Finding an addition formula without trigonometry I'm trying to understand better the following addition formula: $$\int_0^a \frac{\mathrm{d}x}{\sqrt{1-x^2}} + \int_0^b \frac{\mathrm{d}x}{\sqrt{1-x^2}} = \int_0^{a\sqrt{1-b^2}+b\sqrt{1-a^2}} \frac{\mathrm{d}x}{\sqrt{1-x^2}}$$
The term $a\sqrt{1-b^2}+b\sqrt{1-a^2}$ can be derived from trigonometry (since $\sin(t) = \sqrt{1 - \cos^2(t)}$) but I have not been able to find any way to derive this formula without trigonometry, how could it be done?
edit: fixed a mistake in my formula.

The formula is proved easily by assuming $f(a) =\int_{0}^{a}(1-x^2)^{-1/2}\,dx$ and then setting $$u=f(a)+f(b), v=a\sqrt{1-b^2}+b\sqrt{1-a^2}$$ and showing that $u, v$ are functionally dependent so that $u=g(v) $ for function $g$. Putting $b=0$ we get $v=a$ and $u=f(a)=f(v) $ so that $f=g$ and hence $u=f(v) $ as desired.
The functional dependence between $u, v$ is proved by noting that $$\frac{\partial u} {\partial a} \frac{\partial v} {\partial b} =\frac{\partial u} {\partial b} \frac{\partial v} {\partial a} $$ Using same technique one can prove the more difficult formula $$\int_{0}^{a}\frac{dx}{\sqrt {1-x^4}}+\int_{0}^{b}\frac{dx}{\sqrt{1-x^4}}=\int_{0}^{c}\frac{dx}{\sqrt{1-x^4}}$$ where $$c=\frac{a\sqrt{1-b^4}+b\sqrt{1-a^4}} {1+a^2b^2} $$ (Euler and Fagnano established this and it was one of the key results in early development of elliptic function theory). 
