# Complex Integration, pole of order 3

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

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

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

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

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.

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

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

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

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

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

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

• Thanks I got it👍 Apr 28, 2020 at 0:41