This document contains a chapter from a textbook on integral calculus. It lists 32 problems involving evaluating indefinite integrals using substitution and other integral formulas. The problems cover a wide range of integral formulas and techniques including trigonometric substitutions, hyperbolic substitutions, and inverse trigonometric substitutions.
Le tramway à Amiens, florilège de l'absurdeamiens2014
La construction d'un tramway à Amiens ruinerait la Ville. Quel sera le montant de la facture globale ? Quelle durée pour la construction ? Tout est possible : retards dus à des fouilles archéologiques, des terrains instables, bombes à désamorcer, effondrements d'immeubles, crise financière interdisant la poursuite du projet.
Aujourd'hui, les transports en commun ne sont pas saturés sur Amiens. Pire encore, les habitants les boudent de plus en plus.
Le projet se fait au mépris de toute véritable concertation publique.
Gilles Demailly, maire socialiste d'Amiens, oublie les maux qui frappent la Ville : taux de chômage exponentielle, l'insécurité florissante, la misère galopante, l'illettrisme, désertification urbaine, etc.
L'usine GoodYear vient de fermer ses portes. Arnaud Montebourg a ironisé en qualifiant "Gilles Demailly d'inspecteur des travaux finis." Il faut bien reconnaître que le soutien du maire s'est cantonné à l'affichage du bannière géante. A rien d'autre si ce n'est quelques mots sans intérêt et de fausses colères.
La population doit réagir face à ce projet séduisant mais rétrograde et absolument éco-irresponsable. Le tramway et Amiens n'ont jamais fait bon ménage.
A défaut d'avoir des projets, Demailly vend du rêve mais la folie a un coût impossible à amortir. Se laisser séduire, c'est endosser la faute d'une poignée d'hommes en rupture d'imagination. C'est aussi oblitérer l'avenir de nos enfants.
Après moi, le déluge, telle semble être la devise du maire. Non, le bon sens citoyen doit lui faire barrage.
Le tramway à Amiens, florilège de l'absurdeamiens2014
La construction d'un tramway à Amiens ruinerait la Ville. Quel sera le montant de la facture globale ? Quelle durée pour la construction ? Tout est possible : retards dus à des fouilles archéologiques, des terrains instables, bombes à désamorcer, effondrements d'immeubles, crise financière interdisant la poursuite du projet.
Aujourd'hui, les transports en commun ne sont pas saturés sur Amiens. Pire encore, les habitants les boudent de plus en plus.
Le projet se fait au mépris de toute véritable concertation publique.
Gilles Demailly, maire socialiste d'Amiens, oublie les maux qui frappent la Ville : taux de chômage exponentielle, l'insécurité florissante, la misère galopante, l'illettrisme, désertification urbaine, etc.
L'usine GoodYear vient de fermer ses portes. Arnaud Montebourg a ironisé en qualifiant "Gilles Demailly d'inspecteur des travaux finis." Il faut bien reconnaître que le soutien du maire s'est cantonné à l'affichage du bannière géante. A rien d'autre si ce n'est quelques mots sans intérêt et de fausses colères.
La population doit réagir face à ce projet séduisant mais rétrograde et absolument éco-irresponsable. Le tramway et Amiens n'ont jamais fait bon ménage.
A défaut d'avoir des projets, Demailly vend du rêve mais la folie a un coût impossible à amortir. Se laisser séduire, c'est endosser la faute d'une poignée d'hommes en rupture d'imagination. C'est aussi oblitérer l'avenir de nos enfants.
Après moi, le déluge, telle semble être la devise du maire. Non, le bon sens citoyen doit lui faire barrage.
Une rave-party illégale à dégénérée samedi 27 octobre du côté de Milan en Italie. Les sounds systems français et italiens Nonem, Hazard Unitz, Mechanika Crew et Teknomotive s'étaient réunis dans un hangar désaffecté à l'occasion des 10 ans du collectif Hazard Unitz. Mais la fête a vite pris une tournure de guérilla.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
Chapter 08
1. January 27, 2005 11:45 L24-CH08 Sheet number 1 Page number 337 black
CHAPTER 8
Principles of Integral Valuation
EXERCISE SET 8.1
1 1 1
1. u = 4 − 2x, du = −2dx, − u3 du = − u4 + C = − (4 − 2x)4 + C
2 8 8
3 √
2. u = 4 + 2x, du = 2dx, u du = u3/2 + C = (4 + 2x)3/2 + C
2
1 1 1
3. u = x2 , du = 2xdx, sec2 u du = tan u + C = tan(x2 ) + C
2 2 2
4. u = x2 , du = 2xdx, 2 tan u du = −2 ln | cos u | + C = −2 ln | cos(x2 )| + C
1 du 1 1
5. u = 2 + cos 3x, du = −3 sin 3xdx, − = − ln |u| + C = − ln(2 + cos 3x) + C
3 u 3 3
2 2 1 du 1 1 2
6. u = x, du = dx, = tan−1 u + C = tan−1 x + C
3 3 6 1 + u2 6 6 3
7. u = ex , du = ex dx, sinh u du = cosh u + C = cosh ex + C
1
8. u = ln x, du = dx, sec u tan u du = sec u + C = sec(ln x) + C
x
