we have $\int_{c(0,1)}^{}\sin(\frac{1}{z})dz=\int_{c(0,1)}\sum_{n=0}^{+\infty }\frac{(-1)^n}{(2n+1)!}(\frac{1}{z})^{2n+1}dz=$
$$=\sum_{n=0}^{+\infty }\int_{c(0,1)}\frac{(-1)^n}{(2n+1)!}(\frac{1}{z})^{2n+1}dz= \color{blue}{\underbrace{\sum_{n=0}^{+\infty }\frac{(-1)^n}{(2n+1)!}}_{=\sin(1)}} 2\pi i$$
(the integral is a standar integral equal to $2\pi i$)
so the answer is $\sin(1)2\pi i$ ?
No. The answer is $2\pi i$, since$$\int_{c(0,1)}z^{n}\,\mathrm dz=\begin{cases}2\pi i&\text{ if }n=-1\\0&\text{ otherwise.}\end{cases}$$Anyway, it seemes to be that it's more natural to say that$$\int_{c(0,1)}\sin\left(\frac1z\right)\,\mathrm dz=2\pi i\operatorname{res}_{z=0}\left(\sin\left(\frac1z\right)\right)=2\pi i,$$since$$\sin\left(\frac1z\right)=\frac{\color{red}1}z-\frac1{3!z^3}+\frac1{5!z^5}+\cdots$$