Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

The correct option is A $0$

Let $I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\log \left(\frac{2-\sin \theta }{2+\sin \theta }\right)d\theta$
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\log \left(\frac{2-\sin (-\theta )}{2+\sin (-\theta )}\right)d\theta ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\log \left(\frac{2+\sin \theta }{2-\sin \theta }\right)d\theta$
$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-\log \left(\frac{2-\sin \theta }{2+\sin \theta }\right)d\theta =-I$
$\therefore 2I=0\Rightarrow I=0$