TITLE: Complex Integral using Residues
QUESTION [5 upvotes]: 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 $$4\pi \over 3$$
Can someone please verify it

REPLY [2 votes]: Alternatively, you can write
$$\cos^2{\theta} = \frac14 \left ( e^{i \theta} + e^{-i \theta}\right )^2$$
so that you end up with
$$-i \oint_{|z|=1} \frac{dz}{z} \frac1{1+2 (z^2+2 + z^{-2})} = -i \oint_{|z|=1} dz \frac{z}{2 z^4+5 z^2+2}$$
Now if you want, you can sub $\zeta=z^2$ (which is fine because when $|z|=1$ then $|\zeta|=1$), and get
$$-i \frac12 \oint_{|z|=1} \frac{dz}{2 z^2+5 z+2}$$
so if you did the residue calculation right, you are off by a factor of $1/2$.