9. u = tan x, du = sec2 xdx, eu du = eu + C = etan x + C
1 du 1 1
10. u = x2 , du = 2xdx, √ = sin−1 u + C = sin−1 (x2 ) + C
2 1−u 2 2 2
1 1 6 1
11. u = cos 5x, du = −5 sin 5xdx, − u5 du = − u + C = − cos6 5x + C
5 30 30
√
du 1 + 1 + u2 1 + 1 + sin2 x
12. u = sin x, du = cos x dx, √ = − ln + C = − ln +C
u u2 + 1 u sin x
du
13. u = ex , du = ex dx, √ = ln u + u2 + 4 + C = ln ex + e2x + 4 + C
4 + u2
1 −1
14. u = tan−1 x, du = dx, eu du = eu + C = etan x
+C
1 + x2
√ 1 √
15. u = x − 1, du = √ dx, 2 eu du = 2eu + C = 2e x−1
+C
2 x−1
1 1 1
16. u = x2 + 2x, du = (2x + 2)dx, cot u du = ln | sin u| + C = ln sin |x2 + 2x| + C
2 2 2
√ 1 √
17. u = x, du = √ dx, 2 cosh u du = 2 sinh u + C = 2 sinh x + C
2 x
337
2. January 27, 2005 11:45 L24-CH08 Sheet number 2 Page number 338 black
338 Chapter 8
dx du 1 1
18. u = ln x, du = , 2
=− +C =− +C
x u u ln x
√ 1 2 du 2 −u ln 3 2 −√x
19. u = x, du = √ dx, =2 e−u ln 3 du = − e +C =− 3 +C
2 x 3u ln 3 ln 3
20. u = sin θ, du = cos θdθ, sec u tan u du = sec u + C = sec(sin θ) + C
2 2 1 1 1 2
21. u = , du = − 2 dx, − csch2 u du = coth u + C = coth + C
x x 2 2 2 x
dx
22. √ = ln x + x2 − 4 + C
x2 − 4
du 1 2+u 1 2 + e−x
23. u = e−x , du = −e−x dx, − = − ln + C = − ln +C
4−u 2 4 2−u 4 2 − e−x
1
24. u = ln x, du = dx, cos u du = sin u + C = sin(ln x) + C
x
ex dx du
25. u = ex , du = ex dx, √ = √ = sin−1 u + C = sin−1 ex + C
1 − e2x 1 − u2
1
26. u = x−1/2 , du = − dx, − 2 sinh u du = −2 cosh u + C = −2 cosh(x−1/2 ) + C
2x3/2
1 du 1 1 1
27. u = x2 , du = 2xdx, = sin u du = − cos u + C = − cos(x2 ) + C
2 csc u 2 2 2
2du
28. 2u = ex , 2du = ex dx, √ = sin−1 u + C = sin−1 (ex /2) + C
4 − 4u2
29. 4−x = e−x
2 2
ln 4
, u = −x2 ln 4, du = −2x ln 4 dx = −x ln 16 dx,
1 1 u 1 −x2 ln 4 1 −x2
− eu du = − e +C =− e +C =− 4 +C
ln 16 ln 16 ln 16 ln 16
1 πx ln 2 1
30. 2πx = eπx ln 2 , 2πx dx = e +C = 2πx + C
π ln 2 π ln 2
1 2 1
31. (a) u = sin x, du = cos x dx, u du = u + C = sin2 x + C
2 2
1 1 1
(b) sin x cos x dx = sin 2x dx = − cos 2x + C = − (cos2 x − sin2 x) + C
2 4 4
1 1 1 1
(c) − (cos2 x − sin2 x) + C = − (1 − sin2 x − sin2 x) + C = − + sin2 x + C,
4 4 4 2
and this is the same as the answer in part (a) except for the constants.
1 1
32. (a) sech 2x = = (now multiply top and bottom by sech2 x)
cosh 2x cosh2 x + sinh2 x
sech2 x
=
1 + tanh2 x
3. January 27, 2005 11:45 L24-CH08 Sheet number 3 Page number 339 black
Exercise Set 8.2 339
sech2 x
(b) sech2x dx = dx = tan−1 (tanh x) + C, or, replacing 2x with x,
1 + tanh2 x
sechx dx = tan−1 (tanh(x/2)) + C
1 2 2ex
(c) sech x = = x = 2x
cosh x e + e−x e +1
ex
(d) sech x dx = 2 dx = 2 tan−1 (ex ) + C
e2x+1
sec2 x 1 1
33. (a) = =
tan x cos2 x tan x cos x sin x
1 1 1 sec2 x 1
(b) csc 2x = = = , so csc 2x dx = ln tan x + C
sin 2x 2 sin x cos x 2 tan x 2
1 1
(c) sec x = = = csc(π/2 − x), so
cos x sin(π/2 − x)
1
sec x dx = csc(π/2 − x) dx = − ln tan(π/2 − x) + C
2
EXERCISE SET 8.2
1. u = x, dv = e−2x dx, du = dx, v = − 1 e−2x ;
2
1 1 −2x 1 1
xe−2x dx = − xe−2x + e dx = − xe−2x − e−2x + C
2 2 2 4
1 3x 1 3x 1 1 3x 1 3x
2. u = x, dv = e3x dx, du = dx, v = e ; xe3x dx = xe − e3x dx = xe − e + C
3 3 3 3 9
3. u = x2 , dv = ex dx, du = 2x dx, v = ex ; x2 ex dx = x2 ex − 2 xex dx.
For xex dx use u = x, dv = ex dx, du = dx, v = ex to get
xex dx = xex − ex + C1 so x2 ex dx = x2 ex − 2xex + 2ex + C
1 1
4. u = x2 , dv = e−2x dx, du = 2x dx, v = − e−2x ; x2 e−2x dx = − x2 e−2x + xe−2x dx
2 2
For xe−2x dx use u = x, dv = e−2x dx to get
1 1 1 1
xe−2x dx = − xe−2x + e−2x dx = − xe−2x − e−2x + C
2 2 2 4
1 1 1
so x2 e−2x dx = − x2 e−2x − xe−2x − e−2x + C
2 2 4
1
5. u = x, dv = sin 3x dx, du = dx, v = − cos 3x;
3
1 1 1 1
x sin 3x dx = − x cos 3x + cos 3x dx = − x cos 3x + sin 3x + C
3 3 3 9
4. January 27, 2005 11:45 L24-CH08 Sheet number 4 Page number 340 black
340 Chapter 8
1
6. u = x, dv = cos 2x dx, du = dx, v = sin 2x;
2
1 1 1 1
x cos 2x dx = x sin 2x − sin 2x dx = x sin 2x + cos 2x + C
2 2 2 4
7. u = x2 , dv = cos x dx, du = 2x dx, v = sin x; x2 cos x dx = x2 sin x − 2 x sin x dx
For x sin x dx use u = x, dv = sin x dx to get
x sin x dx = −x cos x + sin x + C1 so x2 cos x dx = x2 sin x + 2x cos x − 2 sin x + C
8. u = x2 , dv = sin x dx, du = 2x dx, v = − cos x;
x2 sin x dx = −x2 cos x + 2 x cos x dx; for x cos x dx use u = x, dv = cos x dx to get
x cos x dx = x sin x + cos x + C1 so x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C
1 1 1 2 1 1 2 1
9. u = ln x, dv = x dx, du = dx, v = x2 ; x ln x dx = x ln x − x dx = x ln x − x2 + C
x 2 2 2 2 4
√ 1 2
10. u = ln x, dv = dx, v = x3/2 ;
x dx, du =
x 3
√ 2 3/2 2 2 4
x ln x dx = x ln x − x1/2 dx = x3/2 ln x − x3/2 + C
3 3 3 9
ln x
11. u = (ln x)2 , dv = dx, du = 2 dx, v = x; (ln x)2 dx = x(ln x)2 − 2 ln x dx.
