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  1. 1. CHAPTER 7 Techniques of Integration 7.1 Concepts Review 1. elementary function 2. ∫u5du 3. ex 4. 2 3 ∫ u du 1 Problem Set 7.1 1. ( – 2)5 1 ( – 2)6 ∫ x dx = x +C 6 2. 3 1 3 3 2 (3 )3/ 2 ∫ x dx = ∫ x ⋅ dx = x +C 3 9 3. u = x2 +1, du = 2x dx When x = 0, u = 1 and when x = 2, u = 5 . ∫ 2 x ( x 2 + 1) 5 dx = 1 ∫ 2 ( x 2 + 1) 5 (2 x dx ) 0 0 2 5 5 1 1 2 = ∫ u du 6 5 6 6 ⎡u ⎤ − = ⎢ ⎥ = ⎢⎣ ⎥⎦ = = 12 12 1 15624 1302 12 5 1 4. u = 1– x2 , du = –2x dx When x = 0, u = 1 and when x = 1, u = 0 . ∫ 1 x 1– x 2 dx = – 1 ∫ 1 1 − x 2 ( − 2 x dx ) 0 0 2 1 2 1 2 0 1/ 2 1 1 1/ 2 0 u du ∫ ∫ u du = − = 1 1 1 3 3 = ⎡⎢ u 3/ 2 ⎤⎥ = ⎣ ⎦ 0 dx = 1 tan ⎛ x ⎞ ⎜ ⎟ + C x 5. –1 ∫ 2 + ⎝ ⎠ 4 2 2 6. u = 2 + ex , du = exdx ∫ ∫ 2 + = ln u +C ln 2 ln(2 ) x x e dx = du e u x e C e C = + + = + x + 7. u = x2 + 4, du = 2x dx x dx du x u ∫ 2 ∫ + 1 ln 2 1 = 4 2 = u + C 1 ln 2 4 2 = x + +C 1 ln( 2 4) 2 = x + + C 8. 2 2 2 2 t dt t dt t t 2 2 + 1 − 1 2 1 2 1 ∫ = ∫ + + – 1 = ∫ dt ∫ dt 2 t 2 + 1 u = 2t, du = 2dt t – 1 dt t – 1 du ∫ = ∫ + + – 1 tan–1( 2 ) 2 2 2 1 2 1 t u = t t + C 2 9. u = 4 + z2 , du = 2z dz ∫6z 4 + z2 dz = 3∫ u du = 2u3/ 2 +C = 2(4 + z2 )3/ 2 +C 412 Section 7.1 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  2. 2. 10. u = 2t +1, du = 2dt dt du 5 5 2 1 2 ∫ = ∫ + = 5 u +C = 5 2t +1 + C t u z dz z z dz z tan tan sec cos 11. 2 ∫ = 2 ∫ u = tan z, du = sec2 z dz ∫ tan z sec2 z dz = ∫u du 1 2 2 = u +C 1 tan2 2 = z +C 12. u = cos z, du = –sin z dz ∫ecos z sin z dz = –∫ecos z (– sin z dz) = −∫eudu = −eu +C = –ecos z +C 13. , 1 u t du dt = = 2 t sin t dt 2 sin u du ∫ = ∫ t = –2 cos u + C = –2cos t +C 14. u = x2 , du = 2x dx x dx du x u 2 1– 1– ∫ = ∫ = sin–1 u +C = sin–1(x2 ) +C 4 2 15. u = sin x, du = cos x dx x dx du x u cos 1 sin 1 π / 4 2 / 2 0 2 0 2 ∫ = ∫ + + − u − 1 2/2 [tan ] tan 2 0 1 = ≈ 0.6155 2 = 16. 1– , – 1 u xdu dx 2 1– x = = x dx u du x sin 1– –2 sin 1– 3/ 4 1/ 2 0 1 ∫ = ∫ 1 1/ 2 = 2∫ sin u du 1 = [ − 2cos u ]1/ 2 = − 2 ⎛ cos1 − cos 1 ⎞ ⎜ ⎟ 2 ⎝ ⎠ ≈ 0.6746 17. 3 2 2 1 (3 –1) x + xdx x dx dx x x ∫ = ∫ + ∫ + + 3 2 – ln 1 2 1 1 = x x + x + +C 18. 3 x xdx x x dx dx x x 7 ( 2 8) 8 1 –1 –1 + ∫ = ∫ + + + ∫ 1 3 1 2 8 8ln –1 3 2 = x + x + x + x +C 19. u ln 4x2 , du 2 dx = = x sin(ln 4 2 ) 1 sin x dx u du x ∫ = ∫ 2 – 1 cos = u +C 2 – 1 cos(ln 4 2 ) 2 = x +C 20. u = ln x, du 1 dx x = 2 sec (ln ) 1 sec2 x dx u du x ∫ = ∫ 2 2 1 tan 2 = u +C 1 tan(ln ) 2 = x +C 21. u = ex , du = exdx x x e dx du du e u 6 = 6 ∫ 1 1 2 2 − − 1 1 u C e C − − 6sin 6sin ( ) = + = x + ∫ Instructor’s Resource Manual Section 7.1 413 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  3. 3. 22. u = x2 , du = 2x dx x dx 1 du x u ∫ = ∫ + + 1 tan 1 4 2 4 2 4 2 4 = − u + C 2 ⎛ ⎞ x C 1 tan–1 4 2 = ⎜⎜ ⎟⎟ + ⎝ ⎠ 23. u = 1– e2x , du = –2e2xdx x x 2 2 e dx du e u 3 – 3 1– 2 ∫ = ∫ = –3 u +C = –3 1– e2x +C 24. 3 3 4 4 x dx 1 4 x dx x x ∫ = ∫ + + 1 ln 4 4 4 4 4 4 = x + +C 1 ln( 4 4) 4 = x + + C 25. 3 1 2 3 1 2 1 2 0 0 ∫ t t dt = ∫ t t dt 2 2 1 ⎡ t ⎤ = ⎢ ⎥ = ⎢ ⎥ ⎣ ⎦ 1 0.9102 ln 3 3 3 – 1 2ln3 2ln3 2ln3 0 = ≈ 26. / 6 cos / 6 cos 2 x sin x dx – 2 x (– sin x dx) π π ∫ = ∫ 0 0 cos / 6 ⎡ π =⎢ – 2 x ⎤ ⎥ ⎢⎣ ln 2 ⎥⎦ 0 – 1 (2 3 / 2 – 2) = ln 2 2 − 2 3 / 2 ln 2 0.2559 = ≈ x x dx x dx 27. sin cos 1 cos − ⎛ ⎞ = ⎜ − ⎟ ∫ ∫ sin x ⎝ sin x ⎠ u = sin x, du = cos x dx sin cos x − x dx x du ∫ = − ∫ sin = x − ln u +C = x − ln sin x +C x u 28. u = cos(4t – 1), du = –4 sin(4t – 1)dt t dt t dt t t sin(4 − 1) sin(4 − 1) 1 sin (4 1) cos (4 1) ∫ = ∫ − 2 − 2 − 1 1 4 = − ∫ du u 2 1 1 1 sec(4 1) 4 4 = u− +C = t − +C 29. u = ex , du = exdx ∫ex sec exdx = ∫secu du = ln secu + tan u + C = ln sec ex + tan ex +C 30. u = ex , du = exdx ∫ex sec2 (ex )dx = ∫sec2 u du = tan u + C = tan(ex ) +C 31. x 3 sin sec (sec2 sin cos ) x + e dx x e x x dx x ∫ = ∫ + sec = tan x + ∫esin x cos x dx u = sin x, du = cos x dx tan x + ∫esin x cos x dx = tan x + ∫eu du = tan x + eu +C = tan x + esin x +C 32. u = 3t2 − t −1 , 1 (3 2 1) 1/ 2 (6 1) 2 du = t − t − − t − dt 2 t t t dt u du (6 − 1)sin 3 − − 1 2 sin ∫ = ∫ 3 t 2 − t − 1 = –2 cos u + C = −2cos 3t2 − t −1 + C 414 Section 7.1 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  4. 4. 33. u = t3 − 2 , du = 3t2dt 2 3 t t dt u du cos( − 2) 1 cos sin ( 2) 3 sin ∫ = ∫ − v = sin u, dv = cos u du 2 3 2 t u u du v dv v C u 1 cos 1 1 3 sin 3 3 2 1 ∫ = ∫ − = − − + 2 1 3sin C = − + u 1 = − + 3 t 3sin( 2) C − . x dx dx x dx x x x 1 cos2 1 cos2 sin 2 sin 2 sin 2 + ∫ = ∫ + ∫ = ∫csc2 2x dx + ∫cot 2x csc 2x dx 34. 2 2 2 1 cot 2 1 csc 2 2 2 = − x − x +C 35. u = t3 − 2 , du = 3t2dt 2 2 3 2 t t dt u du cos ( − 2) 1 cos sin ( 2) 3 sin ∫ = ∫ − 1 cot2 3 2 3 2 t u = ∫ u du 1 (csc2 –1) = ∫ u du 3 = − u − u +C 1 1[ cot ] 3 1[ cot( 3 2) ( 3 2)] 3 = − t − − t − + C 1[cot( 3 2) 3] 1 = − t − + t +C 3 36. u = 1 + cot 2t, du = −2csc2 2t csc2 2 1 1 1 cot 2 2 t dt du t u ∫ = − ∫ + = − u + C = − 1+ cot 2t + C 2 1 4 37. u = tan−1 2t , 2 du dt t = + tan − 1 2 e dt eudu ∫ + 2 ∫ 1 1 tan 1 2 2 2 1 t = 1 4 t 2 − eu C e t C = + = + 38. u = −t2 − 2t − 5 , du = (–2t – 2)dt = –2(t + 1)dt 2 2 5 1 ( 1) ∫ t + e−t − t− = − ∫eudu 2 1 2 = − eu +C 1 2 2 5 2 = − e−t − t− +C 39. u = 3y2 , du = 6y dy y dy 1 1 du y u ∫ = ∫ 1 sin 1 6 4 16 9 6 4 4 2 2 − − = − ⎛ u ⎞ +C ⎜ ⎟ ⎝ ⎠ 2 − ⎛ y ⎞ C 1 sin 1 3 6 4 = ⎜⎜ ⎟⎟ + ⎝ ⎠ 40. u = 3x, du = 3dx x dx u du u C ∫ ∫ cosh 3 1 (cosh ) 1 sinh 3 3 1 sinh 3 3 = = + x C = + 41. u = x3 , du = 3x2dx 2 sinh 3 1 sinh ∫ x x dx = ∫ u du 3 1 cosh 3 = u +C 1 cosh 3 3 = x +C 42. u = 2x, du = 2 dx 5 5 1 9 4 2 3 ∫ ∫ 5 sin 1 2 3 dx = du x 2 2 u 2 − − = − ⎛ u ⎞ +C ⎜ ⎟ ⎝ ⎠ = − ⎛ x ⎞ +C ⎜ ⎟ 5 sin 1 2 2 3 ⎝ ⎠ 43. u = e3t , du = 3e3tdt t t 3 6 2 2 e dt 1 1 du e u ∫ = ∫ 1 sin 1 3 2 4 3 2 − − = − ⎛ u ⎞ +C ⎜ ⎟ ⎝ ⎠ 3 e t − C ⎛ ⎞ 1 sin 1 3 2 = ⎜⎜ ⎟⎟ + ⎝ ⎠ 44. u = 2t, du = 2dt dt 1 1 du t t u u ∫ = ∫ 1 sec 1 2 2 4 2 − 1 2 2 − 1 = ⎣⎡ − u ⎦⎤ +C = 1 sec − 1 2 t +C 2 Instructor’s Resource Manual Section 7.1 415 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  5. 5. 45. u = cos x, du = –sin x dx x dx du x u sin 1 / 2 0 0 2 1 2 ∫ = − ∫ + + 16 cos 16 π 1 16 1 0 2 du u = + ∫ 1 ⎡ = 1 tan − 1 ⎛ u ⎞⎤ ⎢ 4 ⎜ ⎣ ⎝ 4 ⎠⎦ ⎟⎥ 0 1 tan 1 1 1 tan 1 0 4 4 4 = ⎡ − ⎛ ⎞ − − ⎤ ⎢ ⎜ ⎟ ⎥ ⎣ ⎝ ⎠ ⎦ 1 tan 1 1 0.0612 4 4 = − ⎛ ⎞ ≈ ⎜ ⎟ ⎝ ⎠ 46. 2 2 x x u e e− = + , du = (2e2x − 2e−2x )dx = 2(e2x − e−2x )dx 1 2 x − 2 x e 2 + e − 2 0 2 x − 2 x 2 e − e dx 1 1 du e e 2 u ∫ = ∫ + 2 2 2 + − = ⎡⎣ ⎤⎦ 1 ln 2 e e u 1 ln 2 2 1 ln 2 2 2 = e + e− − 4 2 e e 1 + ln 1 1 ln 2 2 2 = − 1 ln( 4 1) 1 ln( 2 ) 1 ln 2 2 2 2 = e + − e − 1 4 1 ln 2 0.6625 2 2 = ⎛⎜ ⎛⎜ e + ⎞⎟ − ⎞⎟ ≈ ⎜ ⎜ ⎟ ⎟ ⎝ ⎝ ⎠ ⎠ 1 1 2 5 2 1 4 + + + + + ∫ ∫ 47. 2 2 dx = dx x x x x 1 ( 1) + + ∫ 1 tan–1 1 2 2 = + 2 2 x ( 1) 2 d x x C ⎛ + ⎞ = ⎜ ⎟ + ⎝ ⎠ 1 1 –4 9 –4 4 5 + + + ∫ ∫ 48. 2 2 dx = dx x x x x 1 ( –2) + ∫ 1 tan–1 – 2 5 5 2 2 x ( –2) ( 5) d x = x C ⎛ ⎞ = ⎜ ⎟ + ⎝ ⎠ dx dx + + + + + ∫ ∫ 49. = 9 x 2 18 x 10 9 x 2 18 x 9 1 dx x = ∫ (3 + 3)2 + 12 u = 3x + 3, du = 3 dx dx 1 du x u ∫ = ∫ + + + 1 tan–1(3 3) 3 2 2 2 2 (3 3) 1 3 1 = x + +C 50. dx dx x x x x ∫ = ∫ 16 + 6 – 2 –( 2 – 6 + 9 – 25) dx x dx x ∫ –( – 3)2 52 52 – ( – 3)2 = + = ∫ = sin–1 ⎛ x – 3 ⎞ ⎜ ⎟ +C 5 ⎝ ⎠ x + 1 dx 1 18 x + 18 dx + + + + ∫ ∫ 51. = 2 2 9 x 18 x 10 18 9 x 18 x 10 1 ln 9 2 18 10 18 1 ln 9 18 10 18 x x C = + + + ( 2 ) x x C = + + + 52. x dx x dx x x x x 3– 1 6 – 2 16 6 – 2 16 6 – ∫ = ∫ 2 2 + + = 16 + 6x – x2 +C 53. u = 2t, du = 2dt dt du ∫ = ∫ 2 2 – 9 2 – 32 t t u u ⎛ ⎞ t –1 1 2 sec 3 3 C = ⎜ ⎟ + ⎜ ⎟ ⎝ ⎠ 54. x dx x x dx x x x tan cos tan sec – 4 cos sec – 4 ∫ = ∫ 2 2 x dx = ∫ 2 u = 2 cos x, du = –2 sin x dx sin 1– 4cos x x dx 1 1 2 1 ∫ 2 2 sin 1 4cos − x du u = − − ∫ = − − u +C – 1 sin–1(2cos ) 1 sin 1 2 = x +C 2 55. The length is given by 2 L dy dx = + ⎛ ⎞ ⎜ ⎟ 1 b a ∫ ⎝ dx ⎠ / 4 2 1 1 ( sin ) 0 π = ∫ + ⎡ ⎢ − ⎤ ⎣ cos ⎥ ⎦ x dx x = ∫ + / 4 2 / 4 2 0 1 tan x dx π sec x dx π = ∫ 0 / 4 0 sec x dx π = ∫ 0= ⎡ ⎣ ln sec x + tan x ⎤ π / 4 ⎦ = ln 2 +1 − ln 1 = ln 2 +1 ≈ 0.881 416 Section 7.1 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  6. 6. x x 1 + 56. sec = = 1 sin x x x cos cos (1 + sin ) sin sin2 cos2 sin (1 sin ) cos2 x + x + x x + x + x = = x x x x cos (1 + sin ) cos (1 + sin ) x x x x sin cos cos 1 sin = + + x x x dx sec sin cos ∫ = ∫ ⎛ ⎞ ⎜ + ⎝ cos x 1 + sin x ⎟ ⎠ x dx x dx x x sin cos cos 1 sin = ∫ + ∫ + For the first integral use u = cos x, du = –sin x dx, and for the second integral use v = 1 + sin x, dv = cos x dx. sin cos – cos 1 sin x dx x dx du dv x x u v ∫ + ∫ = ∫ + ∫ + = – ln u + ln v +C = – ln cos x + ln 1+ sin x +C 1 + ln sin = + cos = ln sec x + tan x + C x C x 57. u = x – π , du = dx x x u u sin ( + π ) sin( + π ) 2 π π 0 2 – 2 + + + π ∫ ∫ dx = du x π u 1 cos 1 cos ( ) u u ( )sin 1 cos π π + π + ∫ – 2 du u = u sin u sin u π π π π π + + ∫ ∫ du du u u = + – 2 – 2 1 cos 1 cos u sin u π π ∫ 0 by symmetry. – + 2 1 cos du u = u u du du u u sin sin 2 π π π π π ∫ = ∫ + + v = cos u, dv = –sin u du – 2 0 2 1 cos 1 cos π 2 2 1 –1 1 1 2 –1 2 + + ∫ ∫ dv dv v v − = π 1 1 = 2 π [tan − 1 v ] 1 ⎡π ⎛ π ⎞⎤ − 1 = 2 π ⎢ − ⎜− ⎣ 4 ⎝ 4 ⎟⎥ ⎠⎦ 2 2 ⎛ π ⎞ = π⎜ ⎟ = π 2 ⎝ ⎠ 58. 