Question

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

Answer 1

$\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\log \left(\frac{2-\sin \theta }{2+\sin \theta }\right)d\theta$

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

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

$=\log \frac{2+\sin \theta }{2-\sin \theta }$

$=-\log \frac{2-\sin \theta }{2+\sin \theta }$

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

So, $f\left(\theta \right)$ is an odd function , therefore $\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\log \left(\frac{2-\sin \theta }{2+\sin \theta }\right)d\theta =0$