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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.

- Thread starter FluxCapacitator
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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.

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Tom Mattson

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At least, that would impress me. :tongue2:

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HallsofIvy

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Thanks for confirming my suspicions that it wasn't integrable.

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Tom Mattson

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Hold the phone...ItFluxCapacitator said:Thanks for confirming my suspicions that it wasn't integrable.

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.

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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:FluxCapacitator said:How would one evaluate [tex] \int \sqrt{1+4\cos^2 2x}[/tex] ?

I actually want to learn how to integrate something like this.

[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?

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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.saltydog said:You?

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saltydog

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

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?

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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.

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saltydog

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

[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.

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

An elliptic integral of the second kind?

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Where did these elliptic integrals come from? Do they have anything to do with elliptic curves?

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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.apmcavoy said:Where did these elliptic integrals come from? Do they have anything to do with elliptic curves?

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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.apmcavoy said:Where did these elliptic integrals come from? Do they have anything to do with elliptic curves?

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