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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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    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$2012-11-13

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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!2012-11-13