I. The history of the integral of the secant and the loxodrome is an excellent approach to this difficult, and historically difficult & important, integral. See Rickey and Tuchinsky, An Application of Geography to Mathematics: History of the Integral of the Secant, Mathematics Magazine
Vol. 53, No. 3 (May, 1980), pp. 162-166, http://www.jstor.org/stable/2690106.
One lesson for students is that when a difficult problem is solved, the method of solution can be shared with others, making things easier for future generations, including them.
II. Once I accidentally stumbled upon the following. I was looking for an example of an integral to differentiate (with respect to the upper limit). I wanted something not too complicated but one that a Calculus I student couldn't antidifferentiate. And then I wanted to put a function in the upper limit and apply the chain rule. I thought it would be nice if it simplified, too. So I hit upon
$$\int_0^x {1 \over 1-t^2} \; dt \quad \hbox{and} \quad \int_0^{\sin x} {1 \over 1 - t^2} \; dt \,.$$
And of course the last one simplifies to $\sec x$ (and is rather similar to Michael Joyce's comment and one of Steven Gubkin's answers).
This occurs shortly before we get to substitution and therefore before the students know the antiderivative of $\sec x$. So I could point to the integral, and say, "Hey, this is an antiderivative of secant? Do we know a function that is antiderivative of secant? That's interesting. I wonder what that integral is. I guess we'll have to wait and see." It is motivating in that it piques their curiosity. Even if we don't get to partial fractions in Calc. I or prepare them for the Calc. I method, it gets the students to want to know the answer.
It's more effective if most of your students haven't already had AP calculus.
III. Interestingly, according to Rickey and Tuchinsky, Isaac Barrow came up with essentially the same method as Joyce/Gubkin and was the first use of partial fractions in integration:
$$\int {dx \over \cos x} =
\int {\cos x \; dx \over \cos^2 x} = \cdots =
{1\over 2} \int {\cos x \over 1 - \sin x} + {\cos x \over 1 + \sin x} \; dx =
\text{etc.}$$
IV. A geometric approach. Let $s = \sec\theta$, $t = \tan\theta$:

Then from the differential triangle and the similar $1$-$s$-$t$ triangle, we have
$$
{s \; d\theta \over 1} = {dt \over s} = {ds \over t} \,.
$$
Okay, maybe it is easier to deduce the proportion from the derivative rules,
given the current conditions in math. ed.
Nevertheless, the symmetry of the last two terms of the proportion suggests combining them. Hence we get
$$
{s \; d\theta \over 1} = {dt + ds \over s + t} = {d(s+t) \over s+t} = d(\ln |u|)\,,
$$
where $u = \sec\theta + \tan\theta$. Integrating gives the result.