$\int_{-\pi}^{\pi}\sinh(x) \sin(nx)dx$, using complex transformation. $$\int_{-\pi}^{\pi}\sinh(x) \sin(nx)dx$$
This integral can be done by using integrate by parts twice. However, I was thinking if this can be done by using some complex transformation (I haven't done any complex analysis yet, that is next term), for example use something like $i\sin(ix)=-\sinh(x)$. I encountered this integral when I was trying to find the Fourier Series for $\sinh(x)$ over $-\pi, \pi$.
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\begin{align}&\color{#66f}{\large%
\int_{-\pi}^{\pi}\sinh\pars{x}\sin\pars{nx}\,\dd x}
=-2\ic\int_{0}^{\pi}\sin\pars{nx}\sin\pars{\ic x}\,\dd x
\\[5mm]&=\ic\int_{0}^{\pi}
\braces{\cos\pars{\bracks{n + \ic}x} - \cos\pars{\bracks{n - \ic}x}}\,\dd x
=\ic\,\bracks{{\sin\pars{n\pi + \ic\pi} \over n + \ic}
- {\sin\pars{n\pi - \ic\pi} \over n - \ic}}
\\[5mm]&=-2\,\Im\bracks{\pars{-1}^{n}\sin\pars{\ic\pi} \over n + \ic}
=2\pars{-1}^{n + 1}\sinh\pars{\pi}\,\Im\bracks{\ic \over n + \ic}
=\color{#66f}{\large 2\,{\pars{-1}^{n + 1}n \over n^{2} + 1}\,\sinh\pars{\pi}}
\end{align}
A: I am not really sure about what your intentions are with the use of the words complex transforms. What is general called a transform contains $e^{-i\text{ something}}$ and a transform tends to be named after people not called complex.

We can re-write $\sin(nx)\sinh(x)$ using exponential to obtain a complex $\cosh$. We know that 
\begin{align}
\sinh(x) &= \frac{e^x-e^{-x}}{2}\\
\sin(nx) &= \frac{e^{inx} - e^{-inx}}{2i}
\end{align}
Therefore, we can multiple the exponentials together.
\begin{align}
\frac{e^x-e^{-x}}{2}\frac{e^{inx} - e^{-inx}}{2i}&= \frac{-i}{4}(\exp(x+ixn)+\exp(-(x+ixn))-\exp(x-ixn)-\exp(-(x-ixn))\\
&=\frac{i}{2}\biggl(\frac{\exp(x+ixn)+\exp(-(x+ixn))}{2}+\frac{\exp(x+ixn)+\exp(-(x+ixn))}{2}\biggr)\\
&=\frac{i}{2}(\cosh(x-inx) - \cosh(x+inx))\\
\end{align}
Now we don't need to use integration by parts.
\begin{align}
\frac{i}{2}\biggl[\int_{-\pi}^{\pi}\cosh(x-inx)dx + \int_{-\pi}^{\pi}\cosh(x+inx)dx\biggr] &= \frac{-\sinh(\pi(1 -in))}{n+i}+\frac{i\sinh((\pi(1 +in))}{1+i n}\\
&= \frac{2n(-1)^{n+1}\sinh(\pi)}{n^2+1}
\end{align}
