Prove that $\int_0^x \ 1/\sqrt{1-x^2} dx$ is equal to length of unit circle arc?

Question

How to prove that $\int_0^x \ 1/\sqrt{1-x^2} dx$ is equal to length of unit circle arc?

I know that the integral is $\arcsin(x)+c$ but really do not see how this is related to arc length.

Answer 1

Let $\vec r=\hat xx+\hat yy$ be the vector that traces the unit circle.

A differential vector segment is given by $d\vec r=\hat xdx+\hat ydy$ and the magnitude of this vector $|d\vec r|=\sqrt{(dx)^2+(dy)^2}$ is the length of the differential segment.

We also have from $x^2+y^2=1$ that $\frac{dy}{dx}=-x/y$. Thus,

$$|d\vec r| =\sqrt{1+(-x/y)^2} dx=\frac{dx}{\sqrt{1-x^2}} $$

on the upper unit circle. If we want the arc length along the upper unit circle, we simply integrate this this differential.