## Calculus: Early Transcendentals (2nd Edition)

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