Answer
$$\frac{2}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {{{\cos }^3}x} dx \cr
& {\text{split off }}{\cos ^3}at \cr
& = \int_0^{\pi /2} {{{\cos }^2}x\cos x} dx \cr
& {\text{identity }}{\sin ^2}x + {\cos ^2}x = 1 \cr
& = \int_0^{\pi /2} {\left( {1 - {{\sin }^2}x} \right)\cos x} dx \cr
& = \int_0^{\pi /2} {\left( {\cos x - {{\sin }^2}x\cos x} \right)} dx \cr
& {\text{find antiderivatives}} \cr
& = \left[ {\sin x - \frac{{{{\sin }^3}x}}{3}} \right]_0^{\pi /2} \cr
& {\text{fundamental theorem of calculus}} \cr
& = \left( {\sin \left( {\pi /2} \right) - \frac{{{{\sin }^3}\left( {\pi /2} \right)}}{3}} \right) - \left( {\sin \left( 0 \right) - \frac{{{{\sin }^3}\left( 0 \right)}}{3}} \right) \cr
& {\text{simplifying}} \cr
& = 1 - \frac{1}{3} \cr
& = \frac{{3 - 1}}{3} \cr
& = \frac{2}{3} \cr} $$
