TITLE: Finding the antiderivative of $\sin^6x\cos^2x$
QUESTION [5 upvotes]: I need to find the antiderivative of 
$$\int\sin^6x\cos^2x \mathrm{d}x.$$ I tried symbolizing $u$ as squared $\sin$ or $\cos$ but that doesn't work. Also I tried using the identity of $1-\cos^2 x = \sin^2 x$ and again if I symbolize $t = \sin^2 x$ I'm stuck with its derivative in the $dt$.
Can I be given a hint?

REPLY [3 votes]: Just adding a good point, however, you got the answer completely $\ddot\smile$ :

Consider $$\int\sin^m(x)\cos^n(x)dx$$ where in $m,n\in\mathbb Q$. Whenever $m+n$ is an even integer, you can use $t=\tan(x)$ or $t=\cos(x)$ as a good substitution.

And here $m+n=8$ is an even integer.