How to compute the area of the shadow? 
If we can not use the integral, then how to compute the area of the shadow?
It seems easy, but actually not?
Thanks!

Assume a square of side length $2$ for simplicity.
Denote the various sub-areas as $A, B, \ldots, H$ from top to bottom, left to right. So, the shaded area is $C$, the small disk is $C+D=\pi$, the big quarter disk is $\pi = A+E+G+H$.
Note that $A+B=F+H=D=G=1-\frac\pi 4$.
We locate the right vertex of $A$, measured from the center by solving $x^2+y^2=1$ and $(1+y)^2+(1+x)^2=4$, which implies 
$$x+y=\frac{(1+y^2)+(1+x)^2-(x^2+y^2)-2}2=\frac12$$
and then $x^2+(\frac12-x)^2=1$, hence $x=\frac{1\pm\sqrt 7}4$ and by sign consideration, 
$$ x=\frac{1-\sqrt 7}4, \quad y=\frac{1+\sqrt 7}4.$$
Now $B$ can be computet as the difference of a rectangle $1\times (1-y) =\frac{3-\sqrt 7}4$ and thow halves of circular segments.
But for these segments (using e.g. the formula $ r^2\arccos(1-\frac hr)-(r-h)\sqrt{(2r-h)h}$ for the area of a segment), we need to compute some angles!
I doubt that anything better than the $\arccos$ expression can be found
for these angles (i.e. they are not a rational multiple of $\pi$).
Having thus obtained an expression 
$$B=a+b\sqrt 7+c\arccos\frac{1+\sqrt 7}4+d\arccos\frac{5+\sqrt 7}8$$ 
with $a,b,c,d$ rational, we finally find
$$C = A+G+H = 3G-2B = 3-\frac34\pi -2B. $$
