Answer
$$\frac{4}{{105}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {{{\cos }^5}2x{{\sin }^2}2x} dx \cr
& {\text{substitute }}u = 2x,{\text{ }}du = 2dx \cr
& = \int {{{\cos }^5}u{{\sin }^2}u} \left( {\frac{1}{2}} \right)du \cr
& = \frac{1}{2}\int {{{\cos }^5}u{{\sin }^2}u} du \cr
& = \frac{1}{2}\int {{{\cos }^4}u{{\sin }^2}u} \cos udu \cr
& = \frac{1}{2}\int {{{\left( {1 - {{\sin }^2}u} \right)}^2}{{\sin }^2}u} \cos udu \cr
& = \frac{1}{2}\int {\left( {1 - 2{{\sin }^2}u + {{\sin }^4}u} \right){{\sin }^2}u} \cos udu \cr
& = \frac{1}{2}\int {\left( {1 - 2{{\sin }^2}u + {{\sin }^4}u} \right){{\sin }^2}u} \cos udu \cr
& = \frac{1}{2}\int {\left( {{{\sin }^2}u - 2{{\sin }^4}u + {{\sin }^6}u} \right)} \cos udu \cr
& {\text{find the antiderivative}} \cr
& = \frac{1}{2}\left[ {\frac{{{{\sin }^3}u}}{3} - \frac{{2{{\sin }^5}u}}{5} + \frac{{{{\sin }^7}u}}{7}} \right] + C \cr
& = \frac{1}{2}\left[ {\frac{{{{\sin }^3}2x}}{3} - \frac{{2{{\sin }^5}2x}}{5} + \frac{{{{\sin }^7}2x}}{7}} \right] + C \cr
& \int_0^{\pi /4} {{{\cos }^5}2x{{\sin }^2}2x} dx = \frac{1}{2}\left[ {\frac{{{{\sin }^3}2x}}{3} - \frac{{2{{\sin }^5}2x}}{5} + \frac{{{{\sin }^7}2x}}{7}} \right]_0^{\pi /4} \cr
& {\text{evaluate limits}} \cr
& = \frac{1}{2}\left[ {\frac{{{{\sin }^3}2\left( {\pi /4} \right)}}{3} - \frac{{2{{\sin }^5}2\left( {\pi /4} \right)}}{5} + \frac{{{{\sin }^7}2\left( {\pi /4} \right)}}{7}} \right] - \frac{1}{2}\left[ {0 - 0 + 0} \right] \cr
& {\text{simplify}} \cr
& = \frac{1}{2}\left[ {\frac{1}{3} - \frac{2}{5} + \frac{1}{7}} \right] \cr
& = \frac{4}{{105}} \cr} $$
