Arc Length (Set up the Integral)

johnhuntsman
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x = t + cot t
y = t - sin t
0 ≤ t ≤ 2π

Somehow the answer is:


∫sqrt(3 - 2*sin t - 2*cos t) dt
0

I'm afraid I don't know where to start on this one. I don't need someone to walk me through it (probably) but a point in the right direction would be appreciated.
 
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Nevermind this post everyone. I made a foolish mistake. I'm supposed to do sqrt[(x')^2 + (y')^2] for anyone who makes the same mistake and comes across this in a Google search. It was in an earlie rpart of the chapter I'm working on even. Sorry everyone.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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