Answer
$$ - \frac{{{{\sin }^4}2x\cos 2x}}{{10}} - \frac{{2{{\sin }^2}2x\cos 2x}}{{15}} - \frac{{4\cos 2x}}{{15}} + C $$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^5}2x} dx \cr
& {\text{Use the reduction formula }}\left( {67} \right) \cr
& \,\,\,\int {{{\sin }^n}ax} dx = - \frac{{{{\sin }^{n - 1}}ax\cos ax}}{{na}} + \frac{{n - 1}}{n}\int {{{\sin }^{n - 2}}ax} dx \cr
& {\text{let }}n = 5{\text{ and }}a = 2 \cr
& \,\,\,\int {{{\sin }^5}2x} dx = - \frac{{{{\sin }^{5 - 1}}2x\cos 2x}}{{5\left( 2 \right)}} + \frac{{5 - 1}}{5}\int {{{\sin }^{5 - 2}}2x} dx \cr
& \,\,\,\int {{{\sin }^5}2x} dx = - \frac{{{{\sin }^4}2x\cos 2x}}{{10}} + \frac{4}{5}\int {{{\sin }^3}2x} dx \cr
& \cr
& {\text{Integrate }}\int {{{\sin }^3}2x} dx\cr
& {\text{using the reduction formula with }}\cr
& n = 3{\text{ and }}a = 2 \cr
& \,\,\,\int {{{\sin }^5}2x} dx = - \frac{{{{\sin }^4}2x\cos 2x}}{{10}} + \frac{4}{5}\left( { - \frac{{{{\sin }^{3 - 1}}2x\cos 2x}}{{3\left( 2 \right)}} + \frac{{3 - 1}}{3}\int {{{\sin }^{3 - 2}}2x} dx} \right) \cr
& \,\,\,\int {{{\sin }^5}2x} dx = - \frac{{{{\sin }^4}2x\cos 2x}}{{10}} + \frac{4}{5}\left( { - \frac{{{{\sin }^2}2x\cos 2x}}{6} + \frac{2}{3}\int {\sin 2x} dx} \right) \cr
& \,\,\,\int {{{\sin }^5}2x} dx = - \frac{{{{\sin }^4}2x\cos 2x}}{{10}} - \frac{{2{{\sin }^2}2x\cos 2x}}{{15}} + \frac{8}{{15}}\int {\sin 2x} dx \cr
& \cr
& {\text{Use }}\int {\sin ax} dx = - \frac{1}{a}\cos ax + C \cr
& \,\,\,\int {{{\sin }^5}2x} dx = - \frac{{{{\sin }^4}2x\cos 2x}}{{10}} - \frac{{2{{\sin }^2}2x\cos 2x}}{{15}} + \frac{8}{{15}}\left( { - \frac{1}{2}\cos 2x} \right) + C \cr
& \,\,\,\int {{{\sin }^5}2x} dx = - \frac{{{{\sin }^4}2x\cos 2x}}{{10}} - \frac{{2{{\sin }^2}2x\cos 2x}}{{15}} - \frac{{4\cos 2x}}{{15}} + C \cr} $$
