$\displaystyle \int sin^{4}x$
so what I did was trying to cut it up
$\displaystyle \int sin^{2}x(1-cos^{2}x)$
eventually I realized this method was wrong because I got a big fat zero
so any ideas???
try the identity
$\displaystyle \sin^2(x)=\frac{1}{2}(1-\cos(2x))$
$\displaystyle \int (sin^2(x))^2dx=\int (\frac{1}{2}(1-\cos(2x)))^2dx$
$\displaystyle \frac{1}{4}\int 1-2\cos(2x)+\cos^2(2x)dx$
Now use the cos version of the above
$\displaystyle \cos^{2}(2x)=\frac{1}{2}(1+\cos(4x))$
Then just integrate.
Good luck.