calculus - Finding the antiderivative of $sin^6xcos^2x$

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?

Answer

Hint We can use double-angle identities to reduce powers. We could use $\cos 2t=2\cos^2 t-1$ and $\cos 2t=1-2\sin^2 t$. We end up with polynomial of degree $4$ in $\cos 2x$. Repeat the idea where needed.

It is more efficient in this case to use $\sin 2t=2\sin t\cos t$, that is, first rewrite our expression as $(\sin x\cos x)^2\sin^4 x$. Then rewrite as $\frac{1}{16}(\sin^2 2x)(1-\cos 2x)^2$. Expand the square. Not quite finished. But we end up having to integrate $\sin^2 2x$ (standard), $\sin^2 2x\cos 2x$ (simple substitution), and $\sin^2 2x\cos^2 2x$, a close relative of $\sin^2 4x$.

Remark: In this problem, like in a number of trigonometric integrations, it is possible to end up with dramatically different-looking answers. They all differ by constants, so we are saved by the $+C$.