Mathematics - Definite Integration Question with Solution | TestHub

${\int }_{-\pi }^{\pi }\frac{2y(1+\sin y)}{1+{\cos }^{2}y}dy$ \(=\underset{\text{(Odd)}}{\int_{-\pi}^\pi \frac{2 y}{1+\cos ^2 y} d y}+\underset{\text{(Even)}}{\int_{-\pi}^\pi \frac{2 y \sin y}{1+\cos ^2 y} d y}\)
$=0+2.2{\int }_0^{\pi }y\left(\frac{\sin y}{1+{\cos }^{2}y}\right)dy$ $\begin{array}{rl}& I=4{\int }_0^{\pi }\frac{y\sin y}{1+{\cos }^{2}y}dy\\ & I=4{\int }_0^{\pi }\frac{(\pi -y)\sin y}{1+{\cos }^{2}y}dy\\ & 2I=4{\int }_0^{\pi }\frac{\pi \sin y}{1+{\cos }^{2}y}dy\\ & I=2\pi {\int }_0^{\pi }\frac{\sin y}{1+{\cos }^{2}y}dy\\ & =2\pi {\left(-{\tan }^{-1}(\cos y)\right)}_0^{\pi }\\ & =-2\pi \left[\left(-\frac{\pi }{4}\right)-\left(\frac{\pi }{4}\right)\right]\\ & =-2\pi \left[-\frac{2\pi }{4}\right]={\pi }^2\end{array}$