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.

Source of wikipedia

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

Could someone offer some demonstration for this statement?

Answer 1

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

Now use the beta function.