400-350= = solution weareestimating...5x4 42x 36. g2+x2)3dx solution letu =x +x2.thendu = (5x4...

SECTION 5.5 || Net Changeasthe Integral ofa Rate 621

o (350 < 1 < 400) For N = 10,400-350 _= =At 19.

Thetable of values {r (t;)};=0,..5 is given below:

li 350 360 370 380 390 400r(tj) 5.11438 5.03215 4.95253 4.87539 4.80061 4.72810

The endpoint approximationsare:Ry 10(5.03215 + 4.95253 + 4.87539 + 4.80061 + 4.72810) = 243.888 familiesLn Il 10(5.11438 + 5.03215 + 4.95253 + 4.87539 + 4.80061) = 247.751 families

The right endpoint approximation estimates 243.888 families became extinct in the period 350 < 1 < 400, the left endpointapproximation estimates 247.751 families becameextinct during this time. The averageof the two is 245.820 families.

28. CAS Estimate the total numberofextinct families from f = 0 to the present, using My with N = 544,SOLUTION Weareestimating

544 3130/ dt0 (t + 262). . 544 0 x . 1using My with N = 544. If N = 544, At = Sa > Land {f7}j=1,..n =fAt (At/2) =i - 35.

N 544 3130MN = At r(t*) =1- = 3517.3021.N run cen:i=l i=l

Thus, we estimate that 3517 families have becomeextinct over the past 544 million years. Further Insights and Challenges29. Show thata particle, located at the origin at 1 = 1 and moving along the x-axis with velocity v(t) = f~, will never pass thepoint x = 2.SOLUTION The particle s velocity is v(t) = s (t) = 17?, an antiderivative for which is F(t) = t~!. Hence, the particle sposition at time 7 is

1 t 1=f

for allt > 1. Thus, the particle will never pass x = 1, which impliesit will never pass x = 2 either.30. Show that a particle, located at the origin at £ = 1 and moving along the x-axis with velocity v(t) = 11/2 moves arbitrarilyfar from the origin after sufficient time has elapsed.SOLUTION Theparticle s velocity is v(t) = s/(t) = t~1/2, an antiderivative for which is F(1) = 211/2. Hence,the particle spposition at time ¢ is

1s(t) = [ s(u)du = F(u)| = F(t) FQ) =2V1t-11

for allt > 1. Let S > 0 denote an arbitrarily large distance from the origin. We see that forS41)?i> 72

the particle will be more than S units from the origin. In other words, the particle movesarbitrarily far fromthe origin after sufficienttime has elapsed.

622 CHAPTER 5 | THE INTEGRAL

5.6 Substitution Method

Preliminary Questions1. Which ofthe following integrals is a candidate for the Substitution Method?

(a) [s sin(x?) dx (b) [sind x cos x dx (c) f+ sinx dx

SOLUTION The function in (c): x° sin x is not of the form g(u(x))u'(x). The function in (a) meets the prescribed pattern withg(u) = sinu and u(x) = x°. Similarly, the function in (b) meets the prescribed pattern with g(u) = u? and u(x) = sinx.2. Find an appropriate choice of u for evaluating the following integrals by substitution:

(a) [se + 9)* dx (b) pa sin(x?) dx (c) [ sinx cos? x dx

SOLUTION(a) x(x? + 9)* = 4 (2x) (x? + 9)4; hence, c = 4, f(u) = ut, and u(x) = x? +9.(b) x? sin(x3) = 1(3x2) sin(x?); hence, c = 3, f(u) = sinu, and u(x) = x?.(c) sinx cos? x = ( sinx) cos x; hence, c = 1, f(u) = u?, and u(x) = cos x.

23. Which ofthe following is equal to / x?(x3 + 1) dx for a suitable substitution?

01 2 9 1 9

(a) 5/ u du | udu ©3] u du3 Jo 0 3 JiSOLUTION With the substitution u = x? + 1, the definite integral Io x?(x3 + 1) dx becomes 4 fe u du. The correct answeris(e).

ExercisesIn Exercises 1-6, calculate du.

1. uw = x3 x?SOLUTION Letu = x3 x?. Then du = (3x? 2x) dx.

2. uw = 2x44 8x7!SOLUTION Letu = 2x4 + 8x1. Then du = (8x? 8x?) dx.

3. u = cos(x?)SOLUTION Letu = cos(x?). Then du = -sin(x?) -2x dx = 2x sin(x?) dx.

4. u=tanxSOLUTION Letu = tanx. Then du = sec? x dx.

