I know the article is about sec(x) but I want to share this tidbit about its cousin, the hyperbolic secant: sech(x) is its own Fourier transform (modulo rescalings). That’s right, exp(-x^2) is not the only one.

If we're playing the map-projection-advocacy game, I'd say the Mollweide projection is underrated among equal-area maps [0]. (For local maps, use whatever you want, appropriately centered.) Sure, it distorts shapes away from the central meridian, but locally it only adds a simple horizontal skew. I'm not a big fan of how many equal-area 'compromise' projections lie about how long the lines of latitude are.

[0] https://en.wikipedia.org/wiki/Mollweide_projection

The most elegant proof IMHO is the one that avoids the original problem entirely.

Int[csc(x) dx] = 2 Int[csc(2u) du]

= 2 Int[du / (2 cos(u) sin(u))]

= Int[sec^2(u) du / tan(u)]

= log(tan(u)) + C

= log(tan(x/2)) + C

Then Int[sec(x)] = Int[csc(u)] = log(tan(u/2)) + C = log(tan(pi/4 - x/2)) + C.

Of course, this was no use to Mercator, because the logarithm hadn't been invented yet. But you aren't just pulling a magic factor out of nowhere. There is definitely a bit of cleverness in rearranging the fraction — you have to be used to trying to find instances of the power rule when dealing with integrals of fractions.

[Math Processing Error]