<s>[INST] 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?
 [/INST] A third (or maybe fourth, depending on how you're counting) method (related to the one in David's answer) which uses a bit of linear algebra is as follows:
\begin{align*}
\left[\begin{array}{c}
e^{ax}\cos(bx)\\
e^{ax}\sin(bx)
\end{array}\right]' &= \left[
\begin{array}{c}
ae^{ax}\cos(bx) - be^{ax}\sin(bx)\\
ae^{ax}\sin(bx) + be^{ax}\cos(bx)
\end{array}
\right]\\
&= \left[
\begin{array}{cc}
a & -b\\
b & a
\end{array}
\right] \left[
\begin{array}{c}
e^{ax}\cos(bx)\\
e^{ax}\sin(bx)
\end{array}
\right]
\end{align*}
Using the adjoint method, and the fact that the integral is (up to a constant) the functional inverse of the derivative on the space of smooth functions, we have:
\begin{align*}
\int \left[\begin{array}{c}
e^{ax}\cos(bx)\\
e^{ax}\sin(bx)
\end{array}\right] \text{d}x &= \frac{1}{a^2+b^2} \left[
\begin{array}{cc}
a & b\\
-b & a
\end{array}
\right] \left[\begin{array}{c}
e^{ax}\cos(bx)\\
e^{ax}\sin(bx)
\end{array}\right] + \left[\begin{array}{c}
C_1\\
C_2
\end{array}\right]\\
&= \left[\begin{array}{c}
\frac{1}{a^2+b^2}\left(ae^{ax}\cos(bx) + be^{ax}\sin(bx)\right) + C_1\\
\frac{1}{a^2+b^2}\left(ae^{ax}\sin(bx)-be^{ax}\cos(bx)\right) + C_2
\end{array}\right]
\end{align*}
Hence we have:
\begin{align*}
\int e^{ax}\cos(bx) \,\text{d}x &= \frac{1}{a^2+b^2}\left(ae^{ax}\cos(bx) + be^{ax}\sin(bx)\right) + C\\
\int e^{ax}\sin(bx) \, \text{d}x &= \frac{1}{a^2+b^2}\left(ae^{ax}\sin(bx)-be^{ax}\cos(bx)\right) + C
\end{align*}


Why the downvote?  This method works quite well computing integrals of this type.
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