Calculating $\int_{\pi/2}^{\pi}\frac{x\sin{x}}{5-4\cos{x}}\,\mathrm dx$ 
Calculate the following integral:$$\int_{\pi/2}^{\pi}\frac{x\sin{x}}{5-4\cos{x}}\,\mathrm dx$$

I can calculate the integral on $[0,\pi]$,but I want to know how to do it on $[\frac{\pi}{2},\pi]$.
 A: $$\begin{align}\int_{\pi/2}^\pi\frac{x\sin x}{5-4\cos x}dx&=\pi\left(\frac{\ln3}2-\frac{\ln2}4-\frac{\ln5}8\right)-\frac12\operatorname{Ti}_2\left(\frac12\right)\\&=\pi\left(\frac{\ln3}2-\frac{\ln2}4-\frac{\ln5}8\right)-\frac12\Im\,\chi_2\left(\frac{\sqrt{-1}}2\right),\end{align}$$
where  $\operatorname{Ti}_2(z)$ is the inverse tangent integral and $\Im\,\chi_\nu(z)$ is the imaginary part of the Legendre chi function.

Hint:
Use the following Fourier series and integrate termwise:
$$\frac{\sin x}{5-4\cos x}=\sum_{n=1}^\infty\frac{\sin n x}{2^{n+1}}.$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{\pi/2}^{\pi}{x\sin\pars{x} \over 5 - 4\cos\pars{x}}\,\dd x:\ {\large ?}}$

\begin{align}
&\color{#c00000}{\int_{\pi/2}^{\pi}{x\sin\pars{x} \over 5 - 4\cos\pars{x}}\,\dd x}
={1 \over 4}\int_{x\ =\ \pi/2}^{x\ =\ \pi}x\,\dd\bracks{\ln\pars{5 - 4\cos\pars{x}}}
\\[3mm]&={\pi\ln\pars{5 - 4\cos\pars{\pi}}
  -\pars{\pi/2}\ln\pars{5 - 4\cos\pars{\pi/2}} \over 4}
-{1 \over 4}\int_{\pi/2}^{\pi}\ln\pars{5 - 4\cos\pars{x}}\,\dd x
\\[3mm]&={1 \over 8}\,\pi\ln\pars{81 \over 5}
-{1 \over 4}\int_{0}^{\pi/2}\ln\pars{5 + 4\sin\pars{x}}\,\dd x
\\[3mm]&={1 \over 8}\,\pi\ln\pars{81 \over 25}
-{1 \over 4}\color{#00f}{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin\pars{x}}\,\dd x}
\ \mbox{where}\ \boxed{\ \alpha \equiv {4 \over 5} < 1\ }\qquad\qquad\qquad\pars{1}
\end{align}

With $\ds{x \equiv 2\arctan\pars{t}\quad\imp\quad t = \tan\pars{x \over 2}}$:
\begin{align}
&\color{#00f}{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin\pars{x}}\,\dd x}
=\int_{0}^{1}\ln\pars{1 + \alpha\,{2t \over 1 + t^{2}}}\,
{2\,\dd t \over 1 + t^{2}}
\\[3mm]&=2\int_{0}^{1}
{\ln\pars{t^{2} + 2\alpha t + 1} \over 1 + t^{2}}\,\dd t
-2\
\overbrace{\int_{0}^{1}{\ln\pars{1 + t^{2}} \over 1 + t^{2}}\,\dd t}
^{\ds{-{\rm G} + \half\,\pi\ln\pars{2} }}
\\[3mm]&=2\int_{0}^{1}
{\ln\pars{\bracks{z - t}\pars{z^{*} - t}} \over t^{2} + 1}\,\dd t
+ 2{\rm G} - \pi\ln\pars{2}\tag{2}
\end{align}
where $\ds{z \equiv -\alpha - \root{1 - \alpha^{2}}\ \ic=-\,{4 \over 5} - {3 \over 5}\,\ic}$ and $\ds{\rm G}$ is the Catalan Constant. Note that
$\ds{z^{*} = {1 \over z}}$.

\begin{align}
&\color{#00f}{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin\pars{x}}\,\dd x}
\\[2mm]&=2{\rm G} - \pi\ln\pars{2}
+2\Im\bracks{\int_{0}^{1}{\ln\pars{1 - t/z} \over t - \ic}\,\dd t
+ \int_{0}^{1}{\ln\pars{1 - zt} \over t - \ic}\,\dd t}
\\[2mm]&=2{\rm G} - \pi\ln\pars{2}
+2\Im\bracks{\int_{0}^{1/z}{\ln\pars{1 - t} \over t - \ic z^{*}}\,\dd t
+ \int_{0}^{z}{\ln\pars{1 - t} \over t - \ic z}\,\dd t}
\\[2mm]&=2{\rm G} - \pi\ln\pars{2}
-2\Im\bracks{\int_{0}^{z^{*}}{{\rm Li}_{1}\pars{t} \over t - \ic z^{*}}\,\dd t
+ \int_{0}^{z}{{\rm Li}_{1}\pars{t} \over t - \ic z}\,\dd t}\tag{3}
\end{align}
  where ${{\rm Li_{s}}\pars{z}}$ is the PolyLogaritm Function.

Also,
$$
\int_{0}^{\xi}{{\rm Li}_{1}\pars{t} \over t - \ic \xi}\,\dd t
=-\ln\pars{1 - \xi}\ln\pars{\bracks{1 + \ic}\xi \over \xi + \ic}
+{\rm Li}_{2}\pars{\ic \over \xi + \ic}
+{\rm Li}_{2}\pars{-\ic\,{\xi - 1 \over \xi + \ic}}\tag{4}
$$

$\ds{\color{#88f}{\mbox{The final result was found by combining}\ \pars{1}, \pars{2}, \pars{3}\ \mbox{and}\ \pars{4}}}$.

