Here it is

$\displaystyle \int_{0}^{\frac{\pi}{4}} \sin^3(2x)\cos^2(2x)\ where\ the\ sustitution\ is\ u = \cos(2x)$

$\displaystyle \frac{du}{dx} = -\frac{1}{2}\sin(2x)$

which means

$\displaystyle -2\sin^2(2x)u^2.du$

I get stuck here

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- Jan 2nd 2009, 01:34 AMBeardIntegration by substitution MkII
Here it is

$\displaystyle \int_{0}^{\frac{\pi}{4}} \sin^3(2x)\cos^2(2x)\ where\ the\ sustitution\ is\ u = \cos(2x)$

$\displaystyle \frac{du}{dx} = -\frac{1}{2}\sin(2x)$

which means

$\displaystyle -2\sin^2(2x)u^2.du$

I get stuck here - Jan 2nd 2009, 01:44 AMflyingsquirrel
- Jan 2nd 2009, 02:09 AMtom@ballooncalculus
Just in case it appeals...

http://www.ballooncalculus.org/cosSub.png

This is only an overview - the lower equality relies on the pythag identity.

Balloon Calculus: worked examples from past papers