3 4 4 – π π ⎛ π ⎞ = π ⎜ + ⎟ ⎝ ⎠ ∫ – , 4 V 2 x sin x – cos x dx 4 π u = x du = dx π π ⎛ π ⎞ ⎛ π ⎞ ⎛ π ⎞ = π ⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ∫ V u u u du 2 2 – 2 sin – cos 2 4 4 π π 2 2 sin 2 cos – 2 cos 2 sin ⎛ π ⎞ = π ⎜ + ⎟ + + ⎝ ⎠ ∫ 2 2 – u u u u udu 2 2 2 2 2 π π π −π −π −π ⎛ π ⎞ = π ⎜ + ⎟ = + 2 2 sin 2 2 sin 2 2 sin ⎝ ⎠ ∫ ∫ ∫ u u du π u u du π u du 2 2 2 2 2 2 2 π −π π∫ = by symmetry. Therefore, 2 2 u sin u du 0 2 2 π π 2 2 2 2 2 = π ∫ = π − = π V 2 2 sin u du 2 2 [ cosu] 2 2 0 0 Instructor’s Resource Manual Section 7.1 417 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  7. 7. 7.2 Concepts Review 1. uv – ∫v du 2. x; sin x dx 3. 1 4. reduction Problem Set 7.2 1. u = x dv = exdx du = dx v = ex ∫ xexdx = xex − ∫exdx = xex − ex +C 2. u = x dv = e3xdx du = dx 1 3 v = e x 3 3 1 3 1 3 ∫ xe xdx = xe x − ∫ e xdx 3 3 1 3 1 3 3 9 = xe x − e x +C 3. u = t dv = e5t+πdt du = dt 1 5 v = e t+π 5 5 1 5 – 1 5 ∫te t+πdt = te t+π ∫ e t+πdt 5 5 1 5 – 1 5 5 25 = te t+π e t+π +C 4. u = t + 7 dv = e2t+3dt du = dt v = 1 e 2 t+ 3 2 ( 7) 2 3 1 ( 7) 2 3 – 1 2 3 ∫ t + e t+ dt = t + e t+ ∫ e t+ dt 2 2 1 ( 7) 2 3 – 1 2 3 2 4 = t + e t+ e t+ +C = t e t+ + e t+ + C 2 3 13 2 3 2 4 5. u = x dv = cos x dx du = dx v = sin x ∫ x cos x dx = x sin x – ∫sin x dx = x sin x + cos x + C 6. u = x dv = sin 2x dx du = dx – 1 cos 2 v = x 2 sin 2 – 1 cos 2 – – 1 cos 2 ∫ x x dx = x x ∫ x dx 2 2 – 1 cos 2 1 sin 2 = x x + x +C 2 4 7. u = t – 3 dv = cos (t – 3)dt du = dt v = sin (t – 3) ∫(t – 3) cos(t – 3)dt = (t – 3)sin(t – 3) – ∫sin(t – 3)dt = (t – 3) sin (t – 3) + cos (t – 3) + C 8. u = x – π dv = sin(x)dx du = dx v = –cos x ∫(x – π)sin(x)dx = –(x – π) cos x + ∫cos x dx = (π – x) cos x + sin x + C 9. u = t dv = t +1 dt du = dt v = 2 ( t + 1)3/ 2 3 1 2 ( 1)3/ 2 – 2 ( 1)3/ 2 ∫t t + dt = t t + ∫ t + dt 3 3 2 ( 1)3/ 2 – 4 ( 1)5/ 2 3 15 = t t + t + +C 10. u = t dv = 3 2t + 7dt du = dt v = 3 (2 t + 7)4 / 3 8 3 2 7 3 (2 7)4 / 3 – 3 (2 7)4 / 3 ∫t t + dt = t t + ∫ t + dt 8 8 3 (2 7)4 / 3 – 9 (2 7)7 / 3 8 112 = t t + t + +C 11. u = ln 3x dv = dx du 1 dx = v = x x ln 3x dx x ln 3x x 1 dx ∫ = −∫ = x ln 3x − x +C x 12. u = ln(7x5 ) dv = dx du 5 dx = v = x x ln(7x5 )dx x ln(7x5 ) – x 5 dx x ∫ = ∫ = x ln(7x5 ) – 5x +C 418 Section 7.2 Instructor Solutions Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  8. 8. 13. u = arctan x dv = dx du dx 2 1 1 x = + v = x x x x x dx + ∫ ∫ 2 arctan arctan 1 x = − x x x dx 2 arctan 1 2 + ∫ 2 1 x = − arctan 1 ln(1 2 ) = x x − + x +C 2 14. u = arctan 5x dv = dx du dx 2 5 1 25 x = + v = x x dx x x x dx 2 arctan 5 arctan 5 – 5 + ∫ ∫ 1 25 x = x x x dx 2 arctan 5 – 1 50 + ∫ 10 1 25 x = arctan 5 – 1 ln(1 25 2 ) = x x + x +C 10 dv dx 15. u = ln x 2 x = du 1 dx = v – 1 x x = ln x dx – ln x – – 1 1 dx x x x x ∫ 2 ∫ ⎝ ⎠ – ln x – 1 C = ⎛ ⎞ ⎜ ⎟ = + x x 16. u = ln 2x5 dv 1 dx 2 x = du 5 dx = v 1 x x = − 3 5 3 3 5 2 2 2 2 2 ln 2x dx 1 ln 2x 5 1 dx x x x = ⎡− ⎤ + ⎢⎣ ⎥⎦ ∫ ∫ 3 1 ln 2 5 5 1 ln(2 3 ) 5 1 ln(2 2 ) 5 3 3 2 2 1 ln 2 5 ln 3 5 3ln 2 5 3 3 3 2 = ⎡− ⎢⎣ x − ⎤ x x ⎥⎦ 2 = ⎛ − ⋅ 5 − ⎞ ⎜ ⎟ − ⎛− ⎜ ⋅ 5 − ⎞ ⎟ ⎝ ⎠ ⎝ ⎠ = − − − + + 8 ln 2 5 ln 3 5 0.8507 3 3 6 = − + ≈ 17. u = ln t dv = t dt du 1dt = 2 3/ 2 t v = t 3 e e ∫ t ln t dt = ⎡⎢ 2 t 3/ 2 ln t⎤⎥ – ∫ e 2 t 1/ 2 dt ⎣ 3 ⎦ 3 1 1 1 2 ln – 2 1ln1 4 3 3 9 = e 3/ 2 e ⋅ − ⎡⎢ t 3/ 2 ⎤⎥ ⎣ ⎦ 2 3/ 2 0 4 3/ 2 4 2 3/ 2 4 1.4404 3 9 9 9 9 e 1 = e − − e + = e + ≈ 18. u = ln x3 dv = 2xdx du 3 dx = 1 (2 )3/ 2 x v = x 3 5 3 1 ∫ 2x ln x dx 5 5 3 2 3 3 2 1 (2 ) ln 2 3 = ⎡⎢ x x ⎤⎥ − x dx ⎣ ⎦ ∫ 1 1 5/ 2 5 = ⎡ 1 ⎢⎣ (2 x ) 3/ 2 ln x 3 − 2 x 3/ 2 ⎤ 3 3 ⎥⎦ 1 5 2 5 2 ⎛ ⎞ 1 (10)3 2 ln 53 2 53/ 2 1 (2)3 2 ln13 2 3 3 3 3 = − − ⎜⎜ − ⎟⎟ ⎝ ⎠ 4 2 53 2 4 2 103 2 ln 5 31.699 3 3 = − + + ≈ 19. u = ln z dv = z3dz du 1 dz = 1 4 z v = z 4 3 ln 1 4 ln 1 4 1 ∫ = − ∫ ⋅ 1 4 ln 1 3 4 4 z zdz z z z dz 4 4 z = z z − ∫ z dz 1 4 ln 1 4 4 16 = z z − z +C 20. u = arctan t dv = t dt du dt 2 1 1 t = + 1 2 2 v = t 2 t arctan t dt 1 t 2 arctan t – 1 t dt 2 + ∫ ∫ 2 21 t = 2 t t t dt 1 2 1 1 + − arctan 1 2 2 1 t 2 = − + ∫ 1 2 arctan 1 ∫ 1 ∫ 1 2 2 2 1 + t t dt dt 2 t = − + 1 2 arctan 1 1 arctan 2 2 2 = t t − t + t +C Instructor's Resource Manual Section 7.2 419 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  9. 9. 21. u arctan 1 = ⎛ ⎞ ⎜ ⎟ t ⎝ ⎠ dv = dt du dt 2 – 1 1 t = + v = t dt t t dt t t t 2 arctan 1 arctan 1 ∫ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ = ⎜ ⎟ + ∫ ⎝ ⎠ ⎝ ⎠ 1 + arctan 1 1 ln(1 2 ) = ⎛ ⎞ + + + ⎜ ⎟ t t C 2 t ⎝ ⎠ 22. u = ln(t7 ) dv = t5dt du 7 dt = 1 6 t v = t 6 5 ln( 7 ) 1 6 ln( 7 ) – 7 5 ∫t t dt = t t ∫t dt 6 6 = 1 t 6 ln( t 7 ) – 7 t 6 +C 6 36 23. u = x dv = csc2 x dx du = dx v = −cot x [ ] / 2 2 / 2 / 2 / 6 / 6 / 6 x csc x dx x cot x cot x dx π π π π π π ∫ = − + ∫ 2 6 cot ln sin x x x π π = ⎡⎣− + ⎤⎦ 0 ln1 3 ln 1 ln 2 1.60 2 6 2 2 3 π π π = − ⋅ + + − = + ≈ 24. u = x dv = sec2 x dx du = dx v = tan x [ ] 4 2 4 4 6 6 6 x sec x dx x tan x tan x dx π π π π π π π ⎛ π ⎞ tan ln cos ln 2 ln 3 ∫ = − ∫ x x π 4 x = ⎡⎣ + ⎤⎦ = + − ⎜⎜ + ⎟⎟ 6 4 2 6 3 2 π ⎝ ⎠ 1 ln 2 0.28 π π = − + ≈ 4 6 3 2 3 25. u = x3 dv = x2 x3 + 4dx du = 3x2dx v = 2 ( x 3 + 4)3/ 2 9 ∫ x 5 x 3 + 4 dx = 2 x 3( x 3 + 4)3/ 2 – ∫ 2 x 2 ( x 3 + 4)3/ 2 dx 2 3( 3 4)3/ 2 – 4 ( 3 4)5 / 2 9 3 = x x + x + +C 9 45 26. u = x7 dv = x6 x7 +1 dx du = 7x6dx v = 2 ( x 7 + 1)3/ 2 21 ∫ x 13 x 7 + 1 dx = 2 x 7 ( x 7 + 1)3/ 2 – ∫ 2 x 6 ( x 7 + 1)3/ 2 dx 2 7 ( 7 1)3/ 2 – 4 ( 7 1)5/ 2 21 3 = x x + x + +C 21 105 27. u = t4 3 dv t dt (7 – 3 t 4 )3/ 2 = du = 4t3 dt 4 1/2 1 t 6(7 – 3 ) v = 7 4 3 4 3/ 2 4 1/ 2 4 1/ 2 t dt t – 2 t dt t t t ∫ = ∫ (7 – 3 ) 6(7 – 3 ) 3 (7 – 3 ) 4 t 1 (7 – 3 t 4 ) 1/2 C t = + + 4 1/2 6(7 – 3 ) 9 420 Section 7.2 Instructor Solutions Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  10. 10. 28. u = x2 dv = x 4 – x2 dx du = 2x dx v = – 1 (4 – x 2 )3/ 2 3 ∫ x 3 4 – x 2 dx = – 1 x 2 (4 – x 2 )3/ 2 + 2 ∫ x (4 – x 2 )3/ 2 dx – 1 2 (4 – 2 )3/ 2 – 2 (4 – 2 )5 / 2 3 3 = x x x +C 3 15 29. u = z4 3 dv z dz (4 – z 4 )2 = du = 4z3dz 4 1 z 4(4 – ) v = 7 4 3 z dz z z dz z z z ∫ = − ∫ (4 – 4 )2 4(4 – 4 ) 4 – 4 4 z z 4 C z = + + 4 1 ln 4 – 4(4 – ) 4 30. u = x dv = cosh x dx du = dx v = sinh x ∫ x cosh x dx = x sinh x – ∫sinh x dx = x sinh x – cosh x + C 31. u = x dv = sinh x dx du = dx v = cosh x ∫ x sinh x dx = x cosh x – ∫cosh x dx = x cosh x – sinh x + C 32. u = ln x dv = x–1/ 2dx du 1 dx = v = 2x1/ 2 x ln x dx 2 x ln x – 2 1 dx x x ∫ = ∫ = 2 x ln x – 4 x +C 1/ 2 33. u = x dv = (3x +10)49 dx du = dx v = 1 (3 x + 10)50 150 ∫ x (3 x + 10)49 dx = x (3 x + 10)50 – 1 ∫ (3 x + 10)50 dx (3 10)50 – 1 (3 10)51 150 150 = x x + x + + C 150 22,950 34. u = t dv = (t −1)12 dt du = dt 1 ( 1)13 v = t − 13 ( ) ( ) 1 1 1 12 13 13 0 0 0 t ( t 1) dt t t 1 1 t 1 dt − = ⎡ − ⎤ − − ⎢⎣ ⎥⎦ ∫ ∫ 13 13 t t t 1 1 1 1 = ⎡ 13 14 ⎤ ⎢⎣ ( − ) − ( − ) ⎥⎦ = 1 13 182 182 0 35. u = x dv = 2x dx du = dx 1 2 v = x ln 2 ∫ x 2 x dx = x 2 x – 1 ∫ 2 x dx ln 2 ln 2 = x 2 x – 1 2 x +C 2 ln 2 (ln 2) 36. u = z dv = azdz du = dz 1 v az ln a = zazdz z az – 1 azdz ∫ = ∫ a a ln ln z az az C a a = + 2 – 1 ln (ln ) 37. u = x2 dv = exdx du = 2x dx v = ex ∫ x2exdx = x2ex − ∫ 2xexdx u = x dv = exdx du = dx v = ex ∫ x2ex dx = x2ex − 2(xex − ∫ex dx) = x2ex − 2xex + 2ex +C Instructor's Resource Manual Section 7.2 421 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  11. 11. 38. u = x4 x2 dv = xe dx du = 4x3dx 1 2 2 v = ex 5 2 1 4 2 3 2 – 2 ∫ x ex dx = x ex ∫ x ex dx 2 2 u = x2 dv = 2xex dx du = 2x dx x2 v = e 5 2 1 4 2 2 2 2 – 2 x ex dx = x ex ⎛ x ex − xex dx ⎞ ⎜ ⎟ ∫ ∫ 2 ⎝ ⎠ 1 4 2 2 2 – 2 2 = x ex x ex + ex +C 39. u = ln2 z dv = dz du 2ln z dz = v = z z ∫ln2 z dz = z ln2 z – 2∫ln z dz u = ln z dv = dz du 1 dz = v = z z ∫ln2 z dz = z ln2 z – 2(z ln z – ∫ dz) = z ln2 z – 2z ln z + 2z +C 40. u = ln2 x20 dv = dx 40ln x20 du dx = v = x x ∫ln2 x20dx = x ln2 x20 – 40∫ln x20dx u = ln x20 dv = dx du 20 dx = v = x x ∫ln2 x20dx = x ln2 x20 – 40(x ln x20 – 20∫ dx) = x ln2 x20 – 40x ln x20 + 800x +C 41. u = et dv = cos t dt du = etdt v = sin t ∫et cos t dt = et sin t − ∫et sin t dt u = et dv = sin t dt du = etdt v = –cos t ∫et cost dt = et sin t − ⎡⎣−et cost + ∫et cos t dt⎤⎦ ∫et cos t dt = et sin t + et cos t − ∫et cos t dt 2∫et cos t dt = et sin t + et cos t +C cos 1 (sin cos ) ∫et t dt = et t + t +C 2 42. u = eat dv = sin t dt du = aeatdt v = –cos t ∫eat sin t dt = –eat cos t + a∫eat cos t dt u = eat dv = cos t dt du = aeatdt v = sin t ∫eat sin t dt = –eat cos t + a (eat sin t – a∫eat sin t dt ) ∫eat sin t dt = –eat cos t + aeat sin t – a2 ∫eat sin t dt (1+ a2 )∫eat sin t dt = –eat cos t + aeat sin t + C at at eat sin t dt – e cos t ae sin t C + + ∫ = + + 2 2 a a 1 1 43. u = x2 dv = cos x dx du = 2x dx v = sin x ∫ x2 cos x dx = x2 sin x − ∫ 2x sin x dx u = 2x dv = sin x dx du = 2dx v = −cos x ∫ x2 cos x dx = x2 sin x − (−2x cos x + ∫ 2cos x dx) = x2 sin x + 2x cos x − 2sin x +C 44. u = r2 dv = sin r dr du = 2r dr v = –cos r ∫ r2 sin r dr = –r2 cos r + 2∫ r cos r dr u = r dv = cos r dr du = dr v = sin r ∫ r2 sin r dr = –r2 cos r + 2(r sin r – ∫sin r dr ) = –r2 cos r + 2r sin r + 2cos r +C 422 Section 7.2 Instructor Solutions Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  12. 12. 