$\displaystyle \int sin^4(2x)dx$
I don't know what theorem to use or if to separate it...
I just need a reference point for where to go.
Use $\displaystyle \sin^2\theta=\frac{1-\cos2\theta}2$ and $\displaystyle \cos^2\theta=\frac{1+\cos2\theta}2\colon$
$\displaystyle \int\sin^42x\,dx$
$\displaystyle =\int\left(\sin^22x\right)^2\,dx$
$\displaystyle =\int\left(\frac{1-\cos4x}2\right)^2\,dx$
$\displaystyle =\frac14\int\left(1-2\cos4x+\cos^24x\right)\,dx$
$\displaystyle =\frac14\left[\int dx-2\int\cos4x\,dx+\int\cos^24x\,dx\right]$
$\displaystyle =\frac14\left[\int dx-2\int\cos4x\,dx+\frac12\int(1+\cos8x)\,dx\right]$
The rest should be easy.