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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 $u=g(x)$


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: – 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
up vote 8 down vote accepted

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.

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According to Eli Maor's Trigonometric Delights, this integral was first evaluated by Isaac Barrow, using the partial fractions method you give here; if memory serves, Maor also writes that this is the first known use of the partial fractions method. – user21467 May 14 '15 at 0:00
Thanks for the reference. Wikipedia's article about Isaac Barrow's career mentions the article on the Integral of the Secant Function, which concurs that Barrow was the first to evaluate the integral, and this was the first known use of partial fractions for integration. – robjohn May 14 '15 at 5:29

The wikipedia page: has quite a bit of historical information.

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The Wiki article does not directly address the OP's question. The link contained therein may, but I can only view the first page of it. – David Mitra Jan 17 '12 at 21:53
The paper is available here: – Mikko Korhonen Jan 17 '12 at 22:06

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