Let $r=\sqrt{2-\sqrt2}$. A large semicircle of radius $1+r$ contains three smaller semicircles on its diameter. The middle semicircle has radius $1$ and is centered at $O$; the other two have radius $r$ and are centered at the endpoints $C$ and $F$ of the diameter of the unit circle centered at $O$, as shown. What is the area inside the large semicircle and outside the three smaller semicircles?

1. Let $L$ be the area where one outer semicircle overlaps the middle semicircle. The corresponding full circles have radii $r$ and $1$ and centers one unit apart. The cosine rule gives the two sector angles $\cos^{-1}(r/2)=3\pi/8$ and $\cos^{-1}(1/\sqrt2)=\pi/4$; the kite area between their radii is $\sqrt2/2$. By symmetry across the diameter, the full-circle lens has area $2L=\frac{3\pi r^2}{8}+\frac\pi4-\frac{\sqrt2}{2}$.
2. Subtracting the three semicircle areas from the large one and adding back the two overlaps gives
   $$[\text{shaded}]=\frac\pi2\big((1+r)^2-(1+2r^2)\big)+2L.$$
3. Using $r^2=2-\sqrt2$ and the expression for $2L$, this simplifies to
   $$[\text{shaded}]=\boxed{\pi\left(\sqrt{2-\sqrt2}+\frac{\sqrt2}{8}\right)-\frac{\sqrt2}{2}}.$$