The value of _- ^ 2y(1+ y)1+ ^2ydy is :

Question

The value of $\int _{-\pi }^{\pi }\frac{2y(1+\sin y)}{1+{\cos }^{2}y}dy$ is :

Accepted Answer

$\begin{gathered} I=\int _{-\pi }^{\pi }\frac{2y(1+\sin y)}{1+{\cos }^{2}y}dy \ =\int _{0}^{\pi }(\frac{2y(1+\sin y)}{1+{\cos }^{2}y}+\frac{-2y(1-\sin y)}{1+{\cos }^{2}y})dy \ =\int _{0}^{\pi }(\frac{2y+2y\sin y-2y+2y\sin y}{1+{\cos }^{2}y})dy \end{gathered}$

$I=4{\int }_{\pi }^{0}(\frac{y\sin }{1+{\cos }^{2}y})dy\,$

$\begin{gathered} I=4\int _{0}^{\pi }(\frac{(\pi -y)\sin (\pi -)}{1+{\cos }^{2}(\pi -y)})dy \ I=4[|\int _{0}(\frac{\pi \sin y}{1+{\cos }^{2}y})dy-\int _{0}^{\pi }\frac{y\sin y}{1+{\cos }^{2}y}dy]|\, \end{gathered}$

Adding equation (1) and (2)

$\begin{gathered} 2I=4\int _{0}^{\pi }(\frac{\pi \sin y}{1+{\cos }^{2}y})dy \ I=2\pi \int _{0}^{\pi }\frac{\sin y}{1+{\cos }^{2}y}dy \ =2\pi 	imes \frac{\pi }{2} \ ={\pi }^{2} \end{gathered}$

Solutions · 1