Why can the trig sub $x=\cosh(\theta)$ be used to solve the integral: $\int \frac{1}{(x^2-1)^{3/2}}dx$? I thought that $\theta$ must be chosen such that $\cosh(\theta)$ has a range that is equal to the domain of $\frac{1}{(x^2-1)^{3/2}}$. But this can't be done since the domain includes negative numbers, and $\cosh(\theta)$ is always positive. If this substitution is made, the answer seems to be valid for all values of x, why does it work? 
 A: If you look for a primitive in $(1,+\infty)$,
put $$x=\cosh(\theta)$$
with $$dx=\sinh(\theta) d\theta$$
but if you want a primitive in $(-\infty,-1)$, you should put
$$x=-\cosh(\theta)$$
with $$dx=-\sinh(\theta)d\theta$$
A: So there are singularities at $\pm 1$, you have to avoid those. So let's consider finding an antiderivative valid on $(-\infty,-1)$.
In fact $x=\cosh(\theta)$ is not valid there. The substitution really being used there is $x=-\cosh(\theta)$, so $dx=-\sinh(\theta) d \theta$, so the integral becomes
$$\int \frac{1}{(\cosh^2(\theta)-1)^{3/2}} (-\sinh(\theta)) d \theta.$$
The catch comes when you rewrite $(\cosh^2(\theta)-1)^{3/2}$ as just $\sinh^3(\theta)$. This is not strictly correct, in fact it is $|\sinh^3(\theta)|$ in general. On $(-\infty,-1)$, this absolute value reduces to $-\sinh^3(\theta)$. The formula collapses back to the formula you would get for an antiderivative on $(1,\infty)$ because this minus sign cancels out with the minus sign that came with $dx$.
This is relatively general, but I'm not sure whether pinning down exactly what happens in general will be helpful to you at this point.
A: Part of it's because we're presumably only integrating over values of $x$ for which the integrand is real, i.e. $|x|\ge 1$. Any work with complex numbers that lets you pose the question otherwise in the first place also lets you get a complex $\operatorname{arcosh}x$ with $|x|<1$.
A: $$I=\int\left(x^2-1\right)^{-3/2}dx$$
we want this term $x^2-1$ to be simplified to a single term, so think of trig identities.:
$$\cos^2x+\sin^2x\equiv1,\,\cosh^2x-\sinh^2x\equiv1$$
are the ones you need to know. since it is of the form $x^2-b^2$ rather than $b^2-x^2$ or $x^2+b^2$ we know to use hyperbolic functions. Manipulating the right rule, we obtain:
$$\sinh^2x=\cosh^2x-1$$
now we can use this. let:
$x=\cosh(t)$ and we get $dx=\sinh(t)$ and rewrite the integral as:
$$I=\int\left(\cosh^2(t)-1\right)^{-3/2}.\sinh(t)dt=\int(sinh(t))^{-3}.\sinh(t)dt=\int\text{csch}^2(t)dt=-\coth(t)+C$$
