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The famously most difficult among completely elementary antiderivatives is that of the secant function.

Has someone tabulated all the ways it can be done, or written a somewhat comprehensive history of them, or an account of logical connections among them?

Does the feeling that this particular antiderivative can be found only by methods that are unexpected except by hindsight correspond in some way to some precisely stateable mathematical fact?

Parenthesis: Look at the tangent half-angle formula in the form $$ f(x)= \tan\left(\frac x 2 + \frac\pi4\right) = \tan x + \sec x. $$ Differentiating both sides yields $$ f'(x) = 2\sec^2\left(\frac x 2 + \frac\pi4\right) = \sec^2 x + \tan^2 x = (\sec x)(\sec x + \tan x) = (\sec x)f(x). $$ So $f$ satisfies a differential equation $$ f'(x) = (\sec x)f(x). $$ $$ \frac{df}{f} = \sec x. $$ Antidifferentiating both sides gives $\log|f(x)|= \text{the thing sought}$. Is this one "out there" somewhere (in published source or on the web)?

Later edit: Perhaps some ways of finding this antiderivative are neither "unexpected" (in the sense of being things that can be seen to work only by hindsight) nor applicable only to this one integral. But a fact persists: Lots of ways of doing this that are out there in the literature do match that description. Probably far more so than with all other elementary antiderivatives. So there's a question of whether there's some mathematical fact that expains why that should be true only of this one integral.

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Why do you think the method is unexpected? It is just another solution. What method would qualify as expected? – Pedro Tamaroff Jun 21 '12 at 17:47
@PeterTamaroff : If you know $(d/dx)x^n$, then you'll know how to find antiderivatives of polynomials without anyone's having told you and without giving it more than an instant's thought. If you're looking for $\int \tan x\,dx$ and you know the chain rule, you'll find it right away almost effortlessly. The same applies to most elementary antiderivatives except when there's some complexity to them, and then usually the complexity is the only difficulty, or else it yields to some method that is applicable very generally. None of that seems to apply to this one. The techniques..... – Michael Hardy Jun 21 '12 at 17:52
.....used are of a sort you wouldn't have anticipated without someone telling you about them. At least that's how I've always seen this one, and no others. – Michael Hardy Jun 21 '12 at 17:52
Well, Barrow's $$\int {\frac{{dx}}{{\cos x}}} = \int {\frac{{\cos x}}{{{{\cos }^2}x}}dx} = \int {\frac{{\cos x}}{{1 - {{\sin }^2}x}}dx} = \int {\frac{{du}}{{1 - {u^2}}}} $$ doesn't seem out of league, IMO. I think it is rather subjective. – Pedro Tamaroff Jun 21 '12 at 17:54
@MichaelHardy: That first step is very natural, since it's what you would use to compute $\int \cos^n x \, dx$ for any odd power $n$. – Hans Lundmark Jun 22 '12 at 8:25

I don't agree that evaluating this integral only involves methods that are unexpected. In particular, it is possible to integrate any rational expression involving trigonometric functions using the substitution $$ u = \tan(x/2). $$ This substitution has the nice property that $$ \sin x \;=\; \frac{2u}{1+u^2},\qquad \cos x \;=\; \frac{1-u^2}{1+u^2}, \qquad\text{and}\qquad dx \;=\; \frac{2\,du}{1+u^2}. $$ This is a perfectly standard technique, though it usually isn't taught in calculus classes anymore.

Applying this to the integral of secant gives $$ \int \sec x\,dx \;=\; \int \frac{dx}{\cos x} \;=\; \int \frac{2\,du}{1-u^2} \;=\; \ln\left|\frac{1+u}{1-u}\right|+C \;=\; \ln\left|\frac{1+\tan(x/2)}{1-\tan(x/2)}\right|+C $$ Most computer algebra systems use this technique as part of their integration algorithm, so this is the answer that you tend to get if you ask a computer for the integral of secant.

By the way, if this trick strikes you as clever, be aware that it works just as well to integrate rational expressions of sine and cosine using Euler's identity and the substitution $u=e^{ix}$. This tends to involve a lot of complex numbers, but it might seem more straightforward than the above substitution.

In any case, the only sense in which the integral of secant is difficult is that it can't be evaluated easily using the bag of tricks that we tend to teach in calculus classes nowadays. However, I don't think there's anything mathematically "natural" about the set of tricks that we teach, so I don't think the difficulty of integrating secant has any real mathematical significance.

Edit: By the way, the substitution $u = \tan(\theta/2)$ corresponds to a certain parameterization of the circle by rational functions. In particular, this is essentially the stereographic projection of the unit circle from the point $(-1,0)$ to $y$-axis, with the $\theta/2$ coming from the fact that an inscribed angle is half of the corresponding central angle. This same parameterization can be used to enumerate all Pythagorean triples.

Edit 2: To illustrate the point further, here's a way of integrating secant that only involves an "obvious" substitution. Let $u = \cos x$. Then $$ du \;=\; - \sin x\,dx \;=\; -\sqrt{1-u^2}\,dx $$ so $$ \int \sec x\,dx \;=\; \int \frac{dx}{\cos x} \;=\; \int -\frac{du}{u\sqrt{1-u^2}} \;=\; \mathrm{sech}^{-1}u+ C \;=\; \mathrm{sech}^{-1}(\cos u) + C. $$ Now, you might object to using the derivative formula for $\mathrm{sech}^{-1}$, on the grounds that this isn't usually covered in a first-year calculus course. But again, it seems arbitrary to me that we cover inverse trig functions but not inverse hyperbolic trig functions in first-year calculus.

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+1, but this won't be an "accepted" answer. Could it be that the secant is in some precise sense the simplest function that won't yield to methods simpler than the Weierstrass substitution, and that's a unique role? At any rate, even if the Weierstass substitution is not "unexpected", many---maybe all---of the more routine methods involve a step that can't be anticipated, but is seen by hindsight to work. – Michael Hardy Jun 21 '12 at 22:54
@MichaelHardy Fair enough. I suppose it is possible that the secant has some special property with respect to the standard bag of tricks, though I'm not sure how you'd define this precisely. One problem is that substitution always seems to involve a little bit of cleverness, so I'm not sure how you would account for that. – Jim Belk Jun 21 '12 at 23:28
I can't see that it always involves cleverness; sometimes it glares at you. – Michael Hardy Jun 21 '12 at 23:29
BTW, I am the creator of both Wikipedia's "Weierstrass substitution" article and its "integral of the secant function" article. And the List of things named after Karl Weierstrass. – Michael Hardy Jun 21 '12 at 23:33
@MichaelHardy Thanks for those articles. I think the Weierstrass substitution article may actually be how I learned this method! – Jim Belk Jun 21 '12 at 23:43

Since you ask for references, the original way this integral was found, via cartography, should be kept in mind. See V. F. Rickey and P. M. Tuchinsky, "An Application of Geography to Mathematics: History of the Integral of the Secant", Mathematics Magazine 53 (1980), 162-166. It is available for free on JSTOR.

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I up-voted this answer, but I actually wondered about an account of the plethora of methods and their history and perhaps an account of why there should be such a plethora of methods. – Michael Hardy Jun 30 '12 at 3:26

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