# Homework Help: Integrate x^2 sin(x^2) dx

1. Sep 16, 2005

### jacy

Hi ,
I am having trouble integrating this problem.

Integrate x^2 sin(x^2) dx

Here is what i did. I used the substitution method.

u = x^2 sqrt u = x
du = 2x dx
du/2x = dx

du/2 sqrt u = dx since sqrt u = x

substituting this in the equation

u sin(u) du/2 sqrt u

2. Sep 16, 2005

### Townsend

I don't think the antiderivative of your integrand is an elementary....but I don't really know for sure.

3. Sep 16, 2005

### amcavoy

I believe you're correct. My computer cannot calculate it.

4. Sep 17, 2005

### jacy

Thanks 4 ur time. Do u think that problem is wrong.

5. Sep 17, 2005

### Townsend

The problem is fine....it's just that you cannot find an elementary antiderivative as a solution.

6. Sep 18, 2005

### GCT

where'd you obtain the problem?

7. Sep 18, 2005

### amcavoy

It would work if the integrand was $x\sin{x^{2}}$.

8. Sep 20, 2005

### jacy

Thanks for looking at the problem. My teacher gave this problem on the exam. How can we solve this.

9. Sep 20, 2005

### Tony11235

Integration by parts might be possible, but my computer and my TI 89 cannot compute it, so there's probably not a simple antiderivative. And you were given this problem on a test?

10. Sep 21, 2005

### jacy

Thanks, do u think the substitution method that i used will work. This wasn't the only one there were 3 more.

Last edited: Sep 21, 2005
11. Sep 21, 2005

### amcavoy

Are you sure it wasn't asking you to evaluate it as a definite integral numerically? That's the only way I can see doing this.

12. Sep 22, 2005

### jacy

No it isn't a definite integral. It's tough one. Hopefully the teacher should provide the solution today. If he does, then can i post the solution in here, thanks.

13. Sep 22, 2005

$\int x^2\sin(x^2) dx [/tex] [itex] \int x^2\sin(x^2) dx = \frac{-x}{2}\cos(x^2) + \frac{1}{2} \int \cos(x^2) dx [/tex] 14. Sep 22, 2005 ### Tom Mattson Staff Emeritus True, but now what's [itex]\frac{1}{2} \int \cos(x^2) dx$?

jacy, the initial response you received was correct. Your integrand doesn't have an elementary antiderivative. 10 bucks says that the problem was supposed to be $\int x\sin(x^2) dx$.

15. Sep 22, 2005

### whozum

Its not elementary i was just trying it out and seeing how far I could get, but figured I'd just post it anyway.

16. Sep 22, 2005

### Jameson

Here's the answer from Mathematica in case anyone was wondering. I agree, it was probably supposed to be $$\int x\sin{(x^2)}dx$$.

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17. Sep 23, 2005

### jacy

I don't think so it was a typo. This problem was on the exam, thanks.

18. Sep 23, 2005

### whozum

Was the problem asking to find the derivative of that function?

19. Sep 27, 2005

### jacy

We have to find the anti derivative of that function. Today the teacher said that he made a typo it should be
integrate x^2 sin^2(x) instead of x^2 sin(x^2)

How did u guys type the sign of integral

20. Sep 28, 2005

### bomba923

$$\int {x^2 \sin ^2 x\,dx}$$
can be evaluated via integration by parts.

jacy, the integral sign is just \int when using LaTeX.
Try clicking on this integral sign: $$\int$$