45. u = sin(ln x) dv = dx du cos(ln x) 1 dx = ⋅ v = x x ∫sin(ln x)dx = x sin(ln x) − ∫cos(ln x) dx u = cos (ln x) dv = dx du sin(ln x) 1 dx = − ⋅ v = x x ∫sin(ln x)dx = x sin(ln x) − ⎡⎣x cos(ln x) − ∫ −sin(ln x)dx⎤⎦ ∫sin(ln x)dx = x sin(ln x) − x cos(ln x) − ∫sin(ln x)dx 2∫sin(ln x)dx = x sin(ln x) − x cos(ln x) +C sin(ln ) [sin(ln ) cos(ln )] ∫ x dx = x x − x +C 2 46. u = cos(ln x) dv = dx du – sin(ln x) 1 dx = v = x x ∫cos(ln x)dx = x cos(ln x) + ∫sin(ln x)dx u = sin(ln x) dv = dx du cos(ln x) 1 dx = v = x x ∫cos(ln x)dx = x cos(ln x) + ⎡⎣x sin(ln x) – ∫cos(ln x)dx⎤⎦ 2∫cos(ln x)dx = x[cos(ln x) + sin(ln x)]+C cos(ln ) [cos(ln ) sin(ln )] ∫ x dx = x x + x + C 2 47. u = (ln x)3 dv = dx 3ln2 x du dx = v = x x ∫(ln x)3dx = x(ln x)3 – 3∫ln2 x dx = x ln3 x – 3(x ln2 x – 2x ln x + 2x +C) = x ln3 x – 3x ln2 x + 6x ln x − 6x +C 48. u = (ln x)4 dv = dx 4ln3 x du dx = v = x x ∫(ln x)4 dx = x(ln x)4 – 4∫ln3 x dx = x ln4 x – 4(x ln3 x – 3x ln2 x + 6x ln x − 6x +C) = x ln4 x – 4x ln3 x +12x ln2 x − 24x ln x + 24x +C 49. u = sin x dv = sin(3x)dx du = cos x dx v = – 1 cos(3 x ) 3 sin sin(3 ) – 1 sin cos(3 ) 1 cos cos(3 ) ∫ x x dx = x x + ∫ x x dx 3 3 u = cos x dv = cos(3x)dx du = –sin x dx v = 1 sin(3 x ) 3 Instructor's Resource Manual Section 7.2 423 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  13. 13. sin sin(3 ) – 1 sin cos(3 ) 1 1 cos sin(3 ) 1 sin sin(3 ) ∫ x x dx = x x + ⎡⎢ x x + ∫ x x dx⎤⎥ 3 33 ⎣ 3 ⎦ – 1 sin cos(3 ) 1 cos sin(3 ) 1 sin sin(3 ) = x x + x x + ∫ x x dx 3 9 9 8 sin sin(3 ) – 1 sin cos(3 ) 1 cos sin(3 ) 9 3 9 ∫ x x dx = x x + x x + C sin sin(3 ) – 3 sin cos(3 ) 1 cos sin(3 ) ∫ x x dx = x x + x x +C 8 8 50. u = cos (5x) dv = sin(7x)dx du = –5 sin(5x)dx v = – 1 cos(7 x ) 7 cos(5 )sin(7 ) – 1 cos(5 ) cos(7 ) – 5 sin(5 ) cos(7 ) ∫ x x dx = x x ∫ x x dx 7 7 u = sin(5x) dv = cos(7x)dx du = 5 cos(5x)dx v = 1 sin(7 x ) 7 cos(5 )sin(7 ) – 1 cos(5 ) cos(7 ) – 5 1 sin(5 )sin(7 ) – 5 cos(5 )sin(7 ) ∫ x x dx = x x ⎡⎢ x x ∫ x x dx⎤⎥ 7 77 ⎣ 7 ⎦ – 1 cos(5 ) cos(7 ) – 5 sin(5 )sin(7 ) 25 cos(5 )sin(7 ) = x x x x + ∫ x x dx 7 49 49 24 cos(5 )sin(7 ) – 1 cos(5 ) cos(7 ) – 5 sin(5 )sin(7 ) 49 7 49 ∫ x x dx = x x x x +C cos(5 )sin(7 ) – 7 cos(5 ) cos(7 ) – 5 sin(5 )sin(7 ) ∫ x x dx = x x x x + C 24 24 51. u = eα z dv = sin βz dz du =αeα zdz v – 1 cosβ z β = eα z sin z dz – 1 eα z cos z α eα z cos z dz ∫ = + ∫ u = eα z dv = cos βz dz du =αeα zdz v 1 sinβ z β β β β β β = ⎡ ⎤ eα z sin z dz 1 eα z cos z α 1 eα z sin z α eα z sin z dz ⎣ ⎦ ∫ ∫ β β β β = − + ⎢ − ⎥ β β β β 2 – 1 eα z cos z α eα z sin z –α eα z sin z dz = + ∫ 2 2 β β β 2 2 β β β β α eα z sin z dz – 1 eα z cos z α eα z sin z C ∫ = + + β β β + 2 2 β β β e α z ( α sin β z – β cos β z) C eα z sin z dz –β eα z cos z α eα z sin z C ∫ β = β + β + 2 + 2 2 + 2 2 2 α β α β = + α + β 424 Section 7.2 Instructor Solutions Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  14. 14. 52. u = eα z dv = cosβ z dz du =αeα zdz v 1 sinβ z β = eα z cos z dz 1 eα z sin z – α eα z sin z dz ∫ = ∫ u = eα z dv = sin βz dz du =αeα zdz v – 1 cosβ z β β β β β β = ⎡ ⎤ eα z cos z dz 1 eα z sin z α 1 eα z cos z α eα z cos z dz ⎣ ⎦ ∫ ∫ β β β β = − ⎢− + ⎥ β β β β 2 1 eα z sin z α eα z cos z –α eα z cos z dz = + ∫ 2 2 β β β 2 2 β β β α β eα z cos z dz α eα z cos z 1 eα z sin z C ∫ = + + β β β + 2 2 β β β z α α z e ( α cos β z + β β e cos z dz sin z ) C + ∫ = + 2 2 β α β 53. u = ln x dv = xα dx du 1 dx x = 1 1 α α v x + = + , α ≠ –1 α + 1 ln ln – 1 x x dx x x x dx α α 1 1 + + x x x C α α + + ∫ ∫ α α 1 1 = α 2 ln – , –1 1 ( 1) = + ≠ α α + + 54. u = (ln x)2 dv = xα dx du 2ln x dx x = 1 1 α α v x + = + , α ≠ –1 1 α + x (ln x )2 dx x (ln x )2 – 2 x ln x dx 1 1 1 + ⎡ + + ⎤ x α α α x x x x C α α α α (ln ) 2 ln 1 1 1 ( 1) α α + + ∫ ∫ α α 1 1 = 2 = − ⎢ − ⎥ + 2 + + ⎢⎣ + + ⎥⎦ 1 1 1 + + + α α α x x 2 x x x C α 2 3 (ln ) – 2 ln 2 , –1 1 ( 1) ( 1) = + + ≠ α + α + α + Problem 53 was used for xα ln x dx. ∫ 55. u = xα dv = eβ x dx du =α xα –1dx v 1 eβ x β = α β x α e β xdx x e α – x α –1 e β xdx x ∫ = ∫ β β 56. u = xα dv = sin βx dx du =α xα –1dx v – 1 cosβ x β = α x sin x dx – x cos x x –1 cos x dx α β α α ∫ = + ∫ β β β β Instructor's Resource Manual Section 7.2 425 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  15. 15. 57. u = xα dv = cos βx dx du =α xα –1dx v 1 sinβ x β = α x cos x dx x sin x – x –1 sin x dx α β α α ∫ = ∫ β β β β 58. u = (ln x)α dv = dx (ln x) –1 du dx α α = v = x x ∫(ln x)α dx = x(ln x)α –α ∫(ln x)α –1dx 59. u = (a2 – x2 )α dv = dx du = –2α x(a2 – x2 )α –1dx v = x ∫(a2 – x2 )α dx = x(a2 – x2 )α + 2α ∫ x2 (a2 – x2 )α –1dx 60. u = cosα –1 x dv = cos x dx du = –(α –1) cosα –2 x sin x dx v = sin x ∫cosα x dx = cosα –1 x sin x + (α –1)∫cosα –2 x sin2 x dx = cosα −1 x sin x + (α −1)∫cosα −2 x(1− cos2 x) dx = cosα –1 x sin x + (α –1)∫cosα –2 x dx – (α –1)∫cosα x dx α ∫cosα x dx = cosα −1 x sin x + (α −1)∫cosα −2 x dx –1 α cos x dx cos x sin x –1 cos –2 x dx α α α ∫ = + ∫ α α 61. u = cosα –1 β x dv = cos βx dx du = –β (α –1) cosα –2 β x sinβ x dx v 1 sinβ x β = –1 α cos x dx cos x sin x ( –1) cos –2 x sin2 x dx α β β α β α β β ∫ = + ∫ β 1 − α cos β x sin β x ( 1) cos α 2 x(1 cos2 x) dx = + − ∫ − − α β β β –1 α cos x sin x ( –1) cos –2 x dx – ( –1) cos x dx β β α α = + ∫ ∫ α β α β β 1 − α cos x cos x sin x ( 1) cos 2 x dx α β β α ∫ = + − ∫ − α β α β β –1 α cos x dx cos x sin x –1 cos –2 x dx α β β α α β β ∫ = + ∫ αβ α 62. 4 3 1 4 3 – 4 3 3 ∫ x e xdx = x e x ∫ x e xdx 1 4 3 – 4 1 3 3 – 2 3 3 3 = x e x ⎡ ⎢⎣ x e x ∫ x e xdx⎤ 3 33 ⎥⎦ = 1 x 4 e 3 x – 4 x 3 e 3 x + 4 ⎡ 1 x 2 e 3 x – 2 ∫ xe 3 xdx⎤ 1 4 3 – 4 3 3 4 2 3 – 8 1 3 – 1 3 3 9 33 ⎢⎣ 3 ⎥⎦ = x e x x e x + x e x ⎡ ⎢⎣ xe x ∫ e xdx⎤ 3 9 9 93 3 ⎥⎦ 1 4 3 – 4 3 3 4 2 3 – 8 3 8 3 3 9 9 27 81 = x e x x e x + x e x xe x + e x +C 426 Section 7.2 Instructor Solutions Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  16. 16. 63. 4 cos3 1 4 sin 3 – 4 3 sin 3 ∫ x x dx = x x ∫ x x dx 1 4 sin 3 – 4 – 1 3 cos3 2 cos3 3 3 = x x ⎡⎢ x x + ∫ x x dx⎤⎥ 3 3 ⎣ 3 ⎦ 1 4 sin 3 4 3 cos3 – 4 1 2 sin 3 2 sin 3 3 9 3 3 3 = x x + x x ⎡⎢ x x − x x dx⎤⎥ ⎣ ⎦ ∫ 1 4 sin 3 4 3 cos3 – 4 2 sin 3 8 – 1 cos3 1 cos3 3 9 9 9 3 3 = x x + x x x x + ⎡⎢ x x + ∫ x dx⎤⎥ ⎣ ⎦ = 1 x 4 sin 3 x + 4 x 3 cos3 x – 4 x 2 sin 3 x – 8 x cos3 x + 8 sin 3 x + C 3 9 9 27 81 64. cos6 3 1 cos5 3 sin 3 5 cos4 3 ∫ x dx = x x + ∫ x dx 1 cos5 3 sin 3 5 1 cos3 3 sin 3 3 cos2 3 18 6 = x x + ⎡⎢ x x + ∫ x dx⎤⎥ 18 6 ⎣ 12 4 ⎦ 1 cos5 3 sin 3 5 cos3 3 sin 3 5 1 cos3 sin 3 1 18 72 8 6 2 = x x + x x + ⎡⎢ x x + ∫ dx⎤⎥ ⎣ ⎦ = 1 cos5 3 x sin 3 x + 5 cos3 3 x sin 3 x + 5 cos3 x sin 3 x + 5 x +C 18 72 48 16 65. First make a sketch. From the sketch, the area is given by ∫ e ln x dx 1 u = ln x dv = dx du 1 dx = v = x x ln ln ∫e x dx = x x e − ∫e dx 1 [ ln ]e [ ]1 1 1 = x x − x = (e – e) – (1 · 0 – 1) = 1 (ln ) e V = ∫ π x dx u = (ln x)2 dv = dx du 2ln x dx 66. 2 1 = v = x x π ∫ e 2 e (ln x ) dx = π⎛ 2 e ⎞ 2 1 ⎜ ⎡ ⎤ ⎝ ⎣ x (ln x ) ⎦ − 2 ∫ ln x dx 1 1 ⎟ ⎠ e ⎡x x x x x ⎤ ⎣ ⎦ = π − − 2 1 (ln ) 2( ln ) 1 [ (ln ) 2 ln 2 ]e = π x x − x x + x = π[(e − 2e + 2e) − (0 − 0 + 2)] = π(e − 2) ≈ 2.26 Instructor's Resource Manual Section 7.2 427 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  17. 17. 67. 3 –9 – 1 e x dx = e x ⎛ dx ⎞ ⎜ ⎟ 9 – / 3 9 – / 3 0 0 –9[e x ] – 9 9 ∫ 0 ∫ – /3 ⎝ 3 ⎠ 9= = + ≈ 8.55 3 e 68. 9 – / 3 2 9 –2 / 3 V = ∫ π(3e x ) dx = 9π∫ e x dx 0 0 9 – 3 – 2 = π⎛ ⎞ e x ⎛ dx ⎞ ⎜ ⎟ ⎜ ⎟ 9 –2 / 3 0 – 27 π [ e x ] – 27 π 27 π 42.31 2 2 2 ∫ –2 / 3 9 ⎝ 2 ⎠ ⎝ 3 ⎠ = = + ≈ 0 6 e 69. / 4 / 4 / 4 (x cos x – x sin x)dx x cos x dx – x sin x dx π π π ∫ = ∫ ∫ 0 0 0 4 4 0 0 xsin x sin x dx ⎛ π π ⎞ = ⎜ ⎡⎣ ⎤⎦ ⎟ ⎝ ⎠ − ∫ [ ] 4 4 xπ −⎜ ⎛ − x cos x π ∫ π + cos x dx ⎞ ⎟ ⎝ 0 0 ⎠ [ x sin x cos x x cos x – sin ] / 4 0 = + + 2 –1 = ≈ 0.11 4 π Use Problems 60 and 61 for ∫ x sin x dx and ∫ x cos x dx. V x x dx π ⎛ ⎞ = π ⎜ ⎟ 70. 2 ⎝ ⎠ ∫ 2 sin2 0 dv = x dx u = x sin 2 v = x du = dx –2cos 2 2 2 0 0 ⎛ ⎡ ⎤ π π ⎞ = π⎜ ⎢ ⎥ + ⎟ ⎜ ⎣ ⎦ ⎟ ⎝ ⎠ V x x x dx ∫ 2 –2 cos 2cos 2 2 2 2 ⎛ π ⎞ = 2 π⎜ 4 π + ⎡ 4sin x ⎤ ⎢ ⎥ ⎟ = 8 π ⎜ ⎝ ⎣ 2 ⎦ ⎟ 0 ⎠ ln 2 ln e e ∫ x dx = ∫ x dx u = ln x dv = dx du 1 dx 71. 2 1 1 = v = x x ( ) 1 1 1 1 2 ln 2 [ ln ] 2 [ ] 2 e x dx ⎛ x x e e dx⎞ e x e ⎜ − ⎟ ∫ = ∫ = − = ⎝ ⎠ ∫ e x ln x 2 dx = 2 ∫ e x ln x dx 1 1 u = ln x dv = x dx du = 1 dx 1 2 x v = x 2 e e e x xdx x x xdx ⎛ ⎡ ⎤ ⎞ = ⎜ ⎢ ⎥ ⎟ ⎜ ⎣ ⎦ ⎟ ⎝ ⎠ 2 ln 2 1 ln – 1 e ⎛ ⎡ ⎤ ⎞ = ⎜ ⎢ ⎥ ⎟ = + ⎜ ⎣ ⎦ ⎟ ⎝ ⎠ 2 1 – 1 1 ( 1) 2 4 2 ∫ 2 ∫ 2 2 2 2 2 1 1 1 e x e 1 428 Section 7.2 Instructor Solutions Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  18. 18. ∫e x 2 dx u = (ln x)2 dv = dx du 2ln x dx 1 (ln ) 2 1 = v = x x 1 (ln ) 1 [ (ln ) ] – 2 ln 2 2 ∫ e x 2 dx = ⎛ ⎜ x x 2 e e 1 ∫ x dx ⎞ ⎟ = 1( e –2) 1 ⎝ 1 ⎠ 2 1 2 2 2( 1) 1 e e x + + = = 2 4 1 ( –2) 2– 2 e e y = = 2 4 72. a. u = cot x dv = csc2 x dx du = – csc2 x dx v = –cot x 2 2 2 ∫ ∫ ∫ ∫ x xdx x x xdx x xdx x C x xdx x C cot csc = − cot − cot csc 2 cot csc 2 = − cot 2 + cot csc 2 = − 1 cot 2 + 2 b. u = csc x dv = cot x csc x dx du = –cot x csc x dx v = –csc x 2 2 2 ∫ ∫ ∫ ∫ x xdx x x xdx x xdx x C x xdx x C cot csc = − csc − cot csc 2 cot csc 2 = − csc 2 + cot csc 2 = − 1 csc 2 + 2 c. – 1 cot2 – 1 (csc2 –1) x = x – 1 csc2 1 2 2 = x + 2 2 73. a. p(x) = x3 − 2x g(x) = ex All antiderivatives of g(x) = ex ∫(x3 − 2x)exdx = (x3 − 2x)ex − (3x2 − 2)ex + 6xex − 6ex + C b. p(x) = x2 − 3x +1 g(x) = sin x G1(x) = −cos x G2 (x) = −sin x G3(x) = cos x ∫(x2 − 3x +1)sin x dx = (x2 − 3x +1)(−cos x) − (2x − 3)(−sin x) + 2cos x + C Instructor's Resource Manual Section 7.2 429 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  19. 