Talk:Integral of secant cubed/Archive 1

Some thoughts
As someone who deliberately tries to avoid accidentally putting this integral on multivariable calculus exams (because no one is ever able to evaluate it), I can say that it seems to arise in many unexpected places. Here are a few of the highlights:
 * This lovely little integral is an unfortunate byproduct of attempting to evaluate the arclength of a parabola y = x2/2 (say, from 0 to 1) by a trigonometric substitution. Indeed, the arclength formula gives
 * $$L = \int_0^1 \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = \int_0^1\sqrt{1+x^2}\,dx.$$
 * At this point, most calculus texts instruct the student to perform a trigonometric substitution x = tan&theta;. (A much easier thing to do would be to make a hyperbolic substitution, x = sinh&theta; where sinh is the hyperbolic sine, but this seems to have fallen out of fashion in the latest run of textbooks.)  After performing some simplifications, one is left with the integral
 * $$\int_0^{\pi/4}\sec^3\theta\, d\theta.$$

I'm sure I will think of some more, but these are two of the most obvious cases. One question is whether there are any applications of this integral which don't come from a substitution where one ultimately needs to integrate $$\scriptstyle{\sqrt{1+x^2}}$$. siℓℓy rabbit (  talk  ) 14:42, 11 July 2008 (UTC)
 * For much the same reason, this integral shows up unexpectedly if one tries to compute the surface area of a helicoid with parametric formula r(u,v) = (ucosv, usinv, v). Once again, a hyperbolic substitution (note: why is there no article for this?) is the easiest way to proceed.

Not as difficult?
I don't understand the sentence which says that the integral of secant cubed is not as difficult as the integral of the secant itself. If the method of the article is followed, then you need the integral of the secant to compute the integral of secant cubed. -- Jitse Niesen (talk) 22:42, 12 July 2008 (UTC)


 * The article presupposes that you've already got the integral of the secant function. Going from there to the integral of secant cubed is easier than getting the antiderivative of secant. Michael Hardy (talk) 02:17, 13 July 2008 (UTC)

Any MAA refs?
It might be that there are some MAA-type references for this integral, if people are interested in doing a bit of legwork. I wasn't able to find anything, but I didn't try for too long. --C S (talk) 15:14, 14 July 2008 (UTC)


 * Those do exist; I'll dig some up. Michael Hardy (talk) 15:50, 14 July 2008 (UTC)

Comments
A few comments on the material you've recently added:

1) Rather than saying "rectifying" and then parenthetically explaining that it means computing arc length, wouldn't it be simpler just to say "compute the arc length"? Do we need a special word for this?

2) You write: "This is one of fairly few integrals usually done in a first-year calculus course in which the most natural way to proceed involves integrating by parts and returning to the same integral one started with (another is the integral of the product of an exponential function with a sine or cosine function)." Fairly few seems a bit weaselly to me, and I don't quite agree; I would say that this phenomenon of integrating by parts twice and then solving the resulting equation for the integral is a fairly common one: I would expect any treatment of integration by parts to include that, even if they don't go into the more complicated reduction formulas.  Also this business of reduction formulas is already required for certain powers of sine and cosine.

I do not object to the article, but reading it still does not make clear to me what is so special about the integral of secant cubed. Plclark (talk) 06:32, 13 July 2008 (UTC)Plclark


 * 1) Well, at least next time the reader sees the word "rectify" used in this sense, they'll know what it means.
 * 2) Deriving those reduction formulas isn't usually required of first-year calculus students, so the ones using this technique that they actually go through are not so numerous. Yes, the account of integration by parts always includes it, but that doesn't contradict the "relatively few" comment.
 * Michael Hardy (talk) 15:33, 14 July 2008 (UTC)
 * Michael Hardy (talk) 15:33, 14 July 2008 (UTC)


 * 1) I don't think that using superfluous terminology just to expose people to it is a good idea. The appropriate place for introducing this terminology is the article on arc length, on which it already appears.  However, after being pointed out in the lead that it is synonymous with computing the arc length, "rectify" is not used any further in the article.  I can't think of any situation in which an article that a calculus student might read and understand would use the term "rectify" without explanation.


 * 2) The reduction formulas are typically not derived, but the use of this technique of integration by parts is still used for integrals involving sine and cosine. I don't know of any calculus textbook that discusses secant cubed but not sin^m x cos^n x: do you?  And I just don't agree with the "relatively few" -- most calculus exams on that I have seen and given have a problem of this type; also many students are familiar with it from high school -- but it's not really worth arguing about.  The article as a whole seems reasonably useful to me.  Plclark (talk) 18:29, 14 July 2008 (UTC)Plclark

Should this be a separate article or a redirect?
Perhaps this should be a redirect to List_of_integrals_of_trigonometric_functions? Gyro Copter (talk) 08:08, 10 July 2008 (UTC)


 * I disagree. This is an unusual special case. Michael Hardy (talk) 16:36, 10 July 2008 (UTC)


 * What makes this case unusual? Without a section on the significance of this derivation, the article is incomplete. If there is no such significance, then the subject is not notable. In that case, the general formula available at List_of_integrals_of_trigonometric_functions is sufficient. Remember WP:NOTTEXTBOOK—perhaps this article would be better in Calculus? Gyro Copter (talk) 07:11, 11 July 2008 (UTC)


 * Well, you've taken me by surprise. I'd have thought the answers to those questions were obvious, but maybe I'll have to remind myself that some people who read Wikipedia are not mathematicians, and add some explanation for them. Michael Hardy (talk) 14:14, 11 July 2008 (UTC)


 * ...and now I've looked at the cited section, and it does not derive the integral at all, but only reports the bottom line, and the derivation really wouldn't fit into that article. Michael Hardy (talk) 14:15, 11 July 2008 (UTC)
 * Could you explain to mathematicians why this integral is interesting? I find this completely non-obvious. Perhaps there are some universities where this integral is famous in calculus classes, but it certainly isn't famous in all of mathematics. Maybe this is a lack of mathematical culture on my part, in which case I kindly ask for you to educate me. Kusma (talk) 14:48, 11 July 2008 (UTC)

Well, first look at the comments by siℓℓy rabbit  below. Up above, someone points to the section titled List_of_integrals_of_trigonometric_functions. That section has a recursion formula for integrals of powers of secant. But it doesn't show where the formula comes from. This article does, and the derivation really doesn't fit in that other article. Michael Hardy (talk) 15:26, 11 July 2008 (UTC)
 * I don't think that we really need recursion formula for integrals of powers of secant as an article, but the case you make for that seems to be much stronger than for the specific case of the third power. Integrals that can be solved elementary (and with a fairly standard trick) just don't strike me as interesting enough for their own articles, unlike, say, exponential integral. Kusma (talk) 15:45, 11 July 2008 (UTC)

i think that philip bliss needs to work this one out —Preceding unsigned comment added by 196.211.60.60 (talk) 07:02, 11 July 2008 (UTC)

GyroCopter said in his/her edit summary that this integral is not the Julius Caesar of antiderivatives. Well, why does that redirect to this article then?? --C S (talk) 04:35, 15 July 2008 (UTC)


 * Because user:C S created that ridiculous redirect. I've deleted it. Michael Hardy (talk) 04:54, 15 July 2008 (UTC)