Why do we state the antiderivative of $\sec x$ as $\ln |\sec x +\tan x|+C$?

Question

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?

Answer 1

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.

Answer 2

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 $\frac{1}{2}\ln\left|\frac{1+\sin x}{1-\sin x}\right|$.