How do you find the antiderivative of int (x^3cosx) dx?
1 Answer
Explanation:
I=intx^3cosxdx
This will require multiple iterations of integration by parts (IBP). Integration by parts takes the form
So, for
{(u=x^3),(dv=cosxdx):}
Now, differentiating
{(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
{(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
I=x^3sinx+3x^2cosx-[6xsinx+6sinx]
I=x^3sinx+3x^2cosx-6xsinx-6cosx+C