# 1 x tan1 2 2 11 1 3x e 3 12 1 lim ln3 2ex 2 lim 3 1

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Unformatted text preview: 307 Chapter 9 2π 53. (a) sin mx cos nx dx = 0 2π 1 2 [sin(m+n)x+sin(m−n)x]dx = − 0 2π 2π 0 2π = 0, cos(m − n)x but cos(m + n)x cos(m + n)x cos(m − n)x − 2(m + n) 2(m − n) 0 = 0. 0 2π (b) 1 2π [cos(m + n)x + cos(m − n)x]dx; 20 0 since m = n, evaluate sin at integer multiples of 2π to get 0. cos mx cos nx dx = 2π (c) 1 2π [cos(m − n)x − cos(m + n)x] dx; 20 0 since m = n, evaluate sin at integer multiples of 2π to get 0. sin mx sin nx dx = 55. y = tan x, 1 + (y )2 = 1 + tan2 x = sec2 x, π /4 L= √ π /4 π /4 sec x dx = ln | sec x + tan x| sec2 x dx = 0 0 0 π /4 π /4 π /4 (1 − tan2 x)dx = π 56. V = π (2 − sec2 x)dx = π (2x − tan x) 0 0 π /4 (cos x − sin x)dx = π 2 2 57. V = π 0 0 π sin2 x dx = 58. V = π 0 π 2 60. (a) D = 1 cos 2x dx = π sin 2x 2 π (1 − cos 2x)dx = 0 59. With 0 &lt; α &lt; β, D = Dβ −Dα = L 2π π 2 x− β sec x dx = α 100 ln(sec 25◦ + tan 25◦ ) = 7.18 cm 2π csc x dx = = 0 π /4 61. (a) √ = ln( 2 + 1) 1 sin 2x 2 π /4 = π/2 0 π = π 2 /2 0 L ln | sec x + tan x| 2π (b) D = 1 π (π − 2) 2 β = α sec β + tan β L ln 2π sec α + tan α sec 50◦ + tan 50◦ 100 ln = 7.34 cm 2π sec 30◦ + tan 30◦ sec(π/2 − x)dx = − ln | sec(π/2 − x) + tan(π/2 − x)| + C = − ln | csc x + cot x| + C (b) − ln | csc x + cot x| = ln | csc x − cot x| 1 = ln | csc x − cot x|, = ln | csc x + cot x| | csc2 x − cot2 x| − ln | csc x + cot x| = − ln = ln cos x sin x 1 + = ln sin x sin x 1 + cos x 2 sin(x/2) cos(x/2) = ln | tan(x/2)| 2 cos2 (x/2) √ √ √ 2[(1/ 2) sin x + (1/ 2) cos x] √ √ = 2[sin x cos(π/4) + cos x sin(π/4)] = 2 sin(x + π/4), 62. sin x + cos x = 1 dx =√ sin x + cos x 2 1 csc(x + π/4)dx = − √ ln | csc(x + π/4) + cot(x + π/4)| + C 2 √ 1 2 + cos x − sin x +C = − √ ln sin x + cos x 2 Exercise Set 9.4 308 b a sin x + √ cos x = a2 + b2 (sin x cos θ + cos x sin θ) 2 2 + b2 +b a √ √ √ where cos θ = a/ a2 + b2 and sin θ = b/ a2 + b2 so a sin x + b cos x = a2 + b2 sin(x + θ) a2 + b2 √ 63. a sin x + b cos x = and a2 1 dx =√ 2 + b2 a sin x + b cos x a csc(x + θ)dx = − √ 1 ln = −√ a2 + b2 π /2 sinn x dx = − 64. (a) 0 1 sinn−1 x cos x n √ 1 ln | csc(x + θ) + cot(x + θ)| + C + b2 a2 a2 + b2 + a cos x − b sin x +C a sin x + b cos x π /2 + 0 n−1 n π /2 sinn−2 x dx = 0 n−1 n π /2 sinn−2 x dx 0 (b) By repeated application