`int e^(-3x)sin5x dx` Find the indefinite integral

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....

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`