I noticed some similarities between these two formulas when I refer to the basic integral table.
$\int \sqrt{x^{2}\pm a^{2} }dx= \frac{1}{2}( \sqrt{a^{2}+x^{2}}\cdot x+a^{2}\operatorname{arcsinh}\frac{x}{a} )+C$
And,
$\int \sqrt{a^{2}-x^{2}}dx= \frac{1}{2}( \sqrt{a^{2}-x^{2}}\cdot x+a^{2}\arcsin\frac{x}{a} )+C$
Is this a coincidence, or is it due to the definition of hyperbolic functions? Similar coincidences also exist in the following two integrals, and perhaps in more integrals.
$\int \frac{dx}{\sqrt{x^{2}\pm a^{2}}} = \operatorname{arsinh}\frac{x}{a}+C$
$\int \frac{dx}{\sqrt{a^{2}-x^{2}}} = \arcsin\frac{x}{a}+C$
$\int \frac{\sqrt{x^{2}\pm a^{2}}}{x} dx$ and $\int \frac{\sqrt{a^{2}-x^{2}}}{x} dx$
$\int x^{2} \sqrt{a^{2}-x^{2}} dx \,\,$ and $\,\,\int x^{2} \sqrt{x^{2}\pm a^{2}} dx$.
This is not a coincidence. For all $z \in \mathbb{C}$, $\sin(z)=\sinh(iz)/i$, and for all $w$ such that $|w|<1$, $\arcsin(w) = \arg\sinh(iw)/i$. This explains the formulas, although one must be very careful when handling expression like $\sqrt{a^2-z^2} = \sqrt{a^2+(iz)^2}$ when $z$ is a complex number: which square root is chosen?