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

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


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

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

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

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 $L(2\pi)$

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


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

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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Haven't you just done so? – Sean Eberhard Sep 12 '12 at 8:52
@SeanEberhard: I proved it via geometric relation. I wonder how to show it via some analytic methods in relation 2. Thanks – Mathlover Sep 12 '12 at 9:02
'answers' ... lol +1 – Santosh Linkha Sep 13 '12 at 8:32

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.

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


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

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

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

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

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

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

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

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


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

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

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

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


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

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

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

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This looks like an enormously complicated way of expressing what I wrote in my answer. Perhaps you could comment on what you found missing in my answer? – joriki Sep 13 '12 at 8:25
@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. – Mathlover Sep 13 '12 at 8:32

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