q7-sol-1042-sp08

q7-sol-1042-sp08 - = Z 4 sin 2 4 cos 2 2 cos d , (As 4-t 2...

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Calculus II—1042 Solution of quiz # 7 Fri., 03/17/08 1. Z 2 2 0 1 (1 + x 2 ) 3 2 dx . Solution: a = 1 , x = tan θ, dx = sec 2 θ dθ Z 2 2 0 1 (1 + x 2 ) 3 / 2 dx = Z 2 2 0 1 ( sec 2 θ ) 3 / 2 · sec 2 θ dθ , (As 1 + x 2 = 1 + tan 2 θ = sec 2 θ ) = Z 2 2 0 sec 2 θ sec 3 θ = Z 2 2 0 1 sec θ = Z 2 2 0 cos θ dθ = sin θ (Write back using tan θ = x 1 , θ 1+x^2 x 1 and sin θ = x 1 + x 2 ) = x 1 + x 2 | 2 2 0 (Now plug in integral limits) = 2 2 1 + 8 - 0 1 + 0 = 2 2 3 2. Z 1 4 x 2 - 9 dx . Solution: Re-write it as Z 1 4 x 2 - 9 dx = Z 1 p (2 x ) 2 - 9 dx . a = 3 , 2 x = 3 sec θ, 2 dx = 3 sec θ tan θ dθ dx = 3 2 sec θ tan θ dθ Z 1 4 x 2 - 9 dx = 3 2 Z 1 9 tan 2 θ · sec θ tan θ dθ , (As 4 x 2 - 9 = 9 sec 2 θ - 9 = 9(sec 2 θ - 1) = 9 tan 2 θ ) = 3 2 Z sec θ tan θ 3 tan θ = 1 2 Z sec θ dθ = ln | sec θ + tan θ | + C (Write back using sec θ = 2 x 3 , and θ 2x 3 4x^2-9 and tan θ = 4 x 2 - 9 3 ) = ln | 2 x 3 + 4 x 2 - 9 3 | + C 3. Z t 2 4 - t 2 dt . Solution: a = 2 , t = 2 sin θ, dt = 2 cos θ dθ Z t 2 4 - t 2 dt
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Unformatted text preview: = Z 4 sin 2 4 cos 2 2 cos d , (As 4-t 2 = 4-4 sin 2 = 4(1-sin 2 ) = 4 cos 2 ) = Z 8 sin 2 cos 2 cos d = 4 Z sin 2 d (The power of sine is EVEN, so break the even power) = 4 Z ( 1-cos 2 2 d = Z (2-2 cos 2 ) d = 2 -2 2 sin 2 + C = 2 -2 sin cos + C (Write back again using sin = t 2 , t 4-t^2 4 and cos = 4-t 2 2 , and = sin-1 t 2 ) = 2 sin-1 t 2 + t 2 1-t 2 2 + C...
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This homework help was uploaded on 04/18/2008 for the course MATH 1042 taught by Professor Dr.z during the Spring '08 term at Temple.

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