Answer
$$\frac{1}{2}x + \frac{1}{4}\sin 2x + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cos }^2}x} dx \cr
& {\text{use the half - angle formula }}{\cos ^2}x = \frac{{1 + \cos 2x}}{2} \cr
& = \int {\left( {\frac{{1 + \cos 2x}}{2}} \right)} dx \cr
& = \int {\left( {\frac{1}{2} + \frac{{\cos 2x}}{2}} \right)} dx \cr
& {\text{slipt the integrand}} \cr
& = \int {\frac{1}{2}} dx + \int {\frac{{\cos 2x}}{2}} dx \cr
& = \frac{1}{2}\int {dx} + \frac{1}{{2\left( 2 \right)}}\int {\cos 2x\left( 2 \right)} dx \cr
& {\text{use }}\int {dx} = x + C{\text{ and }}\int {\cos u} du = \sin u + C \cr
& = \frac{1}{2}x + \frac{1}{4}\sin 2x + C \cr} $$
