$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over 1 + x^{2}}\,\dd x -
\half\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x^{2}} \over 1 + x^{2}}\,\dd x =
\half\,\ln\pars{2}\,\mrm{G}}$.
$\ds{\mrm{G}:\ \mbox{Catalan Constant.}}$
\begin{align}
&\color{#f00}{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over 1 + x^{2}}\,\dd x -
\half\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x^{2}} \over 1 + x^{2}}\,\dd x}
\\[5mm] = &\
\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over 1 + x^{2}}\,\dd x -
\Re\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x\ic} \over 1 + x^{2}}\,\dd x
\label{1}\tag{1}
\end{align}
$\ds{\ln}$-function branch-cut is chosen along the 'negative real axis'. Namely, in $\ds{\left.\vphantom{\large A}\ln\pars{z}\right\vert_{\ z\ \not=\ 0}}$ we have $\ds{-\pi < \mrm{arg}\pars{z} < \pi}$. For instance, when $\ds{x \in \pars{0,1}}$ we have:
\begin{align}
\ln\pars{1 + \ic x} & =
\ln\pars{\root{1 + x^{2}}} + \arctan\pars{x}\ic =
\ol{\bracks{\ln\pars{\root{1 + x^{2}}} - \arctan\pars{x}\ic}}
\\[5mm] & =
\ol{\ln\pars{1 - x\ic}}
\\[5mm] \mbox{and}\ \ln\pars{1 + x^{2}} & =
\ln\pars{1 + x\ic} + \ln\pars{1 - x\ic} = 2\,\Re\ln\pars{1 + x\ic}\quad
\mbox{which we already used in \eqref{1}}.
\end{align}
With the identity
$\ds{ab = \half\,a^{2} + \half\,b^{2} - \half\,\pars{a - b}^{2}}$, the expression \eqref{1} can be rewritten in the form
\begin{align}
&\color{#f00}{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over 1 + x^{2}}\,\dd x -
\half\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x^{2}} \over 1 + x^{2}}\,\dd x}
\\[5mm] = &\
\overbrace{\half\int_{0}^{1}{\ln^{2}\pars{1 - x} \over 1 + x^{2}}\,\dd x}
^{\ds{\color{#f00}{\mc{J}_{1}}}}\
\overbrace{-\,\half\int_{0}^{1}{\ln^{2}\pars{x/\bracks{1 - x}} \over 1 + x^{2}}
\,\dd x}
^{\ds{\color{#f00}{\mc{J}_{2}}}}\
\overbrace{-
\half\,\Re\int_{0}^{1}{\ln^{2}\pars{1 + x\ic} \over 1 + x^{2}}\,\dd x}
^{\ds{\color{#f00}{\mc{J}_{3}}}}
\\[5mm] + &\ \underbrace{%
\half\,\Re\int_{0}^{1}{\ln^{2}\pars{x/\bracks{1 + x\ic}} \over 1 + x^{2}}
\,\dd x}_{\ds{\color{#f00}{\mc{J}_{4}}}}\ =\
\color{#f00}{\mc{J}_{1}} + \color{#f00}{\mc{J}_{2}} +\color{#f00}{\mc{J}_{3}} +\color{#f00}{\mc{J}_{4}}
\end{align}
It turns out that the above integrals can be reduced to the form
$$
\left.\int{\ln^{2}\pars{x} \over a - x}\,\dd x\,\right\vert_{\ a\ \not=\ 0}
\,\,\,\,\,\stackrel{x\ =\ at}{=}\,\,\,\,\,
\int{\ln^{2}\pars{at} \over 1 - t}\,\dd t =
-\int\ln^{2}\pars{at}\,\dd\bracks{\ln\pars{1 - t}}
$$
which can be easily evaluated by successive integration by parts:
\begin{equation}
\int{\ln^{2}\pars{x} \over a - x}\,\dd x =
\left\lbrace\begin{array}{lcl}
\ds{-\ln^{2}\pars{x}\ln\pars{1 - {x \over a}} -
2\ln\pars{x}\Li{2}\pars{x \over a} + 2\Li{3}\pars{x \over a}} & \mbox{if} &
\ds{a \not= 0}
\\
\ds{-\,{1 \over 3}\,\ln^{3}\pars{x}} & \mbox{if} & \ds{a = 0}
\end{array}\right.
