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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.

brbrbrbr
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    Use Residue theorem or Cauchy integral formula. – Nikita Evseev Oct 14 '15 at 11:29
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    Should the condition be $a>b>0$? Otherwise things are going to get extremely messy. – David Oct 14 '15 at 11:32
  • @NikitaEvseev yes I know, is with Residue theorem... but how? – brbrbrbr Oct 14 '15 at 11:32
  • 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. – Ben Grossmann Oct 14 '15 at 11:32
  • In particular: if $a/b < 1$, then both poles lie inside the contour. At $a = b$, something strange happens. If $a/b > 1$, then exactly one pole lies inside the contour. – Ben Grossmann Oct 14 '15 at 11:35
  • Is it really $(a+b\cos^2\theta)^2$ in the denominator? If so, your final integral is wrong, this would have come from $(a+b\cos\theta)^2$ in the denominator. Also there is a factor of $1/b^2$ missing. – David Oct 14 '15 at 11:39
  • @Omnomnomnom thanks a lot. I try that. – brbrbrbr Oct 14 '15 at 11:40
  • @David Yes it is, I think I wrote it wrong. Wait. – brbrbrbr Oct 14 '15 at 11:44
  • Alternatively you could use a weierstrass substitution to bring the integral into a form where it is much clearer which poles you have to pick up and which not... – tired Oct 14 '15 at 11:57
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    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? – mickep Oct 14 '15 at 12:10
  • @ItzelBlum (+1)From a learning perspective, it might be helpful to practice this problem first with (appropriate) values for $a$ and $b$, say pick $a=4$ and $b=3$. Now try the Residue theorem and see what you get. Once you get the drift where the numbers "are going" you may be able to generalize. This always worked well for a person like me who is rather slow in learning new concepts.. – imranfat Oct 14 '15 at 12:30

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