# Inverse Functions: Arcsecant (Part 29)

We now turn to a little-taught and perhaps controversial inverse function: arcsecant. As we’ve seen throughout this series, the domain of this inverse function must be chosen so that the graph of $y = \sec x$ satisfies the horizontal line test. It turns out that the choice of domain has surprising consequences that are almost unforeseeable using only the tools of Precalculus.

The standard definition of $y = \sec^{-1} x$ uses the interval $[0,\pi]$ — or, more precisely, $[0,\pi/2) \cup (\pi/2, \pi]$ to avoid the vertical asymptote at $x = \pi/2$ — in order to approximately match the range of $\cos^{-1} x$. However, when I was a student, I distinctly remember that my textbook chose $[0,\pi/2) \cup [\pi,3\pi/2)$ as the range for $\sec^{-1} x$.

I believe that this definition has fallen out of favor today. However, for the purpose of today’s post, let’s just run with this definition and see what happens. This portion of the graph of $y = \sec x$ is perhaps unorthodox, but it satisfies the horizontal line test so that the inverse function can be defined.

Let’s fast-forward a couple of semesters and use implicit differentiation (see also https://meangreenmath.com/2014/08/08/different-definitions-of-logarithm-part-8/ for how this same logic is used for other inverse functions) to find the derivative of $y = \sec^{-1} x$:

$x = \sec y$

$\displaystyle \frac{d}{dx} (x) = \displaystyle \frac{d}{dx} (\sec y)$

$1 = \sec y \tan y \displaystyle \frac{dy}{dx}$

$\displaystyle \frac{1}{\sec y \tan y} = \displaystyle \frac{dy}{dx}$

At this point, the object is to convert the left-hand side to something involving only $x$. Clearly, we can replace $\sec y$ with $x$. As it turns out, the replacement of $\tan y$ is a lot simpler than we saw in yesterday’s post. Once again, we begin with one of the Pythagorean identities:

$1 + \tan^2 y = \sec^2 y$

$\tan^2 y = \sec^2 y - 1$

$\tan^2 y = x^2 - 1$

$\tan y = \sqrt{x^2 - 1} \qquad \hbox{or} \tan y = -\sqrt{x^2 - 1}$

So which is it, the positive answer or the negative answer? In yesterday’s post, the answer depended on whether $x$ was positive or negative. However, with the current definition of $\sec^{-1} x$, we know for certain that the answer is the positive one! How can we be certain? The angle $y$ must lie in either the interval $[0,\pi/2)$ or else the interval $[\pi,3\pi/2)$. In either interval, $\tan y$ is positive. So, using this definition of $\sec^{-1} x$, we can simply say that

$\displaystyle \frac{d}{dx} \sec^{-1} x = \displaystyle \frac{1}{x \sqrt{x^2-1}}$,

and we don’t have to worry about $|x|$ that appeared in yesterday’s post.

Turning to integration, we now have the simple formula

$\displaystyle \int \frac{dx}{x \sqrt{x^2 -1}} = \sec^{-1} x + C$

that works whether $x$ is positive or negative. For example, the orange area can now be calculated correctly:

$\displaystyle \int_{-2}^{-2\sqrt{3}/3} \frac{dx}{x \sqrt{x^2 -1}} = \sec^{-1} \left( - \displaystyle \frac{2\sqrt{3}}{3} \right) - \sec^{-1} (-2)$

$= \displaystyle \frac{7\pi}{6} - \frac{4\pi}{3}$

$= \displaystyle -\frac{\pi}{6}$

So, unlike yesterday’s post, this definition of $\sec^{-1} x$ produces a simple integration formula.

So why isn’t this the standard definition for $\sec^{-1} x$? I’m afraid the answer is simple: with this definition, the equation

$\sec^{-1} x = \cos^{-1} \left( \displaystyle \frac{1}{x} \right)$

is no longer correct if $x < -1$. Indeed, I distinctly remember thinking, back when I was a student taking trigonometry, that the definition of $\sec^{-1} x$ seemed really odd, and it seemed to me that it would be better if it matched that of $\cos^{-1} x$. Of course, at that time in my mathematical development, it would have been almost hopeless to explain that the range $[0,\pi/2) \cup [\pi,3\pi/2)$ had been chosen to simplify certain integrals from calculus.

So I suppose that The Powers That Be have decided that it’s more important for this identity to hold than to have a simple integration formula for $\displaystyle \int \frac{dx}{x \sqrt{x^2 -1}}$