How to find this rare integral? Let $\Lambda$ be a circle centered at zero on $\mathbb{C}\text{\ }\{0\}$. The value of $$\int_{\Lambda}\left(\sin\frac1z+\sin^2\frac 1z\right)dz\ \ \ \ \ \text{is equal to}$$ 

$(a) \ 0 \ \ \ \ \ \ (b)\ \  \pi i\ \ \  (c) \ \ 2\pi i \ \ \ \ \ \ (d)\ \ 4\pi i$

I am really confused how to find the poles & then apply residue theorem. 
Someone please help me solve this integral.  
 A: Instead of Laurent expansion, another way to evaluate the integral is compute the residue at "infinity".
To achieve this, we change variable to $w = \frac1z$.
For any $r > 0$, let $\Lambda(r)$ be a circular contour of radius $r$ in counterclockwise orientation. Let $-\Lambda(r)$ be the same contour but with
orientation reversed. In terms of $w$, we have.
$$\oint_{\Lambda(r)}\left(\sin\frac1z + \sin^2\frac1z\right) dz
= \oint_{\color{red}{-}\Lambda(r^{-1})}\left(\sin w + \sin^2 w\right)\left(\color{red}{-}\frac{dw}{w^2}\right)\\
= \oint_{\Lambda(r^{-1})} \left[\frac{1}{w}\left(\frac{\sin w}{w}\right)+ \left(\frac{\sin w}{w}\right)^2\right] dw$$
Since $\displaystyle\;\frac{\sin w}{w}$ has a removable singularity at $w = 0$ with limiting value $1$, the residue of what inside the square bracket at $w = 0$ is $1$. This means the original contour integral over $z$ is $2\pi i$.
A: For all $z\ne0$ one has
$$\sin{1\over z}={1\over z}-{1\over 6z^3}+{1\over 120 z^5}-\ldots\ .$$
whereby the series converges uniformly on any circle of radius $\rho>0$ around the origin. It follows that
$$\int_\Lambda\sin{1\over z}\>dz=2\pi i\ .$$
The function $z\mapsto\sin^2{1\over z}$ is even, hence the integral along $\Lambda$ of this function is $0$. It follows that option c) is correct.
