## Calculus: Early Transcendentals (2nd Edition)

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