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Help with Integration

  1. Mar 20, 2005 #1
    How do you integrate cos^3/2(x) or square root of cos^3(x), I know for the integration of cos^3(x) you break it apart to cos^2(x)cos(x). Then chnage that to (1 - sin^2(x))(cos(x) and do a u subsitution, with u set equal to sin. However, because of that square root it completely changes the problem, or at least in my eyes it does. Step by step help would be appreciated.

    Thanks.
     
  2. jcsd
  3. Mar 20, 2005 #2

    HallsofIvy

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    Do you have any reason to believe that this has a simple integral?
     
  4. Mar 20, 2005 #3
    It's from a practice AP booklet, so I am assuming you can do this by hand.
     
  5. Mar 20, 2005 #4

    cepheid

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    If the problem is:

    [tex] \int{\sqrt{\cos^3 x}dx} [/tex]

    Then I can't do it. Mathematica gave an answer involving "EllipticF" of something. I don't know what elliptic functions are.

    =/
     
  6. Mar 20, 2005 #5
    Aren't eliptic funtions x^2/a + y^2/b = c, or something near that?
     
  7. Mar 20, 2005 #6

    cepheid

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    No...that's just the equation of an ellipse in cartesian coordinates. That function, if you solve for y, is quite ordinary: y = square root of a bunch of stuff. Elliptic functions refer to something else entirely (not ordinary functions). I found it here: but don't really understand what it means: http://mathworld.wolfram.com/EllipticFunction.html

    I doubt you are expected to either, so maybe there is some other way of solving this integral. I don't understand why they would give you one that isn't solvable at a high school level. :confused:
     
  8. Mar 20, 2005 #7
    Could it be that the question is [tex]\int \cos^3 x dx/2x[/tex]

    [tex]\cos^3 x = \cos x \cos^2 x = \cos x (1- \sin^2 x) = \cos x - \cos x \sin^2 x[/tex] and now you have to get the derivative of 2x which is 2. So the answer should be: [tex]\frac{\cos x - \cos x \sin^2 x}{2}[/tex]

    Anyway, check out Integration of trigonometric functions

    Elliptic functions are part of complex analysis and are more sophisticated than these ones. This is beyond the first two or three semesters of college calculus.
     
    Last edited: Mar 20, 2005
  9. Mar 20, 2005 #8

    HallsofIvy

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    How does that help one integrate cos3/2x?
     
  10. Mar 20, 2005 #9
    I don't think you can integrate this...
     
  11. Mar 20, 2005 #10
    You can try the simple substitution [tex]u = \sqrt{\cos^3{x}}[/tex]. I don't know how far it will get you (I got it to a reasonably friendly form, but it still required the EllipticF function for Maple to evaluate).

    One of the main problems with finding this antiderivative is, of course, that the integrand is real sometimes and imaginary other times.
     
  12. Mar 20, 2005 #11
    I think it should be possible, it's a practice problem from the ap test for calculus ab. Maybe I don't need to show work for it? Anybody know how the calculus ap exam works?

    That was only part of the problem, and I figured I could just do the rest after this step was accomplished. The whole problem is:

    Consider the graphs of y=2/3x and y=cos^3/2(x). The two curves intersect at point P. Region R is bounded by the two curves and the y-axis. Region S is bounded by the two curves and the x-axis.

    a) Find the slope of the tangent line to y=cos^3/2(x), or y=sqrt(cos^3(x)) at point P. (I already know how to do this.)

    b) Find the area of R.
    c) Find the area of S.

    b) R = int(cos^3/2(x) - 2/3x)dx
    c) S = int(2/3x)dx + int(cos^3/2(x))dx

    Assuming I set those up right, you would need to do an itegration of cos^3/2(x) either way, which is where I get stuck. Maybe I'm just headed in the wrong direction? I can't think of any other way to find the area, besides the use of integrals.
     
  13. Mar 20, 2005 #12

    tony873004

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    In the order of operations, doesn't cos^3/2(x) mean

    [tex]\frac{cos^3}{2(x)}[/tex]

    instead of

    [tex]cos^{3/2}x[/tex]

    since the parenthesis are around the x and not around the 3/2 ?
     
  14. Mar 20, 2005 #13

    cepheid

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    You can't apply order of operations here because cos^3/2(x) is just the way he chose to try and write cos3/2x on the forum. In fact, the reason why he put parentheses around the x was to try and isolate it, to show that it was the argument of cosine, and that it wasn't related to those number 3 and 2 in any way, for they make up the exponent.

    By the way, cos3x is just a shorthand for (cos x)3. So

    cos3 by itself, without an argument, is meaningless. That should have tipped you off that you were barking up the wrong tree. :tongue2:
     
  15. Mar 20, 2005 #14
    LOL, yeah I'm sorry, I don't know how to do your fancy javascript/math notation things. I found this forum by accident, searching google for an answer to my problem. I was just hoping that a large group of math/physics gurus would be able to answer this "simple" problem quick and easily. However, I guess I assumed incorrectly. Once again to clarify y=(cos x)^3/2.
     
  16. Mar 20, 2005 #15

    cepheid

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    For what it's worth, I get:

    Area of R:

    [tex] \int_0^P {\left(\sqrt{\cos^3(x)} - \frac{2}{3}x\right)}dx [/tex]

    Area of S:

    [tex] \int_0^{\pi/2} {[\sqrt{\cos^3(x)}]}dx \ \ - \ \ \text{Area of R} [/tex]
     
    Last edited: Mar 20, 2005
  17. Mar 20, 2005 #16
    Yeah, I got the same. Just how are you suppose to integrate (cos x)^3/2...
     
  18. Mar 20, 2005 #17
    I think that in order to solve [tex]u = \sqrt{\cos^3{x}}[/tex] you need to use this equation [tex]u = \int_0^x \frac{ds}{\sqrt{(1-s^2)(1-k^2 s^2)}}[/tex]
     
    Last edited: Mar 20, 2005
  19. Mar 20, 2005 #18

    tony873004

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    It did, as soon as I hit Submit :grumpy:
     
  20. Mar 20, 2005 #19
    *shrug* X-43D, I've never seen that equation before, or I just don't remember it... I'm so lost on this problem, I just hope my teacher doesn't take a grade on this problem. I've asked two others in my class, and they can't seem to figure it out either.
     
  21. Mar 20, 2005 #20
    I don't get it either but i think it's an elliptic integral.
     
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