Do there exist other tricks for trig with Cauchy's Theorem?

Question

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 $\int_{\theta=0}^{2\pi} \frac{\mathrm{d}\theta}{a+b\sin\theta}$?

Answer 1

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 $\sin$.

Answer 2

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!

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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 $z=-(a/b)+\sqrt{(a/b)^2-1}$.