It is possible to find the "middle" area, and therefore the shaded area, without calculus. Let the top point of the top triangle be $P$, and the bottom point of the bottom triangle be $Q$.
Consider the sector that goes $PAQ$, and then around the circle back to $P$.
We know the angle of that sector (it is twice your given angle), and we know the radius, so we know the area.
Double that. This is the area of our "middle" region, except that the areas of the two triangles have been counted twice.
So calculate twice the area of the sector described earlier, and subtract the sum of the areas of the triangles. These areas are not hard to find.