19. 74. a. We note that the nth arch extends from x = 2π (n −1) to x =π (2n −1) , so the area of the nth arch is ( ) sin n (2 − 1) 2 ( − 1) A n x xdx π = ∫ . Using integration by parts: n π u = x dv = sin xdx du = dx v =− cos x A n x xdx x x xdx x x x n n n n n n n n n n (2 − 1) (2 − 1) (2 − 1) (2 − 1) (2 − 1) 2 ( − 1) 2 ( − 1) 2 ( − 1) 2 ( − 1) 2 ( − 1) π [ ] π π [ ] π [ ] π π π π π π ( ) sin cos cos cos sin = ∫ = − − ∫ − = − + = − − − + − − +[ − − − ] [ ] n n n n n n )) (2 1) cos( (2 1)) 2 ( 1) cos(2 ( 1)) sin( (2 1)) sin(2 ( 1 π π π π π π [ ] = −π (2n −1)(−1) + 2π (n −1)(1) + 0 − 0 =π (2n −1) + (2n − 2) . So A(n) = (4n − 3)π b. 3 2 V 2 x sin x dx π = π∫ 2 π u = x2 dv = sin x dx du = 2x dx v = –cos x 2 3 3 π π π π ⎛ ⎡ ⎤ ⎞ ⎜ ⎣ ⎦ ⎟ ⎝ ⎠ = π + ∫ 2 2 3 V 2 –x cos x 2x cos x dx 2 2 2 2 9 4 2x cos x dx π ⎛ ⎞ ⎜ ⎝ π ⎟ ⎠ = π π + π + ∫ u = 2x dv = cos x dx du = 2 dx v = sin x 2 3 3 2 2 V 2 13 [2xsin x] – 2sin x π π ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = π π + π ∫ π ( 2 3 ) 2 2 13 [2cos x]2 2 (13 – 4) π = π π + π = π π ≈ 781 75. u = f(x) dv = sin nx dx du = f ′(x)dx v 1 cos nx n = − π π −π −π ⎡ ⎡ ⎤ ′ ⎤ = ⎢ ⎢− ⎥ + ⎥ π ⎣ ⎣ ⎦ ⎦ ∫ 1 1 cos( ) ( ) 1 cos( ) ( ) an nx f x nx f x dx n n Term 1 Term 2 Term 1 = 1 cos(n π )( f ( −π ) − f ( π )) =± 1 ( f ( −π ) − f ( π )) n n Since f ′(x) is continuous on [–∞ ,∞ ], it is bounded. Thus, cos(nx) f (x)dx π π ∫ ′ is bounded so – a 1 ⎡± f π lim n = lim ⎢⎣ ( ( −π ) − f ( π )) + ∫ cos( nx ) f ′ ( x ) dx ⎤ ⎥⎦ = 0. n n π n →∞ →∞ −π 76. 1 G n n n n n n [( + 1)( + 2) ⋅⋅⋅ ( + )] n 1 n [ ] n n = n n 1/ 1 1 1 2 1 = ⎡⎛ ⎢⎜ + ⎞⎛ ⎟⎜ + ⎟…⎜ ⎞ ⎛ + ⎞⎤ ⎣⎝ n ⎠⎝ n ⎠ ⎝ n ⎟⎥ ⎠⎦ ln Gn 1 ln 1 1 1 2 1 n ⎛ ⎞ ⎡⎛ ⎞⎛ ⎜ ⎟ = + + ⎞ ⎟…⎜ ⎛ + ⎞⎤ ⎝ n ⎠ n ⎢⎜ n ⎟⎜ ⎣⎝ ⎠⎝ n ⎠ ⎝ n ⎠⎦ ⎟⎥ 1 ln 1 1 ln 1 2 ln 1 n n n n n ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ = ⎢ ⎜ + ⎟ + ⎜ + ⎟ + ⋅⋅⋅+ ⎜ + ⎟⎥ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ G xdx ⎛ ⎞ 2 ⎜ ⎟ = ∫ = ⎝ ⎠ 1 lim ln n ln 2ln 2 –1 n →∞ n G e e lim n 2ln 2–1 4 –1 4 n ⎛ ⎞ = = = ⎜ ⎟ ⎝ ⎠ →∞ n e 430 Section 7.2 Instructor Solutions Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  20. 20. 77. The proof fails to consider the constants when integrating . 1t The symbol ( ) 1 t dt ∫ is a family of functions, all of who whom have derivative . 1t We know that any two of these functions will differ by a constant, so it is perfectly correct (notationally) to write ∫(1 t )dt = ∫(1 t )dt +1 [ ( 1 cos 7 2 sin 7 ) 3] 5 ( 1 cos7 2 sin 7 ) (–7 1 sin 7 7 2 cos 7 ) d e x C x C x C e x C x C x e x C x C x dx 78. 5 5 5 + + = + + + 5 [(5 1 7 2 ) cos 7 (5 2 – 7 1)sin 7 ] = e x C + C x + C C x Thus, 5C1 + 7C2 = 4 and 5C2 – 7C1 = 6. Solving, 1 2 – 11 ; 29 37 37 C = C = 79. u = f(x) dv = dx du = f ′(x)dx v = x ( ) [ ( )] – ( ) b b b a a a ∫ f x dx = xf x ∫ xf ′ x dx Starting with the same integral, u = f(x) dv = dx du = f ′(x)dx v = x – a ( ) [( – ) ( )] – ( – ) ( ) b b b a a a ∫ f x dx = x a f x ∫ x a f ′ x dx 80. u = f ′(x) dv = dx du = f ′′(x)dx v = x – a ( )– ( ) ( ) b f b f a = ∫ f ′ x dx [( – ) ( )] – ( – ) ( ) b b a = x a f ′ x ∫ x a f ′′ x dx ( )( – ) – ( – ) ( ) b a a = f ′ b b a ∫ x a f ′′ x dx a Starting with the same integral, u = f ′(x) dv = dx du = f ′′(x)dx v = x – b ( ) ( ) ( ) [( – ) ( )] – ( – ) ( ) b b b f b − f a = ∫ f ′ x dx = x b f ′ x ∫ x b f ′′ x dx ( )( ) – ( – ) ( ) b a a a = f ′ a b − a ∫ x b f ′′ x dx a 81. Use proof by induction. n = 1: ( ) ( )( – ) ( – ) ( ) ( ) ( )( – ) [ ( )( – )] ( ) t t t f a + f ′ a t a + ∫ t x f ′′ x dx = f a + f ′ a t a + f ′ x t x + ∫ f ′ x dx a a a ( ) ( )( – ) – ( )( – ) [ ( )]t ( ) = f a + f ′ a t a f ′ a t a + f x a = f t Thus, the statement is true for n = 1. Note that integration by parts was used with u = (t – x), dv = f ′′(x)dx. Suppose the statement is true for n. n ( i ) n i t n f ( t ) f ( a ) f ( a ) ( t – a ) ( t – x ) f ( 1) ( x ) dx = +Σ + ∫ 1 i n ! a ! i + = t n n a t x f x dx n Integrate ( – ) ( 1) ( ) + ∫ by parts. ! t x n dv dx u = f (n+1) (x) ( – ) ! n = du = f (n+2) (x) t x n v – ( – ) + 1 ( n 1)! = + 1 1 n n + t t t n + n n n a a + ⎡ + ⎤ + t x f x dx t x f x t x f x dx n n n ( – ) ( 1) ( ) – ( – ) ( 1) ( ) ( – ) ( 2) ( ) ∫ ∫ = ⎢ ⎥ + ⎢⎣ + ⎥⎦ + ! ( 1)! ( 1)! a Instructor's Resource Manual Section 7.2 431 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  21. 21. n + 1 t n + 1 n n t a f a t x f x dx n n ( – ) ( 1) ( ) ( – ) ( 2) ( ) ( 1)! ( 1)! = + + + ∫ + a + Thus n ( i ) n + 1 t n + 1 i n n f ( t ) f ( a ) f ( a ) ( t – a ) ( t – a ) f ( a ) ( t – x ) f ( x ) dx + + Σ ∫ ( 1) ( 2) = + + + 1 + + i n n ! ( 1)! a ( 1)! i = n + 1 ( i ) t n + 1 i n f ( a ) f ( a ) ( t – a ) ( t – x ) f ( 2) ( x ) dx + Σ ∫ = + + 1 i n ! a ( 1)! i + = Thus, the statement is true for n + 1. 82. a. 1 1 1 B(α , β ) = ∫ xα − (1− x)β − dx where α ≥ 1,β ≥ 1 0 x = 1 – u, dx = –du 1 1 1 0 1 1 0 1 ∫ xα − (1− x)β − dx = ∫ (1− u)α − (u)β − (−du) 1 1 1 = ∫ (1− u)α − uβ − du = B(β , α ) 0 Thus, B(α, β) = B(β, α). b. 1 1 1 B(α , β ) = ∫ xα − (1− x)β − dx 0 u = xα −1 dv = (1− x)β −1dx du = (α −1) xα −2dx v = − 1 (1 − x)β β 1 ⎡ − ⎤ − − − − = ⎢− − ⎥ + − = − ⎣ ⎦ ( , ) 1 (1 ) 1 (1 ) 1 (1 ) α β α α β α α β 1 1 2 1 2 ∫ ∫ B x x x x dx x x dx β β 0 β 0 0 1 B ( 1, 1) α β − α = − + α β β (*) Similarly, 1 1 1 0 B(α , β ) = ∫ xα − (1− x)β − dx u = (1− x)β −1 dv = xα −1dx du = −(β −1)(1− x)β −2 dx v 1 xα α = 1 1 1 2 0 0 B( , ) 1 xα (1 x)β β 1 xα (1 x)β dx β − 1 α β − β − x(1 x)dx 1 B( 1, 1) = ⎡ − − ⎤ + − − − ⎢⎣ ⎥⎦ ∫ 1 2 α β α α = ∫ − = + − α 0 α α β c. Assume that n ≤ m. Using part (b) n times, ( ) n n n B n m B n m B n m − 1 − 1 ( − 2) ( , ) = ( − 1, + 1) = ( − 2, + 2) m mm ( 1) ( n ) n ( n ) + 1 ( 2) 3 2 1 − − − …⋅ ⋅ = … = + − ( ) B m n (1, 1). m m m m n ( 1) 2 ( 2) + + … + − 1 2 11 0 0 (1, 1) (1 ) 1 [(1 ) ] 1 B m n x m n dx x m n + − = − + − = − − + − = ∫ + − + − m n m n 1 1 Thus, ( ) ( ) ( ) ( ) ( 1 ( 2) 3 2 1 1)!( 1)! ( 1)!( 1)! B n m ( , ) n n n n m n m − − − …⋅ ⋅ − − − − = = = m m m m n m n m n n m ( + 1) + 2 … ( + − 2) + − 1 ( + − 1)! ( + − 1)! If n m, then ( 1)!( 1)! B n m B m n ( , ) ( , ) n m − − n m ( 1)! = = + − by the above reasoning. 432 Section 7.2 Instructor Solutions Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  22. 22. 83. u = f(t) dv = f ′′(t)dt du = f ′(t)dt v = f ′(t) ( ) ( ) [ ( ) ( )] – [ ( )]2 b b b a a a ∫ f ′′ t f t dt = f t f ′ t ∫ f ′ t dt = f ( b ) f ′ ( b ) − f ( a ) f ′ ( a ) − ∫ b [ f ′ ( t )]2 dt b [ ( )]2 a = −∫ f ′ t dt a [ ( )]2 0, so [ ( )]2 0 b f ′ t ≥ − ∫ f ′ t ≤ . a 84. ∫ x ⎛ ⎜ ∫ t f ( z ) dz ⎞ ⎟ dt 0 ⎝ 0 ⎠ u = ∫ t f ( z ) dz dv = dt 0 du = f(t)dt v = t ∫ x ⎛⎜ x ∫ t f ( z ) dz ⎞⎟ dt = ⎡⎢t ∫ t f ( z ) dz⎤⎥ – ∫ x t f ( t ) dt ⎝ ⎠ ⎣ ⎦ 0 0 0 0 0 0 0 ( ) – ( ) x x = ∫ x f z dz ∫ t f t dt By letting z = t, ( ) ( ) , x x ∫ x f z dz = ∫ x f t dt so 0 0 ∫ x ⎛⎜ ∫ t f ( z ) dz ⎞⎟ dt = ∫ x x f ( t ) dt – ∫ x t f ( t ) dt 0 ⎝ 0 ⎠ 0 0 0 ( – ) ( ) x = ∫ x t f t dt ( ) ... x t tn I f tn dtn dt dt = ∫ ∫ ⋅⋅⋅∫ − be the iterated integral. Note that for i ≥ 2, the limits of integration of the 85. Let 1 1 0 0 0 2 1 integral with respect to ti are 0 to ti−1 so that any change of variables in an outer integral affects the limits, and hence the variables in all interior integrals. We use induction on n, noting that the case n = 2 is solved in the previous problem. Assume we know the formula for n −1, and we want to show it for n. I x t 1 t 2 tn − 1 f ( tn ) dtn ... dt dt dt t 1 t 2 tn − 2 F ( tn ) dtn ... dt dt dt = ∫ ∫ ∫ ⋅⋅⋅∫ = ∫ ∫ ⋅⋅⋅∫ − − where ( ) 1 ( ) 0 0 0 0 3 2 1 0 0 0 1 1 3 2 1 tn − = ∫ . F tn f tn dn − 1 0 By induction, 1 2 ! I x F t x t n dt ( ) ( )( ) 2 − = − − ∫ ( ) 1 ( ) 1 0 0 1 1 1 n t dv x t n− = − u = F t = ∫ f tn dtn , ( ) 2 1 1 1 v x t n − = − − ( ) du = f t1 dt1 , ( ) 1 1 n − t = x ⎧⎪⎡ ⎤ ⎪⎫ = ⎨⎢− − ⎥ + − ⎬ − ⎩⎪⎣ − ⎦ − ⎪⎭ = − 1 1 1 2 ! 1 1 1 ( )( ) ( 1)! ( ) ( ) ( ) ( )( ) 1 n t x n − − 1 1 ∫ 1 ∫ I x t f t dt f t x t dt n n 1 0 0 1 1 1 n n n t 1 0 − 1 x n f t x t dt ∫ 0 1 1 1 n = − . (note: that the quantity in square brackets equals 0 when evaluated at the given limits) Instructor’s Resource Manual Section 7.3 433 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  23. 23. 86. Proof by induction. n = 1: u = P1(x) dv = exdx du x) = dP1(dx v = ex dx exP (x)dx exP (x) – ex dP (x) dx exP (x) – dP (x) exdx ∫ = ∫ 1 1 1 1 dx exP (x) – ex dP (x) = ∫ 1 1 dx 1 dx = Note that dP1(x) dx is a constant. Suppose the formula is true for n. By using integration by parts with u = Pn+1(x) and dv = exdx, dP x 1 x ( ) x ( ) – x n n n + ∫ + = + ∫ Note that n 1( ) dP x e P x dx e P x e dx 1 1 ( ) dx + is a polynomial of degree n, so dx ⎡ ⎛ ⎞⎤ = − ⎢ − ⎜ ⎟⎥ = − − ⎢⎣ ⎝ ⎠⎥⎦ n j n j + 1 ( ) ( ) ( ) e P x dx e P x e d dP ( x ) d P x e P x e x x x j n x x j n n n j n j ∫ Σ Σ 1 1 + + ( ) ( ) ( 1) 1 1 1 1 1 + + + + dx dx dx j j 0 0 = = n j 1 d P x + = + Σ − x x j n n j 1 e P x e 1 ( ) ( 1) 1 ( ) j + dx + = n 1 j x j n d P x 1 + = Σ − 0 ( ) ( 1) + j j e dx = 87. j 4 4 2 x x e dx e d x x (3 2 ) (–1) (3 2 ) ∫ 4 + 2 x = x Σ j = 0 dx = ex[3x4 + 2x2 –12x3 – 4x + 36x2 + 4 – 72x + 72] = ex (3x4 –12x3 + 38x2 – 76x + 76) + j j 7.3 Concepts Review 1. 