x
Use u = ln x, dv = dx to get ln x dx = x ln x − dx = x ln x − x + C1 so
(ln x)2 dx = x(ln x)2 − 2x ln x + 2x + C
1 1 √ ln x √ 1 √ √
12. u = ln x, dv = √ dx, du = dx, v = 2 x; √ dx = 2 x ln x−2 √ dx = 2 x ln x−4 x+C
x x x x
3 3x
13. u = ln(3x − 2), dv = dx, du = dx, v = x; ln(3x − 2)dx = x ln(3x − 2) − dx
3x − 2 3x − 2
3x 2 2
but dx = 1+ dx = x + ln(3x − 2) + C1 so
3x − 2 3x − 2 3
2
ln(3x − 2)dx = x ln(3x − 2) − x − ln(3x − 2) + C
3
2x x2
14. u = ln(x2 + 4), dv = dx, du = 2+4
dx, v = x; ln(x2 + 4)dx = x ln(x2 + 4) − 2 dx
x x2 + 4
x2 4 x
but dx = 1− dx = x − 2 tan−1 + C1 so
x2 +4 x2 + 4 2
x
ln(x2 + 4)dx = x ln(x2 + 4) − 2x + 4 tan−1 +C
2
5. January 27, 2005 11:45 L24-CH08 Sheet number 5 Page number 341 black
Exercise Set 8.2 341
√
15. u = sin−1 x, dv = dx, du = 1/ 1 − x2 dx, v = x;
sin−1 x dx = x sin−1 x − x/ 1 − x2 dx = x sin−1 x + 1 − x2 + C
2
16. u = cos−1 (2x), dv = dx, du = − √ dx, v = x;
1 − 4x2
2x 1
cos−1 (2x)dx = x cos−1 (2x) + √ dx = x cos−1 (2x) − 1 − 4x2 + C
1 − 4x2 2
3
17. u = tan−1 (3x), dv = dx, du = dx, v = x;
1 + 9x2
3x 1
tan−1 (3x)dx = x tan−1 (3x) − dx = x tan−1 (3x) − ln(1 + 9x2 ) + C
1 + 9x2 6
1 1 1 1 x2
18. u = tan−1 x, dv = x dx, du = dx, v = x2 ; x tan−1 x dx = x2 tan−1 x − dx
1 + x2 2 2 2 1 + x2
x2 1
but dx = 1− dx = x − tan−1 x + C1 so
1 + x2 1 + x2
1 1 1
x tan−1 x dx = x2 tan−1 x − x + tan−1 x + C
2 2 2
19. u = ex , dv = sin x dx, du = ex dx, v = − cos x; ex sin x dx = −ex cos x + ex cos x dx.
For ex cos x dx use u = ex , dv = cos x dx to get ex cos x = ex sin x − ex sin x dx so
ex sin x dx = −ex cos x + ex sin x − ex sin x dx,
1 x
2 ex sin x dx = ex (sin x − cos x) + C1 , ex sin x dx = e (sin x − cos x) + C
2
1
20. u = e3x , dv = cos 2x dx, du = 3e3x dx, v = sin 2x;
2
1 3x 3
e3x cos 2x dx = e sin 2x − e3x sin 2x dx. Use u = e3x , dv = sin 2x dx to get
2 2
1 3
e3x sin 2x dx = − e3x cos 2x + e3x cos 2x dx, so
2 2
1 3 9
e3x cos 2x dx = e3x sin 2x + e3x cos 2x − e3x cos 2x dx,
2 4 4
13 1 1 3x
e3x cos 2x dx = e3x (2 sin 2x + 3 cos 2x) + C1 , e3x cos 2x dx = e (2 sin 2x + 3 cos 2x) + C
4 4 13
1
21. u = eax , dv = sin bx dx, du = aeax dx, v = − cos bx (b = 0);
b
1 ax a
eax sin bx dx = − e cos bx + eax cos bx dx. Use u = eax , dv = cos bx dx to get
b b
1 a
eax cos bx dx = eax sin bx − eax sin bx dx so
b b
1 a a2
eax sin bx dx = − eax cos bx + 2 eax sin bx − 2 eax sin bx dx,
b b b
eax
eax sin bx dx = 2 (a sin bx − b cos bx) + C
a + b2
6. January 27, 2005 11:45 L24-CH08 Sheet number 6 Page number 342 black
342 Chapter 8
e−3θ
22. From Exercise 21 with a = −3, b = 5, x = θ, answer = √ (−3 sin 5θ − 5 cos 5θ) + C
34
cos(ln x)
23. u = sin(ln x), dv = dx, du = dx, v = x;
x
sin(ln x)dx = x sin(ln x) − cos(ln x)dx. Use u = cos(ln x), dv = dx to get
cos(ln x)dx = x cos(ln x) + sin(ln x)dx so
sin(ln x)dx = x sin(ln x) − x cos(ln x) − sin(ln x)dx,
1
sin(ln x)dx = x[sin(ln x) − cos(ln x)] + C
2
1
24. u = cos(ln x), dv = dx, du = − sin(ln x)dx, v = x;
x
cos(ln x)dx = x cos(ln x) + sin(ln x)dx. Use u = sin(ln x), dv = dx to get
sin(ln x)dx = x sin(ln x) − cos(ln x)dx so
cos(ln x)dx = x cos(ln x) + x sin(ln x) − cos(ln x)dx,
1
cos(ln x)dx = x[cos(ln x) + sin(ln x)] + C
2
25. u = x, dv = sec2 x dx, du = dx, v = tan x;
sin x
x sec2 x dx = x tan x − tan x dx = x tan x − dx = x tan x + ln | cos x| + C
cos x
26. u = x, dv = tan2 x dx = (sec2 x − 1)dx, du = dx, v = tan x − x;
x tan2 x dx = x tan x − x2 − (tan x − x)dx