5. u = ¿el

SOLUTION Letu =e**+1, Then du = 4e**! dx,6. u = In(x* +1)

4x3 dx.x44] *SOLUTION Letu In(x* + 1). Then du =In Exercises 7-22, write the integral in terms ofu and du. Then evaluate.

7. [=Pas, u=x-71SOLUTION Letu =x 7. Then du = dx. Hence

3 3 La 1 4(x-7)°> dx = Ju du = Zu +C=7@-N +C.

8. fe +29«x u=x+25SOLUTION Letu =x + 25. Then du = dx and

1 +C.sife +29 ax = pra = yt4+C=-

SECTION 5.6 | Substitution Method 623

9. rara, u=12+1SOLUTION Let u = 12? + 1. Then du = 21 dt. Hence,

1 1 1[we +1dt= ¿puras = zu? +C=50?+ 13 +C.10. fe +1)cos(x* +4x)dx, u=x*+4xSOLUTION Letu = x4+ 4x. Then du = (4x3 + 4) dx = 4(x3 + 1) dx and

1 1 1fe + 1) cos(x4 + 4x) dx = 7 / cos u du = 7 sinu +C= gina +4x)+C.31zT di(14-291 ">

SOLUTION Let u = 4 2r4. Then du = 813 dt. Hence,11. u=4- 214

12. [Tax u = 4x 1SOLUTION Letu = 4x 1. Then du = 4dx or tdu = dx. Hence

[ve ¿Ju? ; (n°2) 2x -13. [etn ax, u=x+l1SOLUTION Letu =x+1.Thenx = # 1 and du = dx. Hence

[xe 1)? dx = fu-du = fur ur au1 1 1 1" P+c= Zar! - Zarrc.= U U11 10

14. fria, u=4x-1SOLUTION Letu = 4x 1. Then x = Lu + 1) and du = 4dx or 4 du = dx. Hence,

1xvV4x 1dx= jur Du! du = qrra1/2 sp), 1 2 3j= 0 [zu Hr) re

1 1= x= 192 + 5gte 124 .15. peras, u=x+1SOLUTION Letu = x + 1. Then x =u- 1 and du = dx. Hence

fa=fauRS203? 40!) du

a zu ? de ee +C

= Stes 17/7 = 6 +12 + Za++

ll

716. J sincao 10, u = 40 7SOLUTION Let u 40 7. Then du = 4 d6 and

1 1J sino 7) d0 = 7) sino du = 7 cos u +C= 73 cos(490 7) + C.


17. f sin? 0cos040, u = sind

SOLUTION Letu = sin 9. Then du = cos 6 dé. Hence,1 1[si80050 40 = fu? au EL gute = sinó 0 +C.

18. sec? xtan x dx, u=tanx

SOLUTION Letu =tanx. Then du = sec? x dx. Hence1 1[ sec? xtansx ax = frau = zur +c = pianx $C.

19. fa dx, u=-x?

SOLUTION Letu = x?. Then du = 2x dx or 4 du = x ax. Hence,x2 1 à l y 1 _,2E dx = - du = - C = - C.IE x se u 5° + 7° +

20. fee tye"! di, u=tan!SOLUTION Letw = tant. Then du = sec? t dt and

fienew dt = / ce! du =e 4+ C=e 4,

Inx)? dxuu = InxSOLUTION Letu = In x. Then du = + dx, and

In x)? 1 1[Rae[rane tc = Zinn? +c.x

t 1l, 24 e22. are u=tan!xx2 +1SOLUTION Letu = tan ! x. Then du = dx, andI+x

tan ! xy? 1 1Gany = / u? du = zu +C= zan xy? + C.x2 +1

In Exercises 23-26, evaluate the integral in theform a sin(u(x)) + C for an appropriate choice ofu(x) and constant a.

23. pa cos(x*) dx

SOLUTION Letu = x*. Then du = 4x? dx or 4 du = x3dx. Hence

Je cost) as a Jesu au q sinu qsin@) + C.

24. [* cos(x? + 1) dx

SOLUTION Letu =x?° +1. Then du = 3x? dx or 4 du = x? dx. Hence1 1feos? + 1) dx = 3 [cosuau = y sinu +C.

25. Î x1/2 cos(x 4/2) ax

SOLUTION Letu = x?/2. Then du = 3x1/2 dx or 3 du = x1/2 dx. Hence2 2 2J 1? cose) ax = 3 [ cos au = 3 sinu +C= = sin@®?) +C.

SECTION 5.6 || Substitution Method 625

26. / cos x cos(sin x) dxSOLUTION Let u = sinx. Then du = cos x dx. Hence

[cos cos(sinx) dx = [cos du = sinu +C.

In Exercises 27-72, evaluate the indefinite integral.