1 cos2 2 x dx + ∫ 2. ∫(1– sin2 x) cos x dx 3. ∫sin2 x(1– sin2 x) cos x dx 4. cos mx cos nx = 1 [cos( m+ n ) x + cos( m− n ) x ] 2 Problem Set 7.3 ∫ x dx = ∫ x dx 1 – 1 cos 2 2 2 1. sin2 1– cos 2 2 = ∫ dx ∫ x dx = 1 x – 1 sin 2 x +C 2 4 2. u = 6x, du = 6 dx sin4 6 1 sin4 ∫ x dx = ∫ u du 6 1 1– cos 2 2 6 2 = ⎛ u ⎞ du ⎜ ⎟ ⎝ ⎠ ∫ 1 (1– 2cos 2 cos2 2 ) 24 = ∫ u + u du 1 – 1 2cos 2 1 (1 cos 4 ) 24 24 48 = ∫ du ∫ u du + ∫ + u du 3 – 1 2cos 2 1 4cos 4 48 24 192 = ∫ du ∫ u du + ∫ u du 3 (6 ) – 1 sin12 1 sin 24 48 24 192 = x x + x +C = 3 x – 1 sin12 x + 1 sin 24 x +C 8 24 192 434 Section 7.2 Instructor Solutions Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  24. 24. 3. ∫sin3 x dx = ∫sin x(1− cos2 x)dx = ∫sin x dx − ∫sin x cos2 x dx cos 1 cos3 = − x + x +C 3 4. ∫cos3 x dx = = ∫cos x(1− sin2 x)dx = ∫cos x dx − ∫cos x sin2 x dx = sin x − 1 sin3 x + C 3 5. / 2 5 / 2 2 2 π π ∫ = ∫ cos θ dθ (1– sin θ ) cosθ dθ 0 0 / 2 2 4 0 = ∫ + (1– 2sin θ sin θ ) cosθ dθ π / 2 ⎡ π = sin – 2 sin 3 + 1 sin 5 ⎤ ⎢⎣ ⎥⎦ 1– 2 1 – 0 8 θ θ θ 0 3 5 = ⎛ + ⎞ = ⎜ ⎟ ⎝ ⎠ 3 5 15 6. π / 2 π sin 6 = / 2 ⎛ 1– cos 2 θ ⎞ 3 0 0 ⎜ ⎟ ⎝ ⎠ ∫ ∫ d d θ θ θ 2 1 / 2 (1– 3cos 2 3cos 2 2 – cos 3 2 ) 8 0 = ∫ + θ θ θ dθ π 1 π / 2 3 π / 2 3 π / 2 2 1 π – 2cos 2 cos 2 – / 2 cos 3 2 8 0 16 0 8 0 8 0 = ∫ ∫ + ∫ ∫ dθ θ dθ θ θ dθ 1[ 3 3 1 cos 4 θ ] – [sin 2 ] – 1 (1– sin 2 ) cos 2 8 16 8 2 8 π π π ⎛ + ⎞ π = + ⎜ ⎟ / 2 / 2 / 2 / 2 2 0 0 0 0 ⎝ ⎠ ∫ ∫ d d θ θ θ θ θ θ 1 π 3 π / 2 3 π / 2 π 4cos 4 – 1 / 2 π 2cos 2 1 / 2 sin 2 2 2cos 2 8 2 16 0 64 0 16 0 16 0 = ⋅ + ∫ + ∫ ∫ + ∫ ⋅ dθ θ dθ θ dθ θ θ dθ θ π θ π θ π π π = + + + 5 3 3 [sin 4 ] – 1 [sin 2 ] 1 [sin 2 ] / 2 / 2 3 / 2 0 0 0 16 32 64 16 48 π 32 = 7. ∫sin5 4x cos2 4x dx = ∫(1– cos2 4x)2 cos2 4x sin 4x dx = ∫(1– 2cos2 4x + cos4 4x) cos2 4x sin 4x dx = ∫ x x + x x dx – 1 cos3 4 1 cos5 4 – 1 cos7 4 – 1 (cos2 4 – 2cos4 4 cos6 4 )(–4sin 4 ) 4 = x + x x +C 12 10 28 8. ∫(sin3 2t) cos 2tdt = ∫(1– cos2 2t)(cos 2t)1/ 2 sin 2t dt – 1 [(cos 2 )1/ 2 – (cos 2 )5/ 2 ](–2sin 2 ) = ∫ t t t dt 2 – 1 (cos 2 )3/ 2 1 (cos 2 )7 / 2 3 7 = t + t +C 9. ∫cos3 3θ sin–2 3θ dθ = ∫(1– sin2 3θ )sin–2 3θ cos3θ dθ 1 (sin 2 3 1)3cos3 = ∫ − θ − θ dθ 3 – 1 csc3 – 1 sin 3 3 3 = θ θ +C 10. ∫sin1/ 2 2z cos3 2z dz = ∫(1– sin2 2z)sin1/ 2 2z cos 2z dz = ∫ z z z dz 1 sin3/ 2 2 – 1 sin7 / 2 2 1 (sin1/ 2 2 – sin5/ 2 2 )2cos 2 2 = z z +C 3 7 Instructor’s Resource Manual Section 7.3 435 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  25. 25. 11. 2 2 t t dt t t dt ∫ sin4 + 3 cos4 3 = ∫ ⎛ 1– cos 6 ⎞ ⎛ 1 cos 6 ⎞ 1 ⎜ ⎟ ⎜ ⎟ (1– 2cos2 6 cos4 6 ) ⎝ 2 ⎠ ⎝ 2 ⎠ = ∫ t + t dt 16 = 1 ∫ ⎡⎢ 1– (1 + cos12 t ) + 1 (1 + cos12 t )2 ⎤⎥ dt – 1 cos12 1 (1 2cos12 cos2 12 ) 16 ⎣ 4 ⎦ = ∫ t dt + ∫ + t + t dt 16 64 – 1 12cos12 1 1 12cos12 1 (1 cos 24 ) 192 64 384 128 = ∫ t dt + ∫ dt + ∫ t dt + ∫ + t dt – 1 sin12 1 1 sin12 1 1 sin 24 = t + t + t + t + t +C 3 – 1 sin12 1 sin 24 192 64 384 128 3072 = t t + t +C 128 384 3072 12. 3 ∫ cos6 sin2 d = ∫ ⎛ 1 + cos 2 θ ⎞ ⎛ 1– cos 2 θ ⎞ ⎜ ⎟ ⎜ ⎟ d 1 (1 2cos 2 – 2cos3 2 – cos4 2 ) ⎝ ⎠ ⎝ ⎠ = ∫ + θ θ θ dθ θ θ θ θ 2 2 16 1 1 2cos 2 – 1 (1– sin2 2 ) cos 2 – 1 (1 cos 4 )2 16 16 8 64 = ∫ dθ + ∫ θ dθ ∫ θ θ dθ ∫ + θ dθ 1 1 2cos 2 – 1 2cos 2 1 2sin2 2 cos 2 – 1 (1 2cos 4 cos2 4 ) 16 16 16 16 64 = ∫ dθ + ∫ θ dθ ∫ θ dθ + ∫ θ θ dθ ∫ + θ + θ dθ 1 1 sin2 2 2cos 2 – 1 – 1 4cos 4 – 1 (1 cos8 ) 16 16 64 128 128 = ∫ dθ + ∫ θ ⋅ θ dθ ∫ dθ ∫ θ dθ ∫ + θ dθ = 1 1 sin3 2 1 1 sin 4 1 1 sin8 16 48 64 128 128 1024 θ + θ − θ − θ − θ − θ +C 5 1 sin3 2 – 1 sin 4 – 1 sin8 128 48 128 1024 = θ + θ θ θ + C 13. sin 4 cos5 1 [sin 9 sin( )] 1 (sin 9 sin ) ∫ y y dy = ∫ y + −y dy = ∫ y − y dy 2 2 1 1 cos9 cos 1 cos 1 cos9 2 9 2 18 = ⎛⎜ − y + y ⎞⎟ +C = y − y +C ⎝ ⎠ ∫ y y dy = ∫ y + − y dy 1 sin 5 1 sin( 3 ) 14. cos cos 4 1 [cos5 cos( 3 )] 2 = y − − y +C 1 sin 5 1 sin 3 10 6 = y + y +C 10 6 15. 2 w w dw w w dw ∫ sin4 ⎛ ⎞ cos2 ⎛ ⎞ = ∫ ⎛ 1– cos ⎞ ⎛ 1 + cos ⎞ 1 ⎜ (1– cos – cos2 cos3 ) ⎝ 2 ⎟ ⎜ ⎠ ⎝ 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ = ∫ w w+ w dw 8 = 1 ∫ ⎡⎢ 1– cos w – 1 (1 + cos 2 w ) + (1– sin2 w ) cos w⎤⎥ dw 1 1 – 1 cos 2 – sin2 cos 8 ⎣ 2 ⎦ = ∫ ⎡⎢ w w w⎤⎥dw 8 ⎣ 2 2 ⎦ 1 – 1 sin 2 – 1 sin3 16 32 24 = w w w+C sin 3 sin 1 cos 4 cos 2 16. [ ] ∫ ∫ t t dt = − t − t dt 2 ( ) 1 ∫ cos 4 ∫ cos 2 2 1 1 sin 4 1 sin 2 2 4 2 1 sin 4 1 sin 2 8 4 tdt tdt = − − = − ⎛ − ⎞ + ⎜ ⎟ t t C ⎝ ⎠ t tC = − + + 436 Section 7.3 Instructor's Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  26. 26. 17. 2 ∫ x cos x sin xdx u = x du = 1 dx dv = 2 x x dx v = − x − x dx = − x cos sin (cos ) ( sin ) 1 cos 2 3 cos t = x 3 ∫ Thus 2 ∫ x x xdx x x xdx cos sin = ( 1 cos 3 ) ∫ (1)( 1 cos 3 ) 3 3 1 cos ∫ cos 3 1 cos ∫ cos (1 sin ) 3 − − − = ⎡− x 3 x + 3 ⎣ xdx ⎤ ⎦ = ⎡− x 3 ⎣ x + x − 2 xdx ⎤ ⎦ = ⎡ ⎤ ⎢− + − ⎥ = ⎣ ⎦ ⎡− + − ⎤ + ⎢⎣ ⎥⎦ 1 cos 3 ∫ (cos cos sin 2 ) 3 1 cos sin 1 sin 3 3 x x x x xdx t sin x 3 3 = x x x x C 18. 3 ∫ x sin x cos xdx u = x du = 1 dx dv = sin 3 x cos x dx v = (sin x ) (cos xdx ) = 1 sin x 3 4 sin t = x 4 ∫ Thus 3 ∫ x x xdx x x xdx sin cos = (1 sin 4 ) ∫ (1)(1 sin 4 ) 4 4 1 sin ∫ (sin ) 4 1 sin 1 ∫ (1 cos 2 ) 4 4 − = 4 2 2 ⎡ ⎣ x x − x dx ⎤ ⎦ = ⎡ 4 2 ⎤ ⎢⎣ x x − − x dx ⎥⎦ = 1 sin 1 (1 2cos 2 cos 2 ) 4 4 1 sin 1 1 sin 2 1 (1 cos 4 ) 4 4 4 8 1 sin 3 1 sin 2 1 sin 4 4 8 4 32 ⎡ ⎢⎣ x 4 x − ∫ − x + 2 xdx ⎤ ⎥⎦ = ⎡ ⎢⎣ x 4 x − x + x − ∫ + xdx ⎤ ⎥⎦ = ⎡ 4 ⎤ ⎢⎣ x x − x + x − x ⎥⎦ + C 19. 4 ( 2 )( 2 ) ∫ ∫ x dx x x dx tan tan tan = = ∫ ( 2 ) 2 − = ∫ ( − ) = ∫ − ∫ − = − + + x x dx x x xdx x x dx x dx x x x C tan (sec 1) tan 2 sec 2 tan 2 tan 2 sec 2 (sec 2 1) 1 tan 3 tan 3 20. 4 ( 2 )( 2 ) ∫ ∫ ∫ ( 2 2 2 ) x dx x x dx x x dx cot = cot cot cot (csc 1) ( ) 2 2 = − ∫ ∫ ∫ x x xdx x x dx x dx x x x C cot csc cot cot csc (csc 1) 1 cot cot 3 = − = 2 2 − 2 − = − 3 + + + 21. tan 3 x ( )( 2 ) x x dx x x dx x x C ∫ ∫ tan tan tan sec 1 1 tan ln cos 2 = = − ( )( 2 ) 2 = + + 22. 3 ( )( 2 ) ∫ ∫ ∫ ∫ ∫ t dt t t dt t t dt t tdt tdt cot 2 = cot 2 cot 2 ( cot 2 )( csc 2 2 1 ) cot 2 csc 2 cot 2 1 cot 2 1 ln sin 2 4 2 = − = 2 − = − 2 − + t t C Instructor’s Resource Manual Section 7.3 437 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  27. 27. θ ⎛ ⎞ θ ⎜ ⎟ ⎝ ⎠ ∫ 23. tan5 2 d u du d ⎛θ ⎞ θ = ⎜ 2 ⎟ ; = ⎝ ⎠ 2 θ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 5 5 ∫ ∫ d udu tan 2 tan ( )( ) ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ u u du u udu udu u u du u u du u udu u udu u 2 tan sec 1 2 tan sec 2 tan 2 tan sec 2 tan sec 1 2 tan sec 2 tan sec 2 tan du 1 tan tan 2ln cos 2 2 2 2 = − = − = − ( − ) = − + 3 2 3 2 3 3 2 2 3 2 2 θ θ θ = 4 ⎛ ⎞ − 2 ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ − + 2 C θ ⎝ ⎠ ⎝ ⎠ 24. ∫cot5 2t dt u = 2t;du = 2dt cot 2 1 cot 5 5 ∫ ∫ t dt u du 2 1 ∫ ( cot 3 u )( cot 2 udu ) 1 ∫ ( cot 3 u )( csc 2 1 ) du 2 2 1 ∫ ( cot )( csc ) 1 ∫ cot 2 2 1 ∫ ( cot )( csc ) 1 ∫ ( cot )( csc 1 ) 2 2 1 ∫ ( cot )( csc ) 1 ∫ ( cot )( csc ) 1 ∫ cot 2 2 2 1 cot 1 cot 1 ln sin 8 4 2 1 cot 2 1 c 8 4 = = − 3 2 3 u udu udu = − 3 2 2 u u du u u du = − − 3 2 2 u u du u u du u = − + 4 2 = − + + + 4 u u u C t = = − + ot2 2 1 ln sin 2 t + t +C 2 25. − 3 4 ( − 3 )( 2 )( 2 ) ∫ ∫ x xdx x x x dx tan sec tan sec sec = = + = + = − + + ( 3 )( 2 )( 2 ) ∫ ∫ ∫ x x x dx x x x x dx x x C − tan 1 tan sec tan sec dx tan sec 1 tan ln tan 2 3 2 1 2 − − ( ) 2 − 26. − 3/ 2 4 ( − 3/ 2 )( 2 )( 2 ) ∫ ∫ x xdx x x x tan sec tan sec sec = = + = + = − + + ( 3/ 2 )( 2 )( 2 ) ∫ ∫ ∫ x x x dx x x dx x x dx x x C − tan 1 tan sec tan sec tan sec 2 tan 2 tan 3/2 2 1/2 2 1/ 2 3/ 2 3 − − 438 Section 7.3 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  28. 28. 27. ∫ tan3 x sec2 x dx Let u = tan x . Then du = sec2 x dx . tan3 sec2 3 1 4 1 tan4 ∫ x x dx = ∫ u du = u +C = x +C 4 4 28. 3 − 1/2 2 − 3/2 ∫ ∫ x xdx x x x xdx tan sec tan sec (sec tan ) = = − = − = + + ( 2 )( − 3/ 2 )( ) ∫ ∫ ∫ x x x x dx x x x dx x x x dx x x C sec 1 sec sec tan sec 1/ 2 sec tan sec − 3/ 2 sec tan 2 sec 2sec 3 ( ) ( ) 3/ 2 − 1/ 2 29. ∫ cos cos = 1 ∫ (cos[( + ) ] + cos[( − ) ]) 1 1 sin[( ) ] 1 sin[( ) ] mx nx dx m n x m n x dx π π π π 2 – – = ⎡ m + n x + m − n x ⎤ 2 ⎢⎣ m + n m − n ⎥⎦ π −π = 0 for m ≠ n, since sin kπ = 0 for all integers k. 30. If we let u x π = then du dx L π = . Making the substitution and changing the limits as necessary, we get L m π x n π x dx L mu nu du L L L π cos cos cos cos 0 L π ∫ = ∫ = (See Problem 29 − π − (x sin x) dx π ∫ π + 2 2 31. 2 0 (x 2x sin x sin x) dx π x dx x x dx x dx π π π π = π∫ + + 2 0 = π∫ + π∫ + ∫ − 2 sin (1 cos2) 2 0 0 0 π π 1 2 sin cos 1 sin 2 3 2 2 ⎡ 3 ⎤ π π = π ⎡ ⎤ ⎢⎣ x ⎥⎦ + π [ x − x x ] + x − x 0 ⎢⎣ ⎥⎦ 0 0 = π + π + π − + π − − 1 4 5 2 57.1437 1 4 π 2 (0 0) ( 0 0) 3 2 = π + π ≈ 3 2 Use Formula 40 with u = x for ∫ x sin x dx 32. / 2 2 2 V 2 x sin (x )dx π = π∫ 0 u = x2 , du = 2x dx / 2 / 2 / 2 2 2 0 0 0 π π ⎡ ⎤π π = π = π = π ⎢ ⎥ = ≈ ⎣ ⎦ ∫ ∫ V u du u du u u sin 1– cos 2 1 – 1 sin 2 2.4674 2 2 4 4 33. a. 1 f (x)sin(mx)dx π π −π ∫ N ⎛ ⎞ 1 sin( ) sin( ) a nx mxdx π −π ∫ Σ = ⎜⎜ ⎟⎟ π ⎝ ⎠ 1 n n = N 1 sin( )sin( ) = π Σ ∫ 1 a nx mxdx π n n −π = From Example 6, 0 if ∫ ⎧ ≠ sin( nx )sin( mx ) dx = ⎨so every term in the sum is 0 except for when n = m. ⎩ π if = If m N, there is no term where n = m, while if m ≤ N, then n = m occurs. When n = m an π sin(nx)sin(mx) dx am n m n m π −π ∫ = π so when m ≤ N, −π 1 π f ( x )sin( mx ) dx 1 am am ∫ = ⋅ ⋅π = . π −π π Instructor’s Resource Manual Section 7.3 439 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  29. 