Homework Help: A tricky integral

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:

 

One of the things you should learn about integrals is that the great majority of "simple" functions cannot be integrated exactly!

 

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.

 

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?

 

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.

 

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

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?

 

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.

 

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?

 

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

 

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.

 

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.