Solution:

Given, $\int_0^{2\pi} \frac{cos2\theta}{5+4cos\theta} d\theta$

Put $z=e^{i\theta}$

$dz=ie^{i\theta} d\theta $

$dz=iz d\theta $

$d\theta=\frac{dz} {iz}$

$cos\theta =\frac{e^{i\theta}+e^{-i\theta}}{2}$

$cos\theta =\frac{z+z^{-1}}{2}$

$cos\theta =\frac{z+\frac{1}{z}}{2}$

$\int_0^{2\pi} \frac{cos2\theta}{5+4cos\theta} d\theta$ =$\int_c \frac{z^2}{5+4\frac{z+\frac{1}{z}}{2}} \frac{dz}{iz}$

=$\int_c \frac{z^2}{5+4\frac{z^2+1}{2z}} \frac{dz}{iz}$

=$\int_c \frac{z^2}{\frac{10z+4z^2+4}{2z}} \frac{dz}{iz}$

=$\int_c \frac{2z^3}{4z^2+10z+4} \frac{dz}{iz}$

=$\int_c \frac{2z^3}{4z^2+10z+4} \frac{dz}{iz}$

=$\frac{1}{i}\int_c \frac{2z^2}{2(2z^2+5z+2)} \frac{dz}{iz}$

=$\frac{1}{i}\int_c \frac{z^2}{(2z^2+5z+2)} \frac{dz}{iz}$

=$\frac{1}{i}\int_c \frac{z^2}{(2z+1)(z+2)}$

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

$z=\frac{-1}{2} \ and \ z=-2$

The pole $z=\frac{-1}{2}$ lies inside the unit circle $|z|=1$ and poles $z=-2$ lies outside the circle,

Find residue at $z=\frac{-1}{2}$

Residue at $z=\frac{-1}{2} =lim_z\to_{\frac{-1}{2}} (z+\frac{1}{2}) \frac{z^2}{(2z+1)(z+2)i}$

$lim_z\to_{\frac{-1}{2}} (z+\frac{1}{2}) \frac{z^2}{2(z\frac{1}{2})(z+2)i}$

$lim_z\to_{\frac{-1}{2}} \frac{z^2}{2(z+2)i}$

=$\frac{(\frac{-1}{2})^2}{2((\frac{-1}{2})+2)i}$

=$\frac{\frac{1}{4}}{2(\frac{-1+4}{2})i}$

=$\frac{1}{4(3)i}$

Residue at $z=\frac{-1}{2} =\frac{1}{12i} $

$\int_0^{2\pi} \frac{cos2\theta}{5+4cos\theta} d\theta$ = $2\pi i (Residue \ at \ z=\frac{-1}{2})$

= 2$\pi i (\frac{1}{12i})$

=$\frac{\pi}{6}$