How can we show that $\int_{0}^{\pi/2}{\ln^2\sin(x) \ln\cos (x)\over \sin(x)\cos (x)}\mathrm dx=-{\left(\pi^2\over 4!\right)^2}?$ Given that

$$\int_{0}^{\pi/2}{\ln^2\sin(x) \ln\cos (x)\over \sin(x)\cos (x)}\mathrm dx=-\color{brown}{\left(\pi^2\over 4!\right)^2}\tag1$$

My try:
$u=\sin^2(x)\implies du=2\sin(x)\cos(x) dx$
$(1)$ becomes
$${1\over 16}\int_{0}^{1}{\ln^2(u)\ln(1-u)\over u(1-u)}\mathrm du\tag2$$
$v=1-u$
$$-{1\over 16}\int_{0}^{1}{\ln^2(1+v)\ln(v)\over v(1+v)}\mathrm dv\tag3$$
Recall $$\ln(1+x)=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n}x^n$$
$${1\over 16}\sum_{n=1}^{\infty}{(-1)^n\over n}\int_{0}^{1}v^{n-1}\ln(v)\cdot{\mathrm dv\over 1+v}\tag4$$
How can one prove $(1)?$
 A: Note that $$I=\int_{0}^{1}\frac{\log^{2}\left(u\right)\log\left(1-u\right)}{u\left(1-u\right)}du=\sum_{k\geq0}\int_{0}^{1}u^{k-1}\log^{2}\left(u\right)\log\left(1-u\right)du$$ $$=-\sum_{k\geq0}\sum_{m\geq1}\frac{1}{m}\int_{0}^{1}u^{m+k-1}\log^{2}\left(u\right)du$$ so integrating by parts we get $$I=-2\sum_{k\geq0}\sum_{m\geq1}\frac{1}{m\left(m+k\right)^{3}}=-2\sum_{m\geq1}\frac{\zeta\left(3\right)-H_{m-1}^{\left(3\right)}}{m}.$$ Now we may note that $$\sum_{m=1}^{N}\frac{\zeta\left(3\right)}{m}=\zeta\left(3\right)H_{N}$$ and, using summation by parts, $$ -\sum_{m=1}^{N}\frac{H_{m-1}^{\left(3\right)}}{m}=-H_{N-1}^{\left(3\right)}H_{N}+\sum_{m=1}^{N}\frac{H_{m}}{m^{3}}$$ so $$-2\sum_{m\geq1}\frac{\zeta\left(3\right)-H_{m-1}^{\left(3\right)}}{m}=-2\lim_{N\rightarrow\infty}\left(H_{N}\left(\zeta\left(3\right)-H_{N-1}^{\left(3\right)}\right)+\sum_{m=1}^{N}\frac{H_{m}}{m^{3}}\right)$$ $$=-2\sum_{m\geq1}\frac{H_{m}}{m^{3}}.$$ So we have to evaluate the last sum. We have $$-2\sum_{m\geq1}\frac{H_{m}}{m^{3}}=-2\sum_{m\geq1}\frac{1}{m^{3}}\sum_{k\geq1}\left(\frac{1}{k}-\frac{1}{k+m}\right)$$ $$=-2\sum_{m\geq1}\sum_{k\geq1}\frac{1}{m^{2}k\left(k+m\right)}=-\sum_{m\geq1}\sum_{k\geq1}\frac{1}{m^{2}k\left(k+m\right)}-\sum_{m\geq1}\sum_{k\geq1}\frac{1}{mk^{2}\left(k+m\right)}$$ $$=-\sum_{m\geq1}\sum_{k\geq1}\frac{1}{m^{2}k^{2}}=-\zeta\left(2\right)^{2}=-\frac{\pi^{4}}{36}$$ hence $$I=\color{red}{-\frac{\pi^{4}}{36}}.$$ Another (not very elegant) way is to observe that from the definition of the Beta function $$B\left(a,b\right)=\int_{0}^{1}u^{a-1}\left(1-u\right)^{b-1}du$$ we have $$\frac{\partial^{2}}{\partial a^{2}}\frac{\partial}{\partial b}B\left(a,b\right)=\int_{0}^{1}u^{a-1}\left(1-u\right)^{b-1}\log^{2}\left(u\right)\log\left(1-u\right)du$$ so $$I=\lim_{a\rightarrow0^{+}}\lim_{b\rightarrow0^{+}}\frac{\partial^{2}}{\partial a^{2}}\frac{\partial}{\partial b}B\left(a,b\right)=\color{green}{-\frac{\pi^{4}}{36}}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\int_{0}^{\pi/2}{\ln^{2}\pars{\sin\pars{x}}\ln\pars{\cos\pars{x}}\over \sin\pars{x}\cos\pars{x}}\dd x =
-\pars{\phantom{^{2}}\pi^{2} \over 4!}^{2}:\ {\large ?}}$


\begin{align}
&\int_{0}^{\pi/2}{\ln^{2}\pars{\sin\pars{x}}\ln\pars{\cos\pars{x}}\over \sin\pars{x}\cos\pars{x}}\dd x
\\[5mm] = &\
{1 \over 16}\int_{0}^{\pi/2}{\ln^{2}\pars{\sin^{2}\pars{x}}\ln\pars{\cos^{2}\pars{x}}\over \sin^{2}\pars{x}\cos^{2}\pars{x}}\bracks{2\sin\pars{x}\cos\pars{x}}\,\dd x
\\[5mm] \stackrel{\sin^{2}\pars{x}\ \mapsto\ x}{=}\,\,\,&\
-\,{1 \over 16}\int_{0}^{1}\ln^{2}\pars{x}\
\overbrace{\bracks{-\,{\ln\pars{1 - x} \over x\pars{1 - x}}}}
^{\ds{\sum_{n = 1}^{\infty}H_{n}\,x^{n}}}\
\,\dd x\qquad\pars{~H_{z}:\ Harmonic\ Number~}\label{1}\tag{1}
\\[5mm] = &\
-\,{1 \over 16}\sum_{n = 1}^{\infty}H_{n}\
\overbrace{\int_{0}^{1}\ln^{2}\pars{x}x^{n - 1}\,\dd x}
^{\ds{2 \over \phantom{^{3}}n^{3}}}\ =\
-\,{1 \over 8}\sum_{n = 1}^{\infty}{H_{n} \over n^{3}}
\end{align}


$\ds{\sum_{n = 1}^{\infty}{H_{n} \over n^{3}} = {\phantom{^{4}}\pi^{4} \over 72}}$ is a
  well known identity. See $\ds{\pars{19}}$ in the cited link.

\begin{align}
&\int_{0}^{\pi/2}{\ln^{2}\pars{\sin\pars{x}}\ln\pars{\cos\pars{x}}\over \sin\pars{x}\cos\pars{x}}\dd x =
-\,{1 \over 8}\,{\phantom{^{4}}\pi^{4} \over 72} =
-\,{\phantom{^{4}}\pi^{4} \over 576} =
\bbx{-\,\pars{\phantom{^{2}}\pi^{2} \over 4!}^{2}}
\end{align}
