I am trying to solve this using Cauchy Integral Formula for pole of order $n$ but don't understand how to deal with the multiple denominators. In other questions, I could use Partial Fraction and make it so that only 1 term of $(z-a)^n$ exists in denominator and directly apply the formula of
$$∫\frac{f(z)}{(z-a)^{n+1}} dz = 2\pi i\cdot\frac{f^{(n)}}{n!} $$
Hint:-Note that the poles inside the region are $z=0,2i$ . So write $\frac{1}{z^{2}(z-2i)}$ as partial fraction decomposition and evaluate one by one by taking $f=\frac{\sin(z)}{z+2i}$ .
The partial fraction decomposition is $$\frac{1}{z^{2}(z-2i)}=\frac{1}{2iz}\bigg(\frac{1}{z-2i}-\frac{1}{z}\bigg) = \frac{1}{2iz(z-2i)}-\frac{1}{2iz^{2}}=\frac{1}{4}\bigg(\frac{1}{z}-\frac{1}{z-2i}\bigg)-\frac{1}{2iz^{2}}$$ .
So evaluate one by one by using the Cauchy-Integral formula by taking $f=\frac{\sin(z)}{z+2i}$
Otherwise use Cauchy Residue Theorem to evaluate the integral directly. Note that again the poles are $0$ of order $2$ and $2i$ of order $1$