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I was trying to integrate $\sec^3x$ and discovered that I would have to integrate $\sec x$ in the process. I had not seen the "standard" approach and came up with my own solution, which is apparently quite different: $$\int \sec xdx = \int \frac{dx}{\cos x}$$ I substituted $u = \sin x$ so that $dx = \frac{du}{\cos x}$. Then $$\int \frac{dx}{\cos x} = \int \frac{du}{\cos^2x} = \int \frac{du}{1 - \sin^2x} = \int \frac{du}{1 - u^2}$$ The solution to this is $\tanh^{-1}u + C$. Since $u = \sin x$, this means that $$\int \sec xdx = \tanh^{-1}(\sin x) + C \tag{1}$$

After looking it up, I found out that the standard form of the integral is $$\int\sec x dx = \ln|\text{sec}x + \text{tan}x| + C \tag{2}$$

I couldn't find anything about the alternate form $(1)$ which is, as far as I can tell, equivalent to $(2)$. So, did I make a mistake here? If not, is there a reason to prefer the usual form $(2)$?

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  • $\begingroup$ You might be interested in reading about the Gudermannian function (see here also). See this 8 October 2009 sci.math post and these notes for connections with the integral of secant. $\endgroup$ – Dave L. Renfro Dec 9 '18 at 20:38
  • $\begingroup$ See also matheducators.stackexchange.com/questions/14631/… $\endgroup$ – J.G. Dec 9 '18 at 22:47
  • $\begingroup$ No mistake, you've just come across on of the many amazing relationships that exist between our elementary (and non-elementary) functions. No one form is better or worse than the other. Depending upon the situation they each have their advantages. $\endgroup$ – user150203 Dec 10 '18 at 1:20
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$$\begin{align} \tanh^{-1}(\sin x) &=\frac12\ln\left|\frac{1+\sin x}{1-\sin x}\cdot \frac{1/\cos x}{1/\cos x}\right|\\[4pt] &=\frac12\ln\left|\frac{\sec x+\tan x}{\sec x-\tan x}\right| \\[4pt] &=\frac12\ln\left|\frac{\sec x+\tan x}{\sec x-\tan x}\cdot\frac{\sec x+\tan x}{\sec x+\tan x}\right|\\[4pt] &=\phantom{\frac12}\ln\left|\sec x+\tan x\right| \end{align}$$

I believe most introductory calculus books use the equivalent form because their readers are not aware of hyperbolic trigonometric functions or their inverses.

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I was taught to calculate $\;\int (\sec x) dx\;$ via the substitution $\;u = \tan(x/2).\;$ As the other answer suggested, in math, you don't want to use a steamroller when a flyswatter will do. Problems involving $\;\int (\sec x) dx\;$ (for example $\;\int (\sec x)^3 dx)\;$ can be solved in a straightforward (if somewhat arduous) manner without using hyperbolic functions.

The convention is to avoid advanced topics (e.g. hyperbolic functions), unless the problem requires it.

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