Integration by substitution MkII

**Beard** · January 2nd 2009, 01:34 AM

Here it is

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

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

which means

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

I get stuck here

**flyingsquirrel** · January 2nd 2009, 01:44 AM

Hello,

Originally Posted by Beard

which means

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

I get stuck here

Now you have to find an expression of $\sin^2(2x)$ in terms of $u$ (recall that $\sin^2\theta=1-\cos^2\theta$ ).

**tom@ballooncalculus** · January 2nd 2009, 02:09 AM

Just in case it appeals...

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

Balloon Calculus: worked examples from past papers

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