Question

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

Answer 1

`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`