Ways to evaluate $\int \sec \theta \, \mathrm d \theta$ The standard approach for showing $\int \sec \theta \, \mathrm d \theta = \ln|\sec \theta + \tan \theta| + C$ is to multiply by $\dfrac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}$ and then do a substitution with $u = \sec \theta + \tan \theta$.
I like the fact that this trick leads to a fast and clean derivation, but I also find it unsatisfying: It's not very intuitive, nor does it seem to have applicability to any integration problem other than $\int \csc \theta \,\mathrm d \theta$.  Does anyone know of another way to evaluate $\int \sec \theta \, \mathrm d \theta$?

Instead of presenting another way of evaluating this integral, I justify a more general case in an approach which uses partial fractions and trigonometric identities, at the level of a Calculus class, I think:
$$\int \dfrac{1}{a+b\cos x}dx=\dfrac{1}{\sqrt{b^{2}-a^{2}}}\ln \left\vert 
\dfrac{\sqrt{a+b}+\sqrt{b-a}\tan x/2}{\sqrt{a+b}-\sqrt{b-a}\tan x/2}\right\vert \quad a\lt b.\quad (\ast)$$
Since
$$a+b\cos x=(a-b)+2b\cos ^{2}x/2,$$
we have
$$\dfrac{1}{a+b\cos x}=\dfrac{\sec ^{2}x/2}{(a-b)\sec ^{2}x/2+2b}=\dfrac{\sec
^{2}x/2}{(a-b)\sec ^{2}x/2+2b}=\dfrac{\sec ^{2}x/2}{a+b-(b-a)\tan ^{2}x/2}.$$
But
$$\dfrac{1}{a+b-(b-a)\tan ^{2}x/2}=$$
$$=\dfrac{1}{2\sqrt{a+b}}\left( \dfrac{1}{%
\sqrt{a+b}-\sqrt{b-a}\tan x/2}+\dfrac{1}{\sqrt{a+b}+\sqrt{b-a}\tan x/2}%
\right) .$$
Hence
$$\int \dfrac{1}{a+b\cos x}dx=$$
$$=\dfrac{1}{2\sqrt{a+b}}\int \left( \dfrac{\sec
^{2}x/2}{\sqrt{a+b}-\sqrt{b-a}\tan x/2}+\dfrac{\sec ^{2}x/2}{\sqrt{a+b}+%
\sqrt{b-a}\tan x/2}\right) dx$$
$$=\dfrac{1}{\sqrt{b^{2}-a^{2}}}\ln \left\vert \dfrac{\sqrt{a+b}+\sqrt{b-a}%
\tan x/2}{\sqrt{a+b}-\sqrt{b-a}\tan x/2}\right\vert .$$
Thus, we have your particular case
$$\int \dfrac{1}{\cos x}dx=\int \dfrac{1}{0+1\cos x}dx=\ln \left\vert \dfrac{%
1+\tan x/2}{1-\tan x/2}\right\vert . \qquad (\ast\ast)$$
From $\tan \dfrac{x}{2}=\dfrac{\sin x}{1+\cos x}$ and $\sec x+\tan x=\dfrac{1+\sec x+\tan x}{1+\sec x-\tan x}$ it follows that
$$\dfrac{1+\tan x/2}{1-\tan x/2}=\dfrac{1+\dfrac{\sin x}{1+\cos x}}{1-\dfrac{%
\sin x}{1+\cos x}}=\dfrac{1+\cos x+\sin x}{1+\cos x-\sin x}=\sec x+\tan x$$
and, finally
$$\int \sec x\; dx=\ln \left\vert \sec x+\tan x\right\vert .$$
