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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?2012-06-21
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    @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.....2012-06-21
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    .....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.2012-06-21
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    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.2012-06-21
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    @MichaelHardy: I would agree that it is attractive, and unexpected. In elementary calculus we seldom look for solutions that are cleverly shifted standard trig functions. Certainly not with tan.2012-06-21
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    @PeterTamaroff : Barrow's method is simple, but that first step seems to be something that applies only to this integral rather than like something you'd generally do.2012-06-21
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    Why is the integral in `\displaystyle`? It looks conspicuous on the front page.2012-06-21
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    @RahulNarain I agree. I have removed the \displaystyle from the integral in the title.2012-06-21
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    30 years ago, I asked one of the early symbolic math programs to do this integral (or possibly it was the even tougher $\int\sec^3x\,dx$). It didn't give me the answer I was expecting. When I looked into it, I saw that the answer the software gave me differed from the one I expected by a constant --- and the constant was $\pi i/4$.2012-06-21
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    @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 $n$.2012-06-22
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    My 8 October 2009 sci.math post **The Gudermannian function** (see [Google archive version](https://groups.google.com/group/sci.math/msg/dfb992fe3d16fc49) or [Math Forum archive version](http://mathforum.org/kb/message.jspa?messageID=6864360)) might suggest some useful directions for you to look into.2012-06-22

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