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.

Replace the first integral by the same thing from $-a$ to $0$, and consider the points W,X,Y,Z on the unit circle above $-a,0,b$ and $c = a\sqrt{1-b^2} + b \sqrt{1-a^2}$.  Draw the family of lines parallel to XY (and WZ). This family sets up a map from the circle to itself; through each point, draw a parallel and take the other intersection of that line with the circle.
Your formula says that this map [edit: or rather the map it induces on the $x$-coordinates of points on the circle] is a change of variables converting the integral on $[-a,0]$ to the same integral on $[b,c]$.  Whatever differentiation you perform in the process of proving this, will be the verification that $dx/y$ is a rotation-invariant differential on the circle $x^2 + y^2 = 1$.
[The induced map on x-coordinates is: $x \to$ point on semicircle above $x \to$ corresponding point on line parallel to XY $\to x$-coordinate of the second point.  Here were are just identifying $[-1,1]$ with the semicircle above it.]
