Show that $\int_{0}^{\pi/2}\frac {\log^2\sin x\log^2\cos x}{\cos x\sin x}\mathrm{d}x=\frac14\left( 2\zeta (5)-\zeta(2)\zeta (3)\right)$ 
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?
 A: \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}
A: Related problems: (I), (II), (III), (IV), (V), (6). Use the change of variables $\ln(\cos(x))=t$ to transform the integral to

$$ I = \int_{0}^{\frac{\pi }{2}}{\frac{{{\ln }^{2}}\cos x{{\ln }^{2}}\sin x}{\cos x\sin x}}\text{d}x = \frac{1}{4}\,\int _{-\infty }^{0}\!{\frac {{t}^{2} \left( \ln  \left( 1-{
{\rm e}^{2\,t}} \right)\right) ^{2}}{1-{{\rm e}^{2t}}}}{dt}.$$

Follow it by another change of variables $ 1-e^{2t}=z $ gives
$$\frac{1}{4}\,\int _{-\infty }^{0}\!{\frac {{t}^{2} \left( \ln  \left( 1-{
{\rm e}^{2\,t}} \right)  \right) ^{2}}{1-  {{\rm e}^{2t}}
 }}{dt}= \frac{1}{32}\,\int _{0}^{1}\!{\frac { \left( \ln  \left( 1-z \right) 
 \right) ^{2} \left( \ln  \left( z \right)  \right) ^{2}}{z \left( 1-
z\right) }}{dz}$$ 
$$= \frac{1}{32}\,\int _{0}^{1}\!{\frac { \left( \ln  \left( 1-z \right) 
 \right) ^{2} \left( \ln  \left( z \right)  \right) ^{2}}{z }}{dz}+\frac{1}{32}\,\int _{0}^{1}\!{\frac { \left( \ln\left( 1-z \right) 
 \right) ^{2} \left( \ln  \left( z \right)  \right) ^{2}}{ \left( 1-
z\right) }}{dz} $$

$$ \implies I = \frac{1}{16}\,\int _{0}^{1}\!{\frac { \left( \ln  \left( 1-z \right) 
 \right) ^{2} \left( \ln  \left( z \right)  \right) ^{2}}{z }}{dz}\longrightarrow (1). $$

Getting the exact result: Integral (1) can be evaluated as
$$ \frac{1}{16}\,\int _{0}^{1}\!{\frac { \left( \ln  \left( 1-z \right) 
 \right)^{2} \left( \ln  \left( z \right)  \right)^{2}}{z }}{dz}=\frac{1}{16} \lim_{w\to 0}\lim_{s\to 0^+}\frac{d^2}{dw^2}\frac{d^2}{ds^2}\int_{0}^{1} (1-z)^{w}z^{s-1}dz $$
$$ = \frac{1}{16}\lim_{w\to 0}\lim_{s\to 0^+}\frac{d^2}{dw^2}\frac{d^2}{ds^2}\beta(s,w+1)=\frac{1}{16}\lim_{w\to 0}\lim_{s\to 0^+}\frac{d^2}{dw^2}\frac{d^2}{ds^2}\frac{\Gamma(s)\Gamma(w+1)}{\Gamma(s+w+1)}$$

$$ I=\frac{1}{4}\left( 2\zeta \left( 5 \right)-\zeta \left( 2 \right)\zeta \left( 3 \right) \right) \longrightarrow (*), $$

where $\beta(u,v)$ is the beta function. 
Other forms for the solution 1: Using integration by parts with $u=\ln^2(1-z)$, integral $(1)$ can be written as
$$ \frac{1}{16}\,\int _{0}^{1}\!{\frac { \left( \ln  \left( 1-z \right) 
 \right)^{2} \left( \ln  \left( z \right)\right)^{2}}{z }}{dz}=\frac{1}{24}\,\int _{0}^{1}\!{\frac{ \ln\left( 1-z \right)\left( \ln  \left( z \right) \right)^{3}}{1-z}}{dz} $$
$$ = -\sum_{n=0}^{\infty}(\psi(n+1)+\gamma)\int_{0}^{1}z^n\ln^3(z)dz = \frac{1}{4}\sum_{n=0}^{\infty}\frac{\psi(n+1)+\gamma}{(n+1)^4}. $$

$$ I= \frac{1}{4}\sum_{n=1}^{\infty}\frac{\psi(n)}{n^4}+\frac{\gamma}{4}\zeta(4)\sim 0.02413779000 \longrightarrow (**). $$

You can use the identity $ H_{n-1}=\psi(n)+\gamma $, where $H_n$ are the harmonic numbers, to write the result as

$$ I=\frac{1}{4}\sum_{n=1}^{\infty}\frac{H_{n-1}}{n^4} \longrightarrow (***). $$

Other forms for the solution 2: We can have the following form for the solution

$$ I=\frac{1}{16}\sum_{n=1}^{\infty}\frac{H^2_{n}}{n^3}+\frac{1}{16}\sum_{n=1}^{\infty}\frac{\psi'(n+1)}{n^3}-\frac{1}{16}\zeta(2)\zeta(3)\longrightarrow (****). $$

Note 1: we used the power series expansion of the function $ \frac{\ln(1-z)}{1-z}, $

$$\frac{\ln(1-z)}{1-z}= -\sum _{n=0}^{\infty } \left( \psi \left( n+1 \right) + \gamma \right){x}^{n}=-\sum _{n=0}^{\infty } H_{n}{x}^{n}. $$

