Question:
The value of $\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x$ is:
Correct Option: , 3
Solution:
$I=\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x$
$=\int_{-\pi / 2}^{0} \frac{1}{1+e^{\sin x}} d x+\int_{0}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x$
$=\int_{0}^{\pi / 2}\left(\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{-\sin x}}\right) d x$
$=\int_{0}^{\pi / 2} \frac{1+e^{\sin x}}{1+e^{\sin x}} d x=\frac{\pi}{2}$