Question

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

Answer 1

$$\begin{gathered} \mathrm{Let},I={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{a-\sin \theta }{a+\sin \theta }\right)d\theta \\ \mathrm{Here},f\left(\theta \right)=\log \left(\frac{a-\sin \theta }{a+\sin \theta }\right) \\ \mathrm{Consider},f\left(-\theta \right)=\log \left[\frac{a-\sin \left(-\theta \right)}{a+\sin \left(-\theta \right)}\right]=-\log \left(\frac{a-\sin \theta }{a+\sin \theta }\right)=-f\left(\theta \right) \\ \mathrm{i}.\mathrm{e}.,f\left(\theta \right)\mathrm{is}\mathrm{odd}\mathrm{function}. \\ \mathrm{Therefore},I=0 \end{gathered}$$