TITLE: Show that $\int_{0}^{\pi/2}\frac {\log^2\sin x\log^2\cos x}{\cos x\sin x}\mathrm{d}x=\frac14\left( 2\zeta (5)-\zeta(2)\zeta (3)\right)$
QUESTION [69 upvotes]: Show that :
  $$
\int_{0}^{\Large\frac\pi2}
{\ln^{2}\left(\vphantom{\large A}\cos\left(x\right)\right)
\ln^{2}\left(\vphantom{\large A}\sin\left(x\right)\right)
\over
\cos\left(x\right)\sin\left(x\right)}\,{\rm d}x
={1 \over 4}\,
\bigg[2\,\zeta\left(5\right) - \zeta\left(2\right)\zeta\left(3\right) \bigg]
$$

I can only do non squared one. Anyone has a clue?

REPLY [4 votes]: \begin{align}
I&=\int_0^{\pi/2}\frac{\ln^2\cos x\ln^2\sin x}{\cos x\sin x}\ dx\overset{\sin x=u}{=}\frac1{4}\int_0^1\frac{\ln^2(1-x^2)\ln^2x}{x(1-x^2)}\ dx\\
&=\frac1{32}\int_0^1\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx\\
&=\frac1{32}\int_0^1\frac{\ln^2(1-x)\ln^2x}{x}\ dx+\frac1{32}\underbrace{\int_0^1\frac{\ln^2(1-x)\ln^2x}{1-x}\ dx}_{\large{1-x\ \mapsto\  x}}\\
&=\frac1{16}\int_0^1\frac{\ln^2(1-x)\ln^2x}{x}\ dx\\
&=\frac18\sum_{n=1}^{\infty}\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1x^{n-1}\ln^2x\ dx=\frac14\sum_{n=1}^{\infty}\left(\frac{H_n}{n^4}-\frac{1}{n^5}\right)\\
&=\frac14\left(3\zeta(5)-\zeta(2)\zeta(3)-\zeta(5)\right)\\
&=\frac14\left(2\zeta(5)-\zeta(2)\zeta(3)\right)
\end{align}