 (u + 1) du u  − u du (u + 1)   u3      =   = (u + 1)4  (u + 1)2 

3u 4 + 6u 3 + 3u 2 − 2u 4 − 2u 3

(u + 1)

4

=

u 4 + 4u 3 + 3u 2

(u + 1)

4

[(u + 1) ⋅ 3u ]− [u 2

2

3

⋅ 2(u + 1)

(u + 1)

4

d du

at u = 1

(

] = [(u

2

) ][

+ 2u + 1 ⋅ 3u 2 − 2u 4 + 2u 3

(u + 1)

]

4

( ) ( )

 u3  1 14 + 4 ⋅ 13 + 3 ⋅ 12 8 1+ 4 + 3 = = =   = 2 16 2 24 (1 + 1)4  (u + 1) 

) (

)

(

)

d 2 d    2  2 2 2 3w du w + 1  −  w + 1 du 3w    d w w + 1 w + d 1 d  w w2 +1  /     = [3w ⋅ 2 w] − w + 1 ⋅ 3 = = = f.       dw  3w 2  dw  3w2/ =1  dw  3w  9 w2 (3w)2

(

=

g.

)

(

6 w 2 − 3w 2 − 3 9w

[(

2

=

(

)

)=w

3 w2 − 1

2

9w

3w

) ]

2

−1

) ]

3 1 22 − 1 d  w w2 +1  = = =   2 12 4 dw  3w 2  3⋅ 2

at w = 2

2

[(

( )

)

(

d 5 3 d 5 d 3 d v 2 +1 v3 = v +v = v + v = 5v 4 + 3v 2 dv dv dv dv

at v = −2

[(

) ]

(

)

d v 2 + 1 v 3 = 5 ⋅ (− 2 )4 + 3 ⋅ (− 2 )2 = (5 ⋅ 16 ) + (3 ⋅ 4 ) = 80 + 12 = 92 dv

(

)

d 3  3 d 2  2   x + 1 dx x  −  x dx x + 1  d  x 3      = = h. 2 dx  x 2 +1  x2 + 1

(

at x = 0

)

( ) ( )

[(x + 1)⋅ 3x ]− [x (x + 1) 2

2

2

2

3

⋅ 2x

]

3x 4 + 3x 2 − 2 x 4

=

(x + 1) 2

2

=

x 4 + 3x 2

(x + 1)

2

2

0 04 + 3 ⋅ 02 0+0 d  x 3  = = 2 = = 0 2 2 1 dx  x 2 +1  1 0 +1

d 5  5 d   (1 − u ) du u  − u du (1 − u ) d  3  u 2  d  u 5      = u  = = i. 2 du   1 − u  du  1 − u  ( ) − u 1  

(

[(1 − u )⋅ 5u ]− [(u ⋅ −1)] 4

5

(1 − u )

2

=

5u 4 − 5u 5 + u 5

(1 − u )

2

=

− 4u 5 + 5u 4

(1 − u )2

) ( )

0+0 0 − 4 ⋅ 05 + 5 ⋅ 0 4 d  3  u 2  u  = = = = 0 2 1 1 du   1 − u  (1 − 0)   4. Given the functions below find their derivatives at the specified value.

at u = 0

a.

(

)

(

)

ds 2 ds ds ds ds given s = t 2 − 1 + (3t + 2 )2 , then = t − 1 + (3t + 2 )2 = 2t + 2(3t + 2 )2 −1 ⋅ (3t + 2 ) = 2t + 2(3t + 2 ) ⋅ 3 dt dt dt dt dt

Hamilton Education Guides

411

Calculus I

Chapter 2 Solutions

ds = (20 ⋅ 2 ) + 12 = 40 + 12 = 52 dt d 3 d t + 3t 2 + 1 − t 3 + 3t 2 + 1 2t 2t t 3 + 3t 2 + 1 dy dy 2t ⋅ 3t 2 + 6t − t 3 + 3t 2 + 1 ⋅ 2 dt dt b. given y = , then = = 2 2t dt dt 4t 2 (2t )

= 2t + 18t + 12 = 20t + 12

at t = 2

=

6t 3 + 12t 2 − 2t 3 − 6t 2 − 2 4t

=

2

4t 3 + 6t 2 − 2 4t

(

) = 2t

2/ 2t 3 + 3t 2 − 1

=

2

)

)(

(

4/ t

2

3

(

) (

)

+ 3t 2 − 1 2t 2

2

4 2 + 3 −1 dy 2 ⋅ 13 + 3 ⋅ 12 − 1 = = = = 2 dt 2 2 2 ⋅ 12 2 2 dw dw d 2 c. given w = x 2 + 1 + 3 x , then = x + 1 + 3x = 2 x 2 + 1 dx dx dx

at t = 1

(

)

(

)

(

)

2 −1

(

)

(

)

d 2 d x + 1 + 3x = 2 x 2 + 1 ⋅ 2 x + 3 dx dx

dw = 4 ⋅ (− 1)3 + (4 ⋅ −1) + 3 = −4 − 4 + 3 = −5 dx 2 2 2 dy dy d  2 3 d d  2 3  d. given y = x 2 x3 + 2 x + 1 + 3 x , then = x x + 2 x + 1  + 3x x x + 2 x + 1 + 3x  =   dx dx dx  dx    dx

= 4 x3 + 4 x + 3

at x = −1

(

(

)

= x3 + 2 x + 1

2

(

)

(

)

d 2 d 3 x + x2 x + 2x + 1 dx dx

) [ (

(

2

(

(

)

(

)

)

2    + 3 =  x3 + 2 x + 1 ⋅ 2 x  +  x 2 ⋅ 2 x3 + 2 x + 1   

)]

)(

2

)

= 2 x x3 + 2 x + 1 + 2 x 2 x3 + 2 x + 1 3x 2 + 2 + 3

2 −1

(

)

= 6x2 −

2

d 3  x + 2x + 1  + 3 dx 

⋅

dy = 0+0+3 = 3 dx

at x = 0

Section 2.4 Solutions - The Chain Rule 1. Find the derivative of the following functions. Do not simplify the answer to its lowest term.

( ) ( ) ⋅ 2 x = 6 x (x + 2) b. Given y = (x + 1) , then y ′ = − 2(x + 1) ⋅ 2 x = − 4 x (x + 1) 5−1 c. Given y = (x − 1) , then y ′ = 5(x 3 − 1) ⋅ 3 x 2 = 15 x (x − 1) 3

a. Given y = x 2 + 2 , then y ′ = 3 x 2 + 2 2

−2

3

5

3−1

−2 −1

2

2

1 3x

2

4

 1 + x2   1 + x2  f. Given y =  3  , then y ′ = 4  3   x   x      ⋅

− x 4 − 3x 2 x

6

2 −1

, then y ′ = (2 ⋅ 3)x 3−1 +

3

(

4 −1

−3

2

2

  1  1   , then y ′ = 21 −  d. Given y = 1 − 2  2  x x    

e. Given y = 2 x 3 +

2

2

⋅

2x x4

3

4

 4  1  1  x/ ⋅ = 41 − = 3  1 − 2  2  4/ =3 x x x x    

(0 ⋅ 3x )− (6 x ⋅1) = 6 x + 0 − 6 x = 6 x − 6 x/ 9x 9x (3x ) [2 x ⋅ x ]− [3x (1 + x )] = 4  1 + x  ⋅ 2 x − 3x ⋅ 2

2

2 2

3

2

2

2

)

3

3

4/ =3

4

 x3   

x6

 1 + x2   1 + x 2  − x 2/ x 2 + 3  = 4 3  ⋅ = − 4 / = 6 4 3   x   x    x 

2

4

2

x6

− 3x 4

3x3  1 + x2    x3   

3

= 4

 x2 + 3     x4   

3 2 3−1 3 3   1  x +1  x +1  x +1    x +1  x +1 g. Given y = x 2   + x2   ⋅ ⋅ x2  = 2 x    + 3  , then y ′ = 2 x 2 −1 ⋅  3  3     3   3   3   3   

[

h. Given y = x (x + 1)2 + 2 x

(

) (

] = [x (x 3

⋅ 3x 2 + 4 x + 3 = 3 x 3 + 2 x 2

x  i. Given y =  − 2 x3  3  

−1

) ] = (x + 3 x ) (3 x + 4 x + 3 ) 2

+ 2x + 1 + 2x 2

3

+ 2x2 + x + 2x

) = (x 3

3

)

3

(

+ 2 x 2 + 3 x , then y ′ = 3 x3 + 2 x 2 + 3 x

)

2

x  , then y ′ = −  − 2 x3  3  

Hamilton Education Guides

3

−1−1

−2

x  1  1  ⋅  − (2 ⋅ 3)x3−1  = −  − 2 x 3   − 6 x 2  3 3 3      

412

3−1

Calculus I

Chapter 2 Solutions

(

)

(

4

) ⋅ (3x + (3 ⋅ 2)x + 0) = 4(x + 3 x + 1) (3 x + 6 x ) [2t ⋅ (1 + t )]− [2t ⋅ t ] = 3 t  ⋅ 2t + 2/ t/ − 2/ t/ = 3t ⋅ 2t ⋅ 1+ t  (1 + t ) (1 + t ) (1 + t ) (1 + t )   4 −1

j. Given y = x3 + 3 x 2 + 1 , then y ′ = 4 x3 + 3 x 2 + 1 3

 t2   2   , then y ′ = 3  t  k. Given y =  2 2  1+ t  1+ t    3t 4 ⋅ 2t

=

6t 4 +1

=

3−1

2 2

− 2 −1−1

, then y ′ =

x3

(

⋅ −2 x − 2 −1 = 2 x −3 1 + x − 2

[− 2 (x + 1)

][

⋅ 1 ⋅ x3 − 3 x3−1 ⋅ (x + 1)− 2 x6

(

=

6x2

(x + 2)

2

3

x3

(

6x2 − 2x x3 + 2 2x = 2 1 x3 + 2

−

(

)

3

)

−2

=

2

(

x 3 1 + x −2

4

2 2

2 2

)

2

] = [− 2 x (x + 1) ]− [3 x (x + 1) ] −3

3

(

)

2

3

2

3

)

  2  − 1 =  1  ⋅ 6 x 2  1 − x3  1 − x3 x2  

[3x ⋅ (x + 2)]− [3x ⋅ x ] − 2 x (x + 2) 3−1

− x 2 , then y ′ =

x3 + 2

3/

2 2

−2

2

x6

2   1  2 −1 3x 2 1  1  ′ = 2  n. Given y =  , then y ⋅ +       1 − x3  x  1 − x3  1 − x3 

o. Given y =

3/

2

2 2

− 2 −1

(x + 1)−2

2

2

2

2 4

−2 −1

m. Given y =

2

3

2

6t 5

=

2 2+ 2

3

2 −1

2

(1 + t ) ⋅ (1 + t ) (1 + t ) (1 + t ) l. Given y = (1 + x ) , then y ′ = − (1 + x ) 2 2

3−1

2 −1

=

 6x2 − 1 = 2 x2 1 − x3 

)

( ) (x + 2)

3x 2 x3 + 2 − 3x5 2

3

(

− 2x =

)

3

−

1

x2

3/ x/ 5/ + 6 x 2 − 3/ x/ 5/

(x + 2) 3

2

− 2x

2

2. Find the derivative of the following functions at x = 0 , x = 1 , and x = −1 .

(

)

(

5

)

a. Given y = x 3 + 1 , then y ′ = 5 x3 + 1

( )( ) = 0 ⋅1 y′(1) = (15 ⋅ 1 )(1 + 1) = 15⋅ 2 4

y′(0 ) = 15 ⋅ 02 03 + 1 2

[

4

3

y′(−1) = 15 ⋅ (− 1) 2

(

] [(− 1) + 1] 3

)

5 −1

(

(

(

= 0 ⋅1 = 0

4

= 15⋅ 24 = 15⋅ 16 = 240 and

4

= 15 ⋅ (− 1 + 1) 4 = 15 ⋅ 04 = 15 ⋅ 0 = 0

(

4

)

4

4

) (3x

b. Given y = x3 + 3 x 2 − 1 , then y ′ = 4 x3 + 3 x 2 − 1

) (0 + 2) = 0.(− 1)

y′(0 ) = (12 ⋅ 0 ) 03 + 3 ⋅ 02 − 1

)

⋅ 3 x 2 = 15 x 2 x3 + 1 . Therefore,

3

3

4 −1

2

(

)

) (x + 2) . Therefore,

+ 6 x = 12 x x3 + 3 x 2 − 1

3

⋅2 = 0

3

y′(1) = (12 ⋅ 1) 13 + 3 ⋅ 12 − 1 (1 + 2 ) = 12 ⋅ (1 + 3 − 1) 3 ⋅ 3 = 12 ⋅ 27 ⋅ 3 = 972 and

[

] (− 1 + 2) = − 12 (− 1 + 3 − 1)

y′(−1) = (12 ⋅ −1) (− 1) 3 + 3 ⋅ (− 1)2 − 1 2

3

 x   x  c. Given y =    , then y ′ = 2  x +1  x +1

2⋅0

y′(0 ) = y′(1) =

=

(0 + 1) 3 2 ⋅1

(1 + 1) 3

=

2 ⋅ −1

y′(−1) =

(− 1 + 1) 3

(

0

13 2

23

=

)

2 −1

⋅

=

0 = 0 1

=

2 1 = = 0.25 and 8 4

−2 03

= −

1 ⋅ (x + 1) − 1 ⋅ x

(x + 1)2

(

)

2

(

= − 12 (− 2 + 3) 3 = − 12 ⋅ 13 = −12 ⋅ 1 = −12

1 2x  x   x  x/ + 1 − x/ = 2 = 2 = . Thus, ⋅ ⋅  x + 1  (x + 1)2  x + 1  (x + 1)2 (x + 1)3

2 which is undefined due to division by zero 0

(

)

(

)

2 2   d. Given y = x x 2 + 1 , then y ′ = 1 ⋅ x 2 + 1  + 2 x 2 + 1   

y′(0 ) = 02 + 1

3

)

2 −1

(

)

 ⋅ 2 x = x2 + 1 

2

(

)

+ 4 x x 2 + 1 . Therefore,

+ (4 ⋅ 0 ) 02 + 1 = 12 + 0 = 1

Hamilton Education Guides

413

Calculus I

Chapter 2 Solutions

(

)

y′(1) = 12 + 1

2

y′(−1) =

2

(

)

+ (4 ⋅ 1) 12 + 1 = 22 + 4 ⋅ 2 = 4 + 8 = 12 and

[(− 1) + 1]

2

[

]

+ (4 ⋅ −1) (− 1)2 + 1 = (1 + 1) 2 − 4 ⋅ (1 + 1) = 22 − 4 ⋅ 2 = 4 − 8 = −4

(

)

(

3

)

e. Given y = x3 + 2 x 2 + 1 , then y ′ = 3 x3−1 + 2 ⋅ 3 x 2 + 1

(

)

y′(0 ) = 3 ⋅ 02 + (12 ⋅ 0 ) 02 + 1

(

)

y′(1) = 3 ⋅ 12 + (12 ⋅ 1) 12 + 1

2

2

=

3x 4

[

]

=

(1 + x ) (1 + x ) 6 ⋅ 05

y′(0 ) = y′(1) =

(1 + 0 ) (1 + 1 )

2 4

0

=

2 4

6 ⋅ 15

14 6

=

2

(3 ⋅ 2)x 4+1

6 ⋅ (− 1)

(1 + x )

[1 + (− 1) ]

3−1

(

[2 x ⋅ (1 + x )]− [2 x ⋅ x ] ⋅ (1 + x ) 2

2

2 2

6 x5

(1 + x )

2 4

4+ 2

2

6 ⋅ −1

(1 + 1)

= −

4

6 2

4

= −

3 6 = − = −0.375 8 16

[1 ⋅ (x + 1)]− [2 x ⋅ x] = 5 x  ⋅ x + 1 − 2 x    x +1 (x + 1) (x + 1) 2

5 −1

⋅

4

2

2

y′(1) =

6

6

2

2

(

2

) (

) (3x

2 x2 + 1

=

3

4

2

2

) ( 2

1

2

) = 2⋅2

0

2

)

− x2 − 1

=

(

)

) (2 x − 1)

2 x2 + 1

2

x3

(

)

(

)

3

2

. Therefore,

2

(− 1)3

Hamilton Education Guides

0

0

⋅ (2 − 1) 2 ⋅ 4 ⋅ 1 8 = = = 8 and 1 1 1

] ⋅ [2 ⋅ (− 1) − 1]

2 (− 1) + 1 2

2

(

3   2 ⋅ 2 x ⋅ x 2  − 2 x ⋅ x 2 + 1  2 3 2    = 6x x + 1 − 2x x + 1 4 4 x x

2

2 1 + 1 ⋅ 2 ⋅1 − 1

y′(−1) =

2

) = 2 ⋅1 ⋅ (0 −1) = 2 ⋅ −1 = − 2 which is undefined due to division by zero

2

3

2

x3

2 02 + 1 ⋅ 2 ⋅ 0 2 − 1 0

3−1

2

x2

x2

x 4/ =3

[

(x + 1) (x + 1)

6

3

2

3

) [3x − (x + 1)]

(

1 − x2

2

2

2

⋅

6

2

(

2

5x4

2

4

y′(0 ) =

2

=

6

6

2

(

2

2

2

4

2 x/ x 2 + 1

2

6

6

2

=

)

2

2

4

(

. Therefore,

) = 5x (1 − x ) . Therefore, (x + 1) (x + 1) 0 5 ⋅ 0 ⋅ (1 − 0 ) 0 ⋅1 = = = 0 y′(0 ) = 1 1 (0 + 1) 5⋅0 0 5 ⋅ 1 ⋅ (1 − 1 ) 5 ⋅ (1 − 1) = = = = 0 and y′(1) = 64 64 2 (1 + 1) 0 5 ⋅ 1 ⋅ (1 − 1) 5 ⋅ (− 1) ⋅ (1 − (− 1) ) 5⋅0 = = = = 0 y′(−1) = 64 2 (1 + 1) ((− 1) + 1)  3(x + 1) ( x + 1) 1 h. Given y = (x + 1) ⋅ = y= , then y ′ =  4

2

 x 2  2 x + 2 x3 − 2 x3  ⋅ = 3 2  1 + x2    1 + x2

6 3 = = 0.375 and 16 8

5

5x4 1 − x2

. Therefore,

= 3 ⋅ 1 − 12 ⋅ (1 + 1)2 = 3 − 12 ⋅ 22 = 3 − 12 ⋅ 4 = 3 − 48 = −45

 x   x   , then y ′ = 5 g. Given y =   2  x2 + 1   x +1 

=

2

0 = 0 1

=

=

2 4

=

2 2+ 2

=

4

5

y′(− 1) =

2

3

2 2

)

= 3 + 12 ⋅ 22 = 3 + 12 ⋅ 4 = 3 + 48 = 51 and

 2    , then y ′ = 3 x    1 + x2    

2x

⋅

2 2

(

⋅ 2 x = 3 x 2 + 12 x x 2 + 1

= 3 ⋅ 0 + 0 ⋅12 = 0 + 0 = 0

y′(−1) = 3 ⋅ (− 1) 2 + (12 ⋅ −1) (− 1) 2 + 1

 x2 f. Given y =   1+ x2 

3−1

2

=

2(1 + 1)2 ⋅ (2 ⋅ 1 −1) 8 2 ⋅ 22 ⋅ (2 −1) 2 ⋅ 4 ⋅1 = = = − = −8 −1 −1 −1 1

414

Calculus I

Chapter 2 Solutions

2

 x3   3   + 5 x , then y ′ = 2 x  i. Given y =     x −1  x −1   

[3x

2 −1

⋅

2

] [ ] + 5 = 2

⋅ (x − 1) − 1 ⋅ x3

(x − 1)

x3   3 x3 − 3 x 2 − x3  ⋅ +5  x −1  (x − 1)2    

2

(

)

2 x3 x 2 (2 x − 3) 2 x3 ⋅ x 2 2 x3 − 3x 2 2 x3 2 x3 − 3x 2 2 x3+ 2 (2 x − 3) 2 x5 (2 x − 3) ⋅ +5 = +5 = +5= + 5 . Therefore, +5 = ⋅ 2 2 2 2 1 + x − 1 (x − 1) x − 1 (x − 1) (x − 1) (x − 1) (x − 1) (x − 1) 3

=

2 ⋅ 05 ⋅ (2 ⋅ 0 − 3)

y′(0 ) = y′(1) =

(0 − 1)

2 ⋅ 15 ⋅ (2 ⋅ 1 − 3)

(1 − 1) 3

0 ⋅ (0 − 3)

+5 =

3

+5 =

1

2 ⋅ (2 − 3)

2 ⋅ (− 1) 5 ⋅ [(2 ⋅ −1) − 3]

y′(−1) =

(− 1 − 1)

3

+5 =

03

+5 =

0 +5 = 0+5 = 5 1

+5 =

3

2 2 ⋅ −1 + 5 = − + 5 which is undefined due to division by zero and 0 0

2 ⋅ −1 ⋅ [− 2 − 3]

+5 =

(− 2) 3

−2 ⋅ −5 10 5 +5 = + 5 = + 5 = 1.25 + 5 = 6.25 8 4 −8

3. Use the chain rule to differentiate the following functions. 2 d 3  3 d 2 2 3 2 t ⋅ dt (t + 1)  − (t + 1) ⋅ dt t   (t + 1)3      = t ⋅ 3(t + 1) − (t + 1) ⋅ 2t   = t4 t4  t 2 

[

d a. dt

(t + 1)2 (3t − 2t − 2)

=

t

3

][

= d c. dx = d.

(

)

2

18u 5 u 2 + 1 − 12u 3

d dx

9u

8

3t 2 (t + 1)2 − 2t (t + 1)3 t

4

=

t/ (t + 1)2 [3t − 2(t + 1)] t 4/ =3

t3

) ( ) ( ) (u + 1) = 6u (u + 1) [3u (

)

=

(t + 1) 2 (t − 2)

=

3  3 d  4 d 2  u +1  −  u 2 +1 3u 3u 4  3  2  du du d  u +1      = = b. du  3u 4  4 2 3u  

(

]

2

3

3/

2

2

9u

2

(

(

)

(

(

)(

)

3−1 3   4   2 ⋅ 2u  −  u 2 + 1 ⋅ (3 ⋅ 4 )u 4−1 ⋅1 3u ⋅ 3 u + 1    9u 8

)]

− 2 u2 + 1

8/ =5

2

2 u 2 + 1 3u 2 − 2 u 2 − 2

=

3u

5

) = 2(u + 1) (u − 2) 2

2

3u

2

5

 3 d 2 2 d 3  2 3−1 3 2 −1 (1 − x ) dx (2 x + 1)  − (2 x + 1) dx (1 − x )   (2 x + 1)3   = (1 − x ) ⋅ 3(2 x + 1) ⋅ 2 − (2 x + 1) ⋅ 2(1 − x ) ⋅ −1    =   (1 − x )4 (1 − x )4  (1 − x )2 

][

[

6(1 − x ) 2 (2 x + 1) 2 + 2(2 x + 1) 3 (1 − x )

=

(1 − x )4

(

)

2(1/ − x/ )(2 x + 1) 2 [3(1 − x ) + (2 x + 1) ]

(1 − x ) 4/ =3

(

)

(

=

]

2 (2 x + 1) 2 (4 − x )

(1 − x )3

(

)

)

2  3 2 d 2 2 −1    3  3 d 3 3 3 3 3 ⋅ 3x 2   x − 1 (2 x + 1)  = (2 x + 1) dx x − 1  +  x − 1 dx (2 x + 1)  = (2 x + 1) ⋅ 2 x − 1     

(

[

)

(

)] (

(

)

)[

(

)]

2 2     +  x3 − 1 ⋅ 3(2 x + 1)3−1 ⋅ 2 = 6 x 2 (2 x + 1)3 x3 − 1 + 6 x3 − 1 (2 x + 1)2  = 6(2 x + 1)2 x3 − 1 x 2 (2 x + 1) + x3 − 1    

(

)(

(

)

)(

)

= 6(2 x + 1)2 x3 − 1 2 x3 + x 2 + x3 − 1 = 6(2 x + 1) 2 x 3 − 1 3 x 3 + x 2 − 1 e.

2

d  3 1  1   3 s − 2  = 2 s − 2  ds  s + 6 s + 6 

(

)

(

2 −1

d d 2  2  3−1 s + 6 ⋅ ds (1) − 1 ⋅ ds s + 6 ⋅ 3s − 2  s2 + 6 

(

(

)

) (

) (

⋅

1  d  3 1  1  d 3 d 1   3  3 s − 2  = 2 s − 2  = 2 s − 2   ⋅  s − ds 2 ds  s + 6 s + 6 s + 6 s + 6   ds  

)

  3 1   2 2s 1   2 0 − 2 s   3 = − ⋅ 2 s 3s +  = 2 s − 2    ⋅ 3s − 2 2 2  s + 6  s + 6   s +6  s2 + 6   

( )

(

) (

)

(

)

) (

(

) (

(

)

  2 

)

3  3 d 2 3−1 3 d 2  2   2    t +1 t −1  −  t2 −1 t +1  3 t +1 ⋅ 3 t2 −1 ⋅ 2t  −  t 2 − 1 ⋅ 2t   2   d  t −1  dt dt         f. = = 2 2 2 2 dt  t 2 + 1  t +1 t +1  

(

)

Hamilton Education Guides

)

415

Calculus I

Chapter 2 Solutions

(

)(

)

(

)

) [( ) ( ( )

(

)]

2  3  2 2 2 2 6t t + 1 t − 1  − 2t t − 1  2t t 2 − 1 3 t 2 + 1 − t 2 − 1 = = 2 2 t2 +1 t2 +1

(

)

=

)[ ] = 4t (t − 1) (t (t + 1) (t + 1) d (u + 1)  − (u + 1) dud (u + 1)  du

(

2

2t t 2 − 1 3t 2 + 3 − t 2 + 1

(

)

(

(

)

 2 2 (u + 1) ⋅ 3 u + 1 =

=

)[

(

=

(

2θ 2 + 3

(

3 −1

)

  ⋅ 2u  −  u 2 + 1   (u + 1)4

3

(u + 1)

(θ − 1)

(

3

2

 θ2 +3  = 2  3  (θ − 1) 

(θ − 1)

3

2 −1

] [(

(θ − 1) 2

)]

(θ − 1)

(

(

)

(

(

7 r 6 r 2 + 2r

=

(

)

(

) = 2(u + 1) (2u

3

2

2

+ 3u − u 2 − 1

(

]

=

)

2

)

+ 3u − 1

(u + 1)

3

) (

7

(

)

)(

(

(θ − 1)

(

)

2

6

2

)

(

)

=

[

)

(

3+ 4

) = − 2(θ

2

)(

+ 3 ⋅ θ 2 + 2θ + 9

(θ − 1) 7

)

(

)

(

(

) − [r ⋅ 3r (r + 2)⋅ 2(r + 1) ] = 7r (r (r + 2r )

(

7 r 6 r 2 + 2r

3

)

2 −1

6

6

2

(

)

⋅

d 2  r + 2r  dr 

2

+ 2r

)

7

)

) − 6r (r + 2)(r + 1) (r + 2r ) 3

8

6

2

4. Given the following y functions in terms of u , find y ′ . a. Given y = 2u 2 − 1 and u = x − 1 , then y = 2(x − 1)2 − 1 and y ′ = 2 ⋅ 2(x − 1)2 −1 − 0 = 4( x − 1) b. Given y = =

[

(x − 1) 3

c. Given y =

)] [ ( )

(

3 x3−1 ⋅ x3 − 1 − 3 x3−1 ⋅ x3 x3 u and u = x 3 , then y = 3 and y ′ = 2 u −1 x −1 x3 − 1

3/ x5/ − 3 x 2 − 3/ x5/ 2

u

1 + u2

(

= −

3x2

(x − 1) 3

and u = x + 1 , then y =

) [ ( ( )

)(

)]

x2 + 1

2

3

2

3

3

2

(

(

)

1 + x2 + 1

2

(

)

(

)

(

)

2  2 −1   ⋅ 2x ⋅ x2 + 1  2 x ⋅ 1 + x 2 + 1  − 2 x 2 + 1     and y ′ = 2 2  2 1 + x + 1 

(

2  2 2 2 2 2 2 x + 2 x x + 1  − 4 x x + 1 ⋅ x + 1 2x + 2x x2 + 1 − 4x x2 + 1 = = = 2 2 2 2   2 2 1 + x + 1  1 + x + 1  1 1 d. Given y = u 2 − and u = x 4 , then y = x8 − and y ′ = 8 x8−1 − 0 = 8x 7 2 2

Hamilton Education Guides

] = [3x (x − 1)]− [3x ⋅ x ] (x − 1)

2

2

)]

2/ 2 2 θ 2 + 3 (θ − 1) 2θ (θ − 1) − 3 θ + 3 2 θ 2 + 3 2θ (θ − 1)3 − 3 θ 2 + 3 (θ − 1)2 = ⋅ ⋅ (θ − 1) 6/ = 4 (θ − 1) 3 (θ − 1)3 (θ − 1)6

2 θ 2 + 3 ⋅ 2θ 2 − 2θ − 3θ 2 − 9

) − [r ⋅ 3(r + 2r )⋅ (2r + 2) ] (r + 2r ) 3

2

2  3   2 2 2 6u (u + 1) u + 1  − 2 u + 1 (u + 1) (u + 1)4

3 3 d 7 3  2 7 −1   7 2 d 2 r 2 + 2r r − r7 r + 2r   r + 2r ⋅ 7 r  − r ⋅ 3 r + 2r  dr dr  = = 6 6 3 2 r + 2 r r 2 + 2r 

)

)

(u + 1)4

(

6

+3

3

2

+2

 (θ − 1)3 d θ 2 + 3  −  θ 2 + 3 d (θ − 1)3  2 2      d θ +3 θ +3  dθ dθ    ⋅   = 2 ⋅ 3 dθ  (θ − 1) 3  (θ − 1)6  (θ − 1) 

)

4

2

(u + 1)

⋅ 2θ − θ 2 + 3 ⋅ 3(θ − 1) 2

) ⋅ [2θ (θ − 1) − 3(θ

2θ 2 + 3 3

4/ =3

2

2

2

2

3

2

)

 ⋅ 2(u + 1) 2 − 1   =

)] = 2(u + 1) (3u

(

2

) ⋅ [(θ − 1)

 d  r7 i. dr  r 2 + 2r  =

)

2(u/ + 1/ ) u 2 + 1 3u (u + 1) − u 2 + 1

d θ 2 +3  h.   dθ  (θ − 1)3  =

(

2

2

 2 3 2  (u + 1) 3 1   3 d  u2 + 1  d  2 d  2 1    u +1  g. = =   =  u +1 2 du  u +1   du  (u + 1)2  du   ) ( u 1 +      

2 2

2

(

)

)

(

)

(

)

2

)

2x − 2x x + 1

(

)

2  2 1 + x + 1 

2

2

(

)

2  2 x 1 − x 2 + 1    = 2 2   2 1 + x + 1   

(

)

416

Calculus I

Chapter 2 Solutions

e. Given y = u

=

(

8x 1 − x

)

2 3/

(1 − x )

and u = 8x

=

2 8/ =5

f. Given y =

=

4

1 1− x 2

(

(1 − x )

2 4

 2 0 ⋅ 1 − x and y ′ =

)  − 4(1 − x ) (1 − x )

2 4 −1

4

(u + 1)

3

 ⋅ −2 x ⋅ 1 2  = 0 + 8x 1 − x 8 1 − x2

(

2 8

(1 − x )

u2

(

)

)

3

2 5

and u = x − 1 , then y =

2 x3 (x − 1) − 3 x 2 (x − 1)2 x

, then y =

1

x 2/ (x − 1) [2 x − 3(x − 1) ]

=

6

(x − 1)2 (x − 1 + 1)3 =

x 6/ = 4

=

(x − 1)2 x

3

and y ′ =

(x − 1) (2 x − 3x + 3) x4

=

[2(x − 1)

2 −1

][

⋅ x3 − 3 x 2 ⋅ (x − 1) 2 x

]

6

( x − 1)(− x + 3) x4

Section 2.5 Solutions - Implicit Differentiation Use implicit differentiation method to solve the following functions. a. Given x 2 y + x = y , then ; y′ =

2x y + 1

)

(

(

(

)

d 2 d ( y ) ; 2 x ⋅ y + x 2 ⋅ y′ + 1 = y′ ; 2 x y + 1 = y′ − x 2 y′ ; 2 x y + 1 = y′ 1 − x 2 x y+x = dx dx

1 − x2

(

)

)

d d (0) ; (1 ⋅ y + x ⋅ y′) − 6 x + y′ = 0 ; y − 6 x = − x y′ − y′ ; y − 6 x = − y′(x + 1) x y − 3x 2 + y = dx dx y − 6x y − 6x 6x − y ; ; y′ = = − y′ ; y′ = − x +1 x +1 x +1

b. Given x y − 3 x 2 + y = 0 , then

c. Given x 2 y 2 + y = 3y 3 , then

(

)

( ) (

)

d 2 2 d x y +y = 3y 3 ; 2 x ⋅ y 2 + 2 y y′ ⋅ x 2 + y′ = 9 y 2 y′ ; 2 x y 2 + 2 x 2 y y′ = 9 y 2 y′ − y′ dx dx

(

)

; 2 x y 2 = 9 y 2 y′ − y′ − 2 x 2 y y′ ; 2 x y 2 = y′ 9 y 2 − 1 − 2 x 2 y ; y ′ =

(

(

)

9 y2 − 1 − 2 x2 y

d d (5 x ) ; (1 ⋅ y + y′ ⋅ x ) + 3 y 2 ⋅ y′ = 5 ; y + y′x + 3 y 2 y′ = 5 ; x y′ + 3 y 2 y′ = 5 − y x y + y3 = dx dx 5− y

d. Given x y + y 3 = 5 x , then ; y′ x + 3 y 2 = 5 − y ; y′ =

)

2 x y2

x + 3 y2

e. Given 4 x 4 y 4 + 2 y 2 = y − 1 , then

(

(

)

)

d d ( y − 1) ; 4 4 x3 ⋅ y 4 + 4 y 3 y′ ⋅ x 4 + 4 y y′ = y′ ; 16 x3 y 4 + 16 x 4 y 3 y′ 4x4 y 4 + 2 y 2 = dx dx

(

)

+4 y y′ = y′ ; 16 x3 y 4 = y′ − 16 x 4 y 3 y′ − 4 y y′ ; 16 x3 y 4 = y′ 1 − 16 x 4 y 3 − 4 y ; y′ =

(

(

)

(

)

+ 2 x 2 y y′ = 0 ; y + 2 x y 2 = − y′x − 2 x 2 y y′ ; y + 2 x y 2 = − y′ x + 2 x 2 y ; y′ = −

(

h. Given x y 3 + x 3 y = x , then

( ) (

)

y + 2 x y2 x + 2x2 y

)

(

(

)

) (

)

d d (x ) ; 1 ⋅ y 3 + 3 y 2 y′ ⋅ x + 3 x 2 ⋅ y + y′ ⋅ x 3 = 1 ; y 3 + 3 x y 2 y′ + 3 x 2 y + x 3 y′ = 1 x y 3 + x3 y = dx dx

(

)

; 3 x y 2 y′ + x 3 y′ = 1 − y 3 − 3 x 2 y ; y′ 3 x y 2 + x 3 = 1 − y 3 − 3 x 2 y ; y ′ =

Hamilton Education Guides

)

d 2 d 2 x − y2 x ; 1 ⋅ y 2 + 2 y y′ ⋅ x + y′ = 2 x ; 2 y y′x + y′ = 2 x − y 2 ; y′ = x y2 + y = dx dx 2x y + 1

g. Given x y 2 + y = x 2 , then

1

1 − 16 x 4 y 3 − 4 y

d d (0) ; (1 ⋅ y + y′ ⋅ x ) + 2 x ⋅ y 2 + 2 y y′ ⋅ x 2 = 0 ; y + y′x + 2 x y 2 x y + x 2 y 2 − 10 = dx dx

f. Given x y + x 2 y 2 − 10 = 0 , then

i. Given y 2 + x 2 y = x , then

16 x 3 y 4

(

1 − y3 − 3x2 y 3x y2 + x3

)

 d 1 12 −1 1 − 12 d  12 2  y + x2 y  = ′ ′ ( ) ; ; y y x y y x ⋅ + ⋅ + ⋅ = 2 1 y y′ + 2 x y + x 2 y′ = 1 x  dx 2 2 dx  

417

Calculus I

Chapter 2 Solutions

   1 −1   1  1 − 2 xy ; y′ y 2 + x 2  = 1 − 2 xy ; y′ 1 + x 2  = 1 − 2 xy ; y′ = 1 2  2y 2    + x2   2 y

(

)

(

)

1 −3 1 1 −1 d 2 d  14   y  ; 2 x ⋅ y + y′ ⋅ x 2 + 2 y y′ = y 4 y′ ; 2 x y + x 2 y′ + 2 y y′ = y 4 y′ x y + y2 =   4 4 dx dx  

1

j. Given x 2 y + y 2 = y 4 , then

 1 − 34 1 −3 −2 x y −2 x y y y′ = −2 x y ; y′ x 2 + 2 y − y 4  = −2 x y ; y′ = ; y′ = 3 1 − 1 4 4 2 2   x + 2y − x + 2y − y 4 4 3 4 4 y

; x 2 y′ + 2 y y′ −

k. Given x + y 2 = x 2 − 3 , then l. Given x 4 y 2 + y = −3 , then

(

)

(

)

d d 2 2x − 1 x + y2 = x − 3 ; 1 + 2 y y′ = 2 x ; 2 y y′ = 2 x − 1 ; y ′ = dx dx 2y

(

(

)

)

d 4 2 d (− 3) ; 4 x3 ⋅ y 2 + 2 y y′ ⋅ x 4 + y′ = 0 ; 4 x3 y 2 + 2 x 4 y y′ + y′ = 0 x y +y = dx dx

(

)

