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Calculus II                                                       Quiz #5                                April 6, 2005

Name____________________                 R.  Hammack                            Score ______
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(1)     ∫ x^3/(2 - x^2)^(1/2)dx =


  ∫ (2^(1/2) sin(θ))^3/(2 - 2sin^2(θ))^(1/2)2^(1/2)cos(θ)dθ =
∫ (22^(1/2) sin^3(θ))/(2^(1/2) cos(θ))2^(1/2)cos(θ)dθ = 
 22^(1/2)∫ (sin)^3(θ)dθ =

22^(1/2)∫ (sin)^2(θ)sin(θ)dθ =

22^(1/2)∫(1-cos^2(θ))sin(θ)dθ =

-22^(1/2)∫(1-cos^2(θ))(-sin(θ))dθ =

22^(1/2)(-cos(θ)+cos^3(θ)/3)+C =

-22^(1/2)(2 - ( x^2)/2)^(1/2)+22^(1/2)(2 - x^2/2)^(1/2)^3/3+C



x = 2^(1/2) sin (θ)        dx = 2^(1/2) cos(θ) d& ... ))  cos(θ) = cos(sin^(-1)(x/2^(1/2))) = (2 - x^2)^(1/2)/2^(1/2) = (1 - x^2/2)^(1/2)