Solving Integral Problem: Arc Length of Sine

[PWiz]

Homework Statement

I was trying to find a general expression for the arc length of sine, i.e. $\int \sqrt{1+cos^2 x} dx$, but got stuck. I made this problem up, and with some google searches realized that it uses something known as an elliptical integral. How do I go about using it here?

2. Homework Equations
Trig substitutions

The Attempt at a Solution

By letting $cosx=tanu$ and by differentiating arccos, I came down to the integral $\int \frac{-sec^3 u}{\sqrt{2-sec^2 u}}du$, which seems as unsolvable as the original question.

 

[Ray Vickson]

Homework Statement

I was trying to find a general expression for the arc length of sine, i.e. $\int \sqrt{1+cos^2 x} dx$, but got stuck. I made this problem up, and with some google searches realized that it uses something known as an elliptical integral. How do I go about using it here?

2. Homework Equations
Trig substitutions

The Attempt at a Solution

By letting $cosx=tanu$ and by differentiating arccos, I came down to the integral $\int \frac{-sec^3 u}{\sqrt{2-sec^2 u}}du$, which seems as unsolvable as the original question.

Your integral is non-elementary, so no matter how hard you look and how many millions of pages you allow yourself for writing out a finite formula, you won't be able to do it. That is a genuine theorem---that is, has been proved rigorously. However, you can do it in terms of the non-elementary elliptic functions.

Elliptic functions are widely tabulated and are implemented in many computer algebra systems (Maple, Mathematica, etc.). There are also numerous sources of computer codes for them in various computer languages. You can also look on-line for more details, giving things like series expansions, asymptotic expansions and the like. What you probably will not find is a hand-held calculator with an "elliptic function" button (but I could be wrong).

Another approach would be straight numerical integration, where the actual function details are more-or-less ignored.

 

[Dick]

Homework Statement

I was trying to find a general expression for the arc length of sine, i.e. $\int \sqrt{1+cos^2 x} dx$, but got stuck. I made this problem up, and with some google searches realized that it uses something known as an elliptical integral. How do I go about using it here?

2. Homework Equations
Trig substitutions

The Attempt at a Solution

By letting $cosx=tanu$ and by differentiating arccos, I came down to the integral $\int \frac{-sec^3 u}{\sqrt{2-sec^2 u}}du$, which seems as unsolvable as the original question.

To express the result as an elliptic integral you just need the definition of an elliptic integral of the second kind. $E(x|m)=\int_0^x \sqrt{1-m \sin^2(\theta)} d\theta$. If you can express your integral in terms of that form (find m), then you can express the result in terms of $E$.

 

[PWiz]

Alright, say it was a definite integral:
$$\int_0^4 \sqrt{1+cos^2x} dx$$
What should my procedure then be?

 

[Ray Vickson]

Alright, say it was a definite integral:
$$\int_0^4 \sqrt{1+cos^2x}$$ $$dx$$
What should my procedure then be?

Do you want a numerical answer? If so, either apply an elliptic function software package, or else use a numerical integration scheme such as Simpson's Rule.

 

[PWiz]

Do you want a numerical answer? If so, either apply an elliptic function software package, or else use a numerical integration scheme such as Simpson's Rule.

So you mean there is no way to calculate the exact value of the finite integral as well? Will I always be forced to use numerical approximations with these kind of functions? If yes, then is there any quick way to realize that a function is non-integrable (and hence save myself a lot of effort)?

 

[Dick]

Alright, say it was a definite integral:
$$\int_0^4 \sqrt{1+cos^2x} dx$$
What should my procedure then be?

Use the identity $cos^2 x=1-sin^2 x$ to express it in the elliptic form. If you are expecting to get an 'exact value' out of this you going to be disappointed unless you consider something like $sin(1)$ an 'exact value'.

 

[PWiz]

Use the identity $cos^2 x=1-sin^2 x$ to express it in the elliptic form. If you are expecting to get an 'exact value' out of this you going to be disappointed unless you consider something like $sin(1)$ an 'exact value'.

