<s>[INST] How to calculate this integral using Cauchy's Formula? 
Calculate $\int_{C} \sinh(z+1)/(z^2+1)dz$ where $C$ is $x^{2/3}+y^{2/3}=3^{2/3}$

My try is redefine $\int_{C} \sinh(z+1)/(z^2+1)dz$ as
$
\begin{align}
\int_{C} \sinh(z+1)/(z^2+1)dz &= \int_{C} \sinh(z+1)/((z-i)(z+i))dz \\
&= \int_{C} \sinh(z+1)/(2i(z-i))dz-\int_{C} \sinh(z+1)/(2i(z+i))dz \\
&= \frac{1}{2i} \left( \int_{C} \sinh(z+1)/(z-i)dz-\int_{C} \sinh(z+1)/(z+i)dz \right)
\end{align} $
I know that $\sinh(z+1)$ is analytic inside and on the contour $C$ (because is analytic in $\mathbb{C}$) and $i,-i$ lie in $C$, then using the Cauchy's Formula
$$\int_{C} \sinh(z+1)/(z-i)dz=2 \pi i\cdot \sinh(i+1)$$
and
$$\int_{C} \sinh(z+1)/(z+i)dz=2 \pi i\cdot \sinh(1-i)$$
then
$
\begin{align}
\int_{C} \sinh(z+1)/(z^2+1)dz &= \frac{1}{2i} \left( 2 \pi i\cdot \sinh(i+1)-2 \pi i\cdot \sinh(1-i) \right) \\
&= \pi \cdot \sinh(1+i)- \pi\cdot \sinh(1-i) \\
\end{align} $
I've do it well?. I don't know how to get the answer $i2 \pi \sin(1)\cosh(1)$
 [/INST] Your answer is actually equivalent after applying some identities!

There is a formula for the difference of hyperbolic sines (Wikipedia article)
$$\sinh(x) - \sinh(y) = 2 \cosh \left( \frac{x+y}{2} \right) \sinh \left( \frac{x-y}{2}\right)$$
Another identity of importance is 
$$\sinh(x) = -i \sin(ix) \;\;\; \text{or, equivalently,} \;\;\; \sinh(ix) = -i \sin(-x)$$
Finally, note that, in particular, the above holds if $x=1$, and that $\sin(x)$ is an odd function ($\sin(-x) = -\sin(x)$). You could skip a bit of a middleman by using this in the previous fact to conclude
$$\sinh(ix) =  i \sin(x)$$

Apply all of these facts to your answer:
$$\begin{align}
\pi \sinh(1+i) - \pi \sinh(1-i) &\overset{(1)}{=} \pi \left( \sinh(1+i) - \sinh(1-i) \right) \\
&\overset{(2)}{=}  2\pi \cosh \left( \frac{1+i+1-i}{2} \right) \sinh \left( \frac{1+i-(1-i)}{2}\right) \\
&\overset{(3)}{=} 2\pi \cosh \left( 1 \right) \sinh \left( i \right) \\
&\overset{(4)}{=} 2i\pi \cosh \left( 1 \right) \sin(1) 
\end{align}$$
Each equality follows as:


*

*$(1):$ Factor out $\pi$

*$(2):$ Apply the difference formula for hyperbolic sine

*$(3):$ Simplify

*$(4):$ Apply $\sinh(ix) = i \sin(x)$ for $x=1$
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