27. / (4x +5)? dxSOLUTION Letu = 4x +5. Then du = 4dx and

fer + 9°a4[wau= geto = Geet+6.dx28. / Ey95

SOLUTION Letu = x 9. Then du = dx anddx _5 1 _, 1A =! =- 4C.| 95 /use E gs +

t2.|Vi +12SOLUTION Let u = 1 + 12. Then du = di and

dt -1/2 1/2= u du = 2u +C=2Vt4+124+C.[5 Î30. Î (ot + 2)7/3 dtSOLUTION Letu = 91 + 2. Then du = 9dt and

1 1for +29"at =; [uau=;LigOr +a? +e.9 9 5 1531. x+l

(x?AranSOLUTION Let u = x? + 2x. Then du = (2x + 2) dx or ¿du = (x + 1) dx. Hence

x+1 1 I/ 1. Ll, un idu=-[-- C= - 2 ch,[es24208 =; ] = = ( 2" )+ ARTO aa +32. Ja + 1)(x? + 2x) >14 dxSOLUTION Letu = x? + 2x. Then du = (2x + 2) dx = 2(x + 1) dx and

2 3/4 Lf 3/4 1 4 774(x + Da +2x)"/" dx = 3) du = ¿$e +C2= 3074244 C.

x33. / d)Vx? +9SOLUTION Let u = x? + 9. Then du = 2x dx or Adu = x dx. Hence

1 vitdu = < = Vx2+9+C.les= =31h2x2 + x34. nd(4x3 + 3x2)? *

SOLUTION Let u = 4x? + 3x?. Then du = (12x? + 6x) dx or Ldu = (2x? + x) dx. Hence

[ox + 3x7)? (2x? + x) dx = ¿Ju? zu! ¿e + 3x?)EC.


35. for + 1)(x3 + x)? dx

SOLUTION Letu = x? + x. Then du = (3x? + 1) dx. Hence1 1[oe +1)@&° + x)? dx = fea = zu +C= 36° +x)3+C.

5x4 4 2x36. dxG2 +x2)3SOLUTION Let u = x° + x2. Then du = (5x4 + 2x) dx. Hence

[ze_ lJ = dU = -(x° + x2)3 u3 2 u2 2 (x + x2)2

37. fe +8)dxSOLUTION Letu = 3x + 8. Then du = 3 dx and

fe + fr ag 364: +8)

38. [re +8)dxu 8

31 1[xo +8)" dx = ; [u au" au = ; Ju - 8ul!) du

a _"(yr 3" JT1117

SOLUTION Letu = 3x +8. Then du = 3dx,x = , and

2Gx +5)~ ex +s+.39, [evs +1dx

SOLUTION Letu =x3 + 1. Then du = 3x? dx and

[ever las = 5 [WP au = swt +c SG+ 0924 c.

40. [ve Hra

SOLUTION Letu = x3 + 1. Then du = 3x2 dx, x? = u 1 and

1 1[> Vx3 +1 4x = ; [u Drau = 5Jura

3\5 32,3 5/2 2,3 3/2= (x? +11? - 2x c.ie + 1) ge +1) +

dx41. / aoSOLUTION Letu =x +5. Then du = dx and

dx 23 1 1 _= du=--u C=-- 5) 2+cC.[<p E u ut 5 +5) +x? dx

(x + 5)?SOLUTION Letu = x + 5. Then du = dx, x =u- 5 and

anEAfu 10u + 25u

42.

SECTION 5.6 || Substitution Method 62725= In|u| + 10u7! Zu +C

10 25- +C.x+5 sa? =In|x +5|+

43. fre +1)! dzSOLUTION Let u = z3 + 1. Then du = 32? dz and

= = 1 i/ E 79% 39%

44. fe + 427)(29 + 1)? dzSOLUTION Let u = z? +1. Then du = 32? dz,z° = u- 1land

IlJe + 427)(z3 +1)! dz ; je +3)ul? du = 5 fue + 3412) du

11 14,3 13las = E3 a Fag yt

Il loa 14, L 3 13= 1 _ 1 C.AoA+45. JarnarnasSOLUTION Let u = x + 1. Then x =u4 1, du = dx and

fu + ut au = fost +u/4) duIlfe + 2)(x + DU dx4 4ae + ze +C4 4pet D+etre.

46. feeyaSOLUTION Letuw=x2- 1. Thenu + 1= x? and du = 2x dx or + du = x dx. Hence

feeya =exeya

| 3/2 1 5/2 3/2=> (u + Du da => (ue + ur!) du

1 (2 77% 1 (2 572 loa nv, 1,2 5/2==[= =[= = -(x?-1 ur .5 (Fu E +C qe ) +56 1) +C

47. [ sis 38) doSOLUTION Letu = 8-30. Then du = 3 d$ and

1 1 1Î sn 30) 40 = 2 J sinu du = qos = 3 cos 36) + C.

