Inverse Functions: Arcsecant (Part 27)

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], so that

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

Why would this be controversial? 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. However, doing the same with \tan y is trickier. 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? The answer is, without additional information, we don’t know!

  • If 0 \le y < \pi/2 (so that x = \sec y \ge 1), then \tan y is positive, and so \tan y = \sqrt{\sec^2 y - 1}.
  • If \pi/2 < y \le \pi (so that x = \sec y \le 1), then \tan y is negative, and so \tan y = -\sqrt{\sec^2 y - 1}.

We therefore have two formulas for the derivative of y = \sec^{-1} x:

\displaystyle \frac{d}{dx} sec^{-1} x = \displaystyle \frac{1}{x \sqrt{x^2-1}} \qquad \hbox{if} \qquad x > 1

\displaystyle \frac{d}{dx} sec^{-1} x = \displaystyle -\frac{1}{x \sqrt{x^2-1}} \qquad \hbox{if} \qquad x < 1

These may then be combined into the single formula

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

green lineIt gets better. Let’s now find the integral

\displaystyle \int \frac{dx}{x \sqrt{x^2 -1}}

arcsec2

Several calculus textbooks that I’ve seen will lazily give the answer

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

This answer works as long as x > 1, so that |x| reduces to simply x. For example, the red signed area in the above picture on the interval [2\sqrt{3}/3,2] may be correctly computed as

\displaystyle \int_{2\sqrt{3}/3}^2 \frac{dx}{x \sqrt{x^2 -1}} = \sec^{-1} 2 - \sec^{-1} \displaystyle \frac{2\sqrt{3}}{3}

= \cos^{-1} \left( \displaystyle \frac{1}{2} \right) - \cos^{-1} \left( \displaystyle \frac{\sqrt{3}}{2} \right)

= \displaystyle \frac{\pi}{3} - \frac{\pi}{6}

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

However, the orange signed area on the interval [-2,-2\sqrt{3}/3]  is incorrectly computed using this formula!

\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)

= \cos^{-1} \left( -\displaystyle \frac{\sqrt{3}}{2} \right) -\cos^{-1} \left( \displaystyle \frac{1}{2} \right)

= \displaystyle \frac{5\pi}{6} - \frac{2\pi}{3}

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

This is patently false, as the picture clearly indicates that the above integral has to be negative. For this reason, careful calculus textbooks will often ask students to solve a problem like

\displaystyle \int \frac{dx}{x \sqrt{x^2 -1}}, \qquad x > 1

and the caveat x > 1 is needed to ensure that the correct antiderivative is used. Indeed, a calculus textbook that doesn’t include such caveats are worthy of any scorn that an instructor cares to heap upon it.

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