# 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, 2018 at 13:56
• @DanFox That's interesting. What is the constant curvature $-4$ metric?
– J.G.
Oct 4, 2018 at 13:59
• @J.G.: I just mean a rescaling of the hyperbolic metric. Hopefully I got the constant right. Oct 4, 2018 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, 2018 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, 2018 at 6:30

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, 2018 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, 2018 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, 2018 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, 2018 at 20:59
• maa.org/sites/default/files/pdf/cms_upload/… Feb 6, 2021 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, 2018 at 16:19
• Most Calculus II textbooks I've seen (in Canada) also use this method. Oct 3, 2018 at 16:22
• Link now provided to Stewart's text. This is a very standard Calculus text in the U.S. Oct 3, 2018 at 16:57
• @Aeryl Ah yes, between Examples 6 & 7 in Sec. 7.2.
– J.G.
Oct 4, 2018 at 5:35