Integral of $x\log(\sin x)$ 
This is from an old S level paper. I am struggling with part (ii). Any hints?
 A: $\log_e$ for denoting the natural logarithm? Oh my dear.
$$ \int_{0}^{\pi/2}x\log\sin x\,dx = \int_{0}^{1}\frac{\arcsin(u)\log u}{\sqrt{1-u^2}}\,du\tag{1}$$
and by recalling 
$$ \arcsin(u) = \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(2n+1)}u^{2n} \tag{2} $$
$$ \int_{0}^{1}\frac{u^{2n}\log(u)}{\sqrt{1-u^2}}\,du = \frac{\pi\binom{2n}{n}}{4^{n+1}}\left(H_{n-1/2}-H_n\right)\tag{3} $$
(where $(3)$ follows by differentiating Euler's Beta function) the LHS of $(1)$ is converted into a twisted hypergeometric series, according to the terminology introduced here. On the other hand, by exploiting the Fourier series of $\log\sin$ or Fourier-Chebyshev series expansions, the LHS of $(1)$ turns out to be
$$ \int_{0}^{\pi/2}x\log\sin x\,dx = \color{red}{\frac{7}{16}\,\zeta(3)-\frac{\pi^2}{8}\,\log(2)}.\tag{4}$$
One may tackle the equivalent integral $\int_{0}^{\pi/2}x^2\cot(x)\,dx$ also by recalling that $\cot(x)=\frac{1}{x}+\sum_{n\geq 1}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right)$, but symmetry is definitely not enough to carve the $\zeta(3)$ term out of thin air.
A: From point (i) it can be easily shown that
$$\int_{0}^{\pi/2}\log\sin x\,dx=\int_{0}^{\pi/2}\log\cos x\,dx=-\frac{\pi}{2}\log 2$$
Now consider for point (ii)
$$I=\int_{0}^{\pi/2}x\log\sin x\,dx=\int_{0}^{\pi/2}\left(\frac{\pi}{2}-x\right)\log\cos x\,dx$$
thus
$$2I=\int_{0}^{\pi/2}x\log\sin x\,dx+\int_{0}^{\pi/2}\left(\frac{\pi}{2}-x\right)\log\cos x\,dx=$$
$$=\int_{0}^{\pi/2}x\log\tan x\,dx+\frac{\pi}{2}\int_{0}^{\pi/2}\log\cos x\,dx=\int_{0}^{\pi/2}x\log\tan x\,dx-\frac{\pi^2}{4}\log 2$$
since from the following reference from Paul Enta
$$\int_{0}^{\pi/2}x\log\tan x\,dx=\frac{7}{8}\,\zeta(3)$$
we finally have
$$I=\int_{0}^{\pi/2}x\log\sin x\,dx=\frac{7}{16}\,\zeta(3)-\frac{\pi^2}{8}\log 2$$
A: Using $\text{(1d)}$ from this answer
$$
\log(\sin(x))=-\log(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}k
$$
we get
$$
\begin{align}
\int_0^{\pi/2}x\log(\sin(x))\,\mathrm{d}x
&=-\int_0^{\pi/2}x\left(\log(2)+\sum_{k=1}^\infty\frac{\cos(2kx)}k\right)\,\mathrm{d}x\\
&=-\frac{\pi^2}8\log(2)-\sum_{k=1}^\infty\frac1k\int_0^{\pi/2}x\cos(2kx)\,\mathrm{d}x\\
&=-\frac{\pi^2}8\log(2)-\sum_{k=1}^\infty\frac1k\frac{(-1)^k-1}{4k^2}\\
&=-\frac{\pi^2}8\log(2)+\frac12\sum_{k=1}^\infty\frac1{(2k+1)^3}\\
&=\frac7{16}\zeta(3)-\frac{\pi^2}8\log(2)
\end{align}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\pi/2}x\ln\pars{\sin\pars{x}}\,\dd x} =
\left.\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}
\bracks{-\ic\ln\pars{z}}\ln\pars{z - 1/z \over 2\ic}
\,{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
\left. -\,\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}
\ln\pars{z}\ln\pars{{1 - z^{2} \over 2z}\,\ic}
\,{\dd z \over z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,&
\Re\int_{1}^{\epsilon}
\bracks{\ln\pars{y} + {\pi \over 2}\,\ic}\ln\pars{1 + y^{2} \over 2y}
\,{\dd y \over y}
\\[2mm] + &\
\underbrace{\Re\int_{\pi/2}^{0}\bracks{\ln\pars{\epsilon} + \ic\theta}
\bracks{-\ln\pars{2\epsilon} + \pars{{\pi \over 2} - \theta}\ic}\ic\,\dd\theta}
_{\ds{\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\,-\,{\pi^{2} \over 8}\,\ln\pars{2}}}
\\[2mm] + &\ 
\Re\int_{\epsilon}^{1}
\ln\pars{x}\bracks{\ln\pars{1 - x^{2} \over 2x} + {\pi \over 2}\,\ic}
\,{\dd x \over x}
\\[1cm]  \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{=}\,\,\,&
-\,{\pi^{2} \over 8}\,\ln\pars{2} +
\int_{0}^{1}\ln\pars{x}\ln\pars{1 - x^{2} \over 1 + x^{2}}
\,{\dd x \over x}
\\[5mm] = &\
-\,{\pi^{2} \over 8}\,\ln\pars{2} +
{1 \over 4}\int_{0}^{1}\ln\pars{x}\ln\pars{1 - x \over 1 + x}
{\dd x \over x}
\\[5mm] = &\
-\,{\pi^{2} \over 8}\,\ln\pars{2} +
{1 \over 4}\int_{0}^{1}\ln\pars{x}\,
{\ln\pars{1 - x} \over x}\,\dd x -
{1 \over 4}\int_{0}^{-1}\ln\pars{-x}\,
{\ln\pars{1 - x} \over x}\,\dd x
\\[5mm] = &\
-\,{\pi^{2} \over 8}\,\ln\pars{2} -
{1 \over 4}\int_{0}^{1}\ln\pars{x}\,
\mrm{Li}_{2}'\pars{x}\,\dd x +
{1 \over 4}\int_{0}^{-1}\ln\pars{-x}\,\mrm{Li}_{2}'\pars{x}\,\dd x
\\[5mm] = &\
-\,{\pi^{2} \over 8}\,\ln\pars{2} +
{1 \over 4}\int_{0}^{1}
\mrm{Li}_{3}'\pars{x}\,\dd x -
{1 \over 4}\int_{0}^{-1}\mrm{Li}_{3}'\pars{x}\,\dd x
\\[5mm] = &\
-\,{\pi^{2} \over 8}\,\ln\pars{2} +
{1 \over 4}\sum_{n = 1}^{\infty}{1 - \pars{-1}^{n} \over n^{3}} =
-\,{\pi^{2} \over 8}\,\ln\pars{2} +
{1 \over 2}\sum_{{\large n\ =\ 1} \atop {\large n\ odd}}^{\infty}
{1 \over n^{3}}
\\[5mm] = &\
-\,{\pi^{2} \over 8}\,\ln\pars{2} +
{1 \over 2}\bracks{\sum_{n = 1}^{\infty}{1 \over n^{3}} -
\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{3}}} =
-\,{\pi^{2} \over 8}\,\ln\pars{2} +
{7 \over 16}\sum_{n = 1}^{\infty}{1 \over n^{3}}
\\[5mm] = &\ \bbx{-\,{\pi^{2} \over 8\phantom{^{2}}}\,\ln\pars{2} +
{7 \over 16}\,\zeta\pars{3}} \approx -0.3292
\end{align}
