## An arcsecant/arctangent integral

Recently in my integral calculus class I assigned the problem of evaluating

$\displaystyle \int \frac{dx}{x \sqrt{x^4-1}}$.

I intended for the students to recognize the integrand as similar to the derivative of arcsecant:

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

This leads to the substitution $u = x^2$, with $du = 2x dx$.  The original integral then becomes

$\displaystyle \frac{1}{2} \int \frac{2x dx}{x^2 \sqrt{x^4-1}} = \frac{1}{2} \int \frac{du}{u \sqrt{u^2-1}} = \frac{1}{2} \sec^{-1} u + C = \frac{1}{2} \sec^{-1} (x^2) + C.$

However, one of my students, Max, tried the substitution $u = \sqrt{x^4-1}$ (on the general principle that letting u be the most complicated part of the integral is a good idea).  This gives $du = 2x^3/\sqrt{x^4-1}$.  Max then got stuck with the substitution.  The problem was in the inverse trig functions section of the text, though, and so he guessed $\arctan (\sqrt{x^4-1})$ as an antiderivative.  Checking this, he obtained

$\displaystyle \frac{d}{dx} \arctan (\sqrt{x^4-1}) = \left(\frac{1}{x^4-1+1} \right) \left(\frac{1}{2}\right) \left(\frac{1}{\sqrt{x^4-1}}\right) \left(4x^3\right) = \frac{2}{x \sqrt{x^4-1}}.$

Since this is only off by a factor of 2, he concluded that the answer is

$\displaystyle \int \frac{dx}{x \sqrt{x^4-1}} = \frac{1}{2} \arctan (\sqrt{x^4-1}) + C$!

Why are both of these answers correct?  Let’s complete the evaluation of the integral using Max’s substitution, and then we’ll show that the two answers are actually equal to each other.

With $u = \sqrt{x^4-1}$$du = 2x^3/\sqrt{x^4-1}$, and $x^4 = u^2+1$, we can rewrite the original integral and evaluate it via

$\displaystyle \frac{1}{2} \int \frac{2x^3 dx}{x^4 \sqrt{x^4-1}} = \frac{1}{2} \int \frac{du}{u^2+1} = \frac{1}{2} \arctan u + C = \frac{1}{2} \arctan (\sqrt{x^4-1}) + C$.

It’s easy to get stuck here because the correct evaluation doesn’t actually substitute u into the integrand (at least not by itself).  Instead, the radical becomes part of the du, and the remaining expression in x gets replaced by $u^2+1$.  The rest of the evaluation is surprisingly simple, but getting to here is the trick.

The reason the two answers are equal has to do with trigonometry.  If $\theta = \sec^{-1} (x^2)$, then $\sec \theta = x^2$.  Using the trig identity $\tan^2 \theta + 1 = \sec^2 \theta$, we have $\tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{x^4-1}$.  Thus $\theta$ can also be expressed as $\theta = \arctan (\sqrt{x^4-1})$.  (This can also be seen by setting $x^2$ and 1 as the hypotenuse and adjacent side of a right triangle; the Pythagorean theorem plus the triangle definition of tangent gets you the rest of the way.)

I’ve come across different-looking answers to the same integral evaluation before, but this is probably the most complicated different-looking pair of answers I’ve seen.

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