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Homework Help: Antiderivative of cosh(x^2)

  1. May 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Just trying to figure out the anti-derivative of cosh(x^2).


    2. Relevant equations
    I knowthe antiderivative cannot be expressed as an elementary function but I am pretty clueless of getting the antiderivative though!


    3. The attempt at a solution
    I am baffled by this one :(. Any help and pointers would be much appreciated.

    thanks in advance!
     
  2. jcsd
  3. May 20, 2010 #2
    Sorry but isn't the chain rule for differentiating, not antidifferentiating?
     
  4. May 21, 2010 #3
    I don't think there is an elementary antiderivative for cosh(x2). What's this for?

    EDIT: I know there isn't an elementary antiderivative because it would involve integrating ex2, which doesn't have a simple antiderivative.
     
  5. May 21, 2010 #4
    i am trying to solve a double integral question but I can't figure out how to anti-differentiate cosh(x^2).

    it's quite frustrating!
     
  6. May 21, 2010 #5
    Hmm. Were there prior steps to this problem where you could've made a mistake, giving you the wrong integral to deal with at this point?
     
  7. May 21, 2010 #6
    Good question, dont think I've done anything wrong to that point.

    Basically I am trying to evaluate the following double integral, by changing the order of integration first.

    cosh(x^2).dx.dy for 3y < or equal to: x < or equal to 3
    and 0 < or equal to: y < or equal to 1

    I changed the order of integration to get:

    cosh(x^2).dy.dx for x/3 < or equal to: y < or equal to 1
    and 0 < or equal to: x < or equal to 3

    Then:
    [math]= \int_0^1 (1- x/3)cosh(x^2)dx= \int_{x=0}^1 cosh(x^2)dx- \frac{1}{3}\int_0^1 xcosh(x^2)dx[/math]

    From here I cant figure out how to differentiate cosh(x^2). I hope I made a mistake and there is an easier way!
     
  8. May 21, 2010 #7
    [tex]= \int_0^1 (1- x/3)cosh(x^2)dx= \int_{x=0}^1 cosh(x^2)dx- \frac{1}{3}\int_0^1 xcosh(x^2)dx[/tex]

    I can't see a problem in that line of math, but I also haven't encountered multiple integrals in my studies yet, perhaps someone else will be able to help there.

    The second integral you have there is easy enough (by u-substitution), but the first one is quite troubling.

    EDIT: Just looking at your LaTeX (use tex and /tex in your brackets for the proper coding).
     
  9. May 21, 2010 #8
    Oh yea, antiderivative. I thought I saw derivative. XD

    I guess.. the only way to express the antiderivative of cosh(x^2) would be expressing it as a infinite sum using the maclaurin series?
     
  10. May 21, 2010 #9
    Or you can also use the definition of cosh(x). Which is 1/2 (e^x+e^-x). But that won't help since you have cosh(x^2), not cosh(x), so therefore you'll have a sum of two infinite maclaurin series for e^x^2. XD
     
  11. May 21, 2010 #10
    Oh it's a double integral... I should read ahead before posting. :\
     
  12. May 21, 2010 #11

    gabbagabbahey

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    Homework Helper
    Gold Member

    When you sketch the integration region for

    [tex]\int_0^1\int_{3y}^3\cosh(x^2)dxdy[/tex]

    you should get a triangle with its base along the x-axis and vertices (0,0), (1,0) and (1,1). When you cut that region into vertical slices of thickness [itex]dx[/itex], you should find that [itex]y[/itex] ranges from zero to [itex]\frac{x}{3}[/itex], not x/3 to one.

    [tex]\int_0^1\int_{3y}^3\cosh(x^2)dxdy=\int_0^3\int_0^{x/3}\cosh(x^2)dydx[/tex]
     
  13. May 21, 2010 #12
    And there's where my lack of knowledge on multiple integrals comes in! Good luck on the rest of the problem!
     
  14. May 21, 2010 #13
    And from there is really easy. :]
     
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