Let $I=\int_0^{\pi / 2} \sin ^3 x d x=\int_0^{\pi / 2}\left(1-\cos ^2 x\right) \sin x d x$
$=\int_0^{\pi / 2} \sin x d x-\int_0^{\pi / 2} \cos ^2 x \sin x d x=I_1-I_2$
$I_1=\int_0^{\pi / 2} \sin x d x=-[\cos x]_0^{\pi / 2}=-(0-1)=1$
Put $\cos x=t, \Rightarrow-\sin x d x=d t \therefore I_2=-\int_1^0 t^2 d t$
when $x=0, t=1$ when $x=\frac{\pi}{2}, t=0$
$\therefore \quad I_2=\int_0^1 t^2 d t=\left[\frac{t^3}{3}\right]_0^1=\frac{1}{3}$
$\quad I=I_1-I_2=1-\frac{1}{3}=\frac{2}{3}=R . H . S$