The integral: $\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$ The integral: $$\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$$
has been encountered today while solving a longlish problem at MSE. The question here is: How would one evaluate it?
Addendum
For an interesting use of this integral see my Answer to: Evaluate: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$

Define $u=2\cos^{2}{(x)}-1$, then
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}}{\frac{\sin{(x)}\cos^{5}{(x)}}{(1-2\sin^{2}{(x)}\cos^{2}{(x)})^{2}}dx}&=\frac{1}{4}\left(\int_{-1}^{1}{\frac{du}{u^{2}+1}}+\int_{-1}^{1}{\frac{2u\ du}{(u^{2}+1)^{2}}}\right)\\
\\
&=\frac{\pi}{8}
\end{aligned}
$$
