Using complex exponential to show the indefinite integration of sin(x)sinh(x) dx Use the complex exponential to evaluate the indefinite integral of 
$\sin x \sinh x$.
Express your answer in terms of trigonometric and/or hyperbolic functions
The attached photo is what I have tried so far


Start with $\sin x=\Im e^{ix}.$ Then your integral is the imaginary part of the following integral:
$$\int e^{ix}\cdot\frac{e^x-e^{-x}}2\, dx=\frac 12 e^{ix}\cdot\left(\frac {e^x}{i+1}-\frac{e^{-x}}{i-1}\right)=\frac 12e^{ix}\cdot \frac{-(e^{-x}+e^x) + i(e^x-e^{-x})}{-2}=\\=\frac 12e^{ix}\cdot (\cosh x-i\sinh x).$$
As we are interested in the imaginary part, we can take use of the fact that $\Im ab=\Im a\cdot \Re b+\Re a\cdot \Im b$ and get the final result:
$$\frac 12\left(\sin x\cosh x-\cos x\sinh x\right)$$
It agrees with the result from WolframAlpha.
Disclaimer: I omitted $+C$ for clarity. Many people dislike it, so consider adding it.
