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: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$

$$I=\int_{0}^{\pi/2} \frac{\sin x \cos^5x}{(1-2\sin^2 x \cos^2 x)^2} dx$$
Next using $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx,$
we get $$I=\int_{0}^{\pi/2} \frac{\sin^5 x \cos x}{(1-2\sin^2 x \cos^2 x)^2} dx$$ Adding the  two integrals, we get
So $$2I=\int_{0}^{\pi/2} \frac{\sin x \cos x(\sin^4 x+ \cos^4 x)}{(1-2\sin^2 x \cos^2 x)^2}=\int_{0}^{\pi/2} \frac{\sin x \cos x}{(1-2\sin^2 x \cos^2 x)}dx=\int_{0}^{\pi/2}\frac{1}{2} \frac{\sin 2x dx}{1+\cos^2 2x}$$
$$ \text{domain halved,}~\implies 2 I=\int_{0}^{\pi/4}\frac{2\sin 2x dx}{1+\cos^2 2x}=-\int_{1}^{0}\frac{dt}{1+t^2}=\frac{\pi}{4} \implies I=\frac{\pi} {8} $$
Lastly we have used $\cos 2x=t.$ 
