Find the arc length of the curve y = arcsin x ± sqrt(1

Find the arc length of the curve $y = \arcsin x \pm \sqrt{1 - x^2}$ (over its full domain $x\in[-1,1]$).

We handle the $+$ and $-$ branches together using $\mp$ for the alternate sign.

Step 1: Differentiate

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

Step 2: Simplify $1+(y')^2$

\[1+\left(\frac{1\mp x}{\sqrt{1-x^2}}\right)^2 = \frac{(1-x^2)+(1\mp x)^2}{1-x^2}=\frac{2\mp2x}{1-x^2}=\frac{2(1\mp x)}{(1-x)(1+x)}\]

Step 3: The two cases

For $y=\arcsin x+\sqrt{1-x^2}$ (upper sign): $\sqrt{1+(y')^2}=\sqrt{\dfrac2{1+x}}$.

For $y=\arcsin x-\sqrt{1-x^2}$ (lower sign): $\sqrt{1+(y')^2}=\sqrt{\dfrac2{1-x}}$.

Step 4: Integrate Case 1

With $u=1+x$: \[L_1=\int_{-1}^1\sqrt{\frac2{1+x}}\,dx = \sqrt2\int_0^2u^{-1/2}\,du=\sqrt2\left[2\sqrt u\right]_0^2 = 2\sqrt2\cdot\sqrt2=4\]

Step 5: Integrate Case 2

With $u=1-x$, the same computation gives $L_2=4$.

\[\boxed{L_1=L_2=4}\]

Although the integrand has an (integrable) singularity at one endpoint in each case, the resulting arc length is finite and identical for both branches.