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

Let $\quad I=\int_{-\pi}^\pi\left[\frac{2 x(1+\sin x)}{\left(1+\cos ^2 x\right)}\right] d x$ $=\int_{-\pi}^\pi\left[\frac{2 x}{\left(1+\cos ^2 x\right)}\right] d x+\int_{-\pi}^\pi$ $\left[\frac{2 x(\sin x)}{\left(1+\cos ^2 x\right)}\right] d x$ $I=I_1+I_2$ Take $I_1=\int_{-\pi}^\pi\left[\frac{2 x}{\left(1+\cos ^2 x\right)}\right] d x$ $\begin{aligned} & \Rightarrow \quad f(x)=\left[\frac{2 x}{1+\cos ^2 x}\right] \\ & \Rightarrow f(-x)=\left[\frac{2(-x)}{1+\cos ^2(\pi-\mathrm{x})}\right]\end{aligned}$ $\begin{aligned} & =-\left[\frac{2\left({ }^x\right)}{1+(-\cos x)^2}\right] \\ & =-\left[\frac{2\left({ }^x\right)}{1+\cos ^2 x}\right]\end{aligned}$ $=-f\left({ }^x\right)$ $\Rightarrow \quad I_1=\int_{-\pi}^\pi \frac{2 x}{1+\cos ^2 x} d x$ $=0 \quad($ since $f(x)$ is an odd function) Again consider $\begin{aligned} & \Rightarrow \mathrm{g}(\mathrm{X})=\frac{2 \mathrm{x} \sin \mathrm{x}}{1+\cos ^2 \mathrm{x}} \\ & \Rightarrow \mathrm{g}(-\mathrm{x})=\frac{2(-\mathrm{x}) \sin (-\mathrm{x})}{1+\cos ^2(-\mathrm{x})}\end{aligned}$ $=\frac{2(\mathrm{x}) \sin (\mathrm{x})}{1+\cos ^2(\mathrm{x})}$ $[$ since $\sin (-x)=-\sin x, \cos (-x)=\cos x]$ $=\mathrm{g}\left({ }^{\mathrm{X}}\right)$ $\Rightarrow \quad I_2=\int_{-\pi}^\pi \frac{2 x \sin x}{1+\cos ^2 x} d x$ $=2 \times 2 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$ $=4 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$...(i) [since $\mathrm{g}(\mathrm{x})$ is an even function] Then, $\Rightarrow \quad I_2=4 \int_0^\pi \frac{(\pi-\mathrm{x}) \sin (\pi-\mathrm{x})}{1+\cos ^2(\pi-\mathrm{x})} d x$ $=4 \int_0^\pi \frac{(\pi-\mathrm{x}) \sin (\mathrm{x})}{1+\cos ^2(\mathrm{x})} d x$ ...(ii) $\left[\int_0^a f(x) d x=\int_0^a f(a-x) d x\right]$ Adding (i)and (ii), we get $\Rightarrow \quad 2 I_2=4 \pi \int_0^\pi \frac{\sin \mathrm{x}}{1+\cos ^2 \mathrm{x}} \mathrm{dx}$ Substitute $\cos x=z$ $-\sin x d x=d z$ when $x \rightarrow 0, z \rightarrow 1$ and when $x \rightarrow \pi, \mathrm{z} \rightarrow-1$ $\Rightarrow \quad 2 I_2=-4 \pi \int_1^{-1} \frac{d z}{1+z^2}$ $\begin{array}{ll}\Rightarrow & 2 \mathrm{I}_2=-4 \pi\left[\tan ^{-1} \mathrm{z}\right]_1^{-1} \\ \Rightarrow & 2 \mathrm{I}_2=-4 \pi\left[-\frac{\pi}{4}-\frac{\pi}{4}\right] \\ \Rightarrow & 2 \mathrm{I}_2=2 \pi^2 \\ \Rightarrow & \mathrm{I}_2=\pi^2\end{array}$ $\Rightarrow \quad I=I_1+I_2$ $\begin{aligned} & =0+\pi^2 \\ & =\pi^2\end{aligned}$