Question

How do you integrate `int x^2 cos3 x dx` using integration by parts?

Answer 1

`I = (9x^2sin(3x) +18xcos(3x) -2sin(3x))/9 + c`

Explanation:

`I = intx^2cos(3x)dx`

Say `u = x^2` so `du = 2x` and `dv = cos(3x)` so `v = sin(3x)/3`

`I = (x^2sin(3x))/3 - 2/3intxsin(3x)dx`

Say `u = x` so `du = 1` and `dv = sin(3x)` so `v = -cos(3x)/3`

`I = (x^2sin(3x))/3 - 2/3(-xcos(3x) +1/3intcos(3x)dx)`

`I = (x^2sin(3x))/3 - 2/3(-xcos(3x) +sin(3x)/9) + c`

`I = (9x^2sin(3x) +18xcos(3x) -2sin(3x))/9 + c`