 Solution

First use $\sin \theta /(a-\sin \theta )=a/(a-\sin \theta )-1$. If $J={\int }_{-\pi }^{\pi }(a-\sin \theta {)}^{-1}d\theta$, the required integral is $aJ-2\pi$.

Set $z=e^{i\theta }$. The unit circle is traversed counterclockwise once, $d\theta =dz/(iz)$, and $\sin \theta =(z-z^{-1})/(2i)$. ThusThe function $h(z)/z=-2i/(z^2-2iaz-1)$ is regular at zero. Its only enclosed pole is $z_{-}$ from part (b), and its residue isBy the residue theorem, $J=(1/i)2\pi i/\Delta =2\pi /\Delta$. ThereforeAs a sign and scale check, this is positive and behaves as $\pi /a^2$ for large $a$, in agreement with expanding the integrand and integrating its first nonzero, ${\sin }^2\theta /a^2$, term.