# Complex Integral using Residues

This is the question:

Find the integral using residue theorem.

$$\int_0^{2\pi}{d\theta \over1+8\cos^2\theta}$$

I solved it like this :

$$\int_0^{2\pi}{d\theta \over1+8\cos^2\theta}=\int_0^{2 \pi} {d\phi \over 5+4\cos\phi}$$

using $$2\cos^2\theta=\cos 2\theta+1 \quad\quad and \quad 2\theta=\phi$$

Then i took $z=e^{i\phi}$ , so th integral now becomes :

$$\int_C {1 \over (2z^2+5z+2)} {dz \over iz} \quad c:|z|=1$$

Now using the residue theorem on the obtained poles i get answer as $$4\pi \over 3$$

Could you please add a short description of the main steps of your computation? I've computed this integral in SWP and I got $\int_{0}^{2\pi }\frac{d\theta }{1+8\cos ^{2}\theta }=\frac{2\pi }{3}$ –  Américo Tavares Apr 26 at 13:36
Thanks. ${}{}{}{}{}{}$ –  Américo Tavares Apr 26 at 13:52
@rongrodon i see it now. but the substitution should be made to find the poles easily. what should be done then ? what is the logic for dividing by 2 there so that the limits $0$ to $4\pi$ becomes $0$ to $2 \pi$ –  Aman Mittal Apr 26 at 14:13