lim n n 1 0 converges lim n n n 2 ln 2 2 15

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Unformatted text preview: π/3 ≈ 1.047 +1 +∞ (b) +∞ e−u du = √ π/2 ≈ 0.8862 0 4 +∞ 2 2 e−ax dx = √ a √ 2 2σ du, √ 2πσ e−x dx ≈ 0.8862; 0 61. (a) +∞ a dx, 2 0 e (b) √ 66. −2 = √ 2 + π/2 0 1 sin(1 − u2 )du = 2 1 1 sin(1 − u2 )du ≈ 1.187 0 Chapter 9 Supplementary Exercises 336 CHAPTER 9 SUPPLEMENTARY EXERCISES 1. (a) (c) (e) (g) (i) integration by parts, u = x, dv = sin x dx reduction formula u-substitution: u = x3 + 1 integration by parts: dv = dx, u = tan−1 x u-substitution: u = 4 − x2 (b) (d) (f ) (h) u-substitution: u = sin x u-substitution: u = tan x u-substitution: u = x + 1 trigonometric substitution: x = 2 sin θ 1 3 1 9 2. (a) x = 3 tan θ (d) x = 3 sec θ (b) x = 3 sin θ √ (e) x = 3 tan θ (c) x = (f ) x = 5. (a) #40 (d) #108 (b) #57 (e) #52 (c) #113 (f ) #71 6. (a) u = x2 , dv = √ 1 0 √ x x2 + 1 dx, du = 2x dx, v = 3 x dx = x2 x2 + 1 = √ x2 + 1 1 0 sin θ tan θ √ x2 + 1; 1 −2 x(x2 + 1)1/2 dx 0 2 2 − (x2 + 1)3/2 3 1 = 0 √ √ 2√ 2 − [2 2 − 1] = (2 − 2)/3 3 (b) u2 = x2 + 1, x2 = u2 − 1, 2x dx = 2u du, x dx = u du; 1 0 x3 √ dx = x2 + 1 1 0 x2 √ x dx = x2 + 1 √ 2 1 u2 − 1 u du u √ 2 13 u −u 3 (u − 1)du = 2 = 1 √ 2 = (2 − 1 7. (a) u = 2x, sin4 2x dx = 1 2 sin4 u du = 1 1 3 − sin3 u cos u + 2 4 4 sin2 u du 1 1 3 1 − sin u cos u + = − sin3 u cos u + 8 8 2 2 du 3 3 1 sin u cos u + u + C = − sin3 u cos u − 8 16 16 3 3 1 sin 2x cos 2x + x + C = − sin3 2x cos 2x − 8 16 8 (b) u = x2 , x cos4 (x2 )dx = 1 2 cos4 u du = 11 3 cos3 u sin u + 24 4 cos2 u du = 1 31 1 cos3 u sin u + cos u sin u + 8 82 2 = 3 3 1 cos3 u sin u + cos u sin u + u + C 8 16 16 = 1 3 3 cos3 (x2 ) sin(x2 ) + cos(x2 ) sin(x2 ) + x2 + C 8 16 16 8. (a) With x = sec θ: 1 dx = x3 − x cot θ dθ = ln | sin θ| + C = ln du √ x2 − 1 + C ; valid for |x| > 1. |x| √ 2)/3 337 Chapter 9 (b) With x = sin θ: 1 dx = − x3 − x 1 dθ = − sin θ cos θ 2 csc 2θ dθ √ = − ln | csc 2θ − cot 2θ| + C = ln | cot θ| + C = ln 1 − x2 + C, 0 < |x| < 1. |x| (c) By partial fractions: − 1/2 1/2 1 + + x x+1 x−1 dx = − ln |x| + = ln 1 1 ln |x + 1| + ln |x − 1| + C 2 2 |x2 − 1| + C ; valid for all x except x = 0, ±1. |x| √ 9. (a) With u = x: √ 1 1 dx = 2 √ du = 2 sin−1 (u/ 2) + C = 2 sin−1 ( x/2) + C ; √√ x 2−x 2 − u2 √ with u = 2 − x: √ √ √ 1 1 dx = −2 √ du = −2 sin−1 (u/ 2) + C = −2 sin−1 ( 2 − x/ 2) + C ; √√ 2 x 2−x 2−u completing the square: 1 dx = sin−1 (x − 1) + C . 