Evaluate the integral $ I=\frac{1}{2\pi i}\int_{\vert z \vert =R}(z-3)\sin \left(\frac{1}{z+2}\right)dz$, Evaluate the integral
$$ I=\frac{1}{2\pi i}\int_{\vert z \vert =R}(z-3)\sin\left(\frac{1}{z+2}\right)dz$$ 
where $R \geq 4$
My work: Here the zeros of $\sin\left(\frac{1}{z+2}\right)$ is $\frac{1}{n \pi}-2$, so $-2$ is non-isolated singularity. But after I am stuck, Please help. Thanks.
 A: The integrand function has just ONE singularity at $z=-2$. For any non-zero integer $n$ then by letting $z=\frac{1}{n \pi}-2$, we have that
$$\sin\left(\frac{1}{z+2}\right)=\sin(n\pi)=0.$$
So, for $R>2$, the singularity $z=-2$ is inside the circle $|z|=R$, and, by the Residue Theorem,
$$\begin{align}\frac{1}{2\pi i}\int_{\vert z \vert =R}(z-3)\sin\left(\frac{1}{z+2}\right)dz&=\text{Res}\left((z-3)\sin\left(\frac{1}{z+2}\right), -2\right)\quad(w:=z+2)\\
&=
\text{Res}\left((w-5)\left(\frac{1}{w}-\frac{1}{3! w^3}+\frac{1}{5! w^5}+\dots\right), 0\right)\\
&=
\text{Res}\left(1-\frac{5}{w}-\frac{1}{3! w^2}+\frac{5}{3! w^3}+\dots, 0\right)\\
&=-5.
\end{align}$$
P.S. If $0<R<2$ then the above integral is zero.
A: All of the singularities are included by $|z|=R$. There is a theorem (I forgot the name) implying that in such case the integral would be:$$I=\int_{|z|=R}f(z)dz=2\pi i Rez_{z=0}\dfrac{1}{z^2}f(\frac{1}{z})$$here we have:$$\dfrac{1}{z^2}f(\frac{1}{z})=\frac{1-3z}{z^3}\sin\frac{z}{1+2z}$$obviously$$\lim_{z\to 0}z^2\frac{1-3z}{z^3}\sin\frac{z}{1+2z}=1$$so $z=0$ is a 2nd-order pole and the residue would be:$$b_1=\lim_{z\to 0}\frac{d}{dz}z^2\frac{1-3z}{z^3}\sin\frac{z}{1+2z}=-5$$therefore the integral is $$I=-10\pi i$$
