Really really confused with integration by parts

I try and try, but it never seems to come out correctly. Can someone please help me solve this step by step??

The indefinite integral of (x^2)sin(pix) dx

Please and thank you! I'm so frustrated! ><

First hint I would say is the exponent, in this case the exponent of x is 2 therefore you will have to integrate by parts two times.
Let u=x^2 and dv = sin(pix), therefore du=2xdx, v = -(cos(pix))/pi.
By using uv - [integral] vdu you get :

-(x^2*cos(pix)/pi) + 2/pi [integral] x*cos(pi)xdx

Note: - turns into a + because of the -cos(pix)

Now repeat the process only this time u= x and dv = cos(pix), therefore du =(1)dx and v = (sin(pix))/pi

Using the same formula you get :

-(x^2*cos(pix)/pi + 2/pi [ (xsin(pix))/pi - 1/pi [integral] sin(pix)]

Now you're left with :

-(x^2*cos(pix)/pi + 2/pi { (xsin(pix))/pi + (cos(pix))/pi^2]

Just distribute 2/pi :

-(x^2*cos(pix)/pi + (2xsin(pix))/pi^2 + (2cos(pix))/pi^3

I hope that I explained it well.