How do you find the antiderivative of int (x^3cosx) dx?

1 Answer
Nov 5, 2016

x^3sinx+3x^2cosx-6xsinx-6cosx+C

Explanation:

I=intx^3cosxdx

This will require multiple iterations of integration by parts (IBP). Integration by parts takes the form intudv=uv-intvdu.

So, for intx^3cosxdx as intudv, let:

{(u=x^3),(dv=cosxdx):}

Now, differentiating u and integrating dv, we see that:

{(u=x^3,=>,du=3x^2dx),(dv=cosxdx,=>,v=sinx):}

Now, plugging this into the IBP formula:

I=uv-intvdu=x^3sinx-int3x^2sinxdx

Now, for int3x^2sinxdx, perform IBP again:

{(u=3x^2,=>,du=6xdx),(dv=sinxdx,=>,v=-cosx):}

Thus:

I=x^3sinx-[3x^2(-cosx)-int6x(-cosx)dx]

Pay close attention to sign:

I=x^3sinx+3x^2cosx-int6xcosxdx

IBP again on the integral:

{(u=6x,=>,du=6dx),(dv=cosxdx,=>,v=sinx):}

So:

I=x^3sinx+3x^2cosx-[6xsinx-int6sinxdx]

Since intsinxdx=-cosx:

I=x^3sinx+3x^2cosx-[6xsinx+6sinx]

I=x^3sinx+3x^2cosx-6xsinx-6cosx+C