100 Integrals: Solutions and Engineering Applications 1683929675, 9781683929673

Table of contents :
Cover
Half-Title
Title
Copyright
Dedication
Contents
Preface
About the Author
Part 1: List of Selected Integrals with Their Step-By-Step Solutions
Table 1 List of Selected Integrals for Part 1
Integral 1
Integral 2
Integral 3
Integral 4
Integral 5
Integral 6
Integral 7
Integral 8
Integral 9
Integral 10
Integral 11
Integral 12
Integral 13
Integral 14
Integral 15
Integral 16
Integral 17
Integral 18
Integral 19
Integral 20
Integral 21
Integral 22
Integral 23
Integral 24
Integral 25
Integral 26
Integral 27
Integral 28
Integral 29
Integral 30
Integral 31
Integral 32
Integral 33
Integral 34
Integral 35
Integral 36
Integral 37
Integral 38
Integral 39
Integral 40
Integral 41
Integral 42
Integral 43
Integral 44
Integral 45
Integral 46
Integral 47
Integral 48
Integral 49
Integral 50
Integral 51
Integral 52
Integral 53
Integral 54
Integral 55
Integral 56
Integral 57
Integral 58
Integral 59
Integral 60
Integral 61
Integral 62
Integral 63
Integral 64
Integral 65
Integral 66
Integral 67
Integral 68
Integral 69
Integral 70
Integral 71
Integral 72
Integral 73
Integral 74
Integral 75
Integral 76
Integral 77
Integral 78
Integral 79
Integral 80
Integral 81
Integral 82
Integral 83
Integral 84
Integral 85
Integral 86
Integral 87
Integral 88
Integral 89
Integral 90
Integral 91
Integral 92
Integral 93
Integral 94
Integral 95
Integral 96
Integral 97
Integral 98
Integral 99
Integral 100
Part 2: Examples Applied in Engineering
1 Semi-Circle Shapes
2 Circular Segment Shapes
3 Semi-Ellipse Shapes
4 Two-Degree Polynomial Shape-Quadratic
5 Three-Degree Polynomial Shape-Cubic
6 n-Degree Polynomial Shape-Spandrel
7 Sinusoidal Shapes
8 Triangular Shapes
9 Rectangular Shapes
10 Complex Shapes
11 A Cantilever Beam with Cubic Load Distribution
12 A Cantilever Beam with Quarter-Ellipse Load Distribution
13 A Cantilever Beam with Inverse Cosine Load Distribution
14 A Cantilever Beam with Parabolic Load Distribution
15 A Cantilever Beam with Circular Segment Cross-Section and Quarter-Ellipse Load Distribution
16 Probability Density Functions-PDF
References

Citation preview

100 Integrals

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license, disclaimer of liability, and limited warranty By purchasing or using this book (the “Work”), you agree that this license grants permission to use the contents contained herein, but does not give you the right of ownership to any of the textual content in the book or ownership to any of the information or products contained in it. This license does not permit uploading of the Work onto the Internet or on a network (of any kind) without the written consent of the Publisher. Duplication or dissemination of any text, code, simulations, images, etc. contained herein is limited to and subject to licensing terms for the respective products, and permission must be obtained from the Publisher or the owner of the content, etc., in order to reproduce or network any portion of the textual material (in any media) that is contained in the Work. Mercury Learning and Information (“MLI” or “the Publisher”) and anyone involved in the creation, writing, or production of the companion disc, accompanying algorithms, code, or computer programs (“the software”), and any accompanying Web site or software of the Work, cannot and do not warrant the performance or results that might be obtained by using the contents of the Work. The author, developers, and the Publisher have used their best efforts to insure the accuracy and functionality of the textual material and/or programs contained in this package; we, however, make no warranty of any kind, express or implied, regarding the performance of these contents or programs. The Work is sold “as is” without warranty (except for defective materials used in manufacturing the book or due to faulty workmanship). The author, developers, and the publisher of any accompanying content, and anyone involved in the composition, production, and manufacturing of this work will not be liable for damages of any kind arising out of the use of (or the inability to use) the algorithms, source code, computer programs, or textual material contained in this publication. This includes, but is not limited to, loss of revenue or profit, or other incidental, physical, or consequential damages arising out of the use of this Work. The sole remedy in the event of a claim of any kind is expressly limited to replacement of the book, and only at the discretion of the Publisher. The use of “implied warranty” and certain “exclusions” vary from state to state, and might not apply to the purchaser of this product.

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100 Integrals Solutions and Engineering Applications

Mehrzad Tabatabaian, PhD, PEng

Mercury Learning and Information Boston, Massachusetts

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Copyright ©2023 by Mercury Learning and Information LLC. All rights reserved. An imprint of De Gruyter Inc. This publication, portions of it, or any accompanying software may not be reproduced in any way, stored in a retrieval system of any type, or transmitted by any means, media, electronic display or mechanical display, including, but not limited to, photocopy, recording, Internet postings, or scanning, without prior permission in writing from the publisher. Publisher: David Pallai Mercury Learning and Information 121 High Street, 3rd Floor Boston, MA 02110 [email protected] www.merclearning.com (800) 232-0223 M. Tabatabaian. 100 Integrals: Solutions and Engineering Applications. ISBN: 978-1-68392-967-3 The publisher recognizes and respects all marks used by companies, manufacturers, and developers as a means to distinguish their products. All brand names and product names mentioned in this book are trademarks or service marks of their respective companies. Any omission or misuse (of any kind) of service marks or trademarks, etc. is not an attempt to infringe on the property of others. Library of Congress Control Number: 2023942788 232425321  This book is printed on acid-free paper in the United States of America. Our titles are available for adoption, license, or bulk purchase by institutions, corporations, etc. For additional information, please contact the Customer Service Dept. at 800-232-0223 (toll free). All of our titles are available in digital format at www.academiccourseware.com and other digital vendors. The sole obligation of Mercury Learning and Information to the purchaser is to replace the book, based on defective materials or faulty workmanship, but not based on the operation or functionality of the product.

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To the relentless explorers of integral calculus, may this textbook become your trusted companion, guiding you through the labyrinth of calculations, modelling, and simulations in engineering and technical fields.

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Contents Prefacexiii About the Author

xv

PART 1: LIST OF SELECTED INTEGRALS WITH THEIR STEP-BY-STEP SOLUTIONS Table 1  List of Selected Integrals for Part 1 Integral 1 Integral 2 Integral 3 Integral 4 Integral 5 Integral 6 Integral 7 Integral 8 Integral 9 Integral 10 Integral 11 Integral 12 Integral 13 Integral 14 Integral 15 Integral 16

1 2 5 6 7 8 10 12 14 15 16 17 18 19 20 21 22 25

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viii • Contents



Integral 17 Integral 18 Integral 19 Integral 20 Integral 21 Integral 22 Integral 23 Integral 24 Integral 25 Integral 26 Integral 27 Integral 28 Integral 29 Integral 30 Integral 31 Integral 32 Integral 33 Integral 34 Integral 35 Integral 36 Integral 37 Integral 38 Integral 39 Integral 40 Integral 41 Integral 42 Integral 43 Integral 44 Integral 45 Integral 46 Integral 47 Integral 48

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26 27 28 30 31 33 34 36 37 39 40 42 43 44 46 47 48 49 51 53 54 55 57 59 60 61 63 64 65 66 67 68

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Contents • ix



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Integral 49 Integral 50 Integral 51 Integral 52 Integral 53 Integral 54 Integral 55 Integral 56 Integral 57 Integral 58 Integral 59 Integral 60 Integral 61 Integral 62 Integral 63 Integral 64 Integral 65 Integral 66 Integral 67 Integral 68 Integral 69 Integral 70 Integral 71 Integral 72 Integral 73 Integral 74 Integral 75 Integral 76 Integral 77 Integral 78 Integral 79 Integral 80

69 70 72 73 74 75 77 78 80 82 83 84 86 87 89 91 93 95 97 99 100 102 104 105 106 108 110 111 113 114 115 118

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x • Contents



Integral 81 Integral 82 Integral 83 Integral 84 Integral 85 Integral 86 Integral 87 Integral 88 Integral 89 Integral 90 Integral 91 Integral 92 Integral 93 Integral 94 Integral 95 Integral 96 Integral 97 Integral 98 Integral 99 Integral 100

PART 2: EXAMPLES APPLIED IN ENGINEERING 1 Semi-Circle Shapes 2 Circular Segment Shapes 3 Semi-Ellipse Shapes 4 Two-Degree Polynomial Shape-Quadratic 5 Three-Degree Polynomial Shape-Cubic 6 n-Degree Polynomial Shape-Spandrel 7 Sinusoidal Shapes  8 Triangular Shapes 9 Rectangular Shapes 10 Complex Shapes 11 A Cantilever Beam with Cubic Load Distribution

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119 120 121 122 123 124 126 128 131 132 134 135 136 138 139 140 141 142 143 144 147 148 150 156 159 162 165 168 172 174 176 182

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Contents • xi

12 A Cantilever Beam with Quarter-Ellipse Load Distribution 13 A Cantilever Beam with Inverse Cosine Load Distribution 14 A Cantilever Beam with Parabolic Load Distribution 15 A Cantilever Beam with Circular Segment Cross-Section and Quarter-Ellipse Load Distribution 16 Probability Density Functions-PDF References

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184 188 191

193 196 199

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Preface This monograph contains a collection of integrals, some more challenging than others, with their worked-out solutions as indefinite integrals. The integrals were randomly selected, modified, or designed with the condition of having closed forms solutions with common functions. This list is meant for helping readers in practicing and getting hints for working out solutions to similar integrals that they might encounter. Readers might want to add their own favorite integrals to this list. By no means is this a comprehensive list of integrals, as many authors have created such lists ( [1], [2], [3], [4], [5], [6], [7], [8], [9], [10]). The exercise of integration operation is a mind stimulating activity, as it requires the knowledge of certain mathematical techniques and the discovery of tricks and short cuts while solving them. The latter feature makes integration different and more enjoyable compared to other topics in calculus (e.g., differentiation and algebraic manipulation). In addition, we present the application of some integrals in engineering related topics. For example, nonuniform loading, hydrostatic force, moment of inertia, polar moment of inertia, etc. We introduce an up-to-date online software tool, WolframAlpha1 that can be used for comparing our answers for the integrals listed. However, readers may want to update to WolframAlpha Pro for recovering some of the integrals’ step-by-step solutions. Interested readers might like to try this tool for their selected integrals from the list. However, please note that sometimes equivalent results

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xiv • Preface

are provided by this tool for the same integral as input when compared to the results presented in this volume. In addition, similar CAS (computer algebra system) tools like Maple2, Mathematica3, or Mathcad4 may also be employed. Finally, readers should be aware that some integrals may have alternative equivalent solutions rather than a corresponding unique one. Thus worked-out solutions may be different, but equivalent, to those provided by some online CAS tools. Mehrzad Tabatabaian, PhD, PEng Vancouver, B.C. July 22, 2023

1

https://www.wolframalpha.com/calculators/integral-calculator/

2

https://www.maplesoft.com

3

https://www.wolfram.com/mathematica/

4

https://www.mathcad.com

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About the Author

Dr. Mehrzad Tabatabaian is a faculty member in the Mechanical Engineering Department, School of Energy at the British Columbia Institute of Technology. He has several years of teaching and industry experience. Dr. Tabatabaian is currently Chair of the BCIT School of Energy Research Committee. He has published several papers in scientific journals and conferences, and he has written textbooks on multiphysics and turbulent flow modelling, advanced thermodynamics, tensor analysis, direct energy conversion, and Bond Graph modelling method. He holds several registered patents in the energy field resulting from years of research activities. Dr. Tabatabaian volunteered to help establish the Energy Efficiency and Renewable Energy Division (EERED), a new division at Engineers and Geoscientists British Columbia (EGBC). Mehrzad Tabatabaian received his BEng from Sharif University of Technology (former AUT) and advanced degrees from McGill University (MEng and PhD). He has been an active academic, professor, and engineer in leading alternative energy, oil, and gas industries. Mehrzad has also a Leadership Certificate from the University of Alberta and holds an EGBC P.Eng. License.

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PART

1

List of Selected Integrals with Their Step-by-Step Solutions

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2 • 100 Integrals

TABLE 1  List of selected integrals for Part 1 Integral

Integral

1

∫ xln3 xdx

19

 1   ln  1  1  2  dx x  

2

 e x 1  e2 x dx

20

 csch 1 xdx

3

∫

cos3 x dx sin x

21



4

 ln

22

2

5

∫

23

 ln 1  1  x2 dx

6



24

∫ sin 6 x cos5 xdx

7



25

∫ x m ln n xdx

8



26

 e x sin 1  e x  dx

9



x dx 8  4 x2  x4

27

x

10



4 x2  x  1 dx 4 x3  x

28

 x x tan 1 xdx

11



x4  1 dx x2  2

29



tan 1 x dx  x  1 3

12



1 dx 5  4 cos x

30



sin 1 x dx x2

13

∫ sec 4 xdx

31

 x sec 1 xdx

14



32

 sec 1 xdx

15

∫ tan xdx

33

 x2 tan 1 xdx

16

∫ xe x sin xdx

34



17

∫

ln 3 x dx x3

35



18

 ln 1  x dx

36



100_Integrals.001_3pp.indd 2





x  1  x dx

1 dx cos x x

 sin x  cos x 2 x

dx

 7 x  117 x3

1  x 

2 2

dx

1

1  x 

2 5





dx

dx

1 dx sin x  cos x x

dx





1  x2 dx 1  x2

2x  3 3  6 x  9 x2

x

3x  2

2

dx

 4  x2  4

dx

2 x2  5 x  1 dx x  2 x2  x  2 3

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List of Selected Integrals with Their Step-by-Step Solutions • 3

Integral

Integral

37



38



39



40



41

2 x3  3 x2  4 dx  x  1 4

54

x 1



1 1  x  x 

2

dx

55

 e2 x tan 1  e x  dx

56



57

∫ tan 3 xdx

 1  ln x  ln  ln x  dx

58



1 dx 4  5 cos x

42

 ln  x2  x  1  dx

59



tan 3 1  ln x  dx x

43

 cos x 4  sin 2 xdx

60

 sin 2 x sin 3 x    dx  sin x sin 6 x 

44



61

∫

sin 3 x dx cos x

45



62



5 x  31 dx 3 x2  4 x  11

46



dx

63



47

∫ sin 6 x cos5 xdx, alternative solution

64



48

∫ x2 e x dx

65

 sin 1 x ln xdx

49

∫

tan 3 x dx cos3 x

66

 ln  sin x  1  sin xdx

50



tan x dx  sin x  cos x 2

67



51

1   1   2  dx  ln x ln x 

68



52

 sin  x  sin x   sin  x  sin x   dx

69



53

100_Integrals.001_3pp.indd 3

 x  1 x3  x2  x

dx

x4  4 x3  6 x2  4 x  1 dx x3  3 x2  3 x  1 x



1 4

x 1



10

dx

cos x dx sin 2 x  3 sin x  2



1 1 x



1  2 x2

x 5 1  x

x  x2



2 3

dx

3/2



3

x  2   1  dx  1  5 x2  1  5 x2 

70

1

dx

x 6 x  x2

2

3 x 5  x 4  2 x3  12 x2  2 x  1

x

3

 1

2

dx

4 x3  x  1 dx x3  1

1

x

1 x

2

x3 esin

dx

tan x 1  sec 3 x

dx



x ln x  x2  1

 sin 1

x 1 2



 dx



1  x dx

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4 • 100 Integrals

Integral

Integral

71



1 dx 2  2 sin x  cos x

86



72



sec 2 x dx tan 2 x  2 tan x  2

87



73

 x2  x  1 dx

88



74



89

 sin 2  ln x  dx

75



90

∫ sin x sin 2 x sin 3 x dx

76

 sin x tan 1 sec x  1dx

91

 1  x2 dx

77



92



78

 x3 ln  x ln x6   5 dx

93



79

 tan 1

94

 x4   dx 6  1 x 

80



95



81

 sin 1 x dx

96



82

 tan 1 x dx

97

  sin 1 x  dx

83

 sinh 1 x dx

98

 e x 1  ln x  x2 x dx

84

 tanh 1 x dx

99

 xln x dx

85

1  cos1   dx  x

100



x

 x2  2 x  2  1 x3 x2  9

1 x xx

2

dx

dx

dx



100_Integrals.001_3pp.indd 4

2







x  1  x dx

x9 dx x  48 x10  575 20

1 dx x4  4

4x

x5 2

 4

5/2

dx

1 dx x4  1

1 ln x cos x  sin x x dx  ln x 2 2 e2 x  e x 3 e2 x  6 e x  1

dx

2

x3 e x

2

1  x 

2 2

1 1

dx

dx

x 2

x

ln  sin x  dx 1  sin x

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List of Selected Integrals with Their Step-by-Step Solutions • 5

INTEGRAL 1 Problem

∫ xln xdx 3

Solution: x2  4 ln3 x  6 ln2 x  6 lnx  3   constant 8 Techniques used: Change of variables, Integration by parts Step-by-step solution: Let lnx = z, then we have = dx xdz = , x e z. After rewriting the inte­ gral in terms of variable z, we get  xln3 xdx   e2 z z3 dz. Integrating 1 2z 3 1 2z 2z 3 2 by parts gives,  e z dz  e z   e  3 z  dz. Performing the   2 2 dg f

­integration by parts technique, two times, on the second term gives 1 3 3 1 3 3 3  e2 z z3 dz  e2 z z3  e2 z z2   e2 z zdz  e2 z z3  e2 z z2  e2 z z  e2 z 2 4 2 2 4 4 8 3 3 3 e2 z z3  e2 z z2  e2 z z  e2 z. Factoring out e2 z = x2 and substituting for z = lnx, we can 4 4 8 rewrite the result in terms of the original variable x as 1 2z x2 e  4 z3  6 z2  6 z  3    4 ln3 x  6 ln2 x  6 lnx  3 . 8 8

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6 • 100 Integrals

INTEGRAL 2 Problem

e

x

1  e2 x dx

Solution: 1 sinh 1  e x   e x 1  e2 x   constant  2 Techniques used: Change of variables, Trigonometric identities Step-by-step solution: dz . Rewriting the integral in terms of v­ ariable z dz z gives  e x 1  e2 x dx   z 1  z2   1  z2 dz. Now, let z z  sinh u  dz  cosh u du. Rewriting the new integral in terms of variable u, gives  1  z2 dz   1  sinh2 u cosh u du. But,

Let e x  z  dx 

using the cosh2 u  sinh2 u  1 identity we get  1  sinh2 u 2

cosh u du   cosh2 u cosh u du   cosh2 udu. Now, using the cosh u 1  cosh 2 u = (1 + cosh 2 u) / 2 identity, we get  cosh2 udu   du. 2 Expanding the integrand and integrate each term 1  cosh 2 u 1 1 u sinh 2 u u 2 sinh u cosh u du   du   cosh 2 u du     gives  2 2 2 2 4 2 4 1 u sinh 2 u u 2 sinh u cosh u  cosh 2 u du     . Rewriting this expression in terms of variable z 2 2 4 2 4 u 2 sinh u cosh u sinh 1 z z 1  z2 1 and subsequently x, we get     sinh 1  e x   e x 2 2 2 2 2 sinh 1 z z 1  z2 1 2x . 1 x x   sinh  e   e 1  e 2 2 2



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List of Selected Integrals with Their Step-by-Step Solutions • 7

INTEGRAL 3 Problem

∫

cos3 x dx sin x

Solution: 2 sin x  4  cos2 x   constant 5 Techniques used: Change of variables, Trigonometric identities Step-by-step solution:

Let sin x  z  cos xdx  dz. Rewriting the integral in terms of vari­ cos2 x 1  z2 able z gives  cos xdx   dz. Now, let z  u  dz  2 udu. sin x z Rewriting the new integral in terms of variable u gives 1  z2 1  u4 2  dz  2  udu  2 u  u5. Substituting back and ­rewriting u 5 z the answer in terms of variable x , after some manipulations, gives 5 2 2 1 1    2 u  u5  2 sin x  sin x  2 sin x  1  sin2 x   2 sin x  1  1  cos2 x  5 5 5 5    1 2  1 2    x  1  sin x   2 sin x  1  1  cos2 x    sin x  4  cos2 x  . 5 5     5



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8 • 100 Integrals

INTEGRAL 4 Problem

 ln 



x  1  x dx

Solution: 1 sinh 1 x  x 1  x   2 x ln  2





x  1  x   constant 

Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: We use integration by parts technique with considering  dx  x.

 1  x  dx. Performing the integration gives x ln 

Therefore, we can write  ln





x  1  x  dx   x

d ln dx



x

1

x ln





1 x  1  x   x 2 x 2 1  x  x ln x  1 x







x  1  x dx  ln





x  1  x  dx   x

d ln dx



x  11  x  x  1 x x 2 x x





1 x x  1 x   dx. 2 1 x

x  z  dx  2 zdz. Rewriting the new integral in z2 1 x terms of the variable z, gives   dx    dz. But 2 1 x 1  z2 d z 1  z2  using , and the integration by parts technique dx 1  z2 z2 z   zdz   z 1  z2   1  z2 dz. we can write   2 2 1 z 1 z Now, we let z  sinh u  dz  cosh udu. Hence, the latter integral

Now, we let



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List of Selected Integrals with Their Step-by-Step Solutions • 9

can be written as  1  z2 dz   1  sinh 2 u cosh udu   cosh 2 u du, 1 after using the cosh 2 u  sinh 2 u  1 identity. But cosh 2 u  1  cosh 2 u  2 1 cosh 2 u  1  cosh 2 u  and after rewriting the last integral we have 2 1 1 1 1 1  cosh 2 u du   1  cosh 2 u  du  u  sinh 2 u  u  sinh u cosh u 2 2 4 2 2 1 1 sinh 2 u  u  sinh u cosh u. After collecting all related answers and rewrite them in 2 2 terms of the original variable x, we get the solution as shown.

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10 • 100 Integrals

INTEGRAL 5 Problem

1

∫ cos x dx Solution:   x   1  tan  2   ln    constant  1  tan  x       2  Techniques used: Integration by parts, Partial fractions, Trigonometric identities Step-by-step solution: We rewrite the x cos x  cos2  sin 2 2

integral using the trigonometric identity 1 1 x . Therefore,  dx   dx   x x x x 2 cos x  cos2  sin 2  cos  sin 2 2 2 2  1 1 dx. But the integrand can dx   dx   x x x x  x x  cos2  sin 2  cos sin cos sin     2 2 2 2  2 2  be ­written as, using the partial fractions technique, x  a  b sin   a  b co 1 a b 2    x x x x  x x  x x x x  x   cos  sin   cos  sin  cos 2  sin 2 cos 2  sin 2  cos  sin   cos  si 2 2  2 2 2 2  2   x x  a  b sin   a  b cos b 2 2 . Therefore, to have equality valid,   x  x x x x  x x n cos  sin cos  sin   cos  sin  2 2 2  2 2  2 2 x x the constants a and b should satisfy  a  b  sin   a  b  cos  1. 2 2 x x x x  a  b sin   a  b cos  1. Or, a  b  sin and  a  b   cos . 2 2 2 2

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List of Selected Integrals with Their Step-by-Step Solutions • 11

x x 1 Solving for a and b gives, a   cos  sin  and b = 2 2 2 1 x x  cos  sin . Therefore, the original integral can be 2 2 2 x x x x cos  sin cos  sin 1 1 1 2 2 dx   2 2 dx   written as  x x  x x 2 cos x  sin x 2 cos x  sin x   cos  sin   cos  sin  2 2 2 2 2 2  2 2  x x x x cos  sin cos  sin 1 1 2 2 dx . But d  cos x  sin x     cos x  sin x  and 2 2 dx   x      x x x 2 cos  sin 2 2 2 2 2 cos  sin x dx   2 2 2 2

d x x  x x  cos  sin    cos  sin  . Using these relations and recall­ 2 2  2 2 dx  d dz ing that ln z = , we can arrive at the final answer for the integral dz z x x x x cos  sin cos  sin 1 1 x x x x   2 2 2 2 as  dx   dx   ln  cos  sin   ln  cos  sin x x x x 2 cos  sin 2 cos  sin 2 2 2 2   2 2 2 2 x x  cos  sin   x x x x     2 2 . To simplify, after dividing the x   ln  cos  sin   ln  cos  sin   ln  x x 2 2 2 2    cos  sin  2 2  numerator and denominator of the argument of the logarithm by x   1  tan 2  1 x . dx  ln  cos , we get  x cos x 2  1  tan  2  Short-cut solution: An alternative solution can be obtained as follow: dz . Rewriting the integral gives cos x 1 1 1 1  dx   dz   dz  tanh 1 z, where dz. But  2 2 cos x cos x 1 z 1  z2 in terms of the original variable x the answer reads tanh 1  sin x . Let sin x  z  dx 

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12 • 100 Integrals

INTEGRAL 6 Problem

x

  sin x  cos x 

2

dx

Solution: 2 x cos x  1  x  ln  sin x  cos x     constant sin x  cos x  2 Techniques used: Integration by parts, Partial fractions, Trigonometric identities Step-by-step solution: We apply the integration by parts technique to the integral, Or x dx dx dx  x     x . This 2 2  sin x  cos x   sin x  cos x   sin x  cos x 2 dx requires calculating the  . But, by inspection the  sin x  cos x 2 f  x   derivative of an expression, like   would contain the  sin x  cos x  correct expression as exists in the denominator of the integrands 2 (i.e.,  sin x  cos x  ). Assuming f  x  as a polynomial function, we f  x   f   sin x  cos x   f   sin x  cos x  d then have ,   dx  sin x  cos x   sin x  cos x 2 with prime symbol indicating differentiation operation. To have the numerator equal to 1, the function f  x  should be a sinusoi­ dal one. Here, we let f  x   cos x . Note that selecting f  x   sin x

f  x  d  cos x d   sin x  sin x  coss x      dx  sin x  cos x  dx  sin x  cos x   sin x s x  cos x  sin x  cos x x x  sin sin  cos     d 1 cos x   .   2 dx  sin x  cos x   sin x  cos x   sin x  cos x 2 works as well. Therefore,

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2

List of Selected Integrals with Their Step-by-Step Solutions • 13

Therefore, The

we

can

write



1

 sin x  cos x 

2

dx  

cos x . sin x  cos x

original

integral can be written as  x cos x cos dx dx dx  x     x   2 2 2  sin x  cos x   sin x  cos x   sin x  cos x  sin x  cos x sin x  cos x  x cos x cos x can be worked   dx . But  sin x  cos x sin x  cos x sin x  cos x x

  x

dx

 sin x  cos x 2

out

by first dividing the integrand by cos x / cos x 1  dx   Now, dx. sin x / cos x  cos x / cos x tan x  1 manipulate the integrand in this integral as 

cos x,

or

we can 0   1 1  tan x  tan dx   tan x  1 tan x  1 0 0    1 1  tan x  tan x tan x tan x 1  1 1 1  dx   dx   dx   dx   dx  x   dx  x   x  tan x  1 tan tan x  1 tan x  1 tan x  1 1 t    0  1 tan x 1  1 1 1  tan x dx dx   x x dx . Now we take the   dx  x   tan x  1 x  1 tan x  1 tan x  1 tan          1 1  tan x from RHS to the LHS, to get 2  dx  x   dx . tan x 1 tan x  1    1  tan x cos x  sin x But the last integrand can be w ­ritten as  dx    tan x  1 sin x  cos x 1  tan x cos x  sin x   ln  sin x  cos x . Finally, the original integral can be dx   tan x  1 sin x  cos x x  x cos x x 1 dx  written as    ln  sin x  cos x . 2 sin x  cos x 2 2  sin x  cos x  Factoring ½ and rearranging the terms gives the expression as shown above for the answer.

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14 • 100 Integrals

INTEGRAL 7 Problem

x

  7 x  1

17

dx

Solution: 

1  112 x 11760  7 x  1 

16

 constant

Techniques used: Change of variables, Partial fractions Step-by-step solution:

z 1 . Writing the ­integral in terms 7  z  1 / 7 dz 1 z  1 x 1 of variable z, gives  dx     17 dz   z16 dz   z17 dz 17 17 z 7 49 z 49  7 x  1 Let 7 x  1  z  7 dx  dz, x 



dz 1 z  1 1   17 dz   z16 dz   z17 dz . The integrals in the last expression, can 7 49 z 49





1 15 1 z and  z17 dz   z16. After 15 16 substituting and rewriting the integral in terms of original x, we 1 1  1 1 1   1 7x  1   z16 dz   z17 dz    z15  z16   get   16  49 49  15 16 15   49  7 x  1   16 1 1 11   1 7x  1   1 7x  1  z16      . Further simplification gives 16  16  15  15   49  7 x  1   16 11760  7 49  7 x  1   16 1 1  112 x  1 7x  1   , as shown in the answer.  16  16 15  49  7 x  1   16 11760  7 x  1  be worked to have  z16 dz  



100_Integrals.001_3pp.indd 14



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List of Selected Integrals with Their Step-by-Step Solutions • 15

INTEGRAL 8 Problem

1  x 

2 2

dx

Solution: 1 1  ln 1  x2    constant  2 1  x2  Techniques used: Change of variables, Partial fractions Step-by-step solution:

By inspection and considering the denominator of the integrand, we 2 d have 1  x2   4 x3  4 x. Therefore, we rewrite the integral as  dx 0     2 x 1 4 x3 4 x  4 x 1 4 x3  4 x x3 1      dx   dx  ln 1  x2   dx dx 2 2 2 2 4 1  x 2  4 1  x 2  4 1  x 2  1  x 2 

 4x

x



x3



2 2

dx  

x

1  x 

2 2

dx 

2 1 x ln 1  x2    dx . To calculate the new integral, we let 2 4 1  x 2 

dz . Therefore, rewriting this integral in terms of the 2x x 1 x dz 1 dz 1 variable z we get      2  . Back dx    2 2 2 2 z x 2 z 2z 1  x 

1  x2  z  dx 

substituting, gives the final answer in terms of the variable x as shown 2 x3 1 x 1 1   . dx  ln 1  x2    dx   ln 1  x2   2 2 4 2 1  x2  1  x 2  1  x 2 

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16 • 100 Integrals

INTEGRAL 9 Problem x

 8  4x Solution:

2

 x4

dx

 x2  2  1 tan1    constant 4  2 

Techniques used: Change of variables, Trigonometric identities Step-by-step solution: We rewrite the 2 denominator of the integrand as 8  4 x2  x 4   x2  2   4 . Therefore, the integral reads as x x dz  dx. Now, let x2  2  z  dx  dx   2 2 4 2 8  4x  x 2x  x  2  4 and rewrite the integral in terms of the variable z, or x 1 dz 1 dz z  . Now, we let  u  dz  2 du dx     2 2 2 2  x2  2   4 2 4  z 8 1   z / 2  z  u  dz  2 du and write the new integral in terms of the variable u as 2 1 dz 1 du    . The last integral is readily equal to 2 4 1  u2 8 1   z / 2 tan −1 u, after back substitutions and rewriting the integral in terms 2 x 1 1  x  2  of the original variable x, we get  dx  tan   8  4 x2  x 4 4  2  as shown in the answer.

