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