Question

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

Answer 1

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

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

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

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

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

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

Thus $f\left(\theta \right)$ is an odd function

Therefore,${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\log \left(\frac{2-\sin \theta }{2+\sin \theta }\right)=0$