Do there exist other tricks for trig with Cauchy's Theorem? I have noticed that $$\int_{\theta=0}^{2\pi} \frac{\mathrm{d}\theta}{a+b\cos\theta}=\frac{2\pi}{\sqrt{a^2-b^2}}$$ with $|b|<|a|$
Are there any other tricks like this for say $\int_{\theta=0}^{2\pi} \frac{\mathrm{d}\theta}{a+b\sin\theta}$?
 A: If one wishes to use contour integration to evaluate the integral of interest, then one simply enforces the substitution $z=e^{i\theta}$.  Thus, $\cos (\theta)=\frac{z+z^{-1}}{2}$, $d\theta =\frac{1}{iz}\,dz$ and we have
$$\begin{align}
\int_0^{2\pi}\frac{1}{a+b\cos(\theta)}\,d\theta&=\oint_{|z|=1}\frac{1}{a+b\left(\frac{z+z^{-1}}{2}\right)}\,\left(\frac{1}{iz}\right)\,dz\\\\
&=\frac{2}{ib}\oint_{|z|=1}\frac{1}{\left(z+(a/b)+\sqrt{(a/b)^2-1}\right)\left(z+(a/b)-\sqrt{(a/b)^2-1}\right)}\,dz \tag 1\\\\
&=2\pi i \left(\frac{2}{ib}\frac{1}{2\sqrt{(a/b)^2-1}}\right) \tag 2\\\\
&=\frac{2\pi}{\sqrt{a^2-b^2}}
\end{align}$$
as expected!

NOTE:
In going from $(1)$ to $(2)$ we applied the Residue Theorem.  Note that in proceeding, we recognizing that for $|a|>|b|$, the only pole inside $|z|=1$ is at $z=-(a/b)+\sqrt{(a/b)^2-1}$.  
A: Due to the periodicity of $\cos$, the integrand is periodic with period $2\pi$, so the following integrals integrate over a whole period and thus have the same value.
$$ \int_0^{2\pi} \frac{d\theta}{a + b\cos\theta} = \int_0^{2\pi} \frac{d\theta}{a + b\cos(\theta - \theta_0)}
$$
Now substitute something for $\theta_0$ to change it to $\sin$.
