<s>[INST] Trigonometic Substitution VS Hyperbolic substitution The following tables were taken from University of Pennsylvania's page about Calculus:
Trigonometric Substitution

Hyperbolic Substitution

As you can see, the forms $1+x^2$ and $x^2-1$ are repeated in the tables. How does one know when a trigonometric substitution is more suitable to a problem than a hyperbolic substitution and vice versa?
 [/INST] In my opinion, the following substitutions are enough:





appears in integrand
substitution
convention




1
$\sqrt{x^2+1}$
$x=\sinh\theta$
$\theta\in\mathbb R$


2
$\sqrt{x^2-1}$
$x=\pm\cosh\theta$
$\theta\geq0$


3
$\sqrt{1-x^2}$
$x=\sin\theta$
$-\frac{\pi}2\leq\theta\leq\frac{\pi}2$




$\sinh$ and $\cosh$ are better substitutions than $\tan$ and $\sec,$ respectively, as they are easier to differentiate and integrate, and have nicer principal domains. $\sin$ is a better substitution than $\tanh$ as it is easier to differentiate and integrate. The following examples illustrate this:





integrand
substitution
result




1
$\displaystyle\frac{x^2}{\sqrt{x^2+4}}$
$x=2\sinh\theta\quad\color{green}✔$
$4\int\sinh^2\theta\,\mathrm d\theta$



$\displaystyle\frac{x^2}{\sqrt{x^2+4}}$
$x=2\tan\theta\quad\color{red}✗$
$4\int\tan^2\theta\sec\theta\,\mathrm d\theta$


2
$\displaystyle\frac{\sqrt{x^2-4}}{x^2}\quad(x\leq-2)$
$x=-2\cosh\theta\quad\color{green}✔$
$-\int\tanh^2\theta\,\mathrm d\theta$



$\displaystyle\frac{\sqrt{x^2-4}}{x^2}\quad(x\leq-2)$
$x=2\sec\theta\quad\color{red}✗$
$-\int\tan^2\theta\cos\theta\,\mathrm d\theta$


3
$\sqrt{9-x^2}$
$x=3\sin\theta\quad\color{green}✔$
$9\int\cos^2\theta\,\mathrm d\theta$



$\sqrt{9-x^2}$
$x=3\tanh\theta\quad\color{red}✗$
$9\int\mathrm{sech^3}\,\theta\,\mathrm d\theta$



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