1 − (x − 1)2 (b) In the three results in part (a) the antiderivatives differ by a constant, in particular √ √ 2 sin−1 ( x/2) = π − 2 sin−1 ( 2 − x/ 2) = π/2 + sin−1 (x − 1). 2 10. A = 1 A B 3−x C 3−x = + 2+ ; A = −4, B = 3, C = 4 dx, 2 3 + x2 x x (x + 1) x x x+1 A = −4 ln |x| − 3 + 4 ln |x + 1| x 2 1 3 3 3 3 = (−4 ln 2 − + 4 ln 3) − (−4 ln 1 − 3 + 4 ln 2) = − 8 ln 2 + 4 ln 3 = + 4 ln 2 2 2 4 11. Solve y = 1/(1 + x2 ) for x getting 1−y and integrate with respect to y x= 1 to y to get A = 0 +∞ 12. A = e +∞ but lim e +∞ 0 = 1/e e →+∞ − y = 1 / (1 + x2) x xe−x dx = 2π lim −e−x (x + 1) 0 →+∞ 1 1−y dy (see figure) y ln x ln x − 1 dx = lim − →+∞ x2 x 13. V = 2π 14. y ( + 1) = lim →+∞ = 2π lim 0 →+∞ 1 − e− ( + 1) +1 1 = lim = 0 so V = 2π →+∞ e e dx 1 tan−1 (x/a) = lim →+∞ a x2 + a2 = lim 0 →+∞ π 1 tan−1 ( /a) = = 1, a = π/2 a 2a Chapter 9 Supplementary Exercises 338 2 u1/2 du = − cos3/2 θ + C 3 15. u = cos θ, − 16. Use Endpaper Formula (31) to get 17. u = tan(x2 ), 1 2 1 1 1 tan6 θ − tan4 θ + tan2 θ + ln | cos θ| + C . 6 4 2 tan7 θdθ = 1 tan3 (x2 ) + C 6 u2 du = √ √ 18. x = (1/ 2) sin θ, dx = (1/ 2) cos θ dθ, 1 √ 2 1 cos3 θ sin θ 4 π /2 1 cos4 θ dθ = √ 2 −π/2 3 =√ 42 19. x = √ 3 tan θ, dx = 1 1 dθ = sec θ 3 1 3 π /2 1 cos θ sin θ 2 π /2 + −π/2 + −π/2 x+3 21. (x + 1)2 + 1 cos θ dθ = cos2 θ dθ −π/2 π /2 dθ −π/2 3π 31 = √ π= √ 4 22 82 1 x sin θ + C = √ +C 3 3 3 + x2 √ 1 1 du = √ tan−1 [(sin θ − 3)/ 3] + C u2 + 3 3 dx, let u = x + 1, u+2 √ du = u2 + 1 = 1 2 π /2 √ 3 sec2 θ dθ, cos θ dθ, let u = sin θ − 3, (sin θ − 3)2 + 3 20. 3 4 u(u2 + 1)−1/2 + √ 2 u2 +1 du = u2 + 1 + 2 sinh−1 u + C x2 + 2x + 2 + 2 sinh−1 (x + 1) + C Alternate solution: let x + 1 = tan θ, (tan θ + 2) sec θ dθ = = 22. Let x = tan θ to get sec θ tan θ dθ + 2 sec θ dθ = sec θ + 2 ln | sec θ + tan θ| + C x2 + 2x + 2 + 2 ln( x2 + 2x + 2 + x + 1) + C. x3 1 dx. − x2 A B C 1 = + 2+ ; A = −1, B = −1, C = 1 so x2 (x − 1) x x x−1 1 dx − x − = 23. 1 dx + x2 1 1 dx = − ln |x| + + ln |x − 1| + C x−1 x x−1 tan θ − 1 1 + ln + C = cot θ + ln + C = cot θ + ln |1 − cot θ| + C x x tan θ A B C 1 1 1 1 = + + ;A=− ,B= ,C= so (x − 1)(x + 2)(x − 3) x−1 x+2 x−3 6 15 10 − 1 6 1 1 dx + x−1 15 1 1 dx + x+2 10 1 dx x−3 1 1 1 ln |x + 2| + ln |x − 3| + C = − ln |x − 1| + 6 15 10 339 24. Chapter 9 A Bx + C 1 = +2 ; A = 1, B = C = −1 so + x + 1)...
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