Finding an addition formula without trigonometry

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.]