Answer
$$\frac{{ - 147\sqrt 3 + 256}}{{480}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /6} {{{\sin }^5}\theta } d\theta \cr
& = \int_0^{\pi /6} {{{\sin }^4}\theta \sin \theta } d\theta \cr
& = \int_0^{\pi /6} {{{\left( {{{\sin }^2}\theta } \right)}^2}\sin \theta } d\theta \cr
& = \int_0^{\pi /6} {{{\left( {1 - {{\cos }^2}\theta } \right)}^2}\sin \theta } d\theta \cr
& {\text{expanding}} \cr
& = \int_0^{\pi /6} {\left( {1 - 2{{\cos }^2}\theta + {{\cos }^4}\theta } \right)\sin \theta } d\theta \cr
& {\text{find the antiderivative}} \cr
& = \left( { - \cos \theta + \frac{{2{{\cos }^3}\theta }}{3} - \frac{{{{\cos }^5}\theta }}{5}} \right)_0^{\pi /6} \cr
& {\text{evaluate limits}} \cr
& = \left( { - \cos \left( {\frac{\pi }{6}} \right) + \frac{{2{{\cos }^3}\left( {\frac{\pi }{6}} \right)}}{3} - \frac{{{{\cos }^5}\left( {\frac{\pi }{6}} \right)}}{5}} \right) - \left( { - \cos \left( 0 \right) + \frac{{2{{\cos }^3}\left( 0 \right)}}{3} - \frac{{{{\cos }^5}\left( 0 \right)}}{5}} \right) \cr
& = \left( { - \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{4} - \frac{{9\sqrt 3 }}{{160}}} \right) - \left( { - 1 + \frac{2}{3} - \frac{1}{5}} \right) \cr
& {\text{Simplify}} \cr
& = \frac{{ - 80\sqrt 3 + 40\sqrt 3 - 9\sqrt 3 }}{{160}} + \frac{8}{{15}} \cr
& = - \frac{{49\sqrt 3 }}{{160}} + \frac{8}{{15}} \cr
& = \frac{{ - 147\sqrt 3 + 256}}{{480}} \cr} $$
