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Suppose we have the following smooth curve $\sigma:]0,2\pi[\leftarrow\mathbb{R}^2, \sigma(t) = (t, \sin t)$. I want to find the total curvature $\kappa := \int_0^{2\pi}||\sigma''(t)||dt$, but first I want to re-parametrize $\sigma$ such that its parameterized by arc length, i.e. s.t. $||\sigma'(t)|| = 1$ for all $t\in [0,2\pi]$.

How do I parameterize this specific function by arc length? I know how to do it in general but I am not able to solve the arc length integral $s(t) := \int_0^t||\sigma'(t)||dt = \int_0^t\sqrt{1+\cos^2(t)}dt$.

Could anyone point me in the right direction? Thanks

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You almost never want to (or, indeed, can) reparametrize explicitly by arclength. You can compute curvature (and many other things) by applying the chain rule: $$k=\left\|\frac{dT}{ds}\right\|=\left\|\frac{dT/dt}{ds/dt}\right\|,$$ and remember that $ds/dt = \|\alpha'(t)\|$. Here the total curvature will turn out to be $$\int k\,ds = \int_0^{2\pi} k(t)\frac{ds}{dt}\,dt = \int_0^{2\pi} \frac{\sin t}{1+\cos^2 t}\,dt.$$

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