In deriving the catenary equation, how does integrating $\frac{dy'}{\sqrt{1+(y')^2}}=\frac1a dx$ yield $\sinh^{-1}(y')=\frac{x}{a}$?

Question

Watch, consider and refer the following you tube video named " Catenary equation derivation".

Catenary equation derivation

At time $00:17:07$, author of this video integrates both sides of the equation $$\frac{dy'}{\sqrt{1+(y')^2}} = \frac1a dx \tag{1}$$

and arrived at the following step

$$ \sinh^{-1} {(y')}=\dfrac{x}{a}\tag{2}$$

Now I understood the R.H.S. of equation $(2)$, but I don't understand how did the author(math professor) of video integrate L.H.S of equation $(1)$ and arrived at L.H.S. of equation $(2)$.

Would any member of this math stack exchange, what are the steps involved in this integration of equation $(1)$?

Answer 1

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.