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I need evaluate the following integral using residue theorem: With $a>0$ and $b>0$

$\int_0^{2 \pi} {d \theta \over (a + b \cos^2 \theta)^2}$

I have this:

$\int_0^{2 \pi} {d \theta \over (a + b \cos^2 \theta)^2} = \int_{|z| = 1}{ {16z^4} \over{bz^4 + (2a+b)2z^2 +b}} {1 \over iz} dz$

But I do not know how to follow, I really need help.

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    $\begingroup$ Use Residue theorem or Cauchy integral formula. $\endgroup$ – Nikita Evseev Oct 14 '15 at 11:29
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    $\begingroup$ Should the condition be $a>b>0$? Otherwise things are going to get extremely messy. $\endgroup$ – David Oct 14 '15 at 11:32
  • $\begingroup$ @NikitaEvseev yes I know, is with Residue theorem... but how? $\endgroup$ – brbrbrbr Oct 14 '15 at 11:32
  • $\begingroup$ The poles are at $$ \frac ab \pm \sqrt{\left(\frac{a}{b}\right)^2 - 1} $$ you need to find out for which values of $a/b$ do each of these lie within the relevant contour. $\endgroup$ – Omnomnomnom Oct 14 '15 at 11:32
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    $\begingroup$ One could also note that this integral equals $-\frac{d}{da}\int_0^{2\pi}\frac{1}{a+b \cos^2\theta}\,d\theta$. The latter integral has been taken care of several times on this site. But maybe the purpose is to practice residue calculus? $\endgroup$ – mickep Oct 14 '15 at 12:10

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