\(\int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} d x=\)

Let \(f(x)=\frac{x \sin x}{1+\cos ^2 x}\)
\(f(-x)=\frac{(-x) \sin (-x)}{1+\cos ^2 x}=\frac{x \sin x}{1+\cos ^2 x}\)
\(\therefore \mathrm{f}(\mathrm{x})=\mathrm{f}(-\mathrm{x}) \Rightarrow \mathrm{f}(\mathrm{x})\) is an even function
Let \(I=\int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} d x\)
\(\therefore \mathrm{I}=2 \int_0^\pi \frac{\mathrm{x} \sin \mathrm{x}}{1+\cos ^2 \mathrm{x}} \mathrm{dx}\)
\(=2 \int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+[\cos (\pi-x)]^2} d x=2 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} d x\)
Eq. (1) \(+(2)\) gives,
\(
2 I=2 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 \mathrm{x}}
\)
Put \(\cos x=t \Rightarrow \sin x d x=-d t\). Also when \(x=0, t=1\) and when
\( \mathrm{x}=\pi, \mathrm{t}=-1 \)
\( \therefore 2 \mathrm{I}=2 \pi \int_1^{-1} \frac{-\mathrm{dt}}{1+\mathrm{t}^2} \)
\( =2 \pi \int_{-1}^1 \frac{\mathrm{dt}}{1+\mathrm{t}^2}=4 \pi \int_0^1 \frac{\mathrm{dt}}{1+\mathrm{t}^2}=4 \pi\) \(\left[\tan ^{-1}\right]_0^2=4 \pi\left(\frac{\pi}{4}\right)=\pi^2 \)
\( \therefore \mathrm{I}=\frac{\pi^2}{2}\)