Integrating $\int_0^\pi \frac{x\cos x}{1+\sin^2 x}dx$ 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

Let be
$$\begin{align} I=\int_0^\pi \frac{x\cos x}{1+\sin^2 x}\operatorname{d}x&=-\int_0^\pi \arctan(\sin x)\operatorname{d}x \\
&=-2\int_0^{\pi/2} \arctan(\sin x)\operatorname{d}x\\
&=-2\int_0^1\frac{\arctan t}{\sqrt{1-t^2}}\operatorname{d}t
\end{align}$$
Let $$ \displaystyle I(a) = \int_{0}^{1} \frac{\arctan at}{\sqrt{1-t^{2}}} \ dt.$$
Differentiating under the integral, 
$$ \begin{align}  I'(a) &= \int_{0}^{1} \frac{t}{(1+a^{2}t^{2})\sqrt{1-t^{2}}} \ dt \\ &=  \frac{1}{a \sqrt{1+a^{2}}} \text{arctanh} \left( \frac{a}{\sqrt{1+a^{2}}} \right)\\ 
&= \frac{1}{a \sqrt{1+a^{2}}} \text{arcsinh}(a) . 
\end{align}$$
Integrating back,
$$ \begin{align} I(1)-I(0) = I(1) &= \int_{0}^{1} \frac{\text{arcsinh}(a)}{a \sqrt{1+a^{2}}} \ da \\ &= - \text{arcsinh}(a) \text{arcsinh}(\frac{1}{a}) \Big|^{1}_{0} + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \ da \\ &= - \text{arcsinh}^{2}(1) +  \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \ da \\ &= - \ln^{2}(1+\sqrt{2}) + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \ da . \end{align}$$
Now let $ \displaystyle w = \frac{1}{a}$.
Then
$$ I(1) = - \ln^{2}(1+\sqrt{2}) + \int_{1}^{\infty} \frac{\text{arcsinh}(w)}{w \sqrt{1+w^{2}}}$$
$$ =  - \ln^{2}(1+\sqrt{2}) + I(\infty) - I(1) .$$
Therefore,
$$ \begin{align} I(1) &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{I(\infty)}{2} \\ &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi}{4} \int_{0}^{1} \frac{1}{\sqrt{1-t^{2}}} \ dt \\ &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi^{2}}{8} . \end{align}$$
Finally
$$\color{blue}{\int_0^\pi \frac{x\cos x}{1+\sin^2 x}\operatorname{d}x=-2I(1)=\ln^{2}(1+\sqrt{2})-\frac{\pi^{2}}{4}.}$$