1 1
= x tan x − x2 + ln | cos x| + x2 + C = x tan x − x2 + ln | cos x| + C
2 2
2 1 x2
27. u = x2 , dv = xex dx, du = 2x dx, v = e ;
2
2 1 2 x2 2 1 2 x2 1 x2
x3 ex dx = x e − xex dx = x e − e +C
2 2 2
1 1
28. u = xex , dv = dx, du = (x + 1)ex dx, v = − ;
(x + 1)2 x+1
xex xex xex ex
dx = − + ex dx = − + ex + C = +C
(x + 1)2 x+1 x+1 x+1
1 2x
29. u = x, dv = e2x dx, du = dx, v = e ;
2
2 2 2 2
1 2x 1 1 1
xe2x dx = xe − e2x dx = e4 − e2x = e4 − (e4 − 1) = (3e4 + 1)/4
0 2 0 2 0 4 0 4
7. January 27, 2005 11:45 L24-CH08 Sheet number 7 Page number 343 black
Exercise Set 8.2 343
1
30. u = x, dv = e−5x dx, du = dx, v = − e−5x ;
5
1 1 1
1 1
xe−5x dx = − xe−5x + e−5x dx
0 5 0 5 0
1
1 1 1 1
= − e−5 − e−5x = − e−5 − (e−5 − 1) = (1 − 6e−5 )/25
5 25 0 5 25
1 1
31. u = ln x, dv = x2 dx, du = dx, v = x3 ;
x 3
e e e e
1 3 1 1 3 1 3 1 3 1 3
x2 ln x dx = x ln x − x2 dx = e − x = e − (e − 1) = (2e3 + 1)/9
1 3 1 3 1 3 9 1 3 9
1 1 1
32. u = ln x, dv = 2
dx, du = dx, v = − ;
x x x
e e e
ln x 1 1
√ 2
dx = − ln x √
+ √
dx
e x x e e x2
√
1 1 √ 1
e
1 1 1 1 3 e−4
= − + √ ln e − √
=− + √ − +√ =
e e x e e 2 e e e 2e
1
33. u = ln(x + 2), dv = dx, du = dx, v = x;
x+2
1 1 1 1
x 2
ln(x + 2)dx = x ln(x + 2) − dx = ln 3 + ln 1 − 1− dx
−1 −1 −1 x+2 −1 x+2
1
= ln 3 − [x − 2 ln(x + 2)] = ln 3 − (1 − 2 ln 3) + (−1 − 2 ln 1) = 3 ln 3 − 2
−1
1
34. u = sin−1 x, dv = dx, du = √ dx, v = x;
1 − x2
√
3/2
√
3/2
√
3/2
√ √ √
3/2
−1 −1 x 3 3
sin x dx = x sin x − √ dx = sin−1 + 1 − x2
0 0 0 1 − x2 2 2 0
√ √
3 π 1 π 3 1
= + −1= −
2 3 2 6 2
√ 1
35. u = sec−1 θ, dv = dθ, du = √ dθ, v = θ;
2θ θ − 1
4 √ √1
4
1 √ 4 √ 4
sec−1 θdθ = θ sec−1 dθ = 4 sec−1 2 − 2 sec−1 2 − θ − 1
θ − √
2 2 2 θ−1 2 2
π π √ 5π √
=4 −2 − 3+1= − 3+1
3 4 6
1 1
36. u = sec−1 x, dv = x dx, du = √ dx, v = x2 ;
x x2−1 2
2 2 2
1 2 1 x
x sec−1 x dx = x sec−1 x − √ dx
1 2 1 2 1 x2 − 1
1 1
2 √
= [(4)(π/3) − (1)(0)] − x2 − 1 = 2π/3 − 3/2
2 2 1
8. January 27, 2005 11:45 L24-CH08 Sheet number 8 Page number 344 black
344 Chapter 8
1
37. u = x, dv = sin 2x dx, du = dx, v = − cos 2x;
2
π π π π
1 1 1
x sin 2x dx = − x cos 2x + cos 2x dx = −π/2 + sin 2x = −π/2
0 2 0 2 0 4 0
π π π π
1 2 π2
38. (x + x cos x)dx = x + x cos x dx = + x cos x dx;
0 2 0 0 2 0
u = x, dv = cos x dx, du = dx, v = sin x
π π π π π
x cos x dx = x sin x − sin x dx = cos x = −2 so (x + x cos x)dx = π 2 /2 − 2
0 0 0 0 0
√ √ 1 2
39. u = tan−1 x, dv = xdx, du = √ dx, v = x3/2 ;
2 x(1 + x) 3
3 √ √ 2 3/2 √ 3
1 3
x
x tan−1 xdx = x tan−1 x − dx
1 3 1 3 1 1+x
2 3/2 √ 3
1 3
1
= x tan−1 x − 1− dx
3 1 3 1 1+x
2 3/2 √ 1 1
3 √
= x tan−1 x − x + ln |1 + x| = (2 3π − π/2 − 2 + ln 2)/3
3 3 3 1
2x
40. u = ln(x2 + 1), dv = dx, du = dx, v = x;
x2 + 1
2 2 2 2
2x2 1
ln(x2 + 1)dx = x ln(x2 + 1) − dx = 2 ln 5 − 2 1− dx
0 0 0 x2+1 0 x2 + 1
2
= 2 ln 5 − 2(x − tan−1 x) = 2 ln 5 − 4 + 2 tan−1 2
0
√
41. t = x, t2 = x, dx = 2t dt
√
x
(a) e dx = 2 tet dt; u = t, dv = et dt, du = dt, v = et ,
√ √ √
e x
dx = 2tet − 2 et dt = 2(t − 1)et + C = 2( x − 1)e x + C
√
(b) cos x dx = 2 t cos t dt; u = t, dv = cos tdt, du = dt, v = sin t,
√ √ √ √
cos x dx = 2t sin t − 2 sin tdt = 2t sin t + 2 cos t + C = 2 x sin x + 2 cos x + C
9. January 27, 2005 11:45 L24-CH08 Sheet number 9 Page number 345 black
Exercise Set 8.2 345
42. Let q1 (x), q2 (x), q3 (x) denote successive antiderivatives of q(x),
so that q3 (x) = q2 (x), q2 (x) = q1 (x), q1 (x) = q(x). Let p(x) = ax2 + bx + c.
