Question

Integration by parts?

`int 6x^2sin(3x)dx`

Answer 1

`-2x^2cos(3x)+(4xsin(3x))/3+(4cos(3x))/9+C`

Explanation:

First, let's factor out `6` to leave us with `intx^2sin(3x)dx`

Integration by parts: `intvu'=uv-intuv'`

`u'=sin(3x),u=-cos(3x)/3`
`v=x^2,v'=2x`

`6(-(x^2cos(3x))/3+2/3intxcos(3x)dx)`

`u'=cos(3x),u=sin(3x)/3`
`v=x,v'=1`

`6(-(x^2cos(3x))/3+2/3((xsin(3x))/3-intsin(3x)/3dx))`

`6(-(x^2cos(3x))/3+2/3((xsin(3x))/3+cos(3x)/9))`

`-2x^2cos(3x)+(4xsin(3x))/3+(4cos(3x))/9+C`