Calculus: Early Transcendentals (2nd Edition)

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