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March 8th, 2017, 01:50 PM  #1 
Newbie Joined: Mar 2017 From: Glasgow Posts: 4 Thanks: 0  Contour Integration Problem
I'm struggling to evaluate this integral: $\displaystyle \int_{0}^{2\pi} {\frac{d\theta}{72\cos \theta}}$ I have used the substitution $\displaystyle z=e^{i\theta}$ and obtained $\displaystyle d\theta = i\frac{dz}{z}$ and substituted it in. I'm then left with the integrand equal to $\displaystyle \frac{1}{z^2 7z +1}$ after taking the constant $\displaystyle i$ out the front of the integral. This left me with a pole inside the unit circle equal to $\displaystyle \frac{73\sqrt{5}}{2}$ and I'm not quite sure how to proceed after this. I tried using the residue formula for a simple pole but it's not giving me the correct answer which is given in the question and is $\displaystyle \frac{2\sqrt{5}\pi}{15}$. So I think my method is wrong. Any help would be appreciated. 
March 8th, 2017, 02:40 PM  #2 
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 
First, what is your contour integral and what is your contour?

March 8th, 2017, 02:49 PM  #3 
Newbie Joined: Mar 2017 From: Glasgow Posts: 4 Thanks: 0 
I'm not given any more information. I thought I could use the contour as the unit circle centred at 0 since f(z) equal to the integrand given would be analytic inside the unit circle except at the pole calculated. Just generally a bit confused with the method.

March 8th, 2017, 03:45 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,100 Thanks: 1093 
did you multiply the residue by $2\pi i$ ?

March 8th, 2017, 04:35 PM  #5 
Newbie Joined: Mar 2017 From: Glasgow Posts: 4 Thanks: 0 
I think I am calculating the residue wrongly. I'm using Res(f,c) for the calculated simple pole c is equal to the limit as z tends to c of (zc)f(z) however I am getting 0 as an answer.

March 8th, 2017, 05:09 PM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,100 Thanks: 1093  
March 8th, 2017, 11:27 PM  #7 
Newbie Joined: Mar 2017 From: Glasgow Posts: 4 Thanks: 0 
Thanks a lot!


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