So the definite integral is simply represented by $E(4|cot^2x)$? And what do we do if the lower limit is non-zero?

 

[Dick]

So the definite integral is simply represented by $E(4|cot^2x)$? And what do we do if the lower limit is non-zero?

No, how did you get $cot^2 x$??

 

[PWiz]

Sorry, I meant $-cot^2x$. Since $1-msin^2x=1+cos^2x$, $m=-cot^2x$ right?

 

[Dick]

Sorry, I meant $-cot^2x$. Since $1-msin^2x=1+cos^2x$, $m=-cot^2x$ right?

No, $m$ needs to be a constant! Start from $\sqrt{1+cos^2 x} = \sqrt{1+(1-sin^2 x)}=\sqrt{ 2-sin^2 x}$. Now you are getting close to the elliptic form, except that '2' needs to be a '1'. Do you see what to do?

 

[SteamKing]

So you mean there is no way to calculate the exact value of the finite integral as well? Will I always be forced to use numerical approximations with these kind of functions? If yes, then is there any quick way to realize that a function is non-integrable (and hence save myself a lot of effort)?

There's no quick way to decide whether a given integral is non-elementary, other than to look in a table of integrals and see if it, or one of its forms, appears.

There are some famous integrals which don't have closed-form solutions, but which are studied because of their utility in physics, mechanics, and other sciences. Elliptic integrals are a broad category of such functions, and these arose out of attempts to find a formula for the arc length of an ellipse.

http://en.wikipedia.org/wiki/Jacobi_elliptic_functions

One other famous non-elementary integral which finds daily use is the normal distribution, which appears quite often in statistics.

http://en.wikipedia.org/wiki/Normal_distribution

Other non-elementary functions are the result of trying to solve certain classes of ordinary differential equations. Bessel functions are one such class of solutions:

http://en.wikipedia.org/wiki/Bessel_function

As you go thru your mathematics education, you'll find that there are quite a few integrals and solutions to ODEs which cannot be expressed in terms of elementary functions.

 

[Ray Vickson]

So you mean there is no way to calculate the exact value of the finite integral as well?
Will I always be forced to use numerical approximations with these kind of functions?

**********************************************************
Yes, I thought that is what I said.

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If yes, then is there any quick way to realize that a function is non-integrable (and hence save myself a lot of effort)?

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The (now) free article
http://www4.ncsu.edu/~singer/ma792Kdocs/rosenlicht.pdf
has the basic results and gives proofs. See also
http://www.rangevoting.org/MarchisottoZint.pdf

Just Google 'integration if finite terms' to see more articles.

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Note added in edit: There is an article by Mathew Wiener that goes through all this in a fairly elementary way, taking the time to introduce the tools and prove results in about as simple a way as possible. It also lists a number of provably-non-elementary integrations, which should suit your purposes of knowing when to stop trying to integrate and when to use other methods (such as numerics). Here is the link:

http://www.math.niu.edu/~rusin/known-math/93_back/elementary.int

I recommend you look at it.

 

[PWiz]

@Dick Okay, if m is supposed to be a constant then will the integral resolve out to become $\sqrt{2} E(4|0.5)$?
@Ray Vickson Thanks for the articles, there is definitely some interesting information in them on which I'd like to spend some more time.
@SteamKing I'm familiar with none of those functions except the normal distribution.

 

[Dick]

@Dick Okay, if m is supposed to be a constant then will the integral resolve out to become $\sqrt{2} E(4|0.5)$?

Yes!

 

[PWiz]

Yes!

And what if one of the limits isn't zero, like $\int_{-2}^4 \sqrt{1+cos^2x}dx$?

 

[Dick]

And what if one of the limits isn't zero, like $\int_{-2}^4 \sqrt{1+cos^2x}dx$?

I'm not going to tell you until you think about it.

 

[PWiz]

I'm not going to tell you until you think about it.

I was thinking about splitting the problem into two : $-\int_0^{-2} \sqrt{1+cos^2x}dx$ + $\int_0^4 \sqrt{1+cos^2x}dx = \sqrt{2} (-E(-2|0.5)+E(4|0.5))$, but I still have my doubts.

