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The standard approach for showing $\int \sec \theta \, \mathrm d \theta = \ln |\sec \theta + \tan \theta| + C$ is to multiply by $\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}$ and then do a substitution with $u = \sec \theta + \tan \theta$. I like the fact that this trick leads to a fast and clean derivation, but I also find it unsatisfying: It's not very intuitive, nor does it seem to have applicability to any integration problem other than $\int \csc \theta \mathrm \,d \theta$. Does anyone know of another way to evaluate $\int \sec \theta \, \mathrm d \theta$?

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One can always use the half-angle substitution: $t=\tan(\theta/2)$. – Robin Chapman Oct 13 '10 at 14:13
If you make the substitution $u=\sec(\theta)$, the integrand becomes the derivative of the inverse hyperbolic cosine. – Jonas Meyer Oct 13 '10 at 21:54
While I like (and upvoted) the general approaches, I've decided to accept Derek Jennings' answer because it is the most useful to me: I plan to use it in class this coming week when we discuss integration by partial fractions decomposition! Thanks to everyone for their answers and comments; they greatly exceeded what I was expecting. Go Math Stack Exchange. – Mike Spivey Oct 14 '10 at 4:25
This came up a couple of years ago in an ap-calculus listserv, and I wound up writing an essay about the Gudermannian function that you might find of interest. I reposted the essay in sci.math at <>;. – Dave L. Renfro Jul 15 '11 at 21:18
No matter what anybody says, Wikipedia falls short of being infallible. But there is this: – Michael Hardy Jul 17 '11 at 1:37
up vote 69 down vote accepted

Another way is:

$$\int \sec x dx = \int \frac{\cos x}{1-\sin^2 x} dx = \frac{1}{2} \int \left( \frac{1}{1-\sin x} + \frac{1}{1+\sin x} \right) \cos x dx $$ $$= \frac{1}{2} \log \left| \frac{1+\sin x}{1-\sin x} \right| + C.$$

It's worth noting that the answer can appear in many disguises. Another is $$\log \left| \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right| $$

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Also nice. Plus it has the advantage of using only partial fractions and trig identities my students already know. (Unfortunately, I doubt any of them will remember the tangent half-angle identities - assuming that they have even seen them before.) – Mike Spivey Oct 13 '10 at 15:05
This is the method I use whenever I teach this material. Unfortunately, usually we do not cover the Weierstrass substitution. – Andrés E. Caicedo Oct 14 '10 at 2:05
I always like to present this solution to my students. Also I point to them than $\int \sec(\theta) d\theta$ falls into the cathegory $\cos(\theta)$ to an odd power, so the theory sais that we need to make the substitution $u = \sin(\theta)$.. Once I point that $d u$ needs to be at the top, the solution becomes obvious.... – N. S. Jul 15 '11 at 17:56
This'd be my approach. – ncmathsadist Jul 17 '11 at 1:27
I think this method my be the historically oldest one, due to Isaac Barrow in the 1600s, and the first time partial fractions were used in antidifferentiation. – Michael Hardy Jul 11 '12 at 19:54

A useful technique is to use the half angle formulas in terms of $\tan (\theta/2)$ in order to convert trigonometric (rational) functions into rational functions.

For example if $t = \tan(\theta/2)$ we have that $\sec \theta = \frac{1+t^2}{1-t^2}$

We have $2\,\mathrm dt = (1 + \tan^2(\theta/2))\,\mathrm d\theta$

And so

$$\int \sec \theta \,\mathrm d\theta = \int \frac{2\;\mathrm dt}{1-t^2}$$

Which can easily be evaluated.

Similarly we get

$$\int \csc \theta \,\mathrm d\theta = \int \frac{\mathrm dt}{t}$$

using $\csc \theta = \frac{1+t^2}{2t}$

Check this page out.