− 4 x3 y2

; 2 x 4 y y′ + y′ = −4 x3 y 2 ; y′ 2 x 4 y + 1 = −4 x3 y 2 ; y′ =

(

2 x4 y + 1

)

(

)

d 7 d (8) ; 7 y 6 y′ − 2 x ⋅ y 4 + 4 y 3 y′ ⋅ x 2 − 1 = 0 ; 7 y 6 y′ − 2 x y 4 − 4 x 2 y 3 y′ = 1 y − x2 y 4 − x = dx dx

m. Given y 7 − x 2 y 4 − x = 8 , then

(

)

; 7 y 6 y′ − 4 x 2 y 3 y′ = 1 + 2 x y 4 ; y′ 7 y 6 − 4 x 2 y 3 = 1 + 2 x y 4 ; y′ =

[

]

(

1 + 2 x y4

7 y6 − 4 x 2 y3

)

n. Given (x + 3)2 = y 2 − x , then

d (x + 3)2 = d y 2 − x ; 2(x + 3) = 2 y y′ − 1 ; 2 x + 6 + 1 = 2 y y′ ; y′ = 2 x + 7 dx dx 2y

o. Given 3 x 2 y 5 + y 2 = − x , then

d d (− x ) ; 3 2 x ⋅ y 5 + 5 y 4 y′ ⋅ x 2 + 2 y y′ = −1 ; 6 x y 5 + 15 x 2 y 4 y′ + 2 y y′ = −1 3x 2 y 5 + y 2 = dx dx

(

(

)

(

)

)

; 15 x 2 y 4 y′ + 2 y y′ = −1 − 6 x y 5 ; y′ 15 x 2 y 4 + 2 y = −1 − 6 x y 5 ; y′ = −

1 + 6 x y5

15 x 2 y 4 + 2 y

Section 2.6 Solutions - The Derivative of Functions with Fractional Exponent 1. Find the derivative of the following exponential expressions. 1 1 1 −1 1 1−5 1 −4 a. Given y = x 5 , then y′ = x 5 = x 5 = x 5 5 5 5

( )

b. Given y = 4 x3

1 2

3

= 4 x 2 , then y′ =

2 1− 3 1 1 (2 x + 1)3 −1 ⋅ 2 = 2 (2 x + 1) 3 = 2 (2 x + 1) − 3 3 3 3

1

c. Given y = (2 x + 1)3 , then y′ =

(

)

1

d. Given y = 2 x 2 + 1 8 , then y′ =

(

e. Given y = 2 x 3 + 3 x

(

f. Given y = x 3 + 8

( )

g. Given y = x3 =

1 2

)

2 3

)

3 5

(

)

(

(

3 3 2 x + 3x 5

2 3 x +8 3

, then y′ = 3

(

(

)

)

1−8 1 −1 4x 1 2 x 2x2 + 1 8 = 2x + 1 8 ⋅ 4x = 2x2 + 1 8 2 8

, then y′ =

1

1 3 −1 3 2 3− 2 3 ⋅ 4 x 2 = ⋅ 4/ x 2 = 6 x 2 2/ 2

)

) ⋅ (6 x + 3) = 53 (2 x 3 −1 5

2 −1 3 ⋅ 3x 2 1

2

=

(

2 3 x +8 3/

− (3 x − 1)3 = x 2 − (3 x − 1)3 , then y′ =

)

2 −3 3

3

+ 3x

)

−2 5

−7

8

(

)

⋅ 3 2x2 + 1 =

(

⋅ 3/ x 2 = 2 x 2 x 3 + 8

)

(

)(

9 2x2 + 1 2x3 + 3x 5

)

−2 5

−1

3

1− 3 1 3 3− 2 1 3 32 −1 1 x − (3 x − 1)3 −1 ⋅ 3 = x 2 − (3 x − 1) 3 ⋅ 3/ 2 3 2 3/

2 3 12 x − (3 x − 1) − 3 2 1

1

h. Given y = x 2 (x + 1) 8 , then y′ = 2 x ⋅ (x + 1)8 +

Hamilton Education Guides

2 2 1 7 1−8 1 1 1 (x + 1)8 −1 ⋅ x 2 = 2 x (x + 1)8 + x (x + 1) 8 = 2 x ( x + 1)8 + x ( x + 1) − 8 8 8 8

418

Calculus I

Chapter 2 Solutions

(

)

2 5

i. Given y = x 3 + 1

)

(

)

(

)

(

2 −5 2 −1 6 2 3 1 1 −1 1 1− 2 6 x + 1 5 ⋅ 3x 2 + x 2 = x 2 x3 + 1 5 + x 2 = x 2 x 3 + 1 5 5 2 2 5

1

+ x 2 , then y′ =

−3 5

+

1 − 12 x 2

2    2 2 2 2 −1 2 −1 2 2 −3 1 ⋅ x 3  −  x 3 ⋅ (x + 1) x 3 − x 3 ( x + 1) x 3 − x 3 (x + 1) 3 x +1  =   3 3 j. Given y = 2 , then y′ =  = 4 4 4 3 3 3 x x x3 x

(x + 1) 2

k. Given y = =

(

x3 x2 + 1

)

1 2

x2

−1

(

x4

)

)

(

(

)

(

)

)

1− 2   1  2/ x3 2 x + 1 2  − 2 x ⋅ x 2 + 1 2    2/    =  x4

x4

1 2

1   1 1 1 −1 2 2 −1 1 − 23 3 ( x + 1) − 2(x + 1) ⋅ x 3  −  x 3 ⋅ (x + 1)  2 x x ( x + 1) 2 3     3 , then y′ = = 2 2 x3 x3

1

x3

2. Use the

, then y′ =

− 2x x2 + 1

2

(x + 1)2

l. Given y =

(

1 −1 1 1 2 2  2  x + 1 2 ⋅ 2 x ⋅ x  − 2 x ⋅ x + 1 2     2

d notation to find the derivative of the following exponential expressions. dx

d  15 x a. dx 

2

 1   = 2  x 5    

2 −1

⋅

1 2 1−4 1 1 −1 2 1 −4 d 15 2 −3 x = 2 x 5 ⋅ x 5 = x 5 ⋅ x 5 = x 5 5 = x 5 , or 5 5 5 5 dx

2

 d 52 2 2 −1 2 2 −5 2 −3  = x = x5 = x 5 = x 5 5 5 dx 5  1 1 1 −1 d 1 1 1 d (x − 1) 2 = (x − 1)2 ⋅ (x − 1) = (x − 1) − 2 ⋅ 1 = 1 ( x − 1) − 2 b. dx 2 2 2 dx d  15 x dx 

(

)

(

)

(

)

(

)

(

(

)

)

−2

c.

1− 3 1 −1 1 1 1 d 2 2x 2 d 2 x + 1 3 = x2 + 1 3 ⋅ x +1 x + 1 = x2 + 1 3 ⋅ 2x = dx 3 dx 3 3

d.

−1− 4 − 1 −1 d −1 1 1 3x2 3 d 3 x + 1 4 = − x3 + 1 4 ⋅ x3 + 1 = − x3 + 1 4 ⋅ 3x 2 = − x +1 dx 4 dx 4 4

(

1 d  (x − 1)2 e. dx  x 2 

(

)

1 8

d 3 x + 2x dx

g.

d  3 2  x +1 x dx  = 3x

(

(

1 3 x + 2x 8

=

)

1 −1 8 ⋅

(

1 3

3

2 3

(

)

)

−5 4





 d  3 1 h. x ⋅ dx  x 2 +1 

(

)

)

1 2

  3  = d  x 1  dx  2  x + 1 2 

(

(

)

)

1  1− 2   2 2 4 2 3 x x + 1 2  −  x x + 1 2     = =  x2 + 1

Hamilton Education Guides

(

(

−1

)

(

)

) ⋅ (3x −7 8

) (

(

)

 3x2 + 2  3  x + 2x +2 =    8  

2

)

(

)

−7

8

)

 2 d 3     2   d 2 2 2 −1  x + 1  +  x3 + 1 ⋅ x 3  =  x 3 ⋅ 3x 2  +  x3 + 1 ⋅ x 3   = x 3 ⋅ 3 dx dx         

8 −1 2 3 2x 3 + x +1 ⋅ x 3 = 3x 3 + 3 3

(

(

1 d 3 x + 2 x = x3 + 2 x 8 dx

)( )  = dxd (x + 1)x

2+ 2

)

1  1 d 1 −1   1  2 1  2 d 1 1  2 x2 ( x − 1) − 2 − 2 x ( x − 1) 2   x ⋅ 2 (x − 1)2  − (x − 1)2 ⋅ 2 x   x dx (x − 1)2  − (x − 1)2 dx x        2  = = =  x4 x4 x4 

f.

(

)

3

3

(

)

(

)

⋅ x3 + 1

(

)

(

)

(

)

1 1 1 1 −1  2   2 d 3  3 d 2 2  3 1 2 x +1 2   x + 1 2 ⋅ x  − x ⋅  x + 1 2 ⋅ 3x  −  x ⋅ x + 1 2 ⋅ 2 x   2 dx   dx  =      =  2 2  x +1 x +1 

(

)

(

1   2 2 4 2 3 x x + 1 2  −  x x + 1    x2 + 1

)

−1  2

 

419

Calculus I

Chapter 2 Solutions

 d  x5 i. dx  3  x + 1

)

(

(

=

j.

(

2 3

)

)

(

(

(

)

)

2 2 2 2 −1  3  3 d 5  5 d 3 4  5 2 3 2 x +1 3    x + 1 3 ⋅ x  − x ⋅  x + 1 3 ⋅ 5 x  −  x ⋅ x + 1 3 ⋅ 3x  dx   dx 3  =   =     4 4  3 3 3 3 x +1 x +1 

(

)

(

)

)

2 −3  2   4 3 2 x7 3 x +1 3  5 x x + 1 3  −     3 

(x + 1)

4 3

3

(

(

)

(

)

2   4 3 7 3 5 x x + 1 3  −  2 x x + 1     = 4 x3 + 1 3

(

)

−1  3

)

 

1 1 1 1 1 1 1 1 1 1 d  (x − 1) 2 (x + 1) 3  = (x + 1)3 d (x − 1)2  + (x − 1)2 d (x + 1)3  = (x + 1)3 ⋅ 1 (x − 1)2 −1  + (x − 1)2 ⋅ 1 (x + 1)3 −1   dx dx dx  2 3          2 1 1 1 1− 3  1 1− 2  1 1 1 1 1 =  (x + 1)3 ⋅ (x − 1) 2  +  (x − 1)2 ⋅ (x + 1) 3  =  ( x + 1)3 ( x − 1) − 2  +  ( x − 1) 2 ( x + 1) − 3  2  3   2  3

k.

d dx

(

)

(

)

(

(

)

(

)

(

)

(

)

(

)

(

)

1− 2  1   = 3 x 2 x 2 + 1 2  +  x 4 x 2 + 1 2  = 3 x 2 x 2 + 1    

l.

)

(

)

(

)

1 1 1 −1 1 1      2  2  3 2 2  3 1 2 2 2 d x3 + x3 d x 2 + 1 2  =  x + 1 2 ⋅ 3x  +  x ⋅ x + 1 2 ⋅ 2 x    x x +1  =  x + 1 dx dx 2          

1 2

)

(

+ x4 x2 + 1

(

)

−1

2

(

)

(

)

 2     −1  − 1 d 3  3 d −1  −1 − 1 −1 d  3 2 1 x  + x x 2 + 1 3  =  x 2 + 1 3 ⋅ 3x 2  +  x3 ⋅ − x 2 + 1 3 ⋅ 2 x  x x + 1 3  =  x + 1 3 dx   dx 3 dx        

(

(

)

−1− 3   − 1   2x4 2 = 3 x 2 x 2 + 1 3  + − x + 1 3  = 3x2 x2 + 1     3

)

−1

3

−

(

2 x4 2 x +1 3

)

−4

3

Section 2.7 Solutions - The Derivative of Radical Functions 1. Find the derivative of the following radical expressions. Do not simplify the answer to its lowest term.

(

)

1

a. Given y = x 2 + 1 = x 2 + 1 2 , then y′ =

b. Given y = x 3

(

(

)

)

1 −1 −1 1 2 2/ x 2 x + 1 2 ⋅ 2x = x +1 2 = 2/ 2

x

(x + 1) 1 1 + 3 x − 5 = (x + 3 x − 5) , then y′ = (x + 3 x − 5) ⋅ (3 x + 3) = (x + 3 x − 5) (3 x + 3) = 2 2 1 2

3

1 −1 2

3

2

1

1

1 2

2

−1

3

d. Given y =

e. Given y =

f. Given y =

g. Given y =

(x + 1) x +1 = x x x

2

x 2 −1

=

x3 + 3x 2 x2 + 3 x +1

x

(

1

2

Hamilton Education Guides

)

3

= x 2 + 3 x 2 , then y′ = 1

=

2 x3 + 3x − 5

)

1 2

1

( x − 1) 2

(

)

1 −1 1   1 2 2 2 1 2 x ⋅ x − 1 2  −  x − 1 2 ⋅ 2/ x ⋅ x  2/ 2x x2 − 1 2 − x3 x2 − 1     , then y′ = = x2 − 1 x2 − 1

(x − 1) = (x ) + 3 x 1 2

3 2

(

)

1 −1 1 1   1 1 x ( x + 1) − 2 − ( x + 1) 2  2 (x + 1)2 ⋅ x  − 1 ⋅ (x + 1)2    2 , then y′ = = x2 x2

2

2

(

3 x2 + 1

1

1

c. Given y = x 2 + x − 1 = x 2 + (x − 1)2 , then y′ = 2 x 2 −1 + (x − 1) 2 −1 = 2 x + (x − 1) − 2 = 2 x +

1 2

2

2

(

)

(

)

−1

2

3 1 3 32 − 1 3 3− 2 x + (3 ⋅ 2 )x 2 −1 = x 2 + 6 x = x 2 + 6 x 2 2 2 1 −1

 x2 + 3  2  2 2 x2 + 3  , then y′ = 1  x + 3  =   x +1  x +1 2  x + 1   

⋅

(

) = 1  x

2 x ⋅ (x + 1) − 1 ⋅ x 2 + 3

(x + 1)

2

2

+ 3  2  x + 1 

−1

2

⋅

x2 + 2x − 3

( x + 1) 2

420

Calculus I

Chapter 2 Solutions

4

h. Given y =

i. Given y =

2. Use the

x 3 −1

x2 x

3

1

x

x3

(x − 1) =

1 4

(

)

1 3  x −1 4 , then y′ = 

1 −1 4 ⋅ 3x 2

x2 =

x3/ =1

=

1

x 2/ ⋅ x 2

x 1

= x⋅x

−1

2

1− 1

= x

2

(

)

5

1 1  1 1 −1 3x 2 3 ⋅ x 2  −  x 2 ⋅ x3 − 1 4  x −1 2    = 4 x

1

= x 2 , then y′ =

x2

(

)

−3 4

− x

(

1 − 12 3 x x −1 2

)

1 4

1 1− 2 1 12 − 1 1 1 −1 1 −1 x = x2 1 = x 2 = x 2 2 2 2 2

d notation to find the derivative of the following radical expressions. dx

(0 ⋅ x ) − (1 ⋅ 1) = 1 − 1 d d 1 d  1 d  2 1 x +   = 1+ x+  =  x +  = dx  x dx dx  x  dx  x x2 x2

a.

 1 −1 1 −1 1 (x − 1) d x  −  x d (x − 1)    2 2 1 x  d  x  d  x  dx   dx  = 1  x  2 ⋅ [1 ⋅ (x − 1) ] − [1 ⋅ x ] b. = ⋅    =  2  x −1 2  x −1 dx  x − 1  dx  x −1  (x − 1) 2 (x − 1) 2 1 x  =   2  x −1

1− 2 2

x/ − 1 − x/ 1 x  = −  ⋅  2 2  x −1 (x − 1)

d  x3 d  x 3  c. = dx  (x + 1)12 dx  x +1  

−1

2

⋅

1

( x − 1) 2

1 d 1  (x + 1) 12 ⋅ 3 x3 −1  −  x3 ⋅ 1 (x + 1) 12 −1  3  3 d  (x + 1) 2 dx x  −  x dx (x + 1) 2      2    =    =   x +1 x +1 

3 3(x + 1) 12 x 2  −  x (x + 1) 1−22      2   = = x +1

d  (x + 5) 2 d  x + 5  d. = dx  x dx  x   1

 1   x3  2 −1 3 x ( x + 1) 2  −  2 ( x + 1) 2    x +1

1  1 d 1 1 −1    d   1 1 1  x ( x + 5) − 2 − ( x + 5) 2   x (x + 5) 2  − (x + 5) 2 dx x   x ⋅ 2 (x + 5) 2  − (x + 5) 2 ⋅ 1  =  dx    =   = 2  x2 x2 x2 

1   1 1 −1   −1 d 3 d − 12 d  3 d  3 d  3 x2  d  3 d  3 x   x + x 2 ⋅ x −1  = x + x2  = x + x 2  = = x + x e. x + x +  =         dx dx dx  x  dx  dx  dx  dx  x      

= 3 x3 −1 −

1 − 12 − 1 1 −1− 2 1 −3 = 3x 2 − x 2 = 3 x 2 − x 2 x 2 2 2

d  2 x f. 1+ dx  x3 =

1  2    = d 1 + 2 x 3  dx  x  

 1− 6  1 −3 1   −5 d  d  d  d  2  = 1 + 2 x 2  1 + 2 x 2  = 1 + 2 x 2 x −3  = + 1 2 x  =     dx  dx  dx  dx       

−5 − 2  −5  −7 − 5 −1 d (1) + d  2 x 2  = 0 − 5 x 2 ⋅ 2 = − 5 x 2 ⋅ 2/ = − 5 x 2 2/ 2 dx dx  

3. Find the derivative of the following radical expressions. 3− 2 3 −1 3 1 −1 1 1   3 −1 d  3  d (x ) ; d  x 2 + y 2  = 1 ; d x 2 + d y 2 = 1 ; 3 x 2 + 1 y 2 y′ = 1 ; 3 x 2 + 1 y 2 y′ = 1  x + y= 2 2 dx dx 2 2 dx  dx   dx 

a. ;

b.

1 −1 3 1 3 12 1 − 12 x + y y′ = 1 ; y 2 y′ = 1 − x 2 ; 2 2 2 2 d dx

(

)

x + y3 =

y′ 1 2y 2

= 1−

1 1 1 3 1 3 12 x ; y′ = 2 y 2 1 − x 2  ; y′ = 2 y 2 − 3( xy ) 2 2 2  

1 −1 1  1  −1 −1 d (2) ; d  x 2 + y3  = 0 ; d x 2 + d y 3 = 0 ; 1 x 2 + 3 y 2 y′ = 0 ; 1 x 2 + 3 y 2 y′ = 0 ; 3 y 2 y′ = − 1 x 2 dx 2 2 dx dx 2 dx  

Hamilton Education Guides

421

Calculus I

Chapter 2 Solutions

; 3 y 2 y′ = −

1

c.

d (x y ) = d dx dx

d.

d dx

(

y′

;

1 2y 2

1

6 x 2 y2

( x ) ; dxd (x y ) = dxd x

)

1 2

; y

  d d d 12 1 1 −1 1 −1 1 1  x+x y= x ; y + x y′ = x 2 ; x y′ = x 2 − y ; y′ =  1 − y  2 2 dx dx dx x 2   2x 

 d 12 d 3 1 12 − 1 1 − 12 d  12 2  y + x3  = 0 ; ′ ; ; y + x = 0 y y + 3 x = 0 y y′ = −3 x 2 ;  2 2 dx dx dx  

y + x3 = 0 ;

=

1

; y′ = −

1 2x 2

y′ 1 2y 2

= −3 x 2

1 − 3x 2 ; y′ = −6 x 2 y 2 1

(

)

e.

d  4 d 2 ( x ) ; d x 2 + y 2 = 1 ; d x 2 + d y 2 = 1 ; 2 x + 2 y y′ = 1 ; 2 y y′ = 1 − 2 x ; y′ = 1 − 2 x  x +y = dx dx dx dx  2y  dx

f.

d dx

(

)

( )

1 1 1 d 1 3 y 2 y′ d 3 (x + 1)2 = 3 y 2 d y ; 1 (x + 1) 2 − 1 = 3 y 2 y′ ; 1 (x + 1) − 2 = 3 y 2 y′ ; y ; = 1 dx dx dx 2 2 1 2(x + 1) 2

x +1 =

1

1

; 6 y 2 y′(x + 1) 2 = 1 ; y′ =

(

1

6 y ( x + 1) 2 2

)

1 −1 1   −1 d d (2) ; d  x y 2 + x 2  = 0 ; y 2 d x + x d y 2 + 1 x 2 = 0 ; y 2 ⋅ 1 + x ⋅ 2 y y′ + 1 x 2 = 0 x y2 + x = dx dx 2 dx dx 2 dx  

g.

−1

 1 −1 1 1  1 − 12 y y2 x 2 y2 1  − x − y 2  ; y′ = − ′ ; 2 x y y′ = − x 2 − y 2 ; y′ = ; ; y′ = − 3 − − y = − − 1   1 + 2 2x y  2 2x 4 xy 2 xy 2 xy  4x 2 y 4x 2 y 3 1 1 3 −1 d  3 d (x y ) = 0 ; d x 2 + d (x y ) = 0 ; 3 x 2 +  y d x + x d y  = 0 ; 3 x 2 + ( y + x y′) = 0 ; x y′ = − 3 x 2 − y  x + 2 dx dx 2 2 dx  dx   dx  dx

h.

1

; y′ =

i.

d dx

(

 1  3 12 y 3 3x 2 y 3 y y 3  − x − y  ; y′ = − ; y′ = − 1 − − ; y′ = − − ; y′ = − − 1 1   − − 1 x 2 2 x x x x x  2x ⋅ x 2 2x 2 2x 2

)

x + 3y =

1 −1 1  1  −1 −1 d ( y ) ; d  x 2 + 3 y  = y′ ; d x 2 + d 3 y = y′ ; 1 x 2 + 3 y′ = y′ ; 1 x 2 + 3 y′ − y′ = 0 ; 1 x 2 + 2 y′ = 0 dx dx  2 2 2 dx dx 

1 −1 1 1 ; 2 y′ = − x 2 ; 2 y′ = − 1 ; y′ = − 1 2 2x 2 4x 2 4. Evaluate the derivative of the following radical expressions for the specified value of x . 3

2

a. Given y = 3 x + x =

( )

1

3x3 2

2

+x =

2

+ x , then y′ =

1 32

3

3

3 3 −1 3 2 12 3 2 3−2 2 ⋅ x 2 + 2 x 2 −1 = x + 2x x + 2x = 2 2 2

5.196 32 3 2 12 y′ = + 2 = 4.598 +2 = ⋅ 1 + (2 ⋅ 1) = 2 2 2

at x = 1

)

(

(

b. Given y = x 2 + 1

x2 − 1 4x2

)

1

x = x2 + 1 x 2 = x

5 23 1 − 12 5 3 1 x + x = x2 + 1 2 2 2 2x 2

c. Given y =

3 x2

3

3

=

1 32

=

2+ 1

2

1

+ x2 = x

at x = 0

[ (

4 +1 2

y′ =

)]

[2 x ⋅ 2 x] − 2 ⋅ x 2 − 1 x2 − 1 , then y′ = 2x (2 x )2

Hamilton Education Guides

5

1

1

+ x 2 = x 2 + x 2 , then y′ = 5 32 ⋅0 + 2

=

1 1 2⋅ 02

= 0+

4x2 − 2x2 + 2 4x2

=

1 0

5 5 − 2 1 1− 2 5 52 − 1 1 12 − 1 x + x = x 2 + x 2 2 2 2 2 is undefined due to division by zero

2x2 + 2 4x2

=

(

)=

2 x2 + 1 4x2

x2 + 1 2x2

422

Calculus I

Chapter 2 Solutions 22 + 1

y′ =

at x = 2

d. Given y =

=

2 ⋅ 22

e. Given y =

2

⋅

− 12 + 1

=

11   22

−1

2

⋅

2

1 2

3

(

2

)

3 ⋅ 02 3 ⋅ 0 +1 2

−1

2

2

3

3 ⋅ 02 ⋅ 2

+ 12 ⋅ 02 =

1

(0 + 1) 3

1 2

11   22

=

2

2

2

2

−1+1

3

−1

2

⋅

2 (1 + 1) 1 = (x + 1) + 4 x , then y′ = (x + 1) 2

x3 + 1 + 4 x3 y′ =

at x = 0

−1

[1 ⋅ (x + 1)]− [2x ⋅ x] = 1  x  ⋅ − x + 1 2  x + 1  (x + 1) (x + 1)

1 −1

1

1 x  2  x 2 =  2   , then y′ =  2 2 2  x + 1   x +1 x +1 x

1 1  y′ =  2  2 1 +1

at x = 1

4 +1 5 = = 0.625 2⋅4 8

−1

2

1 −1 2 ⋅ 3x 2

2

2

2

⋅0 = 0

+ (4 ⋅ 3)x3 − 1 = 1

+ 12 ⋅ 02 = 0 ⋅

(

1 12

(

)

3x 2 3 x +1 2

−1

2

+ 12 x 2

+0 = 0+0 = 0

)

3    5 3 3 −1 2 3 1 2 x ⋅ x 2  −  x 2 ⋅ x + 1  2x 2 − x 2 x2 + 1 2 x +1 x +1 x +1     2 f. Given y = = = , then y′ = = 3 1 3 x3 x 3 2 2 x3 x x

2

2

2

(

)

( ) 5

2 ⋅ 32 −

y′ =

at x = 3

(

)

3 12 2 ⋅3 3 +1 (2 ⋅ 15.58) − (1.5 ⋅ 1.732 ⋅ 10) = 31.18 − 25.98 = 0.193 2 = 27 27 33

Section 2.8 Solutions - Higher Order Derivatives 1. Find the second derivative of the following functions. a. Given y = x3 + 3 x 2 + 5 x − 1 , then y ′ = 3 x 3−1 + (3 ⋅ 2 )x 2 −1 + 5 x1−1 − 0 = 3 x 2 + 6 x + 5 x 0 = 3 x 2 + 6 x + 5 and y ′′ = (3 ⋅ 2 )x 2 −1 + 6 x1−1 + 0 = 6 x + 6 x 0 = 6 x + 6

b. Given y = x 2 (x + 1)2 , then y ′ = 2 x ⋅ (x + 1)2 + 2(x + 1)2 −1 ⋅ x 2 = 2 x (x + 1)2 + 2 x 2 (x + 1) = 2 x (x + 1)2 + 2 x3 + 2 x 2 and y ′′ =

[2 ⋅(x + 1) + 2(x + 1)

2 −1

2

]

(

)

⋅ 2 x + (2 ⋅ 3)x3 −1 + (2 ⋅ 2 )x 2 −1 = 2 (x + 1)2 + 4 x(x + 1) + 6 x 2 + 4 x = 2 x 2 + 2 x + 1 + 4 x 2 + 4 xx

(

)

+ 6 x 2 + 4 x = 2 x 2 + 4 x + 2 + 10 x 2 + 8 x = 12 x 2 + 12 x + 2 = 2 6 x 2 + 6 x + 1

c. Given y = 3 x 3 + 50 x , then y′ = (3 ⋅ 3)x3−1 + 50 x1−1 = 9 x 2 + 50 and y ′′ = (9 ⋅ 2 )x 2 −1 + 0 = 18 x d. Given y = x5 +

e. Given y =

y ′′ =

=

[(6x

2

1

x2

= x5 + x −2 , then y′ = 5 x5−1 − 2 x −2 −1 = 5 x 4 − 2 x −3 and y ′′ = (5 ⋅ 4 )x 4 −1 + (− 2 ⋅ −3)x −3−1 = 20 x 3 + 6 x −4

] [ ] − (5 ⋅ 2)x

[

3 x 2 ⋅ (x + 1) − 1 ⋅ x3 x3 − 5 x 2 , then y′ = x +1 (x + 1) 2

][

)

(

+ 6 x ⋅ (x + 1) 2 − 2(x + 1) ⋅ 2 x3 + 3 x 2

(x + 1) 4

(x/ + 1/ )[6 x(x + 1) 2 − 2(2 x3 + 3x 2 ) ] (x + 1) 4/ =3

(

− 10 =

)]

− 10 x1 − 1 =

2 −1

=

3x 3 + 3x 2 − x 3

[6x(x + 1)(x + 1) ]− [2(x + 1)(2x 2

3

( x + 1) 3

2 x 3 + 3x 2

(x + 1)2

+ 3x 2

(x + 1) 4

6 x ( x + 1) 2 − 2 x 2 (2 x + 3 )

)

(x + 1)2

− 10 x =

)]

− 10 x and

− 10

− 10

f. Given y = x 3 x 2 − 1 = x5 − x3 , then y′ = 5 x5−1 − 3 x3−1 = 5 x 4 − 3 x 2 and y ′′ = (5 ⋅ 4 )x 4 −1 − (3 ⋅ 2 )x 2 −1 = 20 x 3 − 6 x g. Given y = x 4 +

(

)

(

)

1 x 8 − 7 x 5 + 5x 1 , then y′ = 4 x 4−1 + 8 x 8−1 + (− 7 ⋅ 5)x 5−1 + 5 x 1−1 = 4 x 3 + 8 x 7 − 35 x 4 + 5 and 10 10 10

y ′′ = (4 ⋅ 3)x3−1 +

[

]

(

)

1 (8 ⋅ 7 )x7 −1 + (− 35 ⋅ 4)x 4−1 + 0 = 12 x 2 + 1 56 x6 − 140 x3 = 5.6 x 6 − 14 x 3 + 12 x 2 10 10

Hamilton Education Guides

423

Calculus I

Chapter 2 Solutions

1 = x 2 − (x + 1)−1 , then y′ = 2 x 2−1 + (x + 1)−1−1 = 2 x + (x + 1)−2 and y ′′ = 2 x1−1 − 2(x + 1) − 2 −1 x +1 2 = 2 x 0 − 2(x + 1)−3 = 2 − 2(x + 1)−3 = 2 − ( x + 1)3 1 i. Given y = 2 − 3 x = x −2 − 3 x , then y′ = − 2 x −2−1 − 3 x 1−1 = − 2 x −3 − 3 and y ′′ = (− 2 ⋅ −3)x −3−1 − 0 = 6 x −4 x

h. Given y = x 2 −

2. Find y′′′ for the following functions. a. Given y = x5 + 6 x3 + 10 , then y ′ = 5 x 5−1 + (6 ⋅ 3)x 3−1 + 0 = 5 x 4 + 18 x 2 , y ′′ = (5 ⋅ 4 )x 4−1 + (18 ⋅ 2 )x 2−1 = 20 x 3 + 36 x , and y ′′′ = (20 ⋅ 3)x 3−1 + 36 x 1−1 = 60 x 2 + 36 b. Given y = x 2 +

1 = x 2 + x −1 , then y ′ = 2 x 2−1 − x −1−1 = 2 x − x −2 , y ′′ = 2 x1−1 + 2 x −2 −1 = 2 + 2 x −3 , and y ′′′ = − 6 x −4 x

c. Given y = 4 x3 (x − 1)2 , then y′ =

[(4 ⋅ 3)x

][

]

(

)

⋅ (x − 1)2 + 2(x − 1)2 −1 ⋅ 4 x3 = 12 x 2 (x − 1)2 + 8 x 3 (x − 1) = 12 x 2 x 2 − 2 x + 1

3−1

+ 8 x 4 − 8 x3 = 12 x 4 − 24 x 3 + 12 x 2 + 8 x 4 − 8 x 3 = 20 x 4 − 32 x 3 + 12 x 2 y ′′ = (20 ⋅ 4 )x 4−1 − (32 ⋅ 3)x 3−1 + (12 ⋅ 2 )x 2−1 = 80 x 3 − 96 x 2 + 24 x and y ′′′ = (80 ⋅ 3)x 3−1 − (96 ⋅ 2 )x 2−1 + 24 x 1−1 = 240 x 2 − 192 x + 24 x 0 = 240 x 2 − 192 x + 24

[1 ⋅ (x + 1) ] − [1 ⋅ x] = x + 1 − x = 1 = (x + 1)−2 , y ′′ = − 2(x + 1)−2−1 = − 2(x + 1) − 3 and x , then y ′ = x +1 (x + 1)2 (x + 1)2 (x + 1)2

d. Given y =

6

y ′′′ = (− 2 ⋅ −3)(x + 1) − 3 −1 = 6 (x + 1) − 4 =

( x + 1)4

e. Given y = x8 − 10 x5 + 5 x − 10 , then y ′ = 8 x8−1 + (− 10 ⋅ 5)x5−1 + 5 x1−1 − 0 = 8 x 7 − 50 x 4 + 5 x 0 = 8 x 7 − 50 x 4 + 5 , y ′′ = (8 ⋅ 7 )x 7 −1 − (50 ⋅ 4 )x 4 −1 + 0 = 56 x 6 − 200 x3 and y ′′′ = (56 ⋅ 6 )x 6 −1 − (200 ⋅ 3)x3−1 = 336 x 5 − 600 x 2 f. Given y =

=

=

=

x −1 x

2

+ 5 x 3 , then y ′ =

[1 ⋅ x ]− [2 x ⋅ (x − 1) ] + (5 ⋅ 3)x 2

x

4

[

3−1

][

− 1 ⋅ x3 − 3 x 2 ⋅ (− x + 2 ) x/ (− x + 2 ) −x + 2 2 2 ′′ = = , y + x 15 + 15 x x 4/ =3 x3 x6

2 x3 − 6 x 2 x

6

+ 30 x =

− 6 x 4 + 24 x3 x

8

x 2/ (2 x − 6 )

+ 30 =

x

6/ = 4

− 6 x3/ (x − 4 ) x

8/ =5

2x − 6

+ 30 x =

x

4

+ 30 = −

x2 − 2x2 + 2x x

x5

4

] + (15 ⋅ 2)x

[2 ⋅ x ]− [4x 4

+ 30 x and y ′′′ =

6( x − 4 )

3. Find f ′′(0 ) and f ′′(1) for the following functions.

=

2 −1

3

x

+ 15 x 2 =

− x2 + 2x x4

− x3 + 3x3 − 6 x 2

=

x6

⋅ (2 x − 6 )

8

] + 30x

1 −1

=

+ 15 x 2

+ 30 x 2 x 4 − 8 x 4 + 24 x3 x8

+ 30

+ 30

a. Given f (x ) = 6 x5 + 3 x3 + 5 , then f ′(x ) = (6 ⋅ 5)x5−1 + (3 ⋅ 3)x3−1 + 0 = 30 x 4 + 9 x 2 and f ′′(x ) = (30 ⋅ 4 )x 4 −1 + (9 ⋅ 2 )x 2 −1 = 120 x3 + 18 x . Therefore, f ′′(0 ) = 120 ⋅ 03 + 18 ⋅ 0 = 0 and f ′′(1) = 120 ⋅ 13 + 18 ⋅ 1 = 120 + 18 = 138 b. Given f (x ) = x3 (x + 1)2 , then f ′(x ) =

[3x

3−1

][

]

(

)

⋅ (x + 1)2 + 2(x + 1)2 −1 ⋅ x3 = 3 x 2 (x + 1)2 + 2 x3 (x + 1) = 3 x 2 x 2 + 2 x + 1

+ 2 x 4 + 2 x3 = 3 x 4 + 6 x3 + 3 x 2 + 2 x 4 + 2 x3 = 5 x 4 + 8 x3 + 3 x 2 and f ′′(x ) = (5 ⋅ 4 )x 4 −1 + (8 ⋅ 3)x3−1 + (3 ⋅ 2 )x 2 −1

(

) (

)

(

) (

)

= 20 x3 + 24 x 2 + 6 x . Therefore, f ′′(0 ) = 20 ⋅ 03 + 24 ⋅ 02 + (6 ⋅ 0 ) = 0 and f ′′(1) = 20 ⋅ 13 + 24 ⋅ 12 + (6 ⋅ 1) = 50

Hamilton Education Guides

424

Calculus I

Chapter 2 Solutions

c. Given f (x ) = x + (x − 1) 2 , then f ′(x ) = 1 + 2(x − 1) 2 −1 = 1 + 2(x − 1) = 1 + 2 x − 2 = 2 x − 1 and f ′′(x ) = 2 x1−1 − 0 = 2 . Therefore, f ′′(0 ) = 2 and f ′′(1) = 2 d. Given f (x ) = (x − 1) − 3 , then f ′(x ) = − 3(x − 1) − 3 −1 = − 3(x − 1) − 4 and f ′′(x ) = (− 3 ⋅ −4 )(x − 1) − 4 −1 = 12 (x − 1) − 5 = Therefore, f ′′(0 ) =

12

(0 − 1)

5

(

=

12

(x − 1) 5

12 12 12 = −12 and f ′′(1) = = which is undefined due to division by zero. 5 −1 0 (1 − 1)

)

e. Given f (x ) = (x − 1) x 2 + 1 , then f ′(x ) =

[1 ⋅ (x + 1)]+ [2 x ⋅ (x − 1) ] = x 2

2

+ 1 + 2 x 2 − 2 x = 3 x 2 − 2 x + 1 and

f ′′(x ) = (3 ⋅ 2 )x 2 −1 − 2 x1−1 + 0 = 6 x − 2 . Therefore, f ′′(0 ) = (6 ⋅ 0 ) − 2 = 0 − 2 = −2 and f ′′(1) = (6 ⋅ 1) − 2 = 6 − 2 = 4

(

)

f. Given f (x ) = x3 − 1

2

(

)

+ 2 x = x3 − 1

[ (

)

2

1

(

)

+ (2 x ) 2 , then f ′(x ) = 2 x3 − 1

2 −1

⋅ 3x 2 +

(

)

1 1 1 (2 x ) 2 − 1 ⋅ 2 = 6 x 2 x3 − 1 + (2 x ) − 2 2

]

