Need help with $\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$ Please help me to evaluate this integral:
$$\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$$
Using substitution $x=2\arctan t$ it can be transformed to:
$$\int_0^\infty\frac{2}{1+t^2}\arctan^2\left(\frac{2t}{3+t^2}\right)dt$$
Then I tried integration by parts, but without any success...
 A: A Fourier analytic approach. If $x\in(0,\pi)$,
$$\begin{eqnarray*}\arctan\left(\frac{\sin x}{2+\cos x}\right) &=& \text{Im}\log(2+e^{ix})\\&=&\text{Im}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 2^n}\,e^{inx}\\&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 2^n}\,\sin(nx),\end{eqnarray*}$$
hence by Parseval's theorem:

$$ \int_{0}^{\pi}\arctan^2\left(\frac{\sin x}{2+\cos x}\right)\,dx=\frac{\pi}{2}\sum_{n\geq 1}\frac{1}{n^2 4^n}=\color{red}{\frac{\pi}{2}\cdot\text{Li}_2\left(\frac{1}{4}\right)}.$$


As a side note, we may notice that $\text{Li}_2\left(\frac{1}{4}\right)$ is quite close to $\frac{1}{4}$. 
By applying summation by parts twice we get:
$$ \sum_{n\geq 1}\frac{1}{n^2 4^n} = \color{red}{\frac{1}{3}-\frac{1}{12}}+\sum_{n\geq 1}\frac{1}{9\cdot 4^n}\left(\frac{1}{n^2}-\frac{2}{(n+1)^2}+\frac{1}{(n+2)^2}\right)$$
and the last sum is positive but less than $\frac{11}{486}$, since $f:n\mapsto \frac{1}{n^2}-\frac{2}{(n+1)^2}+\frac{1}{(n+2)^2}$ is a positive decreasing function on $\mathbb{Z}^+$.
A: Generalizing Jack's answer. Take $a\in\Bbb R\setminus\{0\}$, and write
$$J(a)=\int_0^\pi\arctan^2\left(\frac{\sin x}{a+\cos x}\right)dx.$$
From symmetry, we can write
$$J(a)=\frac12\int_{-\pi}^\pi\arctan^2\left(\frac{\sin x}{a+\cos x}\right)dx$$
Then
$$\arctan\left(\frac{\sin x}{a+\cos x}\right)=\Im\log(a+e^{ix})=\sum_{n\ge1}\frac{(-1)^{n+1}}{na^n}\sin(nx).$$
Then from Parseval's theorem,
$$\frac1\pi\int_{-\pi}^\pi\arctan^2\left(\frac{\sin x}{a+\cos x}\right)dx=\sum_{n\ge1}\left(\frac{(-1)^{n+1}}{na^n}\right)^2,$$
so that
$$J(a)=\frac\pi2\mathrm{Li}_2\left(\tfrac1{a^2}\right).$$
Unfortunately, your integral, given by $J(2)=\frac\pi2\mathrm{Li}_2(1/4)$, does not appear to have a closed form.
