TITLE: How to evaluate $I=\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$
QUESTION [59 upvotes]: Find the value of
$I=\displaystyle\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$
We have the information that 
$J=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)\ \mathrm dx=\dfrac{(\pi\ln{2})^2}{8}-\dfrac{\pi^4}{192}$

REPLY [6 votes]: $\newcommand{\+}{^{\dagger}}%
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$\ds{I \equiv \int_{0}^{\pi/2}x^{2}\ln\pars{\sin\pars{x}}\ln\pars{\cos\pars{x}}
     \,\dd x:\ {\Large ?}}$
Yet another partial idea...

First: Some reductions
  \begin{align}
I&=\int_{0}^{\pi/2}x^{2}\,
{\bracks{\ln\pars{\sin\pars{x}} + \ln\pars{\cos\pars{x}}}^{2}
-\ln^{2}\pars{\sin\pars{x}} -\ln^{2}\pars{\cos\pars{x}} \over 2}\,\dd x
\\[3mm]&=\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\sin\pars{x}\cos\pars{x}}\,\dd x
-\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\sin\pars{x}}\,\dd x
-\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\cos\pars{x}}\,\dd x
\\[3mm]&=\half\int_{0}^{\pi/2}x^{2}\bracks{%
\ln^{2}\pars{\sin\pars{2x}} - 2\ln\pars{2}\ln\pars{\sin\pars{2x}} + \ln^{2}\pars{2}}
\,\dd x\\[3mm]&
-\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\sin\pars{x}}\,\dd x
-\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\cos\pars{x}}
\,\dd x
\\[3mm]&={\pi \over 4}\,\ln^{2}\pars{2}
+ {1 \over 16}\int_{0}^{\pi}x^{2}\ln^{2}\pars{\sin\pars{x}}\,\dd x
- {1 \over 8}\ln\pars{2}\int_{0}^{\pi}x^{2}\ln\pars{\sin\pars{x}}\,\dd x
\\[3mm]&-\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\sin\pars{x}}\,\dd x
-\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\cos\pars{x}}\,\dd x
\qquad\qquad\qquad\qquad\qquad\qquad\pars{1}
\end{align}

Also, with $n = 1,2$:
\begin{align}
&\int_{0}^{\pi}x^{2}\ln^{n}\pars{\sin\pars{x}}\,\dd x=
\int_{-\pi/2}^{\pi/2}\pars{x + {\pi \over 2}}^{2}\ln^{n}\pars{\cos\pars{x}}\,\dd x
\\[3mm]&=\int_{0}^{\pi/2}
\bracks{\pars{x + {\pi \over 2}}^{2} + \pars{-x + {\pi \over 2}}^{2}}
\ln^{n}\pars{\cos\pars{x}}\,\dd x
=\int_{0}^{\pi/2}
\pars{2x^{2} + {\pi^{2} \over 2}}\ln^{n}\pars{\cos\pars{x}}\,\dd x
\\[3mm]&=2\int_{0}^{\pi/2}x^{2}\ln^{n}\pars{\cos\pars{x}}\,\dd x
+
{\pi^{2} \over 2}\int_{0}^{\pi/2}\ln^{n}\pars{\cos\pars{x}}\,\dd x
\end{align}

With this result, $\pars{1}$ is reduced to:
  \begin{align}
I&={\pi \over 4}\,\ln^{2}\pars{2}
-{3 \over 8}\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\cos\pars{x}}\,\dd x
+ {\pi^{2} \over 32}\
\overbrace{\int_{0}^{\pi/2}\ln^{2}\pars{\cos\pars{x}}\,\dd x}
^{\ds{{\pi^{3} \over 24} + {\pi \over 2}\,\ln^{2}\pars{2}}}
\\[3mm]&-{1 \over 4}\,\ln\pars{2}\
\overbrace{\int_{0}^{\pi/2}x^{2}\ln\pars{\cos\pars{x}}\,\dd x}
^{\ds{-\,{\pi^{3} \over 24}\,\ln\pars{2} - {1 \over 4}\,\zeta\pars{3}}}
-{\pi^{2} \over 16}\,\ln\pars{2}\
\overbrace{\int_{0}^{\pi/2}\ln\pars{\cos\pars{x}}\,\dd x}
^{\ds{-\,{\pi \over 2}\,\ln\pars{2}}}
-\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\sin\pars{x}}\,\dd x
\end{align}

$I$ is reduced to:
$$
I = C 
-\half\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\sin\pars{x}}\,\dd x
-{3 \over 8}\int_{0}^{\pi/2}x^{2}\ln^{2}\pars{\cos\pars{x}}\,\dd x
$$
where $C$ is given by:
\begin{align}
C &= {\pi \over 4}\,\ln^{2}\pars{2} + {\pi^{5} \over 728}
+ {\pi^{3} \over 64}\,\ln^{2}\pars{2} + {\pi^{3} \over 96}\,\ln^{2}\pars{2}
+ {1 \over 16}\,\ln\pars{2}\zeta\pars{3} + {\pi^{3} \over 32}\,\ln^{2}\pars{2}
\\[3mm]&=
{\pi \over 4}\,\bracks{\ln\pars{2} + {1 \over 4}\,\zeta\pars{3}}\ln\pars{2} + {\pi^{5} \over 728}
+ {1 \over 16}\,\ln\pars{2}\zeta\pars{3} + {11\pi^{3} \over 192}\,\ln^{2}\pars{2}
\end{align}
$\tt @sos440$  user had already derived results which are related to the remaining integrals.