48. / 9 sin(0?) doSOLUTION Letu = 02. Then du = 26 dO and

1Josinco?) as = 5 [su du = = cosu+C = = c0s(9°) +C.cos 4/1Jt

SOLUTION Letu = ÿ1 = 11/2, Then du = qna dt andcos 4/1Jt

49. dt

di= 2 $ cosuda =2sinu+C =2sinvi +C.


50. IE sin(x? + 1) dx

SOLUTION Letu = x? + 1. Then du = 3x? dx or ¿du = x? dx. Hence

[sine 5) sino T3 T3

51. Junco + 9) d0

SOLUTION Letu = 40 + 9. Then du = 4d0 and1 1 1Juanas + 9) d0 = 3 Jeu au = 3 mlsecul +C = q nl sec(46 +9)| +C.

52. f sin 8 cos 6 40

SOLUTION Letu = sind. Then du = cos 0 d6 and1 1[ sin® 60080 a0 = [v8 au = zw += ¿sin? 0 +C.

53. ES dx

SOLUTION Letu = sin x. Then du = cos x dx, and

d[ea = Î ST dx = Î = =In|ul + C =In|sinx|+C.sin x u54. [rndx

4SOLUTION Letu = x4/5, Then du = ges dx and

5 5 aJe5anx*/s ax = 2 f tanu du = 3 misecul +C = 7 in| secx*/>| +C.

55. [seta + 9) dx

SOLUTION Letu = 4x + 9. Then du = 4dx or t du = dx. Hence

[2x + ax = q) suda q anu +C q ant + 9) +-C.

56. / sec? x tan x dx

SOLUTION Let u = tan x. Then du = sec x dx. Hence1 1sec? xtant x dx = fu du = zu +C = tanda +C,

pps57. x= =. LL = ÀSOLUTION Letu = x. Then du = afi dx or 2du = A dx. Hence,

[ep =2f sec? udx = 2tanu + Cx

s ==ay(1 + sin 2x)?SOLUTION Letu = 1 +sin2x. Then du = 2c052x or +du = cos2x dx. Hence

1 1 1fa + sin 2x)2 cos 2x dx = a / u~? du = zu +C=-351+ sin2x)7? + C.59, Î sin 4xcos 4x + 1 dx

SECTION 5.6 ] Substitution Method 629

SOLUTION Letu = cos4x + 1. Then du = 4sin 4x or 4du = sin 4x. Hence

¡NS =;fut? du =7 (°°) =(cos4x + 124.

60. fe x@sinx 1)dx

SOLUTION Letu = 3sinx 1. Then du = 3cosx dx or 3du = cos x dx. Hence1 1/1 1Jesnx- 1) cos x dx = x [eau =; (5) +C= ¿Osinx 1)? +C.

61. sectn O(sec 6 1) ddSOLUTION Letu = sec@ 1. Then du = sec O tan 0 d0 and

1 1[see tan 6(s000 - 1)d0 = frau = zu? +C= 3 (secó - 1? +6,

62. [ cost cos(sint) dtSOLUTION Letu = sint. Then du = cost dt and

J eos: cos(sin1) dt = ES du = sinu + C =sin(sint) + C.

63. fess dxSOLUTION Letu = 14x 7. Then du = 14 dx or u du = dx. Hence,

1 1 114x-7 u u 14x 7dx = du = es Es ife x 14 e du ia +C 14° +C

64. fe + Lex 2x dx

SOLUTION Letu = x? + 2x. Then du = (2x + 2) dx or à du = (x + 1) dx. Hence,2 1 l 1Jo + Ler+2x dx = 5 fe du = zer +C= zePi +C.

Xx -65. /in(ex + 1)4

SOLUTION Letu = e* + 1. Then du = e* dx, ande* 1 1_dx = [utdu=- 40 = - xCc.la+t f = st en +

66. Î (sec? 6) eagSOLUTION Let u = tan 6. Then du = sec? 9 d0, and

fe 0) end 0 = fe du="+C=eM0 4 Cc.t61. Laell + 2el +1

SOLUTION Letu = e . Then du = e! dt, and

| e! di -| du =| du 1 Cx 1 iceM42e 41) 0242441 Jtl? uti ~ et +] :

dx68. =| x(Inx)?SOLUTION Letu = Inx. Then du = l dx, and

dx 1 1=pu=h+0=-h+c.

