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What is the antiderivative of #x sin(x)#?

Answer 1

#intxsinxdx=-xcosx+sinx+C#

We'll use integration by parts for this integral.

Choose your #u# to be #x#, so that way #(du)/dx=1->du=dx#. That means #dv=sinxdx->intdv=intsinxdx->v=-cosx#.

The integration by parts formula is: #intudv=uv-intvdu#

We have #u=x#, #du=dx#, and #v=-cosx#. Substituting into the formula gives: #intxsinxdx=-xcosx-int(-cosx)dx# #color(white)(XX)=-xcosx+intcosxdx# #color(white)(XX)=-xcosx+sinx+C#

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Answer 2

Ezra Adams

The antiderivative of ( x \sin(x) ) is ( -x \cos(x) + \sin(x) + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.