3 3 1 1 (2 x ) − 2 − 1 ⋅ 2/ = 12 x 4 − 12 x + 18 x 4 − (2 x ) − 2 = 30 x 4 − 12 x − (2 x ) − 2 . Thus, f ′′(0) 2/ 1 1 = − which is undefined, and f ′′(1) = 30 ⋅ 14 − 12 ⋅ 1 − = 30 − 12 − 0.35 = 17.65 3 0 (2 ⋅ 1) 2

and f ′′(x ) = 12 x ⋅ x3 − 1 + 3 x 2 ⋅ 6 x 2 − = 30 ⋅ 04 − 12 ⋅ 0 −

1 3

(2 ⋅ 0) 2

Hamilton Education Guides

425

Chapter 3 Solutions: Section 3.1 Practice Problems – Differentiation of Trigonometric Functions 1. Find the derivative of the following trigonometric functions: dy d [ sin (3x + 1) ] = cos (3x + 1) ⋅ d (3x + 1) = cos (3x + 1) ⋅ 3 = 3 cos (3 x + 1) = dx dx dx

a. Given y = sin (3 x + 1) , then

dy d d 3 = 5 cos x3 = − 5 sin x3 ⋅ x = − 5 sin x3 ⋅ 3 x 2 = − 15 x 2 sin x 3 dx dx dx

b. Given y = 5 cos x 3 , then

c. Given y = x 3 cos x 2 , then

(

)

dy d 3 d d d = x cos x 2 = cos x 2 ⋅ x3 + x3 ⋅ cos x 2 = cos x 2 ⋅ 3 x 2 + x3 ⋅ − sin x 2 ⋅ x 2 dx dx dx dx dx

= cos x 2 ⋅ 3 x 2 − x3 ⋅ sin x 2 ⋅ 2 x = 3 x 2 cos x 2 − 2 x 4 sin x 2 d. Given y = sin 5 x ⋅ tan 3 x , then + sin 5 x ⋅ sec2 3 x ⋅

dy d = ( sin 5 x ⋅ tan 3x ) = tan 3x ⋅ d sin 5 x + sin 5 x ⋅ d tan 3x = tan 3x ⋅ cos 5 x ⋅ d 5 x dx dx dx dx dx

d 3 x = tan 3 x ⋅ cos 5 x ⋅ 5 + sin 5 x ⋅ sec2 3 x ⋅ 3 = 5 tan 3 x cos 5 x + 3 sin 5 x sec 2 3 x dx

(

e. Given y = tan 2 x3 = tan x3

)

2

, then

(

dy d = tan x3 dx dx

= 6 x 2 tan x 3 sec 2 x 3 f. Given y = cot (x + 3) 3 , then

[

)

2

= 2 tan x3 ⋅

d d tan x3 = 2 tan x3 ⋅ sec2 x3 ⋅ x3 = 2 tan x3 ⋅ sec2 x3 ⋅ 3 x 2 dx dx

]

dy d d d = cot (x + 3) 3 = − csc2 (x + 3) 3⋅ (x + 3) 3 = − csc2 (x + 3) 3⋅ 3 (x + 3) 2⋅ (x + 3) dx dx dx dx

= − csc2 (x + 3) 3⋅ 3 (x + 3) 2⋅ 1 = − 3 csc 2 ( x + 3 ) 3 ( x + 3 ) 2 g. Given y = x 5 cos x 7 , then

(

)

dy d d d d 5 = x cos x 7 = cos x 7 ⋅ x5 + x5 ⋅ cos x 7 = cos x 7 ⋅ 5 x 4 + x5 ⋅ − sin x 7 ⋅ x 7 dx dx dx dx dx

= cos x 7 ⋅ 5 x 4 − x5 ⋅ sin x 7 ⋅ 7 x 6 = 5 x 4 cos x 7 − 7 x11 sin x 7 4

4

3

3 3 3 3 3 3 3   dy d  d   sec x 2  = 4  sec x 2  ⋅  sec x 2  = 4 sec3 x 2 ⋅ sec x 2 h. Given y = sec4 x3 = sec4 x 2 =  sec x 2  , then =       dx dx       dx   3 3 3 1 3 3 3 3 3 −1 d 32 x = 4 sec3 x 2 ⋅ sec x 2 tan x 2 ⋅ x 2 = 6 x 2 sec 3 x 2 sec x 2 tan x 2 2 dx dy d d d d 3 i. Given y = sec 3 x 2 + x3 , then = sec 3 x 2 + x3 = sec 3 x 2 + x = sec 3 x 2 ⋅ tan 3 x 2 ⋅ 3x 2 + 3x 2 dx dx dx dx dx

3

× tan x 2 ⋅

(

)

( )

= sec 3 x 2 ⋅ tan 3 x 2 ⋅ 6 x + 3 x 2 = 6 x sec 3 x 2 tan 3 x 2 + 3 x 2

(

)

(

)

( )

j. Given y = tan x 5 , then

dy d d 5 = tan x5 = sec2 x5 ⋅ x = sec2 x5 ⋅ 5 x 4 = 5 x 4 sec2 x 5 dx dx dx

k. Given y = tan 5 x , then

dy d d (tan x ) 5 = 5 tan 4 x ⋅ d tan x = 5 tan 4 x ⋅ sec2 x ⋅ d x = 5 tan4 x sec2 x = tan 5 x = dx dx dx dx dx

(

)

l. Given y = csc x 3 + 1 , then

(

)

(

) (

) (

(

)

) (

dy d 3 d = x + 1 = − 3 x 2 csc x 3 + 1 cot x 3 + 1 csc x3 + 1 = − csc x3 + 1 cot x3 + 1 ⋅ dx dx dx

)

2. Find the derivative of the following trigonometric functions: a. Given y =

tan x dy d  tan  , then = csc x dx dx  csc

Hamilton Education Guides

x  = x 

( csc x ⋅ dxd tan x ) − ( tan x ⋅ dxd csc x ) csc2 x

( csc x ⋅ sec x)− ( tan x ⋅ − csc x cot x ) 2

=

csc2 x

426

Calculus I

=

Chapter 3 Solutions

(

csc x

(

)

sin x3 + 1 x2

2

) ][ (

(

sec x csc x

3

( sec x ⋅ − csc x cot x 3

3

x

⋅ 3x 2

)

2 3

csc x

=

( csc x

3

3

)

] [x

(

)

(

x

)(

4

d sec x − sec x ⋅ d csc x 3 ⋅ dx dx

)

2 3

csc x

) (

sec x tan x + 3 x 2 sec x csc x 3 cot x 3 2

) = 3x

( csc x

=

(

)]

) (

d x3 + 1 ⋅ cos x3 + 1 ⋅ dx

x4

3 x 4 cos x3 + 1 − 2 x sin x3 + 1

=

2

=

x4

) ]

( csc x

)] [ (

(

d sin x 3 + 1 − sin x 3 + 1 ⋅ d x 2 ⋅ dx dx

4

dy d  sec x  = = dx dx  csc x3 

, then

2

⋅ cos x3 + 1 ⋅ 3 x 2 − sin x3 + 1 ⋅ 2 x

4

c. Given y =

) = [x

(

[sin (x + 1)⋅ 2 x] = [ x x

sec 2 x + tan x cot x csc x

3 dy d sin x + 1 , then = dx dx x2

3

−

=

2

b. Given y =

−

)

csc x sec2 x + tan x cot x

3

3

(

)

(

cos x 3 + 1 − 2 sin x 3 + 1 x

⋅ sec x tan x

3

)

2 3

csc x

)

3

csc x dy d d d d 5 3 d. Given y = x tan x , then = x tan x = tan x3 ⋅ x5 + x5 ⋅ tan x3 = tan x3 ⋅ 5 x 4 + x5 ⋅ sec2 x3 ⋅ x3 dx dx dx dx dx 5

)

(

3

(

= tan x3 ⋅ 5 x 4 + x5 ⋅ sec2 x3 ⋅ 3 x 2 = 5 x 4 tan x3 + 3 x 7 sec2 x3 = x 4 5 tan x 3 + 3 x 3 sec 2 x 3 e. Given y = x 5 sin x 2 , then

(

)

dy d d d d 5 = x sin x 2 = sin x 2 ⋅ x5 + x5 ⋅ sin x 2 = sin x 2 ⋅ 5 x 4 + x5 ⋅ cos x 2 ⋅ x 2 dx dx dx dx dx

(

= sin x 2 ⋅ 5 x 4 + x5 ⋅ cos x 2 ⋅ 2 x = 5 x 4 sin x 2 + 2 x 6 cos x 2 = x 4 5 sin x 2 + 2 x 2 cos x 2 f. Given y = (x + 5) 2cos x , then = cos x ⋅ 2(x + 5) ⋅

[

(

)

(

−1

)

dy d d d d = x 2 tan 3 x5 = tan 3 x5 ⋅ x 2 + x 2 ⋅ tan 3 x5 = tan 3 x5 ⋅ 2 x + x 2 ⋅ 3 tan 2 x5 ⋅ tan x5 dx dx dx dx dx

h. Given y = x + sin x 3 , then

(

]

dy d (x + 5) 2cos x = cos x ⋅ d (x + 5) 2+ (x + 5) 2⋅ d cos x = dx dx dx dx

= tan 3 x5 ⋅ 2 x + x 2 ⋅ 3 tan 2 x5 ⋅ sec2 x5 ⋅

1 x + sin x3 2

)

d (x + 5) + (x + 5) 2⋅ − sin x = cos x ⋅ 2(x + 5) ⋅ 1 − (x + 5) 2⋅ sin x = 2( x + 5 ) cos x − ( x + 5 ) 2sin x dx

g. Given y = x 2 tan 3 x5 , then

=

)

d 5 x = 2 x tan 3 x5 + x 2 ⋅ 3 tan 2 x5 ⋅ sec2 x5 ⋅ 5 x 4 = 2 x tan 3 x 5 + 15 x 6 tan 2 x 5 sec2 x 5 dx

(

dy d d  3 = x + sin x3  x + sin x  = dx dx dx  

(

1 d d  x + sin x3 x + sin x3  = 2 dx dx  

2⋅  

)

i. Given y = sin 1 + x5 , then j. Given y = cot 2 x 3 , then

(

)

) ⋅ (1 + 3x −1

2

(

2

)

1 2

=

)

cos x3 =

) (

(

1 x + sin x3 2

)

1 −1 2 ⋅

1 + 3 x 2 cos x3

(

2 x + sin x3

)

(

)

1 2

=

)

(

d x + sin x3 dx

)

1 + 3 x 2 cos x 3 2 x + sin x 3

(

dy d d = sin 1 + x5 = cos 1 + x5 ⋅ 1 + x5 = cos 1 + x5 ⋅ 5 x 4 = 5 x 4 cos 1 + x 5 dx dx dx

(

)

(

dy d d = cot 2 x3 = cot x3 dx dx dx

)

2

= 2 cot x3 ⋅

)

d d cot x3 = 2 cot x3 ⋅ − csc2 x3 ⋅ x3 dx dx

= 2 cot x3 ⋅ − csc2 x3 ⋅ 3 x 2 = − 6 x 2 cot x 3 csc 2 x 3 k. Given y = sin 3 ( 1 + 5 x ) , then

dy d [ sin (1 + 5 x ) ] 3 = 3[ sin (1 + 5 x ) ] 2⋅ d sin (1 + 5 x ) = 3[ sin (1 + 5 x ) ] 2⋅ cos (1 + 5 x ) = dx dx dx

d (1 + 5 x ) = 3[ sin (1 + 5 x ) ] 2⋅ cos (1 + 5 x ) ⋅ 5 = 15 sin 2 (1 + 5 x ) cos (1 + 5 x ) dx dy d d 5 d d l. Given y = x 5 csc x 3 , then = x csc x3 = csc x3 ⋅ x5 + x5 ⋅ csc x3 = csc x3 ⋅ 5 x 4 + x5 ⋅ − csc x3 cot x3 ⋅ x3 dx dx dx dx dx ×

(

)

= csc x3 ⋅ 5 x 4 − x5 ⋅ csc x3 cot x3 ⋅ 3 x 2 = 5 x 4 csc x 3 − 3 x 7 csc x 3 cot x 3

Hamilton Education Guides

427

)

Calculus I

Chapter 3 Solutions

3. Find the derivative of the following trigonometric functions: a. Given y = sin 7 x , then

dy d (sin 7 x ) = cos 7 x ⋅ d (7 x ) = cos 7 x ⋅ 7 = 7 cos 7 x = dx dx dx

b. Given y = cos x 3 , then

dy d d 3 = x = − sin x3 ⋅ 3 x 2 = − 3 x 2 sin x 3 cos x3 = − sin x3 ⋅ dx dx dx

c. Given y = cot x 3 , then

dy d 3 d = cot x3 = − csc2 x3 ⋅ x = − csc2 x3 ⋅ 3 x 2 = − 3 x 2 csc 2 x 3 dx dx dx

(

)

( )

(

)

dy d d d d = x3 tan x 2 = tan x 2 ⋅ x3 + x3 ⋅ tan x 2 = tan x 2 ⋅ 3 x 2 + x3 ⋅ sec2 x 2 ⋅ x 2 dx dx dx dx dx

d. Given y = x 3 tan x 2 , then

(

= tan x 2 ⋅ 3 x 2 + x3 ⋅ sec2 x 2 ⋅ 2 x = 3 x 2 tan x 2 + 2 x 4 sec2 x 2 = x 2 3 tan x 2 + 2 x 2 sec 2 x 2 e. Given y = cot (x + 9 ) , then

(

)

dy d [ cot (x + 9) ] = − csc2 (x + 9) ⋅ d (x + 9) = − csc2 (x + 9) ⋅ 1 = − csc2 ( x + 9) = dx dx dx

f. Given y = sin 2 x3 + 5 x + 2 , then

(

) (

× cos (x + 3) ⋅

)

(

(

dy d d = sin 2 x3 + 5 x + 2 = sin x3 + 5 x + 2 dx dx dx

) dxd (x

= 2 sin x3 + 5 x + 2 ⋅ cos x3 + 5 x + 2 ⋅ g. Given y = sin

)

1

x + 3 = sin (x + 3) 2 , then

3

) (

)

) (

2

(

) dxd sin (x

= 2 sin x3 + 5 x + 2 ⋅

) (

+ 5 x + 2 = 2 3 x 2 + 5 sin x 3 + 5 x + 2 cos x 3 + 5 x + 2

)

3

+ 5x + 2

1 −1 d −1 dy 1 d 1 = sin (x + 3) 2 = sin (x + 3) 2 ⋅ sin (x + 3) = sin (x + 3) 2 dx 2 dx 2 dx

−1 d (x + 3) = 1 sin (x + 3) 2 ⋅ cos (x + 3) ⋅ 1 = 1 cos (x + 3)1 = cos ( x + 3 ) 2 dx 2 2 sin ( x + 3 ) sin (x + 3) 2

h. Given y = sin x 2 + cos x 3 , then

(

)

dy d d d d d = sin x 2 + cos x3 = cos x 2 ⋅ x 2 − sin x3 ⋅ x3 sin x 2 + cos x3 = dx dx dx dx dx dx

= cos x 2 ⋅ 2 x − sin x3 ⋅ 3 x 2 = 2 x cos x 2 − 3 x 2 sin x 3

(

)

dy d 2 d d d 2 = x sin x3 = sin x3 ⋅ x + x 2 ⋅ sin x3 = sin x3 ⋅ 2 x + x 2 ⋅ cos x3 ⋅ x3 dx dx dx dx dx

i. Given y = x 2 sin x 3 , then

= 2 x sin x3 + x 2 ⋅ cos x3 ⋅ 3 x 2 = 2 x sin x 3 + 3 x 4 cos x 3

Section 3.2 Practice Problems – Differentiation of Inverse Trigonometric Functions 1. Find the derivative of the following inverse trigonometric functions: a. Given y = sin −1 3 x , then

b. Given y = x 3 + arc cos 5 x , then

c. Given y = x 5 arc tan x 4 , then

+ x5 ⋅

1 1+ x

8

⋅

1

dy d = arc sin 3 x = dx dx

⋅

1 − ( 3 x )2

(

d 3x = dx

)

1 − 9x2 1

dy d 3 = x + arc cos 5 x = 3 x 2 − dx dx

(

3

1 − (5 x )2

⋅

5

d 5x = 3 x 2 − dx

1 − 25 x 2

)

dy d d d 5 = x arc tan x 4 = arc tan x 4 ⋅ x5 + x5 ⋅ arc tan x 4 = arc tan x 4 ⋅ 5 x 4 dx dx dx dx

4 x8 1 d 4 ⋅ 4 x3 = 5 x 4arc tan x 4 + x = 5 x 4 arc tan x 4 + x5 ⋅ 8 dx 1+ x 1 + x8

(

)

d. Given y = arc sin x 3 + 2 , then

e. Given y = arc cot

1 x3

, then

Hamilton Education Guides

[

(

dy d = arc sin x3 + 2 dx dx

dy d 1 = arc cot 3 = − dx dx x

)] =

(

1 − x3 + 2 1

1 +  

1

1 x3

  

2

⋅

)

⋅ 2

(

)

d 3 x +2 = dx

3x2

(

1 − x3 + 2

3 d 1 1 −3 = − = ⋅ dx x3 4 1 + 16 x 4 x 1 + x 

1 x6

)

2

  

428

)

Calculus I

Chapter 3 Solutions

f. Given y = x2 ⋅ =

arc sin 3 x

1 1−9 x 2

x2

x2

− ( arc sin 3 x ⋅ 2 x ) =

x4

1−9 x 2

)

(

1+ x 2 2

=

=

x

(

)

2

)(

d arc sin 3 x − arc sin 3 x ⋅ d x 2 ⋅ dx dx

x4 d dx

[3 x

2

)(

x − 2arc sin 3 x

=

x3 1 − 9x2

(

)

)

   x ⋅ 1 2  − tan −1 x ⋅ 1 1 + x  =  = x2

x2

1 − 9x2

− tan −1 x

x 1+ x 2

x2

x − 1 + x 2 tan −1 x

(

x2 1 + x2

][ ]−

⋅ (x + 5) − x 3 ⋅ 1

(x + 5)

1 − 9x2

x4 1 − 9x2

d x tan −1 x − tan −1 x ⋅ dx

)

dy x3 d  x3 h. Given y = = − arc sin x , then − arc sin dx x+5 dx  x + 5

=

x 2 − 2 x arc sin 3 x

=

(x ⋅

)

x4

− 2 x arc sin 3 x

dy tan −1 x d  tan −1 x  g. Given y = , then = = dx x dx  x  x − 1+ x 2 tan −1 x

(x

dy d  arc sin 3 x  , then =   = dx dx  x2 

1

2

1 − x2

i. Given y = x + cos −1 x , then

=

3 x3 + 15 x 2 − x3

(x + 5)

(

d x3 ⋅ dx

1

−

2

[

 x =  

=

1 − x2

(x + 5) ]− [ x3 ⋅ dxd (x + 5) ] (x + 5) 2

2 x 3 + 15 x 2

(x + 5) 2

)

1 − x2

1

−

1 − x2 1

dy d d d = x + cos −1 x = x + cos −1 x = 1 − dx dx dx dx

1

−

⋅

1 − x2

1

d x = 1− dx

1 − x2

2. Find the first and second derivative of the following inverse trigonometric functions: a. Given y = x 2 + arc sin ax , then

b. Given y = cos −1 6 x , then

(

)

1

dy d = cos −1 6 x = − dx dx

c. Given y = x + arc sin x 3 , then

d. Given y = arc tan 2 x , then

1

dy d 2 = x + arc sin ax = 2 x + dx dx

1 − (6 x ) 2

(

⋅

d 6x = − dx

)

dy d = x + arc sin x3 = 1 + dx dx

(

)

(

dy d d = arc tan 2 x = tan −1 x dx dx dx

(

1 − ( ax ) 2

1

⋅

1 − x6

)

2

⋅

1 1 − 36 x 2

d 3 x = 1+ dx

= 2 tan −1 x ⋅

a

d ax = 2 x + dx

6

⋅6 = −

1 1 − x6

1 − a2 x2

1 − 36 x 2

⋅ 3x 2 = 1 +

)

dy d d d 2x d 1 = x − tan −1 x 2 = x+ tan −1 x 2 = 1 + ⋅ x2 = 1 + 4 dx dx dx dx dx 1+ x 1 + x4

f. Given y = x 2 − arc sin x , then

dy d 2 d 2 d = x − arc sin x = x − arc sin x = 2 x − dx dx dx dx

(

)

1 1 − x2

dy d 1 = tan −1 x3 = dx dx 1 + x3

h. Given y = tan −1 5 x , then

dy d d 1 5 1 = tan −1 5 x = ⋅ 5x = ⋅5 = dx dx 1 + ( 5 x ) 2 dx 1 + 25 x 2 1 + 25 x 2

( )

Hamilton Education Guides

)

dy d = arc sin 5 x 2 = dx dx

2

⋅

⋅

d x = 2x − dx

1 1 − x2

d 3 3x2 1 2 = x = ⋅ 3 x dx 1 + x6 1 + x6

g. Given y = tan −1 x 3 , then

(

1 − x6

2 tan −1 x d 1 = tan −1 x = 2 tan −1 x ⋅ dx 1+ x 2 1+ x 2

e. Given y = x − arc tan x 2 , then

i. Given y = arc sin 5 x 2 , then

3x2

1 1 − 25 x 4

⋅

d 5x2 = dx

1 1 − 25 x 4

⋅ 10 x =

10 x 1 − 25 x 4

429

Calculus I

Chapter 3 Solutions

Section 3.3 Practice Problems – Differentiation of Logarithmic and Exponential Functions 1. Find the derivative of the following logarithmic functions: a. Given y = ln10 x , then

dy d ( ln10 x ) = 1 ⋅ d (10 x ) = 1 ⋅ 10 = 10 = 1 = dx dx 10 x 10 x x 10 x dx

(

(

)

c. Given y = ln x 2 + 3 , then

[ (

(

)

)

(

(

 x/  1 d  =  x ⋅ ⋅ (x ) + ln x ⋅ 1 − 10 x =  + ln x dx    x/

)

)

dy d d ( x ln x ) − d 5 x 2 =  x ⋅ d ( ln x ) + ln x d (x ) − 10 x = x ln x − 5 x 2 = dx dx dx dx dx   dx

e. Given y = x ln x − 5 x 2 , then

f. Given y = x 3 + 2 ln x 2 , then

+

)]

dy d 1 d 2 2x = ln x 2 + 3 = 2 ⋅ x +3 = 2 dx dx x + 3 dx x +3

dy d 3 1 d 3  3 d   = x ln x =  ln x ⋅ x  +  x ⋅ ln x  = ln x ⋅ 3 x 2 + x3 ⋅ = 3 x 2 ln x + x 2 = x 2 (3 ln x + 1) dx dx x dx   dx  

d. Given y = x3 ln x , then

(

( )

)

30 dy d 10 1 d 150 x 2 = = 10 ln 5 x3 = 10 ⋅ 3 ⋅ 5 x3 = ⋅ 15 x 2 = 3 dx dx x 5 x dx 5 x3 5x

b. Given y = 10 ln 5 x3 , then

 x  − 10 x = 1 + ln x − 10 x = ln x − 10 x + 1 

[(

]

)

(

)

(

( )

)

(

)

dy d 3 d d d 3 d   = x ln x 2 + 2 ln x 2 = ln x 2 ⋅ x3 + 2 ln x 2 = x + x3 ⋅ ln x 2  dx dx dx dx dx dx  

4 2 d 2 1  2 ⋅ 2x  ⋅ x = ln x 2 ⋅ 3 x 2 + x3 ⋅ 2 ⋅ 2 x  + 2 = 3 x 2 ln x 2 + 2 x 2 + x x x x 2 dx  

(

)

g. Given y = sin ln x 2 , then h. Given y = ln csc x , then 3

)

(

j. Given y = cos ln x 2 , then

2

2

(

) x1 ⋅ 2 x =

= cos ln x 2 ⋅

2

(

2 cos ln x 2 x

)

3 dy 3 1 −3 d 3 1−3 3 3 1 1 d 3 1 3 1 ln x 2 = 3 ⋅ x 2 = 3 ⋅ x 2 = x 2 ⋅ x 2 = x 2 2 = x − 1 = ⋅ = dx 2 2 dx 2 x 2 2 dx x2 x2

[ (

dy d = cos ln x 2 dx dx

)] = − sin (ln x )⋅ x1 ⋅ dxd x 2

2

2

(

) x1 ⋅ 2 x = − 2x sin (ln x )

= − sin ln x 2 ⋅

2

2

dy x −1 d x +1 x +1 x − 1 (x − 1) − (x + 1) −2 x −1 2 d  x +1  , then = = = = − ⋅ ⋅ ⋅  ln  = dx x −1 x + 1 dx x − 1 x +1 x + 1 (x − 1)2 dx  x −1  (x − 1) 2 x2 − 1

l. Given y = x3 ln x + 5 x , then

(

2

d ( ln csc x ) = 1 ⋅ d csc x = 1 ⋅ − csc x ⋅ cot x = − csc x ⋅ cot x = − cot x dx csc x dx csc x csc x

x3 = ln x 2 , then

i. Given y = ln

k. Given y = ln

[ ( )] = cos (ln x )⋅ x1 ⋅ dxd x

dy d = sin ln x 2 dx dx

)

[

]

dy d 3 d 3 d d d     = x ln x + 5 x = x ln x + 5 x =  ln x ⋅ x3  +  x3 ⋅ ln x  + 5 dx dx dx dx dx   dx  

1  = ln x ⋅ 3 x 2 +  x3 ⋅  + 5 = 3 x 2 ln x + x 2 + 5 x 

2. Find the derivative of the following exponential functions: a. Given y = x 3 e 2 x , then

(

)

dy d d d d = x3e 2 x = e 2 x ⋅ x3 + x3 ⋅ e 2 x = e 2 x ⋅ 3 x 2 + x3 ⋅ e 2 x ⋅ 2 x = 3 x 2e 2 x + x3 ⋅ e 2 x ⋅ 2 dx dx dx dx dx

= 3 x 2e 2 x + 2 x3e 2 x = x 2e 2 x ( 3 + 2 x )

(

)

(

)

b. Given y = x 2 + 3 e 3 x , then

[(

) ]

(

) (

)

(

)

[

(

)]

(

= e3 x ⋅ 2 x + x 2 + 3 ⋅ e3 x ⋅ 3 = 2 x e3 x + 3e3 x x 2 + 3 = e3 x 2 x + 3 x 2 + 3 = e 3 x 3 x 2 + 2 x + 9 c. Given y = e sin 3 x , then

(

)

dy d d d 2 d     = x 2 + 3 e3 x = e3 x ⋅ x + 3  +  x 2 + 3 ⋅ e3 x  = e3 x ⋅ 2 x + x 2 + 3 ⋅ e3 x ⋅ 3 x dx dx dx dx dx    

(

)

)

dy d d = esin 3 x = esin 3 x ⋅ (sin 3 x ) = esin 3 x ⋅ 3 cos 3 x = 3 cos 3 x e sin 3 x dx dx dx

Hamilton Education Guides

430

Calculus I

Chapter 3 Solutions

2

d. Given y = e ln x , then

2 2 dy 2 d ln x 2 1 d 1 = = eln x ⋅ 2 ⋅ x 2 = eln x ⋅ 2 ⋅ 2 x = e ln e dx dx x x dx x

e. Given y = e 9 x sin 5 x , then

)

(

( ) + e

dy d d  = e9 x sin 5 x = sin 5 x ⋅ e9 x dx dx dx 

[

][

9x

⋅

x2

d (sin 5 x ) = sin 5 x ⋅ e9 x ⋅ d 9 x  dx  dx  

]

d   + e9 x ⋅ cos 5 x ⋅ 5 x  = sin 5 x ⋅ e9 x ⋅ 9 + e9 x ⋅ cos 5 x ⋅ 5 = e 9 x ( 9 sin 5 x + 5 cos 5 x ) dx  

(

= arc sin x ⋅ e 2 x ⋅

1

d 2 x + e2 x ⋅ dx

g. Given y = (x − 5) e − x , then

)

( ) + e

dy d d 2x  = e 2 x arc sin x = arc sin x ⋅ e dx dx dx 

f. Given y = e 2 x arc sin x , then

e2 x

= arc sin x ⋅ e 2 x ⋅ 2 +

1 − x2

1 − x2

2x

⋅

d ( arc sin x ) dx 

  = e 2 x  2arc sin x +  

dy d = ( x − 5)e− x =  (x − 5) ⋅ d e− x  +  e− x ⋅ d (x − 5)  = dx dx dx dx    

1 1 − x2

    

[ ( x − 5) ⋅ −e ] + [ e ⋅ 1 ] −x

−x

= − e − x (x − 5) + e − x = − xe− x + 5e − x + e − x = − xe− x + 6e − x = e − x (− x + 6 ) h. Given y = eln ( x + 1) , then

i. Given y =

dy e ln ( x + 1) d ln ( x + 1) d 1 d = = eln ( x + 1) ⋅ ln (x + 1) = eln ( x + 1) ⋅ e ⋅ (x + 1) = dx dx dx x +1 x + 1 dx

[

( )] [

d e x − e x ⋅ d ( tan x ) tan x ⋅ dx dx dy ex d  e x  , then = = dx tan x dx  tan x  tan 2 x

]

=

tan x ⋅ e x − e x ⋅ sec2 x 2

tan x

=

(

e x tan x − sec2 x

)

2

tan x

dy d x x d d d d = 3 ⋅ e = e x ⋅ 3x + 3x ⋅ e x = e x ⋅ 3x ⋅ ln 3 ⋅ x + 3x ⋅ e x ⋅ x = e x ⋅ 3x ⋅ ln 3 ⋅ 1 + 3x ⋅ e x ⋅ 1 dx dx dx dx dx dx

j. Given y = 3 x ⋅ e x , then

= e x ⋅ 3x ⋅ ln 3 + 3x ⋅ e x = ( ln 3 + 1) 3 x e x k. Given y = x3arc sin x , then

(

)

x3

dy d 3 d d 3 = x arc sin x = arc sin x ⋅ x + x3 ⋅ arc sin x = 3 x 2arc sin x + dx dx dx dx

l. Given y = e 5 x cos (5 x + 1) , then

[

]

( ) + e

dy d 5x d  = e cos (5 x + 1) =  cos (5 x + 1) ⋅ e5 x dx dx dx 

5x

⋅

d  cos (5 x + 1) dx 

][

[

1 − x2

d d     =  cos (5 x + 1) ⋅ e5 x ⋅ 5 x  +  e5 x ⋅ − sin (5 x + 1) ⋅ (5 x + 1) = cos (5 x + 1) ⋅ e5 x ⋅ 5 + e5 x ⋅ − sin (5 x + 1) ⋅ 5 dx   dx  

]

= 5e 5 x [ cos (5 x + 1) − sin (5 x + 1) ]

Section 3.4 Practice Problems – Differentiation of Hyperbolic Functions Find the derivative of the following hyperbolic functions:

(

)

a. Given y = cosh x3 + 5 , then b. Given y = x3 sinh x , then

)

(

) (

)

(

)

(

(

)

[

]

dy d (x + 6)sinh x 2 = sinh x3 ⋅ d (x + 6) + (x + 6) ⋅ d sinh x3 = sinh x3 ⋅ 1 = dx dx dx dx

d 3 x = sinh x3 + (x + 6 ) ⋅ cosh x3 ⋅ 3 x 2 = sinh x 3 + 3 x 2 ( x + 6 ) cosh x 3 dx

d. Given y = ln (sinh x ) , then

Hamilton Education Guides

)

dy d 3 d d = x sinh x = sinh x ⋅ x3 + x3 ⋅ sinh x = sinh x ⋅ 3 x 2 + x3 ⋅ cosh x = 3 x 2 sinh x + x 3 cosh x dx dx dx dx

c. Given y = (x + 6 )sinh x3 , then + (x + 6 ) ⋅ cosh x3 ⋅

(

dy d d = cosh x3 + 5 = sinh x3 + 5 ⋅ x3 + 5 = sinh x3 + 5 ⋅ 3 x 2 = 3 x 2 sinh x 3 + 5 dx dx dx

dy d = ( ln sinh x ) = 1 ⋅ d ( sinh x ) = 1 ⋅ cosh x ⋅ d x = cosh x = coth x dx dx sinh x sinh x dx sinh x dx

431

Calculus I

Chapter 3 Solutions

e. Given y = esinh x , then

(

)

dy d sinh x d d = = esinh x ⋅ ( sinh x ) = esinh x ⋅ cosh x ⋅ x = esinh x ⋅ cosh x ⋅ 1 = e sinh x cosh x e dx dx dx dx

f. Given y = cosh 3 x 5 , then

)

(

dy d d = cosh 3 x5 = sinh 3 x5 ⋅ 3 x5 = sinh 3 x5 ⋅ 15 x 4 = 15 x 4 sinh 3 x 5 dx dx dx

][

[

d tanh 2 x − tanh 2 x ⋅ d x x ⋅ dx dx dy d  tanh 2 x  tanh 2 x g. Given y = , then = = 2   x dx dx  x  x

=

2 x tanh x sec h 2 x − tanh 2 x x

2

h. Given y = coth

1 3

x

(

, then

x

2

x

)

2

x ⋅1

]

2

2

dy 1 3 1 d 1 1 −3 d  1  = = − csc h 2 3 ⋅ 4 = csc h2 3  coth 3  = − csc h 2 3 ⋅ 3 4 dx dx  dx x  x x x x x x

)

i. Given y = x 2 + 9 tanh x , then

(

(

tanh x 2 x sec h2 x − tanh x

=

] [ x ⋅ 2 tanh x ⋅ sec h x ]− [tanh =

[(

(

)

]

)

)

) (

(

dy d d 2 d = x + 9 + x 2 + 9 ⋅ tanh x = tanh x ⋅ 2 x x 2 + 9 tanh x = tanh x ⋅ dx dx dx dx

)

+ x 2 + 9 ⋅ sec h 2 x = 2 x tanh x + x 2 + 9 sec h2 x

j. Given y = sinh 3 x 2 , then

(

)

(

dy d d = sinh 3 x 2 = sinh x 2 dx dx dx

)

3

= 3 sinh 2 x 2 ⋅

d d sinh x 2 = 3 sinh 2 x 2 ⋅ cosh x 2 ⋅ x 2 dx dx

= 3 sinh 2 x 2 ⋅ cosh x 2 ⋅ 2 x = 6 x sinh 2 x 2 cosh x 2

(

)

dy d d = tanh 5 x = 5 tanh 4 x ⋅ tanh x = 5 tanh 4 x sec h2 x dx dx dx

k. Given y = tanh 5 x , then

)

(

l. Given y = x5 coth x3 + 1 , then

[

(

dy d 5 = x coth x3 + 1 dx dx

) dxd ( x + 1 ) = coth ( x + 1 )⋅ 5x

(

+ x5 ⋅ − csc h 2 x3 + 1 ⋅

3

3

4

)]

(

) dxd x

= coth x3 + 1 ⋅

(

5

+ x5 ⋅

(

(

)

(

)

)

d coth x3 + 1 = coth x3 + 1 ⋅ 5 x 4 dx

)

(

− x5 ⋅ csc h 2 x3 + 1 ⋅ 3 x 2 = 5 x 4 coth x 3 + 1 − 3 x 7 csc h2 x 3 + 1

)

Section 3.5 Practice Problems – Differentiation of Inverse Hyperbolic Functions Find the derivative of the following inverse hyperbolic functions: a. Given y = x sinh −1 3 x , then 1

= sinh −1 3 x + x ⋅

(

⋅ 3 = sinh −1 3 x +

1 + 9x2

b. Given y = cosh −1 (x + 5) , then

c. Given y = tanh −1 x , then

d. Given y = x3

=

2

x +1

sinh

−1

x3

=

dy d = cosh −1 (x + 5) = dx dx

(

)

1

( x + 5) 2 − 1

dy d 1 = tanh −1 x = dx dx 1− x

x3 − 3 x 2 x 2 + 1 sinh −1 x x6 x 2 + 1

Hamilton Education Guides

⋅

1 + 9x2

d 3x dx

1 + 9x2

( )

2

⋅

⋅

d (x + 5) = dx

d 1 1 x = ⋅ dx 1− x x

=

1

( x + 5) 2 − 1 =

x3 ⋅

⋅1 =

1

( x + 5) 2 − 1

1 x (1 − x ) 1 2

x +1

− sinh −1 x ⋅ 3 x 2 x6

x − 3 x 2 + 1 sinh −1 x x4 x2 + 1

d sinh −1 x − sinh −1 x ⋅ d cosh 3 x cosh 3 x ⋅ dx dy sinh x d  sinh −1 x  dx e. Given y = , then = = = dx cosh 3 x dx  cosh 3 x  cosh 2 3 x

−1

1

3x

d sinh −1 x − sinh −1 x ⋅ d x 3 x3 ⋅ dx dy x d  sinh −1 x  dx , then = = = dx dx  x3  x6

− 3 x 2 sinh −1 x x6

)

dy d d d = x sinh −1 3 x = sinh −1 3 x ⋅ x + x ⋅ sinh −1 3 x = sinh −1 3 x ⋅1 + x ⋅ dx dx dx dx

cosh 3 x 2

x +1

− 3 sinh −1 x sinh 3 x cosh 2 3 x

432

Calculus I

=

Chapter 3 Solutions

cosh 3 x − 3 x 2 + 1 sinh −1 x sinh 3 x cosh 2 3 x

cosh −1 x

f. Given y = x5

d cosh −1 x − cosh −1 x ⋅ d x 5 x5 ⋅ dx dy d  cosh −1 x  dx , then = = = dx dx  x5  x10

− 5 x 4 cosh −1 x

x 2 −1

=

x5

x2 + 1

x5 − 5 x 4 x 2 − 1 cosh −1 x

=

x10

(

)

(

(

) 1 −1x

= 2 x coth −1 x +

2

+e

1

⋅

= cosh

2

−1

x⋅e

3x

e

3x

= 3e cosh

2

−1

e3 x

x+

2

x −1

d x2 e = dx

1

( )

k. Given y = tanh −1 e 2 x , then

)

