How do you find the antiderivative of #int x^2cosx dx#?

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Answer 1 · 2017-10-22T05:45:44

The answer is #=(x^2-2)sinx+2xcosx+C#

#intuv'dx=uv-intu'v#

Apply the integration by parts

Let #u=x^2#, #=>#, #u'=2x#

#v'=cosx#, #=>#, #v=sinx#

Therefore,

#intx^2cosxdx=x^2sinx-int2xsinxdx#

Apply the integration by parts a second time

Let #u=x#, #=>#, #u'=1#

#v'=sinx#, #=>#, #v=-cosx#

So,

#intx^2cosxdx=x^2sinx-int2xsinxdx#

#=x^2sinx-2(-xcosx-int-cosxdx)#

#=x^2sinx+2xcosx-2sinx+C#

#=(x^2-2)sinx+2xcosx+C#