Answer
$ 4 \over 3$
Work Step by Step
$cos^2x sin x = sin x$
$cos^2x sin x - sin x = 0$
$sin x (cos^2 x -1)=0$
$sin x(-sin^2) = 0$
$sin x = 0$
$x = 0 or \pi$
$A = \int^\pi _0 (sin x - cos^2 x sin x)dx = [-cos x +{{1} \over {3}} cos^2 x]^\pi _0$
$(1-{1 \over 3} )-(-1+{1\over3}) = {4\over3}$