Arc length of sin(x)?

Question

My original objective was to find the arc length of $\sin x$ on the interval $[0, 2\pi ]$. Of course the traditional way to calculate arc length is to use the formula: $$L = \int_a^b \sqrt{1 + (f'(x))^2} dx$$. And so obviously the formula becomes in this case: $$\int_{0}^{2\pi} \sqrt{1 + \cos^2 x} \ dx$$ But I am stuck here. Does anyone know of any analytical methods to anti-differentiate $\sqrt{1 + \cos^2 x}$, if any exist at all?

Answer 1

The antiderivative requires elliptic integrals. In Maple's notation, it's $$ \pm {\it EllipticE} \left( \cos \left( x \right) ,i \right) $$ ($-$ if $0 < x < \pi$, $+$ if $\pi < x < 2 \pi$).

This is not an elementary function.

The definite integral is $4 \sqrt{2} {\it EllipticE}(1/\sqrt{2}) $. It is approximately $7.640395578$.