Answer
\[ = - \cos x + \frac{{2{{\cos }^3}x}}{3} - \frac{{{{\cos }^5}x}}{5} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\sin }^5}\,\,xdx} \hfill \\
\hfill \\
Split\,\,{\sin ^5}x\,\,as\,{\left( {{{\sin }^2}x} \right)^2}\,\,\sin x \hfill \\
\hfill \\
then \hfill \\
= \int_{}^{} {\,{{\left( {{{\sin }^2}x} \right)}^2}\sin xdx} \hfill \\
\hfill \\
use\,\,{\sin ^2}x = 1 - {\cos ^2}x \hfill \\
\hfill \\
= \int_{}^{} {\,{{\left( {1 - {{\cos }^2}x} \right)}^2}\sin xdx} \hfill \\
\hfill \\
expand\,\,{\left( {1 - {{\cos }^2}x} \right)^2} \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {1 - 2{{\cos }^2}x + {{\cos }^4}x} \right)\sin xdx} \hfill \\
\hfill \\
set{\text{ }}u = \cos x \to du = - \sin x \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {1 - 2{u^2} + {u^4}} \right)\left( { - 1} \right)du} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
- u + \frac{{2{u^3}}}{3} - \frac{{{u^5}}}{5} + C \hfill \\
\hfill \\
substitute\,\,back\,\,u = \cos x \hfill \\
\hfill \\
= - \cos x + \frac{{2{{\cos }^3}x}}{3} - \frac{{{{\cos }^5}x}}{5} + C \hfill \\
\end{gathered} \]
