∫_-π^π 2 x(1+sin x) 1+cos ^2 x d x is

$$\begin{aligned} & \int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=\int_{-\pi}^\pi \frac{2 x}{1+\cos ^2 x}+2 \int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} \\ & =0+4 \int_0^\pi \frac{x \sin x d x}{1+\cos ^2 x} I=4 \int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} \\ & I=4 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} \Rightarrow I=4 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x}-4 \pi \int \frac{x \sin x}{1+\cos ^2 x} \Rightarrow 2 I=4 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x \end{aligned}$$​∫−ππ​1+cos2x2x(1+sinx)​dx=∫−ππ​1+cos2x2x​+2∫−ππ​1+cos2xxsinx​=0+4∫0π​1+cos2xxsinxdx​I=4∫0π​1+cos2(π−x)(π−x)sin(π−x)​I=4∫0π​1+cos2x(π−x)sinx​⇒I=4π∫0π​1+cos2xsinx​−4π∫1+cos2xxsinx​⇒2I=4π∫0π​1+cos2xsinx​dx​ put $\cos x=t$cosx=t and solve it.