Π 2cos 3x dx π 2 1 sin 2 x cos x dx sin x 1 3 sin 3

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π/ 2 0 cos 3 x dx = π/ 2 0 (1 sin 2 x ) cos x dx = sin x 1 3 sin 3 x π/ 2 0 = 2 3 18. π/ 2 0 sin 2 ( x/ 2) cos 2 ( x/ 2) dx = 1 4 π/ 2 0 sin 2 x dx = 1 8 π/ 2 0 (1 cos 2 x ) dx = 1 8 x 1 2 sin 2 x π/ 2 0 = π/ 16 19. π/ 3 0 sin 4 3 x cos 3 3 x dx = π/ 3 0 sin 4 3 x (1 sin 2 3 x ) cos 3 x dx = 1 15 sin 5 3 x 1 21 sin 7 3 x π/ 3 0 = 0 20. π π cos 2 5 θ dθ = 1 2 π π (1 + cos 10 θ ) = 1 2 θ + 1 10 sin 10 θ π π = π 21. π/ 6 0 sin 4 x cos 2 x dx = 1 2 π/ 6 0 (sin 2 x + sin 6 x ) dx = 1 4 cos 2 x 1 12 cos 6 x π/ 6 0 = [( 1 / 4)(1 / 2) (1 / 12)( 1)] [ 1 / 4 1 / 12] = 7 / 24 22. 2 π 0 sin 2 kx dx = 1 2 2 π 0 (1 cos 2 kx ) dx = 1 2 x 1 2 k sin 2 kx 2 π 0 = π 1 4 k sin 4 πk ( k = 0) 23. 1 2 tan(2 x 1) + C 24. 1 5 ln | cos 5 x | + C 25. u = e x , du = e x dx ; tan u du = ln | cos u | + C = ln | cos( e x ) | + C 26. 1 3 ln | sin 3 x | + C 27. 1 4 ln | sec 4 x + tan 4 x | + C
January 27, 2005 11:45 L24-CH08 Sheet number 17 Page number 353 black Exercise Set 8.3 353 28. u = x, du = 1 2 x dx ; 2 sec u du = 2 ln | sec u + tan u | + C = 2 ln sec x + tan x + C 29. u = tan x , u 2 du = 1 3 tan 3 x + C 30. tan 5 x (1 + tan 2 x ) sec 2 x dx = (tan 5 x + tan 7 x ) sec 2 x dx = 1 6 tan 6 x + 1 8 tan 8 x + C 31. tan 4 x (1 + tan 2 4 x ) sec 2 4 x dx = (tan 4 x + tan 3 4 x ) sec 2 4 x dx = 1 8 tan 2 4 x + 1 16 tan 4 4 x + C 32. tan 4 θ (1 + tan 2 θ ) sec 2 θ dθ = 1 5 tan 5 θ + 1 7 tan 7 θ + C 33. sec 4 x (sec 2 x 1) sec x tan x dx = (sec 6 x sec 4 x ) sec x tan x dx = 1 7 sec 7 x 1 5 sec 5 x + C 34. (sec 2 θ 1) 2 sec θ tan θdθ = (sec 4 θ 2 sec 2 θ + 1) sec θ tan θdθ = 1 5 sec 5 θ 2 3 sec 3 θ + sec θ + C 35. (sec 2 x 1) 2 sec x dx = (sec 5 x 2 sec 3 x + sec x ) dx = sec 5 x dx 2 sec 3 x dx + sec x dx = 1 4 sec 3 x tan x + 3 4 sec 3 x dx 2 sec 3 x dx + ln | sec x + tan x | = 1 4 sec 3 x tan x 5 4 1 2 sec x tan x + 1 2 ln | sec x + tan x | + ln | sec x + tan x | + C = 1 4 sec 3 x tan x 5 8 sec x tan x + 3 8 ln | sec x + tan x | + C 36. [sec 2 x 1] sec 3 x dx = [sec 5 x sec 3 x ] dx = 1 4 sec 3 x tan x + 3 4 sec 3 x dx sec 3 x dx (equation (20)) = 1 4 sec 3 x tan x 1 4 sec 3 x dx = 1 4 sec 3 x tan x 1 8 sec x tan x 1 8 ln | sec x + tan x | + C (equation (20), (22)) 37. sec 2 t (sec t tan t ) dt = 1 3 sec 3 t + C 38. sec 4 x (sec x tan x ) dx = 1 5 sec 5 x + C 39. sec 4 x dx = (1 + tan 2 x ) sec 2 x dx = (sec 2 x + tan 2 x sec 2 x ) dx = tan x + 1 3 tan 3 x + C 40. Using equation (20), sec 5 x dx = 1 4 sec 3 x tan x + 3 4 sec 3 x dx = 1 4 sec 3 x tan x + 3 8 sec x tan x + 3 8 ln | sec x + tan x | + C
January 27, 2005 11:45 L24-CH08 Sheet number 18 Page number 354 black 354 Chapter 8 41. u = 4 x , use equation (19) to get 1 4 tan 3 u du = 1 4 1 2 tan 2 u + ln | cos u | + C = 1 8 tan 2 4 x + 1 4 ln | cos 4 x | + C 42. Use equation (19) to get tan 4 x dx = 1 3 tan 3 x tan x + x + C 43. tan x (1 + tan 2 x ) sec 2 x dx = 2 3 tan 3 / 2 x + 2 7 tan 7 / 2 x + C 44. sec 1 / 2 x (sec x tan x ) dx = 2 3 sec 3 / 2 x + C 45. π/ 8 0 (sec 2 2 x 1) dx = 1 2 tan 2 x x π/ 8 0 = 1 / 2 π/ 8 46. π/ 6 0 sec 2 2 θ (sec 2 θ tan 2 θ ) = 1 6 sec 3 2 θ π/ 6 0 = (1 / 6)(2) 3 (1 / 6)(1) = 7 / 6 47. u = x/ 2, 2 π/ 4 0 tan 5 u du = 1 2 tan 4 u tan 2 u 2 ln | cos u | π/ 4 0 = 1 / 2 1 2 ln(1 / 2) = 1 / 2 + ln 2 48. u = πx, 1 π π/ 4 0 sec u tan u du = 1 π sec u π/ 4 0 = ( 2 1) 49. (csc 2 x 1) csc 2 x (csc x cot x ) dx = (csc 4 x csc 2 x )(csc x cot x ) dx = 1 5 csc 5 x + 1 3 csc 3 x + C 50. cos 2 3 t sin 2 3 t · 1 cos 3 t dt = csc 3 t cot 3 t dt = 1 3 csc 3 t + C 51. (csc 2 x 1) cot x dx = csc x (csc x cot x ) dx cos x sin x dx = 1 2 csc 2 x ln | sin x | + C 52. (cot 2 x + 1) csc 2 x dx = 1 3 cot 3 x cot x + C 53. (a) 2 π 0 sin mx cos nx dx = 1 2 2 π 0 [sin( m + n ) x + sin( m n ) x ] dx = cos( m + n ) x 2( m + n ) cos( m n ) x 2( m n ) 2 π 0 but cos( m + n ) x 2 π 0 = 0 , cos( m n ) x 2 π 0 = 0.

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