630 CHAPTER 5 || THE INTEGRAL

69. p=dxx

SOLUTION Letu = Inx. Then du = + dx, andInx)* 1 1[ax = fut du = iS +0 = z0an +6.x 5 5

dx70. l=SOLUTION Letu = Inx. Then du = À dx, and

dx d/ = = [E =imtul+C =In]inx|+C. xinx u

71. La

SOLUTION Let u = cos(Inx). Then du = 1 sin(In x) dx or du = L sin(Inx) dx. Hence,tan(In in(In x d[Pa[SOw-- oY = In|u| + C = In| cos(Inx)| + C.x x cos(In x) u

72. fe x)In(sinx) dx

SOLUTION Letu = In(sinx). Then1du = cosx = cotx,sin x

andia L. ,. 3(cotx)In(sinx) dx = u du = zu +C= 3 (In(sin x))* + C.

dx73. Evaluate / - using u = 14+ yx. Hint: Show that dx = 2(u 1)du.G+ ype ine Vx. w-1)SOLUTION Letu = 1 + ./x. Then

1du = 275 dx or dx = 24/x du = 2{u 1) du.

Hence,dx u 1ee3 du =2

|

w?-u)d[> / wo fo ee2 1u+u? +40 =- +5+C.1+ J/x (+)?

74. Can They Both Be Right? Hannah usesthe substitution u = tan x and Akiva uses u = sec x to evaluate [tanx sec? x dx.Show that they obtain different answers, and explain the apparent contradiction.SOLUTION With the substitution u = tan x, Hannahfinds du = sec? x dx and

1 1[ranxsec?xax= [udu =+ Cr = 5 tan? x + C1.

Onthe other hand, with the substitution u = sec x, Akiva finds du = sec x tan x dx and1[ans sec? x dx = Î sec x(tan x sec x) dx = 3 sec? x + C2

Hannah and Akiva have each found a correctantiderivative. To resolve what appears to be a contradiction, recall that any twoantiderivatives of a specified function differ by a constant. To show thatthis is true in their case, note that

(5 ©) (5 a) > seo? x

1 15) + C2 C1 = 5 + C2 C1, a constantIl

Here we used the trigonometric identity tan? x + 1 = sec? x.


75. Evaluate f sin x cos x dx using substitution in two different ways:first using u = sin x and then using u = cos x. Reconcilethe two different answers.SOLUTION First, let = sin x. Then du = cos x dx and

. 1 1.J sinxcosxax= f udu = zu? +C = 5 sin? x + Cy.

Next, let uy = cos x. Then du = sinx dx or du = sin x dx. Hence,. 1, 1 4sinxcosxdx = udu= zu + C2 = => cos x + C2.

To reconcile these two seemingly different answers, recall that any two antiderivativesof a specified function differ by a constant.To show that this is true here, note that (4 sin? x + C1) (3 cos? x + Cz) = + + C1 C2, a constant. Here we used thetrigonometric identity sin? x + cos? x = 1.76. Some Choices Are Better Than Others Evaluate

/ sin x cos? x dx

twice. First use u = sin x to show that

ES cos? xdx= uvV1 u? du

and evaluate the integral on the right by a further substitution. Then show that u = cos x is a better choice.SOLUTION Considerthe integral f sin x cos? x dx. If we letu = sin x, then cos x = 4/1 2 and du = cos x dx. Hence

[sx cos? x dx = u 1-2 du.

Now let w = 1 u2. Then dw = 2u du or +dw = u du. Therefore,

/ 1 1/2 1/2 3/2u 1 w? du = > w do ==> 3 +C

1 3/2 1 2,3/2=-sw!*+C=--(l-u*)y/" +C3 31 2 13/2 13=-z(0-sin x) +C = 7 cos x+C.

A better substitution choice is u = cos x. Then du = sinx dx or du = sinx dx. Hence1 1] sinxcos?x dx = =fa = sl +C= 7 cos? x +C.

77. What are the new limits of integration if we apply the substitution u = 3x + to theintegral Io sin(3x + x) dx?SOLUTION The new limits of integration are (0) = 3-0 + x = x and u(x) = 37 + x = 4x.78. Which of the following is the result of applying the substitution 4x 9 to theintegral a (4x 9)?9 dx?

8 1 8(a) / y?du (b) / u2 du2 4 Jo23 1 23

(c) 4f u2dy (d) :/ u?dy-1 4J-1SOLUTION Letu = 4x 9. Then du = 4dx or + du = dx. Furthermore, when x = 2, u = 1, and when x = 8, u = 23.Hence

8 1 23/ (4x 9? dx = if u20 qu.2 4 J-1

The answeris therefore (d).In Exercises 79-90, use the Change-of- Variables Formula to evaluate the definite integral.

