Integration of trigs

**Misa-Campo** · July 30th 2008, 04:53 AM

sorry if you cant see but its between pie/2 and 0, thanks.

**mr fantastic** · July 30th 2008, 04:58 AM

Originally Posted by Misa-Campo

sorry if you cant see but its between pie/2 and 0, thanks.

Double angle formula: $\cos(2 \theta) = 1 - 2 \sin^2 \theta$ .

In your problem, $\theta = 2x \Rightarrow \sin^2 (2x) = \, ......$ .

**Soroban** · July 30th 2008, 09:14 AM

Hello, Misa-Campo!

$\int^{\frac{\pi}{2}}_0 \sin^2(2x)\,dx$

Use the identity: . $\sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}$

And we have: . $\int^{\frac{\pi}{2}}_0 \sin^2(2x)\,dx \;=\;\frac{1}{2}\int^{\frac{\pi}{2}}_0(1 - \cos 4x)\,dx$

. . Got it?

**Misa-Campo** · August 1st 2008, 11:34 PM

Thx alot guys

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