# Total curvature of a parametrized-by-arc-length curve

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

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 $$\int k\,ds = \int_0^{2\pi} k(t)\frac{ds}{dt}\,dt = \int_0^{2\pi} \frac{\sin t}{1+\cos^2 t}\,dt.$$