Đặt $x=\sin t \Rightarrow dx =\cos t dt$
$I=\int_0^{\frac{\pi}{2}} \sin^2 t .\sqrt{1-\sin^2 t} .\cos t dt =\int \sin^2 t \cos^2 t dt=\dfrac{1}{4}\int \sin^2 2t dt$
$=\dfrac{1}{8}\int (1-\cos 4t) dt =(\dfrac{1}{8}t -\dfrac{1}{32}\sin 4t ) \bigg |_0^{\frac{\pi}{2}} =\dfrac{\pi}{16}$