Answer
$$\frac{{3\pi }}{{16}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {{{\cos }^4}x} dx \cr
& {\text{Write the integrand as }}{\left( {{{\cos }^2}x} \right)^2} \cr
& = \int_0^{\pi /2} {{{\left( {{{\cos }^2}x} \right)}^2}} dx \cr
& {\text{Use the identity co}}{{\text{s}}^2}x = \frac{{1 + \cos 2x}}{2} \cr
& = \int_0^{\pi /2} {{{\left( {\frac{{1 + \cos 2x}}{2}} \right)}^2}} dx \cr
& {\text{Expand }} \cr
& = \frac{1}{4}\int_0^{\pi /2} {\left( {1 + 2\cos 2x + {{\cos }^2}2x} \right)} dx \cr
& = \frac{1}{4}\int_0^{\pi /2} {\left( {1 + 2\cos 2x + \frac{{1 + \cos 4x}}{2}} \right)} dx \cr
& = \frac{1}{4}\int_0^{\pi /2} {\left( {1 + 2\cos 2x + \frac{1}{2} + \frac{1}{2}\cos 4x} \right)} dx \cr
& = \frac{1}{4}\int_0^{\pi /2} {\left( {\frac{3}{2} + 2\cos 2x + \frac{1}{2}\cos 4x} \right)} dx \cr
& {\text{Integrate}} \cr
& = \frac{1}{4}\left[ {\frac{3}{2}x + \sin 2x + \frac{1}{8}\sin 4x} \right]_0^{\pi /2} \cr
& {\text{Evaluate}} \cr
& = \frac{1}{4}\left[ {\frac{3}{2}\left( {\frac{\pi }{2}} \right) + \sin 2\left( {\frac{\pi }{2}} \right) + \frac{1}{8}\sin 4\left( {\frac{\pi }{2}} \right)} \right] - \frac{1}{4}\left[ {\frac{3}{2}\left( 0 \right) + \sin 0 + \frac{1}{8}\sin 0} \right] \cr
& = \frac{1}{4}\left[ {\frac{{3\pi }}{4} + 0} \right] - \frac{1}{4}\left[ 0 \right] \cr
& = \frac{{3\pi }}{{16}} \cr} $$
