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

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Answer 1 · 2016-12-29T14:29:18

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.

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

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

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

Hopefully this helps!

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