Show the integrals are alike without explicit computation. I saw a couple of striking integrals which are $\hspace{5em} \displaystyle \color{black}{\displaystyle \int_a^b \frac{x\,\mathrm{d}x}{\sqrt{(x^2-a^2)(b^2-x^2)}} }$$\displaystyle\ =\ $$\displaystyle \color{black}{ \frac{1}{2}\int_a^b \frac{\mathrm{d}x}{\sqrt{(x-a)(b-x)}}}$ On the site the other day. The soltions to these integrals are rather elementary (click on the integrals to find their solutions), but the symmetry about them is striking. Is it a mere coincidence that the integrals are alike, and indeed constant for all $a$, $b$, or is there something deeeper luring behind? Is there some intrinsic property with the integrals such they are alike, and or constant? My question is: **What is the connection between the two integrals, and can they be shown to be alike; without rewriting the integrals into trivialities?**

The identity $$\int_a^b \frac{x\,\mathrm{d}x}{\sqrt{(x^2-a^2)(b^2-x^2)}} =G(a^2,b^2),\qquad G(u,v)= \frac{1}{2}\int_u^v \frac{\mathrm{d}x}{\sqrt{(x-u)(v-x)}} $$ follows from the change of variable $x\to x^2$. Now, a second, unrelated, fact is that $G$ is constant, in particular $G(a^2,b^2)=G(a,b)$ (and $G(a^2,b^2)=G(42a,42b)$ as well, by the way). But I would say the first identity (the change of variable $x\to x^2$) and the second identity (that $G$ is constant) are different in nature.