Find the curve whose arc length between points

Find the curve whose arc length between points with abscissas $x=0$ and $x=x$ is $s=\displaystyle\int_0^x\sqrt{\dfrac{2-t^2}{1-t^2}}\,dt$.

We work backward from the arc-length integrand to the differential equation for $y$.

Step 1: Match integrands

\[\sqrt{1+(y')^2}=\sqrt{\frac{2-x^2}{1-x^2}} \implies 1+(y')^2=\frac{2-x^2}{1-x^2}\]

Step 2: Solve for $(y')^2$

\[(y')^2=\frac{2-x^2}{1-x^2}-1=\frac{(2-x^2)-(1-x^2)}{1-x^2}=\frac1{1-x^2}\]

Step 3: Solve for $y'$ and integrate

\[y'=\pm\frac1{\sqrt{1-x^2}} \implies y=\pm\arcsin x+C\]

Step 4: Fix the constant

Taking the curve through the origin (a natural choice, since $s=0$ at $x=0$): $C=0$.

\[\boxed{y=\pm\arcsin x}\]

Step 5: Verify

For $y=\arcsin x$: $y'=\frac1{\sqrt{1-x^2}}$, so $1+(y')^2=1+\frac1{1-x^2}=\frac{2-x^2}{1-x^2}$, matching the given integrand exactly. ✓