=

  e 3 x  3 x 2 − 1 cosh −1 x + 1    x2 − 1

x −1

(

)

dy d 3 d 3 d 5 x4 1 d = x + tanh −1 x5 = x + tanh −1 x5 = 3 x 2 + ⋅ x5 = 3 x 2 + 10 dx dx dx dx dx 1− x 1 − x 10 2 dy d d d  x2  −1 = sinh −1 7 x + ln e x =  sinh 7 x + ln e  = dx dx dx dx  

1

⋅7 +

1 + 49 x 2

) (

)

e3 x

⋅3+

2

⋅

(

1 − x2

j. Given y = sinh −1 7 x + ln e x , then

x2

)

(

i. Given y = x3 + tanh −1 x5 , then

1

x6 x2 − 1

x2 + 3

x −1

+

x10

dy d d 3x d d = e cosh −1 x = cosh −1 x ⋅ e3 x + e3 x ⋅ cosh −1 x = cosh −1 x ⋅ e3 x ⋅ 3 x dx dx dx dx dx

h. Given y = e3 x cosh −1 x , then

3x

x −1

dy d 2 d d 2 = x + 3 + x 2 + 3 ⋅ coth −1 x = coth −1 x ⋅ 2 x x + 3 coth −1 x = coth −1 x ⋅ dx dx dx dx

g. Given y = x 2 + 3 coth −1 x , then

+ x2 + 3 ⋅

− cosh −1 x ⋅ 5 x 4

1 2

x − 5 x 2 − 1 cosh −1 x

=

x10 x 2 − 1

x5 ⋅

e

x2

2

⋅ ex ⋅

7

d 2 x = dx

+

1 + 49 x 2

1 e

2

x2

⋅ ex ⋅ 2x =

7 1 + 49 x 2

1 1 + (7 x )

⋅

2

d 7x dx

+ 2x

( )

dy d 1 d 1 d 1 2e 2 x = tanh −1 e 2 x = ⋅ e2 x = ⋅ e2 x ⋅ 2 x = ⋅ e2 x ⋅ 2 = 4 x 4 x 4 x dx dx dx dx 1− e 1− e 1− e 1 − e4 x

l. Given y = coth −1 (3 x + 5) , then

[

]

dy 3 d 1 d = coth −1 (3 x + 5) = ⋅ (3 x + 5) = dx dx 1 − (3 x + 5) 2 dx 1 − (3 x + 5 ) 2

Section 3.6 Practice Problems - Evaluation of Indeterminate Forms Using L’Hopital’s Rule Evaluate the limit for the following functions by applying the L’Hopital’s rule, if needed: a. lim x → ∞

ln x x

= lim x → ∞ b. lim x → 0

1 x

⋅1

2x

sin x ex − 1

= lim x → 0 c. lim x → 0

=

2

ln ∞ ∞

2

= lim x → ∞ =

sin 0

e0 − 1

d sin x dx d ex − d 1 dx dx

1 − cos x x

3

∞ ln x Apply the L’Hopital’s rule lim x → ∞ 2 = lim x → ∞ ∞ x

=

1 x

2x

=

= lim x → ∞

2x

=

2

1 2⋅∞

2

=

= lim x → ∞

= lim x → 0

cos x ex − 0

1 − cos x x

3

= lim x → 0 =

1 − cos 0 0

3

cos x ex =

1 x

d x ⋅ dx

2x

1 = 0 ∞

0 sin x 0 = Apply the L’Hopital’s rule lim x → 0 x = lim x → 0 0 1−1 e −1

= lim x → 0

Hamilton Education Guides

1

d ln x dx d x2 dx

=

cos 0 e0

=

d sin x dx d ex − 1 dx

(

)

1 = 1 1

1−1 1 − cos x 0 = Apply the L’Hopital’s rule lim x → 0 0 0 x3

433

Calculus I

Chapter 3 Solutions d dx

= lim x → 0

(1 − cos x ) d x3 dx

rule again lim x → 0 d. limt → 2

2

= lim x → 0

d 1 − d cos dx dx d x3 dx

= lim x → 0

d sin x dx d 3x 2 dx

sin x 3x

t−2

=

t + 2t − 1

2

2−2

2

2 + 2 ⋅ 2 −1

=

x

= lim x → 0

= lim x → 0

0 + sin x 3x 2

= lim x → 0

sin x 3x 2

=

sin 0

3 ⋅ 02

2

0 Apply the L’Hopital’s 0

cos 0 cos x 1 = = = ∞ 6x 6⋅0 0

2−2 0 = = 0 4 + 4 −1 7

cos π2 cos x cos x 0 = = Apply the L’Hopital’s rule lim π = lim π lim π x→ x − π x→ π − π x→ 0 x − π2 2 2 2 2 2 2

e. lim π x→

=

d cos x dx d x−π 2 dx

(

= lim π x→

)

2

− sin x 1

= − lim π sin x = − sin π2 = −1 x→ 2

f. limt → 0

d t cos t 0 ⋅1 t cos t t cos t 0 ⋅ cos 0 0 = = = Apply the L’Hopital’s rule limt → 0 = limt → 0 dt d 0 2⋅0 2 sin t 2 sin t 2 sin 0 2 ⋅ dx sin t

= limt → 0

g. lim x → 8

=

d x− dx d x2 − dx

82 − 64

d 8 dx d 64 dx

=

= limt → 0

1− 0⋅0 cos t ⋅ 1 + t ⋅ − sin t cos t − t sin t cos 0 − 0 ⋅ sin 0 1 = limt → 0 = = = 2 2 ⋅1 2 ⋅ cos t 2 cos t 2 cos 0

8−8 x −8 0 = Apply the L’Hopital’s rule lim x → 8 2 = lim x →8 0 64 − 64 x − 64

= lim x →8

d sin dx

d π dx

x

−

d dx

x

= lim x → π

d 3x dx

d dx

d sin x dx d π −x dx

(

( x − 8)

(x

2

− 64

)

)

cos x cos x = lim x → π = − lim x → π cos x = − cos π = 1 0 −1 −1

sin (8 ⋅ 0 ) sin 8 x sin 8 x sin 0 0 = = = Apply the L’Hopital’s rule lim x → 0 = lim x → 0 0 3x 3⋅ 0 0 3x d 8x cos 8 x ⋅ dx

d dx

1 1 1− 0 1 = lim x →8 = = 16 2⋅8 2x − 0 2x

sin 0 sin x 0 sin x 0 = = = Apply the L’Hopital’s rule lim x → π = lim x → π 0 π −π π −x 0−0 π −x

= lim x → 0 j. limt → ∞

8−8

=

x 2 − 64

= lim x → π i. lim x → 0

d sin t 2 ⋅ dt

x −8

= lim x →8

h. lim x → π

d t + t ⋅ d cos t cos t ⋅ dt dt

= lim x → 0

d sin 8 x dx d 3x dx

cos 8 x ⋅ 8 8 ⋅1 8 cos 8 x 8 ⋅ cos 0 8 ⋅ cos (8 ⋅ 0 ) 8 = lim x → 0 = lim x → 0 = lim x → 0 = = 3 3 3 3 3 3

8t + 3 ∞+3 ∞ 8t + 3 = = Apply the L’Hopital’s rule limt → ∞ = limt → ∞ 4t − 2 ∞−2 ∞ 4t − 2

d dt d dt

( 8t + 3) ( 4t − 2)

= limt → ∞

d 8t dt d 4t dt

+ −

d dt d dt

3 2

8+0 8 = = 2 4 4−0

k. lim x → 0

cos x − 1 x

2

=

cos 0 − 1 0

2

= lim x → 0

d cos x − d 1 dx dx d x2 dx

− lim x → 0

sin x = − lim x → 0 2x

Hamilton Education Guides

=

1−1 0 cos x − 1 = Apply the L’Hopital’s rule lim x → 0 = lim x → 0 0 0 x2

= lim x → 0

d dx

( cos x − 1) d dx

x2

sin 0 0 − sin x − 0 sin x = − lim x → 0 = − = − Apply the L’Hopital’s rule again 2x 2x 2⋅0 0

d sin x dx d 2x dx

= − lim x → 0

cos x cos 0 1 = − = − 2 2 2

434

Calculus I

Chapter 3 Solutions

l. lim π x→

2

1 − sin π2 1 − sin π2 0 1−1 1 − sin x 1 − sin x = = = = Apply the L’Hopital’s rule lim π π x → 1 + cos 2 x 1 + cos π 0 1−1 1 + cos 2 x 1 + cos 2 ⋅ 2 2

(

)

d 1 − sin x dx d 1 + cos 2 x 2 dx

= lim π x→

(

)

d 1 − d sin x dx dx d 1 + d cos 2 x 2 dx dx

= lim π x→

= lim π x→

2

π

π 1 cos x 1 cos 2 0 − cos x = lim π = ⋅ → x 2 2 sin 2 ⋅ π 0 − 2 sin 2 x 2 sin 2 x 2 d cos x dx d sin 2 x dx

=

0 1 0 1 1 1 cos 2 cos x = ⋅ = Apply the L’Hopital’s rule again lim π = lim π ⋅ x→ x → sin 2 x 0 2 0 2 2 sin π 2 2 2

=

π π 1 1 1 sin x 1 − sin x 1 sin 2 1 sin 2 1 = − lim π = − ⋅ = − ⋅ = − ⋅ = lim x → π π x → 4 −1 4 2 4 4 cos 2 ⋅ 4 cos π 2 2 cos 2 x 2 cos 2 x 2

Hamilton Education Guides

435

Chapter 4 Solutions: Section 4.1 Practice Problems – Integration Using the Basic Integration Formulas 1. Evaluate the following indefinite integrals: a.

∫ − 3 dx

∫

∫

= − 3 dx = − 3 x 0 dx = −

3 0 +1 x + c = −3 x + c 0 +1

Check: Let y = −3 x + c , then y ′ = − 3 x1−1 + 0 = − 3x 0 = −3 ⋅ 1 = −3 b.

∫ 5k dx = 5k ∫ dx

5k 0 +1 x + c = 5k x + c 0 +1

∫

= 5k x 0 dx =

Check: Let y = 5k x + c , then y ′ = 5k x1−1 + 0 = 5k x 0 = 5k ⋅ 1 = 5k c.

2

2

2

Check: Let y = d.

∫ x dx = 5

f.

1 2 2 1 x + c , then y ′ = ⋅ 2 x 2 −1 + 0 = x 3 3 3

1 1 6 x + c , then y ′ = ⋅ 6 x 6 −1 + 0 = x5 6 6

1

∫ ax dx = a ∫ x dx = a ⋅ 7 + 1 x 7

7

7 +1

a 8 x +c 8

+c =

Check: Let y =

a 8 1 x + c , then y ′ = a ⋅ ⋅ 8 x8−1 + 0 = a x 7 8 8

∫(x

∫ x dx + ∫ x dx = 1 + 4 x

)

4

+ x3 dx =

Check: Let y = g.

1

1 1 5+1 x + c = x6 + c 5 +1 6

Check: Let y = e.

2 1 1+1 2 1 1 ⋅ x + c = ⋅ x2 + c = x 2 + c 3 1+1 3 2 3

∫ 3 x dx = 3 ∫ x dx = 3 ∫ x dx =

dx

∫ x4 = ∫ x

−4

4

1

3

1+ 4

+

1 1 1 3+1 x + c = x5 + x4 + c 3 +1 5 4

1 5 1 4 1 1 x + x + c , then y ′ = ⋅ 5 x5−1 + ⋅ 4 x 4 −1 + 0 = x 4 + x3 5 4 5 4

dx =

1 1 1 x − 4 +1 + c = − x −3 + c = − +c 3 − 4 +1 3x3

1 1 1 Check: Let y = − x −3 + c , then y ′ = − ⋅ −3 x −3−1 + 0 = x −4 = 4 3 3 x

h.

 1

1 

1

1

∫  x4 − x2  dx = ∫ x4 dx − ∫ x2 d x = ∫ x

−4

∫

dx − x −2 dx =

1 1 1 1 1 x − 4 +1 − x − 2 +1 = − x −3 + x −1 + c = − + +c 3 3 − 2 +1 − 4 +1 x 3x

1 1 1 1 Check: Let y = − x −3 + x −1 + c , then y ′ = − ⋅ −3 x −3−1 − x −1−1 + 0 = x −4 − x −2 = 4 − 2 3 3 x x

i.

∫

3

x 6 dx =

∫

6

x 3 dx =

Check: Let y =

1

∫ x dx = 1 + 2 x 2

2 +1

+c =

1 3 x +c 3

1 3 1 x + c , then y ′ = ⋅ 3 x3−1 + 0 = x 2 3 3

2. Evaluate the following indefinite integrals: a.

3

∫ 

 x 2 + x  dx = 

∫x

2 3

∫

dx + x dx =

Hamilton Education Guides

2 +1 1 1 1+1 x +c = x3 + 2 1+1 1+ 3

1

3+ 2 3

x

2+3 3

+

3 5 1 1 2 x + c = x 3 + x2 + c 5 2 2

436

Calculus I

Chapter 4 Solutions 5−3 2 3 53 1 2 3 5 5 −1 1 x + x + c , then y ′ = ⋅ x 3 + ⋅ 2 x 2 −1 + 0 = x 3 + x = x 3 + x = 5 2 5 3 2

Check: Let y = b.

∫

−

1

dt = −

2 t

Check: Let y =

c.

1

∫ 7 x2

dx =

∫

∫ (1 + x ) 1 2

2 x7

2

1 2

+ c , then y ′ = −

dx =

∫x

−2

7 dx

=

1 −1 t2

2

1 2

+0 = −

1 1 − 72 +c = x 1 − 72

∫(

)

x + x x dx =

1

x

= −

7−2 7

1 2

−1 t 2

∫

∫

x dx + x x dx =

2

+c =

∫x

1 2

1

= −

1 t2

= −

2 2 2 2 2 2 2 x ⋅x + x ⋅ x ⋅ x + c = x x + x2 x + c 3 5 3 5

t

7 75 77 5 x +c = x +c 5 5

∫

1 7

x2

1

dx + x x 2 dx =

∫x

1 2

∫

1+ 1

dx + x

2 dx

3− 2 5− 2 3 1 2 32 2 52 2 3 3 −1 2 5 5 −1 x + x + c , then y ′ = ⋅ x 2 + ⋅ x 2 + 0 = x 2 + x 2 = x 2 + x 2 3 5 3 2 5 2

∫ (2x

2

= 2⋅

1 2 1 1 3+1 1 2 +1 1 1+1 1 0 +1 x − 2⋅ x + x − x + c = x4 − x3 + x2 − x + c 1+ 3 1+ 2 1+1 1+ 0 2 3 2

+ 1 (x − 1) dx =

∫ (2 x

3

)

∫ 2 x dx − ∫ 2 x dx + ∫ x dx − ∫ dx = 2∫ x dx − 2∫ x dx + ∫ x dx − ∫ x dx

− 2 x 2 + x − 1 dx =

3

3

2

2

0

1 2 1 1 4 2 3 1 2 x − x + x − x + c , then y ′ = ⋅ 4 x 4 −1 − ⋅ 3 x3−1 + ⋅ 2 x 2 −1 − 1 + 0 = 2 x3 − 2 x 2 + x − 1 2 3 2 2 2 3

∫ x ( 3x − 1 ) dx = ∫ x (9 x 2

2

)

− 6 x + 1 dx =

∫ (9 x

3

)

− 6 x 2 + x dx =

9

∫ 9 x dx − ∫ 6 x dx + ∫ x dx = 3 + 1 x 3

2

3+1

−

6 2 +1 x 2 +1

1 9 9 6 1 1 1+1 x + c = x 4 − x3 + x 2 + c = x 4 − 2 x 3 + x 2 + c 1+1 4 3 2 2 4

Check: Let y =

9 4 1 9 1 x − 2 x3 + x 2 + c , then y ′ = ⋅ 4 x 4 −1 − 2 ⋅ 3 x3−1 + ⋅ 2 x 2 −1 + 0 = 9 x3 − 6 x 2 + x 4 2 4 2

∫ ( 2 + x ) dx = ∫ ( 2 3

=

1 2

3 +1 1 +1 2 3 2 5 2 3 2 5 1 1 1 1+ 2 1 3+ 2 x + x +c x2 + x 2 + c = 2 +1 x 2 + 2 + 3 x 2 + c = x 2 + x 2 + c = 1 3 3 5 3 5 1+ 2 1+ 2 2 2

3

dx + x 2 dx =

)

1− 2 t 2

7−2 7

=

+

g.

∫

∫x

Check: Let y = f.

1

t2

x2 + x

1 1 − 12 1 1 1 1 − 12 1 1 22−1 2 +c = − t +c t dt = − ⋅ − ⋅ 2 t = = t + c − ⋅ t + c 2 2 2 1− 1 2 2 −1

=

Check: Let y = e.

dt = −

5− 7 −2 7 5 5 −1 7 75 1 x + c , then y ′ = ⋅ x 7 + 0 = x 7 = x 7 = 2 = 5 5 7 x7

x dx =

∫

∫

1

1 −t 2

1

Check: Let y = d.

1 2

3

3

)

+ x3 + 3 ⋅ 22 x + 3 ⋅ 2 x 2 dx =

∫(x

3

)

+ 6 x 2 + 12 x + 8 dx =

∫ x dx + ∫ 6 x dx + ∫12 x dx + ∫ 8 dx 3

2

1 1 3+1 6 2 +1 12 1+1 8 0 +1 x + x + x + x + c = x4 + 2 x3 + 6 x2 + 8 x + c 3 +1 2 +1 1+1 0 +1 4

Check: Let y =

1 4 1 x + 2 x3 + 6 x 2 + 8 x + c , then y ′ = ⋅ 4 x 4 −1 + 2 ⋅ 3 x3−1 + 6 ⋅ 2 x 2 −1 + 8 x1−1 + 0 = x3 + 6 x 2 + 12 x + 8 x 0 4 4

= x3 + 6 x 2 + 12 x + 8 = (2 + x ) 3 h.

∫ ( 2 − x ) dx = ∫ ( 2 3

= −

3

)

− x3 + 3 ⋅ 22 x − 3 ⋅ 2 x 2 dx =

∫ (− x

3

)

∫

∫

∫

∫

− 6 x 2 + 12 x + 8 dx = − x3dx − 6 x 2 dx + 12 x dx + 8 dx

1 1 3+1 6 2 +1 12 1+1 8 0 +1 x − x + x + x + c = − x4 − 2 x3 + 6 x2 + 8 x + c 3 +1 2 +1 1+1 0 +1 4

1 1 Check: Let y = − x 4 − 2 x3 + 6 x 2 + 8 x + c , then y ′ = − ⋅ 4 x 4 −1 − 2 ⋅ 3 x3−1 + 6 ⋅ 2 x 2 −1 + 8 x1−1 + 0 = − x3 − 6 x 2 + 12 x 4 4

+ 8x 0 = − x3 − 6 x 2 + 12 x + 8 = (2 − x ) 3

Hamilton Education Guides

437

Calculus I

i.

∫

Chapter 4 Solutions

y5 + 4 y 2 y

 y5

∫(y

4 y 2  dy = y 2 

∫  y 2 +

dy =

2

)

3

∫ y dy + ∫ 4 dy = 3

+ 4 dy =

1 1 3+1 4 0 +1 y + y + c = y4 + 4 y + c 3 +1 0 +1 4

1 1 4 y + 4 y + c , then w ′ = ⋅ 4 y 4 −1 + 4 ⋅ 1 y1−1 + 0 = y 3 + 4 y 0 = y 3 + 4 4 4

Check: Let w =

Section 4.2 Practice Problems – Integration Using the Substitution Method 1. Evaluate the following indefinite integrals: a. Given

3 2 ∫ t (1 + t ) dt

(

)

d du du = 1 + t 3 = 3t 2 which implies that du = 3t 2 dt and dt = 2 . dt dt 3t

let u = 1 + t 3 , then

Substituting the u and dx values back into the original integral we obtain

du 2 2 3 ∫ t (1 + t ) dt = ∫ t ⋅ u ⋅ 3t 2 Check: Let y = b. Given

dx =

∫ 18x du

2

∫

+ c , then y ′ =

∫

Check: Let y =

∫

(

1 ⋅ 2 1 + t3 6

6 x 3 − 5 dx let u = 6 x3 − 5 , then

2

6 x 3 − 5 dx = 18 x 2 ⋅ u ⋅

2

c. Given

)

(

1 1 1+1 1 1 1 1 u + c = ⋅ u2 + c = u du = ⋅ 1 + t3 3 1+1 3 3 2 6

)

2 −1

⋅ 3t 2 + 0 =

(

(

)

2

+c

(

)

1 1 + t 3 ⋅ 3t 2 = t 2 1 + t 3 3

)

)

d du = 6 x3 − 5 = 18x 2 which implies that du = 18 x 2 dx and dx dx

. Substituting the u and dx values back into the original integral we obtain

18x 2

∫ 18x

(

1 1 + t3 6

=

(

2 6 x3 − 5 3

5

)

3 2

du 18 x

+ c , then y ′ =

dx let u = x + 5 , then

x+5

=

2

∫u

1 2 du

=

1 2

(

(

1 +1 2 3 2 3 2 1 6x3 − 5 u2 +c = u2 +c = u2 +c = 3 3 3 +1

2 3 ⋅ 6 x3 − 5 3 2

)

3 −1 2 ⋅ 18 x 2

(

)

)

3 2

+c

1

+ 0 = 6 x3 − 5 2 ⋅ 18 x 2 = 18 x 2 6 x3 − 5

d du = ( x + 5) = 1 which implies that du = dx . Substituting the u and dx dx dx

values back into the original integral we obtain

∫

5 x+5

dx =

5

∫

∫

⋅ du = 5 u

u

−1

2 du

= 5⋅

1

1 1 1 1 − 12 1 2 −1 u + c = 5 ⋅ 2 −1 u 2 + c = 10 u 2 + c = 10 ( x + 5 ) 2 + c 1 1− 2 2

Check: Let y = 10 (x + 5) 2 + c , then y ′ = 10 ⋅ d. Given

∫ x(x

2

−2

)

1 5

dx let u = x 2 − 2 , then

1 1 1 ( x + 5) 2 −1 + 0 = 5 ( x + 5) − 2 = 2

(

5

( x + 5)

1 2

5

=

x+5

)

du d 2 du = . x − 2 = 2 x which implies that du = 2 x dx and dx = dx dx 2x

Substituting the u and dx values back into the original integral we obtain

∫ x(x

2

−2

)

Check: Let e. Given

∫

1 5

dx =

∫ x ⋅u

(

5 2 x −2 12 3x

x2 + 3

)

6 5

1 5

⋅

(

1 +1 1 5 6 du 5 1 15 1 1 x2 − 2 = u du = ⋅ 1 u5 + c = ⋅ u5 + c = 2 6 2x 2 12 2 1 + 5

∫

+ c , then y ′ =

(

5 6 2 ⋅ x −2 12 5

dx let u = x 2 + 3 , then

)

(

6 −1 5 ⋅ 2x

(

+ 0 = x x2 − 2

)

)

6 5

+c

1 5

)

du d 2 du = . x + 3 = 2 x which implies that du = 2 x dx and dx = dx 2x dx

Substituting the u and dx values back into the original integral we obtain

∫

3x 2

x +3

dx =

∫

3x u

⋅

Hamilton Education Guides

(

1 1 3 3 − 12 du 3 1 1− 12 3 1 2 −1 = u du = ⋅ u + c = ⋅ 2 −1 u 2 + c = ⋅ 2 u 2 + c = 3u 2 + c = 3 x 2 + 3 1 2 2 2x 2 1− 2

∫

2

)

1 2

2

438

+c

Calculus I

Chapter 4 Solutions

(

)

Check: Let y = 3 x 2 + 3 t

∫2

f. Given

1 2

(

)

3 2 x +3 2

+ c , then y ′ =

1 −1 2 ⋅ 2x

(

+0 =

(

3 2 x +3 2

)

−1

2

3x

⋅ 2x =

( x + 3) 2

3x

=

1 2

x2 + 3

)

du d du = . 1 − t 2 = −2t which implies that du = −2t dt and dt = dt dt − 2t

1 − t 2 dt let u = 1 − t 2 , then

Substituting the u and dt values back into the original integral we obtain

∫

t 2

1 − t 2 dt =

Check: Let y = − g. Given

(

1 +1 1 12 1 2 3 1 3 1 t du 1 1 = − u du = − ⋅ 1 ⋅ u⋅ u 2 + c = − ⋅ u 2 + c = − u 2 + c = − 1 − t2 4 4 3 6 − 2t 2 6 4 +1 2

∫

∫

(

1 1− t2 6

)

3 2

+ c , then y ′ = −

2 3 2 3 ∫ x (1 − x ) dx let u = 1 − x , then

(

1 3 ⋅ 1− t2 6 2

)

3 −1 2 ⋅ −2t

(

+0 =

(

1 ⋅ t 1− t2 2

)

1 2

=

)

3 2

+c

t 1− t2 2

)

d du du = . 1 − x3 = − 3x 2 which implies that du = −3 x 2 dx and dx = dx dx − 3x 2

Substituting the u and dx values back into the original integral we obtain

du 2 2 2 3 2 ∫ x (1 − x ) dx = ∫ x ⋅ u ⋅ − 3x2 Check: Let y = − h. Given

(

1 1 − x3 9

x3

∫

)

3

= −

(

1 1 1 2 +1 1 2 1 1 u du = − ⋅ u + c = − ⋅ u3 + c = − 1 − x 3 3 3 3 3 2 +1 9

∫

+ c , then y ′ = −

)

3−1

(

⋅ −3 x 2 + 0 =

(

3 1 − x3 3

)

3−1

3

(

⋅ x 2 = x 2 1 − x3

+c

)

2

)

d 4 du du = x + 3 = 4x3 which implies that du = 4 x3dx and dx = 3 . dx dx 4x

dx let u = x 4 + 3 , then

4

(

1 ⋅ 3 1 − x3 9

)

x +3

Substituting the u and dx values back into the original integral we obtain

∫

x3

dx =

4

x +3

Check: Let y =

i. Given

∫ x (2x 8

9

x3

∫

⋅

u

(

1 4 x +3 2

)

)

du 4x 1 2

3

=

(

1 1 1 1 1 − 12 1 4 1 1 1 − 12 u du = ⋅ x +3 u +c = ⋅2 u2 +c = u2 +c = 1 4 2 4 2 4 1−

∫

)

1 2

+c

2

(

1 4 x +3 4

+ c , then y ′ =

2

+ 1 dx let u = 2 x9 + 1 , then

)

1 −1 2 ⋅ 4 x3

(

+0 =

(

1 4 x +3 4

)

−1

2

⋅ 4 x3 =

x3

(x + 3) 4

1 2

x3

=

x4 + 3

)

d du du = . 2 x9 + 1 = 18x8 which implies that du = 18 x8dx and dx = dx dx 18x8

Substituting the u and dx values back into the original integral we obtain

∫ x (2x 8

9

)

2

+ 1 dx =

Check: Let y =

∫x

8

⋅ u2 ⋅

(

)

du 18x

8

=

(

1 1 3 1 1 2 +1 1 u +c = u 2 du = ⋅ u +c = 2 x9 + 1 54 18 2 + 1 18 54

∫

(

)

3 1 1 ⋅ 3 2 x9 + 1 2 x9 + 1 + c , then y ′ = 54 54

3−1

(

)

(

2

)

3

+c

)

⋅ 18 x8 + 0 = 2 x9 + 1 ⋅ x8 = x8 2 x9 + 1

2

2. Evaluate the following indefinite integrals: a. Given

3 2 3 ∫ 6 x (2 x − 1)dx let u = 2 x − 1 , then

(

)

d du du = 2 x3 − 1 = 6x 2 which implies that du = 6 x 2 dx and dx = 2 . dx dx 6x

Substituting the u and dx values back into the original integral we obtain 1 1+1 du 2 2 3 ∫ 6 x (2 x − 1)dx = ∫ 6 x ⋅ u ⋅ 6 x2 = ∫ u du = 1 + 1 u + c

Check: Let y = b. Given

∫x

(

)

1 2 x3 − 1 2

2

+ c , then y ′ =

1 + x 2 dx let u = 1 + x 2 , then

Hamilton Education Guides

(

)

1 ⋅ 2 2 x3 − 1 2

(

=

2 −1

(

1 2 1 u +c = 2x3 − 1 2 2

(

)

2

+c

)

⋅ 6 x 2 + 0 = 6 x 2 2 x3 − 1

)

du d du = . 1 + x 2 = 2 x which implies that du = 2 x dx and dx = dx 2x dx

439

Calculus I

Chapter 4 Solutions

Substituting the u and dx values back into the original integral we obtain

∫x

∫x⋅

1 + x 2 dx =

Check: Let y = c. Given

∫

5

(

)

1 1 + x2 3

(

1 +1 1 3 1 1 12 1 2 3 du 1 1 = u du = ⋅ 1 1 + x2 u2 +c = ⋅ u2 +c = u2 +c = 2 2 3 3 2x 3 2 1 + 2

∫

u⋅ 3 2

(

1 3 ⋅ 1 + x2 3 2

+ c , then y ′ =

7 x + 1 dx let u = 7 x + 1 , then

)

3 −1 2 ⋅ 2x

+0 =

(

)

1 ⋅ 2x 1 + x2 2

1 2

)

3 2

+c

= x 1 + x2

du d du = (7 x + 1) = 7 which implies that du = 7 dx and dx = . Substituting dx 7 d

the u and dx values back into the original integral we obtain

∫

5

7 x + 1 dx =

∫

∫x

3

u⋅

1 +1 6 1 15 1 5 6 5 65 5 du 1 1 = u du = ⋅ 1 u +c = ( 7 x + 1) 5 + c u5 + c = ⋅ u5 + c = 7 7 6 42 7 42 7 +1 5

∫

6 −1 1 1 6 5 ( 7 x + 1) 5 + c , then y ′ = 5 ⋅ 6 ( 7 x + 1) 5 ⋅ 7 + 0 = 1 ( 7 x + 1) 5 ⋅ 7 = ( 7 x + 1) 5 = 42 5 7 42

Check: Let y = d. Given

5

(

5

7x + 1

)

d du du = . 3 x 2 − 1 = 6 x which implies that du = 6 x dx and dx = 6x dx dx

3 x 2 − 1 dx let u = 3 x 2 − 1 then

Substituting the u and dx values back into the original integral we obtain

∫

3

x 3 x 2 − 1 dx =

Check: Let y = e. Given

∫

∫

(

)

1 3x 2 − 1 8

x

(

1 4 1 13 1 3 4 1 du 1 1 13 +1 = u du = ⋅ 1 3x2 − 1 u +c = ⋅ u3 +c = u3 +c = 6 4 6 8 8 6x 6 1 + 3

∫

x⋅3 u ⋅

4 3

+ c , then y ′ =

)

4 −1 3 ⋅ 6x

(

+0 =

(

)

(

)

1 1 3x 2 − 1 3 ⋅ 6 x = x 3x 2 − 1 6

1 3

4 3

+c 3

= x 3x 2 − 1

)

d 2 du du = . x + 1 = 2 x which implies that du = 2 x dx and dx = dx 2x dx

dx let u = x 2 + 1 , then

2

(

1 4 ⋅ 3x 2 − 1 8 3

)

x +1

Substituting the u and dx values back into the original integral we obtain

∫

x x2 + 1

dx =

x

∫

⋅

u

(

)

Check: Let y = x 2 + 1

1 2

(

1 1 1 1 − 12 du 1 1 1 − 12 1 1 22−1 2 + c = u 2 + c = x2 + 1 ⋅ u = u du = ⋅ = = 2 u + c ⋅ u + c 1 − 2 1 2 2 2x 2 1− 2

∫

2

+ c , then y ′ =

(

2 2 2 ∫ x (1 − x ) dx let u = 1 − x , then

f. Given

)

1 2 x +1 2

)

1 2

+c

2

1 −1 2 ⋅ 2x

+0 =

(

(x + 1) 2

−1

2

x

⋅ 2x =

(x + 1) 2

)

1 2

x

=

2

x +1

d du du = . 1 − x 2 = −2 x which implies that du = −2 x dx and dx = dx dx − 2x

Substituting the u and dx values back into the original integral we obtain 2 du 2 2 ∫ x (1 − x ) dx = ∫ x ⋅ u ⋅ − 2 x

Check: Let y = − g. Given

∫

x5 6

(

1 1 − x2 6

)

3

= −

(

1 2 1 1 2 +1 1 1 u du = − ⋅ u + c = − u3 + c = − 1 − x 2 6 2 2 2 +1 6

∫

+ c , then y ′ = −

dx let u = x 6 + 3 , then

x +3

(

1 ⋅ 3 1 − x2 6

(

)

3 −1

⋅ −2 x + 0 =

(

6 1 − x2 6

)

2

)

3

+c

(

⋅ x = x 1 − x2

)

2

)

d 6 du du = x + 3 = 6x5 which implies that du = 6 x5dx and dx = 5 . dx dx 6x

Substituting the u and dx values back into the original integral we obtain

∫

x5 6

x +3

dx =

∫

x5 u

⋅

Hamilton Education Guides

du 6x

5

=

1 6

∫

1 u

⋅ du =

(

1 − 12 1 6 1 1 1 1 1 − 12 u du = ⋅ x +3 u +c = u2 +c = 1 6 3 3 6 1−

∫

)

1 2

+c

2

440

Calculus I

Chapter 4 Solutions

(

3x

∫

h. Given

)

1 6 x +3 3

Check: Let y =

1 2

+ c , then y ′ =

dx let u = x 2 − 1 , then

x2 − 1

(

)

1 1 6 ⋅ x +3 3 2

1 −1 2 ⋅ 6 x5

(

(

+ 0 = x5 x 6 + 3

)

−1

2

x5

=

x6 + 3

)

du d 2 du = . x − 1 = 2 x which implies that du = 2 x dx and dx = dx 2x dx

Substituting the u and dx values back into the original integral we obtain

∫

3x x2 − 1

dx =

∫

3x u

(

)

Check: Let y = 3 x 2 − 1

i. Given

5x4 + 6x

∫

∫

1 2

+ c , then y ′ =

)

du = 5 x 4 + 6 x dx and dx =

∫

5x4 + 6x

dx =

x5 + 3x 2 + 1

)

(

= 2 x5 + 3x 2 + 1

1 2

du

5x4 + 6x

5x4 + 6x

∫

u

(

(x

5x4 + 6x 5

)

1 −1 2 ⋅ 2x

⋅

(

)

+ 0 = 3x x 2 − 1

(

−1

2

=

3x

(x − 1) 2

1 2

=

)

1 2

+c

3x x2 − 1

)

d 5 du = x + 3 x 2 + 1 = 5 x 4 + 6 x which implies that dx dx

. Substituting the u and dx values back into the original integral we obtain du

4

5x + 6x

=

∫

du

=

u

∫

u

−1

2 du

=

1 1 1 − 12 1 22−1 u + c = 2u 2 + c u +c = 1 2 −1 1− 2 2

+ c = 2 x5 + 3 x2 + 1 + c

)

Check: Let y = 2 x5 + 3 x 2 + 1

=

(

3 2 x −1 2

dx let u = x5 + 3 x 2 + 1 , then

x5 + 3x 2 + 1

(

(

1 3 − 12 3 1 1 − 12 du 3 1 22−1 2 + c = 3 x2 − 1 = = = u du = ⋅ + u c u c 3 u + ⋅ 2 2 1− 1 2x 2 2 −1 2 2

⋅

)

+ 3x 2 + 1

1 2

=

1 2

+ c , then y ′ = 2 ⋅

(

) (5x

1 5 x + 3x 2 + 1 2

1 −1 2

4

)

(

) (5x

+ 6 x + 0 = x5 + 3 x 2 + 1

−1

2

4

+ 6x

)

5x4 + 6x x5 + 3x 2 + 1

Section 4.3 Solutions – Integration of Trigonometric Functions 1. Evaluate the following integrals. a. Given

∫ sin

∫ sin

5 du d 3x 3 3x 3x , then = which implies dx = du . Therefore, = dx let u = dx dx 5 5 5 5 3

3x dx = 5

5

5

5

5

∫ sin u ⋅ 3 du = 3 ∫ sin u du = − 3 cos u + c = − 3 cos

3x +c 5

5 d 3x d 3x d 3x 3x 5 5 3x 3 5 3x Check: Let y = − cos + c = − ⋅ − sin + c , then y ′ = − ⋅ cos ⋅ + 0 = − ⋅ − sin ⋅ = sin 3 5 dx 5 3 5 5 5 3 5 3 dx 5 dx

b. Given

∫ 3 cos 2 x dx let u = 2 x , then

∫ 3 cos 2 x dx = 3∫ cos u ⋅ Check: Let y = c. Given

du du d . Therefore, = 2 x = 2 which implies dx = dx dx 2

3 3 du 3 = cos u du = sin u + c = sin 2 x + c 2 2 2 2

∫

3 d d 6 3 3 d 3 sin 2 x + c , then y ′ = ⋅ sin 2 x + c = ⋅ cos 2 x ⋅ 2 x + 0 = ⋅ cos 2 x ⋅ 2 = ⋅ cos 2 x = 3 cos 2 x 2 2 2 dx 2 dx dx 2

∫ (sin 5x − cos 7 x ) dx = ∫ sin 5x dx −∫ cos 7 x dx let:

a. u = 5 x , then

du du du d and = 5 ; du = 5dx ; dx = = 5x ; 5 dx dx dx

Hamilton Education Guides

441

Calculus I

Chapter 4 Solutions dv dv d dv . = 7 ; dv = 7 dx ; dx = = 7x ; 7 dx dx dx

b. v = 7 x , then Therefore,

∫ sin 5x dx −∫ cos 7 x dx = ∫ sin u ⋅

du dv 1 1 1 1 = − cos v ⋅ cos v dv = − cos u + c1 − sin v + c2 sin u du − 5 7 5 7 7 5

∫

∫

∫

1 1 1 1 = − cos 5 x − sin 7 x + c1 + c2 = − cos 5 x − sin 7 x + c 5 7 5 7 1 d 1 d d 1 d 1 1 Check: Let y = − cos 5 x − sin 7 x + c then y ′ = − ⋅ cos 5 x − ⋅ sin 7 x + c = ⋅ sin 5 x ⋅ 5x 5 7 5 dx 7 dx dx 5 dx 5 7 d 1 ⋅ cos 7 x ⋅ 7 x + 0 = ⋅ sin 5 x − ⋅ cos 7 x = sin 5 x − cos 7 x dx 7 5 7

−

d. Given

∫ 2 csc

∫ 2 csc 2

2

du du du d . Therefore, = 3x ; = 3 ; du = 3dx ; dx = dx dx 3 dx

3 x dx let u = 3 x , then

∫

3 x dx = 2 csc2 u ⋅

2 2 2 du = csc2 u du = − cot u + c = − cot 3 x + c 3 3 3 3

∫

2 2 d 2 2 d d Check: Let y = − cot 3 x + c , then y ′ = − ⋅ cot 3 x + c = − ⋅ − csc2 3 x ⋅ 3 x + 0 = ⋅ csc2 3 x ⋅ 3 = 2 csc2 3 x 3 3 3 dx dx 3 dx

∫ x sec

e. Given

∫ x sec

2

2

x 2 dx let u = x 2 , then

x 2 dx =

Check: Let y =

∫ x sec

2

Check: Let y =

g. Given

∫ sin

3

∫

∫ cos

du d du du . Therefore, = cos x ; du = cos x ⋅ dx ; dx = = sin x ; dx dx dx cos x

x cos x dx let u = sin x , then

Check: Let y =

3

8 8 du 8 = sec u du = ln sec u + tan u + c = ln sec 5 x + tan 5 x + c 5 5 5 5

8 8 1 ln sec 5 x + tan 5 x + c , then y ′ = ⋅ ⋅ 5 sec 5 x tan 5 x + 5 sec2 5 x + 0 5 5 sec 5 x + tan 5 x

x cos x dx =

h. Given

du du d du . Therefore, = 5 ; du = 5 dx ; dx = = 5x ; 5 dx dx dx

8 5 sec 5 x (sec 5 x + tan 5 x ) = 8 sec 5 x ⋅ 5 sec 5 x + tan 5 x

=

∫ cos

∫

1 d 1 d 1 d 2 1 tan x 2 + c , then y ′ = ⋅ tan x 2 + c = ⋅ sec2 x 2 ⋅ x + 0 = ⋅ sec2 x 2 ⋅ 2 x = x sec2 x 2 2 dx 2 dx 2 dx 2

∫ 8 sec 5x dx = 8∫ sec u ⋅

3

du 1 1 1 = sec2 u du = tan u + c = tan x 2 + c 2 2x 2 2

∫ 8 sec 5x dx let u = 5x , then

f. Given

∫ sin

u⋅

du du d 2 du . Therefore, = 2 x ; du = 2 x dx ; dx = = x ; dx 2x dx dx

3

∫u

3

cos x ⋅

du = cos x

∫ u du = 3

1 1 4 u + c = sin 4 x + c 4 4

1 4 1 sin x + c , then y ′ = ⋅ 4 sin 3 x ⋅ cos x + 0 = sin 3 x cos x 4 4 du d du du . Therefore, = cos x ; = − sin x ; dx = dx dx dx − sin x

x sin x dx let u = cos x , then

x sin x dx =

∫u

3

sin x ⋅

1 1 du = − u 3du = − u 4 + c = − cos4 x + c 4 4 − sin x

∫

1 1 4 Check: Let y = − cos 4 x + c , then y ′ = − ⋅ 4 cos3 x ⋅ − sin x + 0 = cos3 x sin x = cos3 x sin x 4 4 4

i. Given x

x

∫ 2 cos x

∫ 2 cos x

2

dx =

2

dx let u = x 2 , then x

du

∫ 2 cos u ⋅ 2 x

Hamilton Education Guides

=

du d 2 du du . Therefore, x ; = 2 x ; du = 2 x ⋅ dx ; dx = = dx 2x dx dx

1 1 1 cos u du = ⋅ sin u + c = sin x 2 + c 4 4 4

∫

442

Calculus I

Chapter 4 Solutions

Check: Let y =

x 1 1 2 sin x 2 + c , then y ′ = ⋅ cos x 2 ⋅ 2 x + c = x cos x 2 + 0 = cos x 2 4 4 4 2

2. Evaluate the following integrals. a. Given

∫e

3x

∫e

3x

Check: Let y =

b.

