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I'm trying to find the integral of $y = x\sin^2 (x^2)$. Can someone help please? I've tried converting it to $x(\frac{1}{2} -\frac{1}{2}\cos(2x^2))$ and using integration by parts but it doesn't seem to help.

Thanks.

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Hint: First let $u=x^{2}$ and notice that $\sin^{2}(u)=\frac{1-\cos(2u)}{2}$

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  • $\begingroup$ Thanks! Turns out it's just a one trick that makes the question so much simpler. $\endgroup$ – inggumnator Aug 3 '14 at 18:31
  • $\begingroup$ You're welcome! $\endgroup$ – user71352 Aug 3 '14 at 18:32
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$x\left(\dfrac12−\dfrac12\cos(2x^2)\right)=\dfrac{x}2-\dfrac18\times4x\cos(2x^2)$

From there $4x\cos(2x^2)$ is of the form $u'\cos u$, and therefore is easy to integrate

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1) $x \sin^2(x^2)dx = \frac{1}{2}\sin^2(x^2)d x^2$

2) $\sin^2 t = \frac{1 - \cos 2t}{2}$

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  • $\begingroup$ What does $dx^2$ mean? $\endgroup$ – Rory Daulton Aug 3 '14 at 18:26
  • $\begingroup$ Yeah I didn't get that either $\endgroup$ – inggumnator Aug 3 '14 at 18:32
  • $\begingroup$ Poor notation. He made the substitution $u=x^2$ so that he has $\frac{1}{2}\sin^2 x^2 \, \mathrm{d}(x^2)$ or $\frac{1}{2}\sin^2 u \, \mathrm{d}u$. $\endgroup$ – Gahawar Aug 3 '14 at 19:19
  • $\begingroup$ yes that's exactly what I did: $\frac{1}{2} d(x^2) =x dx$. What's wrong with this notation? $\endgroup$ – Alex Aug 3 '14 at 19:26

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