Evaluate a complex integral using power series expansions 
Using power series expansions, evaluate the integral 
  $$\int_{\gamma_r}\sin\left(\frac{1}{z}\right)dz.$$
  where $\gamma_r:[0,2\pi]\rightarrow \mathbb C$ is given by $\gamma_r(t)=r(\cos t + i\sin t)$ for some $r>0$.

I know that the answer is $2i\pi$ but I don't really understand how to get to the answer as I keep going wrong somewhere! 
Any help would be greatly appreciated!

Hint: $\,z=re^{it}\Longrightarrow dz=rie^{it}\,,\,\,z^{-1}=r^{-1}e^{-it}$
$$\int\limits_{\gamma_r}\sin\frac{1}{z}dz=ri\int\limits_0^{2\pi}\sin\left(r^{-1}e^{-it}\right)e^{it}dt=i\int\limits_0^{2\pi}re^{it}\left(\frac{1}{re^{it}}-\frac{1}{6r^3e^{3it}}+\ldots\right)dt=$$
$$=i\int\limits_0^{2\pi}\left(1-\frac{1}{r^2e^{2it}}+\ldots\right)dt=2\pi i+0+0+\ldots$$
I'll leave it to you to find out why all those zeros. If you have problems write back a comment.
Alternative way:
$$\int\limits_{\gamma_r}\sin\frac{1}{z}\;dz=\int\limits_{\gamma_r}\left(\frac{1}{z}-\frac{1}{6z^3}+\ldots\right)dz=\int\limits_{\gamma_r}\frac{dz}{z}-\frac{1}{6}\int\limits_{\gamma_r}\frac{dz}{z^3}+\ldots$$
But
$$\int\limits_{\gamma_r}\frac{dz}{z^n}=\begin{cases}i\int\limits_0^{2\pi}dt=2\pi i&,\;\;\;n=1\\{}\\\frac{i}{r^{n-1}}\int\limits_0^{2\pi}\left(\cos(n-1)t-i\sin(n-1)t\right)dt=0&,\;\;\;n>1\end{cases}$$
Since
$$\int\limits_0^{2\pi}\cos mt\;dt=\int\limits_0^{2\pi}\sin mt\;dt=0\;,\;\;m\in\Bbb N$$
