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 $(2)$?
 A: $$\begin{align}
\tanh^{-1}(\sin x) &=\frac12\ln\left|\frac{1+\sin x}{1-\sin x}\cdot \frac{1/\cos x}{1/\cos x}\right|\\[4pt]
&=\frac12\ln\left|\frac{\sec x+\tan x}{\sec x-\tan x}\right| \\[4pt]
&=\frac12\ln\left|\frac{\sec x+\tan x}{\sec x-\tan x}\cdot\frac{\sec x+\tan x}{\sec x+\tan x}\right|\\[4pt]
&=\phantom{\frac12}\ln\left|\sec x+\tan x\right|
\end{align}$$
I believe most introductory calculus books use the equivalent form because their readers are not aware of hyperbolic trigonometric functions or their inverses. 
A: 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.