∫ tan ∫ tan

9

9

c. Given 5

3x

sec u ⋅

du 3e

1 1 1 sec u du = ln sec u + tan u + c = ln sec e 3 x + tan e 3 x + c 3 3 3

∫

=

3x

1 1 1 ln sec e3 x + tan e3 x + c , then y ′ = ⋅ ⋅ 3e3 x sec e3 x tan e3 x + 3e3 x sec2 e3 x + 0 3 x 3 3 sec e + tan e3 x

(

)

1 3e3 x sec e3 x sec e3 x + tan e3 x = e3 x sec e3 x ⋅ 3 sec e3 x + tan e3 x

du d du du . Therefore, = tan x ; = sec 2 x ; du = sec 2 x dx ; dx = dx dx dx sec 2 x

x sec 2 x dx let u = tan x , then x sec 2 x dx =

Check: Let y =

∫ cot

∫e

sec e 3 x dx =

=

du d 3 x du du = 3e3 x ; du = 3e3 x ⋅ dx ; dx = 3 x . Therefore, = e ; dx dx dx 3e

sec e 3 x dx let u = e3 x , then

∫ cot

5

∫u

9

⋅ sec2 x ⋅

du

sec2 x

∫ u du 9

=

=

1 1 10 1 9 +1 u +c = u +c = tan10 x + c 10 9 +1 10

1 1 ⋅ 10 (tan x )10 −1 ⋅ sec2 x + 0 = (tan x )9 sec2 x = tan 9 x sec2 x tan10 x + c , then y ′ = 10 10

x csc 2 x dx let u = cot x , then

x csc 2 x dx =

∫u

5

⋅ csc2 x ⋅

− du 2

csc x

du d du du . Thus, = cot x ; = − csc 2 x c ; du = − csc 2 x dx ; dx = − dx dx dx csc 2 x

∫

= − u 5du =

−1 5+1 1 1 u + c = − u 6 + c = − cot 6 x + c 5 +1 6 6

1 1 Check: Let y = − cot 6 x + c , then y ′ = − ⋅ 6 (cot x ) 6 −1 ⋅ − csc2 x + 0 = cot 5 x csc2 x 6 6

d. Given

∫ sec 2 x tan 2 x dx let u = 2 x , then

∫ sec 2 x tan 2 x dx = ∫ sec u ⋅ tan u ⋅ Check: Let y = e. Given

∫ x cot x

∫ x cot x

2

dx =

Check: Let y = f. Given 1

∫3x

2

1

∫3x

2

=

∫

1 2 1 sec 2 x + c then y ′ = ⋅ sec 2 x tan 2 x ⋅ 2 + 0 = ⋅ sec 2 x tan 2 x = sec 2 x tan 2 x 2 2 2 2

dx let u = x 2 , then du

∫ x ⋅ cot u ⋅ 2 x

=

du du d 2 du . Therefore, = 2 x ; du = 2 x dx ; dx = = x ; dx dx dx 2x

1 1 1 cot u ⋅ du = ln sin u + c = ln sin x 2 + c 2 2 2

∫

1 2 x cos x 2 1 1 2 = = x cot x 2 ln sin x 2 + c , then y ′ = ⋅ ⋅ ⋅ x ⋅ x + cos 2 0 2 2 sin x 2 2 sin x 2

csc x 3 dx =

2

du 1 1 1 = sec u tan u du = sec u + c = sec 2 x + c 2 2 2 2

csc x 3 dx let u = x 3 , then

Check: Let y =

du d du du . Therefore, = 2x ; = 2 ; du = 2dx ; dx = dx dx dx 2

du d 3 du du . Therefore, = 3x 2 ; du = 3 x 2 dx ; dx = = x ; dx dx dx 3x 2

1 1 1 1 2 du csc u ⋅ du = ln csc u − cot u + c = ln csc x 3 − cot x 3 + c x ⋅ csc u ⋅ 2 = 9 9 9 3 3x

∫

∫

(

1 3 x 2 csc x3 csc x3 − cot x3 − csc x3 cot x3 ⋅ 3 x 2 + csc2 x3 ⋅ 3 x 2 ln csc x3 − cot x3 + c , then y ′ = +0 = 9 9 csc x3 − cot x3 9 csc x3 − cot x3

3 x csc x 9

(

3

=

)

(

)

)

2

x 1 ⋅ csc x3 = x 2 csc x3 3 3

Hamilton Education Guides

443

Calculus I

g.

Chapter 4 Solutions sin 5 x

1

∫ (sin 5x csc 5x ) dx = ∫ sin 5x ⋅ sin 5x dx = ∫ sin 5x dx = ∫ dx =

x+c

Check: Let y = x + c , then y ′ = x1−1 + 0 = x 0 = 1 h.

∫ ( cos 5t sec 5t + 3t

2

Check: Let y = t 3 +

∫e

i. Given

∫e

cot x

cot x

)

+ t dt =

1

∫ cos 5t ⋅ cos 5t dt + 3∫ t dt + ∫ t dt = ∫ dt + 3∫ t dt + ∫ t dt 2

= t3 +

1 2 t +t+c 2

1 2 1 t + t + c , then y ′ = 3 t 3−1 + ⋅ 2t 2 −1 + 1 + 0 = 3 t 2 + t + 1 2 2

csc2 x dx let u = cot x , then

∫e

csc2 x dx =

2

u

csc2 x ⋅

− du

csc2 x

du du d du . Therefore, = cot x ; = − csc2 x ; du = − csc2 x dx ; dx = − dx dx dx csc2 x

∫

= − eu du = − eu + c = − e cot

x

+c

Check: Let y = −ecot x + c , then y ′ = − ecot x ⋅ − csc2 x ⋅ 1 + 0 = ecot x csc2 x 3. Evaluate the following integrals. a.

=

∫

∫

=

1 1 2 8 1 1 sin 8 x − sin 2 x cos 8 x + cos 2 x + c , then y ′ = − ⋅ −8 sin 8 x + ⋅ −2 sin 2 x + 0 = 16 4 4 16 4 16

1 1 1 1 1 1 sin 8 x − sin 2 x = sin 8 x + sin (− 2 x ) = sin (3 + 5)x + sin (3 − 5)x = sin 3 x cos 5 x 2 2 2 2 2 2 1

1

1

1

∫ cos 6 x cos 4 x dx = ∫ 2 [ cos (6 − 4)x + cos (6 + 4)x ] dx = ∫ 2 (cos 2 x + cos10 x ) dx = 2 ∫ cos 2 x dx + 2 ∫ cos10 x dx =

1 1 1 1 1 1 sin 10 x + c ⋅ ⋅ sin 2 x + ⋅ sin 10 x + c = sin 2 x + 4 20 2 2 2 10 1 1 1 1 1 1 ⋅ 10 cos 10 x + 0 = cos 2 x + cos 10 x sin 10 x + c , then y ′ = ⋅ 2 cos 2 x + sin 2 x + 2 2 20 4 4 20

Check: Let y = = c.

1

1 1 1 1 1 1 1 1 sin 8 x dx − sin 2 x dx = ⋅ − cos 8 x + ⋅ cos 2 x + c = − cos 8 x + cos 2 x + c 2 8 2 2 2 2 4 16

Check: Let y = −

b.

1

1

∫ sin 3x cos 5x dx = ∫ 2 [ sin (3 − 5)x + sin (3 + 5)x ] dx = ∫ 2 [ sin (− 2 x ) + sin (8x ) ] dx = 2 ∫ (sin 8x − sin 2 x ) dx

∫ sin

5

1 1 1 1 cos 2 x + cos 10 x = cos (6 − 4 )x + cos (6 + 4 )x = cos 6 x cos 4 x 2 2 2 2

∫ sin

3 x dx =

4

3 x sin 3 x dx =

2 2 2 2 ∫ (sin 3x ) sin 3x dx = ∫ (1 − cos 3x ) sin 3x dx . Let u = cos 3x , then

du d du du . Therefore, = cos 3 x ; = −3 sin 3 x ; du = −3 sin 3 x dx ; dx = − dx dx dx 3 sin 3 x

∫ sin = −

5

3 x dx =

2 du 4 2 2 2 2 ∫ (1 − cos 3x ) sin 3x dx = ∫ (1 − u ) sin 3x dx = ∫ (u − 2u + 1)sin 3x ⋅ − 3sin 3x

= −

1 3

4

)

− 2u 2 + 1 du

1 2 1 1 4 2 2 1 1 1 2 1 1 u du + u du − du = − ⋅ u 5 + ⋅ u 3 − u + c = − cos5 3 x + cos 3 3 x − cos 3 x + c 3 3 3 3 5 3 3 3 15 9 3

∫

∫

Check: Let y = −

∫

1 2 1 2 1 1 cos5 3 x + cos3 3 x − cos 3 x + c , then y ′ = − ⋅ 15 cos 4 3 x ⋅ − sin 3 x + ⋅ 9 cos 2 3 x ⋅ − sin 3 x + ⋅ 3 sin 3 x 3 15 9 3 15 9

(

)

(

= sin 3 x cos 4 3 x − 2 sin 3 x cos 2 3 x + sin 3 x = sin 3 x cos 4 3 x − 2 cos 2 3 x + 1 = sin 3 x 1 − cos 2 3 x

(

= sin 3 x sin 2 3 x d.

∫ (u

∫ tan

4

2 x dx =

∫ tan

2

)

2

2

= sin 3 x sin 4 3 x = sin 5 3 x

2 x tan 2 2 x dx =

Hamilton Education Guides

)

∫ tan

2

(

)

2 x sec2 2 x − 1 dx =

∫ tan

2

∫

2 x sec2 2 x dx − tan 2 2x dx

444

Calculus I

=

Chapter 4 Solutions

∫ tan

2

2 x sec2 2 x dx −

2

∫ tan

4

2 x sec2 2 x dx =

∫ tan

2 x dx =

2

)

∫ tan

2 x − 1 dx =

2

∫

∫

2 x sec2 2 x dx − sec2 2 x dx + dx . To solve the first integral let

du d du du . Therefore, = tan 2 x ; = 2 sec2 2 x ; du = 2 sec2 2 x dx ; dx = dx dx dx 2 sec2 2 x

u = tan 2 x , then

∫ tan

∫ (sec

∫u

2

du

sec2 2 x ⋅

=

2

2 sec 2 x

1 2 1 1 1 1 u du = ⋅ u 3 = u 3 = tan 3 2 x . Grouping the terms we obtain 2 2 3 6 6

∫

∫

2

1 1 1 tan 3 2 x − sec2 2 x dx + dx = tan 3 2 x − tan 2 x + x + c 6 6 2

∫

∫

2 x sec2 2 x dx − sec2 2 x dx + dx =

∫

1 1 6 2 tan 3 2 x − tan 2 x + x + c , then y ′ = tan 2 2 x ⋅ sec2 2 x − sec2 2 x + 1 + 0 = tan 2 2 x ⋅ sec2 2 x − sec2 2 x + 1 6 2 6 2

Check: Let y =

(

) ( )( ) 1 1 1 ax )dx = ∫ dx − ∫ sin ax dx = ∫ dx − ∫ (1 − cos 2ax )dx = x − ∫ dx + ∫ cos 2ax dx 2 2 2

= sec2 2 x tan 2 2 x − 1 + 1 = tan 2 2 x + 1 tan 2 2 x − 1 + 1 = tan 4 2 x − tan 2 2 x + tan 2 2 x − 1 + 1 = tan 4 2 x e.

∫ cos

2

∫ (1 − sin

ax dx =

2

2

x sin 2ax x 1 1  1 1 + + ⋅ sin 2ax + c = x1 −  + +c sin 2ax + c = 2 2 2a 2 4a 2  4a 

= x−

1 1 x sin 2ax 1 1 1 2a + + c , then y ′ = + ⋅ cos 2ax ⋅ 2a + 0 = + ⋅ cos 2ax = + cos 2ax 4a 2 2 2 2 4a 2 4a

Check: Let y =

1  1 1 1 1  = 1 −  + cos 2ax = 1 −  − cos 2ax  = 1 − (1 − cos 2ax ) = 1 − sin 2 ax = cos 2 ax 2  2 2 2 2 

f.

∫ tan

3

∫ tan

5 x dx =

2

2

∫ tan

∫ sec

5 x tan 5 x dx =

3

∫ tan

5 x dx =

2

)

5 x − 1 tan 5 x dx =

∫ sec

2

∫

5 x tan 5 x dx − tan 5 x dx . To solve the first

2

5x ⋅ u ⋅

du

5 sec 5 x

5 x tan 5 x dx =

∫ (sec

1 1 1 2 1 1 u = u du = ⋅ u 2 = tan 2 5 x . Combining the term we obtain 5 5 2 10 10

∫

=

2

2

)

5 x − 1 tan 5 x dx =

∫ (sec

2

)

5 x tan 5 x − tan 5 x dx =

∫ sec

2

5 x tan 5 x dx

− tan 5 x dx =

1 1 1 tan 2 5 x − tan 5 x dx = tan 2 5 x − ln sec 5 x + c 10 10 5

Check: Let y =

1 1 1 1 1 tan 2 5 x − ln sec 5 x + c , then y ′ = ⋅ 10 tan 5 x ⋅ sec2 5 x − ⋅ ⋅ 5 sec 5 x tan 5 x + 0 10 5 10 5 sec 5 x

∫

∫

= tan 5 x sec2 5 x − g.

2

du d du du . Therefore, = tan 5 x ; = 5 sec2 5 x ; du = 5 sec2 5 x dx ; dx = dx dx dx 5 sec2 5 x

integral let u = tan 5 x , then

∫ sec

∫ (sec

5 x tan 5 x dx =

(

)

sec 5 x tan 5 x = tan 5 x sec2 5 x − tan 5 x = tan 5 x sec2 5 x − 1 = tan 5 x tan 2 5 x = tan 3 5 x sec 5 x

∫ cot 3x (csc 3x − 1) dx = ∫ cot 3x csc 3x dx − ∫ cot 3x dx = ∫ cot 2 3 x csc2 3 x dx − ∫ ( csc2 3 x − 1) dx = ∫ cot 2 3 x csc2 3 x dx − ∫ csc2 3 x dx + ∫ dx . To solve the first integral let ∫ cot

4

3 x dx =

∫ cot

2

3 x cot 2 3 x dx =

2

2

2

2

2

du d du du . Therefore, = cot 3 x ; = −3 csc2 3 x ; du = −3 csc2 3 x dx ; dx = − dx dx dx 3 csc2 3 x

u = cot 3 x , then

∫ cot

2

3 x csc2 3 x dx =

∫ cot

4

3 x dx =

∫ cot

2

∫u

2

csc2 3 x ⋅ −

du

3 csc2 3 x

= −

1 1 1 2 u du = − u 3 = − cot 3 3 x . Grouping the terms we obtain 9 9 3

∫

1 1 1 3 x csc2 3 x dx − csc2 3 x dx + dx = − cot 3 3 x − csc2 3 x dx + dx = − cot 3 3 x + cot 3 x + x + c 9 9 3

∫

∫

∫

∫

9 3 1 1 Check: Let y = − cot 3 3 x + cot 3 x + x + c , then y ′ = − cot 2 3 x ⋅ − csc2 3 x − csc2 3 x + 1 = cot 2 3 x csc2 3 x − csc2 3 x + 1 3 9 9 3

(

)

(

)(

)

= csc2 3 x cot 2 3 x − 1 + 1 = cot 2 3 x + 1 cot 2 3 x − 1 + 1 = cot 4 3 x − cot 2 3 x + cot 2 3 x − 1 + 1 = cot 4 3 x

Hamilton Education Guides

445

Calculus I

h.

∫ cot

Chapter 4 Solutions

3

2 x dx =

∫ cot

2

integral let u = cot 2 x , then

∫ csc

2

∫ cot

3

∫ csc

2 x cot 2 x dx = 2 x dx =

∫ cot

∫

− cot 2 x dx = −

∫ (csc

2 x cot 2 x dx =

2

2

2

)

2 x − 1 cot 2 x dx =

∫ csc

2

∫

2 x cot 2 x dx − cot 2 x dx . To solve the first

du du d du . Therefore, = = −2 csc2 2 x ; du = −2 csc2 2 x dx ; dx = − cot 2 x ; dx dx dx 2 csc2 2 x

2x ⋅ u ⋅ −

du

2

2 csc 2 x

2 x cot 2 x dx =

∫ (csc

2

= −

)

1 1 1 u du = − u 2 = − cot 2 2 x . Combining the term we obtain 4 4 2

∫

2 x − 1 cot 2 x dx =

∫ (csc

)

2

2 x ⋅ cot 2 x − cot 2 x dx =

∫ csc

2

2 x cot 2 x dx

1 1 1 cot 2 2 x − cot 2 x dx = − cot 2 2 x − ln sin 2 x + c 4 2 4

∫

1 1 1 1 1 Check: Let y = − cot 2 2 x − ln sin 2 x + c , then y ′ = − ⋅ 4 cot 2 x ⋅ − csc2 2 x − ⋅ ⋅ 2 cos 2 x + 0 4 2 4 2 sin 2 x

= cot 2 x csc2 2 x − i.

∫ cot

6

3 x dx =

∫ cot

4

3 x cot 2 3 x dx =

∫ cot

4

(

)

3 x csc2 3 x − 1 dx =

∫

− cot 4 3x dx . From the problem g above we know that

∫ cot =

6

3 x dx =

∫ cot

4

∫ cot

4

)

(

cos 2 x = cot 2 x csc2 2 x − cot 2 x = cot 2 x csc2 2 x − 1 = cot 2 x cot 2 2 x = cot 3 2 x sin 2 x

3 x cot 2 3 x dx =

∫ cot

4

(

∫ cot

4

)

∫ (cot

4

)

3 x csc2 3 x − cot 4 3 x dx =

∫ cot

4

3 x csc2 3 x dx

1 1 3x dx = − cot 3 3 x + cot 3 x + x + c . Therefore, 3 9

3x csc2 3x − 1 dx =

∫ cot

4

∫

3 x csc2 3 x dx − cot 4 3x dx

1 1 1 1  1  3 x csc2 3 x dx −  − cot 3 3 x + cot 3 x + x + c  = − cot 5 3 x + cot 3 3 x − cot 3 x − x + c 15 9 3 3  9 

Check: Let y = − +

15 ⋅ cot 4 3 x ⋅ − csc2 3 x 9 ⋅ cot 2 3 x ⋅ − csc2 3 x 1 1 1 + cot 5 3 x + cot 3 3 x − cot 3 x − x + c , then y ′ = − 9 15 15 9 3

(

)

3 csc2 3 x − 1 = cot 4 3 x ⋅ csc2 3 x − cot 2 3 x ⋅ csc2 3 x + csc2 3 x − 1 = csc2 3 x cot 4 3 x − cot 2 3 x + 1 − 1 3

(

)(

)

= 1 + cot 2 3 x cot 4 3 x − cot 2 3 x + 1 − 1 = cot 4 3 x − cot 2 3 x + 1 + cot 6 3 x − cot 4 3 x + cot 2 3 x − 1 = cot 6 3 x

Section 4.4 Solutions – Integration of Expressions Resulting in Inverse Trigonometric Functions 1. Evaluate the following indefinite integrals. a. First - Write the given integral in its standard form

∫

dx a2 − x2

= arc sin

x + c , i.e., a

∫

dx

=

25 − 9 x 2

∫

dx 52 − (3 x ) 2

du d du . = 3 x = 3 x which implies du = 3 x dx ; dx = dx dx 3 1 u 1 du du dx 1 Therefore, = = = arc sin ⋅ 3 5 3 3 25 − 9 x 2 52 − u 2 52 − u 2 1 u 1 3x Third - Write the answer in terms of the original variable, i.e., x . arc sin + c = arc sin +c 5 3 3 5 1 3x Fourth - Check the answer by differentiating the solution, i.e., let y = arc sin + c then 5 3 1 5 1 d 3x 3 1 1 1 1 1 3 = = y′ = ⋅ +0 = ⋅ = 5 dx 5 3 3 5 15 2 2 2 2 x 25 −9 x 25 − 9 x 25 − 9 x 2 1 − 925 1 − 3x 25

Second - Use substitution method by letting u = 3 x , then

∫

∫

∫

(5 )

b. First - Write the given integral in its standard form

Hamilton Education Guides

∫

dx a2 − x2

= arc sin

x + c , i.e., a

∫

dx 4 − x2

=

∫

dx 22 − x 2

446

Calculus I

Chapter 4 Solutions du d = x = 1 which implies du = dx . Therefore, dx dx

Second - Use substitution method by letting u = x , then

∫

dx 4 − x2

du

∫

=

u +c 2

= arc sin

22 − u 2

u x + c = arc sin + c 2 2 x Fourth - Check the answer by differentiating the solution, i.e., let y = arc sin + c then 2 1 2 1 1 d x 1 1 1 1 = = y′ = ⋅ = ⋅ +0 = 2 2 2 dx 2 2 2 2 2 4− x 4− x 4 − x2 1− x 1− x

Third - Write the answer in terms of the original variable, i.e., x . arc sin

(2 )

4

4

c. First - Write the given integral in its standard form

dx

∫

a2 − x2

∫

=

25 − x 6

x 2 dx

∫

( )

52 − x 3

=

2

x2

∫

⋅

52 − u 2

du 3x

=

2

1 3

∫

du

1

d x3 1 +0 = dx 5 3

⋅

3 1 −  x5   

2

1 1−

⋅ x6 25

1 3x 2 = 5 5

d. First - Write the given integral in its standard form =

dx

∫ 9 (16 + x2 ) 9

=

1 9

dx

∫

16 9

+x

1 9

=

2

∫

x 2 dx

=

25 − x 6

x 2 dx

∫

( )

52 − x 3

2

x2

x3 1 arc sin + c then 5 3

=

25 − x 6 25

1

dx

u 1 arc sin + c 3 5

1 x3 1 u arc sin + c = arc sin +c 3 5 3 5

Fourth - Check the answer by differentiating the solution, i.e., let y = 1 3

=

52 − u 2

Third - Write the answer in terms of the original variable, i.e., x .

y′ =

x + c , i.e., a

du d 3 du = x = 3x 2 which implies du = 3 x 2 dx ; dx = 2 . Thus, dx dx 3x

Second - Use substitution method by letting u = x3 , then

x 2 dx

= arc sin

5x2

1 5

25 − x 6

x

=

∫ a 2 + x2 = a arc tan a + c , i.e., ∫

x2 25 − x 6

dx

9 x 2 + 16

=

dx

∫ 16 + 9 x2

dx

∫ (4 ) 2 + x 2 3

du d = x = 1 which implies du = dx . Therefore, dx dx 1 3 3u 1 3u 1 1 u = ⋅ 4 arc tan 4 + c = ⋅ arc tan +c = arc tan +c 12 4 9 4 4 9

Second - Use substitution method by letting u = x , then 1 9

dx

∫ (4 ) 2 + x 2 3

=

1 9

dx

∫ (4 ) 2 + u 2 3

3

3

Third - Write the answer in terms of the original variable, i.e., x .

1 3x 1 3u arc tan +c arc tan +c = 12 4 12 4

Fourth - Check the answer by differentiating the solution, i.e., let y = y′ =

1 1 ⋅ 12 1 + 3 x

(4 )

2

⋅

1 3x arc tan + c then 12 4

1 1 16 d 3x 1 1 3 3 1 = = ⋅ +0 = ⋅ ⋅ ⋅ = 16 16 + 9 x 2 dx 4 12 1 + 9 x 2 4 48 16 + 9 x 2 16 + 9 x 2 16

16

Note that another way of solving this class of problems is by rewriting the integral in the following way:

∫

dx

2

9 x + 16

Thus,

∫

= dx

dx

∫ 16 + 9 x2 = ∫

42 + (3 x )2

=

∫

Hamilton Education Guides

1

42 + u 2

dx

4 + (3 x ) 2

⋅

2

du 1 = 3 3

. Now, let u = 3 x , then

∫

du

42 + u 2

=

du d du . = 3 x = 3 which implies du = 3dx ; dx = 3 dx dx

1 3x 1 1 u 1 u +c ⋅ arc tan + c = arc tan + c = arc tan 12 12 4 4 3 4 4

447

Calculus I

Chapter 4 Solutions

e. First - Write the given integral in its standard form

=

1 9

∫

x 2 dx 7 9

+x

1 9

=

6

1 9

∫

  

7 3

( )

2

 + x3  

x 2 dx   

2

=

( )

3 2

1 9

∫

7 3

x2   

⋅

2

du

=

3x 2

 + u2   3u +c arc tan 7 7 3

1 27

1 9 7

du

∫

  

7 3

9 7

1

2 3 1 +  3 x   7

⋅

=

2

 + u2  

arc tan

Fourth - Check the answer by differentiating the solution, i.e., let y = ⋅

x 2 dx

∫ 9 (7 + x 6 ) 9

du d 3 du = x = 3x 2 which implies du = 3 x 2 dx ; dx = 2 . dx dx 3x

Third - Write the answer in terms of the x variable, i.e.,

1

7 + 9x

=

6

2

 + x   1 3u 1 3 = arc tan ⋅ +c = 27 7 9 7 7

y′ =

x 2 dx

x

x 2 dx

∫

Second - Use substitution method by letting u = x3 , then Therefore,

1

dx

∫ a 2 + x2 = a arc tan a + c , i.e., ∫

3u 7 1

9 7

u 1 1 ⋅ arc tan +c 27 7 7 3

+c = arc tan

3

1

3x3

arc tan

9 7 3x3

7

+c

+ c then

7

d 3x3 1 1 1 7 x2 x2 1 9x2 x2 = ⋅ = ⋅ = +0 = ⋅ ⋅ 6 6 6 dx 7 7 7 +9 x 7 7 + 9x 7 + 9 x6 9 7 1 + 9 7x 7 7

or, the alternative approach would be to rearrange the integral in the following way:

∫

x 2 dx 7 + 9x

6

Therefore,

=

x 2 dx

∫ ( 7 )2 + (3x3 )2 . Now, let u = 3x x 2 dx

x2

∫ ( 7 )2 + (3x3 )2 ∫ ( 7 )2 + u 2 =

⋅

3

du

1 9

=

9x2

f. First - Write the given integral in its standard form

du d 3 du = 3 x = 9 x 2 which implies du = 9 x 2 dx ; dx = 2 . dx dx 9x

, then

∫

du

∫ ( 7 )2 + u 2 dx x2 − a2

x

∫

dx x

(x )

2 2

= − 52

∫

1 2

x u −5

⋅ 2

du 1 = 2x 2

1 x arc sec + c , i.e., a a

1

⋅ 2

d x2 1 +0 = ⋅ dx 5 10

 x2  − 1  5    2. Evaluate the following indefinite integrals. x2 5

1 x2 5

a. First - Write the given integral in its standard form

x4 25

∫

⋅ −1

Second - Use substitution method by letting u = x , then

Hamilton Education Guides

∫

1 9 7

dx x

x 4 − 25

=

arc tan

∫

3x3 7

+c

dx

(x )

2 2

x

− 52

1 x2 arc sec + c then 10 5

2x 2 = ⋅ 5 50

dx x

7

+c =

x2 1 u 1 arc sec +c arc sec + c = 10 5 5 10

Fourth - Check the answer by differentiating the solution, i.e., let y = 1 ⋅ 10

u

∫

∫

Third - Write the answer in terms of the original variable, i.e., x .

y′ =

1 1 ⋅ arc tan 9 7

du d 2 du . = x = 2 x which implies du = 2 x dx ; dx = 2x dx dx 1 u 1 du du = = arc sec + c 10 5 2 2 2 u u −5 x 2 u 2 − 52

Second - Use substitution method by letting u = x 2 , then Therefore,

=

=

x2 − a2

=

x x2 5

x 4 − 25 25

=

1 ⋅ 25

1 x arc sec + c , i.e., a a

∫

25 x x 2 x 4 − 25

dx x

x 2 −16

1

= x

=

∫

x 4 − 25

dx x x 2 − 42

du d = x = 1 which implies du = dx . Therefore, dx dx

448

Calculus I

∫

Chapter 4 Solutions dx

x x 2 − 42

du

∫

=

=

u u 2 − 42

1 u arc sec + c 4 4

1 u 1 x arc sec + c = arc sec + c 4 4 4 4 1 x Fourth - Check the answer by differentiating the solution, i.e., let y = arc sec + c then 4 4 1 1 16 1 1 1 1 4 1 1 d x = = y′ = ⋅ = ⋅ +0 = ⋅ 16 4 4 16 4 dx 4 2 2 2 2 2 x x −1 x x x x − 16 x x − 16 x x −16 −1

Third - Write the answer in terms of the original variable, i.e., x .

(4 )

4

4

16

16

∫

b. First - Write the given integral in its standard form

=

∫

dx  7 x x 2 −  

2 7

   

1

=

2

x x2 − a2

 x x 2 −  

   

2

7

∫

dx  x x 2 −  

   

2 7

1

=

2

∫

7

7

du  u u 2 −  

2 7

   

1

=

2

7

1

⋅

arc sec

2

7

1

⋅ 2

d 7x 1 +0 = ⋅ dx 2 2

1

1 arc sec 2

+c =

2

∫

dx x 7 x2 − 4

Therefore,

=

=

∫

∫

dx

( 7 x)

x

2

dx x

( 7 x)

2

. Now, let u = 7 x , then − 22 1

∫

= − 22

⋅

u 2 − 22

u 7

1 1 u arc sec + c = arc sec 2 2 2

e x dx

∫ 4 (9 + e 2 x )

=

4

1 4

e x dx

∫ 9 + e2 x

du

1 4

=

4

e x dx

∫ (3 ) 2 + ( e x ) 2 2

Hamilton Education Guides

=

7 x x2 −

4 7

7

1 arc sec 2

7u +c 2

7x +c 2

1

=

2

7 ⋅ 4

4

=

7 x 7 x2 − 4

1 x 7 x2 − 4

du du d . = 7 x = 7 which implies du = 7 dx ; dx = dx dx 7

1

=

∫

7

7 du u u 2 − 22

=

∫

du u u 2 − 22

1 4

1

dx

1 4

ex

x

ex

∫ a 2 + x2 = a arc tan a + c , i.e., ∫ 4 e2 x + 9 dx = ∫ 9 + 4 e2 x dx

e x dx

∫ (3 ) 2 + ( e x ) 2 2

Second - Use substitution method by letting u = e x , then Therefore,

dx

∫

7x +c 2

c. First - Write the given integral in its standard form =

=

7x + c then y ′ 2

7x 7x 7 x 7 x −4  7x  −1 −1 2 4 2 4  2    or, the alternative approach would be to rearrange the integral in the following way: 7x 2

x 7 x2 − 4

7u 1 + c = arc sec 2 2

1 arc sec 2

7 7 = ⋅ 2 4

⋅ 2

u 7

Fourth - Check the answer by differentiating the solution, i.e., let y = 1 2

dx

du d = x = 1 which implies du = dx . Therefore, dx dx

Third - Write the answer in terms of the original variable, i.e., x .

=

∫

2

Second - Use substitution method by letting u = x , then 1

x 1 arc sec + c , i.e., a a

=

dx

∫

7

dx

ex

du

∫ (3 ) 2 + u 2 ⋅ e x 2

=

1 4

du

du d x du = e = e x which implies du = e x dx ; dx = x . dx dx e

∫ (3 ) 2 + u 2 2

=

1 2u 1 1 u +c ⋅ arc tan 3 + c = arc tan 6 3 4 3 2

2

449

Calculus I

Chapter 4 Solutions 1 2e x 1 2u +c arc tan + c = arc tan 6 3 6 3

Third - Write the answer in terms of the original variable, i.e., x .