Repeated Repeated
Differentiation Antidifferentiation
ax2 + bx + c q(x)
+
2ax + b q1 (x)
−
2a q2 (x)
+
0 q3 (x)
Then p(x)q(x) dx = (ax2 + bx + c)q1 (x) − (2ax + b)q2 (x) + 2aq3 (x) + C. Check:
d
[(ax2 +bx + c)q1 (x) − (2ax + b)q2 (x) + 2aq3 (x)]
dx
= (2ax + b)q1 (x) + (ax2 + bx + c)q(x) − 2aq2 (x) − (2ax + b)q1 (x) + 2aq2 (x) = p(x)q(x)
43. Repeated Repeated
Differentiation Antidifferentiation
3x2 − x + 2 e−x
+
6x − 1 −e−x
−
6 e−x
+
0 −e−x
(3x2 − x + 2)e−x = −(3x2 − x + 2)e−x − (6x − 1)e−x − 6e−x + C = −e−x [3x2 + 5x + 7] + C
44. Repeated Repeated
Differentiation Antidifferentiation
x2 + x + 1 sin x
+
2x + 1 − cos x
−
2 − sin x
+
0 cos x
(x2 + x + 1) sin x dx = −(x2 + x + 1) cos x + (2x + 1) sin x + 2 cos x + C
= −(x2 + x − 1) cos x + (2x + 1) sin x + C
10. January 27, 2005 11:45 L24-CH08 Sheet number 10 Page number 346 black
346 Chapter 8
45. Repeated Repeated
Differentiation Antidifferentiation
4x4 sin 2x
+
1
16x3 − cos 2x
2
−
1
48x 2
− sin 2x
+ 4
1
96x cos 2x
8
−
1
96 sin 2x
16
+
1
0 − cos 2x
32
4x4 sin 2x dx = (−2x4 + 6x2 − 3) cos 2x + −(4x3 + 6x) sin 2x + C
46. Repeated Repeated
Differentiation Antidifferentiation
√
x3 2x + 1
+
1
3x 2
(2x + 1)3/2
3
−
1
6x (2x + 1)5/2
15
+
1
6 (2x + 1)7/2
105
−
1
0 (2x + 1)9/2
945
√ 1 1 2 2
x3 2x + 1 dx = x3 (2x + 1)3/2 − x2 (2x + 1)5/2 + x(2x + 1)7/2 − (2x + 1)9/2 + C
3 5 35 315
47. (a) We perform a single integration by parts:
u = cos x, dv = sin x dx, du = − sin x dx, v = − cos x,
sin x cos x dx = − cos2 x − sin x cos x dx. Thus
1
2 sin x cos x dx = − cos2 x + C, sin x cos x dx = − cos2 x + C
2
1 2 1
(b) u = sin x, du = cos x dx, sin x cos x dx = u du = u + C = sin2 x + C
2 2
x
48. (a) u = x2 , dv = √ , du = 2x dx, v = x2 + 1,
x2 +1
1
x3
1 1 √ 2
1
1√ 2
√ dx = x2 x2 + 1 − 2x x2 + 1 dx = 2 − (x2 + 1)3/2 =− 2+
0 x2 + 1 0 0 3 0 3 3
11. January 27, 2005 11:45 L24-CH08 Sheet number 11 Page number 347 black
Exercise Set 8.2 347
√ √
2 2
x 1 3
(b) u = x2 + 1, du = √ dx, (u − 1) du =
2
u −u
x2 + 1 1 3 1
2√ √ 1 1√ 2
= 2− 2− +1=− 2+ .
