Posted by Anonymous on Wednesday, November 7, 2012 at 11:58am.
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 + (2-x^2)cosx + C
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