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List of Selected Integrals with Their Step-by-Step Solutions • 17

INTEGRAL 10 Problem 4 x2  x  1  4 x3  x dx Solution: 1 ln x  tan 1  2 x   constant 2 Techniques used: Change of variables, Partial fractions Step-by-step solution:

2 2 x We rewrite the integral as 4 x  x  1 dx  4 x  1 dx   4 x3  x  4 x3  x  4 x3  x dx 4 x2  1 x x 4 x2  1 dx dx  3 dx   3 dx. But  dx    ln x and  dx   2 . 2 2 4x  x 4x  x x(4 x  1) x(4 x  1) x 4x  1 dz Now, let 2 x  z  dx  and rewrite the new integral in terms 2 1 1 1 dx   dz  tan 1 z. After colof the variable z to get  2 2 4x  1 2 1  z 2 lecting both integrals results and write them in terms of the original 4 x2  x  1 1 ­variable x we get  dx  ln x  tan 1  2 x . 3 4x  x 2

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18 • 100 Integrals

INTEGRAL 11 Problem



x4  1 dx x2  2

Solution: x3 5  x   2x  tan 1    constant 3 2  2 Techniques used: Change of variables, Partial fractions Step-by-step solution:

We perform the division operation for the integrand to get x4  1 5 x4  1 1 2  x  2  . Therefore, the integral reads dx    x2  2  dx  5 2 2 2 2  x x 2 x 2 x 2 3 x4  1 1 x 1 2  x2  2 dx    x  2  dx  5 x2  2 dx  3  2 x  5 x2  2 dx. But the new x 5 1  z  dx  2 dz dx. Now, let integ­ral can be written as  2 2 2  x  1   2 x  z  dx  2 dz and rewrite the last integral in terms of the variable z to 2 5 1 5 2 1 5 2 tan 1 z. After collectget  dx  dz  2 2  z  2 2 1 2  x  1   2 ing both integrals results and write them in terms of the original x4  1 x3 5  x  dx   2 x  tan 1  variable x we get  2 . x 2 3 2  2

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List of Selected Integrals with Their Step-by-Step Solutions • 19

INTEGRAL 12 Problem 1

 5  4 cos x dx Solution:

2 1  x  tan 1  tan     constant 3  2  3

Techniques used: Change of variables, Trigonometric identities Step-by-step solution:

1 dx   5  4 cos x   5  4  cos2  1 1 x 1 dx    5  4 cos x dx    2  x  2  x   dx. Let tan 2  z  dz  2 2  x  . cos   5  4  cos    sin    2 2  2   We rewrite the integral in terms of half-angle to have 

z2 1 2 x  x sin  Therefore, cos    , and   2 . Rewriting the 2 2 1 z 2 1 z integral in terms of the variable z, we get, after some simplifications, 2 1 1 2   2  x  2  x   dx   9  z2 dz  9   z 2 . Now let 5  4  cos    sin    1  2  2   3 z  u  dz  3 du and rewrite the integral in terms of the 3 2 du 2  tan 1 u. After variable u. Therefore, we have 3  1  u2 3 ­rewriting this integral in terms of the original variable x, we get 1 2  x  1  1  5  4 cos x dx  3 tan  3 tan  2  as shown in the answer. 2

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20 • 100 Integrals

INTEGRAL 13 Problem

∫ sec Solution:

4

x dx

1 tan 3 x + tan x + constant 3

Techniques used: Integration by parts, Trigonometric identities Step-by-step solution:

1   2 sin x  dx Rewrite the integral using trigonometric identities as sec 4 xdx   4   cos x cos4 1     dx sin 2 x  cos2 x 4 sec xdx   the integrand by   cos4 x  cos4 x dx. Now, manipulate 1    1 sin 2 x  cos2 x sin 2 x dx writing is as follows,  dx  dx  dx   4 4 4     cos x cos x cos x cos2 x 2 2 x sin x dx 1 dx d d  sin x  . But   tan x, since tan x  . For integ­ dx   dx     the other 4 2 2 cos x cos x cos x dx dx  cos x  cos2 x

 tan

ral, we write  2

sin 2 x sin 2 x  1   1 2  d 3 dx    dx   tan x  tan x   tan x. 4 2 2  dx cos x cos x  cos x  3  

 d  1 x  tan x   tan 3 x. After collecting all terms, we get the answer as  dx  3 1 4 3  sec xdx  3 tan x  tan x.

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List of Selected Integrals with Their Step-by-Step Solutions • 21

INTEGRAL 14 Problem:



1

1  x 

2 5

3 x  5 x3  2 x 5

Solution:

3

1  x 

2 5

dx

 constant

Techniques used: change of variables, Integration by parts, trigonometric identities Step-by-step solution: d 1 , 1  x2  1  tan 2   . Now, 2 cos  cos2  rewriting the integral in terms of the variable α gives 1 1 d dx     cos3 d. Now we can write 2   5 10 cos 2  cos  1  x  x  tan   dx 

Let

 cos  d   cos  cos  d   1  sin   cos  d   cos  d   sin  cos d  d   cos  d   cos  d   sin  cos d. But  cos d  sin  and   sin  cos d    sin   si  d 3

2

2

2

2

2

2

1  d    sin 2  cos d    sin 2   sin     sin 3 . Collecting all related terms 3  d  1 gives the answer as  cos3 d  sin   sin 3 . Rewriting the 3 1 result in terms of the original variable x, having   tan x , gives 1 1 dx  sin  tan 1 x   sin 3  tan 1 x . An alternative form  2 5 3 1 x





of the answer, in terms of polynomial functions of x, can be obtained x 1 by substituting sin   into the expression sin   sin 3  3 1  x2 3 5 3x  5x  2x to get . 5 3 1  x 2 

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22 • 100 Integrals

INTEGRAL 15 Problem:

∫

tan x dx

Solution: 2  1  tan 2  



2 tan x  1  tan 1





2 tan x  1  ln





2 tan x  1  tan 1





2 tan x  1  ln

1  tan x  2 tan x    con 1  tan x  2 tan x 

1  tan x  2 tan x    constant 1  tan x  2 tan x 

Techniques used: change of variables, partial fractions, trigonometric identities Step-by-step solution: dx dz  dz  dx  . We rewrite the integral 2 cos x 1  z2 z dz. in terms of the variable z, to get  tan xdx   1  z2 Now, let z  u  dz  2 udu and the latter integral can be written Let tan x  z 

as 

u2 z dz  2  1  u4 du. The denominator of the integrand can be 1  z2







written as 1  u4  1  u2   2 u2  1  u2  2 u 1  u2  2 u . 2

Therefore, using the partial fractions technique, we can write the u2 u2 1 u du  2  du  du  integral as 2  4  2 2 2 1 u 2 1  u  2u 1  u  2u 1  u  2u



u  2u 2



du 





1 u 1 u du. Now we have two new du    2 2 2 1  u  2u 2 1  u  2u integrals to calculate. Working on the latter, we can write the

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List of Selected Integrals with Their Step-by-Step Solutions • 23

2

 2 1 1 denominator of the integrand as 1  u  2 u   u     1 2  2 2   2 2  2 1 1 dy  u2  2 u   u  Now let    1  2 u  1 . 2 u  1  y  du  ,   2 2 2 2   and write the latter integral in terms of variable y as 2





2u



1 u 2  y  1  / 2 dy 1 1y  dy.   du   22   22 1 y 2  1  y22 2 1  u  2u 2 2 2y 1 1y 1 1 1 But dy  dy  dy. The former 2 2 2    2 1 y 2 1 y 2 2 1 y 2y 1 1 1 1 integral reads tan 1 y and the latter  dy  dy   2 2   2 1 y 2 2 2 1 y 2 2y 1 1  dy   ln 1  y2 . Writing the results in terms of varia2 2 2 1 y 2 2 1 1 1 1 tan 1 y  ln 1  y2   tan 1 2 u  1  ln 1  2 u  1 ble x, gives 2 2 2 2 2 2 2 2 1 1 1 1 tan 1 2 u  1  ln 1  2 u  1  tan 1 2 z  1  ln 1  2 z  1  tan 1 2 tan 2 2 2 2 2 2 2 2 2 1 1 1 ln 1  2 z  1  tan 1 2 tan x  1  ln 1  2 tan x  1 . 2 2 2 2 

































2 tan x  1 



2







2 tan x  1

2  1 tan 2 



2 tan x  1 

ln 1   2 tan x  1  . Therefore, solution reads as  2

2  1 tan xdx  tan 2 





Similarly, the remaining integral can be worked out and 1 u 1 1 du  tan 1 2 tan x  1  ln 1  reads as  2 2 1  u  2u 2 2 2



1











2



100_Integrals.002_3pp.indd 23



2 tan x  1  tan

1



tan xdx 

 2 1  2 tan x  1   ln  4 1  







 

 

2 2 tan x  1   2 2 tan x  1  

27-07-2023 15:57:04





2

24 • 100 Integrals

 2 1    ln 2 tan x  1  4 1  



 

 

2 2 tan x  1  . 2 2 tan x  1  

 2 1  ln 1  4 

 

The

last

term

simplifies

to

 

2 2 tan x  1    2 ln  1  2 tan x  1  2 2 tan x   2 ln 2  2 ln  1  tan    2 4  1  tan  1  2 tan x  1  2 2 tan x  4 2 tan x  1  4 

2  1  tan x  2 tan x  2 1  2 2 tan x  2  1 ln  ln 2   tan 2 tan x  1   . Simplified ­solution reads as 4 1  2 2 tan x  4 2   1  tan x  2 tan x    1  tan x  2 tan x   2 2  1  tan 2 tan x  1  tan 1 2 tan x  1  ln  ln 2.  2  1  tan x  2 tan x   4     tan x  2 tan x   2 ln 2.   tan x  2 tan x   4 





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List of Selected Integrals with Their Step-by-Step Solutions • 25

INTEGRAL 16 Problem:

∫ xe

x

sin x dx

Solution: ex  x sin x  1  x  cos x   constant 2  Techniques used: Integration by parts, trigonometric identities Step-by-step solution:

d xe x   e x  x  1  and xe x dx  e x  x  1 , we  dx apply the integration by parts technique to the integral to get  xe x sin xdx   xe x cos x  e x  x  1  cos xdx   xe x cos x  xe x cos xdx  e x cos

Knowing that

 f

 dg

os xdx   xe x cos x  xe x cos xdx  e x cos xdx. But  xe x cos x dx  xe x sin x  e x  x  1  sin x dx  f

xe

 dg

and Therefore, after substitution, we get xe x sin xdx   xe x cos x  xe x sin x  e x

x

sin xdx   xe x cos x  xe x sin x  e x  x  1  sin x dx  e x cos xdx. Or, after rearranging the involved terms, we get xe x sin xdx 

1  xe x cos x  xe x sin x  e x sin x dx  e x 2

1 x x x  xe x cos x  xe x sin x  e x sin x dx  e x cos xdx  . But  e sin xdx  e cos x  e cos x dx.   2 1 Therefore, after substitution, we get xe x sin xdx    xe x cos x  xe x sin x  e x cos 2 1 1 sin xdx    xe x cos x  xe x sin x  e x cos x  e x cos x dx  e x cos xdx     xe x cos x  xe x sin x  e x cos x   2 2 1 ex dx     xe x cos x  xe x sin x  e x cos x . Or xe x sin xdx   x sin x  1  x  cos x  .  2 2

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2

26 • 100 Integrals

INTEGRAL 17 Problem: ln 3 x ∫ x3 dx Solution: 

1  4 ln 3 x  6 ln 2 x  6 ln x  3   constant 8 x2 

Techniques used: Integration by parts Step-by-step solution: Rewrite the integrand and apply the integration by parts. Therefore, ln 3 x 1 3 3 ln 2 x 3  dx  we get  3 dx   ln x  3   2 ln x   3 dx. Applying the   x  x 2x 2 x f  dg

integration by parts successively twice again to the last integral, gives 3 ln 2 x 3  1 ln x  3  1 1 1 1  3  1 dx   2 ln 2 x   3 dx    2 ln 2 x  2 ln x   3 dx    2 ln 2 x 3  2 x 2  2x x 2x 2 x  2  2x  2  2x 1 1 1 3  1 1 1    ln 2 x  2 ln x   3 dx    2 ln 2 x  2 ln x  2  . After collecting all terms, we x x x 2x 2 x 2 2 2 4    3  2 ln 3 x   receive the answer as  3 dx  2 1  2 ln x  1  ln x  ln 2 x  . x 8x  3  

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List of Selected Integrals with Their Step-by-Step Solutions • 27

INTEGRAL 18 Problem:

 ln 1 

Solution:



x dx

 x  1 ln 1  x   x   constant x 2

Techniques used: Change of variables, Integration by parts, Partial fractions Step-by-step solution:

We use the integration by parts technique and consider dx  x. 1 Therefore, we can write  ln 1  x dx  x ln 1  x  x 2 x dx  x ln 1  x  1 1 x 1 x n 1  x  x 2 x dx  x ln 1  x   dx. Now, let x  z  dx  2 zdz and 2 x x 1 x write the new integral in terms of variable z as z3 z2 1 x   dx    2 dz    dz. Using partial fractions, z z z1 2 x x z2 1   we can write   dz     z  1   dz . Performing the z1 z1  1  1 2  integration, gives    z  1   dz   z  z  ln 1  z . After z1 2  collecting all terms and write them in term of the original variable 1 x, we get the answer as  x  1  ln 1  x  x  x . 2



















100_Integrals.002_3pp.indd 27





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28 • 100 Integrals

INTEGRAL 19 Problem: 

 ln  1 

1

1   dx x2 

Solution:   1  1  x  ln  1  1  2   1  1  2   constant x  x    Techniques used: Change of variables, Integration by parts Step-by-step solution: We use the integration by parts technique along with considering

dx  x.

Therefore, we can write ’

 1  1 1  2  dx  x ln  1  1   x x   dg 

 ln  1 

f  1  1  1  2  x    1  1   x ln  1  1  2   x  dx,  ln  1  1  x2  dx  x 1   1 1 2  dg f x with the prime symbol indicating differentiation. Therefore, ’  1  1  1  2  x  dx  x  dx   . After simplifying the 1 1 2 2 x 1 2  x 1 1 1 2 x x dx dx  integrand, we get  . Let 2 2 1    1 x x 1 x 2 2 x 1 2  x 1 x zdz 1  x2  z  dx  . Therefor, the latter integral in z2  1 1 1 dz   terms of variable z can be written as  2 2 z z 1  z 1 z2  1 z  z2



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List of Selected Integrals with Their Step-by-Step Solutions • 29

z

1 z 1  z 1 2

2

dz  

du z

z  z2  1  u  dz 



1

z 1 z  z 1 2

z2  1

1

u, we have

2



Now,

dz.

let

z  z2  1  u  dz 

du z z2  1

1

. Rewriting the latter integral in terms of the variable





1

z 1 z  z 1 2

2



dz  

du   z  1  z2  1 u 2 z  1    



1 du  . 2 u u

u

1 du   2   . After back substituting for u in terms of z and x we get u u   z 1 1 u  1  z2  1 the results as  1    . Multiplying 2 2 z  1  u   z  z 1 1  x2  x  u the last expression by its conjugates ratio gives, 1 1  x2  x x  1  x2    x  1  x2 . After collect­2 2 2 2   1 x x 1 x  x 1 x  x ing all related terms, we get the final answer as   1  1  2  ln  1  1  x2  dx  x ln  1  1  x2   x  1  x . Please note 1 that 1  x2  x 1  2 . x du

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30 • 100 Integrals

INTEGRAL 20 Problem:

 csch

1

x dx

Solution: xcsch 1 x  sinh 1 x  constant Techniques used: Change of variables, Integration by parts Step-by-step solution: Let csch 1 x    csch 

1 1  x. Therefore, sinh   and sinh  x

dx . Now we can write the integral in terms of varix2 1  cosh   able α, as csch 1 xdx   sinh 1        d. Using the 2  x  sinh    d  cosh   integration by parts, we have    .   d  2 sinh  sinh   sinh   But writing the latter integral in terms of original variable x, we get d dx 1 dx x dx    .   sinh 1 x . 2 2 2 sinh  cosh . x sinh  x 1 x 1 x Now collecting all related terms, we get the solution as   sinh 1 x  xcsch 1 x  sinh 1 x, for x > 0. sinh 

cosh d  

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List of Selected Integrals with Their Step-by-Step Solutions • 31

INTEGRAL 21 Problem: 1

 sin x  cos x dx Solution:    x tan    1  2  2  2   constant ln  2  tan  x   1  2      2 Techniques used: Change of variables, Partial fractions, Trigonometric identities Step-by-step solution: x = z, then we have dx  2 dz / 1  z2 .Using the trigo­ 2  x 2 tan    2   2 z , and nometric identities, we have sin x  1  z2 2 x  1 tan    x 2 1  tan 2   2 2 1 z   cos x  . Therefore, the integral can be written 1  z2 2 x 1  tan   2 1 1 dx  2  dz. in terms of the variable z as  sin x  cos x 1  2 z  z2 But the denominator of the integrand can be written as 2 1  2 z  z2    z2  2 z  1     z  1   2    z  1  2 z  1  2 Let tan









2    z  1  2 z  1  2 . Now we rewrite the integral as 2 

100_Integrals.003_3pp.indd 31





1 1 dz  2  2 1  2z  z z 1 2



27-07-2023 15:45:53



32 • 100 Integrals

2

1 1 dz. dz  2  1  2 z  z2 z 1 2 z 1 2





technique, 1



2 z 1 2



dz  



we

can

write

Using 2 

the

partial 1

fractions

 z  1  2  z  1  2 

dz  

2 dz   2 z 1 2

2 dz 2 dz . Finally, we can write the    2 z 1 2 2 z 1 2

solutions by integrating each term, as 









2 2 ln z  1  2  ln z  1  2  2 2

2 2  z 1 2  2 ln z  1  2  ln  ln z  1  2  . Therefore, the answer in 2 2  z 1 2  2









terms of the original variable x is

  x tan    1  2 2  z 1 2  2  2 ln  ln   2  z 1 2  2  tan  x   1  2    2

   x tan    1  2   z 1 2  2  2 ln  . Please note that the absolute value of the n  2 z  1  2  tan  x   1  2        2 argument of the logarithm should be considered.

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List of Selected Integrals with Their Step-by-Step Solutions • 33

INTEGRAL 22 Problem:

2 Solution:

x

dx

21   x 1   x log 2  constant log 2 2





Techniques used: Change of variables, Logarithm identities Step-by-step solution:

log z = i x log 2 log z 2(log z / log 2) dz idx 2 x 2 x log =  2  . Therefore, dx  = = i x , or  2 z log 2 iz log 2 i z log 2  z log 2 z log 2 2 x 2 log z 2(log z / log 2) 2 x 2 x dx   2  . Rewriting the integral in terms of the var­  iz log 2 i z log 2  z log 2 z log 2 2 2 iable z, we get  2  x dx   2  log z dz. But  log z dz  z log z  z, log 2 and after rewriting this expression in terms of the original x varaiable we 2 21   x get  2  x dx   2 2  x log 2  x  2  x  1   x log 2 . log 2 log 2 2 Let

2

x

 z  log 2 z   x  i x .



100_Integrals.003_3pp.indd 33

We



can

write



log = 2 z



27-07-2023 15:48:34

34 • 100 Integrals

INTEGRAL 23 Problem:

 ln 1 

Solution:





1  x2 dx



x ln 1  1  x2  sinh 1 x  x  constant Techniques used: Change of variables, Trigonometric identities Step-by-step solution: Using the integration by parts technique, we can write the integral as ln 1  1  x2   x ln 1  1  x2  x f  dx .   dx dg













f

d x But f   ln 1  1  x2  . Therefore, the latter 2 dx 1  x  1  x2 dz x2 integral reads xf ’ dx    dx. Let 1  x2  z  dx  z  2 2 x 1 x  1 x dz zdz 1  x2  z  dx  z  and write the latter integral in terms of variable x z2  1

zdz z2  1

z2  1 zdz z2  1    z  1 dz. Now, z2  z z2  1 1  x2  1  x2 let z  cosh u  dz  sinh udu and write the latter integral in terms z2  1 sinh 2 u cosh 2 u  1 of variable u as   dz    du    du     cosh u  1  du z1 1  cosh u 1  cosh u z, as  

100_Integrals.003_3pp.indd 34

x2

dx   

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List of Selected Integrals with Their Step-by-Step Solutions • 35

cosh 2 u  1 du     cosh u  1  du  u  sinh u. This expression in terms of the u 1  cosh u x original variable reads u  sinh u  cosh 1 z  z2  1  cosh 1 1  x2 nh u  cosh 1 z  z2  1  cosh 1 1  x2  1  x2  1  cosh 1 1  x2  x, for x > 0. Collecting all related terms, we get the solution as du   





x ln 1  1  x2  cosh 1 1  x2  x. But cosh 1 1  x2  sinh 1 x.





Hence, the solution simplifies to x ln 1  1  x2  sinh 1 x  x.

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36 • 100 Integrals

INTEGRAL 24 Problem:

∫ sin

6

x cos5 xdx

Solution: 8 4 1 sin11 x + sin 9 x cos2 x + sin 7 x cos4 x + constant 693 63 7 Techniques used: Integration by parts Step-by-step solution: Rewrite the integral as

 sin

6

x cos5 xdx  (sin 6 x cos x)cos4 xdx.

Using the integration by parts, we get

1

x cos x  cos x dx  sin  sin   7 6

4

dg

7

x cos4 x 

f

4 s 7

1 4 4 x cos x  cos x dx  sin 7 x cos4 x   sin 8 x cos3 x dx.Repeating the similar operation     f 7 7

6

dg

1 2 on the latter integral, we get  sin 8 x cos3 x dx   sin 8 x cos x cos2 x dx  sin 9 x cos2 x   s 9 9 1 9 2 8 2 2 10 n x cos x cos x dx  sin x cos x   sin x cos x dx. But the latter integral reads 9 9 1 10 11  sin x cos x dx  11 sin x. Collecting all related results, we get 8 4 1 the solution as sin11 x + sin 9 x cos2 x + sin 7 x cos4 x. 693 63 7

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List of Selected Integrals with Their Step-by-Step Solutions • 37

INTEGRAL 25 Problem:

∫x

m

ln n xdx

m and n are positive integers. Solution:

x m 1 ln n x n  x m ln n1 xdx  constant m 1 m 1  For m= n= 3 4

x  32 ln3 x  24 ln2 x  12 ln x  3  128 Techniques used: Integration by parts, Recursive relation Step-by-step solution: The solution provides a recursive relation applicable to integer values of m and n. We work out the general solution and apply it to a numerical example. Applying the integration by parts technique, we can write x m 1 n n m n x ln xdx  ln x  x m ln n1 xdx. For example, for   m 1 m 1 3 x4 3 3 m= n= 3, we have x ln xdx  ln 3 x  x3 ln 2 xdx. But, using 4 4 the general relation again for = m 3= , n 2, we get 4 2 3 x 2 3 2 x ln xdx  4 ln x  4 x ln xdx. Using the general formula for the 1 x4 m 3= , n 1, we get x3 ln xdx  ln x  x3 dx, latter integral, for = 4 4

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38 • 100 Integrals

x4 . After collecting all related terms, we get 4 x4 3 3  x4 2 2  x4 1 x4   3 3 x xdx  x  x  x  ln ln ln ln    . Or, after  4 4 4 4 4 4 4  x4 simplification, we get the final answer as x3 ln 3 xdx   32 ln3 x  24 ln2 x  128 with

3 3 x ln xdx 

x dx  3

x4  32 ln3 x  24 ln2 x  12 ln x  3 . 128

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List of Selected Integrals with Their Step-by-Step Solutions • 39

INTEGRAL 26 Problem:

e

x

sin 1  e x  dx

Solution: e x sin 1  e x   1  e2 x  constant Techniques used: Change of variables, Integration by parts Step-by-step solution: Let e x  z  dx  dz / z. Rewriting the integral in terms of the variable z, we get

e

x

sin 1  e x  dx   sin 1 z dz. Now, let

sin 1 z    sin   z  cos  d  dz. After back substituting, we get  sin 1 zdz   cos  d. Using the integration by parts technique, we get the answer,    d    sin    sin  d   sin   cos  .  cos   f

dg

Rewriting the results in terms of variables z, and then x, we get  sin   cos   z sin 1 z  1  z2  e x sin 1  e x   1  e2 x .

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40 • 100 Integrals

INTEGRAL 27 Problem:

x Solution:

1  x2 dx 1  x2

1  x2 1  1  x 4  2 tan 1 2  1  x2

   constant  

Techniques used: Change of variables, Partial fractions, Trigonometric identities Step-by-step solution: Let x  tan  z / 2   dx 

dz . Therefore, the integral 2 cos  z / 2  2

can be written in terms of the variable z as x

x

2 1  x2 1 tan  z / 2  1  tan  z / 2  dx dz. But  1  x2 2  cos2  z / 2  1  tan 2  z / 2 

hence we get dz 

1  x2 1 tan  z / 2  1  dx   2 1 x 2 cos2  z / 2  1 

1  tan 2  z / 2   cos z, 1  tan 2  z / 2 

2 1 tan  z / 2  1  tan  z / 2  1 sin  z / 2  cos zdz dz   2 2  2 cos  z / 2  1  tan  z / 2  2 cos3  z / 2 

1 sin  z / 2  cos  z / 2  cos zdz. Now, after multiplying the integrand by , we 2  cos3  z / 2  cos  z / 2  sin z / 2  cos cos z / 2  1 sin 1 sin z have  cos cos zdz   cos zdz. zdz Let cos 44 cos44  z // 2  cos  z // 2  cos 2 4 cos cos z  u   sin z dz  du and have

4 u cos z  . Therefore, we cos  z / 2  1  u 2 4

u 1 sin z du du. Now let u  t  du  2 tdt cos zdz    4  4 cos  z / 2   1  u 2

u  t  du  2 tdt and writing the latter integral in terms of the variable t, we

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List of Selected Integrals with Their Step-by-Step Solutions • 41

get 2 

t2

1  t 

2 2

dt. After using the partial fractions techniques,

  1 1  .  2  dt dt dt   2 2 2 2 2   1  t   1  t   1  t   1 dt  2 tan 1 t . But for calcu­ The latter integral reads 2  2 1  t   we can write 2 

t2

lating the integral 2 

1

1  t 

2 2

dt we let t  tan y  dt  dy / cos2 y 2

 cos2 y  sin 2 y  1 and 1  t   1  tan y    . After   4 2 cos y   cos y cos4 y 1 substation, we get 2   2 dt dy  2  cos2 y dy. But, 2  2 2 cos y 1  t  2 2

2

2

using trigonometric identity cos2 y 

1  cos 2 y , we get 2  cos2 y dy   1  cos 2 y  d 2

1 2  cos2 y dy   1  cos 2 y  dy  y  sin 2 y. Now, after a series of 2 back substitutions and collecting related terms, we get cos  2 tan 1 x 

1  cos  2 tan 1 x 

 tan 1 cos  2 tan 1 x  . This expression simpli­

1  x2 1 fies to  1  x 4  2 tan 1 1  x2 2 

100_Integrals.003_3pp.indd 41

 .  

