<s>[INST] 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: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$
 [/INST] Amazing : the tangent half angle substitution works. Let $x=2\tan^{-1}(t)$ to get
$$I=-4\int\frac{ t \left(t^2-1\right)^5 \left(t^2+1\right)}{\left(t^8-4 t^6+22 t^4-4
   t^2+1\right)^2}\,dt$$  Now, a non-trivial substitution
$$t=\sqrt{1+\frac{2 \left(\sqrt{z+1}\right)}{z}}\implies dt=-\frac{1}{2} \sqrt{\frac{z+ 2 \left(\sqrt{z+1}\right)}{z^3(z+1)}}\,dz$$
$$I=\frac{1}{2} \int \frac{dz}{ \left(z^2+1\right)^2}=\frac{1}{4} \left(\frac{z}{z^2+1}+\tan ^{-1}(z)\right)$$
Back to $t$
$$I=\frac{t^2 \left(t^2-1\right)^2}{t^8-4 t^6+22 t^4-4 t^2+1}+\frac{1}{4} \tan ^{-1}\left(\frac{4 t^2}{\left(1-t^2\right)^2}\right)$$ 
Expanded as a series around $t=1$ gives for the definite integral
$$\frac{\pi }{8}-\frac{1}{6} (1-t)^6+O\left((1-t)^7\right)$$
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