Calculating $$\oint\limits_{|z|=4} \frac{\sin^2(z)}{(z-\frac{\pi}{6})^2(z+\frac{\pi}{6})}dz$$
Due to the fact of the denominator I've looked for the partial fractions: $$1=A\left(z-\frac{\pi}{6}\right)^3+B\left(z+\frac{\pi}{6}\right)\left(z-\frac{\pi}{6}\right)+C\left(z^2-\left(\frac{\pi}{6}\right)^2\right)$$ $$\Leftrightarrow 1=z^3(A+B)+z^2\left(-\frac{\pi}{2}A-\frac{\pi}{6}B+C\right)+z\left(\frac{\pi^2}{12}A-\frac{\pi^2}{36}B\right)-\frac{\pi^3}{216}A+\frac{\pi^2}{36}C$$ $$\begin{align}\Rightarrow 0&=A+B\\ 0&=\frac{\pi^2}{12}A-\frac{\pi^2}{36}B\\ &... \end{align}$$
Unfortunately there I made a mistake. It's a quite heavy equation and maybe there is a more efficient way to solve this integral. Any hints? Thank you!
It happens that$$\frac1{\left(z-\frac\pi6\right)^2\left(z+\frac\pi6\right)}=\frac9{\pi ^2\left(z+\frac\pi6\right)}-\frac9{\pi ^2\left(z-\frac\pi6\right)}+\frac3{\pi\left(z-\frac\pi6\right)^2}$$Therefore\begin{multline*}\oint_{|z|=4}\frac{\sin^2(z)}{\left(z-\frac\pi6\right)^2\left(z+\frac\pi6\right)}dz=\\=\frac9{\pi^2}\oint_{|z|=4}\frac{\sin^2(z)}{z+\frac\pi6}dz-\frac9{\pi^2}\oint_{|z|=4}\frac{\sin^2(z)}{z-\frac\pi6}dz+\frac3\pi\oint_{|z|=4}\frac{\sin^2(z)}{\left(z-\frac\pi6\right)^2}dz.\end{multline*}
Added note: Here's how to obtain the partial fraction decomposition. You want numbers $A$, $B$, and $C$ such that$$\frac1{\left(z-\frac\pi6\right)^2\left(z+\frac\pi6\right)}=\frac A{z+\frac\pi6}+\frac B{z-\frac\pi6}+\frac C{\left(z-\frac\pi6\right)^2}.$$Putting together these three fractions on the right hand side, you get:$$\frac{A z^2-\frac\pi3 A z+\frac{\pi^2}{36}A+B z^2-\frac{\pi ^2}{36} B+C z+\frac\pi6C}{\left(z+\frac\pi6\right)\left(z-\frac\pi6\right)^2.}$$This is equal to $\frac1{\left(z-\frac\pi6\right)^2\left(z+\frac\pi6\right)}$ if and only if$$\left\{\begin{array}{l}\frac{\pi^2}{36}A-\frac{\pi^2}{36}B+\frac\pi6C=1\\-\frac\pi3A+C=0\\A+B=0.\end{array}\right.$$The solution of this system is $A=\frac9{\pi^2}$, $B=-\frac9{\pi^2}$ and $C=\frac3\pi$.