Correct option is (1) \(\pi ^2\)
\(\int\limits_{-\pi}^{\pi} \frac{2 y(1+\sin y)}{1+\cos ^{2} y} d y\)
\(=\underset{\text{(Odd)}}{\int\limits_{-\pi}^{\pi} \frac{2 y}{1+\cos ^{2} y} d y}+=\underset{\text{(Even)}}{\int\limits_{-\pi}^{\pi} \frac{2 y \sin y}{1+\cos ^{2} y} d y }\)
\(=0+2.2 \int\limits_{0}^{\pi} y\left(\frac{\sin y}{1+\cos ^{2} y}\right) d y\)
\(I=4 \int\limits_{0}^{\pi} \frac{y \sin y}{1+\cos ^{2} y} d y\)
\(I=4 \int\limits_{0}^{\pi} \frac{(\pi-y) \sin y}{1+\cos ^{2} y} d y\)
\(2 I=4 \int\limits_{0}^{\pi} \frac{\pi \sin y}{1+\cos ^{2} y} d y\)
\(I=2 \pi \int\limits_{0}^{\pi} \frac{\sin y}{1+\cos ^{2} y} d y\)
\(=2 \pi\left(-\tan ^{-1}(\cos \mathrm{y})\right)_{0}^{\pi}\)
\(=-2 \pi\left[\left(-\frac{\pi}{4}\right)-\left(\frac{\pi}{4}\right)\right]\)
\(=-2 \pi\left[-\frac{2 \pi}{4}\right]\)
\(=\pi^{2}\)
