I would like to know how to prove following integral equality :
$$\int_{0}^{\frac{\pi}{2}}\frac{1-k^2}{(1-k^2\sin^2\theta)^\frac{3}{2}}d\theta = \int_{0}^{\frac{\pi}{2}}\sqrt{1-k^2\sin^2\theta}d\theta \quad \mathrm{where} \quad k^2<1 $$
I would be glad for any suggestions :)
One can confirm (say, by hand or by a CAS) that two integrands differ by a total derivative:
$$\sqrt{1-k^2\sin^2 x}=\frac{1-k^2}{(1-k^2\sin^2x)^{3/2}}+\dfrac{d}{dx}\frac{k^2 \sin x\cos x}{\sqrt{1-k^2\sin^2 x}}$$
But the function inside the derivative vanishes at the limits of integration, so integrating both sides from $x=0$ to $\pi/2$ obtains the desired equality.