Fourth - Check the answer by differentiating the solution, i.e., let y = y′ =

1 ⋅ 6

1

⋅

2 2e x  3 

d 2e x 1 1 2e x 2 = +0 = ⋅ ⋅ ⋅ 2x dx 3 6 1 + 4e 3 18 9

ex 2x

2e x 1 arc tan + c then 3 6

=

1 9e x ex ex = = ⋅ 2 x 2 x 2 9 9 + 4e 9 + 4e 4e x + 9

9 + 4e 1 +  9  or, the alternative approach would be to rearrange the integral in the following way:

ex

∫ 9 + 4 e2 x Thus,

dx =

e x dx

∫ 32 + ( 2e x )2 . Now, let u = 2e

e x dx

ex

∫ 32 + ( 2e x )2 ∫ 32 + u 2 =

d. Write the given integral dx

∫

=

25 − (x − 3) 2

∫

Check: Let y = arc sin

=

1 5

5 25 − (x − 3) 2

⋅

du 2e x

=

dx

∫

25 − (x − 3) 2 dx

52 − u 2

x

, then

du d du = 2e x = 2e x which implies du = 2e x dx ; dx = x . dx dx 2e

2e x 1 1 u 1 1 u 1 du = = = arc +c tan arc tan + c ⋅ arc tan + c 3 6 2 3 3 6 3 2 32 + u 2

∫

in its standard form

= arc sin

1

x−3 + c then y ′ = 5

1−

⋅

( x5−3 ) 2

=

a2 − x2

= arc sin

x + c by letting u = x − 3 . Thus, a

d x−3 +0 = dx 5

1 1−

⋅

( x −3)2

1 1 = 5 5

25

1 25 − ( x −3)2 25

1

=

25 − (x − 3) 2 dx

du

∫ 52 + u 2

=

1

x

dx

∫ a 2 + x2 = a arc tan a + c , i.e., ∫ 25 + (x + 4)2

Second - Use substitution method by letting u = x + 4 , then dx

dx

u x−3 + c = arc sin +c 5 5

e. First - Write the given integral in its standard form

∫ 25 + (x + 4)2

∫

du d (x + 4) = 1 which implies du = dx . Therefore, = dx dx

1 u arc tan + c 5 5

1 u x+4 1 arc tan + c = arc tan +c 5 5 5 5 1 x+4 Fourth - Check the answer by differentiating the solution, i.e., let y = arc tan + c then 5 5 1 25 1 1 1 1 1 1 d  x+4 = y′ = ⋅ ⋅ ⋅ = ⋅  +0 = ⋅ 25 25 + (x + 4 )2 5 1 + ( x + 4 )2 5 5 1 + x + 4 2 dx  5  25 + (x + 4 )2

Third - Write the answer in terms of the x variable, i.e.,

(5)

25

f. First - Write the integral in its standard form =

dx

dx

1

x

dx

dx

∫ a 2 + x2 = a arc tan a + c , i.e., ∫ x2 − 10 x + 26 = ∫ (x2 − 10 x + 25)+ 1

dx

∫ (x − 5) 2 + 1 = ∫ 1 + (x − 5) 2

Second - Use substitution method by letting u = x − 5 , then dx

du

∫ 1 + (x − 5) 2 = ∫ 1 + u 2

du d (x − 5) = 1 which implies du = dx . Therefore, = dx dx

= arc tan u + c

Third - Write the answer in terms of the x variable, i.e., arc tan u + c = arc tan ( x − 5 ) + c Fourth - Check the answer by differentiating the solution, i.e., let y = arc tan (x − 5) + c then

Hamilton Education Guides

450

Calculus I

y′ =

Chapter 4 Solutions 1 1 1 d ⋅1 = ⋅ (x − 5) + 0 = 1 + (x − 5) 2 1 + (x − 5) 2 1 + (x − 5) 2 dx

Section 4.5 Solutions – Integration of Expressions Resulting in Exponential or Logarithmic Functions 1. Evaluate the following indefinite integrals. dx

∫ 2x + 1 let u = 2 x + 1 , then

a. Given

du d (2 x + 1) ; du = 2 ; du = 2 dx ; dx = du . Therefore, = dx 2 dx dx

1 1 1 1 1 du = du = ln u + c = ln 2 x + 1 + c 2 2 2 u 2

dx

∫ 2x + 1 = ∫ u ⋅ Check: Let y =

∫

1 1 1 2 1 1 = ln 2 x + 1 + c , then y ′ = ⋅ ⋅2+0 = ⋅ 2 2 2x + 1 2x + 1 2 2x + 1

x

∫ x 2 − a dx let u = x

b. Given x

x du

∫ x2 − a dx = ∫ u ⋅ 2 x Check: Let y =

=

1 2

x3

x3

x3

du

∫ x 4 − 1 dx = ∫ u ⋅ 4x3 Check: Let y =

1

∫ u du

1 1 ln u + c = ln x 2 − a + c 2 2

=

=

4

(

)

d 4 du du du . Therefore, = x −1 ; = x 4 − 1 ; du = 4 x 3 dx ; dx = dx dx dx 4x 3

− 1 , then

1 1 1 1 du = ln u + c = ln x 4 − 1 + c 4 4 u 4

∫

1 1 1 x3 ln x 4 − 1 + c , then y ′ = ⋅ 4 ⋅ 4 x3 + 0 = 4 4 4 x −1 x −1 1

3 

 1

3

∫  x + 3 + x − 5  dx = ∫ x + 3 dx + ∫ x − 5 dx

d. Given 1

1

3

1

∫ x + 3 dx + ∫ x − 5 dx = ∫ u1 du1 + 3∫ u2 du2 ∫ xe

∫ xe

3x2

3x2

dx =

dx let u = 3x 2 , then

∫ xe

Check: Let y = f. Given

e 5x

e 5x

∫ 1 − e 5x

∫ 1 − e 5x

dx =

u

⋅

let u1 = x + 3 and u2 = x − 5 respectively. Therefore,

= ln u1 + 3 ln u2 + c = ln x + 3 + 3 ln x − 5 + c

Check: Let y = ln x + 3 + 3 ln x − 5 + c , then y ′ = e. Given

)

(

du d 2 du du . Therefore, = x −a ; = 2 x ; du = 2 x dx ; dx = 2x dx dx dx

− a , then

1 1 1 2 x x = 2 ln x 2 − a + c , then y ′ = ⋅ 2 ⋅ 2x + 0 = ⋅ 2 2 2 x −a 2 x −a x −a

∫ x 4 − 1 dx let u = x

c. Given

2

1 3 1 3 + +0 = + x+3 x−5 x+3 x−5

du d 2 du du . Therefore, = 6 x ; du = 6 x dx ; dx = = 3x ; dx 6x dx dx

2 1 1 u 1 du = e du = eu + c = e 3 x + c 6 6 6x 6

∫

2 2 2 6 1 1 3x2 + c , then y ′ = ⋅ e3 x ⋅ 6 x + 0 = ⋅ x e3 x = x e3x e 6 6 6

dx let u = 1 − e5 x , then

e5 x

du

∫ 1 − e5 x ⋅ − 5e5 x

= −

(

)

du d du du = −5e5 x ; du = −5e5 x dx ; dx = − 5 x . Therefore, = 1 − e5 x ; dx dx dx 5e

1 1 1 1 du = − ln u + c = − ln 1 − e 5 x + c 5 u 5 5

∫

1 1 1 e5 x 5x Check: Let y = − ln 1 − e5 x + c , then y ′ = − ⋅ = e ⋅ − + 5 0 5 5 1 − e5 x 1 − e5 x

g. Given

∫ 3e

− ax

dx let u = − ax , then

Hamilton Education Guides

du d (− ax ) ; du = −a ; du = −a dx ; dx = − du . Therefore, = dx a dx dx

451

Calculus I

∫ 3e

Chapter 4 Solutions

∫

dx = 3 eu ⋅ −

− ax

3 u 3 du 3 = − e du = − eu + c = − e − ax + c a a a a

∫

3 3 3a − ax = 3e − ax Check: Let y = − e − ax + c , then y ′ = − e − ax ⋅ − a + 0 = e a a a

h.

∫(x

3

)

+ e 2 x − e −5 x dx =

Check: Let y =

i.

∫ x dx + ∫ e 3

1 3

1

∫ 5x4 dx let u = x3 , then 1 3

∫ 5x4

eu

∫ 5x4

dx =

1 1 1 1 4 1 2 x 1 −5 x x + e − e + c = x 4 + e 2 x + e −5 x + c 4 2 −5 4 2 5

∫

dx − e −5 x dx =

1 4 1 2 x 1 −5 x 1 1 1 x + e + e + c then y ′ = ⋅ 4 x3 + ⋅ e 2 x ⋅ 2 + ⋅ e −5 x ⋅ −5 + 0 = x3 + e 2 x − e −5 x 4 2 5 4 2 5

ex

ex

2x

⋅−

x4 du d 1 du 3 ; = = − 4 ; x 4 du = −3dx ; dx = − du . Therefore, 3 3 dx dx x dx x 1

1 3 1 1 u x4 e du = − eu + c = − e x + c du = − 15 15 15 3

∫

1

1 3

1

1

1 3 1 3 3 x3 1 ex 3 Check: Let y = − e x + c , then y ′ = − ⋅ e x ⋅ − 4 + 0 = ⋅e ⋅ 4 = 15 15 15 x 5x4 x 2. Evaluate the following indefinite integrals. a. Given

∫ (3e

∫ (3e

5x

)

+ 5 e5 x dx let u = 3e5 x + 5 , then

)

5x

+ 5 e5 x dx =

Check: Let y = b. Given

∫ (e

∫ (e

)

x

x

)

(

⋅

)

du 15e

2

∫ 1 + x3

5

∫u

5

⋅ ex ⋅

(

)

e

=

x

dx =

∫

x4

∫ 1 + x5

Check: Let y = x+7

3

∫ x + 6 dx = ∫

2

+c

(

)

(

)

)

30 5 x 1 3e + 5 e5 x = 3e5 x + 5 e5 x ⋅ 2 3e5 x + 5 ⋅ 15e5 x + 0 = 30 30

(

)

du d x du du . Thus, e −1 ; = = e x ; du = e x dx ; dx = dx dx dx ex

(

1 6 1 x u +c = e −1 6 6

(

)

(

)

)

6

+c

(

(

)

)

, then

du du d du = 1 + x3 ; = 3x 2 ; dx = 2 . Therefore, dx dx dx 3x

1 1 1 1 x 2 du = du = ln u + c = ln 1 + x 3 + c ⋅ 3 3 u 3 u 3x 2

∫

1 1 1 3 x2 x2 2 = = ln 1 + x3 + c , then y ′ = ⋅ ⋅ ⋅ 3 x + 0 3 3 1 + x3 3 1 + x3 1 + x3

∫ 1 + x5 dx let u = 1 + x ∫

(

)

6 5 5 5 1 x 6 x 1 e − 1 + c , then y ′ = ⋅ 6 e x − 1 ⋅ e x + 0 = e −1 ex = ex −1 ex 6 6 6

x4

dx =

(

∫

∫ u du = 5

)

1 1 1 2 1 u du = 3e 5 x + 5 ⋅ u +c = 15 15 2 30

+ c , then y ′ =

du

x2

Check: Let y = d. Given

=

5

∫ 1 + x3 dx let u = 1 + x

x2

5x

− 1 e x dx let u = e x − 1 , then

Check: Let y =

e.

5x

1 3e5 x + 5 30

− 1 e x dx =

c. Given

∫u ⋅ e

(

du d du du . Thus, = 3e5 x + 5 ; = 15e5 x ; du = 15e5 x dx ; dx = dx dx dx 15e5 x

5

, then

(

)

du du d du = 5x 4 ; dx = 4 . Therefore, = 1 + x5 ; dx dx dx 5x

1 1 1 1 x 4 du = du = ln u + c = ln 1 + x 5 + c ⋅ 5 5 u 5 u 5x 4

∫

1 1 1 5 x4 x4 4 = = ln 1 + x5 + c , then y ′ = ⋅ ⋅ ⋅ 5 x + 0 5 5 1 + x5 5 1 + x5 1 + x5

(x + 6) + 1 dx x+6

Hamilton Education Guides

=

x+6

1

1

∫ x + 6 dx + ∫ x + 6 dx = ∫ dx + ∫ x + 6 dx =

x + ln x + 6 + c

452

Calculus I

Chapter 4 Solutions

Check: Let y = x + ln x + 6 + c , then y ′ = 1 + f.

(x + 5) + 4 dx

x+9

∫ x + 5 dx = ∫

x+5

=

x+5

1 1 x+7 x + 6 +1 = = ⋅1 + 0 = 1 + x+6 x+6 x+6 x+6

Check: Let y = x + 4 ln x + 5 + c , then y ′ = 1 +

∫a

g. Given

∫

ax

2

+k

x2 +k

∫

au ⋅ x ⋅

= x + 4 ln x + 5 + c

x+9 4 4 x+5+4 = = ⋅1 + 0 = 1 + x+5 x+5 x+5 x+5

(

)

du d 2 du du . Therefore, = x +k ; = 2 x ; du = 2 x dx ; dx = 2x dx dx dx

x dx let u = x 2 + k , then

x dx =

4

4

∫ x + 5 dx + ∫ x + 5 dx = ∫ dx + ∫ x + 5 dx

2

du 1 au 1 1 a x +k = a u du = +c = +c 2 2x 2 ln a 2 ln a

∫

2

Check: Let y = h. Given

∫

2

∫3a

2 2 2 ln a x 2 + k 1 1 a x +k ⋅ a x + k ln a ⋅ 2 x + 0 = ⋅a ⋅ x = a x +k x + c , then y ′ = 2 ln a 2 ln a 2 ln a

(

)

du d 3 du du . Therefore, = 3x 2 ; du = 3 x 2 dx ; dx = x +5 ; = dx dx dx 3x 2

x3 +5 2

x dx let u = x3 + 5 , then

3

2 u 2 x3 +5 2 2 au 2 u 2 du 2 a x +5 a du = ⋅ a x dx = +c = ⋅ +c a ⋅x ⋅ 2 = 9 3 9 ln a 9 ln a 3 3x

∫

∫

3

Check: Let y = i. Given

∫ (e

2x

∫ (e

)

2x

)

3

+ 3 e 2 x dx =

⋅ e2 x ⋅

(

)

1 2x e +3 8

4

du 2e

2x

=

(

)

(

)

4

+c

(

)

)

3 3 3 1 8 2x ⋅ 4 e 2 x + 3 ⋅ 2e 2 x + 0 = e + 3 e2 x = e2 x + 3 e2 x 8 8

1 ln 2⋅5

(x + 1) − 5 (x + 1) + 5

+c =

1 ln 10

x−4 +c x+6

1 x+6 x+6− x+4 1 1 1 ⋅ (x + 6 ) − 1 ⋅ (x − 4 ) 1 x−4 ⋅ ⋅ ln + c then y ′ = ⋅ ⋅ +0 = 2 10 x−4 10 x+6 10 x − 4 (x + 6 ) 2 (x + 6 ) x+6

1 1 10 1 1 1 1 = = 2 = 2 = ⋅ ⋅ 10 x − 4 x + 6 (x + 1) 2− 25 x + 6 x − 4 x − 24 x + 2 x − 24 (x − 4)(x + 6)

1

1

1

=

(x + 2)(x + 4)

1

1

x2 + 4x + 2x + 8 1

Check: Let y =

1

∫ (x + 3) 2 − 1 dx =

dx =

1 ln 2 ⋅1

(x + 3) − 1 (x + 3) + 1

+c =

x+2 1 +c ln x+4 2

1 1 2 1 x+4 x+4− x−2 1 1 1 ⋅ (x + 4 ) − 1 ⋅ (x + 2 ) 1 x+2 = ⋅ ⋅ ⋅ ln + c then y ′ = ⋅ x + 2 ⋅ +0 = ⋅ 2 2 2 x + 2 x + 4 2 x + 2 2 2 x+4 (x + 4 ) (x + 4 ) x+4

∫ 9 − (x − 1) 2 dx = ∫ 32 − (x − 1) 2 dx =

=

(

∫

+ c , then y ′ =

∫ x2 + 6 x + 8 dx = ∫ (x2 + 6 x + 9)− 1

=

)

1 3 1 1 1 1 2x u du = ⋅ u 4 + c = u 4 + c = e +3 2 2 4 8 8

1

Check: Let y =

l.

3

1

Check: Let y =

k.

∫u

∫ (x + 1) 2 − 25 dx = ∫ (x + 1) 2 − 52 dx =

=

(

d 2x du du du = e +3 ; = 2e 2 x ; du = 2e 2 x dx ; dx = 2 x . Therefore, dx dx dx 2e

3

+ 3 e 2 x dx let u = e 2 x + 3 , then

Check: Let y = j.

3 2 3 6 ln a x 3 + 5 2 2 1 2 a x +5 ⋅ a x + 5 ln a ⋅ 3 x 2 + 0 = ⋅ ⋅a ⋅ x = a x +5 x 2 ⋅ + c , then y ′ = ⋅ 3 9 ln a 9 ln a 9 ln a

=

1

x2 + 6x + 8

1 3 + (x − 1) 1 2+ x 3 + x −1 1 ln + c = ln +c + c = ln 2⋅3 3 − (x − 1) 6 4− x 3 − x +1 6

1 1 6 1 4−x 4−x+2+ x 1 1 1 ⋅ (4 − x ) + 1 ⋅ (2 + x ) 1 2+ x = ⋅ ⋅ ⋅ ln + c then y ′ = ⋅ 2 + x ⋅ +0 = ⋅ 2 2 6 2 + x 4 − x 6 4− x 6 2 + x 6 (4 − x ) (4 − x )

1

(2 + x )(4 − x )

=

Hamilton Education Guides

1 8 − 2x + 4x − x

4− x

2

=

1

8 + 2x − x

2

=

1

9 − (x − 1)2

453

Chapter 5 Solutions: Section 5.1 Practice Problems – Integration by Parts 1. Evaluate the following integrals using the integration by parts method.

∫ xe

a. Given

4x

dx let u = x and dv = e 4 x dx then du = dx and

4x

e4 x e4 x 4e 4 x  1 1 1 1 1 1 4x   = xe4 x + ⋅  x −  + 1 ⋅ e 4 x + 0 = e 4 x ⋅  x −  + e 4 x = xe4 x − e  x −  + c , then y ′ = 4 4 4  4 4 4 4 4 4  

x

∫ 2 cos x dx let u = x and dv = cos x dx then du = dx

integration by parts formula x

∫ 2 cos x dx

=

Check: Let y =

∫ u dv = u v − ∫ v du

∫ dv = ∫ cos x dx which implies v = sin x . Using the

we obtain

∫

1 1 1 1 1 1 1 1 x sin x + cos x + c , then y ′ = (1 ⋅ sin x + cos x ⋅ x ) − sin x + 0 = sin x + x cos x − sin x = x cos x 2 2 2 2 2 2 2 2 5x

dx let u = 5 − x and dv = e5 x dx then du = − dx and

the integration by parts formula 5x

and

1 1 1 1 x ⋅ sin x − sin x dx = x sin x + cos x + c 2 2 2 2

∫ ( 5 − x )e

∫ ( 5 − x )e

1 4x e . Using the integration 4

∫

Check: Let y =

c. Given

dx which implies v =

1 1 1 4x 1 4x 1 1  xe − e dx = xe4 x − e 4 x + c = e 4 x  x −  + c 4 16 4 4 4 4 

dx =

b. Given

4x

∫ u dv = u v − ∫ v du we obtain

by parts formula

∫ xe

∫ dv = ∫ e

dx = (5 − x )

Check: Let y = e5 x −

∫ dv = ∫ e

5x

dx which implies v =

e5x . Using 5

∫ u dv = u v − ∫ v du we obtain

1 1 5x e5 x e5 x + dx = e 5 x − xe 5 x + e +c 5 5 5 25

∫

(

)

1 1 1 5x 1 5x 1 5 5x xe + e + c , then y ′ = 5e5 x − e5 x + 5 xe5 x + ⋅ e + 0 = 5e5 x − e5 x − xe5 x + e5 x 5 5 5 25 5 25

= 5e5 x − xe5 x = (5 − x )e5 x d. Given

∫ x sin 5x dx

let u = x and dv = sin 5 x dx then du = dx and

Using the integration by parts formula

∫ x sin 5x dx

∫ u dv = u v − ∫ v du

1

∫ dv = ∫ sin 5x dx which implies v = − 5 cos 5x .

we obtain

1 1 1 1 = x ⋅ − cos 5 x + cos 5 x dx = − x cos 5 x + sin 5 x + c 5 5 5 5

∫

1 5 1 1 1 1 Check: Let y = − x cos 5 x + sin 5 x + c , then y ′ = − (1 ⋅ cos 5 x − sin 5 x ⋅ 5 ⋅ x ) + cos 5 x + 0 = − cos 5 x + x sin 5 x 5 5 5 5 5 5 1 5 + cos 5 x = x sin 5 x = x sin 5 x 5 5

e. Given

∫x

3 − x dx let u = x and dv = 3 − x dx then du = dx and

Using the integration by parts formula

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∫ dv = ∫

3 − x dx which implies v = −

3 2 (3 − x )2 . 3

∫ u dv = u v − ∫ v du we obtain

454

Calculus I

∫x = −

Chapter 5 Solutions

3 − x dx = x ⋅ −

3 5 3 3 3 3 2 (3 − x ) 2 + 2 (3 − x ) 2 dx = − 2 x (3 − x ) 2 − 2 ⋅ 1 3 (3 − x ) 2 +1 + c = − 2 x (3 − x ) 2 − 2 ⋅ 2 (3 − x ) 2 + c 3 3 5 3 3 3 3 1+

∫

2

2 4 x (3 − x ) − (3 − x ) + c 3 15 3 2

5 2

3 5 3 3 1 4 2 2 2 3 4 5 Check: Let y = − x (3 − x ) 2 − (3 − x ) 2 + c , then y ′ = − (3 − x ) 2 + ⋅ x(3 − x ) 2 + ⋅ (3 − x ) 2 + 0 3 15 3 3 2 15 2

3 3 1 1 2 (3 − x ) 2 + x(3 − x ) 2 + 2 (3 − x ) 2 = x(3 − x ) 2 = x 3 − x 3 3

= − f. Given

∫x e

3 3x

dx let u = x3 and dv = e 3 x dx then du = 3 x 2 dx and

3 3x

+

∫x

2 3x

e dx =

1 2 3x 2 3x 2 3x x e − xe + e + c . Therefore, 3 9 27

x 3e 3 x 1 2 3 x 2 3 x 2 3x x3e3 x  1 2 3 x 2 3 x 2 3 x  e +c − x e + xe − −  x e − xe + e +c = 27 9 3 3 27 3 3 9  

Check: Let y =

(

(

) (

x3e3 x 1 2 3 x 2 3 x 2 3 x 1 1 e + c , then y ′ = − x e + xe − 3 x 2 ⋅ e3 x + 3e3 x ⋅ x3 − 2 x ⋅ e3 x + 3e3 x ⋅ x 2 3 3 27 9 3 3

)

)

2 2 2 2 2 2 ⋅ 3e3 x + 0 = x 2e3 x + x3e3 x − xe3 x − x 2e3 x + e3 x + xe3 x − e3 x = x3e3 x 1 ⋅ e3 x + 3e3 x ⋅ x − 9 27 3 9 9 3

∫ cos ( ln x ) dx

let u = cos ( ln x ) and dv = dx then du =

Using the integration by parts formula

∫ cos ( ln x ) dx To integrate du =

1 3x e . Using the 3

∫

∫

x 3 e 3 x dx =

g. Given

dx which implies v =

x3e3 x 1 1 3x − x 2e3 x dx e ⋅ 3 x 2 dx = dx = x3 ⋅ e3 x − 3 3 3

In example 5.1-1, problem letter b, we showed that

∫

3x

∫ u dv = u v − ∫ v du we obtain

integration by parts formula

∫x e

∫ dv = ∫ e

= cos ( ln x ) ⋅ x + x ⋅

∫

− sin ( ln x ) dx and x

∫ dv = ∫ dx which implies v = x .

∫ u dv = u v − ∫ v du we obtain

sin ( ln x ) dx = x cos ( ln x ) + sin ( ln x ) dx x

∫

(1 )

∫ sin ( ln x ) dx use the integration by parts formula again, i.e., let u = sin ( ln x ) and dv = dx then

cos ( ln x ) dx and x

∫ dv = ∫ dx which implies v = x . Therefore,

∫ sin ( ln x ) dx = sin ( ln x ) ⋅ x − ∫ x ⋅

cos ( ln x ) dx = x sin ( ln x ) − cos ( ln x ) dx x

∫

(2 )

Combining equations ( 1 ) and ( 2 ) together we have

∫ cos ( ln x ) dx

= x cos ( ln x ) − sin ( ln x ) dx = x cos ( ln x ) + x sin ( ln x ) − cos ( ln x ) dx

∫

∫

Taking the integral − cos ( ln x ) dx from the right hand side of the equation to the left hand side

∫

we obtain

∫ cos ( ln x ) dx + ∫ cos ( ln x ) dx = x cos ( ln x ) + x sin ( ln x ) Therefore,

2 cos ( ln x ) dx = x cos ( ln x ) + x sin ( ln x ) + c and thus

∫

Check: Let y = =

∫ cos ( ln x ) dx

=

x x cos ( ln x ) + sin ( ln x ) + c 2 2

x x cos ( ln x ) x sin ( ln x ) sin ( ln x ) x cos ( ln x ) cos ( ln x ) + sin ( ln x ) + c , then y ′ = − + + +0 2 2 2 2x 2 2x

cos ( ln x ) cos ( ln x ) cos ( ln x ) sin ( ln x ) sin ( ln x ) cos ( ln x ) = = cos ( ln x ) + + + − 2 2 2 2 2 2

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455

Calculus I

Chapter 5 Solutions x

∫ 3 tan

h. Given

−1

x dx let u = tan −1 x and dv =

Using the integration by parts formula x

∫ 3 tan =

−1

2 x dx = tan −1 x ⋅

x 1 dx then du = dx and 3 1+ x 2

1

x

∫ dv = ∫ 3 dx which implies v = 6 x

2

.

∫ u dv = u v − ∫ v du we obtain

x 2 1 2 dx 1 1 x2 1 1 = x 2 tan −1 x − x ⋅ − dx = x 2 tan −1 x − 2 6 6 6 6 6 6 1 + x2 1+ x

∫

∫



1



∫ 1 − 1 + x2  dx

1 1 1 1 1 1 1 2 x tan −1 x − dx + dx = x 2 tan −1 x − x + tan −1 x + c 2 6 6 6 6 6 6 1+ x

∫

∫

1 1 1 2 1 1 x2 1 1 1 x tan −1 x − x + tan −1 x + c , then y ′ = ⋅ 2 x ⋅ tan −1 x + − + ⋅ +0 2 6 6 6 6 6 1+ x 6 6 1 + x2

Check: Let y =

1 1 1 1 1 1 x2 + 1 1 1 1 x2 1 1 1 −1 = x x tan ⋅ x tan −1 x + ⋅ + ⋅ − + ⋅ − = x tan −1 x + − = x tan −1 x 2 2 2 3 6 6 3 3 6 1+ x 6 1+ x 6 3 6 1+ x 6

i. Given

∫ ln x dx let u = ln x 5

integration by parts formula

∫ ln x dx = ln x 5

5

5

and dv = dx then du =

1

⋅ 5 x 4 dx =

x5

5 dx and x

∫ dv = ∫ dx which implies v = x . Using the

∫ u dv = u v − ∫ v du we obtain

5 ⋅ x − x ⋅ dx = x ln x5 − 5 dx = x ln x 5 − 5 x + c x

∫

∫

5 x5 1   Check: Let y = x ln x5 − 5 x + c , then y ′ = 1 ⋅ ln x5 + 5 ⋅ 5 x 4 ⋅ x  − 5 + 0 = ln x5 + 5 − 5 = ln x5 + 5 − 5 = ln x5 x x   j. Given

∫x e

− ax

dx let u = x and dv = e − ax dx then du = dx and

integration by parts formula

∫x e

− ax

∫

∫e

a a x

2

(

⋅ e − ax = −

1 − ax 1 + xe− ax + e − ax = xe− ax e a a

sin 3 x dx let u = e x and dv = sin 3 x dx then du = e x dx and

sin 3 x dx = e x ⋅ −

To integrate v=

)

1 axe− ax 1 − ax 1 − ax 1 1 − 2e + c , then y ′ = − 1 ⋅ e − ax − ae − ax ⋅ x − 2 ⋅ −2e − ax + 0 = − e − ax + xe a a a a a a

Using the integration by parts formula x

1 dx which implies v = − e − ax . Using the a

we obtain

∫

+

∫e

− ax

1 1 − ax 1 1 − ax 1 1 e dx = − x e − ax − 2 e − ax + c e dx = − x e − ax + dx = x ⋅ − e − ax + a a a a a a

Check: Let y = −

k. Given

∫ u dv = u v − ∫ v du

∫ dv = ∫ e

∫e

x

∫ u dv = u v − ∫ v du

∫ dv = ∫ sin 3x dx which implies v = −

we obtain

cos 3x cos 3 x x 1 1 x e cos 3 x dx − − ⋅ e dx = − e x cos 3 x + 3 3 3 3

∫

∫

cos 3 x dx let u = e x and dv = cos 3 x dx then du = e x dx and

sin 3 x . Thus, 3

∫e

x

cos 3 x dx = e x ⋅

(1 )

∫ dv = ∫ cos 3x dx which implies

sin 3 x sin 3 x x 1 1 x − ⋅ e dx = e x sin 3 x − e sin 3 x dx 3 3 3 3

∫

cos 3 x . 3

∫

(2)

Combining equations (1 ) and ( 2 ) together we obtain:

∫e

x

1 1 x 1 1 1 x e cos 3 x dx = − e x cos 3 x + e x sin 3 x − e sin 3 x dx sin 3 x dx = − e x cos 3 x + 3 3 9 3 9

Taking the −

∫

∫

1 x e sin 3 x dx from the right hand side of the equation to the left hand side we obtain 9

∫

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456

Calculus I

∫e

x

and

Chapter 5 Solutions 1 1 1 x 1 1 10 x e sin 3 x dx = − e x cos 3 x + e x sin 3 x which implies e sin 3 x dx = − e x cos 3 x + e x sin 3 x 3 9 9 3 9 9

∫e

x

sin 3 x dx =

Check: Let y = − = −

∫e

l. Given

∫

∫

sin 3x dx +

x

3 1 x 1 9 1 x  e sin 3 x  − e cos 3 x + e x sin 3 x  = − e x cos 3 x + 10  3 9 10 10 

3 x 1 1 1 3 3 e cos 3 x + e x sin 3 x , then y ′ = − e x ⋅ cos 3 x + sin 3 x ⋅ 3 ⋅ e x + e x ⋅ sin 3 x + cos 3 x ⋅ 3 ⋅ e x 10 10 10 10 10 10

9 x 1 3 x 9 1 3 10 x e sin 3 x + e x sin 3 x = e cos 3 x + e x sin 3 x + e x sin 3 x + e x cos 3 x = e sin 3 x = e x sin 3 x 10 10 10 10 10 10 10

cos 5 x dx let u = e x and dv = cos 5 x dx then du = e x dx and

Using the integration by parts formula

∫

∫ u dv = u v − ∫ v du

1

∫ dv = ∫ cos 5x dx which implies v = 5 sin 5x .

we obtain

e x sin 5 x 1 x 1 1 − e sin 5 x dx sin 5 x ⋅ e x dx = e x cos 5 x dx = e x ⋅ sin 5 x − 5 5 5 5

To integrate

∫

Thus,

∫e

x

(1 )

∫

∫

1

∫ dv = ∫ sin 5x dx which implies v = − 5 cos 5x .

sin 5 x dx let u = e x and dv = sin 5 x dx then du = e x dx and

e x cos 5 x 1 1 1 + cos 5 x ⋅ e x dx cos 5 x ⋅ e x dx = − e x sin 5 x dx = e x ⋅ − cos 5 x + 5 5 5 5

∫

∫

(2)

Combining equations (1 ) and ( 2 ) together we obtain:

 e x sin 5 x e x cos 5 x e x sin 5 x 1 x e x sin 5 x 1  e x cos 5 x 1 x − + e sin 5 x dx = − − + e cos 5 x dx  =  5 25 5 5 5 5  5 5 

∫

∫

e x cos 5 x dx =

−

1 1 e x cos 5 x dx . Taking the − e x cos 5 x dx from the right hand side of the equation to the left hand side we obtain 25 25

∫

∫

∫

e x sin 5 x e x cos 5 x e x sin 5 x e x cos 5 x 1 26 x which implies and + + e x cos 5 x dx = e cos 5 x dx = 25 25 5 5 25 25

∫e

x

cos 5 x dx +

∫e

x

cos 5 x dx =

Check: Let y = =

∫

∫

5 x 1 x 25  e x sin 5 x e x cos 5 x  = e sin 5 x + e cos 5 x +  26 26 26  5 25 

5 x 1 x 1 x 1 5 x 5 e sin 5 x + e cos 5 x , then y ′ = cos 5 x ⋅ 5 ⋅ e x + e ⋅ cos 5 x − sin 5 x ⋅ 5 ⋅ e x + 0 e ⋅ sin 5 x + 26 26 26 26 26 26

26 x 5 x 25 x 1 x 5 x 25 x 1 x e sin 5 x + e cos 5 x + e cos 5 x − e sin 5 x = e cos 5 x = e x cos 5 x e cos 5 x + e cos 5 x = 26 26 26 26 26 26 26

2. Evaluate the following integrals using the integration by parts method. a. Given

∫ x sec

2

the integration by parts formula

∫ x sec

2

∫ dv = ∫ sec

x dx let u = x and dv = sec2 x dx then du = dx and

∫ u dv = u v − ∫ v du

x dx which implies v = tan x . Using

we obtain

∫

x dx = x ⋅ tan x − tan x dx = x tan x − ln sec x + c

(

)

Check: Let y = x tan x − ln sec x + c , then y ′ = 1 ⋅ tan x + sec2 x ⋅ x − b. Given

2

∫ arc sin 3 y dy

let u = arc sin 3 y and dv = dy then du =

the integration by parts formula

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∫ u dv = u v − ∫ v du

sec x tan x + 0 = tan x + x sec2 x − tan x = x sec2 x sec x

3dy 1 − 9 y2

and

∫ dv = ∫ dy which implies v = y . Using

we obtain

457

Calculus I

Chapter 5 Solutions

∫

= arc sin 3 y ⋅ y − y ⋅ 3 y dy

∫

To integrate 3 y dy

∫

3y

⋅

w

(

1− 9y

2

∫

y dy

= y arc sin 3 y − 3

1− 9y

(1 )

2

use the substitution method by letting w = 1 − 9 y 2 then

1 − 9 y2

=

1 − 9 y2

3dy

∫

∫ arc sin 3 y dy

1 1 1 = − w 2 = − 1 − 9 y2 3 3

1 dw = − − 18 y 6

)

dw

∫

w

= −

dw dw . Therefore, = −18 y and dy = − dy 18 y

−1 1− 1 1 1 2 1 1 1 2 −1 1 1 w 2 dw = − ⋅ w 2 = − ⋅ 2 −1 w 2 = − ⋅ w 2 1 6 6 1 6 6 1−

∫

2

2

1 2

(2 )

Combining equations (1 ) and ( 2 ) together we obtain:

∫

arc sin 3 y dy = y arc sin 3 y −

Check: Let w = y arc sin 3 y + 3y

+

−

1 − 9 y2

∫ arc tan x dx

c. Given

1 ⋅ 6

(

1 1 − 9 y2 3

x dw

∫ 1 + x2 = ∫ w ⋅ 2x

)

1 2

= y arc sin 3 y +

∫ u dv = u v − ∫ v du dx

)

+c

3y

3y

3y

−

1 − 9 y2

1 − 9 y2

dx

1 + x2

and

− 2

1 1 ⋅ ⋅ 3 2

18 y 1− 9y

2

+ 0 = arc sin 3 y

= arc sin 3 y

∫ dv = ∫ dx which implies v = x . Using the

we obtain

= x arc tan x −

x dx

(1 )

∫ 1 + x2

use the substitution method by letting w = 1 + x 2 then =

1 2

1− 9y

let u = arc tan x and dv = dx then du =

1 + x2

(

1 1 − 9 y2 3

+ c , then w ′ = arc sin 3 y +

= arc sin 3 y +

1 − 9 y2

∫

x dx

x dx

1 − 9 y2

= arc tan x ⋅ x − x ⋅

∫ 1 + x2

To integrate

∫

18 y

integration by parts formula

∫ arc tan x dx

3 y dy

dw dw . Therefore, = 2 x And dx = 2x dx

1 1 1 dw = ln w = ln 1 + x 2 2 2 w 2

(2)

∫

Combining equations (1 ) and ( 2 ) together we obtain:

∫ arc tan x dx

= x arc tan x −

Check: Let y = x arc tan x − d. Given

∫ sin 5x dx = ∫ sin 3

2

x dx

∫ 1 + x2

= x arc tan x −

1 ln 1 + x 2 + c 2

1 x x 1 2x x = arc tan x ln 1 + x 2 + c , then y ′ = arc tan x + − − + 0 arc tan x + 2 1 + x2 2 1 + x2 1 + x2 1 + x2

5 x ⋅ sin 5 x dx let u = sin 2 5 x and dv = sin 5 x dx then du = 10 sin 5 x cos 5 x dx and 1

∫ dv = ∫ sin 5x dx ∫ dv = ∫ sin x dx which implies v = − 5 cos 5x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain

∫ sin 5x dx = sin

To integrate

3

∫ cos

2

2

5x ⋅ −

2

∫

∫

(1 )

5 x sin 5 x dx use the integration by parts method again, i.e., let u = cos 2 5 x and dv = sin 5 x then

du = −10 sin 5 x cos 5 x dx and

∫ cos

1 cos 5 x 1 + cos 5 x ⋅ 10 sin 5 x cos 5 x dx = − sin 2 5 x cos 5 x + 2 cos 2 5 x sin 5 x dx 5 5 5

1

∫ dv = ∫ sin 5x dx which implies v = − 5 cos 5x . Therefore,

1 1 1 cos 5 x ⋅ 10 sin 5 x cos 5 x dx = − cos3 5 x − 2 cos 2 5 x sin 5 x dx . Taking the 5 x sin 5 x dx = cos 2 5 x ⋅ − cos 5 x − 5 5 5

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∫

∫

458

Calculus I

Chapter 5 Solutions

∫

integral − 2 cos 2 5 x sin 5 x dx from the right hand side of the equation to the left side we obtain

∫ cos

2

1 5 x sin 5 x dx + 2 cos 2 5 x sin 5 x dx = − cos3 5 x . Therefore, 5

∫

∫ cos

2

5 x sin 5 x dx = −

Combining equations (1 ) and ( 2 ) together we have 1

∫ sin 5x dx = − 5 sin 3

−

(

2

1 cos3 5 x 15

(2)

sin 2 5 x cos 5 x 2 1 1 − cos3 5 x 5 x cos 5 x + 2 cos 2 5 x sin 5 x dx = − sin 2 5 x cos 5 x + 2 ⋅ − cos3 5 x = − 5 15 15 5

∫

)

1 1 2 1 1 2 1  cos 3 5 x − cos 5 x + c 1 − cos 2 5 x cos 5 x − cos3 5 x + c = − cos 5 x +  cos3 5 x − cos3 5 x  + c = 15 5 15 5 5 5 15  

Note that another method of solving the above problem is in the following way:

∫ sin 5x dx = ∫ sin 3

1 5

5 x ⋅ sin 5 x dx =

∫ sin 5x dx = ∫ sin 3

Therefore, = −

2

2

5 x ⋅ sin 5 x dx =

1 2 1 2 ∫ (1 − u ) du = ∫  5 u − 5  du

Check: Let y =

2 ∫ (1 − cos 5x )⋅ sin 5x dx let u = cos 5x , then

=

du du . = −5 sin 5 x and dx = − dx 5 sin 5 x

du 2 2 ∫ (1 − cos 5x )⋅ sin 5x dx = ∫ (1 − u )⋅ sin 5x ⋅ − 5 sin 5x

1 3 1 1 1 cos 3 5 x − cos 5 x + c u − u+c = 15 5 15 5

1 1 1 1 cos3 5 x − cos 5 x + c , then y ′ = ⋅ 3 cos 2 5 x ⋅ − sin 5 x ⋅ 5 + ⋅ 5 sin 5 x + 0 = − cos 2 5 x ⋅ sin 5 x + sin 5 x 5 5 15 15

