$\int_{0}^{\pi}{x\log(\sin(x))}dx$ Compute $\displaystyle \int_{0}^{\pi}{x\log(\sin(x))}dx$
I know how to compute the integral $\displaystyle \int_{0}^{\pi}{\log(\sin(x))}dx$. 
Maybe I can reduce to this case by the use of some chan of variable trick or by integration by parts. Please help me )=
 A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\int_{0}^{\pi}x\,\ln\pars{\sin\pars{x}}\,\dd x:\ {\large ?}}$

$$
\color{#c00000}{\int_{0}^{\pi}x\,\ln\pars{\sin\pars{x}}\,\dd x}
=\int_{-\pi/2}^{\pi/2}\pars{x + {\pi \over 2}}\ln\pars{\cos\pars{x}}\,\dd x
=\pi\color{#00f}{\int_{0}^{\pi/2}\ln\pars{\cos\pars{x}}\,\dd x}\quad\pars{1}
$$

\begin{align}
&\color{#00f}{\int_{0}^{\pi/2}\ln\pars{\cos\pars{x}}\,\dd x}
=\half\bracks{%
\int_{0}^{\pi/2}\ln\pars{\cos\pars{x}}\,\dd x
+\int_{0}^{\pi/2}\ln\pars{\cos\pars{{\pi \over 2} - x}}\,\dd x}
\\[3mm]&=\half\int_{0}^{\pi/2}\ln\pars{\sin\pars{2x} \over 2}\,\dd x
=-\,{1 \over 4}\,\pi\ln\pars{2}
+{1 \over 4}\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\,\dd x
\\[3mm]&=-\,{1 \over 4}\,\pi\ln\pars{2}
+\half\color{#00f}{\int_{0}^{\pi/2}\ln\pars{\cos\pars{x}}\,\dd x}
\quad\imp\quad\color{#00f}{\int_{0}^{\pi/2}\ln\pars{\cos\pars{x}}\,\dd x}= -\,\half\,\pi\ln\pars{2}
\end{align}

By replacing this result in expression $\pars{1}$, we'll find:
  $$
\color{#00f}{\large\int_{0}^{\pi}x\,\ln\pars{\sin\pars{x}}\,\dd x}
=\color{#00f}{\large-\,\half\,\pi^{2}\ln\pars{2}} \approx -3.4205
$$

A: There is a well-known trick for functions of this form. Here $f(x)$ is a function of $\sin x$ only.
$$ \begin{align} I :=\\
\int_0^{\pi} x f(x)  dx\\
=\int_{\pi}^{0} (\pi-u) f(\pi - u)  (-du) \\
=\pi \int_{0}^{\pi}  f(u)  du - \int_0^{\pi} u f(u)  du \\
=\pi \int_{0}^{\pi}  f(x)  dx- I.\end{align}$$
Here I used the substitution $u = \pi - x$, and the fact that $\sin(\pi - x) = \sin(x)$.
Hence $$ I = \frac{\pi}{2} \int_{0}^{\pi}  f(x)  dx$$
A similar trick also works if the integral runs to $\pi/2$, except then you need to swap all sines and cosines since $\sin(\pi/2 -x)=\cos x$, $\cos(\pi/2 -x)=\sin x$.