Note 2: Try to tackle integral $(1)$ using the technique used in solving your previous question. 
A: Here is another way to solve the integral. Let
$$
\mathcal{I}=\int_0^{\Large\frac\pi2}\frac{\ln^2(\cos x)\ln^2(\sin x)}{\cos x\sin x}\ dx.
$$
Multiplying $\,\mathcal{I}\,$ by $\,\dfrac{2\sin x\cos x}{2\sin x\cos x}\,$ and setting $\,t=\sin^2x\,$ yield
\begin{align}
\frac1{32}\int_0^1\frac{\ln^2(1-t)\ln^2t}{(1-t)\ t}\ dt&=\frac1{32}\left[\int_0^1\frac{\ln^2(1-t)\ln^2t}{t}\ dt+\color{blue}{\underbrace{\color{black}{\int_0^1\frac{\ln^2(1-t)\ln^2t}{1-t}\ dt}}_{\color{red}{x\ \mapsto\ 1-x}}}\right]\\
&=\frac1{16}\int_0^1\frac{\ln^2(1-t)\ln^2t}{t}\ dt.
\end{align}
The latter integral can be evaluated using IBP by setting $$u=\ln^2(1-t)\ \color{red}{\Rightarrow}\ du=-\dfrac{2\ln(1-t)}{1-t}\quad \text{and}\quad dv=\dfrac{\ln^2t}{t}\ dt\ \color{red}{\Rightarrow}\ v=\dfrac13\ln^3t.$$
Hence
\begin{align}
\frac1{16}\int_0^1\frac{\ln^2(1-t)\ln^2t}{t}\ dt&=\frac1{16}\left[\left.\frac13\ln^3t\ln^2(1-t)\right|_{t=0}^1+\frac23\int_0^1\frac{\ln(1-t)\ln^3t}{1-t}\ dt\right]\\
&=\frac1{24}\int_0^1\frac{\ln(1-t)\ln^3t}{1-t}\ dt.
\end{align}
The latter integral has been evaluated in my other answer (click the link below).
\begin{align}
\color{blue}{\int\frac{\ln^3x\ln (1-x)}{1-x}\ dx}=&\ -\mathbf{H}_{1}(x)\ln^3x+\operatorname{Li}_2(x)\ln^3x+3\,\mathbf{H}_{2}(x)\ln^2x-3\operatorname{Li}_3(x)\ln^2x\\&\ -6\,\mathbf{H}_{3}(x)\ln x+6\operatorname{Li}_4(x)\ln x+6\,\mathbf{H}_{4}(x)-6\operatorname{Li}_5(x),
\end{align}
where $\displaystyle\mathbf{H}_{k}(x)=\sum_{n=0}^\infty\frac{H_nx^n}{n^k}$ and
$$
\mathbf{H}_{k}(1)=\frac{(k+2)}2\zeta(k+1)-\frac12\sum_{m=1}^{k-2}\zeta(k-m)\zeta(m+1)\quad;\quad\text{for}\ k\in\mathbb{Z}\ge2.
$$
Therefore
\begin{align}
\int_0^1\frac{\ln^3x\ln (1-x)}{1-x}\ dx=6\,\mathbf{H}_{4}(1)-6\operatorname{Li}_5(1)=12\zeta(5)-6\zeta(2)\zeta(3).
\end{align}
Alternatively, we can also use the following technique
\begin{align}
\int_0^1\frac{\ln^3x\ln (1-x)}{1-x}\ dx&=-\int_0^1\sum_{n=1}^\infty H_nx^n\ln^3x\ dx\\
&=-\sum_{n=1}^\infty H_n\int_0^1x^n\ln^3x\ dx\\
&=\sum_{n=1}^\infty\frac{3!\ H_n}{(n+1)^4}\tag1\\
&=6\sum_{n=1}^\infty\left[\frac{H_n}{n^4}-\frac1{n^5}\right]\tag2\\
&=6\bigg[3\zeta(5)-\zeta(2)\zeta(3)-\zeta(5)\bigg]\\
&=6\bigg[2\zeta(5)-\zeta(2)\zeta(3)\bigg].\\
\end{align}
Thus
$$
I=\frac1{24}\int_0^1\frac{\ln(1-t)\ln^3t}{1-t}\ dt=\color{blue}{\frac14\bigg[2\zeta(5)-\zeta(2)\zeta(3)\bigg]}.\tag{Q.E.D.}
$$

Notes :
$\displaystyle[1]\ \ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\  n=0,1,2,\ldots$
$\displaystyle[2]\ \ H_{n+1}-H_n=\frac1{n+1}$
A: Let  $x=\sin ^2 \theta$, then
$$
\begin{aligned}
\int_0^1 \frac{\ln ^2(1-x) \ln ^2 x}{x} d x&=\int_0^1 \frac{\ln ^2\left(\cos ^2 \theta\right) \ln ^2\left(\sin ^2 \theta\right)}{\sin ^2 \theta} 2 \sin \theta d \theta\\
&=32 \int_0^{\frac{\pi}{2}} \frac{\ln ^2(\cos \theta) \ln ^2(\sin \theta)}{\tan \theta} d \theta\\
&=\left.32 \cdot \frac{1}{2} \frac{\partial^4}{\partial a^2 \partial b^2} B\left(\frac{a}{2}, \frac{b}{2}+1\right)\right| _{{a \rightarrow 0}, {b \rightarrow 0}}\\
&=\lim _{a \rightarrow 0} \lim _{b \rightarrow 1}\left[\frac{\partial^{4}}{\partial a^2\partial b^2} B(a, b)\right]
\end{aligned}
$$
