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#