Yikes! Integration is some high powered machinery for a geometry problem. Insight, is usually a better tool for these problems.

I got 2pi.

Let T be the area of the equilateral triangle of side 2cm made by joining the centers of the circles. Let x be the small region trapped between one side of the triangle and one of the shaded regions. (Thus the middle white area is T+3x.)

T+x is 1/6th of a circle of radius 2, or (2/3)*pi.

Next, consider the top left shaded region and make the equilateral triangle by connecting the three points of intersection of the arcs that make up the region. It should be clear that this is also an equilateral triangle of side 2cm. Even skipping the proof of congruent regions, it should be clear the shaded region is T+x.

Thus, the three shaded regions equal 3(T+x) = 2pi.