Inverse Functions: Arcsecant (Part 28)

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? Yesterday, we saw that \tan x is both positive and negative on the interval [0,\pi], and so great care has to be used to calculate the integral:

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

Here’s another example: let’s use trigonometric substitution to calculate

\displaystyle \int_{-6}^{-3} \frac{ \sqrt{x^2-9} }{x} dx

The standard trick is to use the substitution x = 3 \sec \theta. With this substitution:

  • x^2 - 9 = 9 \sec^2 \theta - 9 = 9 \tan^2 \theta, and
  • dx = 3 \sec \theta \tan \theta \, d\theta
  • x = -3 \quad \Longrightarrow \quad \sec \theta = -1 \quad \Longrightarrow \quad \theta = \sec^{-1} (-1) = \pi
  • x = -6 \quad \Longrightarrow \quad \sec \theta = -2 \quad \Longrightarrow \quad \theta = \sec^{-1} (-2) = \displaystyle \frac{2\pi}{3}

Therefore,

\displaystyle \int_{-6}^{-3} \frac{ \sqrt{x^2-9} }{x} dx = \displaystyle \int_{2\pi/3}^{\pi} \frac{ \sqrt{9 \tan^2 \theta}} { 3 \sec \theta} 3 \sec \theta \tan \theta \, d\theta

At this point, the common mistake would be to replace \sqrt{9 \tan^2 \theta} with 3 \tan \theta. This is a mistake because

\sqrt{9 \tan^2 \theta} = |3 \tan \theta|

Furthermore, for this particular problem, \tan \theta is negative on the interval [2\pi/3,\pi]. Therefore, for this problem, we should replace \sqrt{9 \tan^2 \theta} with -3 \tan \theta.

Continuing the calculation,

\displaystyle \int_{-6}^{-3} \frac{ \sqrt{x^2-9} }{x} dx = \displaystyle \int_{2\pi/3}^{\pi} \frac{ -3 \tan \theta} { 3 \sec \theta} 3 \sec \theta \tan \theta

= \displaystyle \int_{2\pi/3}^{\pi} -3\tan^2 \theta \, d\theta

= \displaystyle \int_{2\pi/3}^{\pi} -3(\sec^2 \theta-1) \, d\theta

= \displaystyle \int_{2\pi/3}^{\pi} (3-3 \sec^2 \theta) \, d\theta

= \displaystyle \bigg[ 3 \theta - 3 \tan \theta \bigg]_{2\pi/3}^{\pi}

= \displaystyle \left[ 3 \pi - 3 \tan \pi \right] - \left[ 3 \left( \frac{2\pi}{3} \right) - 3 \tan \left( \frac{2\pi}{3} \right) \right]

= \displaystyle 3\pi - 0 - 2\pi + 3(-\sqrt{3})

= \pi - 3\sqrt{3}

So if great care wasn’t used to correctly simplify \sqrt{9 \tan^2 \theta}, one would instead obtain the incorrect answer 3\sqrt{3} - \pi.

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by John Quintanilla on October 28, 2014  •  Permalink
Posted in Calculus, Precalculus
Tagged , , , , ,

Posted by John Quintanilla on October 28, 2014

https://meangreenmath.com/2014/10/28/inverse-functions-arcsecant-part-28/

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