1. Help (trig intergration)

$\int$ (o to pie/2) $sin^2xcos^2x dx$

please if your gonna help dont jump steps like crazy. cause i probably wont understand..

this problem i believe u have to use $sin^2x = \frac{1-cos2x}{2}$ or $cos^2x= \frac{1+cos2x}{2}$

2. Hi

Using what you've written :

$
I=\int_0^{\frac{\pi}{2}}\sin^2x\cos^2x\,\mathrm{d} x=
\int_0^{\frac{\pi}{2}}\frac{1-\cos (2x)}{2}\cdot \frac{1+\cos (2x)}{2}\,\mathrm{d}x = \int_0^{\frac{\pi}{2}}\frac{1-\cos^2(2x)}{4}\,\mathrm{d}x$
because $(a-b)(a+b)=a^2-b^2$
$I=\frac{1}{4}\int_0^{\frac{\pi}{2}}1-\cos^2(2x)\,\mathrm{d}x$

I let you take it from here : you have to find a way of getting rid off the remaining $\cos^2$.

Hope that helps.

3. Hello, Legendsn3verdie!

$\int^{\frac{\pi}{2}}_0\sin^2\!x\cos^2\!x\,dx$
I think it's easier if we use: . $2\sin\theta \cos\theta \;=\;\sin2\theta\quad\text{ and }\quad\sin^2\!\theta \:=\:\frac{1-\cos2\theta}{2}
$

We have: . $\sin^2\!x\cos^2\!x \;=\;\frac{1}{4}(4\sin^2\!x\cos^2\!x) \;=\;\frac{1}{4}(2\sin x\cos x)^2 \;=\;\frac{1}{4}(\sin2x)^2$

. . $= \;\frac{1}{4}\sin^2\!2x \;=\;\frac{1}{4}\left(\frac{1-\cos4x}{2}\right) \;=\;\frac{1}{8}(1 - \cos4x)$

Then we have: . $\frac{1}{8}\int^{\frac{\pi}{2}}_0(1 - \cos4x)\,dx \quad\hdots\quad \text{okay?}$