When teaching calculus, I often present the indefinite integral $$ \int \sec x \, dx = \ln | \sec x + \tan x | + C. $$

I am looking for ways to motivate this anti-derivative rather than just tell the students the answer and show them how to verify it.

Here's the best I have come up with:

We are trying to find a function whose derivative is $\sec x$, so let's think of all the functions we know whose derivative involves $\sec x$. Well, $\frac{d}{dx} \sec x = \sec x \tan x$ and $\frac{d}{dx} \tan x = \sec^2 x$. If we play around a bit, we might stumble upon noticing that $\frac{d}{dx} (\sec x + \tan x) = \sec x (\sec x + \tan x)$ and if we are clever enough, we might realize how that can help us, since if $$ \sec x = \frac{\frac{d}{dx} (\sec x + \tan x)}{\sec x + \tan x},$$ then since we know that $\frac{d}{dx} \ln |f(x)| = \frac{f'(x)}{f(x)}$, we can realize that $$ \sec x = \frac{d}{dx} \ln |\sec x + \tan x|.$$

Is there a simpler way to motivate the result? Is there a better explanation or reason for taking the linear combination $\sec x + \tan x$ than just trying simple combinations in the hope that something good will happen?

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    $\begingroup$ Perhaps first do antiderivative for $\sin^n x \cos^m x$ where at least one of $n,m$ is odd. Then recognize this as a special case. $\endgroup$ Jan 17, 2016 at 23:34
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    $\begingroup$ @GeraldEdgar I think most textbooks treat these inductively, and the methods used only apply to positive $n$ and $m$ $\endgroup$ Jan 18, 2016 at 0:34
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    $\begingroup$ You can rewrite the integral as $\int \frac{\cos x}{\cos^2 x} \, dx$ and use the substitution $u = \sin x$, but then the integral requires partial fractions to evaluate and a little bit of algebra to simplify. So the approach works directly, with a cost of more techniques required but less unmotivated guess-and-verify steps. $\endgroup$ Jan 18, 2016 at 0:36
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    $\begingroup$ You can even do this in the lecture with $\sin^n x \cos^m x$ as an example to illustrate that it works for negative exponents. As long as one of them is odd. $\endgroup$ Jan 18, 2016 at 14:02
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    $\begingroup$ This is just a quibble, but I don't think it's accurate to say that the integral is $\ln|\sec x+\tan x|+C$. This function consists of infinitely many continuous pieces, each of which has its own constant of integration. There is not just one constant of integration $C$ as implied by your notation. $\endgroup$
    – user507
    Jan 19, 2016 at 16:19

3 Answers 3


To expand on Gerald Edgar's comments above:

Assume that you have already covered partial fractions.

Develop a general strategy for integrals of powers of trig functions: If a power is odd, you can use pythagorean identity and a substitution. If both powers are even, use half angle reduction formulas.

For instance

$$ \begin{align*} \int\sin^{2k+1}\cos^m(x) dx &= \int (1-\cos^2(x))^k \cos^m(x) \sin(x) dx\\ &=\int -(1-u^2)^k u^m du \end{align*} $$

Where we made the substitution $u = \cos(x)$.

Start with some positive integer examples. Here, you just get a polynomial, so you can integrate by expanding and integrating termwise. Then observe the same strategy works for all integers, you just get rational functions of $u$ if an exponent is negative, which might have to be handled using partial fraction decomposition.

An example of this general strategy is integrating secant!

$$ \begin{align*} \int \sec(x) dx &= \int \cos(x)^{-1} dx \textrm{ This is an odd power of cosine}\\ &= \int \cos(x)^{-2}\cos(x) dx\\ &=\int (1-\sin^2(x))^{-1}\cos(x) dx\\ &=\int \frac{1}{1-u^2} du\\ &=\frac{1}{2}\int \frac{1}{1-u}+\frac{1}{1+u} du\\ &=\frac{1}{2}\left( \ln(1+u) - \ln(1-u)\right)+C\\ &=\frac{1}{2}\ln(\frac{1+\sin(x)}{1-\sin(x)})+C\\ &=\frac{1}{2}\ln(\frac{(1+\sin(x))^2}{\cos^2(x)})+C\\ &=\ln\left| \frac{1+\sin(x)}{\cos(x)} \right|+C\\ &=\ln\left| \sec(x) +\tan(x)\right|+C \end{align*} $$

So now the integral of secant fits into a more general story: it is an example of the general strategy we have for integrating any function of the form $\sin^n(x)\cos^m(x)$

  • $\begingroup$ Thanks! Both of your suggestions are good ones, but I've accepted this one as it is a bit more concrete in answering my question. $\endgroup$ Feb 4, 2016 at 13:33

In my opinion, a good story to tell about techniques of integration in general goes something like this:

We know that some functions (like $x^2$) have elementary antiderivatives, and some (like $e^{x^2}$) do not. This raises an obvious question: which functions do have elementary antiderivatives? We cannot answer the question in general in this course, but something we can do is tackle some large families of functions. For instance, we can anti differentiate all polynomials. Can we find elementary antiderivatives for all rational functions?

