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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

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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 '13 at 13:36
@Américo Tavares i have added my steps – Aman Mittal Apr 26 '13 at 13:50
Thanks. ${}{}{}{}{}{}$ – Américo Tavares Apr 26 '13 at 13:52
It makes sense that your result is twice the result that @Americo produced: your substitution has you going around the circle twice. – Ron Gordon Apr 26 '13 at 14:02
@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 '13 at 14:13

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 $1/2$.

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