# Cauchy Residue Theorem Integral

I have been given the integral $$\int_0^ {2\pi} \frac{sin^2\theta} {2 - cos\theta} d\theta$$ I have use the substitutions $z=e^{i\theta}$ |$d\theta = \frac{1}{iz}dz$ and a lot of algebra to transform the integral into this $$\frac{-i}{2} \oint \frac{1}{z^2}\frac{(z-1)^2}{z^2-4z+1}dz$$ In order to find the residues i further broke the integral into $$\frac{-i}{2} \oint \frac{1}{z^2}\frac{(z-1)^2}{(z+-r_1)(z-r_2)}dz$$ where $r_1 = 2+\sqrt{3}$ and $r_2 = 2-\sqrt{3}$ giving me three residues at $z=0|z=r_1|z=r_2$

My question is where do I go from here? Thanks.

• What I got is $$f=\frac{(z^2-1)^2}{2z(z^2-4z+1)}$$. And notice that you dont have to care about $r_1$. As I remember, residue for $f$ at $0$ is $\lim_{z\rightarrow 0} f z$. Feb 19 '16 at 22:37
• Let's say you chose to do a contour integral of that integrand, over a quarter circle in the first quadrant with a radius of $2\pi$. You would split that up into three separate contour integrals, one of which being the integral you're trying to find. You can use residue theorem to evaluate that quarter circle. Feb 19 '16 at 22:40
• Do we not care about $r_1$ because it is outside of the unit circle? Feb 19 '16 at 22:49

There was an error in the original post. We have

$$\int_0^{2\pi}\frac{\sin^2(\theta)}{2-\cos(\theta)}d\theta=-\frac i2\oint_{|z|=1}\frac{(z^2-1)^2}{z^2(z^2-4z+1)}\,dz$$

There are two poles inside $|z|=1$. The first is a second order pole at $z=0$ and the second is a first order pole at $z=r_2$.

To find the reside of the first pole we use the general expression for the residue of a pole of order $n$

$$\text{Res}\{f(z), z= z_0\}=\frac{1}{(n-1)!}\lim_{z\to z_0}\left(\frac{d^{n-1}}{dz^{n-1}}\left((z-z_0)^nf(z)\right)\right)$$

Here, we have

\begin{align} \text{Res}\left(\frac{-i(z^2-1)^2}{2z^2(z^2-4z+1)}, z= 0\right)&=\frac{1}{(2-1)!}\lim_{z\to 0}\left(\frac{d^{2-1}}{dz^{2-1}}\left((z-0)^2\frac{-i(z^2-1)^2}{2z^2(z^2-4z+1)}\right)\right)\\\\ \end{align}

To find the residue at $z=r_2$ we have simply

$$\text{Res}\left(\frac{-i(z^2-1)^2}{2z^2(z^2-4z+1)}, z= r_2\right)=\lim_{z\to r_2}\frac{-i(z^2-1)^2}{2z^2(z-r_1)}=-\frac{i}{2}\frac{(r_2^2-1)^2}{r_2^2(r_2-r_1)}$$

The rest is left as an exercise for the reader.