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42 • 100 Integrals

INTEGRAL 28 Problem:

x

x tan 1 xdx

Solution: 1  4 x2 x tan 1 x  2 ln 1  x   x  x  2    constant  10  Techniques used: Change of variables, Integration by parts, Partial fractions Step-by-step solution: Let

x  z  dx  2 zdz, rewriting the integral in terms of the vari­

able z gives x x tan 1 xdx  2 z4 tan 1 z dz. Using the integration by 1  1 z5 parts technique we get 2 z4 tan 1 z dz  2  z5 tan 1 z   dz . 2 5 1 z 5  Note that we used the derivative of inverse tangent func­ d 1 tion of z, or tan 1 z   . Now, we work out the latter  dz 1  z2 integral by using the partial fractions technique to get z  z 1 z5 1  3 1  z 4 z2  1 dz z z dz dz.          2 2    5  1 z  5  4 2  5 1  z2 5 1 z 1 z 1 But  dz  ln 1  z2 . Collecting all related terms, we get 2 5 1 z 10   1 1  z 4 z2  1 2 z4 tan 1 z dz  2  z5 tan 1 z      ln 1  z2  . After 5  4 2  10 5  simplification and rewriting this expression in terms of the original 1 variable x, we get  4 x2 x tan 1 x  2 ln 1  x   x  x  2   . 10

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List of Selected Integrals with Their Step-by-Step Solutions • 43

INTEGRAL 29 Problem:

tan 1 x

  x  1

3

dx

Solution: 2 1   x  1 4 2   tan 1 x   ln   constant 2 2 x 1 8   x  1  x  1  

Techniques used: Integration by parts, Partial fractions Step-by-step solution: Using integration by parts technique we get,

1

f

dx

x  tan  x  1 1

f

3

    dg



dx

x  tan  x  1

3



   

1 2  x  1

2

tan 

dg

1

tan 1 x 

1 1 1 . dx. But using the par­ 2  2  x  1  1  x2

2  x  1 tial fraction technique we can write the latter integral as 1 1 1 1 x 1 1 1 1 dx   dx.   dx   dx   2 2 2  4 1 x 4 x 1 4  x  1 2 2  x  1 1  x 2

1 x 1 1 1 1 dx  ln 1  x2 , and  dx  ln  x  1  . For 2  4 1 x 8 4 x 1 4 the third integral in the last expression, we let x  1  z  dx  dz. 1 1 1 dz 1 1 Therefore, dx   2     . Collecting 2  4  x  1 4 z 4z 4  x  1 But

1 1 1 all the related terms we get the answer as  tan 1 x  ln 1  x2   2 8 4 2  x  1 1 1 1 1 2 1 x x  tan  ln 1   ln x   . 1     2 4 8 4  x  1 2  x  1

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44 • 100 Integrals

INTEGRAL 30 Problem: sin 1 x  x2 dx Solution:  x ln  2 1 1 x

 sin 1 x  constant  x 

Techniques used: Integration by parts, Change of variables, Partial fractions. Step-by-step solution: Using

the integration by parts technique, we can sin 1 x sin 1 x 1  dx  1 write   sin    dx . Note dx x  2    2  x x  x  x 1  x2 f dg

d 1 sin 1 x  . Let x  cos z  dx   sin z dz, then dx 1  x2 1 sin z dz dz  x 1  x2 dx    cos z sin z    cos z   sec z dz. We calculated the integral of sec z in the previous section (see Integral 5). Therefore,  cos  z / 2   sin  z / 2   dz   ln  we have   . But we have cos z  cos  z / 2   sin  z / 2   x  cos z  2 cos2  z / 2   1  1  2 sin 2  z / 2 . Therefore, after some that

manipulations, we get cos  z / 2  

100_Integrals.003_3pp.indd 44

1 x 1 x and sin  z / 2   . 2 2

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List of Selected Integrals with Their Step-by-Step Solutions • 45

  z  z cos    sin     dz 2  2     ln 1  x  1   ln  After back substituting, we get   z z cos z    1 x  1  cos  sin         z  z 2 2       2  cos  2   sin  2          ln 1  x  1  x   ln  1  1  x . Final answer, after n   x 1 x  1 x  cos  z   sin  z          2  2   writing the results in terms of the original variable x, is   1  1  x2  sin 1 x sin 1 x    dx ln .   x2   x x  

100_Integrals.003_3pp.indd 45

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46 • 100 Integrals

INTEGRAL 31 Problem:

x sec

1

xdx

Solution:





1 2 x sec 1 x  x2  1  constant 2 Techniques used: Integration by parts, Trigonometric identities Step-by-step solution: Using the trigonometric identity sec 1 x  cos1 1 / x  and d 1 cos1 1 / x    , we can rewrite the integral by  dx x x2  1

1 applying the integration by parts technique as x sec 1 xdx   cos1    xdx  x 2 2   1 2 1 x  1dg 1 x  1 1  1  1 x sec xdx  cos  cos /  . But xdx x x 1 dx      dx f      x  dg 2 2  x x2  1 2  x x2  1 4  



 

f

1 x2 1 2x 1 2 dx     x  1 . Collecting the related  2 2 2 x x 1 4 2 x 1 1 terms, we get the answer as x sec 1 xdx   x2 cos1 1 / x   x2  1   2 1 x   x2 cos1 1 / x   x2  1 .  2 

100_Integrals.004_3pp.indd 46

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List of Selected Integrals with Their Step-by-Step Solutions • 47

INTEGRAL 32 Problem:

 sec Solution:

 x sec

1

1

xdx



x  x  1  constant

Techniques used: Integration by parts, Trigonometric identities Step-by-step solution:



 sec

1





Using the trigonometric identity sec 1 x  cos1 1 / x and d 1 cos1 1 / x  , we can rewrite the integral by dx 2x x  1  1  1 xdx   cos1  applying the integration by parts technique as  sec   d x    d 1 1 1  1  1 1  1  1 f xdx   cos  dx  x cos    x cos 1 / x     dx   x  1  x sec   2  x x  1 x   dg     









f

1  1  1 dx  x cos1  x  x  1.   x  1  x sec 1  x

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48 • 100 Integrals

INTEGRAL 33 Problem:

x

2

tan 1 xdx

Solution: 1 2 x3 tan 1 x  x2  ln 1  x2    constant  6 Techniques used: Integration by parts, Partial fractions Step-by-step solution:

d 1 tan 1 x   , we can rewrite the integral  dx 1  x2 1 x3 x3 by applying the integration by parts technique as x2 tan 1 xdx  tan 1 x   3 3 1 x x3 x  1 1  x2 1 x3 1 x3 1  2 1 1 x xdx  x  . But  dx x dx    tan tan    dx ln 1  x    3 3  1  x2 3  1  x2 3  1  x2  3  2 2  1  x  1  x2 1     ln 1  x2   . Collecting all related terms, we get the x dx    2  3 2 2 3  1 x   1 answer as x2 tan 1 xdx  2 x3 tan 1 x  x2  ln 1  x2  . 6 Using the relation

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List of Selected Integrals with Their Step-by-Step Solutions • 49

INTEGRAL 34 Problem:



2x  3 3  6 x  9 x2

dx

Solution: 1  1  3x    2 3  6 x  9 x2  11 sin 1     constant 9  2  Techniques used: Change of variables, Trigonometric identities Step-by-step solution: The expression under the radical could be written in

the form of an expression containing the term 1  u2 . d 1 Knowing that sin 1 u  , we can write 3  6 x  9 x2  4  1  6 x  9 x du 1  u2   1  3 x 2  2 3  6 x  9 x2  4  1  6 x  9 x2   4  1  3 x   4 1     . Therefore, let   2   1  3x 2  u  dx   du. After writing the integral in terms of 2 3  1  2u  2 3 1 11  4 u 2x  3 2  3  u, we get the variable  u 1 2 dx   du        2 2  2 3 3 9 1  u2 3  6x  9x 2 1 u 2  3  1 11  4 u du. Further, this integral can be written as   du    3 9 1  u2 2 1  u2 1 11  4 u 11 1 4 u 11 1   du    du   du   du. But   9 1  u2 9 9 1  u2 9 1  u2 1  u2

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50 • 100 Integrals



11 1 11 4 u 4 du   sin 1 u and du   1  u2 . 9  1  u2 9 9  1  u2 9 Collecting all related terms and rewriting the result in terms of the original variable x, we get 

1

2

u

11 1 4 11  1  3x  sin u   1  u2   sin 1   9 9 9  2 

4 11  1  3x  4  1  3x  1 1  u2   sin 1    . After simplification, we get the 9 9  2  9  2  2x  3 11  1  3x  2 answer as  dx   sin 1  3  6 x  9 x2 .  2 9  2  9 3  6x  9x

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List of Selected Integrals with Their Step-by-Step Solutions • 51

INTEGRAL 35 Problem:

 x

3x  2

2

 4  x2  4

dx

Solution: x6 2 x2  4

 constant

Techniques used: Change of variables, Trigonometric identities Step-by-step solution: We rewrite the integral as  dx  3 

2

x

x

x

2

 4

3/2

1

2

 4

3/2

dx  2 

x

1

2

 4

3/2

dx.

But

x

3x  2

2

 4  x2  4 3

x

dx  3 

x

2

 4

3/2

x

x

2

 4

dx  3 /

3/2

x

2

dx  2 

x

1

2

 4

3/2

dx

 4 .

To calculate the latter integral, we get a hint from the relation   x 2  2 1 d x 2 tan u  and rewrite x  4  4 1   2  . Now let  tan u  dx  2     co 1 u dx 2 x 2 du 1 2d  tan u  dx  . Therefore, we have 2  dx  2  3/2 2 2 cos2 u 2 cos u  4 1   x  4 2 du 1 du   . dx  2  3/2 2 2 2 2 cos u 1  tan 2 u 3 / 2 cos u  4 1  tan u   The denominator of the integrand can be simplified as

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52 • 100 Integrals

we

have

sin u 

 x

x

x 4 3x  2

2

2

.

 4  x2  4

100_Integrals.004_3pp.indd 52

3/2

 cos2 u  sin 2 u  1  cos2 u  . Therefore,   2 cos u cos u   du 1 1 1   cos udu  sin u. But 3/2  2 2 2 cos u 1  tan u  2 2

cos2 u 1  tan 2 u 

3/2

Collecting dx  

x

all

3 2

 4



related 1 2

x x2  4

terms, 

we

x6

2

x

2

 4

get

.

27-07-2023 15:52:13

List of Selected Integrals with Their Step-by-Step Solutions • 53

INTEGRAL 36 Problem: 2 x2  5 x  1  x3  2 x2  x  2 dx Solution: ln

 x  1   x  1 2  constant  x  2

Techniques used: Partial fractions, Integration by parts Step-by-step solution:

1

1

Using the partial fractions method, the integrand can be written 2 x2  5 x  1 2 x2  5 x  1 A B C as    .   3 2 x  2 x  x  2  x  2   x  1  x  1 x  2 x  1 x  1 Therefore, we get A  1, B = 1, and C = 2. Now rewrite the inte2 x2  5 x  1 1 1 1 gral as 3  x  2 x2  x  2 dx   x  2 dx   x  1 dx  2  x  1 dx   ln  x  2   lnn 

 x  1 dx  2  x  1 dx   ln  x  2   lnn  x  1  2 ln  x  1. tion rule, we get the answer as ln

100_Integrals.004_3pp.indd 53

Using the logarithm summa-

 x  1   x  1 2 .  x  2

27-07-2023 15:53:28

54 • 100 Integrals

INTEGRAL 37 Problem:

4

dx

Solution: 2 ln  x  1  

9 x2  18 x  4 3  x  1

3

 constant

Techniques used: Partial fractions, Integration by parts, Change of variables Step-by-step solution: Let x  1  z  dx  dz. Therefore, the integral can be written in terms 3 2 2  z  1  3  z  1  4 2 x3  3 x2  4 dx of the variable z as   dz.  z4  x  1 4 After expanding the terms in the nominator of the integrand, 3 2 2  z  1  3  z  1  4 2 z3  3 z2  5 dz  dz.   we get   z4 z4 Using the partial fractions technique, we can write the integral 2 z3  3 z2  5 dz dz dz 3 5 as dz  2   3  2  5  4  2 ln z   3 . 4  z 3z z z z z Rewriting the results in terms of the original variable x, we get 3 5 9 x2  18 x  4 3 5 2 ln z   3  2 ln  x  1     2 ln x  1    3 z 3z x  1 3  x  1 3 3  x  1

5

 1

2 x3  3 x2  4

  x  1

3

 2 ln  x  1  

9 x2  18 x  4 3  x  1

100_Integrals.004_3pp.indd 54

3

.

27-07-2023 15:54:58

List of Selected Integrals with Their Step-by-Step Solutions • 55

INTEGRAL 38 Problem:

  x  1

x 1 x3  x2  x

dx

Solution:   x 2 tan 1    constant 2  x  x 1  Techniques used: Change of Variables, Trigonometric identities Step-by-step solution: Let x  z2  dx   2 z3 dz. Writing the integral in terms of the variable z2  1  z3  x 1 z gives,  dz. dx  2   x  1  x3  x2  x  z2  1 z6  z4  z2 After 2 

multiplying

the

integrand

z5  z2  1  z3

z2  z2  1   z3 

z6  z4  z2

dz  2 

by

z

z5 z5 z2  1

2

we

 1  z 4  z2  1

get dz.

z6

Now we let tan u 

z z  z2  1 4

and differentiate both sides

 4 z3  2 z  z 4  z2  1  z   4 2 1  2 z  z  1  dz. Having to get du  cos2 u z 4  z2  1  z4  2 z2  1 z4  z2  1 du. z 4  2 z2  1 1  , we can write dz  cos2 u z4  z2  1 1  z4

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56 • 100 Integrals

After substituting for dz, we can write the integral as     z 4  2 z2  1  z 4  z2  1  2 z 1  z2  1  du  dz  2   2    z2  1  z 4  z2  1   1  z4  z2  1  z 4  z2  1       

   z  2 z  1 z  z  1    du  2 du  2 u.  2  1  z4  1  z4  z2  1      z2  1

4

2

4

1

2

1

  z 1 But u  tan 1  . Therefore, we get the  and z = 4 2 x  z  z 1    z answer in terms of the original variable x as 2 u  2 tan 1  4   2 2 z  z  1     1       x z x 1   2 tan 1  2 u  2 tan 1  .   2 tan  2 4 2  1 1  x  x  1  z  z 1     2  1  x  x 

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List of Selected Integrals with Their Step-by-Step Solutions • 57

INTEGRAL 39 Problem: x 4  4 x3  6 x2  4 x  1  x3  3 x2  3 x  1 dx Solution: 24 ln  x  1  

x2  14 x 8  4 x  3    constant 2  x  1 2

Techniques used: Partial fractions, Change of variables, Integration by parts. Step-by-step solution: Using the partial fractions technique, we can write the integral x 4  4 x3  6 x2  4 x  1 24 x2  16 x  8 dx  x  7 dx      x3  3 x2  3xx  1 dx. as  x3  3 x2  3 x  1 1 Therefore, we get   x  7  dx  x2  7 x for the former integral. 2

3 x2  24 x2  16 x  8 dx  8  3 3 2 x  3x  3x  1 x  3 x2 0 0       3 x2  6 x  3 3 x2  2 x  1  (4 x  4 x)  (2  2) 3 x2  6 x  24 x2  16 x  8  8 dx  8 dx  8 dx  x3  3 x2  3 x  1  x3  3 x2  3 x  1   x3  3 x2  3 x3  3 x2  3 x  1 0 0     2x  1  (4 x  4 x)  (2  2) 3 x2  6 x  3 dx  16  3 dx. But   dx  8 2 3 2  x  3x  3x  1 x  3 x2  3 x  1 3x  3x  1 3 x2  6 x  3 8 3 dx  8 ln  x3  3 x2  3 x  1   24 ln  x  1  . For    x  3 x2  3 x  1 For the latter integral, we rewrite this integral as 

 x  1 3

the other integral, we have 16 

100_Integrals.004_3pp.indd 57

2x  1 2x  1 dx dx  16  2 x  3x  3x  1  x  1 3 3

27-07-2023 15:59:11

58 • 100 Integrals

we let x  1  z  dx  dz and rewrite the integral in terms of the 2x  1 2z  1 dx  16  3 dz. Using the partial fracvariable z as 16  3 z  x  1 2z  1 dz dz 32 8 tion technique, we get 16  3 dz  32  2  16  3    2 . z z z z z Rewriting the obtained results in terms of the original variable x, 32 8 1 we get the answer as   24 ln  x  1   x2  7 x  2 x  1  x  1 2

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List of Selected Integrals with Their Step-by-Step Solutions • 59

INTEGRAL 40 Problem:



Solution: 

x



1 4

x 1

1  94 x 18



4

x 1



9



10

dx

 constant

Techniques used: Partial fractions, Change of variables Step-by-step solution: Let x  z  dx  2 zdz and the integral in terms of the variable z, after 1 1 dz. Now, let simplifications, reads  dx  2  10 10 4 x x 1 z 1









z  1  u  dz  2  u  1  du and the integral in terms of variable u 1 u 1 can be written as 2  dz  4  10 du  4 u9 du  4 u10 du. 10 u z 1 1 8 4 9 Performing the integrations gives  u  u . Therefore, the 2 9 final solution in terms of the original variable x is 8 9 8 9 1 4 1 4  z 1  z  1   4 x  1  4 x  1 . Or, after 2 9 2 9 1  94 x simplifications, we get the answer as  . 9 18 4 x  1



















100_Integrals.004_3pp.indd 59







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60 • 100 Integrals

INTEGRAL 41 Problem:

 1  ln x  ln  ln x  dx Solution: x  ln x ln  ln x   1  constant Techniques used: Change of variables, Integration by parts Step-by-step solution: Let ln x  z  dx  e z dz. Therefore, the ­ integral can be written in terms of the variable z as

 1  ln x  ln  ln x  dx   1  z  ln z  e dz  z

 ln x  ln  ln x  dx   1  z  ln z  e z dz   ln z e z dz  (z ln z) e z dz.

But, using the integration by parts technique, we have

 ln z e dz   z ln z  z  e    z ln z  z  e dz  z ln z e  ze    z ln z  e dz  ze dz    z ln z  e dz  ze dz. After rearranging the terms, z ln z e  ze    z ln z  e dz  ze d  n z e ze ze    z ln z  e dz  ze dz   ( z ln z ) e dz  z l   dz . The    z

dz  z ln z e z  ze z

z ln z e z  ze z

z

z

z

z

z

z

z

z

z

z

z

z

z

z

z

z



z





new integral can be worked out using the integration by parts

ze dz  ze  e . Therefore, the answer reads  ze dz  z ln z e  ze  ze  e  e  z ln z  1  .

technique as

z

z

z

as

z z z z z z z ln z e z  ze z Rewriting the solution in terms of the original variable x, we get e z  z ln z  1   x  ln x ln  ln x   1 .

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z

List of Selected Integrals with Their Step-by-Step Solutions • 61

INTEGRAL 42 Problem:

 ln  x

2

 x  1  dx

Solution: 1  2 1  2 x  1   x   ln  x  x  1   2 x  3 tan    constant 2   3  Techniques used: Change of variables, Integration by parts, Partial fractions Step-by-step solution: Using the integration by parts technique, we can write x  2 x  1 2 x2  x x  2 x  1 2 2 . But  2   1    1  ln x x dx x ln x x dx     x2  x  1   x2  x  1 x2  x  1  dg f

x  2 x  1 2x  x x2 2 2 , by using the partial fractions technique.  2 2 x  x 1 x  x 1 x  x1 x  2 x  1 x2 Therefore,   2 dx  2 x   2 dx. Now we rewrite x  x 1 x  x1 x2 1 2x  1 1 3 this integral as  2 dx   2 dx   2 dx. But x  x1 2 x  x1 2 x  x1 1 2x  1 1 dx  ln  x2  x  1 . Now, for performing the integral 2  2 x  x 1 2 1 2 3 3  4 1 3 2 dx, we rewrite the denominator as x  x  1  ( x  )   1   2  2 4 4  3  2 x  x 1 2 2 1 3 3  4 1  3  2 1   2x  1 3 x2  x  1  ( x  )2   1   x     1   x  z  dx  dz   . Now, let 2 4 4  3  2   4   3 3   2 3 2x  1 3 dz . Therefore, the integral an be written in  z  dx  2 3 2

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62 • 100 Integrals

1 3 1 3 3 dz  3 tan dx   dz  3  2  1  z2 2 x  x 1 2 3 2    1  z2  4 dz  2x  1  dz  3   3 tan 1 z  3 tan 1  . Collecting all the   2 1 z  3  terms of the z variable as

3 3 3 2    1  z2  4

1 calculated related terms, we get the answer as x ln  x2  x  1   2 x  ln  x2  x  1  2 1 2 2 1  2 x  1  x ln  x  x  1   2 x  ln  x  x  1   3 tan  . 2  3 

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List of Selected Integrals with Their Step-by-Step Solutions • 63

INTEGRAL 43 Problem:

 cos x

4  sin 2 xdx

Solution:    sin x  sin x 2 sin 1  4  sin 2 x   constant  4  2    Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Let sin x  2 sin z  cos x dx  2 cos z dz. We can rewrite the integral in terms of the variable z as  cos x 4  sin 2 xdx  4  cos z 1  sin 2 z dz  4  cos2 zdz 1  cos 2 z . Therefore, the integral can be 2 written as 4  cos2 zdz  2  1  cos 2 z  dz  2 z  sin 2 z. Now writing

x  4  cos z 1  sin 2 z dz  4  cos2 zdz . But cos2 z 

sin x  this result in terms of the original variable x, we get 2 z  sin 2 z  2 sin 1    2  2  sin x   sin x  1  sin x  2 z  sin 2 z  2 sin 1    sin x 1    . Note that z  sin   and 2 2      2  sin 2 z = 2 sin z cos z.

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64 • 100 Integrals

INTEGRAL 44 Problem:

 sin

2

cos x dx x  3 sin x  2

Solution:  sin x  2  ln    constant  sin x  1  Techniques used: Change of variables, Integration by parts, Partial fractions Step-by-step solution: Let sin x  z  cos xdx  dz and rewrite the integral in terms cos x 1 of the variable z to get  2 dx   2 dz. sin x  3 sin x  2 z  3z  2 But z2  3 z  2   z  2   z  1 . Therefore, the integral can be written as, using the partial fractions technique,

z

2

1 1 dz   dz    3z  2 z 2    z  1

1 1 1 1 dz   dz   dz   dz.Performing the integration  3z  2 z2 z 1  z  2   z  1 1 1 z2 of the integrals gives  dz   dz  ln  z  2   ln  z  1   ln z2 z 1 z 1 z2 n  z  2   ln  z  1   ln . After writing the results in terms of the original variable x z 1  sin x  2  we get ln  .  sin x  1 

z

2

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List of Selected Integrals with Their Step-by-Step Solutions • 65

INTEGRAL 45 Problem:



1

1  x 

x  x2

dx

Solution: 2

1 x 1 x

 constant

Techniques used: Change of variables Step-by-step solution:

x  z  dx  2 zdz. Rewriting the integral in terms of the 1 1  1  x x  x2 dx  2  1  z  1  z variable z we get, after some simplifications, 2 1 1  1  x x  x2 dx  2  1  z  1  z2 dz. Now, let 1  z  u  dz   1  z  1  z du. 1 z 1 z Substituting into the integral (writing the integrand in 1 1 terms of both z and u variables), we have 2  dz  2  2 1  z  1  z 1  z  1   2 1  z  1  z 1 1 2 dz  2  du  2u. Therefore, the answer 2 2 1 z 1  z 1  z 1  z  1  z   Let









1

in terms of the original variable x can be written as 2 u  2

100_Integrals.005_3pp.indd 65

1 z 1 x  2 . 1 z 1 x

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66 • 100 Integrals

INTEGRAL 46 Problem:

x

5

1  x 

2 3

dx

Solution: 1

4 x 1  x 2  4

2

 constant

Techniques used: Change of variables, Integration by parts Step-by-step solution: The denominator of the integrand can be written as x 5 1  x2   3

x 5 1  x 2   3

3 3 1 6 1 d x  1  x2    x2  x 4  . Noticing that  x2  x 4   2 x  4 x3,  x x dx a multiple of the numerator after multiplying it by x. Now let dz x2  x 4  z  dx  . Therefore, the integral can be written 2  x  2 x3 

in terms of the variable z as 

 2x

x

1  2 x2

3



4 3

1  2 x2

x 1  x 5



2 3

dx  

x  2 x3

x

2

x



4 3

3 1 6 1 x  1  x 2     x x

1 dz 1 1   3 dz  2 3 4z 2 x  2x  2 z

1 1 1 dz   3 dz  2 . Or, in terms of the original variable x we get 3 4z 2 x  2x  2 z 1 1 1   2 2 2 4 2 4 4z 4 x  x  4 x 1  x 2  . Please note that it is possible to calculate this integral without using the tip mentioned above. But it will be a lengthier operation.

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List of Selected Integrals with Their Step-by-Step Solutions • 67

INTEGRAL 47 Problem:

Solution:

∫ sin

6

x cos5 xdx

sin 7 x  63 sin 4 x  154 sin2 x  99   constant 693

Techniques used: Change of variables, Trigonometric identities Step-by-step solution: The integrand can be written, using trigonometric identities, as

sin 6 x cos5 x  sin 6 x 1  sin 2 x  cos x . Now let sin x  z  cos xdx  dz  2

cos4 x

sin x  z  cos xdx  dz. Therefore, the integral can be written in terms of variable z as

 sin

6

x cos5 x dx   sin 6 x 1  sin 2 x  cos x dx  z6 1  z2  dz .After 2

2

expanding the integral we get z6 1  z2  dz  z10 dz  2 z8 dz  z6 dz. 2

2

dz  z10 dz  2 z8 dz  z6 dz. Performing the integration operation for each term gives, 1 11 2 9 1 7 z  z  z . Rewriting the results in terms of the original 11 9 7 1 2 1 variable x gives sin11 x  sin 9 x  sin 7 x. Or simplify to have 11 9 7 sin 7 x  63 sin 4 x  154 sin2 x  99 . 693 Note that this integral is a version of the general recursive formula, given as: m n  sin ax cos ax dx 

sin m 1 ax cos n1 ax n  1  sin m ax cos n 2 ax dx a  m  n mn

We calculated this integral as given by Integral 24, using only integration by parts technique. Readers can verify that the result given in this section and that of section 24 are equivalent.

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68 • 100 Integrals

INTEGRAL 48 Problem:

∫x e

2 x3 / 2

dx

Solution: 2 x3 / 2 3 / 2 e  x  1   constant 3 Techniques used: Change of variables, Integration by parts Step-by-step solution: Let x3 / 2  z  dx 

2

dz. Therefore, the integral can be written 3 x 3/2 2  x2  z 2 z as x2 e x dx     e dz  ze dz . Using the integration by 3  x 3  x1.5  z

2 2 ze z dz   ze z  e z . Or in terms of the  3 3 2 x3 / 2 3 / 2 original variable x, we have e  x  1 . 3

parts technique we get

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List of Selected Integrals with Their Step-by-Step Solutions • 69

INTEGRAL 49 Problem: tan 3 x ∫ cos3 x dx Solution: 3  5 cos2 x  constant 15 cos5 x Techniques used: Change of variables, Integration by parts Step-by-step solution: 1 dz  z  dx  2 . Therefore, the integral can be written cos x z sin x tan 3 x sin 3 x 3  dz  2 2 in terms of the variable z as  dx  z  2   z  z  1  dz 3 3  cos x cos x  z sin x 

Let

sin 3 x 3  dz  2 2 z    z  z  1  dz. Performing the integration operation for the cos3 x  z2 sin x   z 5 z3 latter integral gives z2  z2  1  dz  z4 dz  z2 dz   . 5 3 Writing the results in terms of the original variable x, we get 1 1 1 z 5 z3 1     3  5 cos2 x     3 sin2 x  2 cos2 x  5 3 5 5 3 5 cos x 3 cos x 15 cos x 15 cos5 x 1 3 − 5 cos2 x 2 2 2  cos  3 sin x  2 cos x . This expression simplifies to . 3 5 x   15 cos5 x   15 cos5 x



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70 • 100 Integrals

INTEGRAL 50 Problem: 3

tan x

  sin x  cos x 

2

dx

Solution:  2 3 tan x  1  3 tan x 1 3 tan 1   ln  3 3   1  tan x 6 

3

tan x 1  ln 1  tan x 6



3







3







1 tan 2 x  3 tan x  1  ln 1  3 tan x  constant 3

Techniques used: Change of variables, Integration by parts Step-by-step solution: Let tan x  z  dx  cos2 xdz, and write the integral in terms of 3 3 3 tan x tan x z dx dz.  dx  variable z to get  2 2   2  sin x  cos x  1  tan x  cos x  1  z 2 3

tan x

 tan x  cos x 2

2

dx  

3

z

 1  z 2

dz. Now let z  u3  dz  3 u2 du, and write the integral

in terms of variable u along with applying the integration by parts 3 du z u 3 u2 technique, to get  . dz   u du  2 3  3 2 1 u 1  u3 1  z  1u  f dg

u  z  3 tan x   . For the latter 1  u3 1  z 1  tan x integral we use the partial fractions technique to have du 1 1 1 2u  1  u3  3  1  u du  3  u2  u  1 du. The former integral reads Hence, we can write

100_Integrals.005_3pp.indd 70



1 tan 2 x  3 tan x  1  ln 1  3 tan x  co 3

3

27-07-2023 16:27:41

List of Selected Integrals with Their Step-by-Step Solutions • 71





1 1 1 du 1 3  ln 1  u . Or ln 1  u   ln 1  tan x . Rewrite  3 3 3 1 u 3 2u 1 1  2u 1 1 the latter integral as 1 du   2 du   2 du 2  3 u  u1 6 u  u1 2 u  u1 1 1  2u 1 1 1 1  2u 1 1 du   ln  u2  u  1    ln 3 z2  du   2 du   2 du. But we have  2 6 u  u1 6 6 6 u  u1 2 u  u1 1 1  1 2 3 2 3 2 ln tan x  3 tan x  1 . For du   ln  u  u  1    ln z  3 z  1  1 6 6 6 1 1 1 1 the remaining integral, we rewrite it as du   2  2 u  u1 2  u  1 / 2 2











1 1 1 1 1 4 du   du  .  2 2  2  u  1 / 2  3 / 4 2 3 2 u  u1

Now let

1  2u  1  1  3  

2

du 

2 3

1  2u  1  1  3  

2

du.

2u  1 3  y  du  dy, and write the integral in terms 2 3

2 of variable y. Hence  3

1 2

du 

3 1 dy. But we  3 1  y2

 2u  1  1   3   23 z 3 3 3  2u  1  1 3 1 . Or, have 3  tan 1 y  tan 1  tan 1  tan  y  3 3 3  3  3 3  1  y2 3   23 z 1   2 3 tan x  1  3 3 3  2u  1   y tan 1  tan 1  tan 1  Collecting all  .  3 3  3 3  3  3   





1 nx 1  ln 1  3 tan x  ln 6 nx 3

100_Integrals.005_3pp.indd 71



3

3





tan x 1 1  ln 1  3 tan x  ln 1  tan x 3 6 3  2 tan x  1  3 tan 2 x  3 tan x  1  tan 1  . 3 3  

answers, we get the solutions as 



3

tan 2 x 



27-07-2023 16:29:24

72 • 100 Integrals

INTEGRAL 51 Problem:  1

1   dx 2 x

  ln x  ln Solution:

x + constant ln x Techniques used: Integration by parts Step-by-step solution:

1  dx dx  1 Rewriting the integral as sum of two parts, we get    2  dx    2 ln x ln x  ln x ln x  1  dx dx  1   ln x  ln2 x  dx   ln x   ln2 x . But the latter integral can be calculated, using the intedx xdx x 1  dx  gration by parts as  2    x  dx .   2 ln x x ln 2 x  f  x ln x ln ln x x    dg

d 1  1 Note that  . Now, by collecting all terms we have  dx  ln x  x ln 2 x dx dx 1 x 1 x  ln x   ln2 x   ln x dx  ln x   ln x dx  ln x .

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List of Selected Integrals with Their Step-by-Step Solutions • 73

INTEGRAL 52 Problem:

 sin  x  sin x   sin  x  sin x  dx Solution: 2 cos  sin x   constant Techniques used: Change of variables, Trigonometry identities Step-by-step solution:

Expand the integrand, sin  x  sin x   sin  x  sin x   sin x cos  sin x   cos x sin  sin x  sin x   sin x cos  sin x   cos x sin  sin x   sin x cos  sin x   cos x sin  sin x   2 cos x sin  sin x  cos x sin  sin x   2 cos x sin  sin x . Let sin x  z  cos xdx  dz and rewrite the integral in terms of the variable z, to get 2  cos x sin  sin x  dx  2  sin zdz  2 cos z n  sin x  dx  2  sin zdz  2 cos z. Rewriting the results in terms of the original variable x, we get 2 cos  sin x .

