$E(x)=\int_0^{\frac{\pi}{2}} \sqrt{1-x^2 \sin^2 t}\, dt$ Where $E(x)$ is complete elliptic integral of the second kind.
$u=\sin t$
$E(x)=\int_0^{1} \frac{\sqrt{1-x^2 u^2}}{\sqrt{1-u^2}}\, du$
$\frac{dE(x)}{dx}=-x\int_0^{1} \frac{u^2}{\sqrt{1-u^2}\sqrt{1-x^2 u^2}} \, du$
$\frac{d}{dx}(x\frac{dE(x)}{dx})=-2x\int_0^{1} \frac{u^2}{\sqrt{1-u^2}\sqrt{1-x^2 u^2}} \, du-x^2\int_0^{1} \frac{xu^4}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du$
$\frac{d}{dx}(x\frac{dE(x)}{dx})=\int_0^{1} \frac{-2xu^2(1-x^2 u^2)-x^3u^4}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du$
$(x^2-1)\frac{d}{dx}(x\frac{dE(x)}{dx})=\int_0^{1} \frac{(-2xu^2+x^3u^4)(x^2-1)}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du \tag1$
$xE(x)=\int_0^{1} \frac{x\sqrt{1-x^2 u^2}}{\sqrt{1-u^2}}\, du \tag 2$
According to Wikipedia, Equation 1 and 2 are equal but I could not prove it. Could you please help me to prove that?
$(x^2-1) \frac{d}{dx}(x \frac{dE(x)}{dx})=xE(x)$ http://en.wikipedia.org/wiki/Elliptic_integral
EDIT:
If $(x^2-1) \frac{d}{dx}(x \frac{dE(x)}{dx})=xE(x)$ is true, then
$(x^2-1) \frac{d}{dx}(x \frac{dE(x)}{dx})-xE(x)=0$
$\int_0^{1} \frac{(-2xu^2+x^3u^4)(x^2-1)}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du-\int_0^{1} \frac{x\sqrt{1-x^2 u^2}}{\sqrt{1-u^2}}\, du=0$ must be. And then
$\int_0^{1} \frac{(-2xu^2+x^3u^4)(x^2-1)-x (1-x^2 u^2)^2}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=0$
$-x\int_0^{1} \frac{1-2u^2+x^2u^4}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=0$
If Wikipedia differential equation is true ,
$\int_0^{1} \frac{1-2u^2+x^2u^4}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=0$
$\int_0^{1} \frac{1-u^2}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du-\int_0^{1} \frac{u^2(1-x^2u^2)}{\sqrt{1-u^2}(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=0$
$\int_0^{1} \frac{\sqrt{1-u^2}}{(1-x^2 u^2)\sqrt{1-x^2 u^2}} \, du=\int_0^{1} \frac{u^2}{\sqrt{1-u^2}\sqrt{1-x^2 u^2}} \, du$ must be true too. Now I need to prove that last equation. Any idea how to proceed? Thanks a lot for advice.