Posted by Anonymous on Wednesday, November 7, 2012 at 11:58am.
Integral x^2 sin(x) dx

Maths  Integration by parts  Steve, Wednesday, November 7, 2012 at 2:13pm
Let I = ∫x^2 sinx dx
u = x^2
du = 2x dx
dv = sinx
v = cosx
∫u dv = uv  ∫v du
= x^2 cosx + ∫2x cosx dx
now, for ∫x cosx dx,
u = x
du = dx
dv = cosx dx
v = sinx
I = x^2 cosx + 2(x sinx  ∫sinx dx)
= x^2 cosx + 2x sinx + 2cosx + C
= 2x sinx + (2x^2)cosx + C
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