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.
 A: 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.]
A: 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). 
A: I will try to summarize here what I've found so far on this:
Since the form being integrated is $\frac{\mathrm{d}x}{\sqrt{1 - x^2}} =: \omega$ let $y = \sqrt{1 - x^2}$, we can also write this as $x^2 + y^2 - 1 = 0$ which makes it more obvious that this is a unit circle.
Let $N$ be the point $(1,0)$ and we can define a group structure on the curve as follows. To add $A$ and $B$, fire a ray from $N$ parallel to $AB$ and pick the point it intersects the curve as $A \oplus B$. Symbolically, this means $A \oplus B = N - k(B - A)$ for some nonzero $k$, since $A \oplus B$ lies on the circle we can expand it into the equation of the curve to solve for $k$,
$$\begin{align}
(- k (x_b - x_a))^2 + (1 - k (y_b - y_a))^2 - 1 &= 0 \\\\
k ((x_b - x_a)^2 + (y_b - y_a)^2) - 2 (y_b - y_a) &= 0 \\\\
\frac{2 (y_b - y_a)}{(x_b - x_a)^2 + (y_b - y_a)^2} &= k
\end{align}$$
hence $x_{a \oplus b} = - 2 \frac{(x_b - x_a)(y_b - y_a)}{(x_b - x_a)^2 + (y_b - y_a)^2} = x_a \sqrt{1-x_b^2} + x_b \sqrt{1-x_a^2}$.
Now I think by the theorem, if 


*

*$\omega$ is invariant along the curve

*$\oplus$ is the group law on points of the curve


we can conclude now that
$$\int_0^a \omega + \int_0^b \omega = \int_0^{a \oplus b} \omega$$
Certainly, the previous equation holds - but for this post to count as a proof the theorem is needed. Furthermore I believe this same idea works for any conic section and also for cubic curves and perhaps not any others curves.
