Question

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

Answer 1

Property:
${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_{-\pi /2}^{\pi /2}\log \left(\frac{2-\sin x}{2+\sin x}\right)dx$ (1)
Using above property:
$I={\int }_{-\pi /2}^{\pi /2}\log \left(\frac{2-\sin (-x)}{2+\sin (-x)}\right)dx$
$⟹I={\int }_{-\pi /2}^{\pi /2}\log \left(\frac{2+\sin x}{2-\sin x}\right)dx$ (2)
Adding (1) and (2):
$2I={\int }_{-\pi /2}^{\pi /2}\log 1dx=0$