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,

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

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

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

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

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