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Show that the length of one arc of the cycloid $x=a(\theta-sin\theta)$ and $y=a(1-cos\theta)$ is $8a$ .
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$ x = a(\theta - cos\theta) \Longrightarrow \frac{dx}{d\theta} = a(1-cos\theta) \\ y = a(1-cos\theta ) \Longrightarrow \frac{dy}{d\theta} = asin\theta \\ $

$ S = \int_0^{2\pi} \sqrt{ a^2(1-cos\theta)^2 + a^2 sin^2\theta} d\theta \\ $

$ S = a\int_0^{2\pi} \sqrt{ 1 + cos^2\theta - 2 cos\theta + sin^2\theta} d\theta \\ $

$ S = a\int_0^{2\pi} \sqrt{ 2 - 2 cos\theta } d\theta \\ $

$ S = a\sqrt2 \int_0^{2\pi} \sqrt 2 sin\frac{\theta}{2} d\theta \\ $

$ = 2a \left [\frac{ \frac{-cos\theta}{2} }{\frac{1}{2}} \right]_0^{2\pi} \\ $

$ = 4a [ -cos\pi + cos 0 ] \\ $

$ = 8a \\ $

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