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Compute the arc length of the curve $y = \sqrt{x-x^2}+\sin^{-1}(\sqrt{x})$ from for $0 \leq x \leq 1$

This problem is pretty brutal! I'd appreciate if somebody could hold my hand through this integral and really lay out the details for me... I've been struggling with it for awhile now and can't get it down!!

Basically we know that: $$L = \int_0^1 \sqrt{1+(\frac{dy}{dx})^2}$$

Where $$\frac{dy}{dx}=\frac{1-2x}{2\sqrt{x-x^2}}+\frac{1}{2\sqrt{1-x}\sqrt{x}}$$

If somebody could help me simplify and integrate this that would be great... Thank so you much!!

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2 Answers 2

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You're practically there, just simplify where you left off: $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1-2x}{2\sqrt{x-x^2}}+\frac{1}{2\sqrt{1-x}\sqrt{x}}=\frac{2-2x}{2\sqrt{x-x^2}}=\frac{1-x}{\sqrt{x}\sqrt{1-x}}=\sqrt{\frac{1-x}{x}}$$ Therefore, $$L = \int_0^1 \sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \; \mathrm{d}x= \int_0^1 \sqrt{\frac{1}{x}} \; \mathrm{d}x=2\sqrt{x} \bigg \rvert_0^1 = \boxed{2}$$

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$\frac{dy}{dx}=\frac{1-2x}{2\sqrt{x-x^2}}+\frac{1}{2\sqrt{1-x}\sqrt{x}}=\frac{1-2x}{2\sqrt{x-x^2}}+\frac{1}{2\sqrt{x-x^2}}=\frac{1-x}{\sqrt{x-x^2}}=\frac{1-x}{\sqrt{1-x}\sqrt x}=\sqrt {x^{-1}-1}$

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