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

Let $u= \cos x$, $du = - \sin x \ dx$:
$$\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$$
$$= -\int_{1}^{0} \frac{du \ (u^5)}{(1-2(1-u^2)u^2)^2}$$
$$= \int_{0}^{1} \frac{u^5}{(1-2u^2+2u^4)^2} \ du$$
Then substitute again: let $v = u^2, dv = 2u\ du$:
$$= \frac{1}{2} \int_{0}^{1} \frac{v^2 \ dv}{(1-2v+2v^2)^2}$$
and do (not really) partial fractions:
$$ \frac{v^2 }{(1-2v+2v^2)^2} = \frac{a}{(1-2v+2v^2)} + \frac{b}{(1-2v+2v^2)^2}$$
$$ v^2= a(1-2v+2v^2) + b$$
$$a = \frac{1}{2} \Rightarrow v^2 = \frac{1}{2}-v+v^2+b$$
$$b = v - \frac{1}{2}$$
So we have:
$$\frac{1}{4} \int_{0}^{1} \frac{dv}{(1-2v+2v^2)} + \frac{1}{2} \int_{0}^{1} \frac{v \ dv}{(1-2v+2v^2)^2} - \frac{1}{2} \int_{0}^{1} \frac{\ dv}{(1-2v+2v^2)^2}$$
For the first integral, complete the square which resolves to the standard $\arctan$ integral. For the second integral, let $w = 2v - 1$ (AoPS), and for the third integral, complete the square, then substitute $w = \frac{\tan v}{\sqrt 2}$ where you can use $\tan^2 w + 1 = \sec^2 w$.
