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 $\int \sec \theta \, \mathrm d \theta$?

Here is another variation on a theme. It relies on the following two double angle formulae for sine and cosine, namely
$$\sin 2\theta = 2 \sin \theta \cos \theta \qquad \text{and} \qquad \cos 2 \theta = \cos^2 \theta - \sin^2 \theta,$$
two obvious substitutions, and a simple partial fraction decomposition. It is similar to making the famous $t$-substitution of $t = \tan \frac{x}{2}$ without having to rely on knowing this. 
If the substitution $x = 2u$ is made, we have
\begin{align}
\int \sec x \, dx &= 2 \int \sec 2u \, du\\
&= 2 \int \frac{du}{\cos 2u}\\
&= 2 \int \frac{du}{\cos^2 u - \sin^2 u}\\
&= 2 \int \frac{du}{\cos^2 u(1 - \tan^2 u)}\\
&= 2 \int \frac{\sec^2 u}{1 - \tan^2 u} \, du.
\end{align}
Now let $t = \tan u, dt = \sec^2 u \, du$. Thus
\begin{align}
\int \sec x \, dx &= 2 \int \frac{dt}{1 - t^2}\\
&= 2 \int \frac{dt}{(1 - t)(1 + t)}\\
&= \int \left [\frac{1}{1 - t} + \frac{1}{1 + t} \right ] dt\\
&= \ln \left |\frac{1 + t}{1 - t} \right | + C\\
&= \ln \left |\frac{1 + \tan u}{1 - \tan u} \right | + C \qquad \text{since} \,\, t = \tan u\\
&= \ln \left |\frac{\cos u + \sin u}{\cos u - \sin u} \right | + C\\
&= \ln \left |\frac{\cos u + \sin u}{\cos u - \sin u} \cdot \frac{\cos u + \sin u}{\cos u + \sin u} \right | + C\\
&= \ln \left |\frac{\cos^2 u + \sin^2 u + 2\sin u \cos u}{\cos^2 u - \sin^2 u} \right | + C\\
&= \ln \left |\frac{1 + \sin 2u}{\cos 2u} \right | + C\\
&= \ln \left |\frac{1 + \sin x}{\cos x} \right | + C \qquad \text{since} \,\, x = 2u\\
&= \ln |\sec x + \tan x| + C,
\end{align}
as expected.