3 3 3 3
e e
49. (a) A = ln x dx = (x ln x − x) =1
1 1
e e
(b) V = π (ln x)2 dx = π (x(ln x)2 − 2x ln x + 2x) = π(e − 2)
1 1
π/2 π/2 π/2 π/2
1 2 π2
50. A = (x − x sin x)dx = x − x sin x dx = − (−x cos x + sin x) = π 2 /8 − 1
0 2 0 0 8 0
π π
51. V = 2π x sin x dx = 2π(−x cos x + sin x) = 2π 2
0 0
π/2 π/2
52. V = 2π x cos x dx = 2π(cos x + x sin x) = π(π − 2)
0 0
π
53. distance = t3 sin tdt;
0
Repeated Repeated
Differentiation Antidifferentiation
t3 sin t
+
3t2 − cos t
−
6t − sin t
+
6 cos t
−
0 sin t
π π
t3 sin t dx = [(−t3 cos t + 3t2 sin t + 6t cos t − 6 sin t)] = π 3 − 6π
0 0
1
54. u = 2t, dv = sin(kωt)dt, du = 2dt, v = − cos(kωt); the integrand is an even function of t so
kω
π/ω π/ω π/ω π/ω
2 1
t sin(kωt) dt = 2 t sin(kωt) dt = − t cos(kωt) +2 cos(kωt) dt
−π/ω 0 kω 0 0 kω
k+1 π/ω k+1
2π(−1) 2 2π(−1)
= + sin(kωt) =
kω 2 k2 ω2 0 kω 2
1 3
55. (a) sin4 x dx = − sin3 x cos x + sin2 x dx
4 4
1 3 1 1
= − sin3 x cos x + − sin x cos x + x + C
4 4 2 2
1 3 3
= − sin3 x cos x − sin x cos x + x + C
4 8 8
12. January 27, 2005 11:45 L24-CH08 Sheet number 12 Page number 348 black
348 Chapter 8
π/2 π/2 π/2
1 4
(b) sin5 x dx = − sin4 x cos x + sin3 x dx
0 5 0 5 0
π/2 π/2
4 1 2
= − sin2 x cos x + sin x dx
5 3 0 3 0
π/2
8 8
=− cos x =
15 0 15
1 4 1 4 1 2
56. (a) cos5 x dx = cos4 x sin x + cos3 x dx = cos4 x sin x + cos2 x sin x + sin x + C
5 5 5 5 3 3
1 4 8
= cos4 x sin x + cos2 x sin x + sin x + C
5 15 15
1 5
(b) cos6 x dx = cos5 x sin x + cos4 x dx
6 6
1 5 1 3
= cos5 x sin x + cos3 x sin x + cos2 x dx
6 6 4 4
1 5 5 1 1
= cos5 x sin x + cos3 x sin x + cos x sin x + x + C,
6 24 8 2 2
π/2
1 5 5 5
cos5 x sin x + cos3 x sin x + cos x sin x + x = 5π/32
6 24 16 16 0
57. u = sinn−1 x, dv = sin x dx, du = (n − 1) sinn−2 x cos x dx, v = − cos x;
sinn x dx = − sinn−1 x cos x + (n − 1) sinn−2 x cos2 x dx
= − sinn−1 x cos x + (n − 1) sinn−2 x (1 − sin2 x)dx
= − sinn−1 x cos x + (n − 1) sinn−2 x dx − (n − 1) sinn x dx,
n sinn x dx = − sinn−1 x cos x + (n − 1) sinn−2 x dx,
1 n−1
sinn x dx = − sinn−1 x cos x + sinn−2 x dx
n n
58. (a) u = secn−2 x, dv = sec2 x dx, du = (n − 2) secn−2 x tan x dx, v = tan x;
secn x dx = secn−2 x tan x − (n − 2) secn−2 x tan2 x dx
= secn−2 x tan x − (n − 2) secn−2 x (sec2 x − 1)dx
= secn−2 x tan x − (n − 2) secn x dx + (n − 2) secn−2 x dx,
(n − 1) secn x dx = secn−2 x tan x + (n − 2) secn−2 x dx,
1 n−2
secn x dx = secn−2 x tan x + secn−2 x dx
n−1 n−1
13. January 27, 2005 11:45 L24-CH08 Sheet number 13 Page number 349 black
Exercise Set 8.2 349
(b) tann x dx = tann−2 x (sec2 x − 1) dx = tann−1 x sec2 x dx − tann−2 x dx
1
= tann−1 x − tann−2 x dx
n−1
(c) u = xn , dv = ex dx, du = nxn−1 dx, v = ex ; xn ex dx = xn ex − n xn−1 ex dx
1 1 1
59. (a) tan4 x dx = tan3 x − tan2 x dx = tan3 x − tan x + dx = tan3 x − tan x + x + C
3 3 3
1 2 1 2
(b) sec4 x dx = sec2 x tan x + sec2 x dx = sec2 x tan x + tan x + C
3 3 3 3
(c) x3 ex dx = x3 ex − 3 x2 ex dx = x3 ex − 3 x2 ex − 2 xex dx
= x3 ex − 3x2 ex + 6 xex − ex dx = x3 ex − 3x2 ex + 6xex − 6ex + C
60. (a) u = 3x,
1 1 1 2 u 2
x2 e3x dx = u2 eu du = u2 eu − 2 ueu du = u e − ueu − eu du
27 27 27 27
1 2 u 2 2 1 2 2
= u e − ueu + eu + C = x2 e3x − xe3x + e3x + C
27 27 27 3 9 27
√
(b) u = − x,
1 √ −1
xe− x
dx = 2 u3 eu du,
0 0
u3 eu du = u3 eu − 3 u2 eu du = u3 eu − 3 u2 eu − 2 ueu du
= u3 eu − 3u2 eu + 6 ueu − eu du = u3 eu − 3u2 eu + 6ueu − 6eu + C,
−1 −1
2 u3 eu du = 2(u3 − 3u2 + 6u − 6)eu = 12 − 32e−1
0 0
61. u = x, dv = f (x)dx, du = dx, v = f (x);
1 1 1
x f (x)dx = xf (x) − f (x)dx
−1 −1 −1
1
= f (1) + f (−1) − f (x) = f (1) + f (−1) − f (1) + f (−1)
−1
62. (a) u dv = uv − v du = x(sin x + C1 ) + cos x − C1 x + C2 = x sin x + cos x + C2 ;
the constant C1 cancels out and hence plays no role in the answer.
