Calculus of a Single Variable, Chapter 8, 8.2, Section 8.2, Problem 29

We shall use partial integration:
int u dv=uv-int v du
Therefore, we have
int e^(-3x)sin5xdx=|[u=e^(-3x),dv=sin5xdx],[du=-3e^(-3x)dx,v=-1/5cos5x]|=
-1/5e^(-3x)cos5x-3/5int e^(-3x)cos5xdx=|[u=e^(-3x),dv=cos5xdx],[du=-3e^(-3x)dx,v=1/5sin5x]|=
-1/5e^(-3x)cos5x-3/25e^(-3x)sin5x-9/25inte^(-3x)sin5xdx
We can see that we have the same integral as the one we've started with. In other words we have the following equation
int e^(-3x)sin5xdx=-1/5e^(-3x)cos5x-3/25e^(-3x)sin5x
-9/25inte^(-3x)sin5xdx
Let us add 9/25int e^(-3x)sin5xdx to the whole equation.
34/25int e^(-3x)sin5xdx=-1/5e^(-3x)cos5x-3/25e^(-3x)sin5x
Now we only need to multiply the whole equation by 25/34 to obtain the solution to our starting problem.
int e^(-3x)sin5xdx=-5/34e^(-3x)cos5x-3/34e^(-3x)sin5x+c, c in RR