# Ways to evaluate $\int \sec \theta \, \mathrm d \theta$

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 $\int \sec \theta \, \mathrm d \theta$?

#### Solutions Collecting From Web of "Ways to evaluate $\int \sec \theta \, \mathrm d \theta$"

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

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

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.

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:

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

Since

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

we have

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

But

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

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

Hence

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

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

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

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

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

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

Now: $$\frac1{\sin x} \, \mathrm dx= \frac1{2\sin \frac{x}{2}\cos\frac{x}{2}} \, \mathrm dx=\frac1{2\tan\frac{x}{2}\cos^2\frac{x}{2}} \, \mathrm dx =\frac{\mathrm d\tan\frac{x}{2}}{\tan\frac{x}{2}}=\mathrm d \ln \left | \tan\frac{x}{2} \right |$$

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

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

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

These articles exist:

http://en.wikipedia.org/wiki/Integral_of_the_secant_function

http://en.wikipedia.org/wiki/Weierstrass_substitution

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.

Here is another way to compute $$\int \sec x\,dx$$

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 x\,dx &=&\int \tan x\,dx+\int \frac{\cos x}{1+\sin x}\,dx \\[6pt]
&=&-\ln \left\vert \cos x\right\vert +\ln \left\vert 1+\sin x\right\vert +C
\\[6pt]
&=&\ln \left\vert \frac{1+\sin x}{\cos x}\right\vert +C \\[6pt]
&=&\ln \left\vert \sec x+\tan x\right\vert +C
\end{eqnarray*}

Here’s the argument in my less-than-one-page paper in the Monthly in June 2013:$^\dagger$
\begin{align}
x & = \tan \left( \frac \pi 4 + \frac \theta 2 \right) \\[10pt]
\frac{x^2-1}{x^2+1} & = \sin\theta \quad (\text{But we won’t use this line, so move on to the next.}) \\[10pt]
\frac{2x}{x^2+1} & = \cos\theta \\[10pt]
\frac{2\,dx}{x^2+1} & = d\theta \\[10pt]
\int \sec\theta \, d\theta & = \int \frac{dx} x = \log|x|+\text{constant} = \log\left| \tan\left( \frac \pi 4 + \frac \theta 2 \right) \right| + \text{constant}.
\end{align}

I regret that in that paper I used the term Weierstrass substitution, following Stewart’s calculus text, because, as I later learned, Stewart’s attribution to Karl Weierstrass is almost certainly erroneous. I wrote to Stewart asking about the evidence for the claim. He didn’t have any, but said the term was in widespread use before his book appeared.

(Privately I think of the trigonometric identity $\displaystyle\tan\left(\frac\pi4 \pm \frac\theta2\right) = \sec\theta\pm\tan\theta$ as the
$\text{“}$cartographer’s tangent half-angle formula,$\text{”}$ but I’m not sure how much sense that makes.)

$^\dagger$Michael Hardy, “Efficiency in Antidifferentiation of the Secant Function”, American Mathematical Monthly, June–July 2013, page 580.

Dual way to compute $$\int \csc x\,dx.$$

First, we need a trigonometry identity
\begin{eqnarray*}
\sin^2x &=&(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 x\,dx &=&\int \cot x\,dx+\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*}

Here is a slightly different approach to calculate

$$\int \frac{1}{\cos(x)}\,dx$$

Define $u := \tan(\frac{x}{2})$ so it follows $dx = \frac{2}{1+u^2}\,du$. It follows that $\cos(x) = \frac{1-u^2}{1+u^2}$ under this substitution. Now we can write the integral as:

$$\int \frac{1}{\cos(x)}\,dx = \int \frac{1}{\frac{1-u^2}{1+u^2}} \frac{2}{1+u^2}\,du = 2 \int \frac{1}{1-u^2}\,du$$

We know that $(\tanh^{-1}(x))’ = \frac{1}{1-x^2}$, so the integral becomes

$$\int \frac{1}{\cos(x)}\,dx = 2 \int \frac{1}{1-u^2}\,du = 2 \tanh^{-1}(u) + C = 2 \tanh^{-1}(\tan(\frac{x}{2})) + C$$

The solution looks a bit different than the others posted here but it’s the same. The trick here is to know the substitution and also how to express $\cos(x)$ in terms of $u$ but after that it’s just the basic substitution rule.