Answer
$$\frac{1}{8}x - \frac{1}{{32}}\sin 4x + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^2}x{{\cos }^2}x} dx \cr
& = \frac{1}{4}\int {4{{\sin }^2}x{{\cos }^2}x} dx \cr
& {\text{Use }}{\left( {ab} \right)^n} = {a^n}{b^n} \cr
& = \frac{1}{4}\int {{{\left( {2\sin x\cos x} \right)}^2}} dx \cr
& {\text{By the trigonometric identity }}\sin 2x = 2\sin x\cos x \cr
& = \frac{1}{4}\int {{{\left( {\sin 2x} \right)}^2}} dx \cr
& {\text{Use }}{\sin ^2}\theta = \frac{1}{2} - \frac{1}{2}\cos 2\theta \cr
& = \frac{1}{4}\int {\left( {\frac{1}{2} - \frac{1}{2}\cos 4x} \right)} dx \cr
& = \frac{1}{8}\int {dx} - \frac{1}{8}\int {\cos 4x} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{8}x - \frac{1}{{32}}\sin 4x + C \cr} $$
