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$$x=6t-6sint$$ $$y=6-6cost$$ Find the arc length of the parametric curve $$Arc length = \int_{0}^{2\pi} \sqrt{(6-6cost)^2+(6sint)^2}dt\\ =\int_{0}^{2\pi} \sqrt{36-72cost+36cos^2t+36sin^2t}dt \\ =\int_{0}^{2\pi} 6 \sqrt{1-2cost+cos^2t+sin^2t}dt\\ =\int_{0}^{2\pi} 6\sqrt{2-2cost}dt\\ =\int_{0}^{2\pi} 6\sqrt{2}\sqrt{1-cost}dt\\ = 6\sqrt{2}\int_{0}^{2\pi}\sqrt{1-cost}dt\\ =6\sqrt{2}\int_{0}^{2\pi}\sqrt{1-cost}\frac{\sqrt{(1+cost)}}{\sqrt{(1+cost)}} dt\\ =6\sqrt{2}\int_{0}^{2\pi}\frac{\sqrt{(1-cos^2t)}}{\sqrt{(1+cost)}} dt\\ =6\sqrt{2}\int_{0}^{2\pi}\frac{sint}{\sqrt{(1+cost)}} dt\\$$ Let $$u =1+cost$$ $$du=-sint$$ $$=-6\sqrt{2}\int_{0}^{2\pi}\frac{1}{\sqrt{u}}\\ =-12\sqrt{2}[u^\frac{1}{2}]\\ =0$$ But the answer is 48.

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    $\begingroup$ the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves $\endgroup$ – G Cab Dec 15 '18 at 16:47
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Leaving the coefficient $6$ on the side, we have

$$s=\int_0^{2\pi}\sqrt{(1-\cos t)^2+(\sin t)^2}\,dt=\int_0^{2\pi}\sqrt{2-2\cos t}\,dt=2\int_0^{2\pi}\left|\sin\frac t2\right|\,dt.$$

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