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p. 464 Top of Calculus: Early Transcendentals (6 edn 2007) doesn't explain the discovery of this trick (in this question's title), that thus feels clairvoyant.

I understand, but ask not about, other less tricky methods to calculate $\int \sec x \,dx $.

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    You can divine the trick by calculating $\int \sec x , dx$ in one of those less tricky ways, and noticing that you get a logarithm. So the original integral must have secretly been of the form $\int \frac{du}{u}$. – Micah May 17 '18 at 04:22
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    @Micah: why write an answer in the "comment" box? – Martin Argerami May 17 '18 at 04:48
  • I wonder if knowing the identity $$ \tan\left( \frac x 2 + \frac \pi 4\right) = \tan x + \sec x, $$ combined with other things, might help here. – Michael Hardy May 19 '18 at 19:00
  • @MichaelHardy You ought link to your full answer here that cites your paper. But how can you expect students to remember, or conjure up, $\tan\left( \frac x 2 + \frac \pi 4\right) = \tan x + \sec x,$? Even teachers wouldn't forebode to appeal to this little-known identity. –  Oct 07 '23 at 23:41

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As Micah suggested, use another method first, e.g. the frequently useful $t=\tan\frac{x}{2}$. Then $\int\sec x dx=\int\frac{2dt}{1-t^2}=\ln\frac{1+t}{1-t}+C$. But $$\frac{1+t}{1-t}=\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}=\frac{1+\sin x}{\cos x}.$$

J.G.
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