How do you find the integral of ##e^(7x)*sin(2x)dx##?

By using twice.

Let ##f(x)=e^(7x)## so that ##f'(x)=7e^(7x)##. Let ##g'(x)=sin(2x)## so that ##g(x)=-1/2cos(2x)##.

Hence

##inte^(7x)sin(2x)dx=-1/2e^(7x)cos(2x)+7/2inte^(7x)cos(2x)dx##

Now consider the integral ##inte^(7x)cos(2x)dx##

Let ##f(x)=e^(7x)## so that ##f'(x)=7e^(7x)##. Let ##g'(x)=cos(2x)## so that ##g(x)=1/2sin(2x)##.

Hence

##inte^(7x)cos(2x)dx=1/2e^(7x)sin(2x)-7/2inte^(7x)sin(2x)dx##

Putting these together we get

##inte^(7x)sin(2x)dx=-1/2e^(7x)cos(2x)+7/2[1/2e^(7x)sin(2x)-7/2inte^(7x)sin(2x)dx]##

##inte^(7x)sin(2x)dx=-1/2e^(7x)cos(2x)+7/4e^(7x)sin(2x)-49/4inte^(7x)sin(2x)dx##

##53/4inte^(7x)sin(2x)dx=7/4e^(7x)sin(2x)-1/2e^(7x)cos(2x)+C##

##53/4inte^(7x)sin(2x)dx=1/4e^(7x)(7sin(2x)-2cos(2x))+C##

##inte^(7x)sin(2x)dx=1/53e^(7x)(7sin(2x)-2cos(2x))+C##