 Solution

Use the positively oriented unit circle and $z=e^{i\theta }$. Writing $d=\sqrt{a^2-b^2}$, the contour integral for a sine-squared trigonometric quotient becomesThere is a double pole at zero and simple poles at $r_{\pm }=(-a\pm d)/b$. Since $a>b>0$, $|r_{+}|<1$ and $|r_{-}|>1$. The coefficient of $z^{-1}$ at zero is $a/(i{b}^2)$. At $r=r_{+}$, the residue isbecause $(r-r^{-1}{)}^2=4d^2/b^2$. The residue theorem consequently givesFor the second integral, use $x=\cos (\theta /2)$ with $0\le \theta \le \pi$. The square root is $\sin (\theta /2)$, and reversing the integration limits givesThe last factor uses the symmetry of the integrand about $\pi$ over a full period. Therefore