First learn partial fraction decomposition to reduce rational functions into simpler terms. Those involving linear functions in the denominator can be solved instantly using variable substitution. Those involving quadratic functions in the denominator require a new technique, trig sub.

Conclude that all rational functions can be integrated via this algorithm.

This also means that, if we can make a clever change of variables to yield a rational function, we can be assured of our ability to integrate it.

For any rational function of sine and cosine, using the tangent half angle substitution yields a rational function, and so we can integrate all of these functions as well. One can ominously foreshadow connections with algebraic geometry here, to lend the subject a suitable air of excitement and mystery.

The integral of secant is a corollary of this kind of work, and an illustration of the power of the method. We would have never been able to guess such a bizarre form for the antiderivative, but this grand structure yields the intuition needed to derive the result.

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    $\begingroup$ I thought I remembered expressing this opinion before, but I thought it was in an offline conversation. Apparently I wrote an almost identical response to another question. matheducators.stackexchange.com/questions/9702/… $\endgroup$ Jan 18, 2016 at 0:56
  • $\begingroup$ It's still interesting here :) Incidentally, it finally hit me today teaching integration that not teaching Weierstrauss $t = \tan(x/2)$ actually makes it harder to know what to do. The fact that Weierstrauss subst. works for any rational function of sine and cosine is a nice fall-back if other methods fail. $\endgroup$ Feb 4, 2016 at 16:31

I. The history of the integral of the secant and the loxodrome is an excellent approach to this difficult, and historically difficult & important, integral. See Rickey and Tuchinsky, An Application of Geography to Mathematics: History of the Integral of the Secant, Mathematics Magazine Vol. 53, No. 3 (May, 1980), pp. 162-166, http://www.jstor.org/stable/2690106.

One lesson for students is that when a difficult problem is solved, the method of solution can be shared with others, making things easier for future generations, including them.

II. Once I accidentally stumbled upon the following. I was looking for an example of an integral to differentiate (with respect to the upper limit). I wanted something not too complicated but one that a Calculus I student couldn't antidifferentiate. And then I wanted to put a function in the upper limit and apply the chain rule. I thought it would be nice if it simplified, too. So I hit upon $$\int_0^x {1 \over 1-t^2} \; dt \quad \hbox{and} \quad \int_0^{\sin x} {1 \over 1 - t^2} \; dt \,.$$ And of course the last one simplifies to $\sec x$ (and is rather similar to Michael Joyce's comment and one of Steven Gubkin's answers). This occurs shortly before we get to substitution and therefore before the students know the antiderivative of $\sec x$. So I could point to the integral, and say, "Hey, this is an antiderivative of secant? Do we know a function that is antiderivative of secant? That's interesting. I wonder what that integral is. I guess we'll have to wait and see." It is motivating in that it piques their curiosity. Even if we don't get to partial fractions in Calc. I or prepare them for the Calc. I method, it gets the students to want to know the answer. It's more effective if most of your students haven't already had AP calculus.

III. Interestingly, according to Rickey and Tuchinsky, Isaac Barrow came up with essentially the same method as Joyce/Gubkin and was the first use of partial fractions in integration: $$\int {dx \over \cos x} = \int {\cos x \; dx \over \cos^2 x} = \cdots = {1\over 2} \int {\cos x \over 1 - \sin x} + {\cos x \over 1 + \sin x} \; dx = \text{etc.}$$

IV. A geometric approach. Let $s = \sec\theta$, $t = \tan\theta$:

Mathematica graphics

Then from the differential triangle and the similar $1$-$s$-$t$ triangle, we have $$ {s \; d\theta \over 1} = {dt \over s} = {ds \over t} \,. $$ Okay, maybe it is easier to deduce the proportion from the derivative rules, given the current conditions in math. ed. Nevertheless, the symmetry of the last two terms of the proportion suggests combining them. Hence we get $$ {s \; d\theta \over 1} = {dt + ds \over s + t} = {d(s+t) \over s+t} = d(\ln |u|)\,, $$ where $u = \sec\theta + \tan\theta$. Integrating gives the result.

  • $\begingroup$ Your argument in $IV$ is beautiful! $\endgroup$ Feb 4, 2016 at 22:05
  • $\begingroup$ @StevenGubkin Thanks! I don't recall having seen it before. I'm sure it's known. I'm just surprised it's not widely known. (I say this, so that someone will say, "Oh that's in calculus textbooks A, B & C.") $\endgroup$
    – user1815
    Feb 5, 2016 at 3:46
  • $\begingroup$ I have drawn lots of similar pictures to find the derivatives of trig and inverse trig functions, but I never came across this relationship before. $\endgroup$ Feb 5, 2016 at 7:34
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    $\begingroup$ Also, given current conditions in math. ed., you may have to spend some time justifying that if $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+c}{b+d}$ is also equal to this common value. $\endgroup$ Feb 5, 2016 at 19:29

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