Integral using a parameter $k\in \mathbb Z$. I am trying to solve an exercise that says:

Compute the following integral for $k\in \mathbb Z$: $$\int _{\vert z \vert =1} \! z^k \sin {\frac{1}{z}} \, \mathrm d z$$

The only thing that I can think is making $z=\mathrm e ^{\mathrm i t}$, so the integral would be
$$\mathrm i \int _0 ^{2\pi} \mathrm e ^{\mathrm i kt} \sin ({\mathrm e ^{-\mathrm i t}})  \, \mathrm d t$$
Then, expanding the exponential ($\mathrm e ^{\mathrm i k t}=\cos(kt)+\mathrm i \sin (kt)$):
$$\mathrm i \int _0 ^{2\pi} \cos(kt) \sin ({\mathrm e ^{-\mathrm i t}})  \, \mathrm d t-\int _0 ^{2\pi} \sin(kt) \sin ({\mathrm e ^{-\mathrm i t}})  \, \mathrm d t$$
But I don't know what to do next. I'm afraid that maybe this is not the way to do it.

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\begin{align}
&\color{#f00}{\left.\oint_{\verts{z}\ =\ 1}
\,\,z^{k}\sin\pars{1 \over z}\,\dd z\,\right\vert_{\ k\ \in\ \mathbb{Z}}} 
\,\,\,\stackrel{z\ \mapsto\ 1/z}{=}\,\,\,
\left.\oint_{\verts{z}\ =\ 1}
\,\,{\sin\pars{z} \over z^{k + 2}}\,\dd z\,\right\vert_{\ k\ \in\ \mathbb{Z}}
\\[5mm] & =
2\pi\ic\bracks{k \geq -1}
\braces{\bracks{z^{k + 1}}\sin\pars{z}\vphantom{\Large A}} = 
2\pi\ic\bracks{k \geq -1}
\braces{\ic^{k}\,\,{1 + \pars{-1}^{k} \over 2\pars{k + 1}!}}
\\[5mm] & =
\color{#f00}{\left\{\begin{array}{rcl}
\ds{2\pi\ic\,{\pars{-1}^{k/2} \over \pars{k + 1}!}} & \mbox{if} &
\ds{k}\ \mbox{is $\ul{even}$ and}\ \ds{k \geq 0}
\\[3mm]
\ds{0} & \mbox{if} &
\ds{k}\ \mbox{is odd or}\ \ds{k < 0} 
\end{array}\right.}
\end{align}
