TITLE: Integrating $\int_0^\pi \frac{x\cos x}{1+\sin^2 x}dx$
QUESTION [22 upvotes]: I am working on $\displaystyle\int_0^\pi \frac{x\cos x}{1+\sin^2 x}\,dx$
First: I use integrating by part then get 
$$ x\arctan(\sin x)\Big|_0^\pi-\int_0^\pi \arctan(\sin x)\,dx $$
then I have $\displaystyle -\int_0^\pi \arctan(\sin x)\,dx$ because $x\arctan(\sin x)\Big|_0^\pi$ is equal to $0$
However, I don't know how to integrate $\displaystyle -\int_0^\pi \arctan(\sin x)\,dx$
Can someone give me a hint?
Thanks

REPLY [4 votes]: Here is another approach.
The idea, in order to evaluate
$$
\int_0^{\pi}\frac{x \cos{x}}{1+\sin^2{x}} {\rm d}x ,
$$
is to start with a Fourier series expansion of $\displaystyle \frac{1}{1+\sin^2{x}} $. 
For $0\leq x \leq\pi$, we easily obtain
$$\begin{align}
 \frac{1}{1+\sin^2{x}} &= \Re \frac{1}{1-i\sin{x}}\\\\
&=  \Re \frac{2e^{ix}}{2-(e^{ix}-1)^2}\\\\
 &=  \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\sum_{n=2}^{\infty}\left(\sqrt{2}-1\right)^n(1+(-1)^n)\cos (nx) \tag2
\end{align}
$$
The series in $(2)$ is normally convergent on $[0,\pi]$, we are then allowed to integrate termwise:
$$
\int_0^{\pi}\frac{x \cos{x}}{1+\sin^2{x}} {\rm d}x =\frac{\sqrt{2}}{2}\sum_{n=2}^{\infty}(\sqrt{2}-1)^n(1+(-1)^n)\!\!\int_0^{\pi}\!\! x\cos{x}\cos (nx)x \:{\rm d}x. \tag3
$$
The latter integral is classic, we write $$ 2\cos{x}\cos (nx)=\cos((n+1)x)+\cos((n-1)x)$$ then integrate twice by parts to get
$$
-2\int_0^{\pi} x\cos{x}\cos (nx)x \:{\rm d}x=\frac{1}{(n+1)^2}+\frac{1}{(n-1)^2},\quad n=2,3,\ldots. \tag4
$$
It is easy to see that
$$
\sum_{n=2}^{\infty}(1+(-1)^n)\left(\frac{x^n}{(n+1)^2}+\frac{x^n}{(n-1)^2}\right)=\left(x+\frac1x\right)\left({\rm{Li}}_2(x)-{\rm{Li}}_2(-x)\right), \quad |x|<1,
$$ where ${\rm{Li}}_2(\cdot)$ is the dilogarithm function.
Now, putting $x:= \sqrt{2}-1$ in the preceding identity and using the following closed form  $$\rm{Li}_2{(\sqrt{2}-1)}-\rm{Li}_2{(-(\sqrt{2}-1))} = \frac{\pi^2}{8} - \frac12 \log^2{(\sqrt{2}+1)}, \tag5$$ as previously explaned by Ron Gordon, gives easily

$$
\int_0^{\pi}\frac{x \cos{x}}{1+\sin^2{x}} {\rm d}x =\log^2{(\sqrt{2}+1)}-\frac{\pi^2}{4}.
$$