Jtdylktuy

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?

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.

If you look for a primitive in $(1,+infty)$,

put $$x=cosh(theta)$$

with $$dx=sinh(theta) dtheta$$

but if you want a primitive in $(-infty,-1)$, you should put
$$x=-cosh(theta)$$

with $$dx=-sinh(theta)dtheta$$

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 $|x|<1$.

$$I=intleft(x^2-1right)^{-3/2}dx$$
we want this term $x^2-1$ to be simplified to a single term, so think of trig identities.:
$$cos^2x+sin^2xequiv1,,cosh^2x-sinh^2xequiv1$$
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:
$$I=intleft(cosh^2(t)-1right)^{-3/2}.sinh(t)dt=int(sinh(t))^{-3}.sinh(t)dt=inttext{csch}^2(t)dt=-coth(t)+C$$