Area of ​​the intersection of two discs: Integral solution? 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 ? 

I derived my own proof with simple geometry and trigonometry. I ended up with a slightly different formula:
$$
A=r_1^2 \text{arccos}\left(\frac{d^2-(r_2^2-r_1^2)}{2dr_1}\right) + r_2^2 \text{arccos}\left(\frac{d^2+(r_2^2-r_1^2)}{2dr_2}\right)-d\sqrt{r_1^2-\left(\frac{d^2-(r_2^2-r_1^2)}{2d}\right)^2}.
$$
The last term can be rewritten:
$$
-d\sqrt{r_1^2-\left(\frac{d^2-(r_2^2-r_1^2)}{2d}\right)^2} = -\frac{\sqrt{Y}}{2},
$$
with
$$
Y=(-d+r_1+r_2)(d+r_2-r_1)(d-r_2+r_1)(d+r_1+r_2).
$$
So this is very close to the first submitted equation, apart from a swap between r1 and r2 in the argument of the arccos(). It seems that this corrects the typo pointed out in the comment. For d=3, r1=1, r2=2 it gives 
$$
A=r_1^2 \text{arccos}\left(1\right) + r_2^2 \text{arccos}\left(1\right)+0=0
$$
