Integrating $\int \sec xdx$: Why is $\ln|\text{sec}x + \text{tan}x| + C$ preferred over $\tanh^{-1}(\sin x) + C$?

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 $$(2)$$?

• 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. – Dave L. Renfro Dec 9 '18 at 20:38
• – J.G. Dec 9 '18 at 22:47
• 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. – user150203 Dec 10 '18 at 1:20

\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 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.