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How do you find the arc length of $y = \sin^{-1}x + \sqrt{1-x^2}$?

I got $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{2-2\sin t}dt$ and became stuck. Any hints?

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The arc length of what part of the curve defined by that equation are you looking for? – joriki Nov 13 '12 at 9:54
Question didn't specify the endpoints, I checked the equation using graphing calculator and the graph ends at (1, $\frac{\pi}{2}$) and (-1, $-\frac{\pi}{2}$). So i used 1 and -1 as the limit for the initial integral. – uohzxela Nov 13 '12 at 10:00
The arc is defined only for $-1\leq x\leq 1$ – Dennis Gulko Nov 13 '12 at 10:03
I know but i didn't start with variable $t$, i started with $x$ instead. I have to change the limits coz i changed the variable $x$ to $t$ by substituting $x$ with $\sin t$ – uohzxela Nov 13 '12 at 10:05
up vote 2 down vote accepted

The length of an arc is given by $\int_a^b\sqrt{1+(y')^2}dx$. In this case, $$y'=\frac{1}{\sqrt{1-x^2}}-\frac{2x}{2\sqrt{1-x^2}}=\frac{1-x}{\sqrt{1-x^2}}=\sqrt\frac{1-x}{1+x}$$ Hence $$L=\int_{-1}^1\sqrt{1+(y')^2}dx=\int_{-1}^1\sqrt{1+\frac{1-x}{1+x}}dx=\int_{-1}^1\sqrt{\frac{2}{1+x}}dx$$ From here the computation is straightforward.

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thank you! simplifying $\frac{1-x}{\sqrt{1-x^2}}$ makes all the difference! – uohzxela Nov 13 '12 at 10:16

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