The integral relation between Perimeter of ellipse and Quarter of Perimeter 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
 A: 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.
A: I proved the relation via using analytic way. I would like to share the solution with you.
$$\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 4u}\,du=K$$
$u=\pi/4-z$
$$K=\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (\pi-4z)}\,dz$$
$\sin (\pi-4z)=\sin \pi \cos 4z-\cos \pi \sin 4z=  \sin 4z$
$$\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (\pi-4z)}\,dz=\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz$$
$$\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz=\int_{-\pi/4}^{0}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz+\int_{0}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (4z)}\,dz$$

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

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

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

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