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Topic: Historyu of the integral of secant
Replies: 13   Last Post: Mar 3, 2009 3:11 PM

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Robert Israel

Posts: 3,262
Registered: 2/1/07
Re: History of the integral of secant
Posted: Mar 1, 2009 7:23 PM
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"Stephen J. Herschkorn" <sjherschko@netscape.net> writes:

> Stephen J. Herschkorn wrote:
>

> > What is the history of the trick to find the indefinite integral of
> > the secant function? Who first used it, and when?

>
>
> Thanks to all for the responses to my query. From these, I summarize
> that Mercator (whose work precedes that of Newton and Leibniz by about
> two centuries) showed that integral(x=0..t, sec x) = ln |tan t/2 +
> pi/4|, though not in that notation.


That's not quite right: nobody really knows what Mercator did, but he
certainly didn't have logarithms available. He was about one century
before Newton and Leibniz.
Edward Wright showed how to use what we would call the integral of sec(x) to
produce a Mercator projection.
Henry Bond noticed that Wright's function coincided numerically with
ln(tan(x)/2 + pi/4).
James Gregory, and later Isaac Barrow, proved the formula.

> I suspect that, after the
> development of calculus, if one notices that ln |tan t/2 + pi/4| = ln
> |sec t + tan t|, one can then come up a posteriori with the trick of
> converting sec x to sec x (sec x + tan x) / (sec x + tan x ).


But it's simpler to just differentiate either formula and check that the
result simplifies to sec(x).

> As I noted in another post, this all comes up because I was presenting
> the integral of secant, via the aforementioned trick, in a
> second-semseter calculus class the other day. (I have been teaching
> calculus for only two years now, and this is the first time this topic
> has come up in the syllabus.) The students very naturally asked, "How
> did you know to multiply by sec x + tan x?" My response was, well,
> that's just a trick someone came up with. I wouldn't expect the
> students to come up with it on their own.
>
> Rick Decker, in another post in this thread, posted another derivation.
> Rewriting it a bit,
>
> sec x = 1 / cos x = cos x / cos^2 x = cos x / (1 - sin^2 x).
>
> Now use the substitution u = sin x and apply partial fractions to get
>
> 1/2 [ln(1 - sin x) + ln(1 + sin x)]
>
> as an antiderivative. Now use the properties of logarithms and the
> trigonometric functions to show that this equals ln |sec x + tan x|.
>
> This derivation seems more "natural" to me: Asked to find the
> antiderivative of the secant, I see nothing unusual in coming up with
> it. Once one sees the final result, one can then come up with the
> efficient trick. I wouldn't be surprised if this how it all went down.
>

--
Robert Israel israel@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada



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