(b) u(v + C1 ) − (v + C1 )du = uv + C1 u − v du − C1 u = uv − v du
14. January 27, 2005 11:45 L24-CH08 Sheet number 14 Page number 350 black
350 Chapter 8
dx
63. u = ln(x + 1), dv = dx, du = , v = x + 1;
x+1
ln(x + 1) dx = u dv = uv − v du = (x + 1) ln(x + 1) − dx = (x + 1) ln(x + 1) − x + C
3dx 2
64. u = ln(3x − 2), dv = dx, du = ,v = x − ;
3x − 2 3
2 2 1
ln(3x − 2) dx = u dv = uv − v du = x− ln(3x − 2) − x− dx
3 3 x − 2/3
2 2
= x− ln(3x − 2) − x − +C
3 3
1 1
65. u = tan−1 x, dv = x dx, du = 2
dx, v = (x2 + 1)
1+x 2
1 2 1
x tan−1 x dx = u dv = uv − v du = (x + 1) tan−1 x − dx
2 2
1 2 1
= (x + 1) tan−1 x − x + C
2 2
1
66. u = , dv = x dx, du = − x(ln x)2 dx,
1 1
v = ln x
ln x
1 1
dx = 1 + dx.
x ln x x ln x
This seems to imply that 1 = 0, but recall that both sides represent a function plus an arbitrary
constant; these two arbitrary constants will take care of the 1.
67. (a) u = f (x), dv = dx, du = f (x), v = x;
b b b b
f (x) dx = xf (x) − xf (x) dx = bf (b) − af (a) − xf (x) dx
a a a a
(b) Substitute y = f (x), dy = f (x) dx, x = a when y = f (a), x = b when y = f (b),
b f (b) f (b)
xf (x) dx = x dy = f −1 (y) dy
a f (a) f (a)
y
(c) From a = f −1 (α) and b = f −1 (β) we get b
bf (b) − af (a) = βf −1 (β) − αf −1 (α); then A1
β β f (b)
f −1 (x) dx = f −1 (y) dy = f −1 (y) dy, a A2
α α f (a) x
which, by Part (b), yields a= f –1(a) b=f –1(b)
β b
f −1 (x) dx = bf (b) − af (a) − f (x) dx
α a
f −1 (β)
−1 −1
= βf (β) − αf (α) − f (x) dx
f −1 (α)
β f −1 (β)
Note from the figure that A1 = f −1 (x) dx, A2 = f (x) dx, and
α f −1 (α)
A1 + A2 = βf −1 (β) − αf −1 (α), a “picture proof”.
15. January 27, 2005 11:45 L24-CH08 Sheet number 15 Page number 351 black
Exercise Set 8.3 351
68. (a) Use Exercise 67(c);
1/2 sin−1 (1/2) π/6
1 1 1 1
sin−1 x dx = sin−1 −0·sin−1 0− sin x dx = sin−1 − sin x dx
0 2 2 sin−1 (0) 2 2 0
(b) Use Exercise 67(b);
e2 ln e2 2 2
ln x dx = e2 ln e2 − e ln e − f −1 (y) dy = 2e2 − e − ey dy = 2e2 − e − ex dx
e ln e 1 1
EXERCISE SET 8.3
1
1. u = cos x, − u3 du = − cos4 x + C
4
1 1
2. u = sin 3x, u5 du = sin6 3x + C
3 18
1 1 1
3. sin2 5θ = (1 − cos 10θ) dθ = θ− sin 10θ + C
2 2 20
1 1 1
4. cos2 3x dx = (1 + cos 6x)dx = x+ sin 6x + C
2 2 12
1 1
5. sin3 aθ dθ = sin aθ(1 − cos2 aθ) dθ = − cos aθ − cos3 aθ + C (a = 0)
a 3a
6. cos3 at dt = (1 − sin2 at) cos at dt
1 1
= cos at dt − sin2 at cos at dt = sin at − sin3 at + C (a = 0)
a 3a
1 1
7. u = sin ax, u du = sin2 ax + C, a = 0
a 2a
8. sin3 x cos3 x dx = sin3 x(1 − sin2 x) cos x dx
1 1
= (sin3 x − sin5 x) cos x dx = sin4 x − sin6 x + C
4 6
9. sin2 t cos3 t dt = sin2 t(1 − sin2 t) cos t dt = (sin2 t − sin4 t) cos t dt
1 1
= sin3 t − sin5 t + C
3 5
10. sin3 x cos2 x dx = (1 − cos2 x) cos2 x sin x dx
1 1
= (cos2 x − cos4 x) sin x dx = − cos3 x + cos5 x + C
3 5
1 1 1 1
11. sin2 x cos2 x dx = sin2 2x dx = (1 − cos 4x)dx = x− sin 4x + C
4 8 8 32
16. January 27, 2005 11:45 L24-CH08 Sheet number 16 Page number 352 black
352 Chapter 8
1 1
12. sin2 x cos4 x dx = (1 − cos 2x)(1 + cos 2x)2 dx = (1 − cos2 2x)(1 + cos 2x)dx
8 8
1 1 1 1
= sin2 2x dx + sin2 2x cos 2x dx = (1 − cos 4x)dx + sin3 2x
8 8 16 48
1 1 1
= x− sin 4x + sin3 2x + C
16 64 48
1 1 1
13. sin 2x cos 3x dx = (sin 5x − sin x)dx = − cos 5x + cos x + C
2 10 2
1 1 1
14. sin 3θ cos 2θdθ = (sin 5θ + sin θ)dθ = − cos 5θ − cos θ + C
2 10 2
1 1
15. sin x cos(x/2)dx = [sin(3x/2) + sin(x/2)]dx = − cos(3x/2) − cos(x/2) + C
2 3
3
16. u = cos x, − u1/3 du = − cos4/3 x + C
4
π/2 π/2
17. cos3 x dx = (1 − sin2 x) cos x dx
0 0
π/2
1 2
= sin x − sin3 x =
3 0 3
π/2 π/2 π/2
1 1
18. sin2 (x/2) cos2 (x/2)dx = sin2 x dx = (1 − cos 2x)dx
0 4 0 8 0
π/2
1 1
= x− sin 2x = π/16
8 2 0
π/3 π/3 π/3
1 1
