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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online