How do you integrate x3sinx by parts?

1 Answer
Jan 6, 2018

I=x3cosx+3x2sinx+6xcosx6sinx+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+3x2sinx6xsinxdx

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+6cosxdx}

Since cosxdx=sinx+C

I=x3cosx+3x2sinx{6xcosx+6[sinx]}+C

I=x3cosx+3x2sinx+6xcosx6sinx+C