The Integral of Secant

My first question on math.SE asked for different ways to evaluate \int \sec \theta \, d \theta.  This post will discuss my original method for evaluating \int \sec \theta \, d \theta, the method in my favorite answer from the math.SE question, and why the two are equivalent (even though they do not appear to be at first glance).

One way to evaluate \int \sec \theta \, d \theta is to multiply the integrand by \displaystyle \frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta} (or, as I tell my students, by 1 in the right way).  This yields

\displaystyle \int \sec \theta \, d \theta = \int \frac{\sec^2 \theta + \sec \theta \tan \theta}{\sec \theta + \tan \theta} \, d \theta.

The numerator is now the derivative of the denominator, and so a substitution with u = \sec \theta + \tan \theta produces an integral of the form \displaystyle \int \frac{du}{u}.  Thus we have

\displaystyle \int \sec \theta \, d \theta = \ln |\sec \theta + \tan \theta| + C.

My favorite answer from my math.SE question (although there are multiple good ones) is the one by Derek Jennings.  Rewrite \sec \theta as 1/\cos \theta and multiply by \displaystyle \frac{\cos \theta}{\cos \theta} (again, by 1 in the right way).  With this, we have

\displaystyle \int \sec \theta \, d \theta = \int \frac{\cos \theta \, d\theta}{\cos^2 \theta} = \int \frac{\cos \theta \, d \theta}{1 - \sin^2 \theta} = \int \frac{\cos \theta \, d \theta}{(1+ \sin \theta)(1 - \sin \theta)}.

Partial fractions decomposition on the last integral yields

\displaystyle \int \sec \theta \, d \theta = \frac{1}{2} \int \left(\frac{\cos \theta}{1 + \sin \theta} + \frac{\cos \theta}{1 - \sin \theta} \right)\, d \theta = \frac{1}{2} \ln \left| \frac{1 + \sin \theta}{1 - \sin \theta}\right| + C.

Why are these equal?  They certainly don’t look like the same answer.  They are, though, even up to the same constant C.  Starting with the second answer, multiply the fraction inside the logarithm by \displaystyle \left| \frac{1 + \sin \theta}{1 + \sin \theta} \right| (once again, by 1 in the right way).  This produces

\displaystyle \frac{1}{2} \ln \left| \frac{1 + \sin \theta}{1 - \sin \theta} \right| = \frac{1}{2} \ln \left| \frac{(1 + \sin \theta)^2}{1 - \sin^2 \theta} \right| = \frac{1}{2} \ln \left| \frac{(1 + \sin \theta)^2}{\cos^2 \theta} \right| = \frac{1}{2} \ln \left(\sec \theta + \tan \theta \right)^2 = \ln | \sec \theta + \tan \theta|,

the same expression as in the first answer.

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