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A tricky integral

  1. Sep 20, 2005 #1
    How would one evaluate [tex] \int \sqrt{1+4\cos^2 2x}[/tex] ?

    It's for an arc length problem, and my teacher says to just find an approximate answer using a graphing calculator, but I actually want to learn how to integrate something like this.
     
  2. jcsd
  3. Sep 20, 2005 #2

    Tom Mattson

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    Unless there is some wacked out substitution you can do on that that I'm just not seeing, I don't think that your integrand has an antiderivative in terms of elementary functions. If you are really determined to do this without a graphing calculator and want to really impress your teacher, write a computer program that will approximate the integral to any desired accuracy using either the trapezoidal rule or Simpson's rule.

    At least, that would impress me. :tongue2:
     
  4. Sep 20, 2005 #3

    HallsofIvy

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    One of the things you should learn about integrals is that the great majority of "simple" functions cannot be integrated exactly!
     
  5. Sep 20, 2005 #4
    Thanks for confirming my suspicions that it wasn't integrable.
     
  6. Sep 20, 2005 #5

    Tom Mattson

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    Hold the phone...It is integrable. It just doesn't have an elementary antiderivative.

    The integrability of a function has nothing to do with whether or not you can find an antiderivative for that function. It has to do with whether or not the limit of the Riemann sum exists.
     
  7. Sep 20, 2005 #6

    saltydog

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    Well I'm glad you said that Flux because this integral is expressible in terms of the complete elliptic integral of the second kind. Oh yea, that is:

    [tex] \int \sqrt{1+4\cos^2 2x}=\frac{\sqrt{5}}{2}\text{EllipticE}[2x,4/5][/tex]

    where:

    [tex]\quote{EllipticE}(u,m)=\int_0^u \sqrt{1-mSin^2(\theta)}d\theta[/tex]

    That's what Mathematica says but I can't figure out how that is arrived at. You?
     
  8. Sep 20, 2005 #7

    Tom Mattson

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    I have to run right now, but you can do it with a substitution ([itex]\theta=2x[/itex]) and using the trig identity [itex]sin^2(\theta)=1-cos^2(\theta)[/itex] in the definition of the elliptic integral that you posted. Those steps should enable you to make Flux Cap's integral look like Mathematica's.
     
  9. Sep 21, 2005 #8

    saltydog

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    Well that's was easy. :blushing: See Flux, I have trouble with this stuf too you know. :smile:

    I tell you what though, the proof in the pudding is to actually calculate all this and verify that everthing lines up. I think there is some constraints on the limits of the Elliptical integral so maybe that formula is only valid for a range of x values. Don't know for sure. Really I think Flux ought to do that. Please post a complete report Ok?
     
  10. Sep 22, 2005 #9
    Quite frankly, I'm feeling waaaaaaaaay out of depth with this problem. I can understand that some mathematician found a solution to a similar problem as mine, but it was so complex that he just called it EllipticE. Oh wait, where's my political correctness, it could have been a female mathematician.

    At this point I just thought that I could learn what the function EllipticE was, how to use it, and any limitations of it, then I could finally solve this integral without approximation.

    Of course, it just got confusing when I started researching it.

    I checked a few pages, and wikipedia.org had a relatively clear definition, but the defintion involved even more functions for which I had never heard of. I really hate to do this, but I'm considering temporarily tabling the search to find out how one uses an elliptic integral to solve this type of problem. I'm still a high school student(I post here because I'm taking a college level class), so I have plenty of time to figure out. However, I'm also a senior, so between work and college applications and the typical nuances of life, I'm pretty busy and don't have time to devote to this.

    If anyone else is interested, however, and also has a pretty good calculus background, then I recommend wikipedia's site on it, since it had the most comprehensive list of information.
     
  11. Sep 23, 2005 #10

    saltydog

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    Hey Flux, I was kinda' funnin' about the report. Suppose we have this:

    [tex]\int_a^b \sqrt{1+4Cos^2(2x)}dx[/tex]

    As per Tom, first let [itex]\theta=2x[/itex]. This gives:

    [tex]\frac{1}{2}\int_{2a}^{2b}\sqrt{1+4Cos^2(\theta)}d\theta[/tex]

    Letting:

    [tex]Cos^2(\theta)=1-Sin^2(\theta)[/tex]

    [tex]\frac{1}{2}\int_{2a}^{2b}\sqrt{1+4(1-Sin^2(\theta))}d\theta[/tex]

    or:

    [tex]\frac{1}{2}\int_{2a}^{2b}\sqrt{5-4Sin^2(\theta)}d\theta[/tex]

    Pulling the 5 out of the radical gives us:

    [tex]\frac{\sqrt{5}}{2}\int_{2a}^{2b}\sqrt{1-\frac{4}{5}Sin^2(\theta)}d\theta[/tex]

    But by definition:

    [tex]\int_0^u \sqrt{1-mSin^2(\theta)}d\theta=EllipticE[u,m][/tex]

    we can write the definite integral as:

    [tex]
    \begin{align*}
    \frac{\sqrt{5}}{2}\int_{2a}^{2b}\sqrt{1-\frac{4}{5}Sin^2(\theta)}d\theta &= \frac{\sqrt{5}}{2}\left(
    \int_{0}^{2b}\sqrt{1-\frac{4}{5}Sin^2(\theta)}d\theta-\int_{0}^{2a}\sqrt{1-\frac{4}{5}Sin^2(\theta)}d\theta
    \right) \\
    &= \frac{\sqrt{5}}{2}\left(\text{EllipticE}[2b,4/5]-\text{EllipticE}[2a,4/5]\right)
    \end{align}
    [/tex]

    And so we can say:

    [tex]\int \sqrt{1+4Cos^2(2x)}dx=\frac{\sqrt{5}}{2}\text{EllipticE}[2x,4/5][/tex]

    And just cause I like some concrete evidence of all this I calculated:

    [tex]\int_2^5 \sqrt{1+Cos^2(2x)}dx[/tex]

    both using EllipticE functions and numerically. Both return 5.01464.
     
  12. Sep 23, 2005 #11
    I understood that part just fine, but I couldn't find out what the heck the ellipticE function was. If you could just show me some function, I don't care how complicated it is, E(a,b), then I'd be happy and content. I just can't find that function for some reason.
     
  13. Sep 23, 2005 #12
    Did you mean this?

    [tex] E(\phi , k) = \int_{0}^{\phi} dt \sqrt{1 - k^{2} \sin^{2}{t}} [/tex]

    An elliptic integral of the second kind?
     
  14. Sep 24, 2005 #13
    Where did these elliptic integrals come from? Do they have anything to do with elliptic curves?
     
  15. Sep 24, 2005 #14

    saltydog

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    Well, Mathematica uses EllipticE to denote the complete elliptic integral of the second kind. These functions are used in the solution of second-order non-linear ODEs. Not sure why they're called elliptic functions.
     
  16. Sep 24, 2005 #15

    Tom Mattson

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    Yes indeedy. The analysis of elliptical orbits is replete with elliptic integrals. In fact, I learned about elliptic integrals in physics before I learned about them in math.
     
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