379. | (x + 2)3 dx1


SOLUTION Letu = x +2. Then du = dx. Hence3 13

/ (x + 2)3 dx =| urdu = -u41 3 43 54 34=2_> = 136,3 4 4

680. / Vx +3dx1SOLUTION Letu =x +3. Then du = dx. Hence

385°6 9

/ vit3ax = [ Ya du = zur?1 49 2= 2027-8) =4 3

1 x81. dÎ rnSOLUTION Letu = x? +1. Then du = 2x dx or 1 du = x dx. Hence

1 1/21 1/1ein -:/ me lu?o (x2+1)3 2 J1 u3 2\ 22

82. / V5x + 6 dx-1

SOLUTION Letu = 5x + 6. Then du = 5dx or + du = dx. Hence16 2 42= (4-1)= .; 5 5

2. 1 16 1/2/ Vox + 6dx = = Vu du = = (3?)

1 1 5\3 4

83. [ xVx2 + 9dx0SOLUTION Letu = x? +9. Then du = 2x dx or 4 du = x dx. Hence

4 25 1I x52 +94» = Vudu = 5 (34°)0 2 Jo 2 \3

= 981= (125 27) = .q ) 3 92 4x +1284. Î atde1 (x2 + 6x + 1)?

SOLUTION Let u = x? + 6x + 1. Then du = (2x + 6) dx and2 174 12 2/ _dx=2 | u?dy=-21 (+61 +19 8

185. / (x + 1)(x? + 2x)? dx0SOLUTION Let u = x? + 2x. Then du = (2x + 2) dx = 2(x + 1) dx, and

1 31 1/ (x + 1)(x? + 2x)? dx = =f u° du = u®o 2 16 123 729 243o 2.4

1786. (x 9)72/3 dx10

SOLUTION Letu = x 9. Then du = dx. Hence17 8 8/ (= 9? ax = | u72/3 dy = 3uV?| =302-1)=3.

1 10 1187. / 9 tan(0?) d9

0SOLUTION Letu = cos 92. Then du = 26 sin 62 dO or du = O sin 0? dé. Hence,

1 9 sin(9?) 1 os! du 16 tan(62) dé = =;| a./ an(0") [ cos(92) 2 Ji u 2 ns}cos 1 1==3 [In(cos 1) + ln 1] = x (sec 1). 1


88. [ sec? (2x - =) dx

SOLUTION Let y = 2x z Then du = 2 dx and

61/6 1 1/6 1 n/6

| sec? (2x =) dx = 5/ sec? u du = tanu0 6 2 J-n/6 2 7/6

EN CEE "32213 3) 3

x/289. / cos? x sinx dx0SOLUTION Letu = cosx. Then du = sinx dx. Hence

n/2 0 1 1 1 1 1/ cost xsinx dx = | 8 au = | wdu= -u) =--0=-.0 1 0 4 0 4 4

1/2 .90. / cot 2 esc? 2 dxx/3 2 2

x 1 2xSOLUTION Let u = cot 7 Then du = > csc 3 and

n/2 5 1/ cot? ~ esc? 2 dx = -2 | u? du1/3 2 2 V3

2 el 2= --u| ==6V3-1).3 V3 32

91. Evaluate | ry5 v4-r2dr.0SOLUTION Letu =5 ~V4 r2. Then

d "drdu = TAL4=r2 5=uso that

rdr =(5 u) du.Hence,the integral becomes:

2.| ry5- v4-r?dr0

5ll 35 5

Î Vu(S u) du = / (su! uw") du = (quer - pun)3 3 3 5

(3v5- 1045) (105 - 3%) = Ts Lys92. Find numbers a and b such that

b n/4/ (u? + 1) du =| sec* 6 d6a 11/4

andevaluate. Hint: Use the identity sec? 9 = tan? 6 + 1.SOLUTION Let 4 = tan 0. Then u? + 1 = tan? 6 + 1 = sec? 6 and du = sec 6 dé. Moreover, because

tan (-) =-1 and tan © = À,

it follows that a = 1 and b = 1. Thus,1/4 1 1

/ sect 9 d@ = / (u? +1) du = (qu + u)n/4 -1 3


93. Wind engineers have found that wind speed v (in meters/second) at a given location follows a Rayleigh distribution of thetype

_ 1 02/64W(v) = 3

This means that at a given momentin time,the probability that v lies between a and b is equal to the shaded area in Figure 1.(a) Show that the probability that v [0, b] is 1 e-b?/64,(b) Calculate the probability that v e [2,5].

