Answer:
Option D
Explanation:
We have ,l=$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log\left(\frac{2-\sin x}{2+\sin x}\right)dx$
Let $f(x)=\log\left(\frac{2-\sin x}{2+\sin x}\right)$
Then, $f(-x)=\log\left(\frac{2-\sin (-x)}{2+\sin (-x)}\right)$
$=\log\left(\frac{2+\sin x}{2-\sin x}\right)=\log\left(\frac{2-\sin x}{2+\sin x}\right)^{-1}$
$=-\log\left(\frac{2-\sin x}{2+\sin x}\right)=-f(x)$
then, f(x) is odd function.
$\therefore$ $\int_{-\pi/2}^{\pi/2} f(x) dx=0$
[ $\because $ f(x) is an odd function, then $\int_{-a}^{a} f(x) dx=0$]
