Closed form of the elliptic integral $\int_0^{2\pi} \sqrt{1+\cos^2(x)}\,dx $

Question

I want to prove the closed form shown in Wikipedia for the arc length of one period of the sine function.

$$\int_0^{2\pi} \sqrt{1+\cos^2(x)} \ dx= \frac{3\sqrt{2}\,\pi^{\frac32}}{\Gamma\left(\frac{1}{4}\right)^2}+\frac{\Gamma\left(\frac{1}{4}\right)^2}{\sqrt{2\pi}}$$

Could someone offer some demonstration for this statement?

Possible duplicate of What is the length of a sine wave from $0$ to $2\pi$? — J. M. ain't a mathematician, Oct 13 '17 at 11:57
@J.M.isnotamathematician Not really a duplicate because no one in that post mentioned the evaluation of the integral in terms of gamma function, which OP wants. — pisco, Oct 13 '17 at 13:02

pisco · Answer 1 · 2017-10-13T13:04:37.030

5

$$\int_0^{2\pi} \sqrt{1+\cos^ 2 x} dx = 4 \int_0^1 \frac{\sqrt{1+x^2}}{\sqrt{1-x^2}} dx = 4\int_0^1 \frac{1+x^2}{\sqrt{1-x^4}} dx $$

Now use the beta function.

edited Oct 13 '17 at 13:04

answered Oct 13 '17 at 08:56

pisco

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Closed form of the elliptic integral $\int_0^{2\pi} \sqrt{1+\cos^2(x)}\,dx $

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