)

(

= sin 5 x 1 − cos 2 5 x = sin 5 x sin 2 5 x = sin 3 5 x e. Given

∫x

2

cos x dx let u = x 2 and dv = cos x dx then du = 2 x dx and

the integration by parts formula

∫x

2

∫ u dv = u v − ∫ v du

∫ dv = ∫ cos x dx which implies v = sin x . Using

we obtain

(1 )

∫

∫

cos x dx = x 2 ⋅ sin x − sin x ⋅ 2 x dx = x 2 sin x − 2 x sin x dx

To integrate

∫ x sin x dx

use the integration by parts formula again, i.e., let u = x and dv = sin x dx then du = dx and

∫ dv = ∫ sin x dx which implies v = − cos x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain ∫ x sin x dx = x ⋅ − cos x + ∫ cos x ⋅ dx = − x cos x + sin x

(2)

Combining equations (1 ) and ( 2 ) together we have

∫x

2

cos x dx = x 2 sin x − 2 x sin x dx = x 2 sin x − 2(− x cos x + sin x ) = x 2 sin x + 2 x cos x − 2 sin x

∫

(

)

Check: Let y = x 2 sin x + 2 x cos x − 2 sin x , then y ′ = 2 x sin x + x 2 cos x + 2(cos x − x sin x ) − 2 cos x

2 x sin x + x 2 cos x + 2 cos x − 2 x sin x − 2 cos x = x 2 cos x f. Given

∫e

−2 x

cos 3 x dx let u = cos 3 x and dv = e −2 x dx then du = −3 sin 3 x dx and

1 v = − e − 2 x . Using the integration by parts formula 2

∫e

−2 x

∫ u dv = u v − ∫ v du

∫ dv = ∫ e

−2 x

we obtain

1 3 −2 x 1 3 −2 x e ⋅ sin 3 x dx = − e − 2 x cos 3 x − e sin 3 x dx cos 3 x dx = cos 3x ⋅ − e − 2 x − 2 2 2 2

To integrate

∫

∫e

−2 x

dx which implies

∫

(1 )

sin 3 x dx use the integration by parts formula again, i.e., let u = sin 3 x and dv = e −2 x dx then

du = 3 cos 3 x dx and

∫ dv = ∫ e

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−2 x

1 dx which implies v = − e − 2 x . Therefore, 2

459

Calculus I

∫e

−2 x

Chapter 5 Solutions 1 1 −2 x 1 3 −2 x e cos 3 x dx e ⋅ 3 cos 3 x dx = − e − 2 x sin 3 x + sin 3 x dx = sin 3 x ⋅ − e − 2 x + 2 2 2 2

(2)

∫

∫

Combining equations (1 ) and ( 2 ) together we have

∫e

−2 x

cos 3 x dx = −

∫e

Taking the integral

∫e

we obtain

−2 x

1 3 9 −2 x 1 −2 x 3 −2 x cos 3 x dx cos 3 x − e e e sin 3 x dx = − e − 2 x cos 3 x + e − 2 x sin 3 x − 2 4 4 2 2

∫

∫

−2 x

cos 3 x dx from the right hand side of the equation to the left hand side

cos 3 x dx +

9 −2 x 1 3 e cos 3 x dx = − e − 2 x cos 3 x + e − 2 x sin 3 x . Therefore, 4 2 4

∫

1 3 13 − 2 x e cos 3 x dx = − e − 2 x cos 3 x + e − 2 x sin 3 x and thus 2 4 4

∫

Check: Let y = −

3

3

x ( 5 x − 1) dx = x ⋅

Check: Let y =

∫ x csc

(5 x − 1) 4 − 20

1 20

3

dx then du = dx and

∫ u dv = u v − ∫ v du

4 ∫ (5x − 1) dx

[

2

∫ u dv = u v − ∫ v du

∫ dv = ∫ csc

2

x dx which implies v = − cot x . Using

we obtain

∫

2

∫ 3 cos

−1

∫

1 (5 x − 1)4 . 20

x dx = x ⋅ − cot x + cot x dx = − x cot x + ln sin x + c

−1

)

∫

cos x + 0 = − cot x + x csc2 x + cot x = x csc2 x sin x

5

5 x dx let u = cos −1 5 x and dv = dx then du = −

1 − 25 x

5 x dx =

To integrate

Thus,

dx which implies v =

]

x dx let u = x and dv = csc2 x dx then du = dx and

Using the integration by parts formula 2

3

x (5 x − 1) 4 1 x (5 x − 1) 4 1 1 (5 x − 1) 5 + c − ⋅ (5 x − 1) 4 +1 + c = − 20 20 25 500 20

=

(

∫ 3 cos

∫ dv = ∫ (5x − 1)

we obtain

Check: Let y = − x cot x + ln sin x + c , then y ′ = − cot x − x csc2 x + i. Given

(

x (5 x − 1) 4 1 (5 x − 1) 5 + c , then y ′ = 1 (5 x − 1) 4 + 20 x (5 x − 1) 3 − 25 (5 x − 1) 4 + 0 − 20 500 500 20

the integration by parts formula 2

)

1 (5 x − 1) 4 + x (5 x − 1) 3 − 1 (5 x − 1) 4 = x (5 x − 1) 3 20 20

=

∫ x csc

3 −2 x 2 −2 x e e sin 3 x + c cos 3 x + 13 13

(

∫ x ( 5x − 1) dx let u = x and dv = (5x − 1)

h. Given

cos 3 x dx = −

2e −2 x cos 3 x 3e −2 x sin 3 x 2 3 + + c , then y ′ = − − 2e − 2 x cos 3 x − 3e − 2 x sin 3 x + − 2e − 2 x sin 3 x + 3e − 2 x cos 3 x 13 13 13 13

Using the integration by parts formula

∫

−2 x

4 −2 x 9 4 −2 x 6 6 9 cos 3 x + e − 2 x sin 3 x − e − 2 x sin 3 x + e − 2 x cos 3 x = e e cos 3 x + e − 2 x cos 3 x = e −2 x cos 3 x 13 13 13 13 13 13

= g. Given

∫e

2 2 cos −1 5 x ⋅ x + x⋅ 3 3

∫

5x 1 − 25 x 2

5x 1 − 25 x 2

∫ u dv = u v − ∫ v du

dx =

Hamilton Education Guides

5 dx 1 − 25 x 2

=

2

dx and

∫ dv = ∫ dx which implies v = x .

we obtain

2 2 x cos −1 5 x + 3 3

∫

5x 1 − 25 x 2

dx

dx use the substitution method by letting w = 1 − 25 x 2 then

∫

5x w

⋅−

1 dw = − 10 50 x

∫

dw w

= −

1 10

∫

1 1

w2

dw = −

dw dw . = −50 x which implies dx = − 50 x dx

1 −1 1 1 w 2 dw = − ⋅ 2 w 2 = − 10 10

∫

1 − 25 x 2 5

460

)

Calculus I

and

∫

Chapter 5 Solutions

2 2 2 cos −15 x dx = x cos −1 5 x + 3 3 3

Check: Let y =

∫ sinh

j. Given

−1

10 x

−1

3 1 − 25 x 2

3 1 − 25 x 2

dx

∫

∫

dw . Therefore, 2x

(

=

∫ u dv = u v − ∫ v du

x dx = sinh −1 x ⋅ x − x ⋅

1

10 x

= w 2 = 1 + x2

)

1 2

=

∫

x dx 1+ x

−

1 − 25 x 2

2 ⋅ 15

−50 x 2 1 − 25 x 2

+0

1

dx and

1 + x2

∫ dv = ∫ x dx which implies v = x . Using

we obtain

= x sinh −1 x −

1 + x2

5x

2 cos −1 5 x 3

x dx let u = sinh −1 x and dv = dx then du =

To get the integral of

dx =

1 − 25 x 2

+

the integration by parts formula

∫ sinh

∫

2 1 − 25 x 2 2 x cos −1 5 x − +c 3 15

dx =

2 1 − 25 x 2 2 2 2 x cos −1 5 x − + c , then y ′ = cos −1 5 x − ⋅ 15 3 3 3

2 cos −1 5 x − 3

=

5x

∫

x dx

(1 )

1 + x2

use the substitution method by letting w = 1 + x 2 then dw = 2 x dx which implies

2

x dx

=

1 + x2

∫

x w

⋅

1 dw = 2 2x

∫

1 w

dw =

1 2

∫

1 1

dw =

w2

−1 1− 1 2 12 1 1 1 2 = w w 2 dw = ⋅ w 2 2 2 1− 1

∫

2

(2)

1 + x2

Combining equations (1 ) and ( 2 ) together we have

∫ sinh

−1

x dx = x sinh −1 x −

x dx

∫

1 + x2

(

Check: Let y = x sinh −1 x − 1 + x 2

)

1 2

(

= x sinh −1 x − 1 + x 2

)

1 2

+c x

+ c , then y ′ = sinh −1 x +

−

1+ x

∫ x sec 10 x dx let u = x and dv = sec 10 x dx then du = dx 2

2

k. Given

Using the integration by parts formula

∫ x sec 10 x dx = x ⋅ 2

Check: Let y = = l. Given

2

2x 2 1+ x

and

+ 0 = sinh −1 x

2

∫ dv = ∫ sec 10 x dx which implies v = 2

tan 10 x . 10

∫ u dv = u v − ∫ v du we obtain

tan 10 x 1 1 1 − tan 10 x dx = x tan 10 x − ln sec 10 x + c 10 10 100 10

∫

1 1 1 1 sec 10 x tan 10 x ⋅ 10 x tan 10 x − ln sec 10 x + c , then y ′ = tan 10 x + x sec2 10 x − +0 10 100 10 100 sec 10 x

1 1 tan 10 x + x sec2 10 x − tan 10 x = x sec2 10 x 10 10 x

x

∫ 5 sinh 7 x dx let u = 5

and dv = sinh 7 x dx then du =

Using the integration by parts formula

Hamilton Education Guides

dx and 5

1

∫ dv = ∫ sinh 7 x dx which implies v = 7 cosh 7 x dx .

∫ u dv = u v − ∫ v du we obtain

461

Calculus I

Chapter 5 Solutions

∫ 5 sinh 7 x dx =

1 1 x 1 1 dx 1 1 = ⋅ cosh 7 x − cosh 7 x ⋅ x cosh 7 x − cosh 7 x ⋅ dx = x cosh 7 x − sinh 7 x + c 5 7 35 35 7 5 35 245

Check: Let y =

1 1 1 1 1 1 ⋅ 7 x sinh 7 x − ⋅ 7 cosh 7 x + 0 = cosh 7 x + x cosh 7 x − sinh 7 x + c , then y ′ = cosh 7 x 35 35 245 35 245 35

x

∫

∫

1 1 1 x sinh 7 x − cosh 7 x = x sinh 7 x 5 35 5

+

Section 5.2 Solutions – Integration Using Trigonometric Substitution Evaluate the following indefinite integrals. a. Given

dx x

2

16 − x

16 − x 2 =

let x = 4 sin t , then dx = 4 cos t dt and

2

(

16 − 16 sin 2 t =

16 1 − sin 2 t

)

16 cos 2 t = 4 cos t . Substituting these values back into the original integral we obtain:

=

∫

∫

dx x 2 16 − x 2

= −

16 − x 2 4 x 4

1 cos t 1 +c = − 16 sin t 16

(

16 − x 2 2

= −

16 x

x2

∫

)

−2 x

∫

− x 2 − 16 + x 2 16 x 2 16 − x 2

2 16 − x 2

16 x

=

−2 x2

⋅ x − 1 ⋅ 16 − x 2 +0 = −

2

16 16 x 2 16 − x 2

2 16 − x 2

16 − x 2 1

−

16 x

− x 2 − 16 − x 2 ⋅ 16 − x 2 16 − x 2 2

= −

2

16 x

1

=

x 2 16 − x 2

9 − x2 =

dx let x = 3 sin t , then dx = 3 cos t dt and

9 − x2

1 1 csc2 t dt = − cot t + c 16 16

=

1 4 ⋅ 16 − x 2 16 − x 2 ⋅ +c = − +c 16 4⋅ x 16 x

16 − x + c , then y ′ = − 16 x

− x 2 − 16 − x 2

b. Given

+c = −

2

Check: Let y = −

= −

1

4 cos t

4 cos t dt

∫ (4 sin t ) 2 ⋅ 4 cos t = ∫ 16 sin 2 t ⋅ 4 cos t dt = ∫ 16 sin 2 t dt

=

9 − 9 sin 2 t =

(

9 1 − sin 2 t

)

=

9 cos 2 t

= 3 cos t . Substituting these values back into the original integral we obtain:

∫ =

x2 9− x

∫

dx =

2

9 sin 2 t ⋅ 3 cos t dt = 3 cos t

∫ 9 sin

2

t dt = 9

∫

1 − cos 2t 9 dt = 2 2

∫ (1 − cos 2t ) dt

=

9 1   t − sin 2t  + c 2 2 

9 x 9 x 9 − x2 9 9 x x 9 + c = sin −1 − t − sin t cos t + c = ⋅ sin −1 − ⋅ ⋅ 9 − x2 + c 2 3 3 2 3 2 2 2 3 2 9 −1 x x 9 sin 9 − x 2 + c , then y ′ = − 2 3 2 2

Check: Let y =

=

=

c. Given

∫

9 2

3

⋅

9 − x2

9

−

2 9 − x2 dx x 9 + 4x

2

 x2 1  9 − x2 − − 3  2 2 9 − x2 

(

2 9 − 2 x2 4 9 − x2

let x =

Hamilton Education Guides

)=

9 2 9 − x2

1 1−

⋅ x2 9

  − 2x 1 1 x ⋅  − 9 − x2 + 2 3 2 2 9 − x2  

(

)

      2 9 − x2 − 2x2   18 − 4 x 2 9 9 − −  =  =   4 9 − x2   4 9 − x2 2 9 − x2 2 9 − x2    

−

9 − 2x2 2 9 − x2

=

3 3 tan t , then dx = sec2 t dt and 2 2

9 − 9 + 2x2 2 9 − x2

9 + 4x2 =

=

2 x2 2 9 − x2 9+4

=

(32 tan t ) 2

    

x2 9 − x2 =

9 + 4 ⋅ 94 tan 2 t

462

Calculus I

=

(

9 + 9 tan 2 t =

=

∫

Chapter 5 Solutions

dx

=

x 9 + 4x2

9 1 + tan 2 t

)

3 sec 2 t 2

∫ 32 tan t ⋅ 3sec t

9 sec2 t = 3 sec t . Therefore,

= dt =

1 sec2 t 1 cos t 1 1 1 1 sec t ⋅ ⋅ sec t dt = dt = dt dt = 3 tan t ⋅ 3 sec t 3 sin t cos t 3 tan t 3 tan t

∫

9 + 4x2

1 1 1 1 1 csc t dt = ln csc t − cot t + c = ln dt = 3 3 3 sin t 3

∫

∫

Check: Let y =

2

9 + 4x − 3

1 ln 3

+ c , then y ′ =

2x 8x2

1 = ⋅ 3

=

2x 9 + 4x

1 ⋅ 3

1 = ⋅ 3

9+ 4 x

⋅

2

⋅

9 + 4x2

)

4x2 9 + 4x2

1

2

+c

2

4x2   8x2 − 2 ⋅  9 + 4 x2 − 3 9 + 4 x2   ⋅

2x

4x2 9 + 4x2

9 + 4x2 − 3

1 ⋅ 3

2x

  ⋅ 8x − 2 ⋅  9 + 4 x2 − 3  

1 2 9+ 4 x

⋅

2x

⋅

8 x 2 − 18 + 8 x 2 + 6 9 + 4 x 2

9 + 4x2 − 3

  6  9 + 4 x2 − 3 6  = 1 ⋅ 2x ⋅  ⋅ 1 3 4x2 9 + 4x2 − 3 4x2 9 + 4x2

∫ (49 + x2 )2 dx let x = 7 tan t , then dx = 7 sec

d. Given

=

9 + 4x2 − 3

3 1 + c = ln 2x 3

2x ⋅

1 = ⋅ 3

8x2 − 2 ⋅ 9 + 4x2 + 6 9 + 4x2

9 + 4x2 − 3

−

9 + 4x2 − 3

4x2

(

2x

2x

  − 2 ⋅  9 + 4 x2 − 3  

−3

2x

2x

2

1 ⋅ 3

∫

∫

∫

=

4x2 9 + 4x2 12 x

1

=

12 x 2 9 + 4 x 2

x

9 + 4x2

(

t dt and 49 + x 2 = 49 + (7 tan t ) 2 = 49 + 49 tan 2 t = 49 1 + tan 2 t

)

= 49 sec2 t . Substituting these values back into the original integral we obtain 7 sec2 t dt

1

7 sec2 t dt

∫ (49 + x2 )2 dx = ∫ ( 49 sec2 t )2 = ∫ 2401sec2 t sec2 t =



∫  ∫

1 1 1 7 dt = ⋅ cos 2 t dt = 343 2 343 2401 sec2 t

∫

∫

 1 1  1  (t + sin t cos t ) + c = 1  tan −1 x +  t + sin 2t  + c = 686 686  7 2 686   

Check: Let y =

e.

=

1 343 + 7 x 2 − 14 x 2 1 7 = ⋅ ⋅ 2 2 686 686 49 + x 2 49 + x

+

(49 − x ) 98( 49 + x )

)

(

2

2 2

 + 5 x  dx =  x2 − 1  x2

x2 − 1

dx =

7

⋅

49 + x 2

(

49 + x 2

)

  1  7x  x  tan −1 +  + c +c = 686 7 49 + x 2    

1 49 1 1 1 7 49 + x 2 − 2 x ⋅ 7 x 7x  1  −1 x = ⋅ + ⋅ ⋅ +  + c then y ′ =  tan 2 2 2 2 686 71 + x  686 686 7 49 + x  686  7 49 + x 2 49 + x  49  

+

x2

x

∫ (1 + cos 2t ) dt

1 x 2

=

∫

(

49 + x 2 + 49 − x 2

(

98 49 + x

x2 2

Hamilton Education Guides

=

(

1 1 343 − 7 x 2 = ⋅ 2 2 686 49 + x 98 49 + x 2

98

98 49 + x

(

)

2 2

)

=

(

1

)

(

( ) ) 686( 49 + x ) +

7 49 − x 2

2 2

=

(

)

1

98 49 + x 2

)

( 49 + x )

2 2

∫

dx + 5 x dx . In Example 5.2-1, problem letter e, the solution to the first integral was:

x −1

x2 − 1 +

)

2 2

)

+

(

1 ln x + x 2 − 1 + c . Therefore, combining the two integrals we have 2

463

Calculus I

Chapter 5 Solutions

x2

∫

∫

dx + 5 x dx =

2

x −1

1 x 2

Check: Let y =

x2 − 1 +

  2x × 1 +  2 x2 − 1  = f. Given

∫

∫

1 x 2

1 ln x + 2

x2 − 1 +

x2 − 1 +

5 2 x +c 2

 1 2x2 1 5 ln x + x 2 − 1 + x 2 + c , then y ′ =  x 2 − 1 + 2 2 2 2 x2 − 1 

    2  x + x2 − 1  2  5 1 2 x2 − 1 + 2x2 1 1  + 5x = 1  4x − 2 +  +   + ⋅ 2x = 2 2 2  2 2 x + x2 − 1 2 x2 − 1 2 x2 − 1  2 x −1 

(

)

4x2 1 4x2 − 2 + 2 + 5x = + 5x = 2 4 x2 − 1 2 x2 − 1

x2 x2 − 1

∫ 5 tan t ⋅ 5 sec t tan t dt = ∫ 25 sec t tan

2

   + 5x x 2 − 1  1

+ 5x x 2 − 25 =

x 2 − 25 dx let x = 5 sec t , then dx = 5 sec t tan t dt and

x 2 − 25 dx =

  1 1 +  2 x + x2 − 1 

(

∫

25 tan 2 t = 5 tan t . Thus,

25 sec2 t − 25 =

)

∫

∫

t dt = 25 sec t sec2 t − 1 dt = 25 sec3 t dt − 25 sec t dt 25 tan t sec t − ln sec t + tan t 2

=

25 tan t sec t + ln sec t + tan t 2

=

 x x 2 − 25 25  x − ln + ⋅  5 5 2 5 

 x 2 25 x 2 − 25  x 2 25 x + x 2 − 25 x − 25 − ln x + x 2 − 25 x − 25 − ln +c = +c =  2 2 2 2 5 5  

+

25 x ln 5 + c = 2 2

25 25 ln 5 is a constant which can be included in the constant c . ln x + x 2 − 25 + c Note: 2 2

(

Check: Let y =

 × 1 +   =

x 2 − 25 −

∫ =

∫

2x 2 x 2 − 25

(

2 x 2 − 25

)

2

Check: Let y =

)+ c

=

x 2 − 25

=

2

x − 25

x 2 − 25

x 2 − 25

×

2

2

x − 25

∫ 6 cos t ⋅ 6 cos t dt = ∫ 36 cos

=

2

36 − x 2 − x 2

=

2 36 − x 2 2

Hamilton Education Guides

2

t dt =

36 2

− 25

(x

2

36 − x 2 =

∫ (1 + cos 2t ) dt =

)

x 2 − 25

− 25

)

=

2 x 2 − 50 2 x 2 − 25

x 2 − 25

=

36 − 36 sin 2 t =

36 cos 2 t = 6 cos t . Thus,

36 36  1  ( t + sin t cos t ) + c  t + sin 2t  + c = 2 2  2 

 36 −1 x x 36 − x 2 36 −1 x 36 x 36 − x 2  +c = sin + +c  + c = 2 sin 6 + 2 ⋅ 36 2 6 2 

36

=

36 − 2 x 2

+

2 36 − x 2

36 − x 2

36 − x 2

(x

x − 25

36 −1 x x 36 − x 2 36 sin + + c , then y ′ = 2 6 2 2

(36 − x )

  25 1 ⋅ − 2  x + x 2 − 25 

2 2  25 x 2 − 25 + x 2 1 x + x 2 − 25  + 0 = x − 25 + x − 25 ⋅ = − ⋅  2  2 x 2 − 25 2 x 2 − 25 2 x 2 − 25 x 2 − 25 x + x 2 − 25

 36  −1 x x 36 − x 2 sin + ⋅  2  6 6 6 

=

(

36 − x 2 dx let x = 6 sin t , then dx = 6 cos t dt and

36 − x 2 dx =

+

sec t + tan t + c =

 2x2 1 25 x 2 x − 25 − ln x + x 2 − 25 + c , then y ′ =  x 2 − 25 + 2 2 2 2 x 2 − 25 

2 x − 25 g. Given

) − 25 ln

2 36 − x 2

=

1 1−

( 6x ) 2

72 − 2 x 2 2 36 − x 2

=

⋅

1 1 + 36 − x 2 + 6 2

(

2 36 − x 2 2 36 − x 2

)

=

− 2 x2 2 36 − x 2

36 − x 2 36 − x 2

=

=

36 ⋅ 2

6 6 36 − x 2

36 − x 2 36 − x 2

⋅

36 − x 2 36 − x 2

36 − x 2

464

Calculus I

∫

h. Given

Chapter 5 Solutions

(

dx

=

)

3 9 + 36 x 2 2

∫

(

dx 9 36

let x =

)

3 x2 2

+

3 1 1  1 tan t = tan t , then dx = sec2 t dt and 9 + 36 x 2 = 9 + 36 ⋅  tan t  2 6 2  2

(

2

)

1 = 9 + 36 ⋅ tan 2 t = 9 + 9 tan 2 t = 9 1 + tan 2 t = 9 sec2 t . Therefore, 4

∫ =

(

dx

)

3 9 + 36 x 2 2

=

i. Given

∫

(

∫

=

)

3 9 sec2 2

1 1 sin t + c = 54 54

Check: Let y =

=

1 sec 2 dt 2

∫

=

6x

∫

9 2 sec

81

)

9 − 4x2 x

dx =

∫

2

(

9 1 − sin 2 t

)

3 cos t 3

)

=

)

1

(9 + 36 x )

1 2 1+ 2

9 − 4x2 =

+

Check: Let y = 3 ln

8x

−

2 9 − 4x2

×

∫

3−

 8x2 − 2 ⋅  3 −  ⋅

6x

=

9 − 4x2

3−

9 − 4x

9 − 4x2

=

4x2

)−

2

9 − 4x2 x 9 − 4x2

Hamilton Education Guides

=

)

(

9 + 36 x 2

81 9 + 36 x 2

)

1

=

(9 + 36 x )

3

2 2

(32 sin t ) 2

∫

9 − 4 ⋅ 94 sin 2 t

=

∫

3 − 2x

∫

9 − 4x2

9 − 4x2

+ 3⋅

2x

3

+c

4x 9 − 4x

−

2

4x 9 − 4x2

=

9 − 4 x2 x 9 − 4 x2

2 9− 4 x

⋅

9 − 4x

×

 9 − 4x2  ⋅ 9 − 4x2  − 2 9 − 4x

4x 9 − 4x2

 2 2  3 − 9 − 4x  ⋅ 2x 9 − 4x  

9 − 4 x2

=

 9 − 4x2  

4x2

  18  3 − 9 − 4 x 2   

9 − 4 x2

2

2

(9 − 4 x ) 9 − 4 x x (9 − 4 x ) 2

2

6x

=

9 − 4x2

3−

  6 x  8 x 2 − 6 9 − 4 x 2 + 18 − 8 x 2   − = 2 2 2 3 − 9 − 4x ⋅ 4x 9 − 4x =

 ⋅ 2x − 2 ⋅  3 − 

1

8x ⋅

2x

9 − 4 x 2 + c , then y ′ = 3 ⋅

9 − 4x2 ⋅ 2x 9 − 4x2 4x

81 9 + 36 x 2

9−4

∫

3−

  3  − 6 9 − 4 x 2 + 18   

=

(

9 + 36 x 2

81+ 324 x 2 −324 x 2

1 6 3 cos 2 t 1 sin 2 t 1 − sin 2 t ⋅ dt = 3 dt = 3 dt − 3 dt = 3 dt sin t sin t 3 2 sin t sin t sin t

=

+

(

4x

−

2x

8x2 − 6 9 − 4x2 + 2 9 − 4x2 2

∫

9 − 4x2 + c

9 − 4x2

3−

)

(

∫

2x

∫

9 cos 2 t = 3 cos t . Therefore,

=

∫ 32 sin t ⋅ 2 cos t dt

9 − 4x2

3−

∫

9 9 + 36 x 2 −324 x 2

− 3 sin t dt = 3 csc t dt − 3 sin t dt = 3 ln csc t − cot t + 3 cos t + c = 3 ln

= 3 ln

1 dt 1 dt 1 = = cos t dt 2 27 sec t 54 sec t 54

=

⋅x

2 9 + 36 x 2

=

1 9 + 36 x 2 2

3 3 sin t , then dx = cos t dt and 2 2

dx let x =

∫

(

72 x

81 9 + 36 x 2

(9 + 36 x )⋅ (

dt

∫ 2 729 sec t

+c

1

=

81 9 + 36 x 2 ⋅ 9 + 36 x 2

9 − 9 sin 2 t =

9 9 + 36 x 2

1 2

=

93 sec3 t

+ c , then y ′ =

9 9 + 36 x 2

x

t

∫2

1 sec 2 dt 2

1 ⋅ 9 9 + 36 x 2 − 9 ⋅

x

9 − 4x2

2

= x

+c =

9 + 36 x 2

(

1 sec 2 dt 2 3 2× 3

4x 9 − 4x2 4x

−

9 − 4x2 2

=

=

9 x 9 − 4x2

9 − 4x2 x

465

Calculus I

Chapter 5 Solutions

Section 5.3 Solutions – Integration by Partial Fractions a. Evaluate the integral

dx

∫ x2 + 5x + 6 .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 2 + 5 x + 6 into (x + 2 ) (x + 3) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way: 2

1

=

x + 5x + 6

1

(x + 2) (x + 3)

A B + x+2 x+3

=

Fourth - Solve for the constants A and B by equating coefficients of the like powers.

1

2

x + 5x + 6

=

A (x + 3) + B (x + 2 ) (x + 2) (x + 3)

1 = A (x + 3) + B (x + 2 ) = Ax + 3 A + Bx + 2 B 1 = ( A + B ) x + (3 A + 2 B ) therefore, 3 A + 2B = 1

A+ B = 0 which result in having A = 1 and B = −1

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

dx

1

B

A

1

∫ x2 + 5x + 6 = ∫ x + 2 dx + ∫ x + 3 dx = ∫ x + 2 dx − ∫ x + 3 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1

1

∫ x + 2 dx − ∫ x + 3 dx = ln

x + 2 − ln x + 3 + c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = ln x + 2 − ln x + 3 + c , then y ′ = b. Evaluate the integral

( x + 3) − ( x + 2 ) = x + 3 − x − 2 = 1 1 1 ⋅1 − ⋅1 + 0 = x+2 x+3 ( x + 2 ) ( x + 3) x 2 + 3 x + 2 x + 6 x 2 + 5 x + 6

x 2 +1

∫ x 3 − 4 x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x3 − 4 x into x x 2 − 4 = x(x − 2 ) (x + 2 ) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way:

x2 + 1

=

x3 − 4 x

A B C x2 + 1 = + + x x−2 x+2 x (x − 2 ) (x + 2 )

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.

x2 + 1 3

2

(

x − 4x

=

A (x − 2 ) (x + 2 ) + Bx (x + 2 ) + Cx (x − 2 ) x (x − 2 ) (x + 2 )

) (

) (

)

x + 1 = A x 2 + 2 x − 2 x − 4 + B x 2 + 2 x + C x 2 − 2 x = Ax 2 − 4 A + Bx 2 + 2 Bx + Cx 2 − 2Cx x 2 + 1 = ( A + B + C )x 2 + (2 B − 2C )x − 4 A therefore, A+ B+C =1

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Chapter 5 Solutions

which result in having A = −

1 5 5 , B = , and C = 4 8 8

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

x 2 +1

A

B

C

∫ x 3 − 4 x dx = ∫ x dx + ∫ x − 2 dx + ∫ x + 2 dx

= −

1 1 5 1 5 1 dx + dx + dx 4 x 8 x−2 8 x+2

∫

∫

∫

Sixth - Integrate each integral individually using integration methods learned in previous sections. −

1 5 5 1 1 5 1 5 1 dx + dx + dx = − ln x + ln x − 2 + ln x + 2 + c 4 x 8 x−2 8 x+2 4 8 8

∫

∫

∫

Seventh - Check the answer by differentiating the solution. The result should match the integrand.

5 5 1 −2(x − 2 )(x + 2 ) + 5 x(x + 2 ) + 5 x(x − 2 ) 1 5 5 Let y = − ln x + ln x − 2 + ln x + 2 + c , then y ′ = − = + + 8 4 8 8 x(x − 2 )(x + 2 ) 4 x 8(x − 2 ) 8(x + 2 )

=

− 2 x 2 + 8 + 5 x 2 + 10 x + 5 x 2 − 10 x

(

8x x2 − 4

c. Evaluate the integral

)

=

8x2 + 8

(

8x x2 − 4

=

)

( ) 8 x (x − 4 ) 8 x2 + 1

=

2

x2 + 1

x3 − 4 x

1

∫ 36 − x 2 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator 36 − x 2 into (6 − x )(6 + x ) . Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once, the integrand can be represented in the following way:

1

36 − x 2

=

1

(6 − x ) (6 + x )

=

A B + 6−x 6+ x

Fourth - Solve for the constants A and B by equating coefficients of the like powers. 1

36 − x 2

=

A (6 + x ) + B (6 − x ) (6 − x ) (6 + x )

1 = A (6 + x ) + B (6 − x ) = 6 A + Ax + 6 B − Bx

1 = ( A − B ) x + (6 A + 6 B ) therefore, 6 A + 6B = 1 which result in having A =

A− B = 0

1 1 , and B = 12 12

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

1

A

B

1

1

1

1

∫ 36 − x2 dx = ∫ 6 − x dx + ∫ 6 + x dx = 12 ∫ 6 − x dx + 12 ∫ 6 + x dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1 1 1 1 1 1 dx + dx = ln 6 − x + ln 6 + x + c 12 6 − x 12 6 + x 12 12

∫

∫

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y =

1 1 6+ x+6− x 1 1 1 1 1 = ⋅ ln 6 − x + ln 6 + x + c , then y ′ = + ⋅ +0 = 12(6 − x ) (6 + x ) 12 12 12 6 − x 12 6 + x 36 − x 2

d. Evaluate the integral

x+5

∫ x3 + 2 x2 + x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

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Chapter 5 Solutions

(

)

Second - Factor the denominator x3 + 2 x 2 + x into x x 2 + 2 x + 1 = x(x + 1)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way:

x+5

x3 + 2 x 2 + x

=

(

x+5

=

)

2

x x + 2x + 1

x+5

=

x(x + 1)2

A B C + + x x + 1 (x + 1)2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.

x+5

x3 + 2 x 2 + x

(

A (x + 1)2 + Bx (x + 1) + Cx

=

) (

x(x + 1)(x + 1)2

)

x + 5 = A x 2 + 2 x + 1 + B x 2 + x + Cx = Ax 2 + 2 Ax + A + Bx 2 + Bx + Cx x + 5 = ( A + B )x 2 + (2 A + B + C )x + A therefore, 2A + B + C = 1

A+ B = 0

A=5

which result in having A = 5 , B = −5 , and C = −4 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values. x+5

1

C

B

A

1

1

∫ x3 + 2 x2 + x dx = ∫ x dx + ∫ x + 1 dx + ∫ (x + 1) 2 dx = 5∫ x dx − 5∫ x + 1 dx − 4∫ (x + 1) 2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 5

1

1

1

∫ x dx − 5∫ x + 1 dx − 4∫ (x + 1) 2 dx = 5 ln

x − 5 ln x + 1 +

4 +c x +1

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = 5 ln x − 5 ln x + 1 + =

5(x + 1) 2 − 5 x (x + 1) − 4 x 4 4 1 1 + c , then y ′ = 5 ⋅ − 5 ⋅ +0 = − 2 x +1 x x + 1 (x + 1) x (x + 1)2

5 x 2 + 10 x + 5 − 5 x 2 − 5 x − 4 x 3

2

x + 2x + x

e. Evaluate the integral

=

x+5

3

x + 2x2 + x

1

∫ x3 − 2 x2 + x dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second – Factor the denominator x3 − 2 x 2 + x into x(x − 1)2 . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way: 1

x(x − 1)2

=

A B C + + x x − 1 (x − 1)2

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 3

1

2

x − 2x + x

(

) (

A (x − 1)2 + Bx (x − 1) + Cx

=

)

x(x − 1)2

1 = A x 2 − 2 x + 1 + B x 2 − x + Cx = Ax 2 − 2 Ax + A + Bx 2 − Bx + Cx 1 = ( A + B )x 2 + (− 2 A − B + C )x + A therefore, A+ B = 0

−2 A − B + C = 0

A =1

which result in having A = 1 , B = −1 , and C = 1 Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

Hamilton Education Guides

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Chapter 5 Solutions

dx

B

A

1

C

1

1

∫ x3 − 2 x2 + x = ∫ x dx + ∫ x − 1 dx + ∫ (x − 1)2 dx = ∫ x dx − ∫ x − 1 dx + ∫ (x − 1) 2 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1

1

1

∫ x dx − ∫ x − 1 dx + ∫ (x − 1) 2 dx

1 +c x −1

= ln x − ln x − 1 −

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = ln x − ln x − 1 − =

x2 − x2 − 2x + 2x + 1 3

=

2

x − 2x + x

f. Evaluate the integral

(x − 1)2 − x (x − 1) + x = x 2 − 2 x + 1 − x 2 + x + x 1 1 1 1 + c , then y ′ = − + +0 = 2 x x − 1 (x − 1) x −1 x3 − 2 x 2 + x x(x − 1)2 3

1

x − 2x2 + x

x2 + 3

∫ x2 − 1 dx . x2 + 3

∫ x2 − 1 dx = ∫

First – Rewrite the integral in the following form:

(x − 1)+ 4 dx = 2

2

x −1

 x2 − 1

4





4



∫  x2 − 1 + x2 − 1  dx = ∫ 1 + x 2 − 1  dx .

Then, check to see if the integrand of the second integral is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction. Second - Factor the denominator x 2 − 1 into (x − 1) (x + 1) . Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand can be represented in the following way:

4 2

x −1

4

=

=

(x − 1) (x + 1)

B A + x −1 x +1

Fourth - Solve for the constants A and B by equating coefficients of the like powers.

4 2

x −1

=

A (x + 1) + B (x − 1) (x − 1) (x + 1)

4 = A (x + 1) + B (x − 1) = Ax + A + Bx − B 4 = ( A + B )x + ( A − B ) therefore, A− B = 4

A+ B = 0 which result in having A = 2 and B = −2

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

x2 + 3



4

4



A

B

1

1

∫ x2 − 1 dx = ∫ 1 + x2 − 1  dx = ∫ dx + ∫ x2 − 1 dx = ∫ dx + ∫ x − 1 dx + ∫ x + 1 dx = ∫ dx + 2∫ x − 1 dx − 2∫ x + 1 dx Sixth - Integrate each integral individually using integration methods learned in previous sections. 1

1

∫ dx + 2∫ x − 1 dx − 2∫ x + 1 dx =

x + 2 ln x − 1 − 2 ln x + 1 + c

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = x + 2 ln x − 1 − 2 ln x + 1 + c , then y ′ = 1 + 2 ⋅ =

(x

2

)

+ x − x − 1 + 2x + 2 − 2x + 2

(x

2

)

+ x − x −1

g. Evaluate the integral

=

x2 − 1 + 4 2

x −1

=

(x − 1) (x + 1) + 2(x + 1) − 2(x − 1) 1 1 − 2⋅ +0 = (x − 1) (x + 1) x −1 x +1

x2 + 3 x2 − 1

1

∫ x3 − 1 dx .