of the formula in part (a) π /2 sinn x dx = 0 = π /2 n−3 n−1 sinn−4 x dx n n−2 0 n−1 n−3 n−5 1 ··· n n−2 n−4 2 n−1 n n−3 n−2 n−5 n−4 2 3 ··· π /2 dx, n even 0 π /2 sin x dx, n odd 0 1 · 3 · 5 · · · (n − 1) · π , n even 2 · 4 · 6···n 2 = 2 · 4 · 6 · · · (n − 1) , n odd 3 · 5 · 7···n π /2 sin3 x dx = 65. (a) 0 π /2 sin5 x dx = (c) 0 sin4 x dx = 1·3 π · = 3π/16 2·4 2 sin6 x dx = 1·3·5 π · = 5π/32 2·4·6 2 π /2 2 3 (b) 2·4 = 8/15 3·5 (d) 0 π /2 0 66. Similar to proof in Exercise 64. EXERCISE SET 9.4 1. x = 2 sin θ, dx = 2 cos θ dθ, 4 cos2 θ dθ = 2 (1 + cos 2θ)dθ = 2θ + sin 2θ + C 1 = 2θ + 2 sin θ cos θ + C = 2 sin−1 (x/2) + x 4 − x2 + C 2 2. x = 1 2 1 1 sin θ, dx = cos θ dθ, 2 2 cos2 θ dθ = = 1 4 (1 + cos 2θ)dθ = 1 1 θ + sin 2θ + C 4 8 1 1 1 1 θ + sin θ cos θ + C = sin−1 2x + x 1 − 4x2 + C 4 4 4 2 309 Chapter 9 3. x = 3 sin θ, dx = 3 cos θ dθ, sin2 θ dθ = 9 = 9 2 (1 − cos 2θ)dθ = 9 9 9 9 θ − sin 2θ + C = θ − sin θ cos θ + C 2 4 2 2 1 9 sin−1 (x/3) − x 9 − x2 + C 2 2 4. x = 4 sin θ, dx = 4 cos θ dθ, √ 1 16 − x2 csc θ dθ = − cot θ + C = − +C 16 16x 1 1 2 dθ = 16 sin θ 1 16 2 5. x = 2 tan θ, dx = 2 sec2 θ dθ, 1 1 dθ = 2θ sec 8 1 8 = √ 6. x = cos2 θ dθ = (1 + cos 2θ)dθ = 1 1 θ+ sin 2θ + C 16 32 1 1 x 1 x θ+ sin θ cos θ + C = tan−1 + +C 16 16 16 2 8(4 + x2 ) 5 tan θ, dx = √ 5 sec2 θ dθ, = 1 1 sec θ tan θ − ln | sec θ + tan θ| + C1 2 2 (sec3 θ − sec θ)dθ = 5 tan2 θ sec θ dθ = 5 5 1 16 1 5 x 5 + x2 − ln 2 2 √ 1 5 5 + x2 + x √ + C1 = x 5 + x2 − ln( 5 + x2 + x) + C 2 2 5 7. x = 3 sec θ, dx = 3 sec θ tan θ dθ, tan2 θ dθ = 3 3 (sec2 θ − 1)dθ = 3 tan θ − 3θ + C = x2 − 9 − 3 sec−1 x +C 3 8. x = 4 sec θ, dx = 4 sec θ tan θ dθ, 1 1 dθ = sec θ 16 1 16 9. x = √ 2 sin θ, dx = 1 cos θ dθ = sin θ + C = 16 x2 − 16 +C 16x √ 2 cos θ dθ, √ sin3 θ dθ = 2 2 √ 22 √ 1 − cos2 θ sin θ dθ √ 1 1 = 2 2 − cos θ + cos3 θ + C = −2 2 − x2 + (2 − x2 )3/2 + C 3 3 10. x = √ √ 25 5 5 sin θ, dx = √ 5 cos θ dθ, √ 1 1 5 1 sin3 θ cos2 θ dθ = 25 5 − cos3 θ + cos5 θ + C = − (5 − x2 )3/2 + (5 − x2 )5/2 + C 3 5 3 5 3 2 3 11. x = sec θ, dx = sec θ tan θ dθ, 2 2 9 2 1 dθ = sec θ 9 2 cos θ dθ = sin θ + C = 9 12. t = tan θ, dt = sec2 θ dθ, sec3 θ dθ = tan θ tan2 θ + 1 sec θ dθ = tan θ (sec θ tan θ + csc θ)dθ = sec θ + ln | csc θ − cot θ| + C = √ 1 + t2 + ln 1 + t2 − 1 +C |t| √ 4x2 − 9 +C 9x Exercise Set...
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