Contour Integration with $\cos (n\theta)$: Integral $\int_0^\pi \frac{3\cos(n\theta)}{5+4\cos(n\theta)}d\theta$

Question

How can I calculate this integral using contour integration? $\displaystyle\int_0^\pi \frac{3\cos(n\theta)}{5+4\cos(n\theta)}d\theta$

I know I can start by using that $\cos(n\theta) = Re (e^{in\theta})$, but I get bogged down in the computing.

Answer 1

$$\int_0^{\pi} d\theta \frac{3 \cos{n \theta}}{5+4 \cos{ n\theta}} = \frac{3}{2}\operatorname{Re}{\int_0^{2 \pi} d\theta \frac{e^{i n \theta}}{5+4 \cos{ n\theta}}}$$

The latter integral is equal to

$$-i \oint_{|z|=1} dz \frac{z^{2 n-1}}{2 z^{2 n} + 5 z^n + 2} $$

The poles of the integrand inside the unit circle are at $z^n = -1/2$, or

$$z_k = 2^{-1/n} e^{i (1+2 k) \pi/n} $$

for $k \in \{0, 1, \ldots,n-1\}$.

The residue of each pole then is given by the numerator evaluated at that pole divided by the derivative of the denominator evaluated that that pole. For arbitrary $k$, we have as the residue

$$\frac{z_k^{2 n-1}}{4 n z_k^{2 n-1} + 5 n z_k^{n-1}} = \frac1{n}\frac1{4 + 5 z_k^{-n}} = -\frac1{6 n}$$

independent of $k$. Further, the integral is $i 2 \pi$ times the sum of the residues of the poles inside the unit circle, so we simply multiply this residue by $n$ and we find, for all $n$, that

$$\int_0^{\pi} d\theta \frac{3 \cos{n \theta}}{5+4 \cos{ n\theta}} = -\frac{\pi}{2}$$