Show that $\int_0^\pi\frac{x \sin x}{5-3\cos x}\, \mathrm dx = \frac{2 \pi \log (4/3)} 3$

Question

Show that $$\int_0^\pi\frac{x \sin x}{5-3\cos x}\, \mathrm dx = \frac{2 \pi \log (4/3)} 3$$

I am struggling with this one. I have tried substitutions and the residue theorem but haven't got anywhere. It would be nice to see an 'elementary' solution if there is one, but any will do.

Answer 1

There's a simple trick: $$\sum_{n\geq 1}\frac{\sin(nx)}{3^n} = \frac{3}{2}\cdot\frac{\sin x}{5-3\cos x}\tag{1} $$ and $$ \int_{0}^{\pi} x \sin(nx)\,dx = \frac{\pi(-1)^{n+1}}{n}\tag{2}$$ hence $$ \int_{0}^{\pi}\frac{x\sin(x)}{5-3\cos x}\,dx = \frac{2\pi}{3}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 3^n}=\frac{2\pi}{3}\log\left(1+\frac{1}{3}\right) \tag{3}$$ nice & easy.

Answer 2

First integrate by parts to get: \begin{align} \int^\pi_0\frac{x\sin(x)}{5-3\cos(x)}\,dx= \frac{\pi}{3}\log(8)-\frac{1}{3}\int^\pi_0 \log(5-3\cos(x))\,dx \end{align} The integral on the right, can be done with many methods, for example, notice that we have with substituting $u=\pi-x$ that: \begin{align} \int^\pi_0 \log(5-3\cos(x))\,dx&=\frac{1}{2}\int^\pi_0 \log(25-9\cos^2(x))\,dx\\ &=\frac{\pi}{2}\log(25)+\frac{1}{2}\int^\pi_0\log\left(1-\frac{9}{25}\cos^2(x)\right)\,dx\\ &=\frac{\pi}{2}\log(25)+\int^{\pi/2}_0\log\left(1-\frac{9}{25}\sin^2(x)\right)\,dx\\ \end{align} We know from this link that: $$\int^{\pi/2}_0\log\left(1-\frac{9}{25}\sin^2(x)\right)\,dx=\pi\log(9/10)$$ Putting everything together yields: \begin{align} \int^\pi_0\frac{x\sin(x)}{5-3\cos(x)}\,dx=\frac{\pi}{3}\log(8)-\frac{1}{3}\left( \frac{\pi}{2}\log(25)+\pi\log(9/10)\right)=\frac{2\pi\log(4/3)}{3}\end{align}

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{% \int_{0}^{\pi}{x\sin\pars{x} \over 5 - 3\cos\pars{x}}\,\dd x}} = \left.\Re\int_{x\ =\ 0}^{x\ =\ \pi}{\bracks{-\ic\ln\pars{z}}\bracks{\pars{1 - z^{2}}\ic/\pars{2z}} \over 5 - 3\bracks{\pars{1 + z^{2}}/\pars{2z}}}\, {\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left.-\,{1 \over 3}\,\Im\int_{x\ =\ 0}^{x\ =\ \pi}\ln\pars{z}\, {\pars{1 - z^{2}} \over \pars{z - 1/3}\pars{z - 3}}\, {\dd z \over z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \qquad\pars{\begin{array}{l} \mbox{with} \\ \ds{-\,{\pi \over 2} < \arg\pars{z} < {3\pi \over 2}} \end{array}} \\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\, &\ {1 \over 3}\,\Im\int_{-1}^{-\epsilon}\bracks{\ln\pars{-x} + \ic\pi}\, {\pars{1 - x^{2}} \over \pars{x - 1/3}\pars{x - 3}}\,{\dd x \over x} \\[2mm] + &\ {1 \over 3}\,\Im\int_{\pi}^{0}\bracks{\ln\pars{\epsilon} + \ic\theta}\ \,{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over \epsilon\expo{\ic\theta}} + {1 \over 3}\,\Im\int_{\epsilon}^{1/3-\epsilon} \ln\pars{x}\,{\pars{1 - x^{2}} \over \pars{x - 1/3}\pars{x - 3}}\, {\dd x \over x} \\[2mm] + &\ {1 \over 3}\,\Im\int_{\pi}^{0}\ln\pars{1 \over 3}\ {8/9 \over \pars{\epsilon\expo{\ic\theta}}\pars{-8/3}}\,{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over 1/3} + {1 \over 3}\,\Im\int_{1/3 + \epsilon}^{1} \ln\pars{x}\,{\pars{1 - x^{2}} \over \pars{x - 1/3}\pars{x - 3}}\, {\dd x \over x} \\[1cm] = &\ -\,{1 \over 3}\,\pi\int_{\epsilon}^{1}{1 - x^{2} \over \pars{x + 1/3} \pars{x + 3}}\,{\dd x \over x} - {1 \over 3}\,\pi\ln\pars{\epsilon} - {1 \over 3}\,\pi\ln\pars{3} \\[5mm] = &\ -\,{1 \over 3}\,\pi\int_{\epsilon}^{1}\bracks{{1 - x^{2} \over \pars{x + 1/3} \pars{x + 3}} - 1}\,{\dd x \over x}\ \underbrace{- {1 \over 3}\,\pi\int_{\epsilon}^{1}{\dd x \over x} - {1 \over 3}\,\pi\ln\pars{\epsilon}}_{\ds{=\ 0}}\ - {1 \over 3}\,\pi\ln\pars{3} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\, &\ \pi\ \underbrace{\int_{0}^{1}\pars{{1 \over 1 + 3x} + {1 \over 9 + 3x}}\dd x} _{\ds{{1 \over 3}\,\ln\pars{16 \over 3}}} - {1 \over 3}\,\pi\ln\pars{3} = \bbx{{2 \over 3}\,\pi\ln\pars{4 \over 3}} \end{align}