# Thread:

1. ## Trigonometric Integrals

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

2. $<br /> \sin ^2 x\cos ^2 x = \frac{1}<br /> {4}\sin ^2 2x = \frac{1}<br /> {4}\left( {\frac{{1 - \cos 4x}}<br /> {2}} \right) = \frac{{1 - \cos 4x}}<br /> {8}$

$<br /> <br /> ^* \sin 2x = 2\sin x\cos x<br /> <br /> <br />$

3. 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.

4. Is that an identity? That [1-cos(2x)^2] = sin(2x)^2 ? I'm not questioning you, I'm just making sure I understand what you did. Thanks

5. Yes, that is an identity. Try deriving it.

6. Originally Posted by thejabronisayz
Is that an identity? That [1-cos(2x)^2] = sin(2x)^2 ? I'm not questioning you, I'm just making sure I understand what you did. Thanks
You know that $1 - cos^2(y) = sin^2(y)$ right? Well, let y = 2x ...

-Dan
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