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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.

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    Why? What's wrong with trigonometry? :-)2010-08-04
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    Since $\int_0^c \frac{\mathrm{d}x}{\sqrt{1-x^2}}=\arcsin c$, I'd expect trigonometry to be required, but I don't know for sure.2010-08-04
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    ShreevatsaR, well the trigonometry is a little *too* good, basically it gives me the answer without giving me any better understanding of the integral sum.2010-08-04
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    What is this formula called?2010-08-04
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    TheMachineCharmer, Well the integral defines $\arcsin$ as Isaac said, the addition formula is effectively the identity $\sin(\theta+\phi)=\sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi)$ written in terms of $\sin(\theta),\sin(\phi)$.2010-08-04
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    @muad: that's what I was thinking, except that the upper limit on the right side is more like $\cos(\theta+\phi)$ than $\sin(\theta+\phi)$, hence my answer below.2010-08-04
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    Thanks! I just wanted to know the name of the formula so that I can search some related things. :)2010-08-04
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    The "right" way to understand this formula - e.g. the one that generalizes to, say, elliptic integrals - is to pass to the Riemann surface w^2 = 1 - z^2 and do the integral there instead. But I'm not familiar with these techniques, so hopefully someone who is will elaborate.2010-08-04
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    Thanks Qiaochu Yuan! I've been trying to read about Riemann surfaces but so far the books I found are too advanced (I couldn't keep up), I'll have to try harder but it's good to see this is the same topic!2010-08-04

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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.]

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    +1 but I will get clearer picture if you draw it http://www.livegeometry.com/ :-)2010-08-04
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    @TheMachineCharmer, maybe *you* could draw the picture and ask T if it represents the situation as he intended... Also: that page seems to want me to install something (and, from the looks of it, another operating system with it!)2010-08-04
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    Somewhat shorter version, but might be only valid for the real line: your integrand is an even function so the integral from 0 to a and the integral from -a to 0 is the same. You end up with the integral from -a to b if you add up the two integrals in the left-hand side. Now find a substitution that transforms the integral from -a to b into an integral with 0 lower limit.2010-08-06
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    "Now find a substitution" -- how does one do that? I wrote down the geometry, which ultimately leads to a substitution when unwound. But to start with the integrals and find a substitution would be quite a challenge. The elliptic case was done by Euler and Fagnano.2010-08-06
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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.

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    You don't need the theorem to conclude the argument, you are proving the theorem (at least in the scheme I posted above). The formula is equivalent to two integrals being equal. The involution defined by the family of lines parallel to AB is a change of variables that matches these two integrals. The differential calculation performed in the process says that for infinitesimally close parallels, measures on the two small pieces of circle are equal. This is called "invariance of $ \omega $ ". That the measures and the intervals of integration are identified proves the formula.2010-08-06
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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).

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Here's a differo-algebraic proof...

