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Ellipse Equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$x=a\cos t$ ,$y=b\sin t$

$L(\alpha)=\int_0^{\alpha}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$

$L(\alpha)=\int_0^\alpha\sqrt{a^2\sin^2 t+b^2 \cos^2 t}\,dt $

$L(2\pi)=\int_0^{2\pi}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt \tag{Perimeter of ellipse}$

$L(\pi/2)=\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt \tag {Quarter of Perimeter }$

Geometrically, we can write $L(2\pi)=4L(\pi/2)$

$4\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt=\int_0^{2\pi}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt \tag1$

If I change variable in integral of $L(2\pi)$

$L(2\pi)=\int_0^{2\pi}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt \tag{Perimeter of ellipse}$

$t=4u$

$L(2\pi)=4\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 4u}\,du$

According to result (1),

$L(2\pi)=4\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 4u},du=4\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt$

$\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 4u}\,du=\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 t}\,dt \tag2$

How to prove the relation $(2)$ analytically? Thanks a lot for answers

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    'answers' ... lol +12012-09-13

2 Answers 2

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I'm not sure about the distinction you're making between geometric and analytic methods. In (2), the left-hand side is an integral over four intervals, in each of which the integrand is a compressed or a compressed and reflected version of the integrand on the right-hand side. Essentially you're averaging a periodic function, up to reflection, and it doesn't matter over how many periods you average it.

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I proved the relation via using analytic way. I would like to share the solution with you.

$\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 4u}\,du=K$

$u=\pi/4-z$

$K=\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (\pi-4z)}\,dz$

$\sin (\pi-4z)=\sin \pi \cos 4z-\cos \pi \sin 4z= \sin 4z$

$\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (\pi-4z)}\,dz=\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz$

$\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz=\int_{-\pi/4}^{0}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz+\int_{0}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz$


$\int_{-\pi/4}^{0}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz$ $z=-p$

$\int_{\pi/4}^{0}\sqrt{b^2+(a^2-b^2)\sin^2 (-4p)}\,(-dp)=\int_{0}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4p)}\,dp$


$K=\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz=2\int_{0}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz$

$K=2\int_{0}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz$

$z=\pi/8-v$

$K=2\int_{-\pi/8}^{\pi/8}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv$ $K=2\int_{-\pi/8}^{0}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv+2\int_{0}^{\pi/8}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv$


$\int_{-\pi/8}^{0}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv$ $v=-h$

$\int_{\pi/8}^{0}\sqrt{b^2+(a^2-b^2)\cos^2 (-4h)}\,(-dh)=\int_{0}^{\pi/8}\sqrt{b^2+(a^2-b^2)\cos^2 (4h)}\,dh$


$K=2\int_{-\pi/8}^{0}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv+2\int_{0}^{\pi/8}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv$ $K=4\int_{0}^{\pi/8}\sqrt{b^2+(a^2-b^2)\cos^2 (4v)}\,dv$

$v=\pi/8-t/4$

$K=4\int_{\pi/2}^{0}\sqrt{b^2+(a^2-b^2)\cos^2 (4(\pi/8-t/4))}\,(-dt/4)$

$K=\int_{0}^{\pi/2}\sqrt{b^2+(a^2-b^2)\cos^2 (\pi/2-t)}\,dt$

$K=\int_{0}^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 (t)}\,dt$

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    @joriki: Thank you very much for your answer. It is OK and clear. I know what average means. if I draw the function , I can see geometrically the periodic function relations. I just asked to see the relation proof via analytic way such as variable changing . It was for me a challenge to solve such relation. Thanks a lot for answer again.2012-09-13