$\int_{0}^{2\pi}\dfrac{d\theta}{a+b\sin\theta}$ where $a,b>0$
Supposedly a very simple residue theorem problem but I'm stuck with the pole(s).
Let $z=e^{i\theta}$ and $d\theta=\frac{dz}{iz}$. Then, $$\begin{align*} \int_{\mathbb{D}}\frac{dz}{\left[a+b\left(\frac{z-z^{-1}}{2i}\right)\right]iz}&=\int_{\mathbb{D}}\frac{2idz}{(2ai+bz-bz^{-1})iz}\\ &=\int_{\mathbb{D}}\frac{2dz}{bz^2+2aiz-b} \end{align*}$$
By quadratic formula, we can deduce that we have pole at $$z=\frac{-ai\pm\sqrt{b^2-a^2}}{b}$$ But I'm having trouble figuring out which, of the two is in fact in the unit circle. All we have is $a>0$ and $b>0$ and I feel like it could go so many different ways depending on the actual values that they take. Can anyone help me to proceed?
For $a\leqslant b$ the integral clearly diverges.
Otherwise, as you've shown, the poles are $$z=\frac{-a\pm\sqrt{a^2-b^2}}{b}i,$$ and the pole with "$+$" is inside the unit circle. It remains to compute the residue.