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74 • 100 Integrals

INTEGRAL 53 Problem: x

 1  5x

2

 2   1  dx  2  1  5x 

Solution: 1 sin 1  5 x2   constant 10 Techniques used: Change of variables, Integration by parts Step-by-step solution: Manipulating the integrand gives,

x x  2   1   2 2 2 1  5x  1  5x  1  5x

1  5 x2  1  5 x2

1  5 x2 x . Let 5 x2  z  10 x dx  dz and write  2 2 4 1  5x x 1  25 x x 1 1 the integral in terms of the variable z, to get  dx   dz 4 10 1  z2 1  25 x x 1 1 1 1  1  25 x4 dx  10  1  z2 dz. But  1  z2  sin z. Therefore, the result in 1 1 1 1 terms of the original variable x reads dz  sin 1 z  sin 1  2 10 1  z 10 10 1 1 1 1 dz  sin 1 z  sin 1  5 x2 . 10  1  z2 10 10 x  2   1   2 2  1  5x  1  5x

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27-07-2023 16:15:33



List of Selected Integrals with Their Step-by-Step Solutions • 75

INTEGRAL 54 Problem:



1

1 x  x 

2

dx

Solution: 1  1 x  tan x  2 1  x2

   constant 

Techniques used: Integration by parts, Change of variables, Trigonometric identities Step-by-step solution: Manipulating the integrand gives 0 

x2



x2 1  1

1  x  1  x  2 2

2 2



1 1  2 1  x 1  x 2 

1





0 

x 1  1 2



1  1  x  1  x  x  x   . Therefore, the integral can be written as 2 2

. 1 1   1 2 dx   1  x2 dx   1  x2 2 dx But   x  x  1

x

2 2 2

1

1 x

2

2 2

1 1  2 1  x 1  x 2

dx  tan 1 x .

dz and cos2 z 1 1 rewrite this integral in terms of the variable z to get  dx   2 2 2 1  x  1  tan2 z 

For performing the latter integral, let x  tan z  dx 



1

1  x 

2 2

1  cos  2 z  dz   cos2 zdz. But cos2 z  . There­2 2 1  tan2 z  cos z 1 1 1 fore,  cos2 zdz   1  cos  2 z   dz  z  sin 2 zz. Collecting all 2 2 4 related terms and rewriting the results in terms of the

dx  

100_Integrals.006_3pp.indd 75

1

2

27-07-2023 16:17:08

76 • 100 Integrals

1 1 1 1 1 1 1 1 original variable x gives tan x  tan x  sin  2 tan x   tan x  sin  2 ta 2 4 2 4 1 1 1 x  sin  2 tan 1 x   tan 1 x  sin  2 tan 1 x . But the last term in the latter expres4 2 4 1 1 1  1  sion can be written as sin  2 tan  x    sin 2   sin  cos .   2 4    4 x 1 But, having tan   x , we get sin  cos   . Therefore, the 2 2  2 x2 x  1 1 1 results of the integral can be written as tan 1 x  sin  2 tan 1 x    tan 1 x   2 4 2 1  x2  1 1 x  x  sin  2 tan 1 x    tan 1 x  . 4 2 1  x2 

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List of Selected Integrals with Their Step-by-Step Solutions • 77

INTEGRAL 55 Problem:

e

2x

tan 1  e x  dx

Solution: 1 1  e2 x  tan 1  e x   e x   constant  2 Techniques used: Chane of variables, Integration by parts Step-by-step solution: Let e x  z  dx  dz / z, and rewrite the integral in terms of the variable z to get e2 x tan 1  e x  dx  z tan 1 zdz. Using the integration z2 z2 1 1 1 by parts technique, we get  z tan z dz  z  dz . tan  2 2  1  z2 dg f

1 z2 d 1 tan 1 z   e2 x tan 1 e x . tan 1 z  . Therefore, 2 2 2 dz 1 z To calculate the remaining integral, rewrite it as 1 z2 1 z2  1  1 1 1 1 1 1   dz   dz   dz   dz    z  tan 1 z     e x 2 2 2  2 1 z 2 1 z 2 2 1 z 2 2 1 1 1 1 1   dz   dz    z  tan 1 z     e x  tan 1 e x . After collecting all obtained 2 2 1  z2 2 2 results, as underlined, and simplifying we get the solution as 1 1  e2 x  tan 1 e x  e x  .  2 Note that

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78 • 100 Integrals

INTEGRAL 56 Problem:

x

1 6 x  x2

dx

Solution: 

6 x  x2  constant 3x

Techniques used: Change of variables, Integration by parts Step-by-step solution:

2  x 2   2 6 x  x  9  x  3  9 1   1       The expression 6 x − x , can be written as     3 2  x 1 1 1 2   2  x  9   x  3   9 1    1   . Substituting into the integral, we get  dx   2 3     3 x 6 x  x2 x  x 1    1 1 1 1 x 3  dx. Now, let and rewrite the dx   1  z  dx  3 dz  x 6 x  x2 2 3 3 x  x 1    1 3  1 1 1 1 dx   d integral in terms of the variable z to get  2 2 3 3  z  z 1 1   x   x 1    1 3  1 1 1 dz. Now, Let and 1  z2  cos u. Therefore, dx   2 2 3  z  1 1  z x  x 1    1 3  1 1 1 1 du. Using the trigonometric dz    2 3 1  sin u 3  z  1 1  z 2

identity sin u 

100_Integrals.006_3pp.indd 78

2 tan  u / 2  1 , we get  1  tan 2  u / 2  3

2 1 1 1  tan  u / 2  du   2 tan  u / 2  3 1  tan  u / 2  2 1 1  tan 2  u / 2 

27-07-2023 16:21:32

List of Selected Integrals with Their Step-by-Step Solutions • 79

2 2 1 1 1  tan  u / 2  1 1 / cos  u / 2  du    du.   du 2 tan  u / 2  3 1  tan  u / 2  2 3  1  tan  u / 2  2 1 1  tan 2  u / 2  Let 1  tan  u / 2   y  du  2 cos2  u / 2  dy, and the solution of the integral in terms of the variable y reads, after substitution, 2 2 2 1 / cos  u / 2  2 1 1 / cos  u / 2  2 1 2 du  cos  u / 2  dy   2 dy   2 2   y 3 3 1  tan  u / 2   3 y 3y

2 1 2 dy   . The answer can be written in terms of the original variable 2  3y 3 y 2 2 sin  u / 2  u u, after substitution, as   . But tan     3y   u   2  cos  u / 2  2 c 3  1  tan     2   2 1  cos u sin u sin u 2  u  sin  u / 2  tan      , and   2 3 1  sin u  c   2  cos  u / 2  2 cos  u / 2  1  cos u  u  3  1  tan     2   2 1  cos u  2 z . Therefore, in terms of the variable , having   3 1  sin u  cos u    u  3  1  tan    2 1  1  z2  2  2 1  cos u   z = sin u, we get   . Finally, 3 1  sin u  cos u  3 1  z  1  z2  21  1  2  2 1 1 z   in terms of the original variable x, we get  2  3 1 z 1 z x 2   x 31   1      2 1  1   1  3     2 1  1  z2  3      . This expression simplifies to 2   3 1  z  1  z2 x x   31   1  1    1   3 3    

os2  u / 2  dy 



























x  6 x  x2 1 6 x  x2   . 3x 3 3x

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80 • 100 Integrals

INTEGRAL 57 Problem:

∫ tan

3

xdx

Solution: 1 tan 2 x  ln  cos x   constant 2 Techniques used: Integration by parts, Trigonometric identities Step-by-step solution: d 1  tan x   2 , we use cos x dx nique to get tan 3 xdx   dx     cos2 x     Having

dg

the integration by parts tech-

 sin 3 x   3 cos2 x sin 2 x  sin sin 3 x x  tan x  tan     cos x cos2 x  cos x    f

 sin 3 x   3 cos2 x sin 2 x  sin 4 x  . But the integrand, after expanding n x    tan x   cos2 x  cos x    sin x  3 cos2 x sin 2 x  sin 4 x  sin 3 x and some manipulations, reads  3 cos2 x  sin   3 cos x  cos2 x cos x  2 x sin 2 x  sin 4 x  sin 3 x 2 2 3 2 2 3 3 cos x  sin x   tan x  3 cos x  1  cos x   tan x  2 tan 3 x cos2 x   2 3 cos x  cos x  sin 3 x  1  cos2 x   tan 3 x  2 tan 3 x cos2 x . After back substitution, we get  tan 3 xdx  tan x      ta  cos x   sin 3 x   sin 3 x  3 3 2 3 3 3 2 tan xdx  tan x tan x 2 tan x cos x dx  tan x           tan x  2  tan x cos x    cos x   cos x  3  sin x  3 3 2 an x     tan x  2  tan x cos x. Or, after rearranging the terms in the last  cos x 

100_Integrals.006_3pp.indd 80

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List of Selected Integrals with Their Step-by-Step Solutions • 81

 sin 3 x  1 3 2 3 xdx  x tan tan     tan x cos x. But  2 cos x   sin x sin x 3 2 2   tan x cos x     sin x  dx   cos x dx   sin x cos xdx. cos x 1 sin x dx  ln  cos x  and  sin x cos xdx  sin 2 x . Therefore, But   2 cos x the final answer, after collecting all related terms, reads  sin 3 x  1 1 2 3 tan xdx  tan x    ln  cos x   sin x. The answer can  2 2  cos x  3  sin x  1 1 2 1 2  sin 2 x  be simplified to tan x     ln  cos x   sin x  sin x  1  2 2 2 cos2 x   cos x    1 1 1 sin 2 x  2 cos x   sin 2 x  sin 2 x  1    ln  cos x   tan x  ln  cos x . 2 cos x 2 2 2   equation, we get

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82 • 100 Integrals

INTEGRAL 58 Problem:

1

 4  5cos x dx Solution:

 tan  x / 2   2 tanh 1    constant 3 3  

Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution:  x  x Let tan    z  dx  2 cos2   dz. Now, using the trigonometric 2 2 x 1  tan 2 2 , rewrite the integral in terms of variable z identity cos x  2 x 1  tan 2  x cos2   1 1 2 as dz  2  dz.    4  5 cos x dx  2   9  z2 2 x   tan 1  2 4  5 x  1  tan 2  2  1 2 1 z But 2  dz   dz. Now let  u  dz  3 du, and 2 2 3 9z 9  z 1  3 rewriting the latter integral in terms of the variable u gives 2 2 1 1 1 dz   du. But  du  tanh 1 u. Therefore, 2 2 2  u 3 1 9 1 u  z 1  3   writing the answer in terms of variables z and then x gives  tan  x / 2   2 2  z 2 tanh 1 u  tanh 1    tanh 1  . 3 3 3 3 3  

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List of Selected Integrals with Their Step-by-Step Solutions • 83

INTEGRAL 59 Problem:

tan 3 1  ln x  dx  x

Solution: 1 tan 2 1  ln x   ln cos 1  ln x    constant 2 Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Let ln x  z  dx  e z dz and rewrite the integral in terms of the vari­ tan 3 1  ln x  tan 3 1  z  z able z, to get  dx   e dz  tan 3 1  z  dz. x ez Now using the results from Integral 57, we can write the answer as 1 tan 2 1  z   ln  cos 1  z  . Or in terms of the original variable x, 2 1 we have tan 2 1  ln x   ln  cos 1  ln x  . 2

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84 • 100 Integrals

INTEGRAL 60 Problem: 2

 sin 2 x sin 3 x    sin x sin 6 x  dx Solution:  3   1  3 tan x  4  ln    constant  18   1  3 tan x  1  3 tan 2 x  Techniques used: Integration by parts, Trigonometric identities, Change of variables Step-by-step solution: Simplify the expression inside the bracket of the integrand to sin 2 x sin 3 x 2 sin x cos x sin 3 x cos x 3 = = . But we have cos 3 x  4 cos x  3 cos x sin x sin 6 x 2 sin x sin 3 x cos 3 x cos 3 x cos 3 x  4 cos3 x  3 cos x. Therefore, the integral can be written as 2 1  sin 2 x sin 3 x  dz   sin x sin 6 x  dx   4 cos2 x  3 2 dx. Now, let tan x  z  dx  cos2 xdz  1  z2   dz and write the integral in terms of the variable z as 1  z2 1 1  z2  dx  4 cos2 x  3 2  1  3 z2 2 dz. Using the partial fractions    

tan x  z  dx  cos2 xdz 

technique,

2

dz 

we

get



1 dz 1 dz 1 dz   6  1  3z 6  1  3z 3  1  3z



100_Integrals.006_3pp.indd 84



1  z2

1  3 z 

2 2

2



dz  

1 

1 dz . 3  1  3z 2





1  z2 3z

 1 

The

2

3z

first



2

dz 

1 dz 1    6 1  3z 6 1

two

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List of Selected Integrals with Their Step-by-Step Solutions • 85



1 1 dz dz integrals in the latter expression read 1  ln 1  3 z    6 1  3z 6 1  3z 6 3

 1  3 tan x   1  3z  1 1 1 1 dz dz 1 ln 1  3 z  ln 1  3 z  ln     ln    1  3z 6 1  3z 6 3 6 3 6 3  1  3 z  6 3  1  3 tan x   1  3z   1  3 tan x  1 1 z  ln  ln    du 6 3  1  3 z  6 3  1  3 tan x  . For the remaining integral, let 1  3 z  u  dz  3 du 1  3 z  u  dz  and write the integral in terms of variable 3 1 dz 1 du 1 1 1     u as 2 2   3 1  3z 3 3 u 3 3u 3 3 1  3z 3 3 1  3 ta





















 3 1 1 1 . Or after simplification reads .   9 1  3 tan x 3u 3 3 1  3 tan x 3 3 1  3z











Similarly, the last integral can be worked out to have 1 dz  3 1  3z





2





3

9 1  3 tan x





1 dz  3 1  3z





2





3

9 1  3 tan x

. After collecting all underlined related answers

we get the solutions as

 1  3 tan x  3 3 ln    6 3  1  3 tan x  9 1  3 tan x 9 1  3 tan x 1



 

an x  3 3  . After some simplifications we get the answer as  an x  9 1  3 tan x 9 1  3 tan x



 



 3   1  3 tan x  4  ln   2 . 18   1  3 tan x  1  3 tan x 

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86 • 100 Integrals

INTEGRAL 61 Problem:

∫

sin 3 x dx cos x

Solution: 2 cos x cos2 x  5   constant  5 Techniques used: Change of variables, Trigonometric identities Step-by-step solution:  sin x , rewrite the integral as 2 cos x sin 3 x   sin x  2  cos x dx  2   2 cos x dx  sin x. Now, using the inte­   sin x  2 tion by parts technique, we get 2   dx  sin x  2 2 cos x    f gra

Having

d dx





cos x 



 sin x





cos x sin 2 x  2

dg

 2 2 dx  sin x  2 cos x sin x  2  cos x sin x cos x dx . But the latter integral cos x  f    4 dg can be worked out as 2  cos x sin x cos x dx  2  cos3 / 2 x sin x dx  cos5 / 2 x 5 4 2  cos3 / 2 x sin x dx  cos5 / 2 x. After collecting all related terms, we get the answer as 5 4   2  cos x sin 2 x  cos5 / 2 x . This expression can be simplified to 5   5   2 cos x 4 2 cos x 5 1  cos2 x   4 cos2 x  2  cos x sin 2 x  cos 2 x   cos2 x  5   5 5 5  





cos2 x   4 cos2 x 



2 cos x  cos2 x  5 . 5

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List of Selected Integrals with Their Step-by-Step Solutions • 87

INTEGRAL 62 Problem:

 3x

5 x  31 dx  4 x  11

2

Solution: 5 103  3x  2  ln  3 x2  4 x  11   tan 1    constant 6 3 29  29  Techniques used: Change of variables, Integration by parts Step-by-step solution: d 3 x2  4 x  11   6 x  4 , we can rewrite the numerator  dx 5 103 as 5 x  31   6 x  4   . Therefore, the integral can be written 6 3 5 x  31 5 6x  4 103 1 as  2 dx   2 dx  dx. 3 x  4 x  11 6 3 x  4 x  11 3  3 x2  4 x  11 5 6x  4 5 dx  ln  3 x2  4 x  11 . For calcula­ But 2  6 3 x  4 x  11 6 ting the latter integral, rewrite the denominator as 2 2 2  4 2  29   2 3 x  4 x  11   3 x     11   3 x    3 . Now let 3 3 3   Having

2 3  z  dx  dz and the integral in terms of varia­ 3 3 103 1 103 3 1 103 ble z reads dx  dz  2   29 3 3 x  4 x  11 9 29 3  1   z2 3 29 3 1 1 103  29 2 dz  29 3  1  3 / 29 z 2 dz. Now let 3 / 29 z  u  dz  3 du, thus the z 3 3x 



100_Integrals.007_3pp.indd 87



1 3 / 29 z



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1

88 • 100 Integrals

103 integral in terms of variable u reads 29 3  1



1 3 / 29 z



2

dz 

103 29 3



1 3 / 29 z



2

dz 

103 29 3

29 1 d 2  3 1 u

1 29 1 103 1 du  tan 1 u. du  du. But  2 2 2   1 u 3 1 u 3 29 1  u

 103 1 103 103 103 du  tan 1 u  tan 1 3 / 29 z  tan 1  3 / 2 2  3 29 1  u 3 29 3 29 3 29    2 x 3 2 103 103 103        tan 1  tan 1 3 / 29 z  tan 1  3 / 29  3 x  .   3 29 3 3 29 29    29   5 103  3x  2 Collecting all related terms, we get the answer as ln  3 x2  4 x  11   tan 1  6 3 29  29 5 103  3x  2  ln  3 x2  4 x  11   tan 1  . 6 3 29  29 

Therefore,









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List of Selected Integrals with Their Step-by-Step Solutions • 89

INTEGRAL 63 Problem: 3 x 5  x 4  2 x3  12 x2  2 x  1



x

3

 1

2

dx

Solution: ln  x3  1  

x2  x  3  constant x3  1

Techniques used: Partial fractions, Change of variables, Integration by parts Step-by-step solution: Having the algebraic identity x3  1   x  1   x2  x  1 , the denom­ inator polynomial common factors are  x  1 ,  x  1  ,  x2  x  1 , 2

x

 x  1  . Using the partial fractions technique we form the following equation, which should be satisfied, for specific values of unknown polynomials P  x   P, Q  x   Q, R  x   R, S  x   S: 2

2

3 x 5  x 4  2 x3  12 x2  2 x  1

x

Q

 x  1

2



3

 1

2



Q P R S   2  2 2 x  1  x  1   x  x  1   x  x  1 2

R S  . After some manipulations (i.e., equating equal 2  x  x  1   x  x  1 2 2

power of x coefficients) the expressions for the unknown polynomials can be obtained as P  Q  1, R  2 x  1, and S  4 x  2. Therefore, the integral can be written as 3 x 5  x 4  2 x3  12 x2  2 x  1 1 1 2x  1 dx   dx   dx   2 dx  2  2 2  3 x 1 x  x1  x  1  x  1  x2

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1

90 • 100 Integrals

dx  

1

 x  1

2

dx  

2x  1 2x  1 dx. But dx  2  2 2 x  x 1  x  x  1



2

1

 x  1 2

x

x

For

dx, let x  1  z  dx  dz and rewriting the integral

in terms of variable z gives   For

1

 x  1 dx  ln  x  1.

2x  1 dx,  x 1

2

having

2x  1 dx  ln  x2  x  1  .  x 1

2

1

 x  1

2

dx   

1 1 1 dz   . 2 z z x 1

d 2  x  x  1  2 x  1, we dx 2x  1 For 2 dx 2  x2  x  1 

get let

du and rewrite the integral as 2x  1 2 x  1 du 1 2 2 . dx  2  2  2  2 du     2 u 2x  1 u u x  x 1

x2  x  1  u  dx  2

x

2x  1 2

 x  1

2

After collecting all related terms, we get the answer as 1 2 ln  x  1    ln  x2  x  1   2 . This expression can be x 1 x  x 1 x2  x  3 . simplified to ln  x3  1   x3  1

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List of Selected Integrals with Their Step-by-Step Solutions • 91

INTEGRAL 64 Problem: 4 x3  x  1  x3  1 dx Solution: 1  x2  x  1  4  2x  1  4 x  ln   tan 1    constant 2  3   x  1   3  3  Techniques used: Partial fractions, Change of variables, Integration by parts Step-by-step solution: Having the order of polynomials for both numerator and denom­ inator of the integrand being equal, we apply division operation 4 x3  x  1 x3 to get 4 3 . Therefore, the integral can be writ­ 3 x 1 x 1 4 x3  x  1 x3 dx  4 dx   3 dx. But 4 dx  4 x and for ten as  3 x 1 x 1 calculating the remaining integral, we use partial fractions method; Q x3 x3 P   2 . Therefore, to main­  3 2 x  1  x  1  x  x  1 x  1 x  x  1 tain the equality we must have x  3  P  x2  x  1   Q  x  1  which requires to have Q  Ax  B. Therefore, after equat­ ing the coefficients of equal power of x on both sides of the A  P  0  equality, we have  A  B  P  1 . The solution for this s­ystem B  P  3  of equations is P   A  2 / 3 and B = 7 / 3 . Using these v­alues,

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92 • 100 Integrals

we

can

write



x3 2 1 2x  7 dx . dx    dx   x3  1 3 x 1 3  x2  x  1 

2 1 2 dx   ln  x  1 . For the remaining integral,  3 x 1 3 2x  7 1 2x  7 1 2x  1  6 1 2x  1 dx   2 dx   2 dx   2 rewrite it as  dx  2  2 3 x  x 1 3 x  x 1 3 x  x 1 3  x  x  1 But  

1 2x  1  6 1 2x  1 1 dx. dx  2  2 dx   2 2  3 x  x1 3 x  x 1 x  x 1

But

1 2x  1 1 dx  ln  x2  x  1  2  3 x  x 1 3

2x  1 1 d 2 dx  ln  x2  x  1 . Note that  x  x  1  2 x  1. Now for per­  x 1 3 dx forming integration for the remaining integral, rewrite it as 1 1 1 8 1 2  2 dx  2  dx  2  dx    dx 2 2 x  x 1 3  2 x  1 2 1 1 1 3   x   1 x     1 2 4 2 4    3  1 8 1 2x  1 3 2 dx. Now, let  z  dx  dz and we can dx    2 2 3  2x  1  2 3 1 3  x     1 2 4   3  write the last integral in terms of the variable z as 1 8 1 4 3 1   dz  tan 1 z. dx   dz. But  2 2 2  z 1 z 1 3  2x  1  3   1  3  4 3 1 4 3 4 3  2x  1  dz   tan 1 z   tan 1  Therefore,   2  3 z 1 3 3  3 . Collecting all related terms, we arrive at the solution of the original 1 4 3 2  2x  1  integral as 4 x  ln  x  1   ln  x2  x  1   tan 1  . 3 3 3  3 

x

2

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List of Selected Integrals with Their Step-by-Step Solutions • 93

INTEGRAL 65 Problem:

 sin

1

x ln xdx

Solution: 1  x2  ln x  2   x sin 1 x  ln x  1   tanh 1





1  x2  constant

Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution:

Using integration by parts, we can write the integral as  dx  1 1 sin 1 x dx  . Therefore, we  x ln x dx  ln x  sin x dx     sin  x   dg f 1 1 require to have  sin x dx. Let sin x    sin   x and cos d  dx . Therefore, we can write  sin 1 x dx   cos  d   sin    sin    cos  in 1 x dx   cos  d   sin    sin  d . In terms of the original variable x, we  







 cos 

 sin

x dx  x sin 1 x  1  x2 . Now, back to the original dx  1 problem, we can write  sin 1 x ln xdx  ln x  sin 1 xdx     x  sin x dx   1  dx   ln x  sin 1 xdx    sin 1 x dx   ln x x sin 1 x  1  x2   x sin 1 x  1  x2 dx .  x x   get



1









Expanding the remaining integral, we get



x sin 1 x  1  x 2 dx   sin 1 x dx  



 x  x sin 1



1



x  1  x 2 dx   sin 1 x dx 

1  x2 1  x2 dx  x sin 1 x  1  x 2   dx.   x x

For the latter integral, let x  sin z  dx  cos z dz. Rewriting

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94 • 100 Integrals

1  x2 1  sin 2 z dx   x  sin z cos zdz cos2 z 1  sin 2 z 1 cos2 z 1  sin 2 z dz   dz   dz   sin zdz cos zdz   dz. But  sin z sin z sin z sin z sin z 

the integral in terms of variable z, gives



1  x2 dx   x

1  sin 2 z 1 dz   sin zdz. For the former integral, let cos z  u   sin zdz  du dz   dz   sin z sin z 

 cos z

 cos z

cos z  u   sin zdz  du and rewrite the integral in terms of varia­ 1 1 1 ble u as  dz    2 du    du. But we have sin z sin z 1  u2 1 1  1  u2 du  tanh u and rewriting in terms of the original variable x, we get tanh 1 u  tanh 1  cos z   tanh 1 1  x2 . After collecting all related terms, we get the answer as ln x x sin 1 x  1  x2  x sin 1 x  1  x2  1  x2  tanh 1





1  x2  tanh 1

  1 x  2

 





1  x2  1  x2

1  x2 . This expression simplifies to 1  x2  ln x  2   x sin 1 x  ln x  1   t

1  x2  ln x  2   x sin 1 x  ln x  1   tanh 1

100_Integrals.007_3pp.indd 94









1  x2  1  x2 .

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List of Selected Integrals with Their Step-by-Step Solutions • 95

INTEGRAL 66 Problem:

 ln  sin x 

1  sin xdx

Solution:  x x x x   x x      1   sin  cos  ln 4  2  sin  cos   ln  sin   ln  cos   2   2  tanh  sin 2 2   2  2     2    2

x   x x   x x      os   ln  sin   ln  cos   2   2  tanh 1  sin   tanh 1  cos    constant 2   2  2   2    2  

2

Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution:

After writing the trigonometric functions in half-angle form, we get x x x x   2 x 2 x  ln(sin x) 1  sin x dx   ln  2 sin 2 cos 2  sin 2  cos 2  2 sin 2 cos 2 dx   ln  2 2

sin 2

x x x x x x  x x x x  x x    cos2  2 sin cos dx   ln  2 sin cos   sin cos  dx   ln  2 sin cos   sin cos  dx. 2 2 2 2 2 2 2 2 2 2 2 2       

x x  x x  dx   ln  2 sin cos   sin cos  dx. Expanding the logarithm expression, we get 2 2  2 2  x x  x x  ln  2 sin cos    ln 2  ln sin  ln cos . Therefore, we can 2 2  2 2  write the integral in terms of summation of its parts, as x x  x x x x  ln 2   sin dx   cos dx , ∫ sin ln sin dx, ∫ cos ln sin dx , 2 2  2 2 2 2  x x x x ∫ sin 2 ln cos 2 dx, ∫ cos 2 ln cos 2 dx. Now we perform the integration

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96 • 100 Integrals

operation for new integrals. Therefore, x

x

 cos 2 dx  2 sin 2 .

x

x

 sin 2 dx  2 cos 2 ,

Using the integration by parts technique,

x cos2 x x x x 2 dx. But we have  sin ln sin dx  2 cos ln sin   x 2 2 2 2 sin 2 x x cos2 1  sin 2 1 x 1 2 dx  2 dx  dx, dx   sin dx . But for ∫  x   x x x 2  sin sin sin sin x 2 2 2 2  2 cos 2

1 x x let cos  z   sin dx  dz and rewrite the integral in terms 2 2 2 1 1 1 of variable z, we get  dx  2  dz  2  dz  2 tanh 1 z  2 tanh 2 x  1 z 2 x sin sin 2 2 1 x  dz  2  dz  2 tanh 1 z  2 tanh 1  cos . Similarly, we can perform the inte­ 1  z2 2   gration process for the remaining integrals. These results are tabu­ lated as shown below: Integral

Answer x

x

x x  x x   2   cos ln sin  cos  tanh 1  cos   2 2 2 2   

x

x

x x x  2  sin ln sin  sin  2 2 2 

x

x

x x x  2   cos ln cos  cos  2 2 2 

x

x

x x  x  x  2 sin ln cos  sin  tanh 1  sin   2 2 2    2

∫ sin 2 ln sin 2 dx ∫ cos 2 ln sin 2 dx ∫ sin 2 ln cos 2 dx ∫ cos 2 ln cos 2 dx

After collecting all related terms and simplifying, we receive the answer as shown above, as solution.

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List of Selected Integrals with Their Step-by-Step Solutions • 97

INTEGRAL 67 Problem:



x3 esin

1

x

1  x2

dx

Solution: 1

3 esin 10

x

  x3   1  x2  1  x2   constant x  3  

Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Having to get



d 1 sin 1 x   , use the integration by parts technique  dx 1  x2 x3 esin

1

x

1 1 1 dx dx    x3 esin x  x3 esin x  3 x2 esin x dx . For 2 f 1 x  x  1

2

dg

1

the new integral, let sin x  z  sin z  x, dx  cos zdz, and write the integral in terms of variable z as 3 x2 esin x dx  3  sin 2 z cos ze z dz. 1

After applying the integration by parts technique, we get  sin 3 z  z 3 z 3  sin 2 z cos ze z dz  3   e   sin z e dz. For the new  3  3 1 integral, after substituting sin 3 z  sin z  sin  3 z , we have 4 4 3 z 1 z 3 z  sin z e dz  4 e sin z dz  4 e sin  3 z  dz. Now we have two integrals to calculate. For ∫e z sin z dz, apply the integration by

parts technique twice to get e z sin z dz   e z cos z  e z cos zdz   e z cos z  e z sin z

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98 • 100 Integrals

os z  e z cos zdz   e z cos z  e z sin z  e z sin z. Now, after rearranging the terms and rewriting this expression as 2 e z sin z dz   e z cos z  e z sin z, we get ez z e sin z dz   sin z  cos z . Or in terms of the original variable x we  2 1 ez esin x have  sin z  cos z   x  1  x2 . Similarly, apply the same 2 2 integration process to the remaining integral, e z sin 3 z  dz. Therefore, ez 1 1  ez  1 z z z 1  cos 3 z    sin 3 z  e we get e sin  3 z  dz  e   cos 3 z   e cos 3 z dz  3 3 3 3  3  3 z z  e 1 1e 3 z dz   cos 3 z    sin 3 z  e z sin 3 z dz . Now, after rearranging the terms and 3 3 3 3  ez writing this expression, we get e z sin  3 z  dz   sin 3 z  3 cos 3 z . 10 1 z esin x e Or in terms of the original variable x we have  sin 3 z  3 cos 3 z   sin  3 sin 1 x  10 10 1 esin x 1 1 sin  3 sin x   3 cos  3 sin x  . Collecting all related terms  sin 3 z  3 cos 3 z   10 1 3 1 1 gives the answer as esin x x  1  x2  esin x sin  3 sin 1 x   3 cos  3 sin 1 x   .   8 40













n 1 x



sin  3 sin 1 x   3 cos  3 sin 1 x   . Using the expansion of the trigonometric functions,   we can simplify the answer as



 1  x2 





1 1 3 sin1 x e x  1  x2  esin x   x3  3 1  x2  1  8 40 1

1 sin1 x  3 3 esin e  x  3 1  x 2  1  x 2  3 x 1  x 2   9 x 2 1  x 2     10 40

x2 1  x2   

1

3 esin 10

x

x

1 3  2 2  x  3 x  1  x 1  x 

1 3  2 2   x  3 x  1  x 1  x  .