29. b. 1 f 2 (x)dx π π −π ∫ N N ⎛ ⎞⎛ ⎞ 1 sin( ) sin( ) a nx a mx dx π −π ∫ Σ Σ = π ⎜⎜ n ⎟⎟⎜⎜ m ⎟⎟ ⎝ ⎠⎝ ⎠ n m 1 1 = = N N 1 sin( )sin( ) a a nx mxdx π = Σ n Σ ∫ πm n m 1 1 −π = = From Example 6, the integral is 0 except when m = n. When m = n, we obtain Σ Σ 2 . πN N 1 ( ) a a π = a = = n n n n n 1 1 34. a. Proof by induction x = x n = 1: cos cos 2 2 Assume true for k ≤ n. n ⎡ ⎤ x x x x x x x x + + cos cos cos cos cos 1 cos 3 cos 2 –1 1 cos ⋅ = ⎢ + + + ⎥ n n n n n n n 1 –1 1 2 4 2 2 2 2 2 2 2 ⎢⎣ ⎥⎦ Note that k x x k x k x + + + cos cos 1 1 cos 2 1 cos 2 –1 , 2n 2n 2 2n 2n ⎛ ⎞⎛ ⎞ ⎡ + ⎤ ⎜ ⎟⎜ = 1 ⎟ ⎢ + ⎥ ⎝ ⎠⎝ ⎠ ⎣ 1 1 ⎦ so 1 n n + ⎡ ⎤ ⎛ ⎞ ⎡ ⎤ ⎢ + + + ⎥ ⎜ ⎟ = ⎢ + + + ⎥ ⎢⎣ ⎥⎦ ⎝ ⎠ ⎢⎣ ⎥⎦ cos 1 cos 3 cos 2 –1 cos 1 1 cos 1 cos 3 cos 2 –1 1 x x x x x x x n n n n n n n n n 1 –1 1 1 1 + + + + 2 2 2 2 2 2 2 2 2 n n ⎡ ⎤ ⎡ ⎤ ⎢ + + + ⎥ = ⎢ + + + ⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ x x x x x x x lim cos 1 cos 3 cos 2 –1 1 1 lim cos 1 cos 3 cos 2 –1 b. –1 –1 n n n n n n n n n n →∞ x →∞ 2 2 2 2 2 2 2 2 1 cos x t dt x = ∫ 0 1 ∫ x cos t dt = 1 [sin t ] x = sin x x x x c. 0 0 x + x = we see that since 35. Using the half-angle identity cos 1 cos , 2 2 π π = = cos cos 2 2 4 2 2 2 π π + + = = = 2 2 1 2 2 cos cos , 8 4 2 2 2 2 π π + + + + = = = etc. 2 2 1 2 2 2 cos cos , 16 8 2 2 Thus, 2 2 + 2 2 + 2 + 2 ⎛ π ⎞ ⎛ π ⎞ ⎛ π ⎞ ⋅ ⋅ cos 2 cos 2 cos 2 2 2 2 = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎠ ⎝ 4 ⎠ ⎝ 8 ⎠ ( ) 2 2 2 2 π π π π ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ sin 2 lim cos cos cos n 2 4 2n = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ →∞ π 2 440 Section 7.3 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  30. 30. π π ∫ π − = ∫ − + 36. Since (k − sin x)2 = (sin x − k)2 , the volume of S is 2 2 2 (k sin x) π (k 2k sin x sin x) dx 0 0 k dx k xdx xdx π π π π x k x 2 k cos x x 1 sin 2 π = π π + π π + π ⎡ − ⎤ ⎢⎣ ⎥⎦ 2 = π ∫ − π∫ + ∫ − 2 [ ] [ ] 2 0 0 0 2 sin (1 cos2) 2 0 0 0 2 2 π π 2 2 2 ( 1 1) ( 0) 2 2 4 k k k k = π + π − − + π − = π − π + 2 2 Let 2 π = π − π + then f ′(k) = 2π2k − 4π and f ′(k) = 0 when k = 2 . ( ) 2 2 4 , 2 f k k k π The critical points of f(k) on 0 ≤ k ≤ 1 are 0, 2 , π 1. 2 2 2 (0) 4.93, 2 4 8 0.93, (1) 2 4 2.24 π ⎛ ⎞ π π = ≈ ⎜ ⎟ = − + ≈ = π − π + ≈ ⎝ π ⎠ f f f 2 2 2 a. S has minimum volume when k = 2 . π b. S has maximum volume when k = 0. 7.4 Concepts Review 1. x – 3 2. 2 sin t 3. 2 tan t 4. 2 sec t Problem Set 7.4 1. u = x +1, u2 = x +1, 2u du = dx ∫ x x +1dx = ∫(u2 –1)u(2u du) = ∫(2u4 – 2u2 )du 2 5 – 2 3 = u u +C 5 3 2 ( 1)5 / 2 – 2 ( 1)3/ 2 5 3 = x + x + + C 2. u = 3 x + π, u3 = x + π, 3u2du = dx ∫ x3 x + πdx = ∫(u3 – π)u(3u2du) = ∫(3u6 – 3πu3)du 3 7 – 3 π 4 u u C = + 7 4 3 7 / 3 3 π ( )– ( )4 / 3 7 4 x x C = +π +π + 3. u = 3t + 4, u2 = 3t + 4, 2u du = 3 dt 1 2 2 3 3 2 ( 4) 2 ( –4) t dt u − u du u du t u ∫ = ∫ = ∫ 3 + 4 9 2 3 – 8 27 9 = u u +C 2 (3 4)3/ 2 – 8 (3 4)1/ 2 27 9 = t + t + + C 4. u = x + 4, u2 = x + 4, 2u du = dx 2 2 2 2 3 ( – 4) 3( – 4)2 x + x u + dx u u du x u ∫ = ∫ + 4 = 2∫(u4 – 5u2 + 4)du 2 5 – 10 3 8 = u u + u + C 5 3 2 ( 4)5/ 2 – 10 ( 4)3/ 2 8( 4)1/ 2 5 3 = x + x + + x + +C 5. u = t , u2 = t, 2u du = dt dt 2u du 2 u e e du t e u e u e 2 2 2 1 1 1 + − ∫ = ∫ = ∫ + + + 2 –2 e du du 2 2 1 1 + ∫ ∫ u e = = 2[u] 2 1 – 2e ⎡⎣ln u + e ⎤⎦ 2 1 = 2( 2 –1) – 2e[ln( 2 + e) – ln(1+ e)] ⎛ 2 + ⎞ 2 2 – 2 – 2 ln e e = ⎜⎜ ⎝ 1 + e ⎟⎟ ⎠ 6. u = t , u2 = t, 2u du = dt t dt u u du t u 1 1 0 0 2 (2 ) ∫ = ∫ + + 1 1 1 2 1 2 0 2 0 2 u du u du u u + − 2 2 1 1 + + ∫ ∫ 1 1 0 0 2 = = 1 1 2 – 2 1 ∫ du ∫ du 1 –1 1 + 1 u = = 2[u]0 – 2[tan u]0 π 2 – 2 tan–11 2 – 0.4292 = = ≈ 2 Instructor’s Resource Manual Section 7.4 441 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  31. 31. 7. u = (3t + 2)1/ 2 , u2 = 3t + 2, 2u du = 3dt (3 2)3/ 2 1 ( 2 – 2) 3 2 t t + dt = u u ⎛⎜ u du ⎞⎟ ∫ ∫ 3 ⎝ 3 ⎠ 2 ( 6 – 2 4 ) 2 7 4 5 9 63 45 = ∫ u u du = u − u +C = 2 (3 t + 2)7 / 2 – 4 (3 t + 2)5 / 2 +C 63 45 8. u = (1– x)1/ 3, u3 = 1– x, 3u2du = –dx ∫ x(1– x)2 / 3dx = ∫(1– u3)u2 (–3u2 )du 3 ( 7 – 4 ) 3 8 3 5 = ∫ u u du = u − u +C 8 5 3 (1– )8/ 3 – 3 (1– )5 / 3 8 5 = x x +C 9. x = 2 sin t, dx = 2 cos t dt 4 – 2 2cos (2cos ) x dx t t dt x 2sin t ∫ = ∫ 1– sin2 2 = ∫ = 2∫csct dt – 2∫sin t dt sin = 2ln csct − cot t + 2cos t + C t dt t 2 2ln 2 4 – x 4 – x2 C = + + x − 10. x = 4sin t, dx = 4cos t dt 2 2 x dx t t dt ∫ = ∫ 16 – x 2 cos t = 16∫sin2 t dt = 8∫(1– cos 2t)dt = 8t – 4sin 2t +C = 8t −8sin t cos t +C 16 sin cos 2 = 8sin–1 ⎛ x ⎞ – x 16 – x ⎜ ⎟ +C 4 2 ⎝ ⎠ 11. x = 2 tan t, dx = 2sec2 t dt 2 dx 2sec t dt 1 cos t dt ∫ = = ( x 2 + 4) 3/2 ∫ 2 ∫ (4sec t ) 3/2 4 1 sin 4 = t + C x C x = + 4 2 + 4 12. t = sec x, dt = sec x tan x dx x π Note that 0 . 2 ≤ t2 –1 = tan x = tan x 3 sec–1(3) 2 2 2 / 3 2 dt sec x tan x dx ∫ = ∫ t t –1 π sec x tan x sec–1(3) / 3 = ∫ cos x dx π sec–1(3) 1 x − π [sin ] / 3 sin[sec (3)] sin π 3 = = − sin cos 1 1 3 2 2 – 3 0.0768 ⎡ = − ⎛ ⎞⎤ ⎢ ⎜ − = ≈ ⎣ ⎝ 3 ⎟⎥ ⎠⎦ 2 3 2 13. t = sec x, dt = sec x tan x dx π ≤ π x Note that . 2 t2 –1 = tan x = – tan x –3 2 sec–1(–3) –2 3 2 / 3 3 t –1 dt – tan x sec x tan x dx t π sec x ∫ = ∫ sec–1(–3) sec–1(–3) 2 2 /3 2 /3 – sin 1 cos 2 – 1 = = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ∫ ∫ x dx x dx π π 2 2 sec–1(–3) π 2 /3 1 sin 2 – 1 4 2 = ⎡ ⎤ ⎢⎣ x x ⎥⎦ sec–1(–3) = ⎡ ⎢⎣ x x ⎤ ⎥⎦ 2 /3 – 2 – 1 sec–1(–3) π 3 x0.151252 1 sin cos – 1 2 2 π = + + ≈ 9 2 8 3 14. t = sin x, dt = cos x dx t dt x dx t ∫ = ∫ = –cos x + C 2 sin 1– = – 1– t2 +C 15. z = sin t, dz = cos t dt z dz t dt z 2 –3 (2sin – 3) 1– ∫ = ∫ 2 = –2 cos t – 3t + C = –2 1– z2 – 3sin–1 z +C 442 Section 7.4 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  32. 32. 16. x = π tan t, dx = πsec2 t dt π x –1 dx ( 2 tan t –1)sect dt x ∫ = ∫ π 2 + π 2 = π2 ∫ tan t sect dt – ∫sect dt = π2 sect – ln sect + tan t +C = π x2 + π2 − ln 1 x2 + π2 + x + C π π x 1 dx x π π − ∫ 0 2 2 + π ⎡ π x 2 + π 2 ⎤ = ⎢π x 2 + π 2 – ln + x ⎥ ⎢ π π ⎥ ⎣ ⎦ 0 = [ 2π2 – ln( 2 +1)] – [π2 − ln1] = ( 2 –1)π2 – ln( 2 +1) ≈ 3.207 17. x2 + 2x + 5 = x2 + 2x +1+ 4 = (x +1)2 + 4 u = x + 1, du = dx dx du ∫ = ∫ u = 2 tan t, du = 2sec2 t dt 2 2 5 2 4 x x u + + + du sec t dt ln sec t tan t C u ∫ ∫ = = + + 2 1 4 + 2 u u C 1 + ln 4 = + + 2 2 2 x x x C 1 + 2 + 5 + + ln 1 = + 2 = ln x2 + 2x + 5 + x +1 +C 18. x2 + 4x + 5 = x2 + 4x + 4 +1 = (x + 2)2 +1 u = x + 2, du = dx dx du ∫ = ∫ u = tan t, du = sec2 t dt 2 4 5 2 1 x x u + + + du t dt t t C u ∫ ∫ 2 sec ln sec tan 1 = = + + + dx u 2 u C 2 ln 1 x x 4 5 = + + + + + ∫ = ln x2 + 4x + 5 + x + 2 +C 19. x2 + 2x + 5 = x2 + 2x +1+ 4 = (x +1)2 + 4 u = x + 1, du = dx x dx u du 3 3– 3 2 5 4 ∫ = ∫ 2 2 x x u + + + u du du u u ∫ ∫ 3 –3 2 2 4 4 = + + (Use the result of Problem 17.) = 3 u2 + 4 – 3ln u2 + 4 + u +C = 3 x2 + 2x + 5 – 3ln x2 + 2x + 5 + x +1 +C 20. x2 + 4x + 5 = x2 + 4x + 4 +1 = (x + 2)2 +1 u = x + 2, du = dx 2 x –1 2 u − dx 5 du x 4 x 5 u 1 ∫ = ∫ 2 2 + + + u du du u u 2 – 5 = ∫ ∫ 2 + 1 2 + 1 (Use the result of Problem 18.) = 2 u2 +1 – 5ln u2 +1 + u +C = 2 x2 + 4x + 5 – 5ln x2 + 4x + 5 + x + 2 +C 21. 5 − 4x − x2 = 9 − (4 + 4x + x2 ) = 9 − (x + 2)2 u = x + 2, du = dx ∫ 5 – 4x – x2 dx =∫ 9 – u2 du u = 3 sin t, du = 3 cos t dt 9 2 9 cos2 9 (1 cos 2 ) ∫ − u du = ∫ t dt = ∫ + t dt 2 9 1sin 2 2 2 = ⎛⎜ t + t ⎞⎟ +C ⎝ ⎠ 9( sin cos ) 2 = t + t t +C = ⎛ u ⎞ + u u +C ⎜ ⎟ 9 sin–1 1 9 – 2 2 3 2 ⎝ ⎠ x x x x C 9 sin–1 2 2 5 – 4 – 2 2 3 2 ⎛ + ⎞ + = ⎜ ⎟ + + ⎝ ⎠ 22. 16 + 6x – x2 = 25 − (9 − 6x + x2 ) = 25 – (x – 3)2 u = x – 3, du = dx dx du x x u ∫ = ∫ u = 5 sin t, du = 5 cos t 16 + 6 – 2 25 – 2 du dt t C u = ⎛ u ⎞ +C ⎜ ⎟ ∫ ∫ sin–1 25 2 = = + − 5 ⎝ ⎠ = sin–1 ⎛ x – 3 ⎞ ⎜ ⎟ +C 5 ⎝ ⎠ Instructor’s Resource Manual Section 7.4 443 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  33. 33. 23. 4x – x2 = 4 − (4 − 4x + x2 ) = 4 – (x – 2)2 u = x – 2, du = dx dx du x x u ∫ = ∫ 4 – 2 4 – 2 u = 2 sin t, du = 2 cos t dt du dt t C u = ⎛ u ⎞ +C ⎜ ⎟ ∫ ∫ sin–1 4 2 = = + − 2 ⎝ ⎠ = sin–1 ⎛ x – 2 ⎞ ⎜ ⎟ +C 2 ⎝ ⎠ 24. 4x – x2 = 4 − (4 − 4x + x2 ) = 4 – (x – 2)2 u = x – 2, du = dx x u + dx 2 du x x u ∫ = ∫ 2 2 4 – 4– u du du u u – – 2 = ∫ + ∫ 4 – 2 4 – 2 (Use the result of Problem 23.) – 4 – 2 2sin–1 = u + ⎛ u ⎞ +C ⎜ ⎟ 2 ⎝ ⎠ = – 4 x – x 2 + 2sin–1 ⎛ x – 2 ⎞ ⎜ ⎟ + C 2 ⎝ ⎠ 25. x2 + 2x + 2 = x2 + 2x +1+1 = (x +1)2 +1 u = x + 1, du = dx x dx u du 2 + 1 2 –1 2 2 1 ∫ = 2 ∫ + + 2 + x x u u du du u u 2 – 1 1 = ∫ 2 ∫ + 2 + = ln u2 +1 – tan–1 u +C = ln (x2 + 2x + 2)− tan−1(x +1) +C 26. x2 – 6x +18 = x2 – 6x + 9 + 9 = (x – 3)2 + 9 u = x – 3, du = dx x dx u du 2 –1 2 5 –6 18 9 ∫ = 2 ∫ + 2 + x x u + 2 u du 5 du u u + + ∫ ∫ ln ( 2 9) 5 tan 1 = + 2 2 9 9 = u + + − ⎛ u ⎞ +C ⎜ ⎟ 3 3 ⎝ ⎠ ln ( 2 6 18) 5 tan 1 3 = x − x + + − ⎛ x − ⎞ +C ⎜ ⎟ 3 3 ⎝ ⎠ 27. 