[scythe]

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.

[qbane]

This was the one I was taught in my high school. It has some cleverness (e.g., some trig. transformations) but looks less like coming out of nowhere than the original.