# integrating the secant function, who figured this out?

I was looking at how the secant function is integrated. The process is not obvious, and I don't expect it to be but I wanted to know if anyone knows who figured this out. Here's what I'm talking about:

\begin{align*}\int \sec(x)dx &=\int \sec(x)\cdot \frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}dx\\ &=\int \frac{\sec^2(x)+\tan(x)\sec(x)}{\sec(x)+\tan(x)}dx. \end{align*}

If $f(x) = \frac{1}{x}$, $g(x)=\sec(x)+\tan(x)$, $g'(x)=\sec^2(x)+\tan(x)\sec(x)$

Then $\int \sec(x)dx = \int f(g(x))\cdot g'(x)dx=\int \frac{1}{u}du$, where $u=g(x)$

$=\ln|\sec(x)+\tan(x)|+c$

So my question is, who first realised how to do this? Who figured out step 2? It's clever and not that obvious.

(and my sub-question is: why does Arturo insist on re-formatting my questions so that my first statements are centre aligned?! :)

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You don't have to do it this way. The substitution $t = \tan \frac{x}{2}$ works wonders: en.wikipedia.org/wiki/Weierstrass_substitution – Qiaochu Yuan Jan 17 '12 at 21:39
@Qiaochu: Yes, "Weierstrass's magical $t$ substitution"... provided you know how to solve integrals of rational functions, at any rate. – Arturo Magidin Jan 17 '12 at 21:44

I don't know who might have come up with the method of integration about which you ask first; however, I have always regarded the method about which you are asking as a bit of reverse engineering. I think the following method utilizing a simple trig substitution and partial fractions is much more natural: \begin{align} \int\sec(x)\;\mathrm{d}x &=\int\sec^2(x)\;\mathrm{d}\sin(x)\\ &=\int\frac{1}{1-\sin^2(x)}\;\mathrm{d}\sin(x)\\ &=\int\frac12\left(\frac{1}{1-\sin(x)}+\frac{1}{1+\sin(x)}\right)\;\mathrm{d}\sin(x)\\ &=\frac12(\log(1+\sin(x))-\log(1-\sin(x)))+C\\ &=\frac12\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)+C\\ &=\frac12\log\left(\frac{(1+\sin(x))^2}{\cos^2(x)}\right)+C\\ &=\log\left(\frac{1+\sin(x)}{\cos(x)}\right)+C\\ &=\log(\sec(x)+\tan(x))+C\tag{1} \end{align} Upon being presented with the result in $(1)$ without the derivation, one might differentiate to get \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\log(\sec(x)+\tan(x)) &=\frac{\sec(x)\tan(x)+\sec^2(x)}{\sec(x)+\tan(x)}\\ &=\sec(x)\tag{2} \end{align} and $(2)$ leads logically to the method about which you ask.