## Calculus: Early Transcendentals (2nd Edition)

See the work step by step section for the method used to obtain the answer$= - \cos x + \frac{{{{\cos }^3}x}}{3} + C$
$\begin{gathered} \int_{}^{} {{{\sin }^3}xdx} \hfill \\ \hfill \\ {\text{first}}\,\,{\text{rewrite}}\,\,{\text{the}}\,\,{\text{integrand}} \hfill \\ \hfill \\ {\sin ^3}x = \,\left( {\sin x} \right)\,\left( {{{\sin }^2}x} \right) \hfill \\ \hfill \\ now\,use \hfill \\ \hfill \\ {\sin ^2}x = 1 - {\cos ^2}x \hfill \\ \hfill \\ {\sin ^3}x = \,\left( {\sin x} \right)\,\left( {1 - {{\cos }^2}x} \right) \hfill \\ \hfill \\ Multiply \hfill \\ \hfill \\ {\sin ^3}x = \sin x - \sin x{\cos ^2}x \hfill \\ \hfill \\ integrate\,\, \hfill \\ \hfill \\ = \int_{}^{} {\sin xdx + \int_{}^{} {{{\cos }^2}x\,\left( { - \sin x} \right)dx} } \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = - \cos x + \frac{{{{\cos }^3}x}}{3} + C \hfill \\ \end{gathered}$