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$\int_0^{2π}\frac{\cos3x}{5-4\cos x}dx$

I have to evaluate the following using complex Integration, I replaced $\cos3x$ as $\frac{z^6+1}{2z^3}$ and $\cos x$ as $\frac{z^2+1}{2z}$ also changed $dx$ as $\frac{dz}{iz}$ and on evaluation I got the poles $0 $ of order 3 and $\frac{1}{2}$ of order 1 however solving for the pole at $0$ by residue formula is quite a daunting task is there any trick to this or have I done anything wrong

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  • $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$
    – Shaun
    Apr 27, 2020 at 23:23

1 Answer 1

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$z =e^{i\theta}\\ dz = i e^{i\theta} d\theta\\ d\theta = \frac {1}{iz} \ dz$

$\oint_{|z| = 1} \frac {\frac {z^3 + z^{-3}}{2}}{iz(5 -4\frac {z + z^{-1}}{2})} \ dz$

$\oint_{|z| = 1} \frac {z^6 + 1}{-2iz^3(2z^2 - 5z + 2)}\ dz\\ \oint_{|z| = 1} \frac {z^6 + 1}{-2iz^3(2z - 1)(z - 2)}\ dz$

At $z = \frac 12$

$2\pi i \frac {\frac 1{2^6} + 1}{-4i (\frac 1{2^3})(-\frac 32)}\\ \frac {65}{24}\pi$

at $z = 0$ Hmmm. you are right, taking the second derivative:

$\frac {d^2}{dz^2} \left(\frac {z^6 + 1}{-2i(2z^2 - 5z + 1)}\right)$ does looks daunting.

How about partial fractions?

$\frac {z^6 + 1}{-2iz^3(2z^2 - 5z + 2)} = \frac{Ax^2 + Bx + C}{-2ix^3} + \frac {Dx + E}{2x^2 - 5x + 2}$

$2A = -2iD\\ -5A + 2B = -2i E\\ 2A - 5B + 2C = 0\\ 2B - 5C = 0\\ 2C = 1$

$C = \frac 12\\ B = \frac 54\\ A = \frac {21}{8}$

We don't actually need to solve for $D,E$ as those pertain to the residuals at $\frac 12, 2$

$2\pi i \text{Res}_{z=0}\left[\frac {z^6 + 1}{-2i(2z^2 - 5z + 1)}\right] = -A\pi$

$(\frac {65}{24} - \frac {21}{8})\pi\\ \frac {65 - 63}{24} \pi = \frac{\pi}{12}$

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  • $\begingroup$ Thanks I got it👍 $\endgroup$ Apr 28, 2020 at 0:41

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