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List of Selected Integrals with Their Step-by-Step Solutions • 99

INTEGRAL 68 Problem: tan x



1  sec 3 x

dx

Solution: 2  tanh 1 1  sec 3 x  constant 3 Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Rewrite the integrand as Let

tan x

the variable z to get



sin x / cos x



sin x

. 1  cos3 x cos x and write the integral in terms of

1  sec 3 x

1 dz  z  dx  2 cos x z sin x



sin x

1  1 / cos3 x

dx  

dz 1  dz z  1 / z2 z sin x z 1  z3 sin x

2

1  cos3 x cos x du sin x 1 dz 3 . Now let 1  z  u  dz  2 . Therefore, we  dz  2 z2 sin x 3 3z z1 / z z 1 z 1 1 1 dz   u  y  du  2 ydy du. Let have  3 3  u  1 u z 1 z and rewrite the latter integral in terms of the variable y as, y 1 1 2 2 1 2 dy    dy   tanh 1 y. du   2 2  3  y  1 y 3 1y 3 3  u  1 u

2 tanh 1 3

After several substitutions, we can write the answer in terms of 2 2 2 2 1 1 u   tanh 1 1  z3   tan the original variable x as  tanh y   tanh 3 3 3 3 2 2 1 3 1 3 u   tanh 1  z   tanh 1  sec x. 3 3

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100 • 100 Integrals

INTEGRAL 69 Problem:





x ln x  x2  1 x 1 2

 dx

Solution:





x2  1 ln x  x2  1  x  constant Techniques used: Change of variables, Integration by parts Step-by-step solution:

the











x d x  x2  1 x  x2  1  1   and rewriting dx x2  1 x2  1 x x   1  1, we have the integral as follows: 2 2  x 1 x 1 0

Having

x ln x  x2  1 x2  1



 dx 



 





   1  1  ln x  x2  1 dx    1  x2  1   x









  ln x  x x2  1  x



 x  2 2   ln x  x2  1 dx    1   ln x  x  1 dx   ln x  x  1 dx . 2 x 1   Now we have two integrals to calculate. For the lat­ ter integral, use the integration by parts technique to get  x  1  2  x 1    ln x  x2  1 dx   x ln x  x2  1  x  dx   x ln x  x2  1   2  x  x  x 1 1  2  dz x x 1  2 x dx   x ln x  x2  1   dx. Now let x  1  z  xdx  2 x  x2  1 x2  1 and write the remaining integral in terms of variable z x 1 1 as  dx   dz  z, or x2 − 1 . For the former 2 2 z x 1

 

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List of Selected Integrals with Their Step-by-Step Solutions • 101

integral,

let



 x  x2  1  u   1  



  dx  du. x 1  x

2

Hence



 2    ln x  x  1 dx   ln udu  u ln u  u. x 1  2 Therefore, in terms of the variable x, we get u ln u  u  x  x  1 ln x

  1 



2

  x 1  

  1  x 





u ln u  u  x  x  1 ln x  x  1  1 . Collecting all related terms gives 2



 x ln x 

  ln  x 

 

2



2

x2



 

 



x2  1 ln x  x2  1  1  x2  1 ln x  x

x2  1  1  x2  1 ln x  x2  1  x.

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102 • 100 Integrals

INTEGRAL 70 Problem:

 sin 



1

1  x dx

Solution: 1 1  2 x  sin 1  2





1  x   x 1  x    constant 

Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Let

1  x  z  dx  2 zdz and write the integral in terms of

the variable z to get

 sin  1



1  x dx  2 z sin 1 z dz. Using

1 2 1 the integration by parts technique, we have 2  z sin z dz  z sin z   dg

1 2 1 2  z sin z dz  z sin z   dg

f

z2

1

rewrite it as  

d 1 sin 1 z   . Therefore,  dz 1  z2 1  x . To calculate the remaining integral,

1  z2

dz. Note that

1  z2 z sin z  1  x  sin 1 2

f

z2

z2 1 z

2

z

dz  z

by parts technique to get

 z f

1  z2  z

dz and apply the integration

z 1     2

dz  z 1  z2   1  z2 dz.

dg

Therefore, z 1  z2   x 1  x  . To calculate the latter integral, let z  sin u  dz  cos u du and write the integral in terms of the 1  cos 2 u variable u to get   1  z2 dz    cos2 u du. But cos2 u  . 2

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dz



List of Selected Integrals with Their Step-by-Step Solutions • 103

1 u sin 2 u . Therefore, 1  cos 2 u  du    2 4 2 1 1 u sin 2 u 1      u  sin u cos u    sin 1 z  z 1  z2   sin 1 1  x   x 2 4 2 2 2 1   sin 1 1  x   x 1  x  . After collecting all obtained results, 2 1 1 as underlined, and simplifying we get the solution as 1  2 x  sin 1  x   x 1 2 Hence,   cos2 u du  





1 sin 1 z  z 1  z2 2











1 1  2 x  sin 1 1  x   x 1  x   .  2

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104 • 100 Integrals

INTEGRAL 71 Problem: 1

 2  2 sin x  cos x dx Solution:

 tan  x / 2   1  ln    constant  tan  x / 2   3 

Techniques used: Change of variables, Trigonometric identities, Partial fractions Step-by-step solution: The integral can be written in terms of half-angle using the trigo2 tan x / 2 1  tan 2 x / 2 nometric identities sin x  and cos x  as 1  tan 2 x / 2 1  tan 2 x / 2 1 1 1 dx   2 2  2  2 sin x  cos x dx   1  tan x / 2 4 tan x / 2 tan x / 2  4 tan x / 2  3 2  2 2 1  tan x / 2 1  tan x / 2 dx 1 x x dx   . Now let tan  z  dx  2 cos2 dz 2 2 2 1  tan x / 2 tan x / 2  4 tan x / 2  3 cos x / 2 2 2  x 2x/2 x 1  tan tan  z  dx  2 cos2 dz and write the integral in terms of the variable z 2 2 1 dz. Using the partial fraction technique, we as 2  2 z  4z  3 1 1 1 1 dz  2  dz   dz   dz. get 2  2 z  4z  3 z1 z3  z  3   z  1 Performing the integration for the new integrals gives 1 1  z  1 dz  ln  z  1 and  z  3 dz  ln  z  3 . Collecting results z+1 . Rewriting the answer in terms of the gives the answer as ln z+3 z1  tan x / 2  1  original variable x gives ln  ln  . z3  tan x / 2  3 

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List of Selected Integrals with Their Step-by-Step Solutions • 105

INTEGRAL 72 Problem: sec 2 x  tan2 x  2 tan x  2 dx Solution: tan 1 1  tan x   constant Techniques used: Change of variables, Integration by parts Step-by-step solution: Rewrite the denominator as tan 2 x  2 tan x  2  1  1  tan x  . 2

Now let 1  tan x   z  dx sec 2 x  dz and write the integral in sec 2 x 1 terms of the variable z to get  dx   dz. 2 tan x  2 tan x  2 1  z2 1 But  dz  tan 1 z, and in terms of the original variable x the 1  z2 answer reads tan 1 z  tan 1 1  tan x .

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106 • 100 Integrals

INTEGRAL 73 Problem:



x2  x  1 dx

Solution:  3 2x  1  2x  1 2 x  x  1  tanh 1    constant 2 4 8  2 x  x 1  Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution:

2 2 3  1  3   2x  1     x    1     4 2 4 3        2 2 3  1  3   2x  1   2x  1 3  1    x    1    z  dx  dz and rewrite  . Now, let 4  2 4   3   2 3 2 3   2x  1   2 z the integral in terms of variable , as  x  x  1 dx   1   4   3  

The expression x2 + x + 1 can be written as x2  x  1 



3   2x  1  1  4   3 

 3 2  dx   1  z dz. For calculating the new 4  du integral, let z  tan u  dz  and rewrite the integral in terms cos2 u 3 3 du 3 1 of variable u, as  1  z2 dz   1  tan 2 u   du. 2 cos u 4 cos3 u 4 4 Using the integration by parts technique, we have du 3 1 3 1 3  tan u tan u sin u     du   du  . But 3 2  cos cos u 4  cos u 4 cos u 4  u cos2 u  

x  x  1 dx   2

2

f

dg

3 1 3  tan u  3 1 3 tan u sin u sin u 1  cos u du   du     . Therefore,    3 3 2 3 3 4 cos u 4  cos u  4 cos u 4 cos u cos u cos u 2

100_Integrals.008_3pp.indd 106

2

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List of Selected Integrals with Their Step-by-Step Solutions • 107

1 3  tan u  3 1 3 1 du   du   du. Rearranging terms involved in the  cos3 u 4  cos u  4  cos3 u 4 cos u 3 1 3  tan u  3 1 du   last equation; we get du.   3  4 cos u 8  cos u  8 cos u The solution to the new integral was presented in the Integral 5 3 1 3 section. Or  du  tan 1  sin u . After collecting all related 8 cos u 8 terms, we get the answer in terms of the original variable x as 3  tan u  3 3 1    tan  sin u   z 8  cos u  8 8



 3 z tanh 1  2 8  1 z

 3  2 x  1      8  3 

1 z



 3 z  tanh 1  2 8  1 z

 2x  1  2    2x  1    3 3 1 1     tanh  2 8  3     2x  1    1     3  

expression simplifies to

100_Integrals.008_3pp.indd 107



2

   .   

 3  2 x  1   1        8  3 

This

 2x  1 2 3 2x  1  x  x  1  tanh 1  . 2 4 8  2 x  x 1 

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108 • 100 Integrals

INTEGRAL 74 Problem:



x

x

2

 2x  2

2

dx

Solution: 1 x2    tan 1  x  1   2  constant 2 x  2 x  2  Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution:

Let x  1  z  dx  dz and rewrite the integral in terms of the x z 1 z 1 dx   dz   dz. variable z, to get  2 2 2 2  z  1 2  2  z  1   2  1  z2   x  2x  2    z 1 z 1 z 1 z dz. The new integral can be written as dz   dz   dz   2 2 2  2 2 2  2  z  1   2  1  z  1  z  1  z2  1  z 1 z 1  1  z2 2 dz   1  z2 2 dz   1  z2 2 dz. Now we have two integrals to       calculate. For the first integral we can write





z

1  z 

2 2

1

1  z 

2 2

dz 



z

1  z 

2 2

dz 

1 2z dz    2  1  z2 2 2

1 2z 1 . For the remaining integral, let dz   2  2 2 1  z  2  1  z2 

1 du 1 dz   du 2 2 2  z  tan u  dz  . Therefore, we have  1  z2  1  tan2 u cos cos2 u 1  cos 2 u du 1 2 and the   cos2 u du . But cos u  dz   2 2 2 1  tan2 u cos u

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List of Selected Integrals with Their Step-by-Step Solutions • 109

1 u sin 2 u . After 1  cos 2 u  du   2 4 2 collecting all related answers, in terms of the original variable x,

new integral reads  cos22 udu  we have 

1 tan 1  x  1  sin  2 tan  x  u sin 2 u 1 1       4 4 2 2  1  z2  2 2  x2  2 x  2 

1 tan 1  x  1  sin  2 tan  x  1   . The answer can be simplified 1   4 2 2x  2 tan 1  x  1  x x2    . dx  x2  2 x  2 2 2 2 2  x  2x  2  

100_Integrals.008_3pp.indd 109

to,

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110 • 100 Integrals

INTEGRAL 75 Problem:

x

1 3

x2  9

dx

Solution: 1  1  3  3 cos    2 54   x x

 x2  9   constant 

Techniques used: Change of variables, Trigonometric identities Step-by-step solution:

sin z 3  dx  3 dz and rewrite the integral in cos z cos2 z cos3 z sin z 1 sin z cos2 z terms of the variable z to get, 1 dz  dz  9  9 / cos2 z  9 cos2 z 27  1  cos2 z 1 os3 z sin z 1 sin z cos2 z dz  cos2 zdz. Using trigonometric identity, dz  2   2 2 27 1  cos z 27 cos z  9 cos z 1 1 1 1 1  cos 2 z cos2 zdz  1  cos 2 z  dz   z  sin 2 z  cos2 z  , we get   27 54 54  2  2 1 1 1 1  cos 2 z  dz   z  sin 2 z . Now in terms of the original variable x we have the  54 54  2  1  1 1  1  3  3 9   1 1 2  answer as cos     z  sin 2 z    z  sin z cos z   x  54  54  2  54  x x Let x 

9  1  1  3  3 1  1  3  3  1  2  . This expression simplifies to cos    2 x2  9  cos    x  54   x x 54  x x    1  1  3  3  cos    2 x2  9  , assuming x > 0. This expression simplifies to 54   x x  1  1  3  3  cos    2 x2  9  .  54   x x 

z 

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List of Selected Integrals with Their Step-by-Step Solutions • 111

INTEGRAL 76 Problem:

 sin x tan

1

sec x  1dx

Solution:





1 cos1 cos x  cos x 1  cos x   cos x tan 1 sec x  1  constant 2 Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Using the integration by parts technique we can write d 1 1 1  sin x tan sec x  1 dx   cos x tan sec x  1   cos x dx tan sec x  1 dx dg





f

c x  1   cos x





d tan 1 sec x  1 dx. But letting tan 1 sec x  1  , or tan   sec x  1 , dx we can write, by differentiating both sides, d sin x  dx . Now using tan   sec x  1, we cos2  2 cos2 x sec x  1 sin x dx  have cos2   cos x and d  tan 1 sec x  1  . 2 cos x sec x  1 sin x 1 sin x Therefore, cos x  2 cos x sec x  1 dx  2  sec x  1 dx . For cal 





d

culating the latter 2 cos z  sin x dx  2 sin z cos z dz and write 1 sin x dx   2  sec x  1

100_Integrals.008_3pp.indd 111

2 integral, let cos x  cos z  sin x dx  2 sin z cos z dz the integral in terms of the variable z as sin z cos z dz   cos2 zdz. Now using the 2 1 / cos z  1

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112 • 100 Integrals

trigonometric identity cos2 z 

1  cos 2 z , we have 2

 cos

2

zdz 

1 1  cos 2 z  dz  2

1 z sin 2 z . Therefore, in terms of the orig1  cos 2 z  dz    2 4 2 1 z sin 2 z 1    z  sin z cos z   cos1 cos x  c inal variable x, we can write 2 4 2 2 1 1 1   z  sin z cos z   cos cos x  cos x 1  cos x  . Collecting all related terms, 2 2 1 cos1 cos x  cos x 1  cos x   cos x tan 1 sec x  1 we receive the answer as 2

 cos

2

zdz 













cos x 1  cos x   cos x tan 1 sec x  1.

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List of Selected Integrals with Their Step-by-Step Solutions • 113

INTEGRAL 77 Problem:

 Solution:



1 x x  x2

dx



2 sin 1 2 x  1  constant Techniques used: Change of variables, Integration by parts Step-by-step solution: x  z  dx  2 zdz, and write the integral in terms of the 1 1 z ­variable z as  dx  2  dz  2  dz. Write 3 4 2 z  z2 z z x xx 1 2 2 the denominator as z  z2  1 / 4   z  1 / 2   1   2 z  1 . 2 Now let  2 z  1   u  2 dz  du and write the new integral in terms of 1 1 1 the variable u as 2  dz  4  dz  2  du. 2 z  z2 1  u2 1   2 z  1 Let

But we have 2 

1 1 u

2

du  2 sin 1 u. Or in terms of the original vari-





able x we get 2 sin 1 u  2 sin 1  2 z  1   2 sin 1 2 x  1 .

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114 • 100 Integrals

INTEGRAL 78 Problem:

x  ln  x 3

ln x  6

  5  dx

Solution: x4 8 ln 2 x  52 ln x  53   constant  32 Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution:





Write the integral as a summation of its terms, x3 ln  x ln x  6   5 dx  5 x3 dx  x3 ln 5 4 3 ln x  6 3 3 ln x  6 3 x ln  x   5 dx  5x dx  x ln  x  dx. But 5x dx  4 x . For the remaining integral, rewrite the integrand by using logarithm rule as,





x ln  x 3

ln x  6

 dx  x

3

 ln x  6  ln x dx  x3 ln 2 x dx  6 x3 ln x dx.  

Now, let ln x  z  x  e z, dx  e z dz , and rewrite the new integrals in terms of the variable z. Therefore,

x

3

x

3

ln 2 x dx  e4 z z2 dz

ln 2 x dx  e4 z z2 dz . Using the integration by parts technique twice, we e4 z 2 1 4 z e4 z 2 e4 z e4 z 4z 2 z   e z dz  z  z get  e z dz  . Similarly, 4 2 4 8 32 dg f

 e4 z  1 e4 z e4 z . 6  x3 ln x dx  6   e4 z z dz  6  z  e4 z dz   3 z3 4 2 8 dg f  4  nal Collecting all answers and rewrite them in terms of the origi 5 5 e4 z 2 e4 z e4 z e4 z e4 z e4 z z  z  3 z 3  x4  variable x, we get x 4   8 z2  5 2 8 4 32 4 4 8 32 e4 z e4 z 5 e4 z 5 x4  3 z 3  x4  8 z2  52 z  13   x 4    8 ln2 x  52 ln x  13  .   2 8 4 32 4 32 x4 The answer simplifies to  53  8 ln 2 x  52 ln x . 32

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1

List of Selected Integrals with Their Step-by-Step Solutions • 115

INTEGRAL 79 Problem:

 tan  1



x  1  x dx

Solution: 1 2





x  tan 1 x  x tan 1





x  1  x  constant

Techniques used: Change of variables, Integration by parts, Trigonometric identities, Partial fractions Step-by-step solution: Using the integration by parts technique, we can write the d 1 integral as  tan 1 x  1  x dx x  1  x  x  tan 1   x tan   dg  dx











x  1  x  , or  



x 1 

f





d x  1  x  x  tan 1  dx





 x  1  x  dx . But having tan 1 



tan   x  1  x , we get, after differentiating both side, d 1 1  1  cos2   2   . But we have tan   dx  cos2   2 x  1 2 x  cos2  2 1 1  cos2  2 . Therefore, tan 2    x  1  x . Or cos   2 2 cos  x 1  x 1

an 1







d  1 1    dx  2 x  1 2 x 



1  1   x  1  x  dx        2 x 2 x 1 



1 x 1  x

100_Integrals.008_3pp.indd 115



x x1  x



2

x 1  x



2





2

1

. Now by plugging back

d into the new integral, we get  x  tan 1  dx





1

dx .



Now,

1  1  x  1  x  dx      2 x 2 x



let

x  z  dx  2 zdz

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116 • 100 Integrals

x  z  dx  2 zdz and write the integral in terms of the variable z as 1 x 1 1  z3  1  dz  dx   2   2 x 2 x  1  x  1  x 2  1   z z2  1  2 z 1  z 1 1 1  z3 dz. For performing the latter intedx      2 z2  1  z2  1  z  1 z 1 1  z3 dz  du   2 . Therefore,   z gral, let z  tan u  dz  z2  1  z2  1  z  1 2 cos u





1 1    z  z2  1 

3

3











3

3



3



sin u / coos u sin 2 du  cos u  dz     cos u  . 2  2 cos2 u 1  sin u  1 / cos u  sin u / cos u   1 cos u z2  1  z  1 z



2

sin 2 u du trigonometric half-angle  2 2  cos2 u 1  sin u  du. Using ­  sin u / cos u   1 cos u 4 tan 2 u / 2 sin 2 u du   . identities we can write the integral as  2 2 cos u 1  sin u  1  tan2 u / 2   4 tan 2 u / 2 sin 2 u 1 du. For performing the new . du   2  os2 u 1  sin u  1  tan2 u / 2   2 tan u   2  1   u  1  tan 2  2  2 u integral let tan  y  du  dy, and rewrite the integral 2 1  y2 y2 1  y2  4 tan 2 u / 2 1 in terms of the variable y as   . du 8 2  1  y 2 2 1  y2  1  tan2 u / 2   2 tan u    2 1   u  1  tan 2  2  2 2 2 y 1 y    dy y 1 . dy. Now by  8 du  8  4 2 2 2 2 2 u  1  y  1  y  2 y  1  y 1  y  1  y   2 tan 2  1 using the partial fractions technique, we can write the integrand as u 1  tan 2  y2 1 1 1 1 2     4 2 2 2 3 4 1  y 1  y 16  y  1 16  y  1 4  y  1 4  y  1

n u / coos u

.

and the integral as a combination of its terms 8 

100_Integrals.008_3pp.indd 116

y2

1  y  1  y  4

2

dy 

1 1 dy  2  y  1 2

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List of Selected Integrals with Their Step-by-Step Solutions • 117

8

y2

1  y  1  y  4

2

dy 

1 1 1 1 1 1 dy   dy  2  dy  2  dy. 2 2 3 4  2  y  1 2  y  1  y  1  y  1

The new integrals can be worked out as 

1 1 1 dy  , 2  2  y  1 2  y  1

2

1

 y  1

4

dy  

2

3  y  1

3

2

1 1 1 dy   , 2  2  y  1 2  y  1

1

 y  1

3

dy  

1

 y  1

2

,

and

. by subsequent change of variables

u 1 x 1 = (y tan = , z tan u = x ) we get y  , assuming y > 0. 2 x Therefore, after collecting all results and writing them in terms of the original variable x, we get the answer as 1  x tan 1 x  1  x  x  tan 1 x . 2



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118 • 100 Integrals

INTEGRAL 80 Problem: x9  x20  48 x10  575 dx Solution: 1  x10  25  ln    constant 20  x10  23  Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: The denominator

of

the

integrand

can

be

written

as

x  48 x  575   x  25   x  23 . Therefore, using the partial 20

10

10

10

x9 1 1 x9 x9 dx  dx   x20  48 x10  575 2  x10  25 2  x10  23 dx 1 x9 1 1 x9 1 10 dx ln x 25 and    dx     10 10   2 x  25 20 2 x  23 20

fractions technique we get dx 

1 x9 1 x9 dx  dx. Or 10 10 2  x  25 2  x  23 1 x9 1   10 dx   ln  x10  23  . Collecting the answers, we get the 2 x  23 20 1 x10 − 25 ln 10 . solution as 20 x − 23

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List of Selected Integrals with Their Step-by-Step Solutions • 119

INTEGRAL 81 Problem:

 sin

1

x dx

Solution: x sin 1 x  1  x2  constant Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Let sin 1 x    x  sin  and dx  cos  d. Therefore, we can write the integral in terms of the variable α as  sin 1 x dx   cos  d 1  sin x dx   cos  d . Using the integration by parts technique, we get  d   sin    sin d   sin   cos  .    cos f

The results in

dg

terms of the original variable x reads,  sin   cos   x sin 1 x  1  x2 sin   cos   x sin 1 x  1  x2 .

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120 • 100 Integrals

INTEGRAL 82 Problem:

 tan Solution:

1

x dx

1 x tan 1 x  ln 1  x2   constant 2

Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Let tan 1 x    x  tan  and dx  cos2  d. Therefore, we can write the integral in terms of variable α as

 tan

1

 tan

1

x dx   cos2  d

x dx   cos2  d . Using the integration by parts technique, we get

 d   tan    tan  d . But  tan  d   sin  d  ln cos     cos   2

f

n  d   

dg

cos 

sin  d  ln cos  . The results in terms of the original variable x reads, cos   1  1  x tan 1 x  ln 1  x2  .  tan   ln cos   x tan 1 x  ln   2 2  1 x  1 Note that having x  tan  we can calculate, cos  . 1  x2

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List of Selected Integrals with Their Step-by-Step Solutions • 121

INTEGRAL 83 Problem:

 sinh

1

x dx

Solution: x sinh 1 x  1  x2  constant Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Let sinh 1 x    x  sinh  and dx  cosh d. Therefore, we can write the integral in terms of variable α as

 sinh

1

 sinh

1

x dx    cosh  d

x dx    cosh  d . Using the integration by parts technique, we get

  cosh  d   sinh    sinh d   sinh   cosh . The results  f

 dg

in terms of the original variable x reads,  sinh   cosh   x sinh 1 x  1  x  sinh   cosh   x sinh 1 x  1  x2 . Note that cosh   1  x2 , using the trigono2 metric identity cosh 2   sinh   1.     x2

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122 • 100 Integrals

INTEGRAL 84 Problem:

 tanh

1

x dx

Solution: 1 x tanh 1 x  ln 1  x2   constant 2 Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Let tanh 1 x    x  tan h  and dx  cos h 2  d. Therefore, we can write the integral in terms of variable α as

 tanh

1

 tanh

1

x dx   cosh 2  d

x dx   cosh 2  d. Using the integration by parts tech2  d   tan h    tan h  d. But nique, we get    cosh    f

dg

sinh  d   ln cosh . The results in terms cosh   1 of the original variable x reads,  tanh   ln cosh   x tanh 1 x  ln  2  1 x  1  1 1 1 2 cosh   x tanh x  ln    x tanh x  ln 1  x . Note that having x  tanh  2 2  1 x  1 we can calculate, cosh   . 1  x2   tanh  d   

100_Integrals.009_3pp.indd 122

   x tan 

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List of Selected Integrals with Their Step-by-Step Solutions • 123

INTEGRAL 85 Problem:

 cos Solution:

1

1   dx  x

 x2  1  1 x cos1    tanh 1    constant  x   x 

Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: cos1 1 / x     x  1 / cos  and dx  sin  / cos2  d . Let Therefore, we can write the integral in terms of the variable α as

 cos 1 / x  dx    sin  / cos   d. Using the integration by parts 1 technique, we get   sin  /cos   d   /cos      cos  d. 1

2

2

 f

dg

1 d  tanh 1  sin   (see Integral 5). Therefore, we can But  cos  1 d   / cos   tanh 1  sin  . write the results as  / cos    cos  The results in terms of the original variable x reads,  x2  1   / cos   tanh 1  sin    x cos1 1 / x   tanh 1  .  x  

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124 • 100 Integrals

INTEGRAL 86 Problem:

x

4

1 dx 4

Solution: 1  x2  2 x  2  1 1 1 ln  2    tan  x  1   tan  x  1    constant 16  x  2 x  2  8 Techniques used: Change of variables, Integration by parts, Partial fractions Step-by-step solution:

Write the denominator of the integrand as x 4  4   x2  2   4 x2   x2  2 x  2   x2  2 2

4   x2  2   4 x2   x2  2 x  2   x2  2 x  2 . Using the partial fractions techniques, write 1 1 2 x 1 2x the integral as  4 dx   2 dx   2 dx. x 4 8 x  2x  2 8 x  2x  2 1 2 x 1 11 x 1 1 1 2  2x dx   2 dx   2 dx   2 But dx 2  8 x  2x  2 8 x  2x  2 8 x  2x  2 16 x  2 x  2 1 1 2  2x  x2  2 x  2 dx  16  x2  2 x  2 dx. The latter integral can be worked out, since d 2 1 2  2x 1 x  2 x  2   2 x  2 , as dx  ln  x2  2 x  2 .  2  dx 16 x  2 x  2 16 1 1 1 1 dx dx   2  8  x  1 2  1 For the former integral, write it as 8 x  2 x  2 2

1 1 1 1 dx. Now let x  1  z  dx  dz and write the dx   2  8  x  1 2  1 8 x  2x  2 1 1 1 1 1 1 integral in terms of variable z as  dx   2 dz  tan 1 z  tan 1  x 2 8  x  1  1 8 z 1 8 8

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List of Selected Integrals with Their Step-by-Step Solutions • 125

 1 2

dx 

1 1 1 1 dz  tan 1 z  tan 1  x  1 . 8 8 8  z2  1

Similarly,

The

remaining

integral

1 2x 1 x 11 1 2x  2 dx    2 dx    2 8  x2  2 x  2 8 x  2x  2 16 x  2 x  2 1 x 11 1 2x  2 1 1 1 2x  2 1 dx   2 dx   ln  x2  2 x  dx. But   2   2 dx    2 8 x  2x  2 16 x  2 x  2 8 x  2x  2 16 x  2 x  2 16 1 2x  2 1 1 1 1 1 d   2 dx   dx   ln  x2  2 x  2 . For the former integral  2 8 x  2x  2 16 x  2 x  2 16 8  x  1 2  1 1 1 1 1 1 dx  tan 1  x  1 . Collecting all related terms, dx   2 2  8  x  1  1 8 8 x  2x  2 1 1 1 ln  x2  2 x  2   tan 1  x  1   ln  x2  2 x  we receive the answer as 16 8 16 1 1 1 1 2 1 2 x  2   tan  x  1   ln  x  2 x  2   tan  x  1 . This expression simplifies to 8 16 8 1  x2  2 x  2  1 1 1 ln     tan  x  1   tan  x  1   . 16  x2  2 x  2  8  can be worked out as

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126 • 100 Integrals

INTEGRAL 87 Problem:

 Solution:

4x

x5 2

 4

5/2

dx

1  3 x 4  12 x2  8   constant 12   4 x2  4 3 / 2   

Techniques used: Change of variables, Integration by parts Step-by-step solution: Expand the denominator of the integrand as

4x

2

 4

x5

x2  1  x2  1 

2

4

5/2

the

 1 x5

2

 4

5/2

 4 5 / 2  x2  1 

5/2

 32

 32 x  1  x  1  . Therefore, the integral reads 1 x5  dx dx . Now, using the partial  4 x2  4 5 / 2 32  x2  1  x2  1 2    1 x5 1  x dx   fractions technique1, we get  dx 32  x2  1  x2  1 2 32  x2  1    x 1  x 2x dx dx   d x dx   x2  1 3 / 2  x2  1 5 / 2 .Now,perform 32  x2  1      

5/2

x

4x

2

new

5/2

integrals,

1 x 1 2x dx  dx    2 64 32 x2  1 x 1 1

2

2

2

one

by

one.

For

2x x2  1 . For  1  2 32  x  1 3 / 2 32

1 2x 1 x dx  dx    2 32 64 x 1 x2  1 1 . For dx  16 x2  1

A CAS tool (e.g., Wolfram Alpha) can be employed for this step.

