Question

Evaluate : ${\int }_{-\pi /2}^{\pi /2}\log \left(\frac{2-\sin x}{2+\sin x}\right)dx$.

Answer 1

Let $I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\log \left(\frac{2-\sin x}{2+\sin x}\right)dx$

Here $f\left(x\right)=\log \left(\frac{2-\sin x}{2+\sin x}\right)$

$f(-x)=\log \left(\frac{2-\sin (-x)}{2+\sin (-x)}\right)$

$=\log \left(\frac{2+\sin x}{2-\sin x}\right)$

$=-\log \left(\frac{2-\sin x}{2+\sin x}\right)$

$=-f\left(x\right)$

Hence $f\left(x\right)$ is an odd function.

$\therefore ,I=0$