Methods from Trigonometric and hyperbolic substitutions

Substitutions for $\sqrt{a^2\pm x^2}$ and $\sqrt{x^2-a^2}$

For $\sqrt{a^2-x^2}$ the hypotenuse is $a$ and the side opposite $\theta$ is $x$, so the adjacent side is $\sqrt{a^2-x^2}$. For $\sqrt{a^2+x^2}$ the opposite side is $x$, the adjacent side is $a$ and the hypotenuse is $\sqrt{a^2+x^2}$. For $\sqrt{x^2-a^2}$, this triangle description assumes $x\ge a$: the hypotenuse is $x$, the adjacent side is $a$ and the opposite side is $\sqrt{x^2-a^2}$.

The identity $\operatorname{sech}^2\theta+\tanh^2\theta=1$ is the same relation as $1-\tanh^2\theta=\operatorname{sech}^2\theta$.

Writing $\sqrt{1-\sin^2\theta}=\cos\theta$ without the absolute value is legitimate on the range of $\sin^{-1}$, where $\cos\theta\ge 0$; it fails if $\theta$ is taken outside that range. For $\sqrt{x^2-a^2}$ on $x\le-a$, use a separate branch, such as $x=-a\cosh\theta$ with $\theta\ge0$, or choose a secant range and retain the required signs.