Here is the problem :
We consider two circles that intersect in exactly two points. There $O_1$ the center of the first and $r_1$ its radius. There $O_2$ the center of the second and $r_2$ its radius. We note $d=O_1O_2$. Question: Express the area of the intersection of the two disks with the distances $d$, $r_1$ and $r_2$.
My proof :
I call $ A $ and $ B $ the points of intersection of the two disks and $C$ is the intersection point of $d$ and $AB$ ($O_1C=x$).
I selected a landmark $\mathcal R (O, \vec e_x , \vec e_y )$ such that $O_1$ is centered at $(0,0)$ and $(d, 0)$ for $O_2$
The equation of the two circles are $x^2+y^2=r_1^2$ and $(x-d)^2=y^2=r_2^2$.
We have $x=\frac{d^2+r_1^2-r_2^2}{2d}$, then $$y^2=r_1^2-x^2=r_1^2-\left(\frac{d^2+r_1^2-r_2^2}{2d}\right)^2$$
Writing that $AB=2y$ we have $$AB=\frac{\sqrt{4d^2r_1^2-(d^2-r_2^2+r_1^2)^2}}{d}$$
I am going to calculate the half of the requested area.
Denote $\theta$ the angle of $AO_1B$. The half of the requested area is :
$A_1$=Area (sector)$-$Area (isosceles triangle $O_1$) =$$\frac{\pi r_1^2\theta}{2\pi}-\frac{1}{2}r_1^2\sin(\theta)$$
Furthermore $$\sin\left(\frac{\theta}{2}\right)=\frac{AB}{2r_1}=\frac{\sqrt{4d^2r_1^2-(d^2-r_2^2+r_1^2)^2}}{2dr_1}\qquad \cos\left(\frac{\theta}{2}\right)=\frac{x}{r_1}$$
Therefore $$A_1=r_1^2\arccos\left(\frac{d^2-r_1^2+r_2^2}{2dr_1}\right)-\frac{(d^2-r_1^2+r_2^2)AB}{4d}$$
Using $sin(x)=2\sin(\frac{x}2)\times \cos(\frac{x}2)$.
After some algebra to compute the request area we have $$A=r_1^2\arccos\left(\frac{d^2-r_1^2+r_2^2}{2dr_1}\right)+r_2^2\arccos\left(\frac{d^2-r_2^2+r_1^2}{2dr_2}\right)- \frac{\sqrt{Y}}{2} $$ where $$Y=(-d+r_1+r_2)(d+r_2-r_1)(d-r_2+r_1)(d+r_1+r_2)$$
My Question: Can we proved this result with integrals for exemple ?
Set the center of the first circle in $(0,0)$ and that of the second circle in $(d,0)$. The equations of the circles are $y=\sqrt{r_1^2- x^2}$ and $y=\sqrt{r_2^2-(x-d)^2}$. In the segment connecting the two centers, the first and second circle intersect the x-axis in the points $(r_1, 0)$ and $(d-r_2, 0)$, respectively. From the two equations, we also get that the intesection points between the two circles have an $x$-value of $\displaystyle h=\frac {(d^2+r_1^2-r_2^2)}{2d}$. Then, to calculate the intersection area $A$, we have to estimate the following integrals:
$$A=\displaystyle 2\int_{d-r_2}^{h} \sqrt{r_2^2-(x-d)^2} + 2\int_{h}^{r_1} \sqrt{r_1^2-x^2} $$
The first indefinite integral is $${(x-d)}\sqrt{r_2^2-(x-d)^2} + r_2^2 \tan^{-1}(\frac {x-d}{\sqrt{r_2^2-(x-d)^2}})$$
which calculated between $(d-r_2)$ and $h$ gives
$$(h-d)\sqrt{r_2^2-(h-d)^2} + r_2^2 \tan^{-1}(\frac {h-d}{\sqrt{r_2^2-(h- d) ^2}}) +\frac {\pi}{2} r_2^2 $$
The second indefinite integral is $$x\sqrt{r_1^2-x^2} + r_1^2 \tan^{-1} (\frac {x}{\sqrt{r_1^2-x^2}})$$
which calculated between $h$ and $r_1$ gives
$$-h\sqrt{r_1^2-h^2} - r_1^2 \tan^{-1}(\frac {h}{\sqrt{r_1^2-h^2}})+ \frac {\pi}{2} r_1^2$$
Summing the two results we get
$$A=(h-d)\sqrt{r_2^2-(h-d)^2} + r_2^2 \tan^{-1}(\frac {h-d}{\sqrt{r_2^2-(h- d) ^2}}) +\frac {\pi}{2} r_2^2 - h\sqrt{r_1^2-h^2} - r_1^2 \tan^{-1} (\frac {h}{\sqrt{r_1^2-h^2}})+ \frac {\pi}{2} r_1^2 $$
and substituting $\displaystyle h=\frac {d^2+r_1^2-r_2^2}{2d}$ we obtain
$$A=-\frac {d^2-r_1^2+r_2^2}{2d}\sqrt{r_2^2-(\frac {d^2-r_1^2+r_2^2}{2d}) ^2} - r_2^2 \tan^{-1} \left( \frac { (d^2-r_1^2+r_2^2)/(2d) }{\sqrt{r_2^2- (\frac {d^2-r_1^2+r_2^2}{2d})^2}} \right) +\frac {\pi}{2} r_2^2 - \frac {(d^2+r_1^2-r_2^2)}{2d} \sqrt{r_1^2-(\frac {d^2+r_1^2-r_2^2}{2d})^2} - r_1^2 \tan^{-1} \left( \frac { (d^2+r_1^2-r_2^2)/(2d) }{\sqrt{r_1^2-(\frac {d^2+r_1^2-r_2^2}{2d})^2}} \right) + \frac {\pi}{2} r_1^2 $$
Now if we set $\frac {d^2-r_1^2+r_2^2}{2d}=r_2 \sin \alpha$, the denominator of the first term containing the arctan becomes equal to $r_2 \cos \alpha$. We then can substitute the whole term with $\displaystyle r_2^2\arcsin \frac {d^2-r_1^2+r_2^2}{2dr_2}$. Similarly, we can follow the same procedure to substitute the second term containing the arctan with $\displaystyle r_1^2\arcsin \frac {d^2+r_1^2-r_2^2}{2dr_1}$.
This leads to
$$A=-\frac {(d^2-r_1^2+r_2^2)}{2d}\sqrt{r_2^2-(\frac {d^2-r_1^2+r_2^2}{2d}) ^2} - r_2^2\arcsin \frac {d^2-r_1^2+r_2^2}{2dr_2} +\frac {\pi}{2} r_2^2 - \frac {(d^2+r_1^2-r_2^2)}{2d} \sqrt{r_1^2-(\frac {d^2+r_1^2-r_2^2}{2d})^2} - r_1^2\arcsin \frac {d^2+r_1^2-r_2^2}{2dr_1} + \frac {\pi}{2} r_1^2 $$
Further moving the terms containing $\pi/2$ into those containing the $\arcsin$ we finally get
$$A=-\frac {(d^2-r_1^2+r_2^2)}{2d}\sqrt{r_2^2-(\frac {d^2-r_1^2+r_2^2}{2d}) ^2} + r_2^2\arccos \frac {d^2-r_1^2+r_2^2}{2dr_2} - \frac {(d^2+r_1^2- r_2^2)}{2d} \sqrt{r_1^2-(\frac {d^2+r_1^2-r_2^2}{2d})^2} + r_1^2\arccos \frac {d^2+r_1^2-r_2^2}{2dr_1}$$