2 ⎛ ⎞ 1 0 2 1 2 5 ⎝ + + ⎠ ∫ V dx = π ⎜ ⎟ x x 2 ⎡ ⎤ 1 1 0 2 ∫ = π ⎢ ⎥ x ( 1) 4 dx ⎢⎣ + + ⎥⎦ x + 1 = 2 tan t, dx = 2sec2 t dt 1 2sec 4sec π ⎛ ⎞ / 4 2 tan (1/ 2) 2 ⎝ ⎠ ∫ V tdt = π –1 ⎜ ⎟ 2 t 1 t dt π π π π = ∫ –1 / 4 tan –1 (1/ 2) 2 8 sec dt t / 4 cos 2 tan (1/ 2) = ∫ 8 t dt π π ⎛ ⎞ = ⎜ + ⎟ / 4 tan (1/ 2) ∫ –1 ⎝ ⎠ 1 1cos 2 8 2 2 / 4 –1 π ⎡ ⎤π = ⎢ + ⎥ ⎣ ⎦ tan (1/ 2) 1 1sin 2 8 2 4 t t / 4 t t t π –1 tan (1/ 2) 1 1sin cos 8 2 2 π ⎡ ⎤ = ⎢ + ⎥ ⎣ ⎦ 1 – 1 tan–1 1 1 π π = ⎡⎛ + ⎞ ⎛ ⎢⎜ ⎟ ⎜ + ⎞⎤ 8 ⎣⎝ 8 4 ⎠ ⎝ 2 2 5 ⎠⎦ ⎟⎥ 1 – tan–1 1 0.082811 π ⎛ π ⎞ = ⎜ + ⎟ ≈ 16 10 4 2 ⎝ ⎠ 2 1 28. 1 + + ∫ 1 0 2 V xdx 0 2 x x 2 5 = π 2 x dx ∫ ( x + 1) + 4 1 1 0 2 0 2 = π x dx dx x x + 2 1 – 2 1 + + + + ∫ ∫ = π π ( 1) 4 ( 1) 4 1 1 x x 2 1 ln[( 1) 4] – 2 1 tan 1 ⎡ ⎡ = π + 2 + ⎤ π –1 ⎛ + ⎞⎤ ⎜ ⎣ ⎢ 2 ⎥ ⎦ ⎢ ⎣ 2 ⎝ 2 ⎠⎦ ⎟⎥ [ln8 – ln 5] – tan–11– tan–1 1 0 0 = π π ⎡ ⎤ ⎢⎣ 2 ⎥⎦ ln 8 – tan–1 1 0.465751 5 4 2 ⎛ π ⎞ = π⎜ + ⎟ ≈ ⎝ ⎠ 29. a. u = x2 + 9, du = 2x dx xdx du u C x u ∫ 2 ∫ + 1 ln 2 9 1 ln ( 2 9) 2 2 1 1ln = = + 9 2 2 = x + +C = x + +C b. x = 3 tan t, dx = 3sec2 t dt xdx t dt x ∫ ∫ tan = – ln cost +C 2 + 9 = ⎛ ⎞ ln 3 ln 3 = − + = − ⎜ ⎟ + C C 2 1 2 1 + ⎜ + ⎟ ⎝ ⎠ x x 9 9 = ln ⎛⎜ x 2 + 9 ⎞⎟ − ln 3 + C1 ⎝ ⎠ = ln ((x2 + 9)1/ 2 )+C 1 ln ( 2 9) = x + +C 2 444 Section 7.4 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  34. 34. 30. u = 9 + x2 , u2 = 9 + x2 , 2u du = 2x dx 3 3 3 2 3 2 2 0 2 0 2 3 x dx x u − x dx 9 udu ∫ = ∫ = ∫ x x u 9 9 + + 3 2 ⎡ 3 2 ⎤ 3 2 u du u u 3 ∫ ≈ 5.272 ( 9) – 9 18 – 9 2 = − = ⎢ ⎥ = 3 3 ⎢⎣ ⎥⎦ 31. a. u = 4 – x2 , u2 = 4 – x2 , 2u du = –2x dx 2 2 2 x dx x x dx u du x x u 4 – 4 − – ∫ = ∫ = ∫ 2 2 4 – 2 2 2 u du du du u u 4 4 4 1 4 4 − + − − − ∫ ∫ ∫ 4 1 ln 2 = =− + u + u C u = − ⋅ + + 4 2 − 2 x x 2 C x − + ln 4 2 4 = − + − + 2 4 − − 2 b. x = 2 sin t, dx = 2 cos t dt 4 – 2 cos2 2 x dx t dt x sin t ∫ = ∫ (1– sin2 ) 2 = ∫ sin = 2∫csct dt – 2∫sin t dt = 2ln csct − cot t + 2cos t + C t dt t 2 − 2ln 2 4 x 4 x2 C = − + − + x x 2 − − 2ln 2 4 x 4 x2 C = + − + x To reconcile the answers, note that 2 2 2 2 x x x x − + − − ln 4 2 ln 4 2 − = 4 − − 2 4 − + 2 2 2 x − − ln ( 4 2) ( 4 x 2 2)( 4 x 2 2) = − + − − 2 2 2 2 x x (2 − 4 − ln ) ln (2 − 4 − ) = = 2 2 x x 4 4 − − − 2 ⎛ − − ⎞ − − = ⎜ ⎟ = 2 4 2 2 4 2 ln x 2ln x x x ⎜ ⎟ ⎝ ⎠ 32. The equation of the circle with center (–a, 0) is (x + a)2 + y2 = b2 , so y = ± b2 – (x + a)2 . By symmetry, the area of the overlap is four times the area of the region bounded by x = 0, y = 0, and y = b2 – (x + a)2 dx . A = 4 ∫ b – a b 2 –( x + a ) 2 dx 0 x + a = b sin t, dx = b cos t dt / 2 2 2 sin ( / ) A b tdt π = ∫ 4 cos –1 a b b tdt π 2 / 2 sin ( / ) = ∫ + 2 (1 cos 2 ) –1 a b b t t π / 2 = ⎡ + ⎤ ⎢⎣ ⎥⎦ 2 –1 a b 2 2 1 sin 2 sin ( / ) b t t tπ 2 /2 –1 2 [ sin cos ] a b sin ( / ) = + ⎡π ⎛ ⎛ a ⎞ a b 2 – a 2 ⎞⎤ = 2 b 2 ⎢ – ⎜ sin–1 ⎢ 2 ⎜ ⎜ ⎟ + ⎟⎥ ⎝ ⎝ b ⎠ b b ⎟⎥ ⎣ ⎠⎦ b2 – 2b2 sin–1 a – 2a b2 – a2 = π ⎛ ⎞ ⎜ ⎟ b ⎝ ⎠ 33. a. The coordinate of C is (0, –a). The lower arc of the lune lies on the circle given by the equation x2 + ( y + a)2 = 2a2 or y = ± 2a2 – x2 – a. The upper arc of the lune lies on the circle given by the equation x2 + y2 = a2 or y = ± a2 – x2 . – – 2 – – a a a a A = a 2 x 2 dx ⎛⎜ a 2 x 2 a ⎞⎟ dx ∫ ∫ – – ⎝ ⎠ – – 2 – 2 a a a a 2 2 2 2 2 = ∫ a x dx ∫ a x dx + a – – Note that 2 2 – a a ∫ a x dx is the area of a – semicircle with radius a, so 2 a 2 x 2 dx a a a ∫ = – For 2 2 π – . 2 2 – , a a ∫ a x dx let – x = 2a sin t, dx = 2a cos t dt 2 – 2 cos a a a 2 x 2 dx π / 4 a 2 2 tdt π ∫ = ∫ – –/ 4 2 π / 4 1 π / 4 (1 cos 2 ) 2 sin 2 – π /4 – π / 4 = a ∫ + tdt = a ⎡ ⎢⎣ t + t ⎤ 2 ⎥⎦ 2 2 a a π = + 2 2 2 – 2 2 2 2 2 2 π ⎛ π ⎞ = ⎜⎜ + ⎟⎟ + = A a a a a a ⎝ ⎠ Thus, the area of the lune is equal to the area of the square. Instructor’s Resource Manual Section 7.4 445 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  35. 35. b. Without using calculus, consider the following labels on the figure. Area of the lune = Area of the semicircle of radius a at O + Area (ΔABC) – Area of the sector ABC. 1 2 2 – 1 ( 2 )2 2 2 2 ⎛ π ⎞ = π + ⎜ ⎟ A a a a ⎝ ⎠ 1 2 2 – 1 2 2 2 2 = πa + a πa = a Note that since BC has length 2a, the π measure of angle OCB is , 4 so the measure π of angle ACB is . 2 34. Using reasoning similar to Problem 33 b, the area is a a b a a b 1 1 (2 ) – – 1 2sin 2 2 2 1 – – sin . 2 π + ⎛ ⎞ ⎜ ⎟ 2 2 2 –1 2 ⎝ ⎠ a 2 a b 2 a 2 b 2 –1 a b b = π + 35. 2 – 2 dy – a x dx x = ; 2 – 2 y – a x dx x = ∫ x = a sin t, dx = a cos t dt cos cos2 – cos – y a t a t dt a t dt = ∫ = ∫ a sin t sin t 1– sin2 – (sin – csc ) a t dt a t t dt = ∫ = sin t ∫ = a (– cos t − ln csct − cot t ) +C 2 cos t = a – x 2 2 2 , csct = a , cot t = a – x a x x ⎛ − ⎞ = ⎜ − − ⎟ + 2 – 2 2 2 y a – a x ln a a x C ⎜ a x x ⎟ ⎝ ⎠ 2 2 a2 x2 a − a ln a − x C = − − − + x Since y = 0 when x = a, 0 = 0 – a ln 1 + C, so C = 0. 2 2 y – a2 x2 a ln a a – x x − = − − 7.5 Concepts Review 1. proper 2. –1 5 1 x x + + 3. a = 2; b = 3; c = –1 A B Cx + D x x x 4. + + –1 ( –1)2 2 + 1 Problem Set 7.5 1. 1 A B = + xx x x ( + 1) + 1 1 = A(x + 1) + Bx A = 1, B = –1 1 1 – 1 ( 1) 1 + + ∫ ∫ ∫ = ln x – ln x +1 + C dx = dx dx x x x x 2. 2 A B 2 = 2 = + 3 ( 3) 3 x x x x x x + + + 2 = A(x + 3) + Bx 2 , – 2 3 3 A = B = ∫ 2 ∫ ∫ + + 2 ln – 2 ln 3 3 3 dx dx B dx 2 = 2 1 – 2 3 3 3 3 x x x x = x x + + C A B 3 3 –1 ( 1)( –1) 1 –1 3. 2 = = + x x + x x + x 3 = A(x – 1) + B(x + 1) – 3 , 3 2 2 A = B = 3 – 3 1 3 1 –1 2 1 2 –1 ∫ 2 ∫ ∫ + – 3 ln 1 3 ln –1 dx = dx + dx x x x = x + + x + C 2 2 446 Section 7.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  36. 36. x x 5 5 5 4. 3 2 2 = = x x x x x x 2 + 6 2 ( + 3) 2 ( + 3) 3 A B x x = + + 5 ( 3) 2 = A x + + Bx 5 , – 5 6 6 A = B = x dx dx 5 5 1 – 5 1 ∫ = x 3 + x 2 ∫ x ∫ x + 5 ln – 5 ln 3 6 6 2 6 6 6 3 = x x + +C x x A B x x x x x x 5. 2 11 11 3 4 ( 4)( 1) 4 1 − − = = + + − + − + − x – 11 = A(x – 1) + B(x + 4) A =3, B = –2 x dx dx dx x x x x ∫ 2 ∫ ∫ + − + − = 3ln x + 4 − 2ln x −1 + C 11 3 1 2 1 3 4 4 1 − = − x x A B x x x x x x 6. 2 – 7 = – 7 = + – –12 ( – 4)( + 3) – 4 + 3 x – 7 = A(x + 3) + B(x – 4) – 3 , 10 7 7 A = B = x dx dx dx x x x x ∫ 2 ∫ ∫ + – 3 ln – 4 10 ln 3 – 7 = – 3 1 + 10 1 – –12 7 – 4 7 3 = x + x + +C 7 7 7. 2 x x A B 3 − 13 3 − = 13 = + 3 10 ( 5)( 2) 5 2 x x x x x x + − + − + − 3x −13 = A(x − 2) + B(x + 5) A = 4, B = –1 ∫ x − dx 4 1 1 2 + − 3 13 3 10 x x + − ∫ ∫ dx dx = − x x 5 2 = 4ln x + 5 − ln x − 2 +C x x + π + π 8. = 2 – 3 2 2 ( – 2 )( – ) x x x x A B x x = + π + π π π – 2 – π π x +π = A(x −π ) + B(x − 2π ) A = 3, B = –2 x + π dx dx dx ∫ 2 π + π 2 ∫ ∫ π π = 3ln x – 2π – 2ln x – π + C 3 – 2 = x x x x – 3 2 – 2 – x x x x x x 2 21 2 21 2 9 – 5 (2 –1)( 5) 9. 2 + + = A B x x = + + + 2 –1 + 5 2x + 21 = A(x + 5) + B(2x – 1) A = 4, B = –1 x dx dx dx x x x x 2 21 4 – 1 2 9 – 5 2 –1 5 ∫ 2 ∫ ∫ + + = 2ln 2x –1 – ln x + 5 +C + = 10. 2 2 2 2 x x x x x x x x x 2 − − 20 2( + − 6)− 3 − 8 = 6 6 + − + − x x x 2 3 8 2 6 + = − + − x x x x x x 3 8 3 8 2 + + = A B x x = + + − + − 3 2 6 ( 3)( 2) + − 3x + 8 = A(x – 2) + B(x + 3) 1 , 14 5 5 A = B = 2 2 x x dx x x 2 20 6 − − + − ∫ 2 1 1 14 1 + − ∫ ∫ ∫ 2 1 ln 3 14 ln 2 dx dx dx = − − x x 5 3 5 2 = x − x + − x − +C 5 5 x x x x x x 17 – 3 17 – 3 3 – 2 (3 – 2)( 1) 11. 2 = A B x x = + + + 3 – 2 + 1 17x – 3 = A(x + 1) + B(3x – 2) A = 5, B = 4 x dx dx dx x x x x + + ∫ ∫ ∫ 5 ln 3 – 2 4ln 1 17 – 3 = 5 + 4 3 2 – 2 3 – 2 1 = x + x + +C 3 Instructor’s Resource Manual Section 7.5 447 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  37. 37. 12. 2 x x 5 – 5 – = x x x x – ( 4) 4 ( – )( – 4) A B x x = + π + + π π – π – 4 5 – x = A(x – 4) + B(x – π ) 5 – π , 1 –4 4– A B = = π π ∫ 2 ∫ ∫ 5 – ln – 1 ln – 4 − π + + π π π π x dx dx dx 5 – 5 – π = 1 + 1 1 ( 4) 4 – 4 – 4 – – 4 x x x x x x C π = π + + π π – 4 4 – 13. 2 2 3 2 x x x x x x x x x x 2 + − 4 2 + − 4 = A B C x x x = + + − − + − 1 2 2 ( 1)( 2) + − 2x2 + x − 4 = A(x +1)(x − 2) + Bx(x − 2) + Cx(x +1) A = 2, B = –1, C = 1 2 3 2 2 x + x − 4 dx 2 dx 1 dx 1 dx x x x x x x ∫ = ∫ − ∫ + ∫ = 2ln x − ln x +1 + ln x − 2 + C − − + − 2 1 2 14. 7 2 2 – 3 x + x A B C = + + x x x x x x (2 –1)(3 + 2)( – 3) 2 –1 3 + 2 – 3 7x2 + 2x – 3 = A(3x + 2)(x – 3) + B(2x –1)(x – 3) +C(2x –1)(3x + 2) 1 , – 1 , 6 35 7 5 A = B = C = 7 2 2 – 3 1 1 1 1 6 1 – x + x dx dx dx dx ∫ = ∫ ∫ + ∫ + + 1 ln 2 –1 – 1 ln 3 2 6 ln – 3 70 21 5 x x x x x x (2 –1)(3 2)( – 3) 35 2 –1 7 3 2 5 – 3 = x x + + x +C 15. 2 2 x x x x x x x x x x 6 + 22 − 23 + − = 6 22 23 (2 1)( 2 6) (2 1)( 3)( 2) A B C x x x = + + − + − − + − 2 − 1 + 3 − 2 6x2 + 22x − 23 = A(x + 3)(x − 2) + B(2x −1)(x − 2) + C(2x −1)(x + 3) A = 2, B = –1, C = 3 2 x x dx dx dx dx x x x x x x 6 + 22 − 23 2 1 3 (2 1)( 6) 2 1 3 2 ∫ = ∫ − ∫ + ∫ = ln 2x −1 − ln x + 3 + 3ln x − 2 + C − 2 + − − + − 16. 3 2 3 2 x x x x x x 6 11 6 − + − − + − 4 28 56 32 3 2 3 2 ⎛ − + − ⎞ x x x x x x 1 6 11 6 4 7 14 8 = ⎜⎜ ⎟⎟ ⎝ − + − ⎠ 2 ⎛ − + ⎞ x x 1 1 3 2 4 7 14 8 = ⎜⎜ + ⎝ x 3 ⎟⎟ − x 2 + x − ⎠ ⎛ − − ⎞ x x x x x 1 1 ( 1)( 2) 4 ( 1)( 2)( 4) = ⎜ + ⎟ ⎝ − − − ⎠ 1 1 1 4 x 4 = ⎛ + ⎞ ⎜ − ⎟ ⎝ ⎠ 3 2 3 2 x x x dx x x x ∫ – 6 + 11 – 6 1 1 1 + 4 –28 56 –32 = ∫ dx + ∫ dx 1 1ln – 4 x 4 4 –4 = x + x +C 4 4 448 Section 7.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  38. 