<s>[INST] How to evaluate this expression: $∫\frac{\sin2x}{(1 - \cos2x)^4}\:\mathrm{d}x$? 
How to evaluate: $$∫\frac{\sin2x}{(1 - \cos2x)^4}\:\mathrm{d}x$$ If anyone knows the answer, please help.

 [/INST] Use $1-\cos 2x=2\sin^2 x$ and $\sin 2x=2\sin x\cos x$, then u will get 
$$\int \frac{\sin 2x}{(1-\cos 2x)^4}dx=\int \frac{\cos x}{2^3 \sin ^7x }dx$$
Now letting $u=\sin x\Rightarrow du=\cos x dx$
$$\int \frac{\cos x}{2^3 \sin ^7x }dx= \frac{1}{2^3}\int\frac{du}{u^7}=\frac{1}{2^3}.\frac{-1}{6u^6}+C=\frac{1}{2^3}.\frac{-1}{6\sin^6x}+C=\frac{-1}{2^4.3\sin^6x}+C$$
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