Answer
The result is $$\int e^{-4x}\sin3xdx=-\frac{1}{25}e^{-4x}(3\cos3x+4\sin3x)+c.$$
Work Step by Step
Denote:
$$I_1=\int e^{-4x}\sin3xdx.$$
Now we will integrate by parts using formula $$\int udv=uv-\int vdu.$$
Set $u=e^{-4x}$ and $dv=\sin3xdx=dv$ which gives us $-4e^{-4x}dx=du$ and $v=-\frac{1}{3}\cos3x$. Putting this into the integral we get:
$$\int e^{-4x}\sin3xdx=\int udv=uv-\int vdu=-\frac{1}{3}\cos3xe^{-4x}-\frac{4}{3}\int e^{-4x}\cos3xdx.$$
To solve the last integral we will integrate by parts again using that $u=e^{-4x}$ and $\cos3xdx=dv$ which gives us $-4e^{-4x}dx=du$ and $v=\frac{1}{3}\sin3x$. Substituting all of this we have:
$$\int e^{-4x}\cos3xdx=\frac{1}{3}e^{-4x}\sin3x-\int-\frac{1}{3}\sin3x4e^{-4x}dx=\frac{1}{3}e^{-4x}\sin3x+\frac{4}{3}\int e^{-4x}\sin3xdx.$$ Putting this into the original integral we get:
$$I_1=\int e^{-4x}\sin3xdx=-\frac{1}{3}\cos3xe^{-4x}-\frac{4}{3}\int e^{-4x}\cos3xdx=-\frac{1}{3}\cos3xe^{-4x}-\frac{4}{3}\left(\frac{1}{3}e^{-4x}\sin3x+\frac{4}{3}\int e^{-4x}\sin3xdx\right)=
-\frac{1}{3}\cos3xe^{-4x}-\frac{4}{9}e^{-4x}\sin3x-\frac{16}{9}\int e^{-4x}\sin3xdx=
-\frac{1}{3}\cos3xe^{-4x}-\frac{4}{9}e^{-4x}\sin3x-\frac{16}{9}I_1.$$
We can rewrite this as
$$I_1=-\frac{1}{3}\cos3xe^{-4x}-\frac{4}{9}e^{-4x}\sin3x-\frac{16}{9}I_1\Rightarrow
\frac{25}{9}I_1=-\frac{1}{3}\cos3xe^{-4x}-\frac{4}{9}e^{-4x}\sin3x
\Rightarrow
I_1=-\frac{1}{25}e^{-4x}(3\cos3x+4\sin3x).$$
Finally we have:
$$\int e^{-4x}\sin3xdx=-\frac{1}{25}e^{-4x}(3\cos3x+4\sin3x)+c,$$
where $c$ is arbitrary constant.
