Find the integral $\int\frac{dx}{\sqrt{x^2-a^2}}$ 
Evaluate the integral of $\frac{1}{\sqrt{x^2-a^2}}$

Put $x=a\sec\theta\implies dx=a\sec\theta\tan\theta d\theta$
$$
\begin{align}
& \ \ \ \int \frac{dx}{\sqrt{x^2-a^2}} \\ 
&=\int \frac{a\sec\theta\tan\theta d\theta}{\sqrt{a^2\sec^2\theta-a^2}} \\ 
&=\int \frac{a\sec\theta\tan\theta d\theta}{a\sqrt{\tan^2\theta}}\\ 
&= \int\frac{a\sec\theta\tan\theta d\theta}{a\color{red}{\tan\theta}}\\ 
&=\int\sec\theta d\theta \\ 
&=\log \lvert \sec\theta+\tan\theta \rvert|+C_0 \\ 
& \mbox{where $C_0$ is an arbitrary constant of integration }  \\
&=\log \left\lvert \frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1} \right\rvert + C_0 \\
&= \log \left\lvert \frac{x+\sqrt{x^2-a^2}}{a} \right\rvert +C_0 \\ 
&= \log \left\lvert x+\sqrt{x^2-a^2} \right\rvert -\log|a|+C_0 \\ 
&= \log|x+\sqrt{x^2-a^2}|+ C, 
& \mbox{ where $C = \log \lvert a \rvert + C_0$}.
\end{align}
$$
This is how it is solve in my reference. But, $\sqrt{\tan^2\theta}=|\tan\theta|$ right ?
Then, does that imply
$$
\int\frac{dx}{\sqrt{x^2-a^2}}=\int\frac{a\sec\theta\tan\theta d\theta}{a\color{red}{|\tan\theta|}}=\color{red}{\pm}\int\sec\theta d\theta
$$
Why am I getting this confusion and is the first solution complete ?
 A: Suppose that $a>0$.
The work is just for the case when $x>a$. The case for $x<-a$ is different, but the finals result is the same.
Let $x=a\sec\theta$, where $\theta\in[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$. This is the domain of $\textrm{arcsec}$.
For $x< -a$, $\theta\in(\frac{\pi}{2},\pi]$ and so $\tan\theta\le0$.
$$\sqrt{x^2-a^2}=\sqrt{a^2\tan^2\theta}=-a\tan\theta$$
\begin{align*}
\int\frac{dx}{\sqrt{x^2-a^2}}&=\int\frac{a\sec\theta\tan\theta}{-a\tan\theta}d \theta\\
&=-\int\sec\theta d\theta\\
&=-\ln|\sec\theta+\tan\theta|+C\\
&=\ln|\sec\theta-\tan\theta|+C\\
&=\ln\left|\frac{x}{a}-\frac{-\sqrt{x^2-a^2}}{a}\right|+C\\
&=\ln\left|x+\sqrt{x^2-a^2}\right|-\ln|a|+C
\end{align*}
There are two minus signs and they cancel each other to reach the final result.
A: This is an elementary integral, using an inverse hyperbolic function.
By the formula for inverse functions, $$\cosh'(x)=\sinh(x)=\sqrt{\cosh^2(x)-1}$$ lets you establish
$$\text{arcosh}'(x)=\frac1{\sqrt{x^2-1}}.$$
It is known that
$$\text{arcosh}(x)=\log(x+\sqrt{x^2-1})$$ though the form $\text{arcosh}$ is fine.
