Evaluate this Integral?

Question

∫x²sinx dx

Please do it step by step with details

Thank you jim black

But I think it should be f(x)g(x) - ∫ f'(x)g(x) dx, right ?

It should be -, not +

Answer 1

by parts = -x^2cosx +2Int x*cos x dx
Int xcosx dx = x*sinx -Int sinx dx = x sin x +cos x
so
Int = -x^2cos x+2( x sinx +cos x)+C

Answer 2

Interesting question...

Note I = ∫x²sinx dx
f(x) = x² => f'(x)= 2x
g'(x) = sinx => g(x)= -cosx
you use : ∫ f(x)g'(x) dx = f(x)g(x) + ∫ f'(x)g(x) dx

So I = ∫x²sinx dx= x² (-cosx) + ∫ 2x(- cosx) dx = -2xcosx - 2 ∫ x cosx dx = -2xcosx - 2 I1

Note I1= ∫ x cosx dx
Again...
f(x)= x => f(x) =1
g'(x)= cosx => g(x)= sinx

So I1= x sinx + ∫ sinx dx =x sinx + ( -cosx) + C= x sinx - cosx + C
So I = -2xcosx - 2 (x sinx - cosx) + C

I = I = -2x cosx - 2x sinx - 2cosx + C
Really hope you understood it. Tell me if you did.