>= W(v)

0.05

v (m/s) Pa b 20FIGURE 1 The shadedareais the probability that v lies betweena and b.

SOLUTION(a) The probability that v [0, b] is

[o 22Let u = v?/64. Then du = v/32 dv and

b2/64[ - [o 32 0 0

(b) The probability that v [2, 5] is the probability that v [0, 5] minus the probability that v [0, 2]. By part (a), the probabilitythat v [2, 5] is

(i 6725164) _ ( _ ets) = ¿71/16 _,-25/641/2

94. Evaluate / sin x cos x dx forn > 0.0SOLUTION Letu = sin x. Then du = cos x dx. Hence

1/2 1 ynel : 1Î sin x cos x dx Î u du = = .0 0 n+l), n+1

In Exercises 95-96, use substitution to evaluate the integral in terms off(x).

9s. syruaSOLUTION Letu = f(x). Then du = f (x) dx. Hence

1[ro f'@) dx = fe du = qe +C= 210) + C.

96. / LO gyf(x)?SOLUTION Letu = f(x). Then du = f'(x) dx. Hence

AQ), fia1 _AFords = fu du=-u +C= Tgx/6 1/2 1

97. Show that / F(sind) dé = / Fu) du.0 0 V1 u2


SOLUTION Letu = sind. Then u(m/6) = 1/2 and u(0) = 0, as required. Furthermore, du = cos 6 d@, so thatdudd = |cos à

If sind = u, then u? + cos? 8 = 1, so that cos9 = V1 y2. Therefore d0 = du/v1- u2. This gives1/6 1/2 1

1 d0 = du./ Fin 6) / ft Further Insights and Challenges98. Usethe substitution u = 1 + x!/ to show that

/ Vitx/ndx =n fura 1)""" dyEvaluate for n = 2, 3.

SOLUTION Letu =1+x!/",Thenx = (u 1)" and dx = n(u 1)""! du. Accordingly, / V1+x1/ dx =n /wy _Pl au.

For n = 2, we have

[VRPax = 2feidu =2[0a") du

=2 (0/0 qu) +C= SaHsEG + x1/293/2 4 .For n = 3, we have

[vi Hadx =3 fu?@ 1)? au =3 fw? 2u3?? 4 ull?) a

3 (Gur? (2) (5)> wn?) +ell 7

Il =0 + x1/3)712 _ Za 4013952 421413938 4 .1/2 do

/n/2 do99, Evaluate J = / à 1 + cot6,000 90 1 + tan6;000 91/2

check that 7 + J = | dé.0

SOLUTION Toevaluate1 x/2 dx~ / 1 + tan6000 x

we substitute 1 = 11/2 x. Then dí = dx, x = x/2 1,1(0) = x/2, and 1(x/2) = 0. Hence,=f" dx _ F dt - [ dtdo 1 + tan6000 y, x/2 1+ tan©00(7/2 1) Jo 1-+cot60005°

dtLada Pe 0007: We know 7 = J,so 7 + J = 27. On the otherhand, bythe definitionof 7 and J and thelinearity0 1 + cot6000(7)of the integral,

1+1= [7 Er -{"" rt dxJo 1 + tan6000 x "1 4 cot6000 x Jy 1 + tan6000 x TG 2016000 y *1/2 1 1 d

~ | (: + tan6000 x + 1 + Tam) +

1/2 1 1= dx/ (, + tan6000 x + (tan6000 y + 1)/ tan6000 =) .

636 CHAPTER 5 || THE INTEGRAL

+ dx1 + tan6000 x 1 + tan6000 x

f 1 tan6000 x

0

Il / 000, dx = /0 1 + tan x 0

Hence, J + J = 2] =x/2,s01 = 1/4.a

100. Use substitution to provethat f(x) dx = Oif f is an odd function.-a

SOLUTION We assume that f is continuous. If f(x) is an odd function, then f( x) = f(x). Letu = x. Then x = u anddu = dx or du = dx. Accordingly,

a 0 a 0 a|Pax = "rarasf fe)ax = [ Fed du + [ f(x) dx

[ roax- [| fu) du = 0.

101. Prove that /> 1 dx = faa 4 dx for a, h > 0. Then show that the regions under the hyperbola over the intervals [1, 2], [2, 4],[4, 8],... all have the samearea (Figure 2).

Equal area

T = Y se T x1 2 4 8FIGURE 2 The area under y = 4 over [2 , 2"*1] is the samefor alln = 0,1,2,....

