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 ?

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.