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Chapter 5 Solutions

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x3 − 8 into (x − 1) x 2 + x + 1 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way:

1

=

x3 − 1

(x − 1) (x

1 2

A Bx + C + x − 1 x2 + x + 1

=

)

+ x +1

Fourth - Solve for the constants A , B and C by equating coefficients of the like powers. 1

=

3

x −1

(

) (x − 1) (x

(

A x 2 + x + 1 + (Bx + C ) (x − 1)

)

)

2

+ x +1

1 = A x 2 + x + 1 + (Bx + C ) (x − 1) = Ax 2 + Ax + A + Bx 2 − Bx + Cx − C 1 = ( A + B ) x 2 + ( A − B + C )x + ( A − C ) therefore, A+ B = 0 which result in having A =

A− B+C = 0

A−C =1

1 1 2 , B = − , and C = − 3 3 3

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

1

∫ x3 − 1

dx =

∫

A dx + x −1

Bx + C

∫ x2 + x + 1

dx =

1 1 dx + 3 x −1

∫

− 13 x −

2 3

1

1

1

x+2

∫ x2 + x + 1 dx = 3 ∫ x − 1 dx − 3 ∫ x2 + x + 1 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections. To solve the second integral let u = x 2 + x + 1 , then x+2 =

1 (2 x + 1) + 3 . Therefore, 2 2

1 1 1 x+2 1 1 1 dx − dx = dx − 3 x −1 3 x2 + x + 1 3 x −1 3

∫

=

=

∫

1 1 1 dx − 3 x −1 6

∫

∫

2x + 1

1

∫

1 2

1

(2 x + 1) + 32 x2 + x + 1 1

dx =

1

1 1 1 dx − 3 x −1 6

∫

2x + 1

∫ x2 + x + 1

1 1 2 x + 1 du ⋅ − u 2x + 1 2

∫ x2 + x + 1 dx − 2 ∫ x2 + x + 1 dx = 3 ∫ x − 1 dx − 6 ∫

1 1 1 1 ln x − 1 − ln u − ⋅ 3 6 2 3 2

=

du du . Also, x + 2 can be rewritten as = 2 x + 1 and dx = dx 2x + 1

2 x +1 tan −1 2 3 2

+c =

dx −

3

1 2 dx 3 x2 + x + 1

∫

1

∫ (x + 1 ) 2 + 3 dx 4

2

1 1 1 2 2(2 x + 1) +c ln x − 1 − ln x 2 + x + 1 − ⋅ tan −1 3 6 2 3 2 3

1 1 1 2x + 1 1 1 3 2x + 1 ln x − 1 − ln x 2 + x + 1 − tan −1 + c = ln x − 1 − ln x 2 + x + 1 − tan −1 +c 3 6 3 6 3 3 3 3

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = y′ =

−

1 1 3 2x + 1 ln x − 1 − ln x 2 + x + 1 − tan −1 + c , then 3 6 3 3

3 1 1 1 1 ⋅ − ⋅ ⋅ (2 x + 1) − ⋅ 3 3 x − 1 6 x2 + x + 1

1 1 +  2 x +1   3 

2

⋅

( 2 ⋅ 3 )− 0 ⋅ (2 x + 1) + 0 = 1 − 2 x + 1 3(x − 1) 6(x + x + 1) ( 3) 2

2

2 3 3 3 6 x 2 + 6 x + 6 − (2 x + 1) ⋅ (3 x − 3) 2 1 2x + 1 2 = = ⋅ ⋅ − − − 2 2 2 3 3 + (2 x + 1)2 3 3(x − 1) 6 x 2 + x + 1 3 + 4x + 4x + 1 3 + (2 x + 1) 18(x − 1) x + x + 1

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(

)

(

)

470

Calculus I

Chapter 5 Solutions

6 x 2 + 6 x + 6 − 6 x 2 + 6 x − 3x + 3

=

) (x + 1) (x + x + 1) − (x − 1) = = 2(x − 1)(x + x + 1) (

−

3

18 x − 1

2

=

9x + 9

(

3

)

18 x − 1

x3 + x 2 + x + x 2 + x + 1 − x3 + 1

3

3

2

4x2 + 4x + 4

(x − 1)(2 x 3

2

2

+ 2x + 2

)

−

1

(

=

)

2

4 x + x +1

(

x +1 3

(x − 1)(2 x 3

2

)

2 x −1

2x2 + 2x + 2

=

−

+ 2x + 2

)

(

1 2

)

2 x + x +1

1

=

x3 − 1

1

∫ x4 − 1 dx .

h. Evaluate the integral

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

)

)(

(

(

)

Second - Factor the denominator x 4 − 1 into x 2 − 1 x 2 + 1 = (x − 1) (x + 1) x 2 + 1 . Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way:

1

=

4

x −1

1

(x − 1) (x + 1) (x

2

=

)

+1

A B Cx + D + + x − 1 x + 1 x2 + 1

Fourth - Solve for the constants A , B , C and D by equating coefficients of the like powers. 1

4

x −1

(

)

(

)

A(x + 1) x 2 + 1 + B (x − 1) x 2 + 1 + (x − 1) (x + 1) (Cx + D )

=

(

(x − 1) (x + 1) (x 2 + 1)

)

(

)

1 = A(x + 1) x 2 + 1 + B(x − 1) x 2 + 1 + (x − 1)(x + 1)(Cx + D ) 1 = Ax3 + Ax 2 + Ax + A + Bx3 − Bx 2 + Bx − B + Cx3 + Dx 2 − Cx − D 1 = ( A + B + C )x3 + ( A − B + D )x 2 + ( A + B − C )x + ( A − B − D ) therefore, A+ B+C = 0 which result in having A =

A− B+ D = 0

A− B− D =1

A+ B−C = 0

1 1 1 , B = − , C = 0 , and D = − 4 4 2

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

1

A

B

∫ x4 − 1 dx = ∫ x − 1 dx + ∫ x + 1 dx + ∫

Cx + D 2

x +1

dx =

1 4

1

1

1

1

1

∫ x − 1 dx − 4 ∫ x + 1 dx − 2 ∫ x2 + 1 dx

Sixth - Integrate each integral individually using integration methods learned in previous sections.

1 4

1

1

1

1

1

∫ x − 1 dx − 4 ∫ x + 1 dx − 2 ∫ x2 + 1 dx =

1 1 1 ln x − 1 − ln x + 1 − tan −1 x + c 4 4 2

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y = =

1 1 1 1 1 1 1 1 1 ln x − 1 − ln x + 1 − tan −1 x + c , then y ′ = ⋅ ⋅1 − ⋅ ⋅1 − ⋅ 2 ⋅1 + 0 4 4 2 4 x −1 4 x +1 2 x +1

1 1 1 − − 4(x − 1) 4(x + 1) 2 x 2 + 1

(

2

− 2x − 2x + 2x + 2

(

)(

4 x2 − 1 4 + x2

)

i. Evaluate the integral

=

=

)

(

4

)

4 x4 − 1

=

(x + 1) (x 2 + 1) − (x − 1) (x 2 + 1) − 2(x − 1) (x + 1) 4(x − 1) (x + 1) (x 2 + 1)

=

x3 + x + x 2 + 1 − x3 − x + x 2 + 1

(

)(

4 x2 − 1 4 + x2

)

1

4

x −1

1

∫ x3 + 64 dx .

First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is a rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.

(

)

Second - Factor the denominator x3 + 64 into (x + 4 ) x 2 − 4 x + 16 .

Hamilton Education Guides

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Chapter 5 Solutions

Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the integrand can be represented in the following way:

1

x3 + 64

=

(x + 4) (x

1 2

− 4 x + 16

=

)

A Bx + C + x + 4 x 2 − 4 x + 16

Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers. 3

1

x + 64

(

(

)

A x 2 − 4 x + 16 + (Bx + C ) (x + 4 )

=

)

(x + 4) (x 2 − 4 x + 16)

1 = A x 2 − 4 x + 16 + (Bx + C )(x + 4 ) = Ax 2 − 4 Ax + 16 A + Bx 2 + 4 Bx + Cx + 4C

1 = ( A + B )x 2 + (− 4 A + 4 B + C )x + (16 A + 4C ) therefore, A+ B = 0 which result in having A =

−4 A + 4 B + C = 0

16 A + 4C = 1

1 1 1 , B=− , and C = 48 48 6

Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.

1

∫ x3 + 64

dx =

∫

A dx + x+4

Bx + C

∫ x2 − 4 x + 16

dx =

1 1 dx + 48 x + 4

∫

−

1 48

x+

1 6

1 1 1 x −8 dx − dx 48 x + 4 48 x 2 − 4 x + 16

∫

∫ x2 − 4 x + 16 dx =

∫

Sixth - Integrate each integral individually using integration methods learned in previous du du . Also, x − 8 can be = 2 x − 4 and dx = dx 2x − 4

sections. To solve the second integral let u = x 2 − 4 x + 16 , then rewritten as x − 8 = (x − 2 ) − 6 =

1 (2 x − 4) − 6 . Therefore, 2

(x − 2) − 6 dx = 1 1 dx − 1 1 1 1 1 1 1 x −8 dx − dx = dx − 48 x + 4 48 48 x + 4 48 x + 4 48 x 2 − 4 x + 16 48 x 2 − 4 x + 16

∫

∫

∫

1 1 1 dx − 48 x + 4 96

=

1 1 1 1 1 dx − du + 48 x + 4 96 u 8

=

1 1 12 x−2 12 x−2 1 1 ln x + 4 − ln x 2 − 4 x + 16 + tan −1 +c = ln x 2 − 4 x + 16 − tan −1 +c ln x + 4 − 48 96 8 ⋅ 12 48 96 96 12 12

∫

1

6

(2 x − 4) − 6

∫ x2 − 4 x + 16 dx

=

∫

2x − 4

∫

∫

1 2

6 1 1 1 2 x − 4 du dx − ⋅ + u 96 2 x − 4 48 48 x + 4

∫

∫ x2 − 4 x + 16 dx + 48 ∫ x2 − 4 x + 16 dx = ∫

1

∫ (x − 2)2 + 12 dx =

∫

1

∫ (x − 2)2 + 12 dx

1 1 1 x−2 ln x + 4 − ln u + tan −1 +c 48 96 12 8 12

Seventh - Check the answer by differentiating the solution. The result should match the integrand. Let y =

y′ =

1 1 12 x−2 ln x + 4 − ln x 2 − 4 x + 16 + tan −1 + c , then 48 96 96 12

1 1 1 12 1 ⋅ − ⋅ ⋅ (2 x − 4 ) + ⋅ 48 x + 4 96 x 2 − 4 x + 16 96

1 1 +  x − 2   12 

2

⋅

(1 ⋅ 12 )− 0 ⋅ (x − 2) + 0 = 1 − 48(x + 4 ) 48(x ( 12 ) 2

+

1 6 x−2 12 12 12 = − + ⋅ ⋅ 2 2 2 48(x + 4 ) 48 x − 4 x + 16 96 12 + (x − 2 ) 12 48 x − 4 x + 16

=

−x + 8 1 + 2 48(x + 4 ) 48 x − 4 x + 16

=

(

16 + 32

(

48(x + 4 ) x − 4 x + 16 2

)

Hamilton Education Guides

)

=

=

(x

(

2

) 48(x + 4 ) (x

)

(

− 4 x + 16 + (− x + 8)(x + 4 )

(

2

− 4 x + 16

48

48(x + 4 ) x − 4 x + 16 2

)

=

)

=

)

=

x−2 2

1 −x + 2 + 6 + 48(x + 4 ) 48 x 2 + 4 x + 16

(

− 4 x + 16

)

)

x 2 − 4 x + 16 − x 2 − 4 x + 8 x + 32

(

48(x + 4 ) x 2 − 4 x + 16 1

x3 − 4 x 2 + 16 x + 4 x 2 − 16 x + 64

=

)

1

x3 + 64

472

Calculus I

Chapter 5 Solutions

Section 5.4 Practice Problems – Integration of Hyperbolic Functions 1. Evaluate the following integrals: a. Given

∫ cosh 3x dx let u = 3x , then

∫ cosh 3x dx = ∫ cosh u ⋅

1 1 du 1 = cosh u du = sinh u + c = sinh 3 x + c 3 3 3 3

∫

1 1 d d 1 d 1 sinh 3 x + c , then y ′ = ⋅ sinh 3 x + c = ⋅ cosh 3 x ⋅ 3 x + 0 = ⋅ cosh 3 x ⋅ 3 = cosh 3 x 3 dx dx 3 dx 3 3

Check: Let y = b. Given

du du d . Therefore, = 3 x = 3 which implies dx = dx dx 3

3x 3x ∫ (sinh 2 x − e ) dx = ∫ sinh 2 x dx + ∫ e dx

let:

a. u = 2 x , then

du du du d and = 2 ; du = 2dx ; dx = = 2x ; dx 2 dx dx

b. v = 3 x , then

dv dv d dv . = 3x ; = 3 ; dv = 3dx ; dx = dx dx 3 dx

∫ sinh 2 x dx + ∫ e

Therefore, =

3x

∫ sinh u ⋅

dx =

1 1 du dv 1 1 v = + e3 x ⋅ e dv = cosh u + c1 + ev + c2 sinh u du + 2 3 2 3 2 3

∫

∫

∫

1 1 1 1 cosh 2 x + e3 x + c1 + c2 = cosh 2 x + e 3 x + c 3 2 2 3

d d 1 1 1 1 1 d d 1 d cosh 2 x + e3 x + c then y ′ = ⋅ cosh 2 x + ⋅ e3 x + c = ⋅ sinh 2 x ⋅ 2 x + ⋅ e3 x ⋅ 3 x + 0 2 3 2 dx 3 dx dx dx dx 2 3

Check: Let y = = c. Given

2 3 1 1 ⋅ sinh 2 x ⋅ 2 + ⋅ e3 x ⋅ 3 = ⋅ sinh 2 x + ⋅ e3 x = sinh 2 x + e3 x 2 3 2 3

∫ csc h 5x dx let u = 5x , then

∫ csc h 5x dx = ∫ csc h u ⋅

du 1 1 u 1 5x = csc h u du = ln tanh +c + c = ln tanh 5 5 5 2 5 2

∫

1 d 5x d 5x 1 1 1 + c = ⋅ + c , then y ′ = ⋅ ln tanh ln tanh 2 5 5 dx 2 dx 5 tanh

Check: Let y =

1 1 = ⋅ 5 tanh

5x 2

5x 5  1  ⋅ ⋅  sec h 2 ⋅ +0 = 10 2 2  

cosh 2 5 x − sinh 2 5 x 2

2⋅ d. Given

∫x

2

∫x

2

2 sinh 5 x 2 cosh 5 x 2

Check: Let y = 2

=

1 cosh 2 5 x

2⋅

2 sinh 5 x 2 cosh 5 x 2

=

∫3x

3

∫x

2

sec h 2u ⋅

du 3x

2

=

5 sec h 2 52x tanh 52x

1 = ⋅ 2

sec h 2 52x tanh 52x

1 = 2

5x 2

⋅

d  5x   tanh +0 dx  2 

1 − tanh 2 52x ⋅ tanh 52x

1 = ⋅ 2

1−

sinh 2 5 x 2

cosh 2 5 x

sinh 5 x

2

2 cosh 5 x 2

5x

cosh 2 1 1 1 1 = = = = csc h5 x ⋅ 2 sinh 5 x 2 cosh 5 x ⋅ sinh 5 x sinh 2 ⋅ 52x 2 cosh 52x ⋅ sinh 52x 2

sec h 2 x3dx let u = x3 , then

sec h 2 x3dx =

e. Given

2

cosh 2 5 x

=

du du d . Therefore, = 5 x = 5 which implies du = 5dx ; dx = 5 dx dx

2

du d 3 du du = = 3x 2 ; du = 3 x 2 dx ; dx = 2 . Therefore, x ; dx dx dx 3x

1 1 1 sec h 2u du = tanh u + c = tanh x 3 + c 3 3 3

∫

1 1 d d 1 d 3 1 tanh x3 + c , then y ′ = ⋅ tanh x3 + c = ⋅ sec h 2 x3 ⋅ x + 0 = ⋅ sec h 2 x3 ⋅ 3 x 2 = x 2 sec h 2 x3 3 3 dx dx 3 dx 3

(

)

csc h 2 x 4 + 1 dx let u = x 4 + 1 , then

Hamilton Education Guides

(

)

du d 4 du du = x +1 ; = 4x3 ; du = 4 x3dx ; dx = 3 . Therefore, dx dx dx 4x

473

Calculus I 2

∫3x

3

Chapter 5 Solutions

(

)

(

(

)

(

) (

1 1 2 3 du 1 x csc h 2u ⋅ 3 = csc h 2u du = − coth u + c = − coth x 4 + 1 + c 6 6 6 3 4x

∫

csc h 2 x 4 + 1 dx =

∫

)

(

)

)

1 1 d d 1 d 4 Check: Let y = − coth x 4 + 1 + c , then y ′ = − ⋅ coth x 4 + 1 + c = − ⋅ − csc h 2 x 4 + 1 ⋅ x +1 + 0 6 dx dx 6 dx 6

(

f. Given

∫

∫x

3

)

(

)

csc h 2 x 4 + 5 dx let u = 2 x 4 + 5 , then

(

)

∫

x3 csc h 2 x 4 + 5 dx =

(

)

x3 csc h u ⋅

du

=

8 x3

1 8

∫

(

)

(

)

du du d du = 2x4 + 5 ; = 2 x 4 + 5 ; du = 8 x3dx ; dx = 3 . Therefore, dx dx dx 8x

csc h u du =

u 1 2 x4 + 5 1 ln tanh + c = ln tanh +c 2 8 2 8

1 2x4 + 5 1 d 1 1 d 2x4 + 5 d 2x4 + 5 ⋅ tanh +0 ln tanh + c = ⋅ + c , then y ′ = ⋅ ln tanh 4 2 8 tanh 2 x + 5 dx 8 8 dx 2 dx 2 2

Check: Let y =

=

(

)

4 x3 1 2 ⋅ csc h 2 x 4 + 1 = x3 csc h 2 x 4 + 1 ⋅ csc h 2 x 4 + 1 ⋅ 4 x3 = 6 3 6

=

1 2 x 4 +5 2

8 tanh

sec h 2 2x4 + 5 d 2x4 + 5 ⋅ sec h 2 ⋅ +0 = 2 2 dx 8 tanh

2 x 4 +5 2 2 x 4 +5 2

4

2 2 x +5 8 x3 x3 1 − tanh x3 2 = = ⋅ ⋅ ⋅ 4 2 2 2 tanh 2 x 2 + 5

1−

4 sinh 2 2 x + 5 2

4 cosh 2 2 x + 5 2

4 sinh 2 x + 5 2

4 cosh 2 x + 5 2

x3 = ⋅ 2

4 4 cosh 2 2 x + 5 − sinh 2 2 x + 5 2 2 2 2 x 4 +5

cosh

2 2 x 4 +5

sinh

2

x3 = ⋅ 2

=

sinh 2 ⋅

∫ cosh

g. Given

∫ cosh

x

7

7

=

(

x

4

sinh 2 x + 5

( x + 1) sinh ( x + 1)dx

=

2

4

cosh 2 x 2 + 5 x3 x3 = = ⋅ 4 4 2 cosh 2 2 x 4 + 5 ⋅ sinh 2 x 4 + 5 2 cosh 2 x 2 + 5 ⋅ sinh 2 x 2 + 5 2 2

2

∫u

(

3

( x + 1) sinh ( x + 1)dx

Check: Let y = h. Given

2 x 4 +5 2

sinh

2 2 x 4 +5

4 cosh 2 x + 5

4 cosh 2 x + 5 2

3

1 4 cosh 2 2 x + 5

7

= x3 csc h 2 x 4 + 5

)

) du d du du . Thus, = cosh (x + 1) ; = sinh (x + 1) ; dx = dx dx dx sinh (x + 1)

let u = cosh (x + 1) , then sinh (x + 1) ⋅

du = sinh (x + 1)

∫ u du 7

=

1 1 8 u + c = cosh 8 ( x + 1) + c 8 8

1 1 cosh8 (x + 1) + c , then y ′ = ⋅ 8 cosh 7 (x + 1) ⋅ sinh (x + 1) + 0 = cosh 7 (x + 1) sinh (x + 1) 8 8

∫ csc h (5x + 3) coth (5x + 3) dx let u = 5x + 3 , then

∫ csc h (5x + 3) coth (5x + 3) dx = ∫ csc h u coth u

du d ( 5 x + 3) ; du = 5 ; du = 5dx ; dx = du . Therefore, = dx dx 5 dx

1 du 1 1 = csc h u coth u du = − csc h u + c = − csc h (5 x + 3 ) + c 5 5 5 5

∫

1 1 csc h (5 x + 3) coth (5 x + 3) d Check: Let y = − csc h (5 x + 3) + c , then y ′ = − ⋅ − csc h (5 x + 3) coth (5 x + 3) ⋅ (5 x + 3) + 0 = ⋅5 5 5 dx 5

i. Given

∫e

=

5 csc h (5 x + 3) coth (5 x + 3) = csc h (5 x + 3) coth (5 x + 3) 5

∫e

x +1 sec h e x +1 dx let u = e x +1 , then du = d e x +1 ; du = e x +1 ; du = e x +1 ⋅ dx ; dx = du . Therefore, dx dx dx e x +1

x +1 sec h e x +1 dx =

(

∫e

x +1

sec h u ⋅

)

du e

x +1

=

Check: Let y = sin −1 tanh e x +1 + c , then y ′ =

Hamilton Education Guides

∫ sec h u du = sin 1 1 − tanh 2 e x +1

−1

⋅

(tanh u ) + c

(

)

= sin −1 tanh e x +1 + c

d tanh e x +1 + 0 = dx

sec h 2e x +1 sec h 2e x +1

⋅

d x +1 e dx

474

Calculus I

Chapter 5 Solutions sec h 2e x +1

=

sec h e x +1

⋅ e x +1 = e x +1 sec h e x +1

2. Evaluate the following integrals: a.

∫ tanh ∫ tanh

5

x sec h 2 x dx let u = tanh x , then

5

x sec h 2 x dx =

Check: Let y = b. Given

∫ coth

6

∫u

5

du

⋅ sec h 2 x ⋅

du d du du . Thus, = tanh x ; = sec h 2 x ; du = sec h 2 x dx ; dx = dx dx dx sec h 2 x 2

sec h x

=

∫ u du = 5

1 1 1 5+1 u + c = u 6 + c = tanh 6 x + c 5 +1 6 6

1 1 tanh 6 x + c then y ′ = ⋅ 6 (tanh x ) 6 −1 ⋅ sec h 2 x + 0 = (tanh x ) 5 sec h 2 x = tanh 5 x sec h 2 x 6 6

(x + 1) csc h 2 ( x + 1) dx

; du = − csc h 2 (x + 1) dx ; dx = −

let u = coth (x + 1) , then du

csc h 2 (x + 1)

. Thus,

∫ coth

6

du d du = coth (x + 1) ; = − csc h 2 (x + 1) c dx dx dx

(x + 1) csc h 2 ( x + 1) dx

=

∫u

6

⋅ csc h 2 (x + 1) ⋅

− du

csc h 2 (x + 1)

1 1 = − u 6 du = − u 7 + c = − coth7 ( x + 1) + c 7 7

∫

1 1 Check: Let y = − coth 7 (x + 1) + c then y ′ = − ⋅ 7[ coth (x + 1) ] 7 −1 ⋅ − csc h 2 (x + 1) + 0 = coth 6 (x + 1) csc h 2 (x + 1) 7 7

c. Given

∫e

3x

∫e

3x

tanh e3 x dx let u = e3 x , then

Check: Let y = d. Given

∫x

3

∫e

tanh e3 x dx =

∫x

3

3x

(

)

)

∫x

(

(

3

⋅ sec h u ⋅

1 1 1 tanh u du = ln cosh u + c = ln cosh e 3 x + c 3 3 3

∫

)

[ (

du

4 x3

=

( (

)

[

)] sec h (x + 1) 1 d ⋅ tanh (x + 1) + 0 = dx 1 − tanh (x + 1) 4 sec h (x + 1) (

1 1 1 sec h u ⋅ du = sin −1 (tanh u ) + c = sin −1 tanh x 4 + 1 + c 4 4 4

∫

)]

2

4

4

) )

2

4

2

4

4

sec h 2 x 4 + 1 4 x3 d 4 ⋅ 4 x3 = ⋅ sec h x 4 + 1 = x3 sec h x 4 + 1 x +1 = 4 4 dx 4 sec h x + 1

(

)

(

∫ sec h ( 3x + 2) dx let u = 3x + 2 , then

Check: Let y =

×

(

du d 4 du du = x +1 ; = 4x3 ; du = 4 x3dx ; dx = 3 . Thus, dx dx dx 4x

1 −1 sin tanh x 4 + 1 + c then y ′ = 4

∫ sec h ( 3x + 2) dx = ∫ sec h u ⋅

f. Given

=

sec h x 4 + 1 dx let u = x 4 + 1 , then

Check: Let y =

e. Given

du

3e3 x

1 3 1 1 ln cosh e3 x + c , then y ′ = ⋅ ⋅ sinh e3 x ⋅ 3e3 x + 0 = e3 x tanh e3 x = e3 x tanh e3 x 3 3 3 cosh e3 x

sec h x 4 + 1 dx =

×

tanh u ⋅

du d 3 x du du = e ; = 3e3 x ; du = 3e3 x dx ; dx = 3 x . Therefore, dx dx dx 3e

(

)

)

du d ( 3x + 2) ; du = 3 ; du = 3dx ; dx = du . Thus, = dx dx 3 dx

1 1 1 du = sec h u du = sin −1 (tanh u ) + c = sin −1 [ tanh (3 x + 2 ) ] + c 3 3 3 3

∫

1 −1 sin [ tanh (3 x + 2 ) ] + c , then y ′ = 3

1 3 1 − tanh 2 (3 x + 2 )

⋅

d tanh (3 x + 2 ) + 0 = dx

sec h 2 (3 x + 2 ) 3 sec h 2 (3 x + 2 )

2 2 d (3x + 2) = sec h (3x + 2) ⋅ 3 = 3 ⋅ sec h (3x + 2) = sec h (3x + 2) dx 3 sec h (3 x + 2 ) 3 sec h (3 x + 2 )

∫e

cosh (3 x + 5 )

sinh (3 x + 5) dx let u = cosh (3 x + 5) , then

Hamilton Education Guides

du d du d = cosh (3 x + 5) ; = sinh (3 x + 5) ⋅ (3 x + 5) dx dx dx dx

475

Calculus I

;

Chapter 5 Solutions

du du . Therefore, = sinh (3 x + 5) ⋅ 3 ; dx = dx 3 sinh (3 x + 5)

∫e

cosh (3 x + 5 )

sinh (3 x + 5) dx =

Check: Let y =

∫e

u

sinh (3 x + 5) ⋅

1 u 1 1 du = e du = eu + c = e cosh (3 x +5 ) + c 3 3 3 3 sinh (3 x + 5)

∫

d 1 cosh (3 x +5 ) 1 3 e + c , then y ′ = ⋅ ecosh (3 x + 5 ) ⋅ sinh (3 x + 5) ⋅ (3 x + 5) + 0 = ⋅ ecosh (3 x + 5 ) ⋅ sinh (3 x + 5) 3 dx 3 3

= ecosh (3 x + 5 ) sinh (3 x + 5) g.

∫ tanh

5

x dx =

∫ tanh

3

3 2 3 2 3 ∫ tanh x (1 − sec h x )dx = − ∫ tanh x sec h x dx + ∫ tanh x dx . To solve the first

x tanh 2 x dx =

du d du du . Therefore, = tanh x ; = sec h 2 x ; du = sec h 2 x dx ; dx = dx dx dx sec h 2 x

integral let u = tanh x , then

du

∫

∫

− tanh 3 x sec h 2 x dx = − u 3 sec h 2 x ⋅ letter d, we found that

∫ tanh

∫ tanh

x tanh 2 x dx =

= −

5

x dx =

∫ tanh

3

3

sec h 2 x

x dx = −

1 1 = − u 3du = − u 4 + c = − tanh 4 x + c . In Example 5.4-6, problem 4 4

∫

1 tanh 2 x + ln cosh x + c . Therefore, 2

3 2 3 3 2 ∫ tanh x (1 − sec h x ) dx = − ∫ tanh x sec h x dx + ∫ tanh x dx

1 1 1 1  1  tanh 4 x + tanh 3 x dx = − tanh 4 x +  − tanh 2 x + ln cosh x + c  = − tanh 4 x − tanh 2 x + ln cosh x + c 4 4 2 4  2 

∫

1 1 4 2 sinh x Check: Let y = − tanh 4 x − tanh 2 x + ln cosh x + c , then y ′ = − tanh 3 x ⋅ sec h 2 x − tanh x ⋅ sec h 2 x + +0 4 2 4 2 cosh x

)

(

)(

(

= − tanh 3 x sec h 2 x − tanh x sec h 2 x + tanh x = − sec h 2 x tanh 3 x + tanh x + tanh x = − 1 − tanh 2 x tanh 3 x + tanh x

(

)

)

+ tanh x = − tanh 3 x + tanh x − tanh 5 x − tanh 3 + tanh x = − tanh 3 x − tanh x + tanh 5 x + tanh 3 + tanh x = tanh 5 x

h.

∫ coth

5

x dx =

∫ coth

3

x coth 2 x dx =

d du du du . Thus, = − csc h 2 x ; du = − csc h 2 x dx ; dx = − = coth x ; dx dx dx csc h 2 x

integral let u = coth x , then

∫ coth

3 x csc h 2 x dx =

letter g, we found that

∫ coth = −

5

x dx =

∫ coth

3

∫u

3 2 3 2 3 ∫ coth x (1 + csc h x ) dx = ∫ coth x csc h x dx + ∫ coth x dx . To solve the first

3

csc h 2 x ⋅ −

∫ coth

x dx = −

3

x coth 2 x dx =

du csc h 2 x

1 1 = − u 3du = − u 4 + c = − coth 4 x + c . In example 5.4-6, problem 4 4

∫

1 coth 2 x + ln sinh x + c . Grouping the terms together we have 2

1 4 3 3 2 3 3 2 ∫ coth x (1 + csc h x )dx = ∫ coth x csc h x dx + ∫ coth x dx = − 4 coth x + ∫ coth x dx

1 1 1  1  coth 4 x +  − cot 2 x + ln sinh x + c  = − coth4 x − coth 2 x + ln sinh x + c 4 2 4 2  

1 1 − 4 ⋅ coth 3 x ⋅ − csc h 2 x 2 ⋅ coth x ⋅ − csc h 2 x cosh x Check: Let y = − coth 4 x − coth 2 x + ln sinh x + c , then y ′ = − + +0 4 2 4 2 sinh x

(

(

)

)(

)

= coth 3 x csc h 2 x + coth x csc h 2 x + coth x = csc h 2 x coth 3 x + coth x + coth x = coth 2 x − 1 coth 3 x + coth x + coth x = coth 5 x + coth 3 x − coth 3 x − coth x + coth x = coth 5 x i.

∫ coth

6

x dx =

∫ coth

4

x coth 2 x dx =

∫

4 2 4 2 4 4 2 ∫ coth x (1 + csc h x ) dx = ∫ ( coth x csc h x + coth x ) dx = ∫ coth x csc h x dx .

+ coth x dx . In example 5.4-6, problem letter e, we found that

Hamilton Education Guides

1

∫ coth x dx = − 3 coth 4

3

x − coth x + x + c . Therefore,

476

Calculus I

∫ coth

Chapter 5 Solutions

6

x dx =

∫ coth

4

x coth 2 x dx =

4 2 4 2 4 4 2 ∫ coth x (1 + csc h x ) dx = ∫ coth x csc h x dx + ∫ coth x dx = ∫ coth x csc h x dx

1 1  1  +  − coth 3 x − coth x + x + c  = − coth5 x − coth 3 x − coth x + x + c 5 3 3  

5 ⋅ coth 4 x ⋅ − csc h 2 x 3 ⋅ coth 2 x ⋅ − csc h 2 x 1 1 Check: Let y = − coth 5 x − coth 3 x − coth x + x + c , then y ′ = − − 5 3 5 3

(

)

+ csc h 2 x + 1 + 0 = coth 4 x ⋅ csc h 2 x + coth 2 x ⋅ csc h 2 x + csc h 2 x + 1 = csc h 2 x coth 4 x + coth 2 x + 1 + 1

(

)(

)

= coth 2 x − 1 coth 4 x + coth 2 x + 1 + 1 = coth 6 x + coth 4 x + coth 2 x − coth 4 x − coth 2 x − 1 + 1 = coth 6 x

Hamilton Education Guides

477

Index A Adjacent, 308 Antiderivative, 213 Arithmetic sequence, 13-16 Arithmetic series, 16-19

B Binomial coefficient, 45 Binomial expansion, 47-52

C Chain rule, 82-96 Common difference, 13 Common multiplier, 20 Common ratio, 20 Constant function, 59 Constant of integration, 213 Converge, 31 Cosecant, 233 Cosecant hyperbolic, 351 Cosine, 233 Cosine hyperbolic, 351 Cotangent, 233 Cotangent hyperbolic, 351

D Derivative higher order, 124-138 of fractions with fractional exponents, 102-108 of radical functions, 109-123 Difference quotient, 54-58 Differentiation of hyperbolic functions, 181-186 of indeterminate forms using L’Hopital’s rule, 193-211 of inverse hyperbolic functions, 187-192 of inverse trigonometric functions, 158-165 of logarithmic and exponential functions, 166-180 of trigonometric functions, 140-157 Differentiation rules using dxd notation, 71-81 using prime notation, 59-70

Hamilton Education Guides

Diverge, 31

E Even functions, 233 Even power, 244, 365

F Factorial notation, 42-52 Finite sequence function, 2 Formulas, addition, 232,350 double angle, 232,350 half angle, 232,350 unit, 232,350

G General term of a sequence, 2 Geometric sequence, 20-25 Geometric series, 25-29

H Hyperbolic functions addition formulas for, 350 double angle formulas for, 350 half angle formulas for, 350 unit formulas for, 350 Hypotenuse, 308

I Identity function, 59 Implicit differentiation, 97-101 Indefinite integral, 213 Indeterminate forms, 140, 193, 207, 209 Index of summation, 7 Infinite sequence function, 2 Integral sign, 213 Integrand, 213 Integration by partial fractions, 320-349 by parts, 287-307 of hyperbolic functions, 350-378 of trigonometric functions, 232-258 resulting in exponential or logarithmic functions, 273-285

478

Calculus I

Index

resulting in inverse trigonometric functions, 259-272 using the basic integration formulas, 213221 using the substitution method, 222-231 using trigonometric substitution, 308-319 Inverse trigonometric functions, 259

L L’Hopital’s rule, 193-211 Limits of sequences, 31-36 of series, 36-38 Linear factors distinct, 320 repeated, 327 Long division, 320

M Method difference quotient, 54-58 implicit differentiation, 97-101 the chain rule, 82-96 the L’Hopital’s rule, 193-211

O Odd functions, 233 Odd power, 244, 365 Opposite, 308

P Partial Fractions method, 320-349 Denominator with distinct linear factors, 320-326 Denominator with distinct quadratic factors, 334-343 Denominator with repeated linear factors, 327-333 Denominator with repeated quadratic factors, 344-349 Polynomial, 320 Product rule, 60, 71 Pythagorean theorem, 308

Q Quadratic factor distinct, 334 repeated, 344

Hamilton Education Guides

R Radical expressions, 308 Range of summation, 7 Rational fraction improper, 320 proper, 320 Reference to chapter 1 problems, 1 to chapter 2 problems, 53 to chapter 3 problems, 139 to chapter 4 problems, 212 to chapter 5 problems, 286 Repeating decimals, 38-41 Right angle, 308 Rule product, 60, 71 scalar, 60, 71 summation, 60, 71

S Scalar rule, 60, 71 Secant, 233 Secant hyperbolic, 351 Sequences, 2-6 Series, 7-12 Sigma notation, 7 Sine, 233 Sine hyperbolic, 351 Solutions to chapter 1, 379-400 to chapter 2, 401-425 to chapter 3, 426-435 to chapter 4, 436-453 to chapter 5, 454-477 Summation index of, 7 range of, 7 rule, 60,71 Symbol approximate ≈ , 48 differentiation

d , 71 dx

exponent e , 273 factorial ! , 42 integration

∫

, 213

logarithm ln = log e , 273

479

Calculus I

Index

not equal to ≠ , 25 ,7 summation

∑

Synthetic division, 320

T Table of basic integration formulas, 378 of differentiation formulas for hyperbolic functions, 181, 351 of differentiation formulas for inverse hyperbolic functions, 187 of differentiation formulas for inverse trigonometric functions, 158 of differentiation formulas for trigonometric functions, 140, 233 of hyperbolic formulas, 181 of integration formulas for hyperbolic functions, 350 of integration formulas for trigonometric functions, 232 of inverse hyperbolic functions, 187 Tangent, 233 Tangent hyperbolic, 351 Terms of a sequence, 2 Trigonometric functions addition formulas for, 232 double angle formulas for, 232 half angle formulas for, 232 unit circle formulas for, 232

Hamilton Education Guides

480

About the Author Dan Hamilton received his B.S. degree in Electrical Engineering from Oklahoma State University and Master's degree, also in Electrical Engineering from the University of Texas at Austin. He has taught a number of math and engineering courses as a visiting lecturer at the University of Oklahoma, Department of Mathematics, and as a faculty member at Rose State College, Department of Engineering Technology, at Midwest City, Oklahoma. He is currently working in the field of aerospace technology and has published several books and numerous technical papers.

Hamilton Education Guides

481