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


Method I
Using integration by parts twice,
\begin{align}
  I &=\int \sin ax \sinh bx \, dx \\
  &= \frac{\sin ax \cosh bx}{b}-\int \frac{\cosh bx}{b} d(\sin ax) \\
  &= \frac{\sin ax \cosh bx}{b}-\frac{a}{b} \int \cos ax \cosh bx \, dx \\
  &= \frac{\sin ax \cosh bx}{b}-\frac{a}{b}
     \left[
       \frac{\cos ax \sinh bx}{b}-\int \frac{\sinh bx}{b} d(\cos ax)
     \right] \\
  &= \frac{\sin ax \cosh bx}{b}-
     \frac{a\cos ax \sinh bx}{b^2}-
     \frac{a^2}{b^2} \int \sin ax \sinh bx \, dx \\
 I &=
 \frac{\sin ax \cosh bx}{b}-
 \frac{a\cos ax \sinh bx}{b^2}-\frac{a^2}{b^2}I+C \\
  I &= \frac{b\sin ax \cosh bx-a\cos ax \sinh bx}{a^2+b^2}+C
\end{align}
Method II
\begin{align}
  \cos (u+iv) &= \cos u \cosh v-i\sin u \sinh v \\
  \sin (u+iv) &= \sin u \cosh v+i\cos u \sinh v \\
  \int \cos [(a+ib)x] \, dx &=
  \int \cos ax \cosh bx \, dx-
  i\int \sin ax \sinh bx \, dx \\
  \frac{\sin [(a+ib)x]}{a+ib} &=
  \int \cos ax \cosh bx \, dx-i\int \sin ax \sinh bx \, dx \\
  \frac{(a-ib)(\sin ax \cosh bx+i\cos ax \sinh bx)}{a^2+b^2} &=
  \int \cos ax \cosh bx \, dx-i\int \sin ax \sinh bx \, dx
\end{align}
Equating the imaginary parts,
$$\fbox{$
\int \sin ax \sinh bx \, dx=
\frac{b\sin ax \cosh bx-a\cos ax \sinh bx}{a^2+b^2}+C
$}$$
Method III
By brute force,
\begin{align}
  & \quad \int \sin ax \sinh bx \, dx \\[10pt]
  &= \int
     \left[
       \frac{(e^{iax}-e^{-iax})(e^{bx}-e^{-bx})}{4i}
     \right] \, dx \\[10pt]
  &= \int
  \left[
    \frac{e^{(b+ia)x}+e^{-(b+ia)x}-e^{(b-ia)x}-e^{(-b+ia)x}}{4i}
  \right] \, dx \\[10pt]
  &= \frac{e^{(b+ia)x}-e^{-(b+ia)x}}{4i(b+ia)}
    +\frac{-e^{(b-ia)x}+e^{-(b-ia)x}}{4i(b-ia)} \\[10pt]
  &= \frac{e^{(b+ia)x}-e^{-(b+ia)x}}{4(-a+ib)}
    +\frac{-e^{(b-ia)x}+e^{-(b-ia)x}}{4(a+ib)} \\[10pt]
  &= \frac{(-a-ib)[e^{(b+ia)x}-e^{-(b+ia)x}]}{4(a^2+b^2)}
    +\frac{(a-ib)[-e^{(b-ia)x}+e^{-(b-ia)x}]}{4(a^2+b^2)} \\[10pt]
  &= \frac{bi}{4(a^2+b^2)}
      [-e^{(b+ia)x}+e^{-(b+ia)x}+e^{(b-ia)x}-e^{-(b-ia)x}] \\
  &  \quad +
     \frac{a}{4(a^2+b^2)}
      [-e^{(b+ia)x}+e^{-(b+ia)x}-e^{(b-ia)x}+e^{-(b-ia)x}] \\[10pt]
  &= \frac{bi}{4(a^2+b^2)}
     [e^{bx}(-e^{iax}+e^{-iax})+e^{-bx}(e^{-iax}-e^{iax})] \\
  &  \quad +
     \frac{a}{4(a^2+b^2)}
     [e^{bx}(-e^{iax}-e^{-iax})+e^{-bx}(e^{-iax}+e^{iax})] \\[10pt]
  &= \frac{b}{a^2+b^2} \times
     \frac{e^{bx}+e^{-bx}}{2} \times
     \frac{e^{iax}-e^{-iax}}{2i}-
     \frac{a}{a^2+b^2} \times
     \frac{e^{bx}-e^{-bx}}{2} \times
     \frac{e^{iax}+e^{-iax}}{2} \\[10pt]
  &= \frac{b\sin ax \cosh bx-a\cos ax \sinh bx}{a^2+b^2}
\end{align}
