<s>[INST] Show that $\int_0^\pi\frac{x \sin x}{5-3\cos x}\, \mathrm dx = \frac{2 \pi \log (4/3)} 3$ 
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.
 [/INST] 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}
</s>