Using the residue theorem, Evaluate $\int_{0}^{2\pi}\cfrac{\cos{2 \theta}}{5+4 \cos {\theta}} d\theta$.

Solution:

Consider, $\int_{0}^{2\pi}\cfrac{\cos{2 \theta}}{5+4 \cos {\theta}} d\theta$

Put $z=e^{j \theta}$ -----------(1)

$dz=j e^{j \theta} d\theta$

$d\theta=\cfrac{dz}{jz}$ ---------------(2)

$\cos {\theta} = \cfrac{e^{j \theta} + e^{-j \theta}}{2} =\cfrac{z+z^{-1}}{2}=\cfrac{z+\cfrac{1}{z}}{2} = \cfrac{z^{2}+1}{2z}$

But $\cos {2 \theta}=R.P(e^{2j \theta}) = R.P(e^{j\theta})^{2}=z^{2}$

$\int_{0}^{2\pi}\cfrac{\cos{2 \theta}}{5+4 \cos {\theta}} d\theta= \int_{c}\cfrac{z^{2}}{5+4\left( \cfrac{z^{2}+1}{2z}\right)} \cfrac{dz}{iz}= \int_{c}\cfrac{z}{5+2\left( \cfrac{z^{2}+1}{z}\right)} \cfrac{dz}{i}$

$= \int_{c}\cfrac{z^{2}}{i(5z+2z^{2}+2)} dz$ where c is the circle $|z|=1$

$= \int_{c}\cfrac{z^{2}}{i(2z+1)(z+2)} dz$

Now the poles are given by

$(2z^{2}+5z+2)=0$

$(2z+1)(z+2)=0$

$\therefore \ z=\cfrac{-1}{2} \ \& \ z=-2$

The pole $z=\cfrac{-1}{2}$ lies inside the circle and the pole $z=-2$ lies outside the circle.

$\therefore$ Residue of $f(z)$ at $z=\cfrac{-1}{2} =\lim _{ z\longrightarrow { z }_{ o } }{ (z-{ z }_{ o })f(z) } = \lim _{ z\longrightarrow \frac{-1}{2} }{ (z+\cfrac{1}{2})f(z) }\cfrac{z^{2}}{i} $

$= \lim _{ z\longrightarrow \frac{-1}{2} }{ \left(z+\cfrac{1}{2} \right)}\cfrac{z^{2}}{i \left[ 2 \left(z+\cfrac{1}{2} \right) (z+2) \right]}= \lim _{ z\longrightarrow \frac{-1}{2} } \cfrac{z^{2}}{i \left[ 2 (z+2) \right]} = \cfrac{\left( \cfrac{-1}{2} \right)^{2}}{2i \left[\cfrac{-1}{2}+2\right]} = \cfrac{ \cfrac{1}{4} }{2i \left[\cfrac{3}{2}\right]}$

$=\cfrac{1}{12i}$

$\int_{0}^{2\pi}\cfrac{\cos{2 \theta}}{5+4 \cos {\theta}} d\theta=2\pi i$ (Residue at $\cfrac{-1}{2}$) $=2 \pi i \times \cfrac{1}{12i} = \cfrac{\pi}{6}$