Answer
$$\int e^{-3 x} \sin 5 x d x=\frac{-5}{34}e^{-3x}\cos 5 x -\frac{3}{34}e^{-3x}\sin 5 x $$
Work Step by Step
Let $$
I=\int e^{-3 x} \sin 5 x d x
$$
Integrate by parts , let
\begin{align*}
u&=e^{-3x}\ \ \ \ \ \ \ \ \ \ \ dv=\sin 5 x d x\\
u&=-3e^{-3x}\ \ \ \ \ \ \ \ \ v=\frac{-1}{5}\cos 5 x
\end{align*}
then \begin{align*}
I&=uv-\int vdu\\
&= \frac{-1}{5}e^{-3x}\cos 5 x -\frac{3}{5}\int e^{-3x} \cos 5 x dx\\
&= \frac{-1}{5}e^{-3x}\cos 5 x -\frac{3}{5}J
\end{align*}
where $$J=\int e^{-3x} \cos 5 x dx$$
Integrate by parts , let
\begin{align*}
u&=e^{-3x}\ \ \ \ \ \ \ \ \ \ \ dv=\cos 5 x d x\\
u&=-3e^{-3x}\ \ \ \ \ \ \ \ \ v=\frac{1}{5}\sin 5 x
\end{align*}
then \begin{align*}
J&=uv-\int vdu\\
&= \frac{1}{5}e^{-3x}\sin 5 x +\frac{3}{5}\int e^{-3x} \sin 5 x dx\\
&= \frac{1}{5}e^{-3x}\sin 5 x +\frac{3}{5}I
\end{align*}
Hence
\begin{align*}
I&=\frac{-1}{5}e^{-3x}\cos 5 x -\frac{3}{5}J\\
&=\frac{-1}{5}e^{-3x}\cos 5 x -\frac{3}{5}\left(\frac{1}{5}e^{-3x}\sin 5 x +\frac{3}{5}I\right)\\
\frac{34}{25}I&=\frac{-1}{5}e^{-3x}\cos 5 x -\frac{3}{25}e^{-3x}\sin 5 x
\end{align*}
$$I=\frac{-5}{34}e^{-3x}\cos 5 x -\frac{3}{34}e^{-3x}\sin 5 x $$
