$\int_0^{2π}\frac{\cos3x}{5-4\cos x}dx$
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
$z =e^{i\theta}\\ dz = i e^{i\theta} d\theta\\ d\theta = \frac {1}{iz} \ dz$
$\oint_{|z| = 1} \frac {\frac {z^3 + z^{-3}}{2}}{iz(5 -4\frac {z + z^{-1}}{2})} \ dz$
$\oint_{|z| = 1} \frac {z^6 + 1}{-2iz^3(2z^2 - 5z + 2)}\ dz\\ \oint_{|z| = 1} \frac {z^6 + 1}{-2iz^3(2z - 1)(z - 2)}\ dz$
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 $\frac 12, 2$
$2\pi i \text{Res}_{z=0}\left[\frac {z^6 + 1}{-2i(2z^2 - 5z + 1)}\right] = -A\pi$
$(\frac {65}{24} - \frac {21}{8})\pi\\ \frac {65 - 63}{24} \pi = \frac{\pi}{12}$