SOLUTION Letu = = Then au = x and du = 4 dx or a du = dx. Henceby b/a b/a

/ Lax= | Lau | À ne.a x 1 au 1 u

/ /1 u 1%

The area underthe hyperbola overthe interval[1, 2] is given by the definite integral f2 1 dx. Denote this definite integral by A.Using the result from part (a), we find the area under the hyperbola overthe interval [2, 4] is

4 4/2 2[ <= f wax | =dx=42 x 1 x 1 x

81 8/4 1 24[=] dx [ <a A,4 X 1 x 1 x

In general, the area underthe hyperbola over the interval [2 , an+1) is

ll ll

Il Il

gn+l 1 2n+1/2n 1 2 1dx = dx = dx = À.

an x 1 x 1 X102. Show that the two regions in Figure 3 have the same area. Then use the identity cos? u = 01 + cos 24) to compute thesecond area.

(A) (B)FIGURE 3

SECTION 5.7 | Further Transcendental Functions 637

SOLUTION The area of the region in Figure 3(A) is given by So 1-x?dx.Letx = sinu. Then dx = cosu du and1 x2 = V1 sin?y = cos u. Hence,

1/ V1l x2 dx =]0 0This last integral represents the area of the region in Figure 3(B). The two regions in Figure 3 therefore have the same area.

Let's now focus on the definite integral Le12 cos? y du. Using the trigonometric identity cos? y = 34 + cos 24), we have1/2 1 1/2 1 1 1/2 1/ cou du = > [| 1+ cos2udu= 5 (w+ 5 sin2u)| =5:3-0=7.

1/2 1/2cos u - cos u du =f cos? u du.0

103. Area of an Ellipse Prove the formula A = wabforthe area ofthe ellipse with equation (Figure 4)2 2xI,|

2 2x yoFIGURE 4 Grapha 5 = 1

y257 = 1; here a, b > 0. The area between thepart ofthe ellipse in the upper2SOLUTION Considerthe ellipse with equation + +( 5), and the x-axis is Sa f(x) dx. By symmetry, the part of the elliptical region in the lower

half-plane has the samearea. Accordingly, the area enclosed bytheellipse isa a 2) a

2 [f@jax=2[ 0? (1-5) ax = 29 Vi e/a) dxNow,let u = x/a. Then x = au and a du = dx. Accordingly,

2b (Da= aa[VEdu = 200 (2) = ad

Here we recognizedthatA V1 u? du represents the area of the upper unit semicircular disk, which by Exercise 102 is 2(4) =TED .

5.7 Further Transcendental Functions

Preliminary Questionsb dx1. Find 5 such that / is equal to1 x

(a) In3 (b) 3SOLUTION For b > 0,

bo b/ nz =mb In1 = Ind.1 X 1x

(a) For the value ofthe definite integral to equal In 3, we must have b = 3.(b) Forthe value ofthe definite integral to equal 3, we must have h = e?,

b a2. Find b such at | ax = E,o 1+x2 3


SOLUTION In general,b= tan! -tan!0=tan!b.

0 [ dx -1= tan x0 1 + x2

Forthe value ofthe definite integral to equal +, we must have

tant b = = or b= tan= V3.3. Which integral should be evaluated using substitution?

9dx dxbol ofSOLUTION Usethe substitution u = 3x on the integralin (b).4. Which relation between x and u yields 16 + x2 = 4/1 + u??SOLUTION To transform 416 + x2 into 4/1 + u?, make the substitution x = Au.

ExercisesIn Exercises 1-10, evaluate the definite integral.

f dx1. 1 x

21 9SOLUTION / -dx =in|x|| = 1n9 - In! = 1n9.

1 X 12 [a4 x20 4 20

sorurion | dx =in|x|| = 1n20 -1n4=In5.4 X 4

e3 13. | Lai1 tey e?sorurion | | dt = In|t| = Ine? Inl =3.1 1-eja. | Laie2 te 1 = eSOLUTION | = dt =In|t| = In| e| In| e?| = n = In(1/e) = 1.e2 t ee e

12 dt5. / ne2 31 +4SOLUTION Let u = 3t + 4. Then du = 3 dt and

[ di IN 1 12 31+4 3 10 u 3 = -(In40 In10) = - In4.3 3 10

3e6. / 4e tintSOLUTION Letu = Int. Then du = (1/t)dt and

e 1 34 3/ a= | = Injull) =1n3 int = In3,e 1

tan 8 dx7. / atant X° +1

8 5tan dx -1SOLUTION / = + tan x

tan I+xtan 8

= tan ? (tan 8) tan! (tanl) =8-1=7.tan 1

400-350= = solution weareestimating...5x4 42x 36. g2+x2)3dx solution letu =x +x2.thendu = (5x4...

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