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The half angle formulas give an algorithm that can be used, in principle, to integrate any rational trigonometric function: make the substitution, then use partial fractions. (I say "in principle" because in practice one needs to numerically factor polynomials to do this.) – Qiaochu Yuan Oct 13 '10 at 14:26
I usually call this the "Weierstrass Magic $t$-Substitution" Method because (i) it originated with Weierstrass; and (ii) it works almost like magic. Same substitution will always do the trick. – Arturo Magidin Oct 13 '10 at 16:14
@Arturo, doesn't this, in essence, come down to the down-to-earth fact that one has a rational parametrization of the unit circle? – Mariano Suárez-Alvarez Oct 16 '10 at 0:16
@Mariano: Essentially, yes (that was Weierstrass observation, if I remember the history correctly). It still seems like magic... or at least sleight-of-hand... – Arturo Magidin Oct 16 '10 at 2:56
@Mariano: I'm quite late to the party here, but would you care to elaborate? I'm not quite sure I understand what you mean by "rational parametrization of the unit circle" and how it's related to Weierstrass substitution. – Bruno Stonek Jun 27 '11 at 1:57

Using the definitions $$\sec \theta = 1/\cos \theta \quad \text{and} \quad \cos \theta = (\exp(i \theta) + \exp(-i \theta))/2$$ gives $$\int \sec \theta \, d \theta = \int \frac{2 \, d \theta}{\exp(i \theta) + \exp(-i \theta)}.$$ The only insight needed is to find the substitution $u = \exp( i \theta )$ (what else is there to try?), leading to a multiple of $\int \frac{du}{1+u^2}$, the inverse tangent. Thus, in an essentially mechanical fashion you obtain the generic solution $$-2 i \arctan(\exp(i \theta)).$$ Unwinding this via the usual algebraic identities between exponential and trig functions not only shows it is equal to the usual solutions, but also reveals why half angles might be involved and where an offset of $\pi /4$ might come from (as in @Derek Jennings' answer): it's a constant of integration, of course.

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+1: For another general technique. – Aryabhata Oct 13 '10 at 16:13
Not only that, but the same technique (essentially); u and t = tan theta/2 are related by invertible fractional linear transformations. – Qiaochu Yuan Oct 13 '10 at 16:28
@Qiaochu: Agreed, they are related; that's a good observation. But as a matter of technique there is a substantial difference. The other responses require tricks or auxiliary knowledge, such as the general effectiveness of using tan(t/2) or choosing the right trig formula. From another comment I now realize that using complex variables will not be appropriate for Mike Spivey's audience, but the original purpose of pointing out this method was to address the intent of his message: how can we do this integral without invoking some mysterious insight or relationship? – whuber Oct 13 '10 at 17:53
I really had two reasons for wanting an answer to my question: 1) something I could use in my integral calculus class, and 2) for my own interest. Thanks for a good response with respect to reason 2. – Mike Spivey Oct 13 '10 at 19:38
+1 for whuber's wonderful answer and comment! I've never understood this tan(x/2) stuff: you have to do a complicated computation to find that z is a primitive of dz. – Pierre-Yves Gaillard Oct 14 '10 at 5:49

Instead of presenting another way of evaluating this integral, I justify a more general case in an approach which uses partial fractions and trigonometric identities, at the level of a Calculus class, I think:

$$\int \dfrac{1}{a+b\cos x}dx=\dfrac{1}{\sqrt{b^{2}-a^{2}}}\ln \left\vert \dfrac{\sqrt{a+b}+\sqrt{b-a}\tan x/2}{\sqrt{a+b}-\sqrt{b-a}\tan x/2}\right\vert \quad a\lt b.\quad (\ast)$$


$$a+b\cos x=(a-b)+2b\cos ^{2}x/2,$$

we have

$$\dfrac{1}{a+b\cos x}=\dfrac{\sec ^{2}x/2}{(a-b)\sec ^{2}x/2+2b}=\dfrac{\sec ^{2}x/2}{(a-b)\sec ^{2}x/2+2b}=\dfrac{\sec ^{2}x/2}{a+b-(b-a)\tan ^{2}x/2}.$$


$$\dfrac{1}{a+b-(b-a)\tan ^{2}x/2}=$$

$$=\dfrac{1}{2\sqrt{a+b}}\left( \dfrac{1}{% \sqrt{a+b}-\sqrt{b-a}\tan x/2}+\dfrac{1}{\sqrt{a+b}+\sqrt{b-a}\tan x/2}% \right) .$$


$$\int \dfrac{1}{a+b\cos x}dx=$$

$$=\dfrac{1}{2\sqrt{a+b}}\int \left( \dfrac{\sec ^{2}x/2}{\sqrt{a+b}-\sqrt{b-a}\tan x/2}+\dfrac{\sec ^{2}x/2}{\sqrt{a+b}+% \sqrt{b-a}\tan x/2}\right) dx$$

$$=\dfrac{1}{\sqrt{b^{2}-a^{2}}}\ln \left\vert \dfrac{\sqrt{a+b}+\sqrt{b-a}% \tan x/2}{\sqrt{a+b}-\sqrt{b-a}\tan x/2}\right\vert .$$

