F x cos x f x cos x f 4 x cos x k2 k4

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Unformatted text preview: (1 − cos 10θ)dθ = 1 1 θ− sin 10θ + C 2 20 6. cos3 at dt = = 7. cos5 θdθ = (1 − sin2 at) cos at dt cos at dt − (1 − sin2 θ)2 cos θdθ = = sin θ − 8. sin3 x cos3 x dx = = 9. sin2 at cos at dt = sin2 2t cos3 2t dt = = 1 1 sin at − sin3 at + C (a = 0) a 3a (1 − 2 sin2 θ + sin4 θ) cos θdθ 1 2 sin3 θ + sin5 θ + C 3 5 sin3 x(1 − sin2 x) cos x dx (sin3 x − sin5 x) cos x dx = 1 1 sin4 x − sin6 x + C 4 6 sin2 2t(1 − sin2 2t) cos 2t dt = 1 1 sin3 2t − sin5 2t + C 6 10 (sin2 2t − sin4 2t) cos 2t dt Exercise Set 9.3 304 (1 − cos2 2x) cos2 2x sin 2x dx sin3 2x cos2 2x dx = 10. = 1 1 cos5 2x + C (cos2 2x − cos4 2x) sin 2x dx = − cos3 2x + 6 10 11. sin2 x cos2 x dx = 1 4 sin2 2x dx = 1 8 12. sin2 x cos4 x dx = 1 8 (1 − cos 2x)(1 + cos 2x)2 dx = = 1 8 sin2 2x dx + = 1 1 x− sin 4x + C 8 32 1 1 1 x− sin 4x + sin3 2x + C 16 64 48 (1 − cos 4x)dx = 1 8 1 8 (1 − cos2 2x)(1 + cos 2x)dx sin2 2x cos 2x dx = 1 16 (1 − cos 4x)dx + 13. sin x cos 2x dx = 1 2 1 1 (sin 3x − sin x)dx = − cos 3x + cos x + C 6 2 14. sin 3θ cos 2θdθ = 1 2 (sin 5θ + sin θ)dθ = − 15. sin x cos(x/2)dx = 1 sin3 2x 48 16. u = cos x, − 1 2 1 [sin(3x/2) + sin(x/2)]dx = − cos(3x/2) − cos(x/2) + C 3 5 u1/5 du = − cos6/5 x + C 6 π /4 π /4 (1 − sin2 x) cos x dx cos3 x dx = 17. 1 1 cos 5θ − cos θ + C 10 2 0 0 = sin x − π /2 1 4 sin2 (x/2) cos2 (x/2)dx = 18. 0 1 8 = π /3 0 √ √ 1√ = ( 2/2) − ( 2/2)3 = 5 2/12 3 π /2 sin2 x dx = 0 x− 1 8 π /2 (1 − cos 2x)dx 0 π /2 1 sin 2x 2 = π/16 0 π /3 0 1 1 sin5 3x − sin7 3x 15 21 sin4 3x(1 − sin2 3x) cos 3x dx = sin4 3x cos3 3x dx = 19. π /4 1 sin3 x 3 0 π cos2 5θ dθ = 20. −π 1 2 π (1 + cos 10θ)dθ = −π π /6 21. sin 2x cos 4x dx = 0 1 2 1 2 θ+ 1 sin 10θ 10 π /6 (sin 6x − sin 2x)dx = − 0 π /3 =0 0 π =π −π 1 1 cos 6x + cos 2x 12 4 π /6 0 = [(−1/12)(−1) + (1/4)(1/2)] − [−1/12 + 1/4] = 1/24 2π sin2 kx dx = 22. 0 1 2 2π (1 − cos 2kx)dx = 0 1 2 x− 1 sin 2kx 2k 2π =π− 0 1 sin 4πk (k = 0) 4k 305 23. Chapter 9 1 24. − ln | cos 5x| + C 5 1 tan(3x + 1) + C 3 25. u = e−2x , du = −2e−2x dx; − 26. 1 2 tan u du = 1 1 ln | cos u| + C = ln | cos(e−2x )| + C 2 2 1 ln | sin 3x| + C 3 28. u = √ 27. 