 

[PWiz]

Um so did I get it right?

I was thinking about splitting the problem into two : $-\int_0^{-2} \sqrt{1+cos^2x}dx$ + $\int_0^4 \sqrt{1+cos^2x}dx = \sqrt{2} (-E(-2|0.5)+E(4|0.5))$, but I still have my doubts.

 

[Ray Vickson]

Um so did I get it right?

Look at areas under the curve $y = f(x) = \sqrt{1+\cos^2(x)}$. The quantity, $\int_{-2}^4 f(x) \, dx$, is the area $A(-2 \to +4)$ under $y = f(x)$ between $x = -2$ and $x = +4$. This is $A(-2 \to 0) + A(0 \to +4)$. Look at $f(x)$. For $a > 0$ can you see how to relate $A(-a \to 0)$ to values of $A(0,b)$ for some appropriate $b > 0$?

 

[PWiz]

Look at areas under the curve $y = f(x) = \sqrt{1+\cos^2(x)}$. The quantity, $\int_{-2}^4 f(x) \, dx$, is the area $A(-2 \to +4)$ under $y = f(x)$ between $x = -2$ and $x = +4$. This is $A(-2 \to 0) + A(0 \to +4)$. Look at $f(x)$. For $a > 0$ can you see how to relate $A(-a \to 0)$ to values of $A(0,b)$ for some appropriate $b > 0$?

Yes, and after splitting the problem into those two parts (post 18) and switching the limits for the -ve case (by adding a -ve sign outside the integral) I came down to $\sqrt{2} (E(4|0.5)-E(-2|0.5))$. I can't simplify this any further, so have I done it correctly?

 

[Dick]

Yes, and after splitting the problem into those two parts (post 18) and switching the limits for the -ve case (by adding a -ve sign outside the integral) I came down to $\sqrt{2} (E(4|0.5)-E(-2|0.5))$. I can't simplify this any further, so have I done it correctly?

Sure that's right. You can also check these sorts of thing on Wolfram. Ask it for the value of sqrt(2)*(EllipticE(4,1/2)-EllipticE(-2,1/2)). Compare it with the value it gives for the integral.

 

[PWiz]

Alright, thanks. One final question: are these sort of problems encountered in math during the first two years of undergrad education?

 

[Dick]

Alright, thanks. One final question: are these sort of problems encountered in math during the first two years of undergrad education?

I don't think special functions like that are a routine part of the curriculum. On the other hand you might see them in a course with a title like 'Mathematical Physics'.

 

[Ray Vickson]

Yes, and after splitting the problem into those two parts (post 18) and switching the limits for the -ve case (by adding a -ve sign outside the integral) I came down to $\sqrt{2} (E(4|0.5)-E(-2|0.5))$. I can't simplify this any further, so have I done it correctly?

OK, but if you notice that $f(x) = \sqrt{1+\cos^2(x)}$ is an even function (meaning that $f(-x) = f(x)$) you ought to be able to see that $A(-2 \to 0) = A(0 \to 2)$, so the answer is $E(4) + E(2)$. That would be convenient if you were using tables of $E$ (or software) that could only evaluate it for positive arguments.

That was why I asked you about the relationship between $A(-2 \to 0)$ and $A(0 \to 2)$, but you did not answer my question.

 

[PWiz]

OK, but if you notice that $f(x) = \sqrt{1+\cos^2(x)}$ is an even function (meaning that $f(-x) = f(x)$) you ought to be able to see that $A(-2 \to 0) = A(0 \to 2)$, so the answer is $E(4) + E(2)$. That would be convenient if you were using tables of $E$ (or software) that could only evaluate it for positive arguments.

That was why I asked you about the relationship between $A(-2 \to 0)$ and $A(0 \to 2)$, but you did not answer my question.

Ahhh, I completely ignored the parity of the function (I didn't see that relationship in the first place) Thanks , I'll surely investigate the properties of the function next time I meet an elliptical integral (which I guess will happen a lot if I play around with the curve lengths of different functions).