How do you integrate ∫x3sinx by parts?

1 Answer
Anthony R.
Jan 6, 2018

I=−x3cosx+3x2sinx+6xcosx−6sinx+C

Explanation:

The formula for Integration by Parts (IBP):∫ u dv=uv−∫ v du

Let u1=x3;dv1=sinx

Thus, du1=3x2dx;v1=−cosx

I=[x3][−cosx]−∫[−cosx][3x2dx]

I=−x3cosx+∫3x2cosxdx

Apply IBP again:

Let u2=3x2;dv2=cosx

Thus, du2=6xdx;v2=sinx

I=−x3cosx+{[3x2][sinx]−∫[sinx][6xdx]}

I=−x3cosx+3x2sinx−∫6xsinxdx

Apply IBP once more:

Let u3=6x;dv3=sinx

Thus, du3=6dx;v3=−cosx

I=−x3cosx+3x2sinx−{[6x][−cosx]−∫[−cosx][6dx]}

I=−x3cosx+3x2sinx−{−6xcosx+6∫cosxdx}

Since ∫cosxdx=sinx+C

I=−x3cosx+3x2sinx−{−6xcosx+6[sinx]}+C

I=−x3cosx+3x2sinx+6xcosx−6sinx+C

Related questions
See all questions in Integration by Parts
Impact of this question
5970 views around the world
You can reuse this answer
Creative Commons License