Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /3} {{{\sin }^4}3x{{\cos }^3}3x} dx \cr
& {\text{split off }}{\cos ^3}3t \cr
& = \int_0^{\pi /3} {{{\sin }^4}3x{{\cos }^2}3x} \cos 3xdx \cr
& {\text{identity }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr
& = \int_0^{\pi /3} {{{\sin }^4}3x\left( {1 - {{\sin }^2}3x} \right)} \cos 3xdx \cr
& = \int_0^{\pi /3} {\left( {{{\sin }^4}3x - {{\sin }^6}3x} \right)} \cos 3xdx \cr
& = \frac{1}{3}\int_0^{\pi /3} {\left( {{{\sin }^4}3x - {{\sin }^6}3x} \right)} \cos 3x\left( 3 \right)dx \cr
& {\text{find antiderivatives}} \cr
& = \frac{1}{3}\left[ {\frac{{{{\sin }^5}3x}}{5} - \frac{{{{\sin }^7}3x}}{7}} \right]_0^{\pi /3} \cr
& {\text{fundamental theorem of calculus}} \cr
& = \frac{1}{3}\left( {\frac{{{{\sin }^5}3\left( {\pi /3} \right)}}{5} - \frac{{{{\sin }^7}3\left( {\pi /3} \right)}}{7}} \right) - \frac{1}{3}\left( {\frac{{{{\sin }^5}3\left( 0 \right)}}{5} - \frac{{{{\sin }^7}3\left( 0 \right)}}{7}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{3}\left( {0 - 0} \right) - \frac{1}{3}\left( 0 \right) \cr
& = 0 \cr} $$
