Question

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

• A. $1$
• B. $3$
• C. $2$
• D. $0$

Answer 1

The correct option is D $0$

We need to evaluate ${\int }_{-\pi /2}^{\pi /2}\log \left(\frac{2-\sin x}{2+\sin x}\right)dx$

$={\int }_{-\pi /2}^0\log \left(\frac{2-\sin x}{2+\sin x}\right)dx+{\int }_0^{\pi /2}\log \left(\frac{2-\sin x}{2+\sin x}\right)dx$

$=-{\int }_0^{-\pi /2}\log \left(\frac{2-\sin x}{2+\sin x}\right)dx+{\int }_0^{\pi /2}\log \left(\frac{2-\sin x}{2+\sin x}\right)dx$

$={\int }_0^{\pi /2}\log \left(\frac{2-\sin x}{2+\sin x}\right)dx+{\int }_0^{\pi /2}\log \left(\frac{2+\sin x}{2-\sin x}\right)dx$

$={\int }_0^{\pi /2}\log \left(1\right)dx=0$