Answer
$$\frac{1}{{16}}x - \frac{1}{{64}}\sin 4x + \frac{1}{{48}}{\sin ^3}2x + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^2}x} {\cos ^4}xdx \cr
& {\text{split co}}{{\text{s}}^4}x \cr
& = \int {{{\sin }^2}x} {\cos ^2}x{\cos ^2}xdx \cr
& = \int {{{\sin }^2}x} {\cos ^2}x{\cos ^2}xdx \cr
& {\text{using identities}} \cr
& = \int {\left( {\frac{{1 - \cos 2x}}{2}} \right)} \left( {\frac{{1 + \cos 2x}}{2}} \right)\left( {\frac{{1 + \cos 2x}}{2}} \right)dx \cr
& {\text{simplify}} \cr
& = \int {\left( {\frac{{1 - {{\cos }^2}2x}}{4}} \right)} \left( {\frac{{1 + \cos 2x}}{2}} \right)dx \cr
& = \frac{1}{8}\int {\left( {1 - {{\cos }^2}2x} \right)} \left( {1 + \cos 2x} \right)dx \cr
& {\text{use }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr
& = \frac{1}{8}\int {\left( {{{\sin }^2}2x} \right)} \left( {1 + \cos 2x} \right)dx \cr
& = \frac{1}{8}\int {\left( {{{\sin }^2}2x + {{\sin }^2}2x\cos 2x} \right)} dx \cr
& {\text{split integrand}} \cr
& = \frac{1}{8}\int {{{\sin }^2}2x} dx + \frac{1}{8}\int {{{\sin }^2}2x\cos 2x} dx \cr
& {\text{identity si}}{{\text{n}}^2}\theta = \frac{{1 - \cos 2\theta }}{2} \cr
& = \frac{1}{8}\int {\left( {\frac{{1 - \cos 4x}}{2}} \right)} dx + \frac{1}{8}\int {{{\sin }^2}2x\cos 2x} dx \cr
& = \frac{1}{8}\int {\left( {\frac{1}{2} - \frac{{\cos 4x}}{2}} \right)} dx + \frac{1}{8}\int {{{\sin }^2}2x\cos 2x} dx \cr
& = \frac{1}{8}\int {\left( {\frac{1}{2}} \right)} dx - \frac{1}{8}\int {\frac{{\cos 4x}}{2}} dx + \frac{1}{8}\int {{{\sin }^2}2x\cos 2x} dx \cr
& = \frac{1}{{16}}\int {dx} - \frac{1}{{16}}\int {\cos 4x} dx + \frac{1}{8}\int {{{\sin }^2}2x\cos 2x} dx \cr
& {\text{find the antiderivatives}} \cr
& = \frac{1}{{16}}x - \frac{1}{{16\left( 4 \right)}}\sin 4x + \frac{1}{{16\left( 3 \right)}}{\sin ^3}2x + C \cr
& = \frac{1}{{16}}x - \frac{1}{{64}}\sin 4x + \frac{1}{{48}}{\sin ^3}2x + C \cr} $$
