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 + (2-x^2)cosx + C

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