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List of Selected Integrals with Their Step-by-Step Solutions • 127

1 x 1 2x 1 . Collecting dx  dx   3/2 2 64   x2  1 5 / 2 32   x2  1 5 / 2 96  x  1  all related results, we receive the answer as 2 x 1 1 1 . This expression simplifies to   3/2 32 16 x2  1 96  x2  1  3 x 4  12 x2  8

96  x2  1  x2  1

100_Integrals.009_3pp.indd 127



1  3 x 4  12 x2  8  . 12   4 x2  4 3 / 2   

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128 • 100 Integrals

INTEGRAL 88 Problem:

x

4

1 dx 1

Solution: 2   x2  2 x  1  1   2  tan ln  2 8   x  2 x  1 

  x2  2 x  1  1   2  tan ln  2   x  2 x  1 





2 x  1  tan 1







2 x  1  tan 1



 2 x  1   constant  



 2 x  1   constant  



Techniques used: Change of variables, Integration by parts, Partial fractions Step-by-step solution: Write the expression in the denominator of the integral as a prod-





uct of two terms. Therefore, x 4  1   x2  1   2 x2  x2  2 x  1 x2  2 x  1 2









2 x2  x2  2 x  1 x2  2 x  1 . But, using the partial fractions technique, we can x 2 1 1 2 write the integral as  4 dx  dx   2  2 2 x 1 4 x  2x  1 x  2x  1 x  2x  1



1



x  1 x  2x  1 2



dx 





2 2 x 2 x 2 dx   2 dx. Now we perform the  2 4 x  2x  1 4 x  2x  1

­integration of the new integrals. Rewrite the former integral as 2 x 2 2 2x  2  2 2 2x  2 2 dx  dx  dx   2 dx  2   2 2 4 x  2x  1 8 8  x  2x  1 x  2x  1 x  2x  1  2 2x  2 2 dx . Therefore, dx   2  2 8  x  2x  1 x  2x  1 

100_Integrals.009_3pp.indd 128

2 2x  2 2 dx  ln x2  2 x  1  2 8 x  2x  1 8





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List of Selected Integrals with Their Step-by-Step Solutions • 129

d 2 2x  2 2 x  2 x  1  2 x  2 . Now, after dx  ln x2  2 x  1 . Note that dx 8 x  2x  1 manipulating the expression in the denominator of the latter 2









2

2  2 1 1  integral, we have x  2 x  1   x     1  2 x  1 .   2 2 2   2 2 1 1 Therefore, we have dx. dx    2 8 x  2x  1 2 1  2x  1 2



2







Let 2 x  1  z  2 dx  dz and write the integral in terms of the 1 1 1 1 2 variable z, as  tan 1 z. dx  dz  2 2  2 1  2x  1 4 2 2 1 z





2 tan 1 4

Or in terms of the original variable, x we get





2x  1 .

Similarly, for the remaining integral, we have 2 x 2 2 2x  2  2 2 2x  2 2  dx   dx  dx   2   2   2 2 4 x  2x  1 8 8  x  2x  1 x  2x  1 x   2x  2  2 2 2x  2 2  x2  2 x  1 dx  8   x2  2 x  1 dx   x2  2 x  1 dx . Therefore, 



2 2x  2 2 dx   ln x2  2 x  1 . Note 8  x2  2 x  1 8







d 2 x  2 x  1  2 x  2 . Now, after manipulating the dx expression in the denominator of the new integral, we have that

2

2  2 1 1  x  2x  1   x     1  2 x  1  . Therefore, we   2 2 2   2 2 1 1 have dx   dx. Let 2 x  1  u  2 dx  du  2 8 x  2x  1 2 1  2x  1 2



2







2 x  1  u  2 dx  du and write the integral in terms of the variable u, as 1 1 1 1 2 tan 1 u. Or in terms dx  du  2   u 2  1  2x  1 2 1 4 2 2



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130 • 100 Integrals

2 tan 1 4 related results, we get the answer as

of the original variable x, we get









2 x  1 . Collecting all















2 2 2 2 ln x2  2 x  1  tan 1 2 x  1  ln x2  2 x  1  tan 1 8 4 8 4 2 2 ln x2  2 x  1  tan 1 2 x  1 . This expression simplifies to 4 2  2  x  2 x  1   2  tan 1 2 x  1  tan 1 2 x  1  . ln 2   8  x  2x  1  







 

100_Integrals.009_3pp.indd 130



2x  1



 



27-07-2023 16:43:21



List of Selected Integrals with Their Step-by-Step Solutions • 131

INTEGRAL 89 Problem:

 sin  ln x  dx 2

Solution: x  5  cos  2 ln x   2 sin  2 ln x    constant 10  Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Let ln x  z  dx  e z dz. Now write the integral in terms of variable z, as

 sin  ln x  dx  e 2

z

sin 2 z dz . Using the trigo-

1  cos 2 z rewrite the integral as 2 1 z 1 1 e 1  cos 2 z  dz   e z dz   e z cos 2 z dz. For the new integrals,  2 2 2 1 z 1 z we get e dz  e and, using the integration by parts technique, 2 2 1 1 z 1 1 1 1   e cos 2 z dz   e z sin 2 z   e z sin 2 z dz   e z sin 2 z   e z cos 2 z    8 2 f dg 4 4 4 8 nometric identity, sin 2 z 

1 1 1 n 2 z dz   e z sin 2 z   e z cos 2 z   e z cos 2 z dz. Therefore, rearranging terms in the 4 8 8 1 last expression gives e z cos 2 z dz  e z  cos 2 z  2 sin 2 z . Now, 5 1 z 1 z z 2 we can write e sin z dz  e  e  cos 2 z  2 sin 2 z . This 2 10 expression written in the original variable x gives the answer as 1 z 1 z 1 1  e  e  cos 2 z  2 sin 2 z   x    cos  2 ln x   2 sin  2 ln x    . 2 10   2 10

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132 • 100 Integrals

INTEGRAL 90 Problem:

∫ sin x sin 2 x sin 3 x dx Solution: 1 6 sin 2 x  3 sin 2 2 x  2 sin 2 3 x   constant  24 Or 

1 6 cos2 x  3 cos2 2 x  cos2 3 x   constant  24

Techniques used: Integration by parts, Trigonometric identities Step-by-step solution:

1 cos  a  b   cos  a  b   2 1 1 1 n b  cos  a  b   cos  a  b   , rewrite the integrand as (sin x sin 2 x)sin 3 x   cos x  cos 3 x  sin 3 x   co   2 2 2 b a Using the trigonometric identity, sin a sin b 

1 1 cos x  cos 3 x sin 3 x  cos x sin 3 x  cos 3 x sin 3x. Now, after substitution, 2 2 1 1 rewrite the integral as  sin x sin 2 x sin 3 x dx   cos x sin 3 x dx   cos 3 x sin 3 x dx 2 2 1 1 cos x sin 3 x dx  cos 3 x sin 3 x dx . For the latter new integral, using the inte2 2 1 1 gration by parts, we get   cos 3 x sin 3 xdx  cos2 3 x. 2 12 For the former integral, using the trigonometric identity, 1 1 1 sin c cos d  sin  c  d   sin  c  d  , we have  cos x sin 3x dx    sin 4 x  sin 2 x  2 4 d c 2

x)sin 3 x 

 a

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List of Selected Integrals with Their Step-by-Step Solutions • 133

1 1 1 1 cos x sin 3x dx    sin 4 x  sin 2 x  dx   sin 2 x cos 2 x dx   sin x cos x dx.   2 4 2 2 d c 1 1 The new integrals are worked out as  sin 2 x cos 2 xdx  sin 2 2 x, 2 8 1 1 2 and  sin x cos xdx  sin x. Collecting all related terms, the 2 4 1 1 1 2 answer reads cos 3 x + sin 2 2 x + sin 2 x. This expression simpli12 8 4 1 1 fies to   6 sin 2 x  3 sin 2 2 x  2 sin 2 3 x . Or the answer in terms 12 24 3 1 of the cosine function reads   6 cos2 x  3 cos2 2 x  cos2 3 x . 8 24

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134 • 100 Integrals

INTEGRAL 91 Problem:



1  x2 dx

Solution:





1 sinh 1 x  x 1  x2  constant 2 Techniques used: Change of variables, Trigonometric identities Step-by-step solution: Let x  sinh z  dx  cosh z dz, and write the integral in terms of the .

variable z as



1  x2 dx   1  sinh 2 z cosh z dz   cosh 2 z dz.

Note that cosh 2 z  sinh 2 z  1. Rewriting the latter inte1 gral using the trigonometric identity, cosh 2 z   cosh 2 z  1 , 2 1 1 z 2 as  cosh z dz    cosh 2 z  1  dz   cosh 2 z dz  . But  2 2 2 1 1 cosh 2 z dz  sinh 2 z. Collecting related terms, the results read  2 4 z 1 + sinh 2 z. This expression in terms of the original variable x 2 4 1 sinh 1 x 1 reads  sinh z cosh z  sinh 1 x  x 1  x2 . 2 2 2



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List of Selected Integrals with Their Step-by-Step Solutions • 135

INTEGRAL 92 Problem:

1 ln x cos x  sin x x dx 2  ln x  

Solution: sin x + constant ln x Techniques used: Integration by parts Step-by-step solution: 1 ln x cos x  sin x cos x sin x . Therefore, x Expand the integrand,   2 ln x x ln 2 x  ln x  1 ln x cos x  sin x cos x sin x x dx   dx   ormer dx . Integrate f­ 2  ln x x ln 2 x  ln x  integral by using the integration by parts technique, to get sin x sin x  1   cos x  ln x  dx  ln x   x ln2 x dx. Collecting all related terms dg  f

gives sin x  sin x dx  sin x dx  sin x .  x ln2 x  ln x ln x x ln 2 x  0

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136 • 100 Integrals

INTEGRAL 93 Problem:



2 e2 x  e x 3 e2 x  6 e x  1

dx

Solution:  3ex  3  1 2x x 1 2 3 e  6 e  1  3 cosh     constant 3  2    Techniques used: Integration by parts, Trigonometric identities Step-by-step solution:

Let e x  z  dx  dz / z and rewrite the integral in terms of the 2 e2 x  e x 2 z2  z 2z  1 dx   dz   variable z, to get  dz 2x x 2 z 3z  6z  1 3e  6e  1 3 z2  6 z  1 2 z2  z 2z  1 dz   dz. The expression in the denominator can be writ2 z 3z  6z  1 3 z2  6 z  1 2  3  z  1  2 2 2 ten as 3 z  6 z  1  3 z  2 z  1 / 3  3  z  1   4 / 3  2    2   2  3  z  1  2 3  z  1  4 / 3  2  Therefore, the integral reads   1. 2   2z  1 1 2z  1 3  z  1 dz. Let 2  3 z2  6 z  1 dz  2  2  u  dz  du  3  z  1  2 3   1 2   3  z  1 2 2  u  dz  du, z  u  1. Now, write the latter integral in 2 3 3













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List of Selected Integrals with Their Step-by-Step Solutions • 137

2z  1 1 4u  3 1 dz   du  2 2 3 u2  1  3  z  1    1 2   2z  1 1 4u  3 1 4u  3 4 u dz   du. Expand the new integral, as  du   du  2 2 2 2 3 3 3  u 1 1 u 1   u  3  z  1 12 4 u 31 4 u 3 1  du  du   du. For the former integral, we   3 3 3 u2  1 u2  1 u2  1 u 4 4 2 have du  u  1 . For the latter integral, let  2 3 3 u 1 terms

of

the

variable

u,

as

u  cosh y  du  sinh ydy and write the integral in terms of the variable y, as

sinh y

osh y  1 2

dy 

sinh y 3 1 3 3 3 3 y cosh du  dy  dy     2 2 3 3 3 3 3 u 1 cosh y  1

3 3 3 cosh 1 u. Collecting all related terms and write them in dy  y 3  3 3 3 4 2 4 u 1  cosh 1 u  terms of the original variable x, gives 3 3 3

3 4 u2  1  cosh 1 u  3 3

2

 3  z  1   3  z  1  4 3 cosh 1    1      2 3 2     3

x  3  ex  1  4  3  e  1  3 1  1   .  cosh     2 3  2 3    

 3  ex  1      2  

2

2

This

expression

simplifies

to

 3  ex  1  1 2 . 3 e2 x  6 e x  1  cosh 1    2 3 3  

100_Integrals.010_3pp.indd 137

2

 3  z  1    2    3 1  cosh 1   3 

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138 • 100 Integrals

INTEGRAL 94 Problem: 2

 x4    1  x6  dx Solution: 1  1 3 x3 tan  x  6 1  x6

   constant 

Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Let x3  z  3 x2 dx  dz and write the integral in terms of the 2

variable z, as

2

2

 x4   x 4  dz 1  z  1 z2    dx dz dz   1  x6    1  z2  3 x2 3   1  z2  3   1  z2 2

2



1  z  1 z2 du dz. Now let z  tan u  dz  and write the new dz    3   1  z2  3   1  z2 2 cos2 u tan 2 u 1 1 integral in terms of the variable u, as  du   sin 2 udu 2 2 2 3 cos u 1  tan u  3 tan 2 u

u 1  tan 2 u 

2

du 

1 1 1 1  cos 2 u 2 sin 2 udu. Butsin 2 u  . Therefore, we have  sin udu   1  cos 2 u  du   3 6 3 2

1 1 u 1 sin 2 udu   1  cos 2 u  du   sin 2 u. Therefore, the answer in  6 12 3 6 u 1 tan 1 z 1 t terms of the original variable x reads  sin 2 u   sin  2 tan 1 z   6 12 6 12 1 3 tan 1 z 1 1 1 tan x 2 x3 sin 2 u   sin  2 tan 1 z    sin  2 tan 1 x3  . But sin  2 tan 1 x3   12 6 12 6 12 1  x6 3 2x sin  2 tan 1 x3   , using tan 1 x3  , or tan   x3, and calculating sin 2α. The 1  x6 tan 1 x3 x3  answer simplifies to . 6 6 1  x 6 

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List of Selected Integrals with Their Step-by-Step Solutions • 139

INTEGRAL 95 Problem:



x3 e x

2

1  x 

2 2

dx

Solution: 2

ex  constant 2 1  x 2  Techniques used: Change of variables, Integration by parts Step-by-step solution: Let x2  z  2 xdx  dz and write the integral in terms of 2 1 x3 e x ze z  dz. Now, using the integravariable z, as  dx 2 2   1  z 2 1  x 2  z dz 1 1 z  1  1 e 1  z  z ze   ze  dz tion by parts technique we get     2 f  1  z 2 1 z 2 1 z 2     dg

z dz 1 ze z 1 z 1 z  1  1 z  1  1 e 1  z  ze    dz    e  e     . This expression in 2  1 z 2 1 z 2 2 1 z 2 1 z 2  z   terms of the original variable x gives the answer as dg 2 ex 1 z 1  e  .  2  1  z  2 1  x 2 

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140 • 100 Integrals

INTEGRAL 96 Problem:

1

1 x

dx

Solution: 4 4 x3  2 x  4 4 x  4 ln 1  4 x  constant 3





Techniques used: Change of variables, Integration by parts Step-by-step solution: x  z  dx  2 zdz and write the integral in terms of the var1 z iable z, as  dx  2  dz. Now, let 1  z  u  dz  2  u  1  du 1 z x 1

Let

1  z  u  dz  2  u  1  du and the new integral in terms of the variable u reads  u  1 3 z 2 dz  4  nomial, du. After expanding the poly­ u 1 z  u  1 3 u3  3 u2  3 u  1 3 du  4  we have  u  1   u3  3 u2  3 u  1. Therefore, 4  u u 3 3 2  u  1 1 u  3u  3u  1 4 du  4 u2 du  12 u du  12 du  4  du.   du  4  u u u 4 3 2 Performing the new integrals, gives u  6 u  12 u  4 ln u. Write 3 this expression in terms of the original variable x gives the answer as 3 2 4 4 3 4 u  6 u2  12 u  4 ln u  1  z  6 1  z  12 1  z  4 ln 1  z  1  3 3 3 3 2 2 4 z  12 1  z  4 ln 1  z  1  x 6 1 x  12 1  x  4 ln 1  x 3 2 4 14 x  12 1  x  4 ln 1  x . This expression simplifies to 4 x3  2 x  4 4 x  4 ln 1  4 x  3 3 14 4 3 x  2 x  4 4 x  4 ln 1  4 x  . 3





















100_Integrals.010_3pp.indd 140







 

































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List of Selected Integrals with Their Step-by-Step Solutions • 141

INTEGRAL 97 Problem:

  sin x  1

2

dx

Solution: x  sin 1 x   2 1  x2 sin 1 x  2 x  constant 2

Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution: Let sin 1 x  z  sin z  x and dx = cos z dz. Write the integral in terms of the variable z, as

  sin x  1

2

dx  z2 cos z dz.

Using the integration by parts technique, we get 2 2 z2 cos z dz  z sin z  2 z sin z dz  z sin z  2 z cos z  2  cos z dz .    f dg sin z

Writing the answer in terms of the original variable x, gives 2 z2 sin z  2 z cos z  2 sin z  x  sin 1 x   2 1  x2 sin 1 x  2 x .

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ln x  z2

142 • 100 Integrals

INTEGRAL 98 Problem:

e 1  ln x  x xx

2x

dx

Solution: e x  x x  1   constant x

Techniques used: Change of variables, Integration by parts Step-by-step solution: Let x x = z . Taking logarithm of both sides gives ln x x = ln z, or dz x ln x = ln z and after differentiating we have (1  ln x)dx  . z dz Therefore, dx  x . Now, write the integral in terms of x 1  ln x  x dz the variable z, as e x 1  ln x  x2 x dx  e z 1  ln x  z2  ze z dz. z 1  ln x  

dz  ze z dz. Using the integration by parts technique, the latter integral z 1  ln x   reads ze z dz  ze z  e z. This expression in terms of the original variable x gives the answer as ze z  e z  e x  x x  1 . x

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List of Selected Integrals with Their Step-by-Step Solutions • 143

INTEGRAL 99 Problem:

 x

Solution:

ln x

dx

x2 ln x  constant 2  ln 

Techniques used: Change of variables, Integration by parts, Logarithmic identities Step-by-step solution: Let ln x  z  dx  e z dz, and write the integral in terms of the variable z, as xln x dx  e2 z  z dz. Using the integration by parts, gives

2z z  e  dz   dg

 f

e2 z z 1 2 z  d z  .   e    dz For calculating 2 2  dz 

d z π , let  z  u and take the logarithm of both sides. Therefore, dz ln  z  z ln   ln u. After differentiating both sides, we d du d get  z ln    . Or du  u ln    z ln . Note that du   z. dz u dz 1 2z  d z  1 Back substituting into the latter integral gives  e    dz   e2 z   z ln  2 2  dz  1 2z  d z  1 2z z ln  2 z z  e    dz   e   ln   dz   e  dz. Now, plugin back into the for2 2 2   dz  e2 z z ln  2 z z mer expression, gives e2 z  z dz    e  dz. Rearranging 2 2  ln   2 z z e2 z z e2 z  z  2z z the terms, gives  1  e dz . Or e  dz  .      2  2 2  ln   Write this expression in terms of the original variable x, to get the e2 z  z e2 ln x ln x x2 ln x answer as   . Note that we used the log2  ln  2  ln  2  ln  arithmic identity, e2 ln x = x2.

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144 • 100 Integrals

INTEGRAL 100 Problem:

ln  sin x 

 1  sin x dx Solution: 2

ln(sin x)   x   2 ln  cos     x  constant 1  cot  x / 2   2  

Techniques used: Change of variables, Integration by parts, Trigonometric identities Step-by-step solution:

x x Write the sine function as half-angle, using the identity sin x = 2 sin cos x x ln  2 sin 2 cos  ln  sin x  x x 2 2 d dx    sin x = 2 sin cos , and expand the integrand to get  1  sin x 1  sin x 2 2 x x  ln  2 sin cos  ln  cos  x / 2   ln  sin  x / 2   ln  sin x  1 2 2  dx   ln 2   dx   dx   dx    dx 1  sin x 1  sin x 1  sin x 1  sin x 1  sin x ln  sin  x / 2   ln  cos  x / 2   dx. Now, we have three new integrals to calculate. dx   1  sin x 1  sin x 1 dx, write it in terms of half-angle using the identity 1  sin x 2 tan  x / 2  sin x  . Therefore, after some manipulations we get 1  tan 2  x / 2  For 

1  tan 2  x / 2  2 1  x 2  1  sin x dx   1  tan  x / 2  2 dx. Let tan  2   z  dx  2 cos  x / 2  dz  1  z2 dz

2 x 2 dz, and write the integral in terms of the variable z   z  dx  2 cos  x / 2  dz  2 1  z2

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List of Selected Integrals with Their Step-by-Step Solutions • 145

. Thus 

1 1 2 . Therefore, the answer in dx  2  dz  2 1  sin x 1 z 1  z 

terms of variable x reads  ln 2  

For



1 2 ln 2 ln 1 / 4 dx    . x x 1  sin x 1  tan 1  tan 2 2

ln  sin x / 2 

dx, using integration by parts we get 1  sin x dx dx    dx  sin x / 2    ln  sin x / 2      ln  sin x / 2       dx   ln   1  sin x  1  sin x  1  sin x   f

dg

x  d ln  sin   x   dx  dx    2   cos x / 2 and we    ln  sin x / 2      dx. But  ln  sin    2 sin x / 2 dx  sin x  1  sin x    2  dx 2 have  . Therefore,  2 ln  sin x / 2  ln  sin x / 2  x 1  sin x  1  sin x dx  1  tan  x / 2  1  tan 2 2 ln  sin x / 2  ln  sin x / 2  cos  x / 2   1  sin x dx  1  tan  x / 2    sin  x / 2  1  tan  x / 2   dx. Write the later

z

terms of variable z (recall tan  x / 2   z ) cos  x / 2  1 as  dz. For the dx  2  sin  x / 2  1  tan  x / 2   z  1  z   1  z2  integral

in

latter

integral, use partial fraction technique to get 1 dz dz dz z 2 dz  2      dz. The 2 2    1 1 1 z z z z2 z 1  z  1  z  first two new integrals from the latter expression read dz dz 2   2 ln z  2 ln  tan  x / 2  , and     ln 1  z    ln 1  tan  x / 2   z 1 z   ln 1  z    ln 1  tan  x / 2   . The remaining integrals can be worked out as

z 1 2z 1 dz dz    dz     tan 1 z   tan 1  tan  x / 2  , and   2 2 2 1 z 2 1 z 2 1 z 1 2z 1 z 1  dz   ln 1  z2    ln 1  tan 2  x / 2  . dz    1  z2 2 1  z2 2 2 

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146 • 100 Integrals

ln  cos x / 2  dx, using integration by parts we get 1  sin x dx dx    dx   ln  cos x / 2    ln  cos x / 2      ln  cos x / 2      dx   1  sin x  1  sin x  1  sin x   f For 

dg

dx    d ln  cos x / 2    sin x / 2    ln  cos x / 2      dx. But  ln  cos x / 2    dx 2  cos x / 2  nx  1  sin x  ln  cos x / 2  2 ln  cos x / 2  dx 2   dx  and we have  . Therefore,  x 1  sin x 1  sin x  1 tan x / 2    1  tan 2 2 ln  cos x / 2  ln  cos x / 2  sin  x / 2    dx . Write the latter dx  1  tan  x / 2  1  sin x  cos x / 2  1  tan  x / 2   integral in terms of variable z (recall tan  x / 2   z ) as 

sin  x / 2  z dx  2  dz. For the cos x / 2 1  tan x / 2        1  z   1  z2 

latter integral, we use partial fraction technique to get z dz z 1 2   dz. The first two dz   dz   2 2 1 z 1 z 1  z2 1  z  1  z  dz  ln 1  z   ln 1  ta new integrals from the latter expression read  1 z dz z   ln 1  z   ln 1  tan  x / 2  , and   dz   1 / 2  ln 1  z2    1 / 2  ln 1  tan 2  1  z2 1 z

 1 / 2  ln 1  z2    1 / 2  ln 1  tan 2  x / 2  . The remaining integral can be worked out as  

dz   tan 1 z   tan 1  tan  x / 2     x / 2 . 2 1 z

Collection all related underlined answers, we get the soluln(sin x)   x   2 ln  cos     x tion to the integral, after some simplifications, as 2 1  cot  x / 2   2   ln(sin x)   x  2  2 ln  cos     x. 1  cot  x / 2   2  

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PART

2

Examples Applied in Engineering In this part of the book, we present some integrals along with their solutions related to engineering topics. The list is not exclusive but meant to help readers with their learning from Part 1 with some application examples in technical computations. The examples will focus on area properties of sections, structural beams, and a couple of probability distributions commonly used in engineering fields.

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148 • 100 Integrals

1  SEMI-CIRCLE SHAPES We consider a half-circle shape with radius R as shown in Figure 1. Its area, centroid, and moment of inertia is calculated using related integrals and the worked-out solutions. A half-circle shape can be the cross section of a beam, rod, or shape of hydraulic gate, for example. y

dθ

dr

r

dA

C θ

y

x 0

R

Figure 1  A Semi-circle shape and differential area element

πR 2 , from geometry. We calculate the area using 2 integration of area differential element dA  rdrd, where r is the radial disctance from the center and θ is polar angle w.r.t x-axis. R  R    R2  R2 R 2 Therefore, A  rdrd  drdr    d   d   .  2  2 0 2 00 0 0 0 Area is equal to

Centroid is the first moment of area divided by its total area and

is located at  0, y  when symmetry about y-axis exists as shown in ydA  ydA , where y  r sin . Note that y is Figure 1. Or y  dA A

R



measured from the centroid of the differential element. But ydA   r 2 sin drd   si R  R   R3  R3 R03 0 2R3 0  2 2 ydA  r sin  drd   sin  d  r dr  sin  d   cos      1 1    0  0 0 0 0 0  3  3 3 3

 d  

2R3 R3 R3  ydA  2R3 2  4R . cos 0    1  1   . Therefore, y  3 3 3 R 2 / 2 3 R 2 3 

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Examples Applied in Engineering • 149

Moment of inertia is the second moment of area with reference to a desired axis. Considering x-axis as reference, ­ R  R  R 4  1  cos 2  2 3 2 2 3 I x  y dA  r sin drd   sin dr dr    d. 4 0  2  00 0 0 Note that y is measured from the centroid of the differential  R 4  1  cos 2  R 4  sin 2 element. Performing the integrations gives d         4 0 2 8  2  



R 4  1  cos 2  R 4  sin 2   R 4 R 4  . Note that I  I  . We can calcud         y x   4 0  2 8  2 0 8 8  R

late Iy  x dA  r 3 cos2 drd, directly as well. 2

00

Using the parallel axis theorem, we can calculate the moment of inertia with respect to the axis at the centroid, Ic. Or I x  Ic  Ad 2 , where d is the normal distance between x-axis and that parallel passing through the point C at the centroid, or d = y. Therefore 2 R 4 R 2  4 R   9 2  64  4 Ic  I x  Ad 2      R . 8 2  3    72   Polar moment of inertia is the second moment of area with reference to a desired point. Considering the origin as the reference, R  R  R4 R 4 Jo  r 2 dA  r 3 drd  dr 3 dr  d   . Note that 4 0 4 00 0 0 Jo  I x  Iy. Using the parallel axis theorem, we can calculate the polar moment of inertia with respect to the centroid, Jc. Or Jo  Jc  Ad 2, where d is the distance between origin and the centroid. Therefore, 2 R 4 R 2  4 R   9 2  32  4 Jc  Jo  Ad 2      R . 4 2  3    36   Table 2 lists the results for semi-circle shape. TABLE 2  Results for Semi-circle

Shape Semi-circle

100_Integrals.011_3pp.indd 149

Area Centroid πR 2 2

4R 3π

Ix

Ic

Jo

2 πR 4  9   64  4 πR 4 R  8  72   4

Jc  9 2  32  4 R   36  

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150 • 100 Integrals

2  CIRCULAR SEGMENT SHAPES We consider a segmental shape of a half circle with radius R as shown in Figure 2, shaded area. Its area, centroid, and moment of inertia is calculated using related integrals and the worked-out solutions. A circular segmental shape can be the cross section of a beam, rod, or shape of hydraulic gate, for example. y

C a b

R

a

θ2

b

y

θ2 β

x

Figure 2  A Circular segment area shape

 Assuming total angle θ, from geometry we have a  R sin , 2   b  R cos , and   . The equation of a circle reads x2  y2  R 2. 2 2 x2 Therefore, for the sector we can write y  R 1  2 . R

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Examples Applied in Engineering • 151

Area: We calculate the area using the integration of area differential y

a

b

a

element dA = dxdy. Therefore, A   dxdy  dy  dx  y

a

b

a

dy  dy  dx  a

a

a

a

a

0 2

0

a

a

0

0

  y  b dx  2   y  b dx  2 ydx  2 bdx  2 ydx  2 ab. a

a

a

a

  y  b dx  2   y  b d

a

a

But 2 ydx  2 R  0

0

0

x2 1  2 dx R

x 2 ydx  2 R  1  2 dx . Let x  cos   dx  R sin  d. Therefore, the R R 0 0     limits of the integral reads x   0, a  R sin      ,  2  2 2  and write the latter integral in terms of the variable β to get a

 2

 2



 sin 2  2 2R  sin  d   R  1  cos 2  d  R      2    0 2 2 2 sin        2  2 2 sin    s in 2   sin 2     2 2  2  1  cos 2  d  R  2    . After applying the limits, we get R  2     R  2   2 2 2  sin        sin         sin   R 2  sin 2  2 R2     R2         sin  . Collecting all 2 2 2 2 2  2   2 

2   R R2    sin    2 ab     sin    2R 2 sin cos related answers, we get A  2 2 2 2  2 2 2 R sin  R R    2 ab     sin    2R 2 sin cos . Hence, A  2    sin  . 2 2 2  2

x 1  2 dx  2 R 2 R

2

2

2

R 2 sin 

Centroid is the first moment of area divided by its total area and is located at C  0, y  when symmetry about y-axis exists as ydA  ydA , where y is measured shown in Figure 2. Or y  dA A from the x-axis to the centroid of the differential element. But y a a a 1 2 2 ydA  ydy dx  y  b dx   y2 dx  ab2 .  b a 2 a  0

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152 • 100 Integrals

a

a  2 x3  a3 2 2 2 2 y dx  R  x dx  R x   aR    0 0  3  0 3  3 3  a R sin 3 3 a 3    dx  R2 x  x3   aR2  a3 . Therefore, y2 dx  ab2  aR2  a3  ab2  R3 sin 2  3 2  R3 sin 2 0  0 2 3 3 3 3  4R s R sin R sin 3 a   2 3 3  . Now, we have y  3 2  3 2  2 3 2  ab  R sin   R sin cos  R sin R2 3  3 2  2 3 3 2 2   sin     2 1  si n 2

x2  2 R

a

The latter integral reads

2

 2 3 3 4 R sin 3   R sin  2 . 2  y  32 R 3   sin      sin    2 Moment of inertia is the second moment of area with reference to a desired axis. Considering x-axis as reference, y a a a 2 2 2 2 I x  y dA  y2 dy  dx    y3  b3  dx  y3 dx  ab3 . But  30 30 3 b a 4 2 2   R    ab3   R 4 sin cos3   sin   1  sin 2  . Now, after sub2 3 3 2 2 3   x2  1  0  R2  After rewriting the latter integral in terms of the a

2 2R3 stituting for y, the latter integral reads y3 dx  30 3

a

3

2 4 ble β(recall x  R cos), we get 2 R  1  x  dx   2 R 3 0  3 R 2    3 a

3

 2R 4  dx    3 

3

  dx.   varia-

2

 1  cos   2

 2

3

sin  d  

2R 4 3

trigonometric relation, sin 4   R4  12

 2

  2

100_Integrals.012_3pp.indd 152

2



sin 4  d.

