Trigonometric Integrals

**thejabronisayz** · February 16th 2008, 12:33 PM

Can anyone help me or give me a good idea of how to approach the definite integral ([sinx]^2)*([cosx]^2) with the limits of integration from 0 to pi/2? I'm pretty sure I should somehow be able to simplify it using half-angle identities but I'm stuck. Thanks

**Peritus** · February 16th 2008, 12:41 PM

$ \sin ^2 x\cos ^2 x = \frac{1} {4}\sin ^2 2x = \frac{1} {4}\left( {\frac{{1 - \cos 4x}} {2}} \right) = \frac{{1 - \cos 4x}} {8}$

$ ^* \sin 2x = 2\sin x\cos x $

**galactus** · February 16th 2008, 12:42 PM

Use the identities $sin^{2}(x)=\frac{1-cos(2x)}{2}$ and $cos^{2}(x)=\frac{(1+cos(2x)}{2}$

Then, $\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-cos(2x))(1+cos(2x))dx$

$=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-cos^{2}(2x))dx$ $=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}sin^{2}(2x)dx$ $=\frac{1}{8}\int_{0}^{\frac{\pi}{2}}(1-cos(4x))dx$

Now continue.