Evaluation the integral in complex analysis $$\int_0^{2\pi}\frac{\cos^23\theta\, d\theta}{5-4\cos2\theta}.$$
In complex analysis, I want to solve the below problem. I know already the answer. But it is not easy to verify the answer. The function in the integral has a pole of order $5$ at $z=0$. it is strange.... The calculation is too hard. Please, help me...
Hint: Using the change of variable $z=\mathrm e^{\mathrm i\theta}$, this is, up to multiplicative constants I will let you discover, the complex integral on the unit circle of the function $f$ defined by
$$
f(z)=\frac{(z^3+z^{-3})^2}{5-2z^2-2z^{-2}}\frac1z\propto\frac1{z^5}g(z),
$$
where
$$
g(z)=\frac{(1+z^6)^2}{1-\frac52z^2+z^4},
$$
and the question is to determine the coefficient of $z^4$ in the expansion of $g(z)$. 

To do so, first note that the numerator is $1+O(z^6)$ hence it does not contribute. Using $w=\frac52z^2-z^4$, the expansion $\frac1{1-w}=1+w+w^2+O(w^3)$ and the fact that $O(w^3)=O(z^6)$, one sees that
$$
g(z)=\frac1{1-w}+O(z^6)=1+\left(\frac52z^2-z^4\right)+\left(\frac52z^2\right)^2+O(z^6),
$$
hence the $z^4$ term in $g(z)$ is $-z^4+\left(\frac52z^2\right)^2$, from which the answer follows.
Hint: Using the change of variable $z=\mathrm e^{\mathrm i\theta}$, this is, up to multiplicative constants I will let you discover, the complex integral on the unit circle of the function $f$ defined by
$$
f(z)=\frac{(z^3+z^{-3})^2}{5-2z^2-2z^{-2}}\frac1z\propto\frac1{z^5}g(z),
$$
where
$$
g(z)=\frac{(1+z^6)^2}{1-\frac52z^2+z^4},
$$
and the question is to determine the coefficient of $z^4$ in the expansion of $g(z)$. 

To do so, first note that the numerator is $1+O(z^6)$ hence it does not contribute. Using $w=\frac52z^2-z^4$, the expansion $\frac1{1-w}=1+w+w^2+O(w^3)$ and the fact that $O(w^3)=O(z^6)$, one sees that
$$
g(z)=\frac1{1-w}+O(z^6)=1+\left(\frac52z^2-z^4\right)+\left(\frac52z^2\right)^2+O(z^6),
$$
hence the $z^4$ term in $g(z)$ is $-z^4+\left(\frac52z^2\right)^2$, from which the answer follows.