show that, if $C$ is the circle $|z|=1$, $\int_\mathcal{C}\frac{d{z}}{z^2\sinh(z)}=-\frac{1}{3}\pi i$ Show that, if $C$ is the circle $|z|=1$,
$$\int_\mathcal{C}\dfrac{d{z}}{z^2 \sinh(z)}=-\frac{1}{3}\pi i$$
I am confused how to do it.
 A: Using The Residue Theorem
There is only one singularity inside $\mathcal{C}$, so we just need to compute the residue of $\dfrac1{z^2\sinh(z)}$ at $z=0$.
$$
\begin{align}
\frac1{z^2\sinh(z)}
&=\frac1{z^2\left(z+z^3/6+O\left(z^5\right)\right)}\\
&=\frac1{z^3}\frac1{1+z^2/6+O\left(z^4\right)}\\
&=\frac1{z^3}\left(1-z^2/6+O\left(z^4\right)\right)\\
&=\frac1{z^3}-\frac1{6z}+O\left(z\right)\tag{1}
\end{align}
$$
Thus, the residue is $-\dfrac16$. Assuming the contour is counter-clockwise, the integral by the Residue Theorem is 
$$
\begin{align}
\int_\mathcal{C}\frac{\mathrm{d}z}{z^2\sinh(z)}
&=2\pi i\left(-\frac16\right)\\[6pt]
&=-\frac{\pi i}{3}\tag{2}
\end{align}
$$

Using Only Cauchy's Theorem
Note that
$$
\int\frac1{z^3}\,\mathrm{d}z=-\frac1{2z^2}+C
$$
Therefore, integrating around any closed contour (which starts and stops at the same point) yields
$$
\int_\mathcal{C}\frac1{z^3}\,\mathrm{d}z=0\tag{3}
$$
Next, parametrize $\mathcal{C}$ by $z=e^{it}$ for $t\in[0,2\pi]$.
$$
\begin{align}
\int_\mathcal{C}\frac1z\,\mathrm{d}z
&=\int_0^{2\pi}e^{-it}\,ie^{it}\,\mathrm{d}t\\[6pt]
&=2\pi i\tag{4}
\end{align}
$$
The integral of the analytic portion $(1)$ is $0$ by Cauchy's Theorem. That is,
$$
\int_\mathcal{C}\left(\frac1{z^2\sinh(z)}-\frac1{z^3}+\frac1{6z}\right)\,\mathrm{d}z=0\tag{5}
$$
Combining $(3)$, $(4)$, and $(5)$ yields
$$
\begin{align}
\int_\mathcal{C}\frac1{z^2\sinh(z)}\,\mathrm{d}z
&=\int_\mathcal{C}\frac1{z^3}\,\mathrm{d}z-\int_\mathcal{C}\frac1{6z}\,\mathrm{d}z\\[6pt]
&=0-\frac{2\pi i}{6}\\[6pt]
&=-\frac{\pi i}{3}\tag{6}
\end{align}
$$
