<s>[INST] Evaluating $\int_c\frac{1}{\sin\frac{1}{z}}\text{d}z$ over $C= \{z\big\vert|z|=\frac{1}{5}\}$ Evaluate 

$$\int\limits_{|z|=\frac{1}{5}} \frac{1}{\sin\frac{1}{z}}\text{d}z$$

My attempt: I know that this function has non isolated singularity at $0$, and simple poles at $\frac{1}{n \pi}$. Which ever methods I know are applicable only when all the singularities are isolated. So I tried with Residue theorem but then I realised that the theorem is not applicable here. So now I am stuck. Can any one help please?
I can observe that there are only two poles outside the curve c, one at $1/\pi$ and the other at infinity. Can this be used?
Thanks in advance.
 [/INST] $$f(z) = \frac1{\sin{\left ( \frac1{\zeta+\frac1{n \pi}} \right )}} $$
where $\zeta = z-\frac1{n \pi}$.  Then
$$\begin{align}f(z) &= \frac1{\sin{\left ( \frac{n \pi}{1+ n \pi \zeta} \right )}} \\ &= \frac1{\sin{\left [ n \pi \left (1-n \pi \zeta + n^2 \pi^2 \zeta^2-\cdots \right )\right  ]}}\\ &= \frac1{\sin{n \pi} \cos{(n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)} - \cos{n \pi} \sin{(n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)}}\\ &= \frac{(-1)^{n+1}}{\sin{(n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)}}\\ &= \frac{(-1)^{n+1}}{(n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)-\frac16 (n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)^3+\cdots}\end{align}$$
The residue is the coefficient of $\frac1{\zeta}$, which is $(-1)^{n+1}/(n^2 \pi^2)$.  Within $|z|=1/5$, we only consider $|n| \ge 2$. Thus away from $z=0$ the contribution to the integral is
$$i 2 \pi (2) \sum_{n=2}^{\infty} \frac{(-1)^{n+1}}{n^2 \pi^2} = -i \frac{4}{\pi} \left (1-\frac{\pi^2}{12} \right)$$
There is also an essential singularity at $z=0$.  $f$ may be written as
$$f(z) = \frac1{\frac1z - \frac16 \frac1{z^3} + \frac1{120} \frac1{z^5} - \cdots} $$
and thus has no residue here.  Thus, the integral is 
$$-i \left (\frac{4}{\pi} - \frac{\pi}{3} \right ) $$
(Thanks to @achillehui for pointing out a crucial mistake.)
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