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

1 Answer
Sep 20, 2015

1/53e^(2x)(2sin 7x-7cos 7x)+C

Explanation:

Integration by parts:

intudv=uv-intvdu

I=inte^(2x)sin 7xdx

e^(2x)=u => 2e^(2x)dx=du
dv=sin 7xdx => v=intsin 7xdx=-1/7cos 7x

I=-1/7e^(2x)cos 7x+2/7inte^(2x)cos 7xdx

Again:

e^(2x)=u => 2e^(2x)dx=du

dv=cos 7xdx => v=intcos 7xdx=1/7sin 7x

I=-1/7e^(2x)cos 7x+2/7[1/7e^(2x)sin7x-2/7inte^(2x)sin7xdx]

I=-1/7e^(2x)cos 7x+2/49e^(2x)sin7x-4/49inte^(2x)sin7xdx

I=-1/7e^(2x)cos 7x+2/49e^(2x)sin7x-4/49I

I+4/49I=-1/7e^(2x)cos 7x+2/49e^(2x)sin7x

53/49I=-1/7e^(2x)cos 7x+2/49e^(2x)sin7x

I=-7/53e^(2x)cos 7x+2/53e^(2x)sin7x+C

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