Let's try to calculate the integral $$I=\int \sqrt{x^2-a^2} dx, x>a>0$$ I think a suitable substitution would be $x=\dfrac{a}{\sin t}$. Then we would have $$dx=d\dfrac{a}{\sin t}=ad\frac{1}{\sin t}=ad\sin^{-1}t=-1a\sin^{-2}t\cos tdt=-\dfrac{a\cos t}{\sin^2 t}dt$$
Note that $$\sqrt{x^2-a^2}=\sqrt{\frac{a^2}{\sin^2 t}-a^2}=\sqrt\frac{a^2-a^2\sin^2 t}{\sin^2 t}=\sqrt\frac{a^2\cos^2 t}{\sin^2 t}=\dfrac{a\cos t}{\sin t}$$
Note that for the last simplification to be true we must have $0<t<\dfrac{\pi}{2}$. Can we actually conclude that, why (why not), as I cannot find a valid reason to do so?
Note that as $x$ and $a$ are both positive, then we must also have $\sin t>0$ as $x=\dfrac{a}{\sin t}$.
Maybe there are some specifics (that I'm not familiar with) in the formulation of the theorems for change of variables with integrals. Thanks!
$x$ can be negative, but the equation requires $|x|>a$
I would say $x = a\sec t$ rather than $x = a\csc = \frac {a}{\sin t}.$ The negative signs confuse me with the cosecant and cotangent functions.
$x = a\sec t\\ dx = a\sec t\cot t$
Yes, $-\frac {\pi}{2}<t<\frac {\pi}{2}$ I am not sure that that is particularly relevant. Our substitution is a mapping from the real line excluding the interval $|x|<a$ to $(-\frac {\pi}{2}, \frac {\pi}{2})$ but nothing has been left out.
$\int \sqrt {a^2 (\sec^2 t - 1)} a\sec t\tan t\ dt\\ \int a^2 \sqrt{\tan^2 t}\sec t\tan t\ dt$
If we are going to be precise about this. $\sqrt{\tan^2 t} = |\tan t|$ rather than just $\tan t$ that is the radical always returns the principal root.
But, I will work with $a^2 \int \sec t\tan^2 t\ dt$ with a note that should we be working with negative values of $x$ and $\tan t$ this is going to flip the sign of the entire integral.
This integral is not obvious, but here is the trick.
Integration by parts:
$u = \tan t, dv = \sec t\tan t\\ du = \sec^2 t, v = \sec t$
$a^2 (\sec t\tan t - \int \sec^3 t\ dt)\\ a^2\sec t\tan t - a^2\int \sec t(\tan^2 t + 1)\ dt\\ a^2\sec t\tan t - a^2\int \sec t\ dt - a^2\int \sec t\tan^2 t\ dt$
But $a^2\int \sec t\tan^2 t\ dt = I$
$I = a^2\sec t\tan t - a^2\int \sec t\ dt - I\\ I = \frac {a^2\sec t\tan t - a^2\ln |\sec t + \tan t|}{2} + C$
Revisit our note that when $\tan t$ is less than zero, the sign flips.
$t< 0$ implies
$I = \frac {-a^2\sec t\tan t + a^2\ln |\sec t + \tan t|}{2} + C\\ \frac {a^2\sec t(-\tan t) - a^2\ln |\frac {1}{\sec t + \tan t}|}{2} + C\\ \frac {a^2\sec t(-\tan t) - a^2\ln |\frac {\sec t - \tan t}{\sec^2 t - \tan^2 t}|}{2} + C\\ \frac {a^2\sec t(-\tan t) - a^2\ln |\sec t +(-\tan t)|}{2} + C$
If we have $-\tan t$ factors/terms when $t<0$ and $\tan t$ when $t\ge 0$ then that is $|\tan t|$
$I = \frac {a^2\sec t|\tan t| - a^2\ln |\sec t + |\tan t||}{2} + C$
Reverse the substitution.
$\sec t = \frac {x}{a}\\ |\tan t| = \sqrt{\frac {x^2}{a^2} - 1} = \frac {\sqrt{x^2 - a^2}}{a}$
$\frac {x\sqrt {x^2-a^2} - a^2\ln |x + \sqrt {x^2- a^2}|}{2} + C$