With the substitution $y=\sqrt\frac{1-x}{1+x}\rightarrow x+1=\frac2{1+y^2}$, you can transform the integrals into $$\int_1^{\sqrt{(1-a)/(1+a)}}d\bigg(\frac2{1+y^2}-1\bigg)\frac{\sqrt{1+x}}{(1+x)\sqrt{1-x}}+\int_1^{\sqrt{(1-b)/(1+b)}}d\bigg(\frac2{1+y^2}-1\bigg)\frac{\sqrt{1+x}}{(1+x)\sqrt{1-x}}$$ \begin{align}&=\int_1^{\sqrt{(1-a)/(1+a)}}\frac{-4y\; dy}{(1+y^2)^2}\frac{1+y^2}{2y}+\int_1^{\sqrt{(1-b)/(1+b)}}\frac{-4y\; dy}{(1+y^2)^2}\frac{1+y^2}{2y} \\[3ex] &=-2\int_1^{\sqrt{(1-a)/(1+a)}}\frac{dy}{1+y^2}-2\int_1^{\sqrt{(1+b)/(1-b)}}\frac{-dz/z^2}{1+z^{-2}}\tag{$z=1/y\,$}\\[3ex] &=2\int_{\sqrt{(1-a)/(1+a)}}^{\sqrt{(1+b)/(1-b)}}\frac{dx}{1+x^2}=2\Bigg[\int_0^{\sqrt{(1+b)/(1-b)}}-\int_0^{\sqrt{(1-a)/(1+a)}}\Bigg]\frac{dx}{1+x^2}. \end{align} Now I'll employ some shorthand here and rescale the bounds... $$2\int_0^{BA}\frac{dy/A}{1+y^2/A^2}-2\int_0^{AB}\frac{dy/B}{1+y^2/B^2}\tag{$A:=\sqrt{\tfrac{1-a}{1+a}},B:=\sqrt{\tfrac{1+b}{1-b}}$}$$ $$=2\int_0^{AB}\frac{(1/A-1/B+y^2\color{purple}[1/(AB^2)-1/(A^2B)\color{purple}])\:dy}{(1+y^2/A^2)(1+y^2/B^2)}$$ $$=\frac{2(A-B)}{AB}\int_0^{AB}\frac{[-1+y^2/(AB)]\:dy}{1+y^2(A^{-2}+B^{-2})+y^4/(A^2B^2)}$$To bring the integral into a form with an $f^2(u)+g^2(u)$ expression in its denominator, just complete the square (with $1$ as the constant): $$=\frac{2(A-B)}{AB}\int_0^{AB}\frac{[-1+y^2/(AB)]\:dy}{(1+y^2/(AB))^2+y^2(A^{-2}-2/(AB)+B^{-2})}.$$And now the rest of the integrand must equal either $\frac{g'}f-\frac{f'g}{f^2}$ or $\frac{f'}g-\frac{fg'}{g^2}$ to create $du/(1+u^2)$; using $u=\frac{1+y^2/(AB)}{y(1/A-1/B)}$ or $u=\frac{y(1/A-1/B)}{1+y^2/(AB)}$ will work, but the second one will keep the lower bound $0$. Thus, $$=-2\int_0^{(B-A)/(1+AB)}\frac{-du}{1+u^2}.$$Now you can reverse the first substitution! $$=2\int_0^{(B-A)/(1+AB)}\frac{-2u\;du}{(1+u^2)^2}\frac{1+u^2}{2u}=\int_1^{-1+\tfrac{2(1+AB)^2}{(1+AB)^2+(B-A)^2}}\frac{-dw}{\sqrt{1-w^2}}\tag{$u=\sqrt{\tfrac{1-w}{1+w}}$}$$ $$=-\int_1^{\tfrac{(1+AB)^2-(A-B)^2}{1+A^2B^2+A^2+B^2}}\frac{dw}{\sqrt{1-w^2}}=-\int_1^{\tfrac{(1-A^2)(1-B^2)+4AB}{[1+(1-a)/(1+a)][1+(1+b)/(1-b)]}}\frac{dw}{\sqrt{1-w^2}}$$ $$=-\int_1^{(4\sqrt{1-a^2}\sqrt{1-b^2}+2a\cdot-2b)/4}\frac{dw}{\sqrt{1-w^2}}.$$But now there's no $0$ bound! So it looks like you'll need one more substitution: $$=\int_0^{\sqrt{1-(1-a^2)(1-b^2)-a^2b^2+2ab\sqrt{(1-a^2)(1-b^2)}}}\frac1z\frac{z\;dz}{\sqrt{1-z^2}}\tag{$z=\sqrt{1-w^2}$}$$$$=\int_0^{\sqrt{a^2(1-b^2)+b^2(1-a^2)+2ab\sqrt{(1-a^2)(1-b^2)}}}\frac{dz}{\sqrt{1-z^2}}$$ $$=\int_0^{a\sqrt{1-b^2}+b\sqrt{1-a^2}}\frac{dx}{\sqrt{1-x^2}}\qquad\square$$