Question

How do you integrate `int e^(2x)sin(3x)` by integration by parts method?

Answer 1

`inte^(2x)sin(3x)dx=-3/13e^(2x)cos(3x)+2/13e^(2x)sin(3x)`

Explanation:

`inte^(2x)sin(3x)dx`

if `u=e^(2x)` and `dv=sin(3x)dx`, then `du=2e^(2x)dx` and `v=-1/3cos(3x)`.

using integration by parts, the integral becomes:

`-1/3e^(2x)cos(3x)-int-1/3*2e^(2x)cos(3x)dx`
`=-1/3e^(2x)cos(3x)+2/3inte^(2x)cos(3x)dx`

integrate by parts again:

`u=e^(2x)` and `dv=cos(3x)dx`, which means `du=2e^(2x)dx` and `v=1/3sin(3x)`

the integral is now:
`=-1/3e^(2x)cos(3x)+2/3(1/3e^(2x)sin(3x)-int2*1/3e^(2x)sin(3x)dx)`
`=-1/3e^(2x)cos(3x)+2/9e^(2x)sin(3x)-2/3int2/3e^(2x)sin(3x)dx`
`=-1/3e^(2x)cos(3x)+2/9e^(2x)sin(3x)-4/9inte^(2x)sin(3x)dx`

remember that this expression is equal to the original integral, meaning:

`inte^(2x)sin(3x)dx=-1/3e^(2x)cos(3x)+2/9e^(2x)sin(3x)-4/9inte^(2x)sin(3x)dx`

solve for `inte^(2x)sin(3x)dx`:

`13/9inte^(2x)sin(3x)dx=-1/3e^(2x)cos(3x)+2/9e^(2x)sin(3x)`
`inte^(2x)sin(3x)dx=9/13(-1/3e^(2x)cos(3x)+2/9e^(2x)sin(3x))`
`inte^(2x)sin(3x)dx=-3/13e^(2x)cos(3x)+2/13e^(2x)sin(3x)`