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What is the antiderivative of #x^2(sinx)dx#?

Answer 1

Aurora Alvarado

Make two uses of integration by parts.

Let #u = x^2# and #dv = sinxdx#. Then #du = 2xdx# and #v = -cosx#.

#int(udv) = uv - int(vdu)#

#int(x^2sinx) = -x^2cosx - int(-cosx * 2xdx)#

#int(x^2sinx) = -x^2cosx + 2int(xcosxdx)#

Now, let #u = x# and #dv = cosxdx#. Then #du = dx# and #v = sinx#. Now, use integration by parts again.

#int(x^2sinx) = -x^2cosx + 2(xsinx - int(sinx)) + C#

#int(x^2sinx) = -x^2cosx + 2xsinx + 2cosx + C#

#int(x^2sinx) = (2 - x^2)cosx + 2xsinx + C#

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Answer 2

Ezra Adams

The antiderivative of ( x^2 \sin(x) ) with respect to ( x ) is ( -x^2 \cos(x) + 2x \sin(x) + 2 \cos(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.