\end{equation}
$$
\begin{array}{|c|}\hline\mbox{}\\
\quad\mbox{Hereafter, we'll use this result to evaluate}\
\ds{\braces{\vphantom{\large A}\color{#f00}{\mc{J}_{k}}\,,\ k = 1,2,3,4}}
\quad
\\ \mbox{}\\ \hline
\end{array}
$$
With $\ds{r \equiv 1 + \ic}$:
- $\ds{\large\color{#f00}{\mc{J}_{1}}:\ ?}$.
\begin{align}
\color{#f00}{\mc{J}_{1}} & \equiv
\half\int_{0}^{1}{\ln^{2}\pars{1 - x} \over 1 + x^{2}}\,\dd x\,\,\,\,\,
\stackrel{x\ \mapsto\ \pars{1 - x}}{=}\,\,\,\,\,
\half\int_{0}^{1}{\ln^{2}\pars{x} \over x^{2} - 2x + 2}\,\dd x
\\[5mm] & =
\half\int_{0}^{1}{\ln^{2}\pars{x} \over \pars{x - r}\pars{x - \ol{r}}}\,\dd x =
-\,\half\,\Im\int_{0}^{1}{\ln^{2}\pars{x} \over r - x}\,\dd x =
\color{#f00}{\Im\Li{3}\pars{\half\,r}}\label{J1}\tag{J1}
\end{align}
- $\ds{\large\color{#f00}{\mc{J}_{2}}:\ ?}$.
\begin{align}
\color{#f00}{\mc{J}_{2}} & \equiv
-\,\half\int_{0}^{1}{\ln^{2}\pars{x/\bracks{1 - x}} \over 1 + x^{2}}\,\dd x
\,\,\,\,\,\stackrel{x/\pars{1 - x}\ \mapsto\ x}{=}\,\,\,\,\,
-\,\half\int_{0}^{\infty}{\ln^{2}\pars{x} \over 2x^{2} + 2x + 1}\,\dd x
\\[5mm] & =
-\,\half\int_{0}^{1}{\ln^{2}\pars{x} \over 2x^{2} + 2x + 1}\,\dd x -
\half\int_{0}^{1}{\ln^{2}\pars{x} \over x^{2} + 2x + 2}\,\dd x
\\[5mm] & =
-\,{1 \over 4}\int_{0}^{1}
{\ln^{2}\pars{x} \over \pars{x + r/2}\pars{x + \ol{r}/2}}\,\dd x - \half\int_{0}^{1}{\ln^{2}\pars{x} \over \pars{x + r}\pars{x + \ol{r}}}\,\dd x
\\[5mm] & =
-\,\half\,\Im\int_{0}^{1}{\ln^{2}\pars{x} \over -r/2 - x}\,\dd x -
\half\,\Im\int_{0}^{1}{\ln^{2}\pars{x} \over -r - x}\,\dd x
\end{align}
However,
$$
\left\lbrace\begin{array}{rcl}
\ds{-\,\half\,\Im\int_{0}^{1}{\ln^{2}\pars{x} \over -r/2 - x}\,\dd x} & \ds{=} & \ds{-\,{5 \over 128}\,\pi^{3} - {1 \over 32}\,\ln^{2}\pars{2}\pi -
\Im\Li{3}\pars{-\,{r \over 2}}}
\\[3mm]
\ds{-\,\half\,\Im\int_{0}^{1}{\ln^{2}\pars{x} \over -r - x}\,\dd x} & \ds{=} & \ds{-\Im\Li{3}\pars{-\,{\ol{r} \over 2}}}
\end{array}\right.
$$
Then,
\begin{equation}
\color{#f00}{\mc{J}_{2}} =
\color{#f00}{-\,{5 \over 128}\,\pi^{3} - {1 \over 32}\,\ln^{2}\pars{2}\pi}
\label{J2}\tag{J2}
\end{equation}
- $\ds{\large\color{#f00}{\mc{J}_{3}}:\ ?}$.
\begin{align}
\color{#f00}{\mc{J}_{3}} & \equiv
-\,\half\,\Re\int_{0}^{1}{\ln^{2}\pars{1 + x\ic} \over 1 + x^{2}}\,\dd x
\,\,\,\,\,\stackrel{\pars{1 + x\ic}\ \mapsto\ x}{=}
-\,\half\,\Im\int_{1}^{r}{\ln^{2}\pars{x} \over \pars{2 - x}x}\,\dd x
\\[5mm] & =
-\,{1 \over 4}\,\Im\int_{1}^{r}{\ln^{2}\pars{x} \over 2 - x}\,\dd x -
{1 \over 4}\,\Im\int_{1}^{r}{\ln^{2}\pars{x} \over x}\,\dd x
\end{align}
The remaining integrals are given by:
$$
\left\lbrace\begin{array}{rcl}
\ds{-\,{1 \over 4}\,\Im\int_{1}^{r}{\ln^{2}\pars{x} \over 2 - x}\,\dd x} & \ds{=} &
\ds{{1 \over 96}\,\pi^{3} - {3 \over 32}\,\ln^{2}\pars{2}\pi +
{1 \over 4}\,\ln\pars{2}\,\mrm{G} - \half\,\Im\Li{3}\pars{r \over 2}}
\\[3mm]
\ds{-\,{1 \over 4}\,\Im\int_{1}^{r}{\ln^{2}\pars{x} \over x}\,\dd x}
& \ds{=} &
\ds{{1 \over 768}\,\pi^{3} - {1 \over 64}\,\ln^{2}\pars{2}\,\pi}
\end{array}\right.
