Hyperbolic substitution in $\int\frac{dx}{\sqrt{x^2-a^2}}$ for $x\le -a$ When trying to solve $\int\frac{dx}{\sqrt{x^2-a^2}}$ we can substitue $x$ in two ways:
METHOD 1: $x=a\sec(\theta)$ and $dx=a\sec(\theta)\tan(\theta) d\theta$ and $\theta=\sec^{-1}(\frac{x}{a})$. 
METHOD 2: $x=a\cosh(u)$ and $dx=a\sinh(u)du$ and $u=\cosh^{-1}(\frac{x}{a})$
My calculus-book states that we can use a similar method when we assume $x\le -a$. 
I can see how that would work with method 1 because $\sec^{-1}(\frac{x}{a})$ is indeed defined for $x\le -a$. 
QUESTION: But how would that work in method 2? $\cosh^{-1}(\frac{x}{a})$ is only defined for $\frac{x}{a}\ge1$ and if $x\le-a$ then $\frac{x}{a}$ is negative. So in this case $\cosh^{-1}(\frac{x}{a})$ is not defined!!! 
What is happening here? Thanks!
 A: The method $2$ only works for $x > a$. That's why your book suggests there is a similar method for $x < -a$. As we shall see below, since the integrand's maximal real domain consists of the union of two disjoint intervals, it is necessary to separately consider each interval together with two different(but indeed similar) variable substitutions, one for each.
What does actually mean to "solve" $\int f(x) \,dx$? It means finding an "integral-free expression" for the function
$$ F(y) = \int_r^y f(x) \,dx $$
where $r$ and $y$ are elements of $f$'s domain and $r$ is arbitrarily fixed.
The change of variables theorem says that if $\varphi : [p,q] \to [r,y]$ is a continuously differentiable bijective function and $f : [r,y] \to \mathbb{R}$ is a continuous function then
$$ \int_{r}^{y} f(x) \,dx = \int_{\varphi{-1}(r)}^{\varphi^{-1}(y)} f(\varphi(u)) \varphi'(u) \,du \tag{1} $$
In this case we have
$$ f : \left]-\infty, -a \right[ \cup \left]a, \infty \right[ \to \mathbb{R} : x \mapsto \frac{1}{\sqrt{x^2-a^2}} $$
where $a > 0$ and $\sqrt{\cdot}$ is the nonnegative real square root.
Since $f$'s domain is the union of two open intervals, at some moment we must choose whether $r, y \in \left]-\infty, -a \right[$ or $r, y \in \left]a, \infty \right[$ for we can't integrate over the gap between these two intervals because $f$ can't be defined or assumes complex values there.
By wanting to subtitute
$$ \varphi : \left]-\infty, \infty \right[ \to \left[a, \infty \right[ : u \mapsto a \cosh(u) $$
we are forced to assume $r, y \in \left]a, \infty \right[$. Therefore consider $f$'s domain to be only the right interval: $\left]a, \infty \right[$. Furthermore this implies that at a first moment $F$ can only be defined on $\left]a, \infty \right[$.
To make $\varphi$ invertible, it is necessary to restrict its domain to either $\left]-\infty, 0 \right]$ or $\left[0, \infty \right[$. Without loss of generality choose $\left[0, \infty \right[$. Then its inverse is
$$ \varphi^{-1} : \left[a, \infty \right[ \to \left[0, \infty \right[ : x \mapsto \cosh^{-1}\left(\frac{x}{a}\right) $$
Now that all the hypothesis of the theorem are satisfied, we can use $(1)$ to compute the integral on the right interval, $\left]a, \infty \right[$.
The similar method your book states to compute $F$ when $x < -a$ is to integrate over the left interval, namely $\left] -\infty, -a \right]$, and subtitute $-\varphi$ instead of $\varphi$ using the same reasoning as above, with the suitable modifications.
Let $\psi = -\varphi$. That is
$$ \psi : \left]-\infty, \infty \right[ \to \left] -\infty, -a \right] : u \mapsto - a \cosh(u) $$
If we want to substitute $\psi$ we must assume that $s, y \in \left]-\infty, -a \right[$ instead of $\left]a, \infty \right[$, where $s$ is playing the role of $r$. Note that $s \neq r$ since they belong to diferent intervals. So in this case we restrict $f$'s domain to $\left]-\infty, -a \right[$ instead. To make $\psi$ invertible we also restrict its domain to $\left[0, \infty \right[$. So
$$ \psi^{-1} : \left] -\infty, -a \right] \to \left[0, \infty \right[ : x \mapsto \cosh^{-1}\left(-\frac{x}{a}\right) $$
And thus by $(1)$ we are able to compute the integral
$$ \int_s^y f(x) \,dx = \int_{\psi{-1}(p)}^{\psi^{-1}(q)} f(\psi(u)) \psi'(u) \,du $$
this time on the left interval $\left]-\infty, -a \right[$.
