I want to use the residue theorem to solve this integral: $$\int_0^{2\pi} \frac{1}{13+5\sin(\theta)}d\theta.$$ When applying the residue theorem, $\sin(\theta)$ becomes $\frac{z-\frac{1}z}{2i}$. When plugging this substitution into my integrand, the integrand simplifies as: $$\frac{2}i\int_C \frac{dz}{-5iz^2+26z-5i}.$$ However, I am having difficulty finding where the singularity is so that I can apply the residue theorem to this problem. Is there an easier way?

  • $\begingroup$ This is a quadratic polynomial. Its zeros are the singularities. BTW, I get $+5i$ instead of $-5i$. $\endgroup$ – Friedrich Philipp Mar 24 '17 at 5:28
  • $\begingroup$ My result is $\pi/6$. $\endgroup$ – Friedrich Philipp Mar 24 '17 at 5:36
  • $\begingroup$ I understand that it's considerably easier to just do the integral in the way it's presented, but I'm trying to apply the residue theorem. $\endgroup$ – Evan Mar 24 '17 at 5:38
  • $\begingroup$ I got my result using the residue theorem (or rather Cauchy's formula). $\endgroup$ – Friedrich Philipp Mar 24 '17 at 5:41

First, $$\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$$Now, let $z=e^{i\theta}$, then $\text{d}z=iz\text{d}\theta$, i.e the integral becoms $$I=\frac{1}{i}\int\limits_{|z|=1}\frac{1}{z\left(13+5\dfrac{z-z^{-1}}{2i}\right)}\text{d}z=\int\limits_{|z|=1}\underbrace{\frac{2}{5z^2+26iz-5}}_{f(z)}\text{d}z$$The poles are $$z_{1,2}=\dfrac{-26i\pm\sqrt{-676+100}}{10}=\dfrac{-26i\pm 24i}{10}$$ and only $z=-\dfrac{i}{5}$ is in the unit circle, i.e $$I=2\pi i\cdot\text{res}\left(f(z),-\dfrac{i}{5}\right)=\frac{\pi}{6}$$


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