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

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?

• 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 $\tan x$. Oct 4 '18 at 13:56
• @DanFox That's interesting. What is the constant curvature $-4$ metric?
– J.G.
Oct 4 '18 at 13:59
• @J.G.: I just mean a rescaling of the hyperbolic metric. Hopefully I got the constant right. Oct 4 '18 at 14:52
• 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). Oct 5 '18 at 18:47
• 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. Oct 6 '18 at 6:30

2 Answers

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.

• +1 for pointing out that the sec+tan form is more helpful at the end of a trig sub problem. Oct 3 '18 at 18:31
• 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. Oct 3 '18 at 19:55
• 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 Oct 4 '18 at 15:56
• 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.) Oct 4 '18 at 20:59
• maa.org/sites/default/files/pdf/cms_upload/… Feb 6 at 14:14

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

• Stewart's text?
– J.G.
Oct 3 '18 at 16:19
• Most Calculus II textbooks I've seen (in Canada) also use this method. Oct 3 '18 at 16:22
• Link now provided to Stewart's text. This is a very standard Calculus text in the U.S. Oct 3 '18 at 16:57
• @Aeryl Ah yes, between Examples 6 & 7 in Sec. 7.2.
– J.G.
Oct 4 '18 at 5:35