Finding the antiderivative of $\sin^6x\cos^2x$ 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?
 A: 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 $+C$. 
A: $$\int \sin^6x\cos^2xdx=\int \sin^6x(1-\sin^2x)dx=\int \sin^6xdx-\int \sin^8xdx$$
$$=I_6-I_8 \text{ where }I_n=\int\sin^nxdx$$
$$\text{Now, }I_{n+2}=\int\sin^{n+2}xdx=\int\sin^{n+1}x\cdot \sin xdx$$
$$=\sin^{n+1}x\int \sin xdx-\int\left(\frac{d \sin^{n+1}x}{dx}  \int \sin xdx\right)dx$$ (using Integration by parts)
$$=\sin^{n+1}x(-\cos x)-\int(n+1) \sin^nx\cos x(-\cos x)dx$$
$$=-\sin^{n+1}x\cos x+(n+1)\int \sin^nx(1-\sin^2x)dx$$
$$=-\sin^{n+1}x\cos x+(n+1)(I_n-I_{n+2})$$
$$\implies I_{n+2}=-\frac{\sin^{n+1}x\cos x}{(n+2)}+\frac{n+1}{n+2}I_n$$
Now, $I_0=\int \sin^0xdx=\int dx=x$
Put $n=0,2,4,6$ respectively to get the values of $I_2,I_4,I_6,I_8$
A: 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.
A: $$\text{ As }\cos2y=2\cos^2y-1=1-2\sin^2y$$
$$\sin^6x\cos^2x=\left(\frac{1-\cos2x}2\right)^3\left(\frac{1+\cos2x}2\right)$$
$$16\sin^6x\cos^2x=(1-3\cos2x+3\cos^2x-\cos^32x)(1+\cos2x)$$
$$=\left(1-3\cos2x+3\frac{(1+\cos4x)}2-\frac{(\cos6x+3\cos2x)}4\right)(1+\cos2x)$$ (applying  $\cos3y=4\cos^3y-3\cos y$)
$$64\sin^6x\cos^2x=(10-15\cos2x+6\cos4x-\cos6x)(1+\cos2x)$$
$$=10-15\cos2x+6\cos4x-\cos6x+10\cos2x-15\cos^22x+6\cos4x\cos2x-\cos6x\cos2x$$
$$=10-5\cos2x+6\cos4x-\cos6x+10\cos2x-15\frac{(1+\cos4x)}2+6\frac{(\cos2x+\cos6x)}2-\frac{(\cos4x+\cos8x)}2$$ (applying $2\cos A\cos B=\cos(A-B)+\cos(A+B)$)
$$\text{ So, }128\sin^6x\cos^2x=5-4\cos2x-4\cos4x+4\cos6x-\cos8x$$

Alternatively 
as we know, $e^{ix}=\cos x+i\sin x,e^{-ix}=\cos x-i\sin x\implies \cos x=\frac{e^{ix}+e^{-ix}}2,\sin x=\frac{e^{ix}-e^{-ix}}{2i}$
So, $$\sin^6x\cos^2x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^6\left(\frac{e^{ix}+e^{-ix}}2\right)^2$$
$$=\frac{\left(e^{6ix}+e^{-6ix}-\binom61(e^{4ix}+e^{-4ix})+\binom62(e^{2ix}+e^{-2ix})-\binom63\right)}{-2^6}$$
$$\cdot\frac{\left(e^{2ix}+e^{-2ix}+2\right)}{2^2}$$
$$=\frac{e^{8ix}+e^{-8ix}-(6-2)(e^{6ix}+e^{-6ix})+(1+\binom62-2\cdot\binom61)(e^{4ix}+e^{-4ix})-(\binom63-2\cdot\binom62+\binom61)(e^{2ix}+e^{-2ix})+2\binom62-2\binom63}{-2^8}$$
$$=\frac{2\cos8x-4\cdot2\cos6x+4\cdot2\cos4x+4\cdot2\cos2x-10}{-256}\text{ as }e^{ix}+e^{-ix}=2\cos x$$
Now, simplify and use $\int\cos mxdx=\frac{\sin mx}m+C$
A: If you only want a hint that will be simple:
Whenever we have even powers of sin and cos multiplied then we must convert the integral into higher angles.
