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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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Dann Corbit

Posts: 1,424
Registered: 12/8/04
Re: History of the integral of secant
Posted: Mar 2, 2009 7:11 PM
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On Mar 1, 4:23 pm, Robert Israel
<> wrote:
> "Stephen J. Herschkorn" <> 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.

This reference:
cited above, in footnote 6 on page 7 shows that Thomas Harriot
(1560-1621) evaluated {the integral from zero to theta of secant
(theta) dtheta} in 1594.

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