# Trigonometric Integrals

• Feb 16th 2008, 12:33 PM
thejabronisayz
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
• Feb 16th 2008, 12:41 PM
Peritus
$
\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

$
• Feb 16th 2008, 12:42 PM
galactus
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.
• Feb 16th 2008, 02:32 PM
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
• Feb 16th 2008, 02:39 PM
galactus
Yes, that is an identity. Try deriving it.
• Feb 16th 2008, 05:18 PM
topsquark
Quote:

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