But

using

 2

 1  cos   2

3

sin

 2

the

 2

1  3  4 cos 2  cos 4  we get 8

R4  3  4 cos 2  cos 4  d   12



1 R4  3   2   sin   sin   3 2 2 4    2 sin      4 12  2 2

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Examples Applied in Engineering • 153



1 1 1 1 R4  3 1   2

2 sin 2  sin 4      2 sin   sin 2 . But we have sin 2  sin  cos   sin   1  2 sin  4 2 2 4 12  2 4   2

 1 1 1 1 1  R4  3  sin 2  sin  cos   sin   1  2 sin 2   sin   sin  sin 2 . Therefore,    2 sin   sin 2 2 2 2 12  2 4 4 2 2  4 3 5 R4  3 1 R     2    2 sin   sin 2      sin   sin  sin . Collecting the 12  2 2 2 4  12  2 R4  3 R4 5   2  results, we get I x  sin   1  sin 2     sin   sin  sin   12  2 2 2 3 2  4 4 R   R       2  sin 2   sin   1  sin 2 . After simplification, we get I x     sin   2 sin  sin . 2 3 2 8 2     y

a

b

a

a

Similarly, considering y-axis as reference, Iy  x2 dA  dy  x2 dx  2   y  b  x2 dx  y

y

a

a

a

a

a

 x2 dA  dy  x2 dx  2   y  b  x2 dx  2 yx2 dx  2 bx2 dx. b

a

a

0

0

2 bx2 dx  

But

0

0

0

2a b   2   R 4 sin 3 cos 3 3 2 2 3

2 a3 b R4 2    2 bx dx     R 4 sin 3 cos   sin  sin 2 . Now, after substituting for y, 3 3 2 2 3 2 0 2

a

a

x2 dx. After rewritR2 0 0 ing the latter integral in terms of variable β (recall x  R cos), the latter integral reads 2 yx2 dx  2 R x2 1 

a

we get 2 R x2 1  0

2

x dx  2 R 4 R2

 2



sin 2  cos2 d. But using

 2

the trigonometric relation, sin 2  cos2   2 R

4

 2

  2

R4  cos 4  d  4

R4 sin 2  cos2 d   4

 2

  2

1 1  cos 4  we get 8

R4 1  cos 4  d  4



1 R4   1   2

   sin 4     sin 2 .    4 2 4 4 2

But

we



1 R4   1   2    sin 4        4 2 4 4

have

2

1 1 1 sin 2  sin  cos   4 2 2

1 1 1  1   sin 2  sin  cos   sin   1  2 sin 2   sin   sin  sin 2 . Therefore, 2 2 2 4 2 2 

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154 • 100 Integrals

4 R4   1  R  1 2    sin 2     sin   sin  sin . Collecting the 4 2 4 2  4 2 2 R4  2 results, after some simplifications, we get Iy   3  3 sin   2 sin  sin 24  R4  2    3  3 sin   2 sin  sin . 24  2

  2

Using the parallel axis theorem, we can calculate the moment of inertia with respect to the centroid, Ic. Or I x  Ic  Ad 2 , where d 2 is the distance between x-axis and the centroid, or y. Therefore  3    R 4 sin    R4  R2  2  2  Ic  I x  Ad 2     sin       sin   2 sin  sin   8  2 2  3    sin    2      3    4 R sin   4  R2  2   . Or, after simplification, I  R   sin   2 sin  sin 2     64 2     sin        c 2 2 8  2 9  3    sin           sin 6    R4   64    2  Ic    sin   2 sin  sin 2    . 8   2  9   sin     Polar moment of inertia is the second moment of area with reference to a desired point. Considering the origin as the reference, Jo  r 2 dA    x2  y2  dA  I x  Iy.

Therefore, using previously obtained results, we have R4  R4  2  2  Jo     sin   2 sin  sin    3  3 sin   2 sin  sin  . 8  2  24  2 R4  2 2 After some simplifications, we have Jo     sin   sin  sin 4  3

R4  2 2    sin   sin  sin 4  3

  2

 . 2

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Examples Applied in Engineering • 155

Using the parallel axis theorem, we can calculate the polar moment of inertia with respect to the centroid, Jc. Or Jo  Jc  Ad 2, where 2 d is the distance between origin and the centroid, or y. Therefore  3    R sin 4    R4  2 R2  2  2  Jc  Jo  Ad 2    sin   sin  sin    sin       4  3 2 2  3    sin    2    3    4 R sin      R2  2  .  sin 2      sin    2 2  3    sin      Table 3 lists some of the results for circular segment shape. TABLE 3  Results for circular segment

Shape

Area

Centroid

Ix

Jo

 4 4 R sin 3   R 4  Circular R 2 2 2  R  2  2    sin   2 sin  sin     sin   sin  sin  segment 2    sin   8  2 4  3 2 3    sin  

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156 • 100 Integrals

3  SEMI-ELLIPSE SHAPES We consider a half-ellipse shape with major and minor radii a and b, respectively. Its area, centroid, and moment of inertia is calculated using integrals. A half-ellipse shape can be the cross-section shape of beams, rods, or hydraulic gates for example. y

dA dy

b

x x

a Figure 3  A Semi ellipse area shape and differential element

πab , from geometry. We calculate the area using 2 integration of the area differential element dA = 2 xdy as shown x2 y2 in Figure 3. Using the equation of ellipse, 2  2  1 we have a b Area is equal to

b

A  2 xdy  2 a   2

0

y y2 1  2 dy. Let  sin   dy  b cos  d  and b b  2

 2

0

0

A  2 ab 1  sin 2  cos  d   2 ab cos2  d   ab 1  cos 2  d . 0

/2

sin 2  ab  Performing the integration gives ab   . Note   2 0 2  that due to symmetry, area of an ellipse is πab. Centroid is the first moment of area divided by its total area and is located at

100_Integrals.012_3pp.indd 156

 0, y 

when symmetry about y-axis exists.

27-07-2023 16:54:04

Examples Applied in Engineering • 157

Or

ydA  ydA . y dA A

y2 ydA  2 0 xydy  2 a0 y 1  b2 dy. b

But

b

Performing the integration, after writing it in terms of   the variable θ, for values y   0, b      0,  gives,  2 



2 y2  cos3   2 2 ab2  2 a y 1  2 dy  2 ab2  sin  cos2 d  2 ab2   . b 3  0 3  0 0 b

Therefore, y 

2 ab2 / 3 4 b  . ab / 2 3 

Moment of inertia is the second moment of area with reference to a desired axis. Considering the x-axis as reference, 

b

b

0

0

I x  y2 dA  2 y2 xdy  2 a y2 Performing  3 2

ab 2

2  sin 2d  0

 3 2

ab 4

the

2 y2 1  2 dy  2 ab3  sin 2  cos2 d. b 0

integrations



gives

ab3 2 2 ab3 sin 2    d 2 0 4

 2

 1  cos 4  d 0

 2

ab3  sin 4   ab3     I x . Similarly, we   4  4   0 8 0 a3 b can calculate Iy  x2 dA  . Note that x is measured from the 8 centroid of the differential element.

 1  cos 4  d 

Using the parallel axis theorem, we can calculate the moment of inertia with respect to the axis at the centroid, Icx. Or I x  Icx  Ad 2 , where d is the normal distance between axis x and that parallel passing through the centroid, or y. Therefore 2 ab3 ab  4 b   9 2  64  3 2 Icx  I x  Ad       ab . Similarly,  8 2  3    72   2 a3 b ab  4 b   9 2  64  3 2 Icy  Iy  Ad       a b. 8 2  3    72  

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158 • 100 Integrals

Polar moment of inertia is the second moment of area with ­reference to a desired point. Considering the origin as the reference, ab3 a3 b ab 2 Jo  r 2 dA    x2  y2  dA  I x  Iy    a  b2 .  8 8 8 we can calculate the moment of inertia with respect to the cen 9 2  64  3 3 troid, Jc. Or Jc  Jcx  Jcy.Therefore, Jo     a b  ab .  72   Table 4 lists the results for semi-ellipse shape. Table 4  Results for semi-ellipse

Shape Area Centroid Semiellipse

100_Integrals.012_3pp.indd 158

πab 2

4b 3π

Ix

Icx

Jo

Jc

2  9 2  64  3 πab3  9   64  3 ab 2 3 a  b2     ab   a b  ab   8 72  8    72  

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Examples Applied in Engineering • 159

4  TWO-DEGREE POLYNOMIAL SHAPE-QUADRATIC  x2  Consider a quadratic polynomial shape of degree two, y  b  1  2  a   which is bounded between coordinate axes in the first quadrant (i.e., x > 0 and y > 0) as shown in Figure 4. y 3.0 2.5

b=3

2.0

dx

1.5 1.0 0.5

dy dA 0.5

1.0

1.5

2.0

x

a=2 Figure 4  A quadratic area shape with a=2 and b=3.

Area, we calculate the area using integration of the differential area element dA = dxdy. Using the equation of polynomial, we y a a a  x2  have A  dA  dydx  ydx  b  1  2  dx. Performing the a  0 0 0 0 a

 x3  2 ab . Simply, subtracting this integration gives A  b  x  2   3a 0 3  area from that of the enclosing rectangle, ab we get the value of ab the area over the polynomial, or . 3

Centroid is the first moment of area divided by its total 2 y a a a ydA  ydA . But 1 2 b2  x2  area. Or yc  ydA  0 ydy0 dx  2 0 y dx  2 0  1  a2  dx  dA A

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160 • 100 Integrals

a

ydx  0

2

a a a b2  x2  b2  x4 1 2 x2  y dx  1  dx 1  2       dx. Performing the integration, a2  2 0  a4 a2  2 0 2 0  a 4 ab2 b2  x 5 2 x3  4 ab2 / 15 2 b gives,  x  4  2   . Therefore, = yc = . 2  5a 3a 0 15 2 ab / 3 5

Similarly, we can calculate the x-coordinate of the centroid, y a a a a xdA xdA   x2  x3   xc   . But xdA  dyxdx  yxdx  bx  1  2  dx  b  x  2  dx a  a  0 0 0 0  0 dA A a

a a  x2 x4  a2 b   x2  x3   bx  1  2  dx  b  x  2  dx. Performing the integration, gives, b   2   4 a  a   2 4a 0 0  0 a a2 b / 4 3 a  x2 x4  a2 b. Therefore, = xc = . b  2   2 ab / 3 8 4  2 4a 0

Moment of inertia is the second moment of area with reference to a desired axis. Considering x-axis as reference, y

3

a a a b3  x2  1 b3  x6 3 x 4 3 I x  y dA   y dxdy  y dydx  y3 dx    1  2  dx    1  6  4  30 3 0 a  3 0 a a a 0 0 3 a a  x2  b3  x6 3 x 4 3 x2   dx 1   2  dx. Performing the integrations gives 1      0  a2  a6 a 4 3 0  a  a 3 7 3 x 5 x3  16 ab3 b  x Ix   x  6  4  2   . Similarly, we can calcu3  7a 5a 105 a 0 2

2

2

a

y

a a a  x4  late Iy  x dA   x dxdy  dyx2 dx  yx2 dx  b  x2  2  dx. a  0 0 0 0 a 3 5 3 x x  2a b Performing the integrations gives Iy  b   2   . a 3 5 15  0 2

2

Using the parallel axis theorem, we can calculate the moment of inertia with respect to the axis at the centroid, Icx. Or I x  Icx  Ad 2 , where d is the normal distance between axis x and that parallel passing through the centroid, or yc.

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Examples Applied in Engineering • 161

2

16 ab3 2 ab  2 b  8 ab3 Therefore Icx  I x  Ayc   . Similarly,    105 3  5  175 2 2 a3 b 2 ab  3 a  19 3 2 Icy  Iy  Axc   a b.    15 3  8  480 2

Polar moment of inertia is the second moment of area with reference to a desired point. Considering the origin, 16 ab3 2 a3 b 2 ab Jo  r 2 dA    x2  y2  dA  I x  Iy     7 a2  8 b2 . 105 15 105 we can calculate the polar moment of inertia with respect to the 8 ab3 19 a3 b ab centroid, Jc. Or Jc     665a2  768 b2 . 175 480 16800 Table 5 lists the results for a quadratic/parabolic polynomial shape. TABLE 5  Results for a parabolic shape

Shape

Area

Centroid, yc

Semiellipse

2 ab 3

2b 5

100_Integrals.012_3pp.indd 161

Ix

Icx

Jo

Jc

ab 16 ab3 8 ab3 2 ab 7 a2  8 b2  a 16800  665a2  768 b2  105 175 105

27-07-2023 17:03:39

162 • 100 Integrals

5  THREE-DEGREE POLYNOMIAL SHAPE-CUBIC  x3  Consider a cubic polynomial shape of degree three, y b  1  3  a   which is bounded between coordinate axes in the first quadrant (i.e., x > 0 and y > 0) as shown in Figure 5 y 4

3

2

1

0.5

1.0

1.5

2.0

x

Figure 5  A cubic area shape with a=2 and b=4

Area, we calculate the area using integration of the differential area element dA = dxdy. Using the equation of polynomial, we y a a a  x3  have A  dxdy  dydx  ydx  b  1  3  dx . Performing the a  0 0 0 0 a

 x4  3 ab integration gives A  b  x  3   . Simply, subtracting this a 4 4  0 area from that of the enclosing rectangle, ab we get the value of ab the area over the polynomial, or . 4 Centroid is the first moment of area divided by its total 2 y a a a ydA  ydA . But 1 2 b2  x3  b2 area. Or yc  ydA  ydy dx  y dx    1 dx    0 0 2 0 2 0  2 a3  dA A

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a



0

Examples Applied in Engineering • 163

2

a

ydydx  0

a a b2  x3  1 2 y dx  1    a3  2 0 2 0  b2 gives, 2

Similarly, centroid,

a b2  x6 x3  1  2    dx. Performing the integration, 2 0  a6 a3  a  9 ab2 x7 2 x 4  9 ab2 / 28 3 b  .   x = y = . Therefore,  c 7 a6 4 a3  0 28 3 ab / 4 7 

dx 

we

can xdA

calculate the x-coordinate of the y a a a xdA  x3   xc   . But xdA  dyxdx  yxdx  bx  1  3  dx a  0 0 0 0  dA A

y

a a a a   x4  x3  dA  dyxdx  yxdx  bx  1  3  dx  b  x  3  dx. Performing the integration, gives, a  a  0 0 0 0  0 a  x2 3 a2 b x5  3 a2 b / 10 2 a b  3   . Therefore, = xc = . 10 3 ab / 4 5  2 5a  0

Moment of inertia is the second moment of area with reference to a desired axis. Considering x-axis as reference, 3 y a a a a b3  x3  1 3 b3  x9 3 2 2 2 I x  y dA    y dxdy  y dy  dx  y dx    1  3  dx    1  9  30 3 0 a  3 0 a a 0 0 3 a 9 6 3 3 3 a 3 3 b  x  3x 3x  b  x dx    1  3  dx    1  9  6  3  dx. Performing the integrations gives a  3 0 a a a  3 0 a

3 x7 3 x 4  81ab3 b3  x10 Ix   x   6  3  . Similarly, we can calcu9 3  10 a 7 a 4a 0 140 y a a a  x5  late Iy  x2 dA    x2 dxdy  dyx2 dx  yx2 dx  b  x2  3  dx. a  0 0 0 0 a  x3 x6  a3 b Performing the integrations gives Iy  b   3   . 6  3 6a 0

Using the parallel axis theorem, we can calculate the moment of inertia with respect to the axis at the centroid, Icx. Or I x  Icx  Ad 2 , where d is the normal distance between axis x and that parallel passing through the centroid, or yc. Therefore 2 a3 b 3 a 81ab3 3 ab  3 b  108 3 2 Icx  I x  Ayc 2   ab . Similarly, I  I  Ax      cy y c 140 4  7  245 6 4 2 3 a b 3 ab  2 a  7 3 Icy  Iy  Axc 2   a b.    6 4  5  150

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164 • 100 Integrals

Polar moment of inertia is the second moment of area with reference to a desired point. Considering the origin, 81ab3 a3 b ab Jo  r 2 dA    x2  y2  dA  I x  Iy    70 a2  243 b2   140 6 420 a3 b ab 2 2   a  243 b . 70   6 420

we can calculate the polar moment of inertia with respect to the ab 108 ab3 7 a3 b centroid, Jc  Icx  Icy. Or Jc     343 a2  3240 b2  3 245 150 7350 a b ab   343 a2  3240 b2 . 50 7350 Table 6 lists the results for a cubic polynomial shape. Table 6  Results for cubic polynomial shape Shape

Area

Centroid, yc

Cubic

ab 4

3b 7

100_Integrals.012_3pp.indd 164

Ix

I cx

Jo

81ab3 108 3 ab ab 70 a2  243 b2  420 140 245

Jc 7 a3 b ab  150 7350  343 a2  3240 b2 

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Examples Applied in Engineering • 165

6  n-DEGREE POLYNOMIAL SHAPE-SPANDREL As shown in Figure 6, consider a polynomial of degree n with its n  x vertex at the origin. The polynomial equation reads y  b   . The a area under the polynomial, Spandrel is of interest in engineering. A differential element dA = dxdy is used for the following calculation. y

4

3

2

1

0.5

1.0

1.5

2.0

x

Figure 6  A spandrel area shape with a=2 and b=4

Area, we calculate the area using integration of the area differential element dA = dxdy. Using the equation of polynomial, we have ay

y

n

a

b  x n 1  ab  x  A  dxdy  dx dy  ydx  b   dx  n  .  a a  n  1 0 n  1 00 0 0 0 0  Simply, subtracting this area from that of the enclosing rectangle, nab hb we get the value of the area over the polynomial, or . Note n+1 that for n = 2, previous results for a parabolic are obtained. a

a

a

Centroid is the first moment of area scaled by total area. Or ydA  ydA . But ydA  a yydxdy  adx yydy  a  y2  dx. yc    0 0 0  2  00 dA A After substituting for y and performing the integration, gives,

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166 • 100 Integrals

 y2 0  2

a  b2 b2 2n dx  x dx   2 a2 n 0 2 a2 n  ab2 2  2 n  1  n  1 b yc   . ab 4n  2 n1 a

a

 x 2 n 1  ab2  .  2n  1    0 2  2 n  1

Therefore,

Similarly, we can calculate the x-coordinate of the centroid, y ay a a xdA xdA  xc   . But xdA  xdxdy  xdx dy  yxdx .  00 0 0 0 dA A

After substituting for y and performing the integration, a

a b n 1 b  x n 2  a2 b  gives, yxdx  n x dx  n  . Therefore,   a 0 a  n  2 0 n  2 0 a2 b  n  1 a xc  n  2  . ab n2 n1 a

Moment of inertia is the second moment of area with reference to a desired axis. Considering x-axis as reference, y a a a  y3  b3  x3 n  b3  a3 n1  I x  y2 dA  dx y2 dy     dx    3 n  dx  3 n  . 3  3 0 a  3a  3n  1  0 0 0 ab3 . Similarly, we 9n  3 y a a a  x n 2  b  a n  3  a3 b 2 2 2 can calculate Iy  x dA  x dx dy  x ydx  b  n  dx  n   a  a  n3 n3 0 0 0 0 Performing the integrations gives I x 

a  x n 2  b  n a 0

 b  a n  3  a3 b dx  .    an  n  3  n  3  Using the parallel axis theorem, we can calculate the moment of inertia with respect to the axis at the centroid, Icx. Or I x  Icx  Ad 2 , where d is the normal distance between axis x and that parallel passing through the centroid, or yc. Therefore 2  3 ab3 ab   n  1  b   7 n2  4 n  1 2 ab . Icx  I x  Ayc       2  9 n  3 n  1  4 n  2   12  3 n  1   2 n  1  

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Examples Applied in Engineering • 167

Similarly, Icy  Iy  Axc 2 

2  3 b ab   n  1  a   1  a b.    2   3 n  1  n  2    n  3   n  2  

2  3 a3 b ab   n  1  a   1 a b     2  n  3 n  1  n  2    n  3   n  2  

Polar moment of inertia is the second moment of area with reference to a desired point. Considering the origin, ab3 b2  a3 b ab  a2 Jo  r 2 dA    x2  y2  dA  I x  Iy       9n  3 n  3 2  n  3 9n  3  3 a3 b ab  a2 b2      .  3 n  3 2  n  3 9n  3 

We can also calculate the polar moment of inertia with respect to the centroid, Jc. Or Jc  Icx  Icy.Therefore   3   3  7n a2 7 n2  4 n  1 1 ab  Jc   a b  ab      2 2 2   n  3   n  2  12  3  12  3 n  1   2 n  1     n  3   n  2     3  7 n2  4 n  1 b2  . a2 1  a b ab   2 2  2 n  3   n  2     n  3   n  2  12  3 n  1   2 n  1   Table 7 lists the results for a parabolic shape. TABLE 7  Results for Spandrel Shape

Spandrel, order n

Area

ab n+1

Centroid, yc

 n  1 b 4n  2

Ix b3 3 a3 n  a3 n1     3n  1 

Icx

Jo

Jc

  a2  a2     2  7 n2  4 n  1    n  3 n  2       2  ab n  3  ab   12  3 n  1   2 n  1   2  7 n2  4 n  1  b2  b2       ab3  12  3 n  1   2 n  1 2   9n  3   

These results are comparable to those obtained in the previous sections for two- and three-degree polynomials.

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168 • 100 Integrals

7  SINUSOIDAL SHAPES  x  Consider a sinusoidal shape, y b sin   which is bounded  a  between coordinate axes in the first quadrant (i.e., x > 0 and y > 0) as shown in Figure 7, for example for a = 2 and b = 3. y

3.0 2.5 2.0 1.5 dx

1.0

dy dA

0.5

0.5

1.0

1.5

2.0

x

 x  Figure 7  A sinusoidal area shape, 3 sin  .  2 

Area, we calculate the area using integration of the differential area element dA = dxdy. Using the equation of the profile, we y a a a  x  have A  dxdy  dydx  ydx  b sin   dx. Performing the  a  0 0 0 0 a

x  2 ab  a integration gives A  b   cos   . Simply, subtracting this a 0    area from that of the enclosing rectangle, ab we get the value of the 2  area over the sinusoidal curve, or ab  1    0.363 ab.   

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Examples Applied in Engineering • 169

Centroid is the first moment of area scaled by total area. ydA  ydA . The y-coordinate of the centroid is yc  dA A y

y

a a a  y2  But ydA  ydxdy  dx ydy     dx. After substituting  2  00 0 0 0 for y and performing the integration, we get a a 2 2 a 2 2 a y  b  2 x  2 x  b b  a ab2 2  x  d x  x  sin 1 dx  sin dx   cos      0  2  2 0 4 0  4  2 4 a  a  0  a  2 ab a 2 2 b  a 2 x  ab . Therefore, y  4  b. 2 x  sin  c  dx   x   2 ab 8 4  2 a 0 a  4 

Similarly, we can calculate the x-coordinate of the centroid, y ay a a xdA xdA  xc   . But xdA  xdxdy  xdx dy  yxdx . After 00 0 0 0 dA A

substituting for y and performing the integration, we get a a a a ab x  xx a2 b  x   ab yxdx  b x sin dx   x cos  cos  . dx 0 0  a    a  0  0  a  0 2 a b a Therefore, xc    . This result confirms that from the shape 2 ab 2  a symmetry (i.e., the shape is symmetric about the line at = ). 2 Moment of inertia is the second moment of area with reference to a desired axis. Considering x-axis as reference, y a a a  y3  b3  x  2 I x  y dA  dx y2 dy     dx   sin 3   dx.  3  3 0  a  0 0 0 a

Rewrite the latter integral as

100_Integrals.013_3pp.indd 169

a

x 2  x  x  b3 b3  x   sin sin dx  sin  1  cos2        3 0 a 3 0 a  a   a 

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170 • 100 Integrals

a

x  b3 x  2  x    dx   sin  1  cos    dx and performing the integration, we get 3 0 a  a  a  a

a

a

2 ab3 2 ab3 2 ab3 b3  a x  b3 x  a 3  x   2  x     cos sin cos  dx cos          3  3  3 9 a  0 3 0 a 3  a 0  a  a a 3 3 3 3 3 a 3 3 b  a x  b x 4 ab 2 ab 2 ab 2 ab  x    a  x     0.1415 a3 b  cos    sin cos2   dx     cos3      3   3 3  3 3 9 9 a 0 3 0 a  a   a 0 b3 3

a

2 ab3 2 ab3 4 ab3 x     0.1415 a3 b.   3 9 9 a 0

y

y

a

a

a

 x  Similarly, we can calculate Iy  x2 dA  dyx2 dx  yx2 dx  bx2 sin   dx  a  0 0 0 0 a

a

a

 x  2 2 2 0 dy0 x dx  0 yx dx  b0 x sin  a  dx. Performing the integration gives a a a a3 b 2 ab x  2 ab x x  ab 2 Iy    x cos   cos dx x cos dx.   x   a 0  0 a   0 a   a

The

latter

integral

reads

x cos 0

a

a  a2  x x  a x  a dx   x sin    sin dx   2 cos a a 0  0  a       0

2 3 a a  a2 a x  a3 b 2 ab 2 a2    4  a b x  x 2 a2  a . 2    0.189   x sin    sin dx   2 cos    2 . Therefore, Iy     3 a 0  0 a 0  a       2 3 0 a3 b 2 ab 2 a2    4  a b    0.1893 a3 b. . 2     3 a

Using the parallel axis theorem, we can calculate the moment of inertia with respect to the axis at the centroid parallel to the x-axis, Icx. Or I x  Icx  Ad 2 , where d is the normal distance between axis x and that parallel passing through the centroid, or yc. Therefore 2 3 2 4 ab3  2 ab   b  128  9   ab Icx  I x  Ayc 2   . Similarly,    9     8  288  2  4  a3 b  2 ab   a 2  2  8  a3 b  2 Icy  Iy  Axc    0.03 a3 b.    3 3  2    2  Polar moment of inertia is the second moment of area with reference to a desired point. Considering the origin,

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Examples Applied in Engineering • 171

Jo  r 2 dA    x2  y2  dA  I x  Iy 

2 3 2 3 4 ab3    4  a b 13   36  a b    0.3308 a3 b. 9 3 9 3

2 2  9 4  1152  a3 b 288 

3

2 3 2 3 4 ab3    4  a b 13   36  a b    0.330 9 3 9 3

We can also calculate the polar moment of inertia with respect to the centroid, Jc. Or Jc  Icx  Icy.Therefore 128  9 2  ab3  2  8  a3 b  272 2  9 4  1152  a3 b  Jc     0.0734 a3 b 3 3 288  2 288   0.0734 a3 b. Table 8 lists the results for the sinusoidal shape area: TABLE 8  Some results for sinusoidal shape

Shape sinusoidal

100_Integrals.013_3pp.indd 171

Area

Centroid, yc

2ab π

πb 8

Ix 4 ab3 9π

Icx

Jo

128  9  ab 13 2

288 

3

Jc 2

 36  a3 b  272 2  9 4  1152  a3 b 9 3

288 3

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172 • 100 Integrals

8  TRIANGULAR SHAPES We consider an equilateral triangle shape with side a. Its area, centroid, and moment of inertia are calculated using related ­ integrals. A triangular shape can be the cross-section shape of beams, rods, or hydraulic gates for example.

y

a

a

h

dx dy 60°

60° a/2

a/2

x

Figure 8  A triangular area shape

a2 3 , from geometry. We calculate the area using 4 the integration of the area differential element dA = dxdy as shown in Figure 8. Using the geometrical properties of the triangle we 2x   a 3 have y  h  1   , where h = is the height of the triangle. a   2 a/2 y a/2 a/2  2x  x2  a  Therefore, we have A   dxdy  dy  dx  2 h   1   dx  a 3  x    a  a 0  0 a/2 0  Area is equal to

a/2

 2x  x2  a2 3   2 h   1   dx  a 3  x    . a  a 0 4  0  a/2

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Examples Applied in Engineering • 173

Centroid is the first moment of area divided by its total area and is ydA  ydA located at  0, y  when symmetry about y-axis exists. Or y  A dA y 2  a/2 a/2 a/2 3 ydA ydA    . But ydA  2 ydy dx  y2 dx  h2  1  2 x  dx  a . y  0 0 0 0  a  8 A dA



Therefore, y =

a3 / 8 a 3 . = 2 6 a 3 /4

Moment of inertia is the second moment of area with reference to a desired axis. Considering x-axis as reference, y 3 a/2 a/2 a/2 2 2x  2 h3  2 2 3 I x  y dA  2 y dy  dx   y dx   1   dx. 3 0 3 0  a  0 0 Performing the integral gives 3 a/2 2x  2 h3 2 h3  Ix  1  dx    3 0  a  3

a/2

3 2 4 3 2 4 3 4   x  a x  a2 x  a3 x   32 a . 0

Similarly, we can calculate the moment of inertia with respect to the y a/2 a/2 a/2 2x   2 y axis. Or Iy  x dA  2 dy  x2 dx  2  yx2 dx  2 h  x2  1   dx . a   0 0 0 0 Performing the integral gives a/2

 x3 x 4  2x  3 4  Iy  2 h  x  1   dx  2 h     a a  96   3 2a 0 0 a/2

2

Using the parallel axis theorem, we can calculate the moment of inertia with respect to the parallel axis at the centroid, Icx. Or I x  Icx  Ad 2 , where d is the normal distance between axis x and the parallel axis passing through the centroid, or y. Therefore 2

3 4 a2 3  a 3  3 4 Icx  I x  Ad  a  a.    32 4  6  96 2

Polar moment of inertia is the second moment of area with reference to a desired point. Considering the origin as the reference, 3 4 3 4 3 4 Jo  r 2 dA    x2  y2  dA  I x  Iy  a  a  a. 32 96 24

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174 • 100 Integrals

9  RECTANGULAR SHAPES We consider a rectangular shape with base b and height h. The area, centroid, and moment of inertia are calculated using related integrals. A rectangular shape can be the cross-section shape of a beam, rods, or hydraulic gates for example.

y

dy

h dx

x

b Figure 9  A rectangular area shape

Area is equal to bh, from geometry. We can calculate the area using integration of the differential area element dA = dxdy as shown in Figure 9. Therefore, considering a coordinate system, x − y with its origin at the down left vertex, we have h

b

b

0

0

0

A   dxdy  dydx  h dx  bh. Centroid is the first moment of area divided by total area. We h

can write

b

ydA   ydy dx . y 0

A

0

bh

Therefore. After performing h

b

0

0

 ydy dx  b  y the integration operation, we get y 

2

h

/ 2  0

h  . 2

bh bh b xdA   0xdx 0dy  h  x / 2  0  b . Therefore, the Similarly, x  A bh bh 2 centroid is at C  b / 2, h / 2 . Theses results are consistent with those b

h

2

obtained from the symmetry property of the rectangular shape.

