# Calculus

posted by Justin

Evaluate ∫ (cos(x))^(1/2)sin(x)dx

Let u = cos(x)?

∫ (u)^(1/2)sin(x)dx = ∫ [2u^(3/2)/3]sin(x)dx
∫ [2cos(x)^(3/2)/3] (-cos(x)) dx?

I thought this involved the FTC, but now I'm thinking that's false.

1. Damon

ok, let u = cos x
then du = -sin x dx

then you have

∫ (u)^(1/2)(-du)

=-(2/3) u^(3/2) + c

2. Damon

of course that is

-(2/3) (cos x)^3/2 + c

Your problem seems to be not replacing dx with a calculated du

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