19. sin4 3x cos3 3x dx = sin4 3x(1 − sin2 3x) cos 3x dx = sin5 3x − sin7 3x =0
0 0 15 21 0
π π π
1 1 1
20. cos2 5θ dθ = (1 + cos 10θ)dθ = θ+ sin 10θ =π
−π 2 −π 2 10 −π
π/6 π/6 π/6
1 1 1
21. sin 4x cos 2x dx = (sin 2x + sin 6x)dx = − cos 2x − cos 6x
0 2 0 4 12 0
= [(−1/4)(1/2) − (1/12)(−1)] − [−1/4 − 1/12] = 7/24
2π 2π 2π
1 1 1 1
22. sin2 kx dx = (1 − cos 2kx)dx = x− sin 2kx =π− sin 4πk (k = 0)
0 2 0 2 2k 0 4k
1 1
23. tan(2x − 1) + C 24. − ln | cos 5x| + C
2 5
25. u = e−x , du = −e−x dx; − tan u du = ln | cos u| + C = ln | cos(e−x )| + C
1 1
26. ln | sin 3x| + C 27. ln | sec 4x + tan 4x| + C
3 4
17. January 27, 2005 11:45 L24-CH08 Sheet number 17 Page number 353 black
Exercise Set 8.3 353
√ 1 √ √
28. u = x, du = √ dx; 2 sec u du = 2 ln | sec u + tan u| + C = 2 ln sec x + tan x + C
2 x
1
29. u = tan x, u2 du = tan3 x + C
3
1 1
30. tan5 x(1 + tan2 x) sec2 x dx = (tan5 x + tan7 x) sec2 x dx = tan6 x + tan8 x + C
6 8
1 1
31. tan 4x(1 + tan2 4x) sec2 4x dx = (tan 4x + tan3 4x) sec2 4x dx = tan2 4x + tan4 4x + C
8 16
1 1
32. tan4 θ(1 + tan2 θ) sec2 θ dθ = tan5 θ + tan7 θ + C
5 7
1 1
33. sec4 x(sec2 x − 1) sec x tan x dx = (sec6 x − sec4 x) sec x tan x dx = sec7 x − sec5 x + C
7 5
1 2
34. (sec2 θ − 1)2 sec θ tan θdθ = (sec4 θ − 2 sec2 θ + 1) sec θ tan θdθ = sec5 θ − sec3 θ + sec θ + C
5 3
35. (sec2 x − 1)2 sec x dx = (sec5 x − 2 sec3 x + sec x)dx = sec5 x dx − 2 sec3 x dx + sec x dx
1 3
= sec3 x tan x + sec3 x dx − 2 sec3 x dx + ln | sec x + tan x|
4 4
1 5 1 1
= sec3 x tan x − sec x tan x + ln | sec x + tan x| + ln | sec x + tan x| + C
4 4 2 2
1 5 3
= sec3 x tan x − sec x tan x + ln | sec x + tan x| + C
4 8 8
36. [sec2 x − 1] sec3 x dx = [sec5 x − sec3 x]dx
1 3
= sec3 x tan x + sec3 x dx − sec3 x dx (equation (20))
4 4
1 1
= sec3 x tan x − sec3 x dx
4 4
1 1 1
= sec3 x tan x − sec x tan x − ln | sec x + tan x| + C (equation (20), (22))
4 8 8
1 1
37. sec2 t(sec t tan t)dt = sec3 t + C 38. sec4 x(sec x tan x)dx = sec5 x + C
3 5
1
39. sec4 x dx = (1 + tan2 x) sec2 x dx = (sec2 x + tan2 x sec2 x)dx = tan x + tan3 x + C
3
40. Using equation (20),
1 3
sec5 x dx = sec3 x tan x + sec3 x dx
4 4
1 3 3
= sec3 x tan x + sec x tan x + ln | sec x + tan x| + C
4 8 8
18. January 27, 2005 11:45 L24-CH08 Sheet number 18 Page number 354 black
354 Chapter 8
41. u = 4x, use equation (19) to get
1 1 1 1 1
tan3 u du = tan2 u + ln | cos u| + C = tan2 4x + ln | cos 4x| + C
4 4 2 8 4
1
42. Use equation (19) to get tan4 x dx = tan3 x − tan x + x + C
3
√ 2 2
43. tan x(1 + tan2 x) sec2 x dx = tan3/2 x + tan7/2 x + C
3 7
2
44. sec1/2 x(sec x tan x)dx = sec3/2 x + C
3
π/8 π/8
1
45. (sec2 2x − 1)dx = tan 2x − x = 1/2 − π/8
0 2 0
π/6 π/6
1
46. sec2 2θ(sec 2θ tan 2θ)dθ = sec3 2θ = (1/6)(2)3 − (1/6)(1) = 7/6
0 6 0
47. u = x/2,
π/4
1
π/4 √
2 tan5 u du = tan4 u − tan2 u − 2 ln | cos u| = 1/2 − 1 − 2 ln(1/ 2) = −1/2 + ln 2
0 2 0
1 π/4
1
π/4 √
48. u = πx, sec u tan u du = sec u = ( 2 − 1)/π
π 0 π 0
1 1
49. (csc2 x − 1) csc2 x(csc x cot x)dx = (csc4 x − csc2 x)(csc x cot x)dx = − csc5 x + csc3 x + C
5 3
cos2 3t 1 1
50. · dt = csc 3t cot 3t dt = − csc 3t + C
sin2 3t cos 3t 3
cos x 1
51. (csc2 x − 1) cot x dx = csc x(csc x cot x)dx − dx = − csc2 x − ln | sin x| + C
sin x 2
1
52. (cot2 x + 1) csc2 x dx = − cot3 x − cot x + C
3
2π 2π
1
53. (a) sin mx cos nx dx = [sin(m + n)x + sin(m − n)x]dx
0 2 0
2π
cos(m + n)x cos(m − n)x
= − −
2(m + n) 2(m − n) 0
2π 2π
but cos(m + n)x = 0, cos(m − n)x = 0.
0 0
2π
1 2π
(b) cos mx cos nx dx = [cos(m + n)x + cos(m − n)x]dx;
0 2 0
since m = n, evaluate sin at integer multiples of 2π to get 0.
2π
1 2π
(c) sin mx sin nx dx = [cos(m − n)x − cos(m + n)x] dx;
0 2 0
since m = n, evaluate sin at integer multiples of 2π to get 0.