evaluate the integral.   (7x^2)(sin pix) dx please explain as you work the problem  

Let I be `I=int(7x^2 sin (pi x) dx`

Integrate it by part

`u(x)=7x^2 u'(x)=14x`

`v'(x)=sin pi x v(x)=-1/(pi)cos pi x`

`I=-7/pi x^2cos (pi x) -int 14x*(-1/(pi)cos pi x) dx`

`I=-7/pi x^2cos (pi x) +14/pi int x*cos pi x) dx`

Integrate it by part again

`u=x u'(x)=1`

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

`I=-7/pi x^2cos (pi x) +14/(pi) (x/pi) sin pi x-int(1/pi) sin pi x dx)`

`I=-7/pi x^2cos (pi x) +14/(pi)^2 x sin pi x- 14/(pi)^2 int sin pi x dx`

Answer: 

`I=-7/pi x^2cos (pi x) +14/(pi)^2 x sin pi x+14/(pi)^3cos pi x +C `