Answer
$$\frac{x}{{16}} - \frac{1}{{64}}\sin 4x + \frac{{{{\sin }^3}2x}}{{48}} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^2}x{{\cos }^4}x} dx \cr
& {\text{use half - angle formulas }} \cr
& = \int {\left( {\frac{{1 - \cos 2x}}{2}} \right)} {\left( {\frac{{1 + \cos 2x}}{2}} \right)^2}dx \cr
& {\text{sxpand}} \cr
& = \int {\left( {\frac{{1 - \cos 2x}}{2}} \right)} \left( {\frac{{1 + 2\cos 2x + {{\cos }^2}2x}}{4}} \right)dx \cr
& = \frac{1}{8}\int {\left( {1 - \cos 2x} \right)} \left( {1 + 2\cos 2x + {{\cos }^2}2x} \right)dx \cr
& = \frac{1}{8}\int {\left( {1 + 2\cos 2x + {{\cos }^2}2x - \cos 2x - 2{{\cos }^2}2x - {{\cos }^3}2x} \right)} dx \cr
& = \frac{1}{8}\int {\left( {1 + \cos 2x - {{\cos }^2}2x - {{\cos }^3}2x} \right)} dx \cr
& {\text{split and rewrite}} \cr
& = \frac{1}{8}\int {\left( {1 + \cos 2x - \frac{{1 + \cos 4x}}{2} - {{\cos }^2}2x\cos 2x} \right)} dx \cr
& = \frac{1}{8}\int {\left( {1 + \cos 2x - \frac{1}{2} - \frac{{\cos 4x}}{2} - \left( {1 - {{\sin }^2}2x} \right)\cos 2x} \right)} dx \cr
& = \frac{1}{8}\int {\left( {\frac{1}{2} + \cos 2x - \frac{{\cos 4x}}{2} - \cos 2x + {{\sin }^2}2x\cos 2x} \right)} dx \cr
& = \frac{1}{8}\int {\left( {\frac{1}{2} - \frac{{\cos 4x}}{2} + {{\sin }^2}2x\cos 2x} \right)} dx \cr
& = \frac{1}{8}\int {\left( {\frac{1}{2} - \frac{{\cos 4x\left( 4 \right)}}{{2\left( 4 \right)}} + \frac{1}{2}{{\sin }^2}2x\left( 2 \right)\cos 2x} \right)} dx \cr
& {\text{evaluate the integral}} \cr
& = \frac{1}{8}\left( {\frac{x}{2} - \frac{1}{8}\sin 4x + \frac{{{{\sin }^3}2x}}{6}} \right) + C \cr
& = \frac{x}{{16}} - \frac{1}{{64}}\sin 4x + \frac{{{{\sin }^3}2x}}{{48}} + C \cr} $$