1 x, du = √ dx; 2x 29. u = tan x, u2 du = 1 ln | sec 2x + tan 2x| + C 2 2 sec u du = 2 ln | sec u + tan u| + C = 2 ln | sec √ √ x + tan x| + C 1 tan3 x + C 3 30. tan5 x(1 + tan2 x) sec2 x dx = (tan5 x + tan7 x) sec2 x dx = 31. tan3 4x(1 + tan2 4x) sec2 4x dx = 32. tan4 θ(1 + tan2 θ) sec2 θ dθ = 33. sec4 x(sec2 x − 1) sec x tan x dx = 34. (sec2 θ − 1)2 sec θ tan θdθ = 35. (sec2 x − 1)2 sec x dx = 1 1 tan6 x + tan8 x + C 6 8 (tan3 4x + tan5 4x) sec2 4x dx = 1 1 tan4 4x + tan6 4x + C 16 24 1 1 tan5 θ + tan7 θ + C 5 7 (sec6 x − sec4 x) sec x tan x dx = (sec4 θ − 2 sec2 θ + 1) sec θ tan θdθ = (sec5 x − 2 sec3 x + sec x)dx = 1 1 sec7 x − sec5 x + C 7 5 1 2 sec5 θ − sec3 θ + sec θ + C 5 3 sec5 x dx − 2 sec3 x dx + = = 1 51 1 sec3 x tan x − sec x tan x + ln | sec x + tan x| + ln | sec x + tan x| + C 4 42 2 = 36. 1 3 sec3 x tan x + 4 4 3 5 1 sec3 x tan x − sec x tan x + ln | sec x + tan x| + C 4 8 8 sec3 x dx − 2 [sec2 (x/2) − 1] sec3 (x/2)dx = =2 =2 sec5 u du − sec3 x dx + ln | sec x + tan x| [sec5 (x/2) − sec3 (x/2)]dx sec3 u du 3 1 sec3 u tan u + 4 4 sec3 u du − (u = x/2) sec3 u du = 1 1 sec3 u tan u − 2 2 = 1 1 1 sec3 u tan u − sec u tan u − ln | sec u + tan u| + C 2 4 4 = x1 x x1 x x 1 x sec3 tan − sec tan − ln sec + tan +C 2 2 24 2 24 2 2 (equation (20)) sec3 u du (equation (20), (22)) sec x dx Exercise Set 9.3 306 37. sec2 2t(sec 2t tan 2t)dt = 39. sec4 x dx = 1 sec3 2t + C 6 sec4 x(sec x tan x)dx = 38. (1 + tan2 x) sec2 x dx = 1 sec5 x + C 5 (sec2 x + tan2 x sec2 x)dx = tan x + 1 tan3 x + C 3 40. Using equation (20), sec5 x dx = = 1 3 sec3 x tan x + 4 4 sec3 x dx 3 3 1 sec3 x tan x + sec x tan x + ln | sec x + tan x| + C 4 8 8 tan4 x dx = 41. Use equation (19) to get 1 tan3 x − tan x + x + C 3 42. u = 4x, use equation (19) to get 1 4 tan3 u du = √ 43. 11 1 1 tan2 u + ln | cos u| + C = tan2 4x + ln | cos 4x| + C 42 8 4 tan x(1 + tan2 x) sec2 x dx = 2 sec3/2 x + C 3 sec1/2 x(sec x tan x)dx = 44. π /6 (sec2 2x − 1)dx = 45. 0 1 tan 2x − x 2 π /6 sec2 θ(sec θ tan θ)dθ = 46. 0 2 2 tan3/2 x + tan7/2 x + C 3 7 1 sec3 θ 3 π /6 = √ 3/2 − π/6 0 π /6 √ √ = (1/3)(2/ 3)3 − 1/3 = 8 3/27 − 1/3 0 47. u = x/2, π /4 tan5 u du = 2 0 48. u = πx, 1 π 1 tan4 u − tan2 u − 2 ln | cos u| 2 π /4 sec u tan u du = 0 1 sec u π π /4 π /4 √ = 1/2 − 1 − 2 ln(1/ 2) = −1/2 + ln 2 0 √ = ( 2 − 1)/π 0 1 1 (csc4 x − csc2 x)(csc x cot x)dx = − csc5 x + csc3 x + C 5 3 49. (csc2 x − 1) csc2 x(csc x cot x)dx = 50. 1 cos2 3t · dt = sin2 3t cos 3t 51. (csc2 x − 1) cot x dx = 52. 1 (cot2 x + 1) csc2 x dx = − cot3 x − cot x + C 3 1 csc 3t cot 3t dt = − csc 3t + C 3 csc x(csc x cot x)dx − 1 cos x dx = − csc2 x − ln | sin x| + C sin x 2...
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