Integral x^2 sin(x) dx

Answer 1

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