Thus, we have your particular case

$$\int \dfrac{1}{\cos x}dx=\int \dfrac{1}{0+1\cos x}dx=\ln \left\vert \dfrac{% 1+\tan x/2}{1-\tan x/2}\right\vert . \qquad (\ast\ast)$$

From $\tan \dfrac{x}{2}=\dfrac{\sin x}{1+\cos x}$ and $\sec x+\tan x=\dfrac{1+\sec x+\tan x}{1+\sec x-\tan x}$ it follows that

$$\dfrac{1+\tan x/2}{1-\tan x/2}=\dfrac{1+\dfrac{\sin x}{1+\cos x}}{1-\dfrac{% \sin x}{1+\cos x}}=\dfrac{1+\cos x+\sin x}{1+\cos x-\sin x}=\sec x+\tan x$$

and, finally

$$\int \sec x\; dx=\ln \left\vert \sec x+\tan x\right\vert .$$

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+1 for Americo. – night owl Apr 22 '11 at 5:18
night owl, Thanks. – Américo Tavares Apr 22 '11 at 11:28

Here is a way an electrician solves the problem. Since $\cos(x)=\sin(\frac{\pi}{2}+x)$ it is easier consider the integral $$ I=\int \csc xdx = \int \dfrac1{\sin x}\mathrm dx$$

Now: $$ \dfrac1{\sin x}\mathrm dx= \dfrac1{2\sin\frac{x}{2}\cos\frac{x}{2}}\mathrm dx=\dfrac1{2\tan\frac{x}{2}\cos^2\frac{x}{2}}\mathrm dx =\frac{\mathrm d\tan\frac{x}{2}}{\tan\frac{x}{2}}=\mathrm d \ln \left | \tan\frac{x}{2} \right | $$

Thus $$I=\ln \left | \tan\frac{x}{2}\right | +C$$

Substituting $x$ with $\frac{\pi}{2}+x$ gives for the original integral:

$$\ln \left| \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right|+C $$

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These articles exist:

V. Frederick Rickey and Philip M. Tuchinsky, An Application of Geography to Mathematics: History of the Integral of the Secant, Mathematics Magazine, volume 53, number 3, May 1980, pages 162–166.

Rickey & Tuchinsky's article tells us that the integral of the secant function was a well known conjecture in the 17th century, that Isaac Barrow solved the problem, and that the original reason for raising the question came from cartography.

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Here is another way to compute $$\int \sec xdx $$

First, we need a trig identity \begin{eqnarray*} \cos ^{2}x &=&(1-\sin x)(1+\sin x) \\ \frac{1-\sin x}{\cos x} &=&\frac{\cos x}{1+\sin x} \\ \sec x &=&\tan x+\frac{\cos x}{1+\sin x} \end{eqnarray*} Next, it suffices to integrate each side \begin{eqnarray*} \int \sec xdx &=&\int \tan xdx+\int \frac{\cos x}{1+\sin x}dx \\ &=&-\ln \left\vert \cos x\right\vert +\ln \left\vert 1+\sin x\right\vert +C \\ &=&\ln \left\vert \frac{1+\sin x}{\cos x}\right\vert +C \\ &=&\ln \left\vert \sec x+\tan x\right\vert +C \end{eqnarray*}

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Dual way to compute $$\int \csc xdx.$$

First, we need a trigonometry identity \begin{eqnarray*} \sin ^{2}x &=&(1-\cos x)(1+\cos x) \\ \frac{1-\cos x}{\sin x} &=&\frac{\sin x}{1+\cos x} \\ \csc x &=&\cot x+\frac{\sin x}{1+\cos x} \end{eqnarray*} Next, it suffices to integrate each side \begin{eqnarray*} \int \csc xdx &=&\int \cot xdx+\int \frac{\sin x}{1+\cos x}dx \\ &=&\ln \left\vert \sin x\right\vert -\ln \left\vert 1+\cos x\right\vert +C \\ &=&-\ln \left\vert \frac{1+\cos x}{\sin x}\right\vert +C \\ &=&-\ln \left\vert \csc x+\cot x\right\vert +C \end{eqnarray*}

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