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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 $[\frac{\pi}{2},\pi]$.

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Some properties of definite integral must be used as the indefinite one is not elementary… – lab bhattacharjee Jun 12 '13 at 6:25
Mathematica gives a closed form for what it's worth: $$\frac{1}{16} \left(-\Im\left(4 \text{Li}_2\left(\frac{i}{2}\right)-4 \text{Li}_2\left(-\frac{i}{2}\right)\right)+\pi \log \left(\frac{6561}{400}\right)-8 \log (2) \tan ^{-1}\left(\frac{1}{3}\right)-\log (16) \tan ^{-1}\left(\frac{1}{2}\right)+\log (16) \tan ^{-1}(2)\right)$$ – Ron Gordon Jun 12 '13 at 6:50
I do not understand your question: if you can compute the integral on $[0,\pi]$, why can't you do the same on the smaller interval $[\frac{\pi}{2},\pi]$? – Avitus Jun 12 '13 at 7:23
First, we have$$\int_{0}^{\pi}\frac{x\sin{x}}{5-4\cos{x}}dx=\frac{\pi}{2}\log{\frac{3}{2}}‌​-\frac{1}{4}\int_{0}^{\pi}\log{(\frac{5}{4}-\cos{x})}dx$$Now let $I(q)=\int_{0}^{\pi}\log{(q-\cos{x})}dx$, we have$$I'(q)=\int_{0}^{\pi}\frac{1}{(q-\cos{x})}dx$$Next,let $t=\tan{\frac{x}{2}}$ and we can work it out:$$I'(q)=\frac{\pi}{\sqrt{q^2-1}}$$Then we integrate it back:$$I(q)=\pi\log{\frac{q+\sqrt{q^2-1}}{2}}$$.Thus,we have$$\int_{0}^{\pi}\frac{x\sin{x}}{5-4\cos{x}}d=\frac{\pi}{2}\log{\frac{3}{2}}-‌​\frac{I(\frac{5}{4})}{4}=\frac{\pi}{2}\log{\frac{3}{2}}$$.@Avitus – YMJou Jun 12 '13 at 8:10
But, when the integral limit becomes $[\frac{\pi}{2},\pi]$,we'll have $$I'(q)=\int_{\pi/2}^{\pi}\frac{1}{(q-\cos{x})}dx=\frac{2}{\sqrt{q^2-1}}(\pi/2-\‌​tan^{-1}{\frac{1}{\sqrt{q^2-1}}})$$,which I failed to integrate back.@Avitus – YMJou Jun 12 '13 at 8:29

$$\begin{align}\int_{\pi/2}^\pi\frac{x\sin x}{5-4\cos x}dx&=\pi\left(\frac{\ln3}2-\frac{\ln2}4-\frac{\ln5}8\right)-\frac12\operatorname{Ti}_2\left(\frac12\right)\\&=\pi\left(\frac{\ln3}2-\frac{\ln2}4-\frac{\ln5}8\right)-\frac12\Im\,\chi_2\left(\frac{\sqrt{-1}}2\right),\end{align}$$ 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: $$\frac{\sin x}{5-4\cos x}=\sum_{n=1}^\infty\frac{\sin n x}{2^{n+1}}.$$

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{\pi/2}^{\pi}{x\sin\pars{x} \over 5 - 4\cos\pars{x}}\,\dd x:\ {\large ?}}$

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

$\ds{\color{#00f}{\large\mbox{The final result is found by combining}\ \pars{1}, \pars{2}, \pars{3}\ \mbox{and}\ \pars{4}}}$.

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