Hence we can extend $F$ to $f$'s entire domain by "gluing" the solutions as
$$ F : \left]-\infty, -a \right[ \cup \left]a, \infty \right[ \to \mathbb{R} : y \mapsto \begin{cases} \int_r^y f(x) \,dx &, y \in \left]a, \infty \right[\\ \int_s^y f(x) \,dx &, y \in \left]-\infty, -a \right[ \end{cases}$$
Computing both integrals we obtain
$$ F(y) = \begin{cases} \log \left(y + \sqrt{y^2-a^2}\right) + C_1(r)  &, y \in \left]a, \infty \right[\\ -\log \left(-y + \sqrt{y^2-a^2}\right) + C_2(s) &, y \in \left]-\infty, -a \right[ \end{cases} $$
where $C_1(r)$ and $C_2(s)$ are constants that respectively depends on $r$ and $s$.
Which can also be written as
$$ F(y) = \text{sgn}(y) \log \left( |y| + \sqrt{y^2-a^2} \right) + \frac{\text{sgn}(y) + 1}{2}C_1(r) + \frac{1 - \text{sgn}(y)}{2}C_2(s) $$
where $\text{sgn}$ is the sign function.
A: You have to make the substitution $t=-x$. Then $\;\mathrm dx=-\mathrm dt$, and
$\DeclareMathOperator{\acosh}{argcosh}$
\begin{alignat}{2} \int\frac{\mathrm dx}{\sqrt{x^2-a^2}}= -\int\frac{\mathrm dt}{\sqrt{t^2-a^2}}&=-\acosh\frac ta=-\acosh\frac{|x|}a \\
&=-\ln\Bigl(|x|+\sqrt{x^2-a^2}\Bigr)&\text{(modulo a constant term).}
\end{alignat}
For method $1$, I don't see where it can lead, since the derivative of $\operatorname{arcsec} x$ is
$$(\operatorname{arcsec})'(x)=\frac1{|x|\sqrt{x^2-1}}\ne\frac1{\sqrt{x^2-1}}.$$
A: Given that $a>0$ you cannot substitute $x=a \cosh u$ when you are interested in $x$-values $<-a$. In order to use the basic idea of method 2 also when $x<-a$ we just set up the substitution
$$x:=-a\cosh u\quad (u>0),\qquad dx=-a\sinh u\>du\ ,$$
and obtain
$$\eqalign{\int{dx\over\sqrt{x^2-a^2}}&=\int{-a\sinh u\>du\over a\sinh u}\Biggr|_{u:={\rm arcosh}(-x/a)}\cr  &=(-u+C)\biggr|_{u:={\rm arcosh}(-x/a)}\cr &=-{\rm arcosh}{-x\over a}\qquad(x<-a)\ .\cr}$$
Note that this is not the same as the answer for $x>a$, since ${\rm arcosh}$ is not an odd function.
A: For method 1 :
Let $x=asec(u)$, $dx=atan(u)sec(u)du$. Then $\sqrt{x^2-a^2}=\sqrt{a^2sec^2(u)-a^2}=atan(u)$ and $u=sec^{-1}(x/a)$. The integral then becomes 
$\int{sec(u)du}$. The trick is then to multiply numerator and denominator by $tan(u)+sec(u)$ and then do the change of variable (#another one) $s=tan(u)+sec(u)$ and $ds=(sec^2(u)+tan(u)sec(u))du$. Then you are left with 
$\int{\frac{1}{s}}ds$. You just need to subtitute back in all the change of variables we did. Note that $tan(arcsec(z))=\sqrt{1-\frac{1}{z^2}}z$.
A: Antiderivative exists also in the case $x<a.$ 
To show this, let us consider the definite integral
$$J=\int\limits_{-3}^{-2}\dfrac{dx}{\sqrt{x^2-1}}.$$
The substitution
$$x= \cosh u, \quad u = \cosh^{-1}x$$
allows to obtain the value
$$J=\int\limits_{\cosh^{-1}(-3)}^{\cosh^{-1}(-2)}\ du = \cosh^{-1}(-2) - \cosh^{-1}(-3),$$
wherein
$$\cosh^{-1}x = \ln(x+\sqrt{x^2-1}).$$
Easy to see that the imaginary phase additive can not change the real difference.
So 
$$\cosh^{-1}(-2) - \cosh^{-1}(-3) = \ln\dfrac{-2+\sqrt{2^2-1}}{-3+\sqrt{3^2-1}}  = \ln\dfrac{2-\sqrt3}{3-\sqrt8} = \ln\dfrac{3+\sqrt8}{2+\sqrt3}\approx 0.445789$$
(see also Wolfram Alpha).
