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Integrate x^2 sin(x^2) dx

  1. Sep 16, 2005 #1
    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

    Now i don't know how to integrate this. Please help, thanks.
  2. jcsd
  3. Sep 16, 2005 #2
    I don't think the antiderivative of your integrand is an elementary....but I don't really know for sure.
  4. Sep 16, 2005 #3
    I believe you're correct. My computer cannot calculate it.
  5. Sep 17, 2005 #4

    Thanks 4 ur time. Do u think that problem is wrong.
  6. Sep 17, 2005 #5
    The problem is fine....it's just that you cannot find an elementary antiderivative as a solution.
  7. Sep 18, 2005 #6


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    where'd you obtain the problem?
  8. Sep 18, 2005 #7
    It would work if the integrand was [itex]x\sin{x^{2}}[/itex].
  9. Sep 20, 2005 #8
    Thanks for looking at the problem. My teacher gave this problem on the exam. How can we solve this.
  10. Sep 20, 2005 #9
    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? :bugeye:
  11. Sep 21, 2005 #10
    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
  12. Sep 21, 2005 #11
    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.
  13. Sep 22, 2005 #12
    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.
  14. Sep 22, 2005 #13
    [itex] \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]
  15. Sep 22, 2005 #14

    Tom Mattson

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    True, but now what's [itex]\frac{1}{2} \int \cos(x^2) dx[/itex]?

    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 [itex]\int x\sin(x^2) dx[/itex].
  16. Sep 22, 2005 #15
    Its not elementary i was just trying it out and seeing how far I could get, but figured I'd just post it anyway.
  17. Sep 22, 2005 #16
    Here's the answer from Mathematica in case anyone was wondering. I agree, it was probably supposed to be [tex]\int x\sin{(x^2)}dx[/tex].

    Attached Files:

  18. Sep 23, 2005 #17
    I don't think so it was a typo. This problem was on the exam, thanks.
  19. Sep 23, 2005 #18
    Was the problem asking to find the derivative of that function?
  20. Sep 27, 2005 #19
    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
  21. Sep 28, 2005 #20
    [tex] \int {x^2 \sin ^2 x\,dx} [/tex]
    can be evaluated via integration by parts.

    jacy, the integral sign is just \int when using LaTeX.
    Try clicking on this integral sign: [tex] \int [/tex]
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