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One easy integration of $\sec x$ substitutes $u=\sin x$, viz.$$\int\frac{\cos x}{1-\sin^2 x}\,\mathrm{d}x=\frac{1}{2}\ln\left|\frac{1+\sin x}{1-\sin x}\right|+C.$$Multiplying top and bottom by $1+\sin x$ and writing $1-\sin^2 x=\cos^2 x$ gives the more familiar $$\ln\left|\sec x+\tan x\right|+C.$$A proof written so as to get to that result first requires more elaborate tricks, such as comparing the derivatives of $\sec x$ and $\tan x$ to spot a solution to $y'=y\sec x$, or to use the fact that $f:=\sec x,\,g:=\tan x$ satisfy$$f^2=g^2+1\implies\frac{f'}{g}=\frac{g'}{f}\implies\frac{d}{dx}\ln\left|f+g\right|=\frac{f'}{g}=\sec x.$$So why is the sec/tan formula for an antiderivative universally taught, instead of the sine-based one?

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  • $\begingroup$ With suitable limits the integral computes the distance from $(0,0)$ to $(\sec x, 0)$ in the constant curvature $-4$ metric on the unit disk. From this point of view the form in terms of $\sin x$ is more natural than the form in terms of $\sec x$ and $\tan x$. $\endgroup$
    – Dan Fox
    Commented Oct 4, 2018 at 13:56
  • $\begingroup$ @DanFox That's interesting. What is the constant curvature $-4$ metric? $\endgroup$
    – J.G.
    Commented Oct 4, 2018 at 13:59
  • $\begingroup$ @J.G.: I just mean a rescaling of the hyperbolic metric. Hopefully I got the constant right. $\endgroup$
    – Dan Fox
    Commented Oct 4, 2018 at 14:52
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    $\begingroup$ My Bronstein gives $\ln (\tan(x/2 +\pi/4))$ as antiderivative. This comes from the nice antiderivative $\ln (\tan(x/2))$ of $\csc$ by using that $\cos$ is just $\sin$ shifted by $\pi/2$. (It also only uses the function mentioned by @BPP). $\endgroup$ Commented Oct 5, 2018 at 18:47
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    $\begingroup$ I guess it was used because it was easier to look up logarithms of tans in tables than to look up sec and tan, add them and then get out a logarithm table. $\endgroup$ Commented Oct 6, 2018 at 6:30

2 Answers 2

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First idea: Maybe just because it's easier to memorize the derivatives and antiderivatives of the trig functions this way: $\sin$ always pairs with $\cos$, $\tan$ always pairs with $\sec$, and $\cot$ always pairs with $\csc$. These are the same pairs you already know from the sum/difference of squares formulas too, if your precalculus trig studies included all* 6 functions.

You may put the dividing line between "stuff to memorize" and "stuff to re-derive when needed" in a different place, but at least some students have been expected to memorize those 6 derivatives and 6 antiderivatives.

Second idea: if you arrived at $\int \sec x\ dx$ through a trig substitution, then an answer involving $\sec x$ and $\tan x$ will be easier to translate back to the original variable than one with a pair of $\sin x$, since you already figured out what $\sec x$ is, and again, the relationship between $\sec x$ and $\tan x$ is one of those easy difference-of-squares-equals-1 deals.

*Sorry, exsecant and haversine.

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    $\begingroup$ +1 for pointing out that the sec+tan form is more helpful at the end of a trig sub problem. $\endgroup$
    – Aeryk
    Commented Oct 3, 2018 at 18:31
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    $\begingroup$ Moral +1 from me. I think having a simpler formula to remember is in fact the reason for why it is written this way. Otherwise leaving it in sin form is more intuitive. IMNSHO. $\endgroup$
    – guest
    Commented Oct 3, 2018 at 19:55
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    $\begingroup$ Not an answer, but I think it is my duty to post a link to this paper whenever this integral is mentioned: math.uconn.edu/~kconrad/math1132s10/secantintegral.pdf $\endgroup$ Commented Oct 4, 2018 at 15:56
  • $\begingroup$ Great article. I have taken navigation class and yes Mercator projection is great because of rhumb lines. (However, you usually still plot a series of segments. Because Great Circle route is not constant bearing. But approximating the trip with set of segments is fine. Hard to really steer a Great Circle.) $\endgroup$
    – guest
    Commented Oct 4, 2018 at 20:59
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    $\begingroup$ maa.org/sites/default/files/pdf/cms_upload/… $\endgroup$ Commented Feb 6, 2021 at 14:14
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Stewart's text (link) derives the result via: $$\int \sec x\ dx = \int \sec x \frac{\sec x +\tan x}{\sec x +\tan x}\ dx = \int \frac{\sec^2 x +\sec x \tan x}{\sec x +\tan x}\ dx$$ and then $u=\sec x +\tan x$. I don't think this is any worse than rewriting $\frac{1}{\cos x} = \frac{\cos x}{1-\sin^2x}$, using a $u$-sub, and then partial fractions.

Also, if you take $\sec$ and $\tan$ to be functions in their own right, then $\ln|\sec x+\tan x|$ requires fewer computational steps than $\frac{1}{2}\ln\left|\frac{1+\sin x}{1-\sin x}\right|$.

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  • $\begingroup$ Stewart's text? $\endgroup$
    – J.G.
    Commented Oct 3, 2018 at 16:19
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    $\begingroup$ Most Calculus II textbooks I've seen (in Canada) also use this method. $\endgroup$
    – orion2112
    Commented Oct 3, 2018 at 16:22
  • $\begingroup$ Link now provided to Stewart's text. This is a very standard Calculus text in the U.S. $\endgroup$
    – Aeryk
    Commented Oct 3, 2018 at 16:57
  • $\begingroup$ @Aeryl Ah yes, between Examples 6 & 7 in Sec. 7.2. $\endgroup$
    – J.G.
    Commented Oct 4, 2018 at 5:35

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