Evaluate $\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x$

Let $I = \int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x$.
By using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have:
$I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} dx = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} dx$.
$I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx - \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx$.
$I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx - I$.
$2I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$.
$I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$.
Let $\cos x = t$,then $-\sin x dx = dt$. When $x=0, t=1$ and when $x=\pi, t=-1$.
$I = \frac{\pi}{2} \int_{1}^{-1} \frac{-dt}{1 + t^2} = \frac{\pi}{2} \int_{-1}^{1} \frac{dt}{1 + t^2}$.
Since $\frac{1}{1+t^2}$ is an even function,$I = \frac{\pi}{2} \times 2 \int_{0}^{1} \frac{dt}{1 + t^2} = \pi [\tan^{-1} t]_{0}^{1}$.
$I = \pi (\tan^{-1}(1) - \tan^{-1}(0)) = \pi (\frac{\pi}{4} - 0) = \frac{\pi^2}{4}$.