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How do I compute the following integral using contour integration: $$\int_0^{2\pi} \frac{\cos^2 \theta}{6-2\cos\theta}d\theta$$

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    $\begingroup$ Parametrize? The integral?...What's the question, anyway? What's your effort, what've you done so far? $\endgroup$ – DonAntonio Mar 18 '13 at 13:33
  • $\begingroup$ I'm supposed to re-express as a paramaerization of a complex line integral of a unit circle. I do not understand how to approach this problem. $\endgroup$ – Tom Mar 18 '13 at 13:35
  • $\begingroup$ Tom, is this $\int_0^{2\pi}\frac{\sin^2\theta}{6-4\cos\theta}\,d\theta$? $\endgroup$ – Ayman Hourieh Mar 18 '13 at 13:36
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    $\begingroup$ @Tom: I edited you question the way I understood it. Feel free to change it. Also, you can learn a little about presenting your questions in $\LaTeX$ by looking at my edit. $\endgroup$ – Dennis Gulko Mar 18 '13 at 13:39
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    $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Mar 18 '13 at 13:42
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Hint: Denote $\gamma(t)=e^{it}$. Then, for $t\in[0,2\pi]$ $\gamma$ is the contour of the unit circle. Use the definition of contour integration in reverse: $$\int_0^{2\pi}f(\cos t,\sin t)dt=\int_\gamma f\left(\frac12(z+z^{-1}),\frac1{2i}(z-z^{-1})\right)\frac{dz}{iz}$$ Now use Cauchy's integral formula.

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  • $\begingroup$ $\sin t=\frac{1}{2i}(z-z^{-1})$. $\endgroup$ – Sungjin Kim Mar 18 '13 at 15:11
  • $\begingroup$ Yeah, now corrected. $\endgroup$ – Dennis Gulko Mar 18 '13 at 16:04

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