1. Integrate sin^2(x)*cos^2(x)

Hi can someone please help me integrate sin^2(x)*cos^2(x). It has something to do with trig identities (obviously). I tried using the double angle identities but just made more of a mess. Even if someone could show me what how to manipulate the trig stuff into something easier to integrate and then I will have a crack at the integration myself.

2. $\cos^{2}x \sin^{2}x = \Big(\frac {1}{2} + \frac {\cos2x}{2}\Big)\Big(\frac {1}{2} - \frac {\cos2x}{2} \Big) = \frac {1}{4} - \frac {1}{4} \cos^{2} 2x = \frac {1}{4} - \Big(\frac {1}{8} - \frac{1}{8} \cos 4x \Big)$ $= \frac {1}{8} - \frac{1}{8} \cos 4x$

3. $sin^2 (x)cos^2(x)=(sin x cos x)^2$

= $(\frac{1}{2}sin (2x))^2$

= $\frac{1}{4}sin^2(2x)$

= $\frac{1}{4}\frac{1-cos (4x)}{2}$

= $\frac{1}{8}(1-cos (4x))$

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

= $\frac {1}{8}(x-\frac{sin (4 x)}{4})$

= $\frac{1}{8}\frac{4x-sin (4x)}{4}$

= $\frac{1}{32} (4x-sin (4x))$

You can verify the answer using Wolfram Mathematica Online Integrator

4. Originally Posted by alexmahone
$sin^2 (x)cos^2(x)=(sin x cos x)^2$

= $(\frac{1}{2}sin (2x))^2$

= $\frac{1}{4}sin^2(2x)$

= $\frac{1}{4}\frac{1-cos (4x)}{2}$

= $\frac{1}{8}(1-cos (4x))$

-------------------------------------------------------------------

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

= $\frac {1}{8}(x-\frac{sin (4 x)}{4})$

= $\frac{1}{8}\frac{4x-sin (4x)}{4}$

= $\frac{1}{32} (4x-sin (4x))$

You can verify the answer using Wolfram Mathematica Online Integrator
Thank you very much for the time you put into that reply. It really makes sense now.