Question

How do you integrate `int x^2 cos^2 3 x dx` using integration by parts?

Answer 1

`x^2(x/2-sin(6x)/12) -x^3/3 - x cos(6x) + sin(6x)/6`

Please check calculation too

Explanation:

`u = x^2`
`cos^2 3x dx= {1 + cos(6x)}/2 dx= dv`

`du = 2x dx`
`v = x/2 - sin(6x)/12`

integral `I = x^2(x/2-sin(6x)/12) -x^3/3 + int (x/6 sin(6x))dx`

integrate by parts again
`u=x`
`du=dx`
`dv = sin(6x)/6 dx`
`v= -cos(6x)`

so `I = x^2(x/2-sin(6x)/12) -x^3/3 - x cos(6x) + sin(6x)/6`