38. 17. 3 x x x –1 3 – 2 = + 2 2 x x x x –2 –2 + + 3 x –2 = 3 x –2 = A + B x 2 x x x x x – 2 ( 2)( –1) 2 –1 + + + 3x – 2 = A(x – 1) + B(x + 2) 8 , 1 3 3 A = B = 3 x dx ∫ ( 1) 8 1 1 1 x 2 + x – 2 ∫ x dx ∫ dx ∫ dx 1 2 – 8 ln 2 1 ln –1 + − = − + + x x 3 2 3 1 = x x + x + + x +C 2 3 3 18. 3 2 2 x x x x x x x x – 4 14 24 + + = + 5 6 ( 3)( 2) + + + + 14 x + 24 = A + B ( x 3)( x 2) x 3 x 2 + + + + 14x + 24 = A(x + 2) + B(x + 3) A = 18, B = –4 3 2 2 5 6 x x dx x x + + + ∫ ( 4) 18 – 4 ∫ x dx ∫ dx ∫ dx 1 2 4 18ln 3 – 4ln 2 + + x x 3 2 = − + = x − x + x + x + +C 2 19. 4 2 2 3 x x x x x x x x x 8 8 12 8 4 ( 2)( – 2) + + + = + − + 12 x 2 + 8 = A + B + C ( 2)( – 2) 2 – 2 x x + x x x + x 12x2 + 8 = A(x + 2)(x – 2) + Bx(x – 2) + Cx(x + 2) A = –2, B = 7, C = 7 4 2 3 x x dx x dx dx dx dx x x x x x + ∫ ∫ ∫ ∫ ∫ 1 2 – 2ln 7ln 2 7 ln – 2 8 8 – 2 1 7 1 7 1 – 4 2 – 2 + + = + + = x x + x + + x +C 2 20. 6 3 2 x x x x x x x x x x 4 4 4 16 68 272 4 – 4 – 4 + + + 3 2 = + + + + 3 2 3 2 2 x A B C 272 + 4 = + + x 2 ( x – 4) x x 2 x – 4 272x2 + 4 = Ax(x – 4) + B(x – 4) +Cx2 – 1 , –1, 1089 4 4 A = B = C = 6 3 3 2 x 4 x 4 dx x – 4 x ( 4 16 68) – 1 1 – 1 1089 1 + + ∫ 3 2 − ∫ ∫ ∫ ∫ x x x dx dx dx dx = + + + + 2 x x x 4 4 4 1 4 4 3 8 2 68 – 1 ln 1 1089 ln – 4 4 3 4 4 x x x x x x C = + + + + + + x x + 1 A B x x x 21. = + 2 2 ( 3) 3 ( 3) − − − x + 1 = A(x – 3) + B A = 1, B = 4 x dx dx dx x x x ∫ + 1 = ∫ 1 + ∫ 4 ln 3 4 − 2 − − 2 ( 3) 3 ( 3) x C = − − + 3 x − Instructor’s Resource Manual Section 7.5 449 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  39. 39. x x A B 5 + 7 5 + 7 4 4 ( 2) 2 ( 2) 22. = = + 2 2 2 x x x x x + + + + + 5x + 7 = A(x + 2) + B A = 5, B = –3 x dx dx dx + + + + ∫ ∫ ∫ 5ln 2 3 5 + 7 = 5 − 3 4 4 2 ( 2) 2 2 x x x x x C = + + + 2 x + x x 3 + 2 3 + 2 3 3 1 ( 1) 23. = 3 2 3 A B C x x x = + + + + + + 1 ( 1)2 ( 1)3 x x x x + + + 3x + 2 = A(x +1)2 + B(x +1) +C A = 0, B = 3, C = –1 x dx dx dx C 3 + 2 3 1 3 1 3 3 1 ( 1) ( 1) 1 2( 1) + + + + + + + ∫ ∫ ∫ = − = − + + 3 2 2 3 2 x x x x x x x 24. 6 x A B C D E F G = + + + + + + ( x – 2)2 (1– x )5 x – 2 ( x – 2)2 1– x (1– x )2 (1– x )3 (1– x )4 (1– x )5 A = 128, B = –64, C = 129, D = –72, E = 30, F = –8, G = 1 6 2 5 2 2 3 4 5 ⎡ ⎤ x dx 128 – 64 129 – 72 30 8 1 dx ∫ ∫ = ⎢ + + − + ⎥ x x x x x x x x x ( – 2) (1– ) – 2 ( – 2) 1– (1– ) (1– ) (1– ) (1– ) ⎢⎣ ⎥⎦ 128ln – 2 64 –129ln 1– 72 15 8 1 x x C = + + − + − + 2 3 4 x x x x x – 2 1– (1– ) 3(1– ) 4(1– ) 25. 2 2 3 2 2 2 x x x x A B C x x x x x x x x 3 − 21 + 32 3 − 21 + 32 = = + + 8 16 ( 4) 4 ( 4) − + − − − 3x2 − 21x + 32 = A(x − 4)2 + Bx(x − 4) + Cx A = 2, B = 1, C = –1 2 3 2 2 x x dx dx dx dx x x x x x ∫ 3 − 21 + 32 = ∫ 2 + ∫ 1 − ∫ 1 2ln ln 4 1 − + − − 8 16 4 ( 4) x x C = + − + + 4 x − 26. 2 2 x x x x x x x x + 19 + 10 + 19 + = 10 2 4 5 3 3 (2 5) A B C D x x x x = + + + + + 2 3 2 + 5 A = –1, B = 3, C = 2, D = 2 2 x + 19 x + 10 ⎛ 1 3 2 2 ⎞ dx – dx 2 x 5 x x x x 2 x 5 – ln x – 3 – 1 ln 2x 5 C ∫ = 4 3 ∫ ⎜ + + + + ⎝ 2 3 + ⎟ ⎠ 2 = + + + x x 27. 2 2 3 2 2 x x x x A BxC x x x x x x 2 –8 2 –8 + + + = = + 4 ( 4) 4 + + + A = –2, B = 4, C = 1 2 3 2 x x dx dx x dx x x x x 2 + – 8 –2 1 4 + 1 dx x dx dx x x x 2 1 2 2 1 + + ∫ ∫ ∫ ∫ = ∫ + ∫ + + 2 2 4 4 = − + + 4 4 = –2ln x + 2ln x 2 + 4 + 1 tan–1 ⎛ x ⎞ ⎜ ⎟ +C 2 2 ⎝ ⎠ 450 Section 7.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  40. 40. x x 3 + 2 3 + 2 ( 2) 16 ( 4 20) 28. = 2 2 A Bx + C x x x = + + + + + 2 4 20 x x x x x x + + 1 , – 1 , 13 10 10 5 A = B = C = x dx 3 + 2 ( 2) 16 ∫ + 2 + xx x 1 x 13 10 5 2 1 1 – + 10 4 20 1 1 14 1 10 5 ( 2) 16 + + ∫ ∫ 2 dx dx x x x = + x dx 1 2 4 20 4 20 + + ∫ ∫ 2 dx dx x x = + + + + ∫ x x − x x 1 ln 7 tan–1 2 10 10 4 ⎛ + ⎞ = + ⎜ ⎟ ⎝ ⎠ – 1 ln 2 4 20 20 x + x + +C 29. 2 x x A Bx C x x x x 2 –3 –36 + = + (2 − 1)( 2 + 9) 2 –1 2 + 9 A = –4, B = 3, C = 0 2 x x dx dx x dx x x x x 2 – 3 – 36 –4 1 3 (2 –1)( 9) 2 –1 9 ∫ = ∫ + ∫ –2ln 2 –1 3 ln 2 9 2 + 2 + = x + x + +C 2 1 1 x –16 (x 2)(x 2)(x 4) 30. = 4 2 − + + A B Cx + D x x x = + + – 2 + 2 2 + 4 1 , – 1 , 0, – 1 32 32 8 A = B = C = D = = x x + ⎛ x ⎞ +C ⎜ ⎟ + + ∫ ∫ ∫ ∫ 1 ln – 2 – 1 ln 2 – 1 tan–1 1 1 1 – 1 1 1 1 –16 32 – 2 32 2 8 4 dx = dx dx − dx 4 2 x x x x 32 32 16 2 ⎝ ⎠ 1 A B C D 31. = + + + 2 2 2 2 x x x x x x ( –1) ( + 4) –1 ( –1) + 4 ( + 4) – 2 , 1 , 2 , 1 125 25 125 25 A = B = C = D = 1 – 2 1 1 1 2 1 1 1 + + + ∫ ∫ ∫ ∫ ∫ – 2 ln –1 – 1 2 ln 4 – 1 dx = dx + dx + dx + dx 2 2 2 2 x x x x x x ( –1) ( 4) 125 –1 25 ( –1) 125 4 25 ( 4) x x C = + + + x x 125 25( –1) 125 25( + 4) 32. 3 2 2 x x x x x x x x x x – 8 –1 + = 1 + –7 7 –16 2 2 ( + 3)( –4 + 5) ( + 3)( − 4 + 5) 2 2 2 x x A Bx C –7 + 7 –16 + = + ( x 3)( x – 4 x 5) x 3 x – 4 x 5 + + + + – 50 , – 41, 14 A = B = C = 13 13 13 3 2 41 14 ⎡ ⎛ ⎞ + ⎤ = ⎢ ⎜ ⎟ + ⎥ + + ⎢⎣ ⎝ + ⎠ + ⎥⎦ x – 8 x –1 – x dx 1– 50 1 13 13 dx x x x x x x ∫ ∫ 2 2 ( 3)( – 4 5) 13 3 – 4 5 dx dx dx x dx 50 1 68 1 41 2 4 13 3 13 ( 2) 1 26 4 5 + − + − + ∫ ∫ ∫ ∫ – 50 ln 3 – 68 tan–1( – 2) – 41 ln 2 – 4 5 = − − − 2 2 − x x x x = x x + x x x + +C 13 13 26 Instructor’s Resource Manual Section 7.5 451 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  41. 41. 33. x = sin t, dx = cos t dt 3 2 3 2 t t t dt x x dx t t t x x x (sin − 8sin − 1) cos − 8 − 1 (sin 3)(sin 4sin 5) ( 3)( 4 5) ∫ = 2 ∫ + − + + 2 − + 50 ln 3 68 tan 1( 2) 41 ln 2 4 5 13 13 26 = x − x + − − x − − x − x + + C which is the result of Problem 32. 3 2 t t t dt t t t t t C t t t (sin – 8sin –1) cos sin – 50 ln sin 3 – 68 tan –1 (sin – 2) – 41 ln sin 2 – 4sin 5 (sin 3)(sin – 4sin 5) 13 13 26 ∫ + 2 + = + + + 34. x = sin t, dx = cos t dt t dt dx x x x C t x cos 1 1 ln 2 1 ln 2 1 tan sin 16 16 32 32 16 2 = = − − + − − 1 ⎛ ⎞ ⎜ ⎟ + ∫ 4 ∫ − 4 − ⎝ ⎠ which is the result of Problem 30. t dt t t t C t cos 1 ln sin – 2 – 1 ln sin 2 – 1 tan sin sin –16 32 32 16 2 = + –1 ⎛ ⎞ ⎜ ⎟ + ∫ 4 ⎝ ⎠ 35. 3 2 2 2 2 2 x – 4 x + + = Ax B + Cx D x x x ( + 1) + 1 ( + 1) A = 1, B = 0, C = –5, D = 0 3 2 2 2 2 2 x – 4 x dx x dx 5 x dx x x x 1 ln 1 5 2 2( 1) ∫ = ∫ − ∫ 2 + + + ( 1) 1 ( 1) x C = + + + 2 x + 36. x = cos t, dx = –sin t dt 2 2 2 4 2 4 t t dt x dx (sin )(4cos –1) – 4 –1 ∫ = ∫ + + + + t t t x x x (cos )(1 2cos cos ) (1 2 ) 2 2 2 4 2 2 2 2 2 x x A BxC DxE 4 − 1 4 − 1 + + = = + + (1 2 ) ( 1) 1 ( 1) x x x x x x x x + + + + + A = –1, B = 1, C = 0, D = 5, E = 0 ⎡ ⎤ − ⎢− + + ⎥ = − + + + ⎢⎣ + + ⎥⎦ + x x dx x x C 1 5 ln 1 ln 1 5 ln cos 1 ln cos 1 5 ∫ 2 2 2 2 2 2 x x x x 1 ( 1) 2 2( 1) t t C = − + + + 2 2 2(cos t + 1) 37. 3 2 2 5 3 4 2 x x x x x x x x x xx x 2 + 5 + 16 (2 + 5 + 16) = 8 16 ( 8 16) + + + + 2 2 2 2 2 2 x x Ax B Cx D x x x 2 + 5 + 16 + + ( 4) 4 ( 4) = = + + + + A = 0, B = 2, C = 5, D = 8 3 2 5 3 2 2 2 x x x dx dx x dx x x x x x 2 + 5 + 16 2 5 + 8 dx x dx dx 2 5 8 4 ( 4) ( 4) ∫ = ∫ + ∫ + + + + 2 2 2 2 2 8 16 4 ( 4) + + + ∫ ∫ ∫ = + + x x x 8 , x + ∫ let x = 2 tan θ, dx = 2sec2θ dθ . To integrate 2 2 ( 4) dx 2 8 16sec θ ∫ dx = ∫ d cos2 1 1 cos 2 2 + 2 4 x ( 4) 16sec θ θ = θ dθ = ⎛⎜ + θ ⎞⎟ dθ ∫ ∫ ⎝ 2 2 ⎠ 1 1 sin 2 1 1 sin cos 2 4 2 2 x x C 1 tan 2 2 4 = θ + θ +C = θ + θ θ +C –1 = + + 2 x + 3 2 x x x dx x x x C x x x x x 2 5 16 tan – 5 1 tan x x C 3 tan 2 – 5 2 2 2( 4) ∫ –1 –1 –1 + + + + + + = + + + 5 3 2 2 8 16 2 2( 4) 2 2 4 = + + 2 x + 452 Section 7.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
  42. 42. x x A B x x x x x x 38. 2 –17 = –17 = + –12 ( 4)( – 3) 4 – 3 + + + A = 3, B = –2 6 6 4 2 4 x dx dx x x x x –17 3 – 2 –12 4 – 3 = ⎛ ⎞ ⎜ + ⎟ + ⎝ ⎠ ∫ ∫ 6 4 = ⎡⎣3ln x + 4 – 2ln x – 3 ⎤⎦ = (3ln10 – 2ln 3) – (3ln8 – 2ln1) = 3ln10 – 2ln 3 – 3ln8 ≈ –1.53 39. u = sin θ, du = cos θ dθ cos θ 1 ∫ / 4 d ∫ 1/ 2 du 1/ 2 0 2 2 + 2 0 2 2 + 2 u u θ = (1– sin θ )(sin θ 1) (1– )( 1) π 1 + + ∫ 0 2 2 u u u (1 – )(1 )( 1) du = 1 A B Cu + D Eu + F = + + + 2 2 2 2 2 2 u u u u u u (1– )( + 1) 1– 1 + + 1 ( + 1) 1 , 1 , 0, 1 , 0, 1 8 8 4 2 A = B = C = D = E = F = 1/ 2 1 1 1/ 2 1 1 1/ 2 1 1 1/ 2 1 1 1/ 2 1 0 2 2 2 0 0 0 2 0 2 2 − + − + + + ∫ ∫ ∫ ∫ ∫ du = du + du + du + du u u u u u u (1 )( 1) 8 1 8 1 4 1 2 ( 1) 1/ 2 ⎡ ⎛ ⎞⎤ = ⎢ – 1 ln 1– u + 1 ln 1 + u + 1 tan –1 u + 1 ⎜ tan –1 u + u ⎣ 8 8 4 4 ⎝ 2 ⎟⎥ + 1 ⎠⎦ 0 u 1/ 2 ⎡ + = 1 ln 1 u 1 ⎢ + tan –1 u + u ⎤ ⎥ ⎢⎣ 8 1 − u 2 4( u 2 + 1) ⎥⎦ 0 1 2 + ln 1 1 tan–1 1 1 0.65 8 2 1 2 2 6 2 = + + ≈ − 1 , u + ∫ let u = tan t.) (To integrate 2 2 ( 1) du x x x x x x 3 13 3 13 4 3 ( 3)( 1) 40. 2 + + = A B x x = + + + + + + 3 + 1 A = –2, B = 5 5 5 1 2 1 x dx x x x x 3 + 13 –2ln 3 5ln 1 4 3 ∫ = ⎡⎣ + + + ⎤⎦ = –2 ln 8 + 5 ln 6 + 2 ln 4 – 5 ln 2 = 5ln 3− 2ln 2 ≈ 4.11 + + 41. dy y(1 y) dt = − so that ∫ dy ∫ dt t C − a. Using partial fractions: 1 1 1 (1 ) y y = = + A B A y By 1 (1 − ) + (1 ) 1 (1 ) = + = ⇒ y y y y y y − − − ( ) 1 0 1, 0 1, 1 1 1 1 + − = + ⇒ = − = ⇒ = = ⇒ = + y (1 y ) y 1 y A B Ay y A B A A B − − ⎛ ⎞ ⎛ ⎞ y y y + = ⎜ + ⎟ = ⎝ − ⎠ ∫ ln ln(1 ) ln t C dy Thus: 1 1 1 y 1 y − − = ⎜ ⎟ ⎝ 1 − y ⎠ so that y e Ce y t C t = 1 = + 1 ( C1 ) C = e − y t e or 1 ( ) C t t e = + (0) 0.5, 0.5 1 or 1 y = = C = Since 1 1 C + y t e ; thus ( ) = + 1 t t e b. 3 3 (3) 0.953 y e = ≈ 1 e + Instructor’s Resource Manual Section 7.5 453 © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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