$$
Then,
\begin{equation}
\color{#f00}{\mc{J}_{3}} =
\color{#f00}{{3 \over 256}\,\pi^{3} - {7 \over 64}\,\ln^{2}\pars{2}\pi +
{1 \over 4}\,\ln\pars{2}\,\mrm{G} - \half\,\Im\Li{3}\pars{r \over 2}}
\label{J3}\tag{J3}
\end{equation}
- $\ds{\large\color{#f00}{\mc{J}_{4}}:\ ?}$.
\begin{align}
\color{#f00}{\mc{J}_{4}} & \equiv
\half\,\Re\int_{0}^{1}{\ln^{2}\pars{x/\bracks{1 + x\ic}} \over 1 + x^{2}}\,\dd x
\,\,\,\,\,\stackrel{x/\pars{1 + x\ic}\ \mapsto\ x}{=}\,\,\,\,\,
{1 \over 4}\,\Im\int_{0}^{\ol{r}/2}{\ln^{2}\pars{x} \over -\ic/2 - x}\,\dd x
\\[5mm] & =
\color{#f00}{{7 \over 256}\,\pi^{3} + {9 \over 64}\,\ln^{2}\pars{2}\pi +
{1 \over 4}\,\ln\pars{2}\,\mrm{G} - \half\,\Im\Li{3}\pars{r \over 2}}
\label{J4}\tag{J4}
\end{align}
Summarising
$\ds{\pars{~\vphantom{\large A}\mbox{see}\ \eqref{J1}, \eqref{J2}, \eqref{J3}\ \mbox{and}\ \eqref{J4}~}}$:
\begin{equation}
\left\lbrace\begin{array}{rcccccccc}
\ds{\color{#f00}{\mc{J}_{1}}} & \ds{=} &&&&&&&\ds{\Im\Li{3}\pars{r \over 2}}
\\[3mm]
\ds{\color{#f00}{\mc{J}_{2}}} & \ds{=} &
\ds{-\,{5 \over 128}\,\pi^{3}} & \ds{-} & \ds{{1 \over 32}\,\ln^{2}\pars{2}\pi}
&&&&
\\[3mm]
\ds{\color{#f00}{\mc{J}_{3}}} & \ds{=} &
\ds{{3 \over 256}\,\pi^{3}} & \ds{-} & \ds{{7 \over 64}\,\ln^{2}\pars{2}\pi} & \ds{+} &
\ds{{1 \over 4}\,\ln\pars{2}\,\mrm{G}} & \ds{-} &
\ds{\half\,\Im\Li{3}\pars{r \over 2}}
\\[3mm]
\ds{\color{#f00}{\mc{J}_{4}}} & \ds{=} &
\ds{{7 \over 256}\,\pi^{3}} & \ds{+} & \ds{{9 \over 64}\,\ln^{2}\pars{2}\pi} & \ds{+} &
\ds{{1 \over 4}\,\ln\pars{2}\,\mrm{G}} & \ds{-} & \ds{\half\,\Im\Li{3}\pars{r \over 2}}
\end{array}\right.
\end{equation}
The $\ds{\quad\ul{final\ result}\quad}$ is given by:
\begin{align}
&\color{#f00}{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over 1 + x^{2}}\,\dd x -
\half\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x^{2}} \over 1 + x^{2}}\,\dd x}
\\[5mm] = &\
\mc{J}_{1} + \mc{J}_{2} + \mc{J}_{3} + \mc{J}_{4} =
\color{#f00}{\half\,\ln\pars{2}\,\mrm{G}}\,,\qquad
\pars{~\mrm{G}:\ \mbox{Catalan Constant}~}
\end{align}
Best Answer
Let's start out with the substitution $ \displaystyle \ln(\sin x) = u $ and get: $$\ln(\cos x)=\frac{\ln(1-e^{2u})}{2}$$ $$\displaystyle\frac{1}{\tan x} \ dx =du$$ that further yields $$\int_0^{\pi/2}\frac{(\ln{\sin x})(\ln{\cos x})}{\tan x}dx= \frac{1}{2} \int_{-\infty}^{0} \ln(1-e^{2u}) u \ du$$
According to Taylor expansion we have $$\ln(1-e^{2u})= \sum_{k=1}^{\infty} (-1)^{2k+1} \frac{e^{2 k u}}{k}$$ then $$\frac{1}{2} \int_{-\infty}^{0} \ln(1-e^{2u}) u \ du=$$ $$\frac{1}{2} \sum_{k=1}^{\infty} \frac{(-1)^{2k+1}}{k} \int_{-\infty}^{0} u e^{2ku} \ du =$$ $$\frac{1}{2} \sum_{k=1}^{\infty} \frac{(-1)^{2k+1}}{k} \frac{-1}{4k^2} = \frac{1}{8} \sum_{k=1}^{\infty} \frac{1}{k^3}=\frac{1}{8} \zeta(3).$$
Remark: the value of $\zeta(3)\approx1.2020569$ is called Apéry's Constant - see here.
Q.E.D. (Chris)