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Examples Applied in Engineering • 175

Moment of inertia is the second moment of area with reference to a desired axis. Considering x-axis as the reference, we can write h b h h bh3 2 2 I x  y dA  y dydx. Therefore, I x  by2 dy  b  y3 / 3  0  . 3 0 0 0 Similarly, moment of inertia about y-axis reads b h 3 hb Iy  x2 dA  x2 dx dy  . 3 0 0 Using parallel axis theorem, we can calculate the moment inertia about a system of coordinates, xc − yc with its origin at 2 bh3 bh3 h 2 the centroid. Or I xc  I x  Ay   bh    , and 3 12 2 2 hb3 hb3  b Iyc  Iy  Ad 2   bh    . 3 12 2 bh 2  h  b2 . 3 Also, the polar moment of inertia about centroid reads bh 2 Jc  I xc  Iyc   h  b2 . 12 Polar moment of inertia about origin O, is Jo  I x  Iy 

We can also calculate Jc using the parallel axis theorem, as  b2 h 2  bh 2 bh 2 Jc  Jo  Ar 2  h  b2   bh       h  b2 . 3 4 4 12  

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176 • 100 Integrals

10  COMPLEX SHAPES A complex shape can be divided into simple shapes. This ­facilitates the calculation of the area properties for complex shapes using those of simple shape components. The only requirement is this that for each simple shape property, for example moment of inertia, we consider a common reference axis. Again, parallel axis theorem can be used for transformation of properties. 10.1  Example: A semi-circle with a semi-elliptical hole We consider a semi-circular shape with radius R and with its center located at the origin. A semi elliptical shape is taken away from the original area with its semi radii being a  R and b  R , respectively (0    1 and 0    1, are constants). Figure 10 shows the complex shape.

y

b

x

a R Figure 10  A complex semi circular area shape with a semi ellipse hole

Area can be obtained by subtracting the area of the semi-ellipse from that of the semi-circle. As shown in the previous sections (see R 2 R 2  1    2 Table 3 and Table 5), we have A    R. 2 2 2

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Examples Applied in Engineering • 177

Centroid is located at y distanced from the x- axis on the y-axis. Using the obtained results from the previous sections, we have  R 2   4 R   R 2   4R       4 1  2  2   3   2   3   y  R.  1    2 3  1    R 2 Moment of inertia with respect to x-axis is simply that of semi-ellipse part subtracted from the semi-circle’s. Therefore, R 4 3 R 4 R 4 Ix    1  3 . But moment of iner8 8 8 tia with respect to the axis at the centroid of the complex shape requires the application of the parallel axis. For the semi-circle part we have, using the previously obtained 2  9 2  64  4  R 2   4 R 4 1    R   . For results, Icx  cirle    R     72    2   3  3  1     the semi-ellipse part we have, using the previously obtained results, 2

Icx  ellipse

 9 2  64  3 4  R 2    R    72    2

2   4R 4 1    R   .     3  3  1     2

As a numerical example, let   0.75, and   0.5. Therefore, we get the following results as shown in Table 9. TABLE 9  Results for a composite shape

Shape

Area

Centroid, y

Ix

Icx _ circle

Icx _ ellipse

composite

0.9817 R 2

0.5093 R

0.3559 R 4

0.39795R 4

0.06228 R 4

10.2  Example: A rectangle with circular segment sides We consider a rectangular shape with base b and height h. The two vertical sides are composed of two circular segments where the corresponding circles’ centers are located at h / 2 with radii R. Figure 11 shows the complex shape.

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178 • 100 Integrals

A

B

4

R θ 2

2

h

-4

-2

2

0

4

θ 2

-2

D

-4 C

b Figure 11  A complex rectangular area shape with circular segment sides

Considering the complex shape composed of two parts: a rectangle and two circular segments. We use the results obtained in previous sections for calculations. Area can be obtained by subtracting the area of the segments from that of the rectangle. As shown in previous sections we R2  have A  bh  2    sin  . But, from geometry, h  2 R sin . 2 2  Therefore, we have A  2 bR sin  R 2    sin  . Note that θ is 2 the angle corresponding to the circular arc. Centroid is simply at the geometrical center of the shape, due to h  symmetry. Therefore, y   R sin . 2 2

Moment of inertia with respect to the axis at the centroid and parallel to the base is simply that of rectangle part subtracted by the circular bh3 R4   segments. Therefore, Icx   2  3  3 sin   2 sin  sin 2 . 12 24  2 3 R  Therefore, after simplifications, we have Icx  8 b sin 3   / 2    12  R3   3 2   Icx  b  R   . Similarly, the 8 sin  / 2 3  3 sin  2 sin  sin     12  2   

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Examples Applied in Engineering • 179

moment of inertia with respect to axis at the centroid and p ­ arallel to the sides is that of the rectangle part subtracted by the circular segments. But the distance from the centroid of the circular arc  4 R sin 3    b  2 . to the y-axis reads, from geometry, d   R cos  2 2 3    sin   Therefore, the moment of inertia of the circular sectors with respect to the y-axis , using the parallel axis theorem, reads

     sin 6    4R 2 b  64  R4  R 2   2 Iys    sin   2 sin  sin    .     sin     R cos  8  2 3  2  9   sin   2 2    2     3    R sin in 6    4   2  2    R   sin   b  R cos    2  .    sin   2 2 3    sin    2    Therefore,

for

hb3 Icy   2 Iys  12

the

complex

 b3 R sin   4 2  R 6 4

shape

we

can

write

    sin 6      64    2    R2   sin   2 sin  sin 2    .   2  9   sin      2

       4 R sin 3    sin 6     b   64    2    R 2   sin    R cos   2  .   2 sin  sin 2    .   sin 2 2 3       2  9   sin       From these results, the polar moment of inertia about centroid can be calculated Jc  Icx  Icy. 10.3  Example: A semi-circle with a triangular shape hole We consider a semi-circle with radius R when a triangular shape with base b and height h. Is subtracted from it, as show in Figure 12. The complex shape is symmetric about the y-axis with conditions b that < R and h < R . 2

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180 • 100 Integrals

y

5

4

3

2

1 D

F X

−4

−3

−2

E

h C

O −1

0

1

2

G 3

4

X

b/2 −1 R −2

Figure 12  A complex semi-circular area shape with a triangular shape hole

Considering the complex shape composed of two parts: a rectangle and two circular segments. We use the results obtained in the previous sections for calculations. Area can be obtained by subtracting the area of the triangle from R 2  bh that of the semi-circle. Or A  . 2 Centroid is simply at the geometrical center of the shape. Therefore, with reference to the base x − x axis we have  R2 / 2   4R / 3    bh / 2  h / 3  . After some simplificay  R2  bh  / 2 tions, we have y 

100_Integrals.013_3pp.indd 180

4 R 3  bh2 . Due to symmetry, x = 0. 3  R 2  bh 

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Examples Applied in Engineering • 181

Moment of inertia with respect to the x-x axis can be πR 4 bh3 . ­calculated as that of triangle, from the semi-circle’s, 8 12 R 4 bh3 Or I x   . After using the parallel axis the 8 12 orem, we can calculate the moment inertia about the centroid

as

 R 4 bh3 I xc    12  8

2

  R 2  bh   4 R 3  bh2  .   2 2     3  R  bh  

 64  9  R 2

Or,

after

 64  9  R  2

Icx

simplifying 6

this

expression

we

 9 bhR 4  32 bh2 R 3  6 bh3 R 2  2 b2 h 4 72  bh  R 2 

get

Icx 

.

Similarly, using symmetry, the moment of inertia with respect to R 4 hb3 the y-axis can be written as Iy  Icy   . 8 12 From these results, the polar moment of inertia about the centroid can be calculated Jc  Icx  Icy.

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6



182 • 100 Integrals

11 A CANTILEVER BEAM WITH CUBIC LOAD DISTRIBUTION A cantilever beam is loaded with a cubically distributed load, as 3 x shown in Figure 13. The load is given as q     where, ω is L the maximum load density per unit length (in N/m), and L beam length. We calculate the equivalent load, its acting location xc, and the distribution of the shear and bending moment for this beam.

ω

A

L

B

X

Z Figure 13  A Cantilever beam with cubic load distribution.

Equivalent load W, is the area under the distribution. Therefore, 3 L L x we can write W  q dx     dx. Performing the integration L 0 0  3 L 4  L  L x gives    dx    3   . L  4L  4 0  Centroid of the load, xc is where W is acting. Therefore, 3   L5  L x L  x  xq dx   0  L   L3  5   4 L . We can calcuxc  0 L L L 5 4 4 4 L late the support reaction force RA  , and the moment 4

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Examples Applied in Engineering • 183

L  L   4 L  MA    , after writing the balance of forces in   5  4  5  the z-direction and the that of the moments of point A. 2

Shear force distribution, V as a function of x, is given as V   q dx  C1 where, C1 is determined by the boundary conditions for the shear force, for example V x = L = 0. Performing the  integration, gives V   q dx  C1   3 x 4  C1 . Applying the 4L  4 L boundary conditions gives  3 L  C1  0  C1  . Hence 4 4L  L V   3 x4  . Note that the shear force equation recovers the 4L 4 L . reaction force at the support A, or V( x  0)  4 Bending moment distribution, M as a function of x, is given as M  V dx  C2 where, C2 is determined by the boundary conditions for the moment, for example M x = L = 0. Performing the integraL   5 L   x  tion, gives M     3 x 4  x  C2 .  dx  C2   4  20 L3 4  4L Applying the boundary conditions gives 2 2  5 L L2  5 L L  L   C  0  C   . Hence M   x   x 2 2 20 L3 4 5 5 20 L3 4 2  5 L L M x  . Note that the bending moment equation recovers the x 20 L3 4 5 L2 reaction moment at the support A, or M( x  0)   . 5

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184 • 100 Integrals

12 A CANTILEVER BEAM WITH QUARTER-ELLIPSE LOAD DISTRIBUTION A cantilever beam is loaded with an elliptical quarter distributed 2

x load, as shown in Figure 14. The load is given as q   1    L where, ω is the maximum load density per unit length (in N/m), and L beam length. We calculate the equivalent load, its acting location xc, and the distribution of the shear and bending moment for this beam.

ω

A

X

B

Z Figure 14  A Cantilever beam with quarter-ellipse load distribution.

Equivalent load W, is the area under the distribution. Therefore,

we

can

write

L

L

2

x W  q dx   1    dx. L 0 0

x  sin   dx  L cos  d. Writing the integral in terms L of the variable α and performing the integration gives

Let



L

 0

2





2 2 L 2 L L 1 x 1    dx  L  cos2  d  1  cos 2  d    sin 2    4 2 0 2  2 L 0 0 

L  1

2 L 2  d    sin 2   .  2  2 4 0

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Examples Applied in Engineering • 185

Centroid of the load, xc is where W is acting. Therefore, 2

L x  x 1    dx 2 L 0 4 L x xc   x 1    dx. Writing the inteL L 0 L 4 gral in terms of the variable α and performing the integration 

L



2

2 2 x 2 L 2 2 21 3 gives x 1    dx  L  sin  cos d   L  cos    . 3 L 3 0 0 0 2  4   L  4L Therefore, xc   .     L   3  3 

We can calculate the support reaction force RA 

L , and the 4

L2  L   4 L  moment M A    , after writing the balance     3  4   3  of forces in the z-direction and the that of the moments about point A . Shear force distribution, V as a function of x, is given as V   q dx  C1 where, C1 is determined by boundary conditions for the shear force, for example V x = L = 0. Performing the inte2

x gration, gives V    1    dx  C1 . Writing the integral L x in terms of variable α (recall  sin   dx  L cos  d) gives L 1 L L   1  cos 2  d  C1      sin 2   C1. Applying the  2 2  2  boundary condition, V x  L  V   / 2  0 gives  L  C1  0  C1  L 4 4 L  1  L L L . Writing the   C1  0  C1  . Hence, V      sin 2   4 2  2  4 4 shear force equation in terms of the original variable x,

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186 • 100 Integrals

1 2x L  L L L   L   1    sin  cos         sin 2   2  2 4 2 4 4 L     2   L 2x L L x x  1  2  2 sin 1   . Note that the shear    sin  cos      2 4 4  L L  L 

 C2 

gives V  

force equation recovers the reaction force at the support A, or L V ( x  0)  . 4 Bending moment distribution, M as a function of x, is given as

M  V dx  C2 where, C2 is determined by boundary conditions for the moment, for example M x = L = 0. writing the integral,  x2 L  2x x2 L  L 1 x   gives M     sin x  x  dx  sin 1  1 2 dx C   2 2 2     L 4 2 4  L L L 2  L  x2 L x x  x 1  2 dx  sin 1 dx  C2. But writing the former integral  4 2 L 2 L x   in terms of the variable   recall,  sin   dx  L cos  d  L   3/2

x2  L2 L2 L2  x2  sin  cos2  d  cos3   reads  x 1  2 dx   1  2   L 2 6 6  2 L  2 2 2 3/2 L L  x   cos3    1  2  . Similarly, the latter integral can be written as L  6 6  L L2 L2 L2  x 1  x  1 x  sin   cos   sin  cos   dx   d        sin    1 2  2  2 2  L L L x2  L2 L2  x 1  x    sin   cos      sin    1  2 . After collecting all related L 2  L L 3/2 L L2  x2  Lx 1  x  L answers, we get the solution as M  1  sin    x   2  L  2 2 4 6  L

x2  L2   1  2  L  6 

3/2



x2 Lx 1  x  L2 sin    1  2  C2 . Applying the boundary con2 2 L L L2 L2 dition, M x = L = 0 gives   C2  0  C2  0 . Hence, 4 4

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Examples Applied in Engineering • 187

the bending moment equation, after simplification, reads   L  x2  x2 1  x  M 3 x  2 L  2  2  1  2  6 x sin   . Note that the 12  L  L  L    bending moment equation recovers the reaction moment at the L2 support A, or M( x  0)   . 3

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188 • 100 Integrals

13 A CANTILEVER BEAM WITH INVERSE COSINE LOAD DISTRIBUTION A cantilever beam is loaded with an inverse cosine distributed load. 2 x The load is given as q  cos1   where, ω is the load density  L per unit length (e.g., N/m), and L beam length. In this section, we present the calculations for the equivalent load, its acting location xc from the support, and the distribution of shear and bending moment along the length of this beam. Equivalent load W, is the area under the distribution. L L 2 x Therefore, we can write W  q dx  cos1   dx. Let   L 0 0 x x cos1      cos   and dx   L sin  d. Note that the limits L L  of integral change to   for x = 0 and   0for x = L. Writing the 2 integral in terms of the variable α and performing the integration gives L 0 2 2 L 2 L 2 L 1  x  cos  sin  d     dx   cos   sin  0       0     2 L 2

 cos   sin  

0  2

2 L  .  Centroid of the load, xc is distance from the support where W is a­ cting. L 2 L x x x cos1   dx  x cos1   dx  0 0  L   L  . Writing the Therefore, xc  2 L L  integral in terms of the variable α and performing the integration gives 

L

2 1  x  2 x dx L  sin cos  d    L2 cos     0  L   0

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 sin 2   0 1 0  L2   2     sin   d       2 2 4        2  2          / 4

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Examples Applied in Engineering • 189

0  sin 2   0 1 0  L2  sin 2   L2  2 d    L   sin  d              . Therefore, 2   2  8 2     4  2 2       2      /2       / 4 2

xc 

L2 L /L . 8 8

We can calculate the reaction force at the support, point A as 2L L2  L   2L  RA  , and the moment M A    , after    4  8    writing the balance of forces in the z-direction and the that of the moments about point A. Shear force distribution, V as a function of x is given by V   q dx  C1 where, C1 is determined by the boundary conditions for shear force, for example V x = L = 0. Performing the 2 x integration gives V   cos1   dx  C1. But the inte  L gral in terms of the variable α reads (see previous paragraph) V    cos   sin    C1. Applying the boundary condition at the tip of the beam, or V x  L  V   0 gives C1 = 0. Hence, shear force as  2 L  x2 x 1  x  a function of x reads V   1  2  cos    . Note that   L L  L   the shear force equation recovers the reaction force at the support 2L A, or V x 0  .  Bending moment distribution, M as a function of x, is given by M  V dx  C2 where, C2 is determined by the boundary condition for moment, for example M x = L = 0. Performing the  2 L  x2 x 1 x integration, gives M 1   cos   dx  C2 .    L2 L L 

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190 • 100 Integrals

But the integral in terms of the variable α reads (recall x 2L2 cos  ), M  sin2    sin  cos   d  C2 . But, L   using the integration by parts technique, we have  2L2 2L2   sin 2  1    sin 2  d   .     sin cos d      2 2  Therefore, after combing the results we have 2 2  2L2  3  si 2 L   sin  1 2 2 M  d    d   C   sin 2  d   sin   sin 2        2 2   2 2  2 2  2 L  3   sin    sin 2  d    d   C2    C2 . The remaining integral can be  2   2   3 3 3 sin 2 2 written as  sin  d     1  cos 2  d      .  2 4 4 8  2L2  3 sin 2  sin 2   Therefore, M      C2 . Applying    4 8 2 

the boundary condition, M x  L  M   0  0, we get C2 = 0 . Now, writing the equation for M in terms of the original variable x, we get 2L2  3 sin 2  sin 2   2L2  3 1 x x2 1 x 1  x  M   cos1        cos  1      2   4 8 2    4 L 2 L L 4 L 3 x2 1 x2   x 1 x  x   cos1    1  2  cos1    1  2  . After collecting similar terms in 4 L 2 L   L 4 L  L  2 x2  1  x  L2  x x2  1 1 this expression and simplifying, we have M        cos   2   L L2  L2  L x2  L2  x 2 x2   x   1  2   1  2  cos1    .  L  2   L L   L  

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Examples Applied in Engineering • 191

14 A CANTILEVER BEAM WITH PARABOLIC LOAD DISTRIBUTION A cantilever beam is loaded with an inverse cosine distributed  x2  load. The load is given as q    1  2  where, ω is the load density L   per unit length (e.g., N/m), and L beam length. In this section, we present the calculations for the equivalent load, its acting location xc from the support, and the distribution of shear and bending moment along the length of this beam. Equivalent load W, is the area under the distribution. Therefore, L L  x2  we can write W  q dx    1  2  dx. Performing the integraL  0 0 L  2 L x3  tion gives W    x  2   . 3L 0 3  Centroid of the load, xc is distance from the support where W is L L   x2 x4  x2   x  1  2  dx 3   2  0 L   2 4L 0 3L    acting. Therefore, xc  . 2 L 2L 8 3 We can calculate the reaction force at the support, point A as 2 L L2  3 L   2 L  RA  , and the moment M A    , after   3 4  8  3  writing the balance of forces in the z-direction and the that of the moments about point A. Shear force distribution, V as a function of x is given by V   q dx  C1 where, C1 is determined by the boundary conditions for shear force, for example V x = L = 0. Performing the integration   x3  x2  gives V    1  2  dx    x  2 . Applying the boundary L  3L    2 L condition at the tip of the beam, or V x = L = 0 gives C1  . Hence, 3 2 L  the shear force as a function of x reads V   x  2 x3. Note 3 3L

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192 • 100 Integrals

that the shear force equation recovers the reaction force at the 2 L ­support A, or V x 0  . 3 Bending moment distribution, M as a function of x, is given by M  V dx  C2 where, C2 is determined by the boundary condition for moment, for example M x = L = 0. Performing the integra 2L x3  tion, gives M     x  2  dx  C2. But the integral reads 3L   3 3  2L  2L x  x2 x4     x  2  dx    x   . Applying the bound3L  2 12 L2   3  3 L2 ary condition, M x = L = 0, we get C2   . Now, writing the equa4  1 4 1 2 2L L2  x  x  x  tion for M we get M    . 2 2 3 4   12 L

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Examples Applied in Engineering • 193

15 A CANTILEVER BEAM WITH CIRCULAR SEGMENT CROSS-SECTION AND QUARTERELLIPSE LOAD DISTRIBUTION We consider a cantilever beam with a circular segment cross-­ section. The beam is under a quarter elliptical distributed load. We calculate the bending and shear stresses at the point of maximum value. In previous sections, we calculated the moment of inertia and the centroid of a circular segment. These results are repeated here for convenience, after some simplifications:  4 R sin 3   θ 2 Centroid yc  R cos , distance of centroid from the 3    sin   2  segment base. y

  64 sin 6     R  2   , with Moment of inertia Ic    sin  cos   8  9    sin      respect to the neutral axis at the centroid. 4

Also, in the previous section, we calculated the maximum bending moment and the shear force due to a quarter-ellipse load distribution. These results are repeated here for convenience: L L2 , Mmax   . where, ω is the maximum load den4 3 sity per unit length measured at the beam support location, and L beam length.

Vmax 

Mmax y. Where, y is the disIc tance measured from the neutral axis in the plane of the cross section. Assuming a bending moment that puts the top fiber of the beam at the tension (i.e., negative moment), we get The bending stress is given as  

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194 • 100 Integrals

     sin 3    4 R sin 3     2   4R  3   2  . Therefore,  ytop  R  y  R  3    sin   3  4   sin           sin 3    64 sin 6    2  4  M 4 RL 3  2   / R   sin  cos   2  top  max ytop     9    sin   Ic 9  4   sin   8        64 sin 6    4  R  2  . After some simplifications, we get    sin  cos   9    sin   8     192    sin    256 sin 3   L2 2  top   3 . . Similarly, 8R 6    64 sin   9    sin      sin  cos 2 the stress due to bending at the bottom fiber of the beam reads     3 64 sin 6    2  4 R sin  4   M L2 L  2   R cos   / R   sin  cos    2   L   bottom  max yc  9    sin   8 R 3 3  3    sin   2 8  Ic   

       64 sin 6   256 sin 3    192    sin   cos 4  2 R L 2  2   L 2 .   sin  cos   . 3 9    sin   8 R 8   9    sin      sin  cos    64 sin 6    2  Note that the ratio of the stress at the top and the bottom of ytop the beam cross section is exactly equal to the value of . Or ybottom  3    sin    4 sin 3    top ytop  2  . These results con    bottom ybottom  4 sin 3    3    sin   cos 2 2   firm that the variation of the stress is linear across the cross section of the beam.

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Examples Applied in Engineering • 195

Exercise problems 1. Calculate area, centroid, moment of inertia, polar moment of inertia for a quarter circle. Consider the axis through centroid for the calculations. 2. Calculate area, centroid, moment of inertia, polar moment of inertia for a half circle with a half circular hole. Consider the axis through centroid for the calculations. 3. Calculate area, centroid, moment of inertia, polar moment of inertia for a quarter ellipse. Consider the axis through centroid for the calculations. 4. Calculate area, centroid, moment of inertia, polar moment of inertia for a half ellipse with a half circle hole. Consider the axis through centroid for the calculations. 5. Calculate area, centroid, moment of inertia, polar moment of inertia for a half ellipse with a half ellipse hole. Consider the axis through centroid for the calculations. 6. Calculate area, centroid, moment of inertia, polar moment of inertia for a half circle with a half ellipse hole. Consider the axis through the centroid for the calculations. 7. Calculate area, centroid, moment of inertia, polar moment of inertia for a quadratic polynomial shape. Consider the axis through centroid for the calculations. 8. Calculate area, centroid, moment of inertia, polar moment of inertia for a half ellipse with a quadratic hole. Consider the axis through centroid for the calculations. 9. Calculate area, centroid, moment of inertia, polar moment of inertia for a half ellipse with a spandrel hole. Consider the axis through centroid for the calculations. 10. Calculate area, centroid, moment of inertia, polar moment of inertia for a half circle with a rectangular add on to it along the diameter. The width of the rectangle is equal to the circle radius divided by two. Consider the axis through the centroid for the calculations.

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y

196 • 100 Integrals

16  PROBABILITY DENSITY FUNCTIONS-PDF Probability density functions/distributions (e.g., Normal, Weibull) are used in engineering and statistics. Usually, the integral of these functions is needed and are referred to as CDF (cumulative distribution function). In practice, engineers use standard tables for related calculations. In this section, we use the integration techniques for calculating the CDF related to Gaussian/Normal and Weibull distributions. 16.1  Normal distribution Basic form of Normal distribution is given as N  x   e x dx, also 2 related to as Error function. Plotting the e− x shows that it has a maximum value of 1 at x = 0 and symmetrically decreases about the vertical axis. This so-called bell-shape can represent many natural and industrial random data probability distributions. In this section, we show how to perform integration this function and calculate N  x  for an infinite range, x  0,  . 2





We can write the integral as N  e x dx  e y dy. Since the answer 0

2

2

0

is equivalent regardless of the variable x or y. Now, we calculate    2 2   x2  y2  N 2  e x dx e y dy  e dxdy. This equation is valid since, 0

0

00

for example, the value of the integral in terms of the variable y is treated as a constant for the integral in terms of x, or vice versa. Now, we transform the integral from Cartesian  x, y  to polar coordinate  r, . We can write x  r cos and y  r sin . Therefore, x2  y2  r 2 and dxdy  rdrd (i.e., the differential area elements defined in each coordinate system). This can be shown as follow: dx 

y y x x dr  d  cos  dr  r sin  d and dy  dr  d   sin  dr  r cos  d  r  r 

y y dr  d   sin  dr  r cos  d  . But the Jacobian of transformation reads r  cos   r sin  J  r , determinant of the Jacobian matrix. sin  r cos 

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Examples Applied in Engineering • 197

Therefore, the differential elements are related through dxdy  Jdrd  rdrd. 

Now we can write

e



 x2  y2



dxdy 

00

/2



0

0

 de

 r2

rdr. Note that the

limits of the integrals read r  0,  and   0,  / 2 , consistent with 

x and y  0, . Now we calculate this integral in terms of r, or e r rdr . 2

0

2 Let r  u  2 rdr  du and write the new integral in terms of   2  1 1 1 1 the variable u as e r rdr  e u du    e u  0    0  1   . 2 2 2 2 0 0 2 Therefore, N 



N  x   e x dx  0

2

/2



0

0

r  d e rdr  2

 1 / 2

1 2

/2

  2  d   . Or N  x   e x dx  0 4 2 0



2  . Note that due to symmetry,  e x dx   . 2 

The normalized form of the integral, π dividing both side with , or Nˆ  x   2 

have

Nˆ  x  can be calculated by  2  x2 e dx  1. Similarly, we  0

2 1 e x dx  1.   

The standard form of the integral, N  x  can be calculated by using z dz the change of the variable technique. Let x   dx  and 2 2 the normalized form of the integral in terms of the variable z reads   z2 z2   2 1 2 2 e dz  e dz  1. as Nˆ  z   . Similarly, we have 1   2 0 2   It is customary to plot the distribution about its mean µ, versus x origin and scale it with its standard deviation σ, or z  . Mean  is defined as the integral of the first moment of distribution. Or

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198 • 100 Integrals



 e x  1   xe dx      . The Variance (or Standard deviation  2  0 2 0 2 square, σ ) is defined as the integral of the second moment of dis 2  2 tribution. Or   x2 e x dx  . 4 0 

2

 x2

16.2  Weibull distribution b 1

t

b

  b t  Weibull density function is given as f  t     e    . It gives  distribution of the function f vs. time for given shape factor, b and scale factor, θ. Here we use two-parameter Weibull for our calculations. For a given set of data, the values of b and θcan be obtained to fit a given set of data. For example, for b = 3 Weibull and Normal distributions are closely alike. t

b

 t The CFD, is F  t    f  t  dt . Let    z  dt  b1 dz.  bt   0 Therefore, writing the integral in terms of z gives b

b

t t  b  t  b 1   b       b1 e z dz  e z dz. 0 0       bt t b b t   t     z    1  e Performing the integration gives F  t    e    e .   0 t

t

b t  0 f  t  dt  0    

b 1

e

t   

Mean and variance can be calculated as well. These quantities involve special function, Gamma [8]. The Gamma func

tion is defined as   x   t x 1 e t dt. It can be shown that 0

  x    x  1    x  1    x  1  ! . The mean for Weibull distri

b 1

t

b

  1 bt  t       1  . bution can be worked out as      e      2 2 2 Similarly, the variance     1  2 b    1  1 / b   .

Weibull distribution has extensive applications in Reliability engineering.

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Examples Applied in Engineering • 199

REFERENCES [1] [Online]. Available: https://en.wikipedia.org/wiki/Lists_of_ integrals. [2] K. Charlwood, “Integration on Computer Algebra Systems,” The Electronic Journal of Mathematics and Technology, vol. 2, no. 3, pp. 291-301, 2008. [3] J. e. Sasikala, “Applications of Integral Calculus in Engineeing,” International Journal of Science, Engineering and Management (IJSEM), vol. 2, no. 11, pp. 112-116, 2017. [4] “University of South Carolina,” [Online]. Available: https:// people.math.sc.edu/girardi/m142/integration/100problems. pdf. [Accessed 2022]. [5] “LibreTexts,” CXone Expert knowledge management, [Online]. Available: https://math.libretexts.org/Courses/Monroe_Community_College/MTH_211_Calculus_II/Chapter_7%3A_Techniques_of_Integration. [Accessed 2023]. [6] “Openstax,” [Online]. Available: https://openstax.org/books/ calculus-volume-1/pages/a-table-of-integrals. [Accessed 2023]. [7] M. Math, “Integration Bee,” [Online]. Available: https://math. mit.edu/~yyao1/integrationbee.html. [Accessed 2023]. [8] Daniel Zwillinger (Editor), CRC Standard Mathematical Tables and Formulas, 33rd Edition, Chapman and Hall/ CRC, 2018. [9] “engineering Fundamentals-eFunda,” eFunda [Online]. Available: efunda.com. [Accessed 2023].

Inc.,

[10] D. Z. (Editor), Table of Integrals, Series, and Products, 8th Edition, Academic Press, 2014.

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