To evaluate $$\int_0^{2\pi}\frac{1}{(1+3\sin^2\theta)^2}d\theta$$ using Cauchy's Residue Theorem, I set $$\sin\theta=\frac{z-z^{-1}}{2i}\qquad\text{and}\qquad\frac{dz}{iz}$$ and obtained $$\int_{C_1^+(0)}\frac{-16z^3i}{(3z^4-10z^2+3)^2}\,dz$$ Then I factored the denominator and obtained double poles at $\pm\sqrt3/3$ and $\pm\sqrt3$. However, only $\pm\sqrt3/3$ lie within the unit circle $C_1^+(0)$. Now, to calculate the residue at $-\sqrt3/3$ I use
$$ \begin{array}{rcl} \text{Res}[f,-\sqrt3/3]&=&\frac{1}{(2-1)!}\lim_{z\to-\sqrt3/3}\frac{d}{dz}(z+\sqrt3/3)^2f(z)\\ \text{Res}[f,-\sqrt3/3]&=&\lim_{z\to-\sqrt3/3}\frac{d}{dz}\left[(z+\sqrt3/3)^2\cdot\frac{-16z^3i}{9(z+\sqrt3/3)^2(z-\sqrt3/3)^2(z+\sqrt3)^2(z-\sqrt3)^2}\right]\\ \text{Res}[f,-\sqrt3/3]&=&\lim_{z\to-\sqrt3/3}\frac{d}{dz}\left[\frac{-16z^3i}{9(z-\sqrt3/3)^2(z+\sqrt3)^2(z-\sqrt3)^2}\right]\\ \text{Res}[f,-\sqrt3/3]&=&\lim_{z\to-\sqrt3/3}\frac{d}{dz}\left[\frac{-16z^3i}{(3z-\sqrt3)^2(z^2-3)^2}\right] \end{array} $$
And here is the beginning of some difficult differentiation.
$$ \begin{array}{rcl} \text{Res}[f,-\sqrt3/3]&=&\lim_{z\to-\sqrt3/3}\frac{(3z-\sqrt3)^2(z^2-3)^2(-48z^2i)+16z^3i[(3z-\sqrt3)^2(2)(z^2-3)(2z)+2(3z-\sqrt3)(3)(z^2-3)^2]}{(3z-\sqrt3)^4(z^2-3)^4} \end{array} $$
Now, this is extremely difficult to continue with and highly unlikely that we obtain a correct answer.
Now, my question. Do instructors share a nice way to simplify this process for this particular problem to ease the work and help their students be successful?
Thanks
The numbers $\pm\frac1{\sqrt3}$ are not simple poles; they are double poles.
Near $\frac1{\sqrt3}$, you have$$-16z^3i=-\frac{16 i}{3 \sqrt{3}}-16 i \left(z-\frac{1}{\sqrt{3}}\right)+\cdots$$and$$\left(3 z^4-10 z^2+3\right)^2=\frac{256}{3} \left(z-\frac{1}{\sqrt{3}}\right)^2+\frac{128}{\sqrt3}\left(z-\frac{1}{\sqrt{3}}\right)^3+\cdots,$$which confirms that $\frac1{\sqrt3}$ is a double pole. So, $\frac{-16z^3i}{\left(3 z^4-10 z^2+3\right)^2}$ can be written as$$\frac{a_{-2}}{\left(z-\frac1{\sqrt3}\right)^2}+\frac{a_{-1}}{z-\frac1{\sqrt3}}+a_0+\cdots$$and so\begin{align}-16z^3i&=\left(\frac{256}{3} \left(z-\frac{1}{\sqrt{3}}\right)^2+\frac{128}{\sqrt3}\left(z-\frac{1}{\sqrt{3}}\right)^3+\cdots\right)\left(\frac{a_{-2}}{\left(z-\frac1{\sqrt3}\right)^2}+\frac{a_{-1}}{z-\frac1{\sqrt3}}+a_0+\cdots\right)\\&=\left(\frac{256}{3}+\frac{128}{\sqrt3}\left(z-\frac{1}{\sqrt{3}}\right)+\cdots\right)\left(a_{-2}+a_{-1}\left(z-\frac1{\sqrt3}\right)+a_0\left(z-\frac1{\sqrt3}\right)^2+\cdots\right).\end{align}So, we have$$\left\{\begin{array}{l}\frac{256}3a_{-2}=-\frac{16i}{3\sqrt3}\\\frac{256}3a_{-1}+\frac{128}{\sqrt3}a_{-2}=-16i.\end{array}\right.$$Therefore,$$\operatorname{res}_{z=1/\sqrt3}\frac{-16z^3i}{\left(3 z^4-10 z^2+3\right)^2}=a_{-1}=-\frac{5i}{32}.$$