Integral of $x\sin^2 (x^2)$

Question

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.

score 4 · Answer 1 · edited Aug 03 '14 at 18:28

4

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

edited Aug 03 '14 at 18:28

2'5 9'2

54,717

answered Aug 03 '14 at 18:23

user71352

13,038

Thanks! Turns out it's just a one trick that makes the question so much simpler. – thbcm Aug 03 '14 at 18:31
You're welcome! – user71352 Aug 03 '14 at 18:32

score 2 · Answer 2 · answered Aug 03 '14 at 18:36

2

$x\left(\dfrac12−\dfrac12\cos(2x^2)\right)=\dfrac{x}2-\dfrac18\times4x\cos(2x^2)$

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

answered Aug 03 '14 at 18:36

imj

1,430

score 0 · Answer 3 · answered Aug 03 '14 at 18:22

0

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

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

answered Aug 03 '14 at 18:22

Alex

19,262

What does $dx^2$ mean? – Rory Daulton Aug 03 '14 at 18:26
Yeah I didn't get that either – thbcm Aug 03 '14 at 18:32
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 $\frac{1}{2}\sin^2 u , \mathrm{d}u$. – Gahawar Aug 03 '14 at 19:19
yes that's exactly what I did: $\frac{1}{2} d(x^2) =x dx$. What's wrong with this notation? – Alex Aug 03 '14 at 19:26

Integral of $x\sin^2 (x^2)$

3 Answers3