1. Integral using Trig Substitution

I'm trying to integrate (1-x^2)^(3/2), however I'm finding it prety difficult. What's the trick? I'm pretty sure you could let x=sint so dx=cost(dt) but I'm not sure if you're allowed to make (1-x^2)^(3/2)=cost and then use the double angle formulas. All help is appreciated!

$(\cos^2{t})^{3/2}=\cos^{6/2}{t}=\cos^3{t}$
Use the substitution x= cos(t) so that, as dwsmith says, $(1- x^2)^{3/2}= (1- cos^2(t))^{3/2}= (sin^2(t))^{3/2}= sin^3(t)$. And, of course, $dx= d(cos(t))= -sin(t)dt$.