Integrate $e^{ax}\sin(bx)?$ Is there a general formula for finding the primitive of $$e^{ax}\sin(bx)?$$
I've done this manually with $a=9$ and $b=4$ using Euler's formulas. But it takes a bit of time. Is there a pattern here?
 A: Notice that $$\sin(bx)=\mathrm{Im}(e^{ibx})$$
so we need just take the imaginary part of this antiderivative
$$\int e^{ax}e^{ibx}dx=\frac{1}{a+ib}e^{(a+ib)x}+C$$
A: Hint 
Integration by parts
$$\int u \ dv =uv-\int v \ du $$
Make substition
$$u=\sin(bx)\ \Rightarrow \ du=b\cos (bx) \ dx$$
and
$$\ dv=e^{ax} \ dx \Rightarrow v= \frac{e^{ax}}{a}$$
So
$$\int e^{ax} \sin(bx) \ dx=\frac{e^{ax}}{a}\sin(bx)-\frac ba\int e^{ax}\cos bx \ dx$$
Then another integration by parts for $\int e^{ax}\cos bx \ dx$ .
I think you can do the rest of it.
A: Try by parts twice (assuming $\;ab\neq 0\;$ to avoid trivialities):
$$u=e^{ax}\;\;,\;\;u'=ae^{ax}\\
v'=\sin bx\;\;,\;\;v=-\frac1b\cos bx$$
and thus
$$I:=\int e^{ax}\sin bx\,dx=-\frac1be^{ax}\cos bx+\frac ab\int e^{ax}\cos bx\,dx=$$
$$=-\frac1be^{ax}\cos bx+\frac a{b^2}e^{ax}\sin bx-\frac{a^2}{b^2}\int e^{ax}\sin bx\,dx$$
Well, now just past the last rightmost summand to the left side (above) and do a little algebra:
$$\left(1+\frac{a^2}{b^2}\right)I=\frac{e^{ax}}b\left(\frac 1b\sin bx-\cos bx\right)\implies I=\ldots$$
A: Write $\sin bx = \Im e^{ibx}$, so that
$$e^{ax} \sin bx = \Im e^{ax}e^{ibx}=e^{(a+ib)x}.$$
Integrate this as a regular exponential and recover the imaginary part:
$$\int e^{ax} \sin (bx) dx = \int \Im e^{ax}e^{ibx}dx= \Im \int e^{(a+ib)x}dx.$$
A: There is a pattern. Differentiating a function of the form $e^{ax}\sin (bx)$ yields a linear combination of a function of the same form, and a function $e^{ax}\cos (bx)$. The analogous property holds for functions $e^{ax}\cos (bx)$. So the primitive of $e^{ax}\sin (bx)$ will be a linear combination of $e^{ax}\sin (bx)$ and $e^{ax}\cos (bx)$ (plus a constant).
It remains to find the coefficients.
$$\begin{align}
\frac{d}{dx}\left(e^{ax}(m\sin (bx) + n\cos (bx)\right) &= e^{ax}\left(a\bigl(m\sin(bx) + n\cos(bx)\bigr) + \bigl(bm\cos(bx) - bn\sin(bx)\bigr)\right)\\
&= e^{ax}\left((am - bn)\sin (bx) + (an+bm)\cos(bx)\right)
\end{align}$$
Now solve the linear system
$$\begin{align}
am - bn &= 1\\
an + bm &= 0.
\end{align}$$
A: $e^{ax}(a\sin(bx)-b\cos(bx))/(a^2+b^2)$
