How should I solve integrals of this type? The general form of the integral I want to solve is:
$$ \int e^{bx}\sin(ax)  dx$$
Euler's formula has a nice connection, but the i makes it too complicated.
Doing it by parts doesn't seem to get me anywhere.
Do you have any tips for how to begin solving this?

By using the exponential version of sine:
\begin{align}
\int e^{bx} \ \sin(ax) \ dx &= \frac{1}{2i} \int \left( e^{(b+ai)x} - e^{(b-ai)x}
\right) \ dx \\
&= \frac{1}{2i} \left[ \frac{e^{(b+ai)x}}{b+ai} - \frac{e^{(b-ai)x}}{b-ai} \right] \\
&= \frac{1}{2i (b+ai)(b-ai)} \left[ (b-ai) e^{(b+ai)x} - (b+ai)e^{(b-ai)x} 
\right] \\
&= \frac{1}{2i(b^{2}+a^{2})} \left[ b \ e^{bx} \left(e^{ai x} - e^{-ai x} \right) - (ai) e^{bx} \left( e^{ai x} + e^{-ai x} \right) \right] \\
&= \frac{e^{bx}}{b^{2} + a^{2}} \left[ b \sin(ax) - a \cos(ax) \right] 
\end{align}
By using integration by parts:
\begin{align}
\int e^{bx} \ \sin(ax) \ dx &= \frac{1}{b} e^{bx} \sin(ax) - \frac{a}{b} \int
e^{bx} \ \cos(ax) \ dx  \\
&= \frac{1}{b} e^{bx} \sin(ax) - \frac{a}{b^{2}} e^{bx} \cos(ax) - \frac{a^{2}}{b^{2}} \int e^{bx} \ \sin(ax) \ dx
\end{align}
which is
\begin{align}
\left( 1 + \frac{a^{2}}{b^{2}} \right) \int e^{bx} \ \sin(ax) \ dx = \frac{1}{b} e^{bx} \sin(ax) - \frac{a}{b^{2}} e^{bx} \cos(ax)
\end{align}
or
\begin{align}
\int e^{bx} \ \sin(ax) \ dx = \frac{e^{bx}}{b^{2} + a^{2}} \left[ b \sin(ax) - a \cos(ax) \right] 
\end{align}
