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

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Answer 1 · 2016-11-05T14:49:45

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

#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#