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.

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.
