In deriving the catenary equation, how does integrating $\frac{dy'}{\sqrt{1+(y')^2}}=\frac1a dx$ yield $\sinh^{-1}(y')=\frac{x}{a}$?
You want to evaluate $\int \frac{1}{\sqrt{1+x^2}}dx,$ so substitute $x=\sinh(u),$ $dx=\cosh(u)du$ and we have
$$\int \frac{1}{\sqrt{1+x^2}}dx=\int\frac{\cosh(u)}{\sqrt{1+\sinh^2(u)}}du=\int\frac{\cosh(u)}{\cosh(u)}du$$ $$=\int 1\ du=u+C=\sinh^{-1}\left(x\right)+C$$
using the hyperbolic identity $\cosh^2(u)=1+\sinh^2(u)$.
See also hyperbolic functions.