## Calculus: Early Transcendentals (2nd Edition)

$= - \frac{2}{5}{\cos ^{\frac{5}{2}}}x + \frac{2}{9}{\cos ^{\frac{9}{2}}}x + C$
$\begin{gathered} \int_{}^{} {{{\sin }^3}x{{\cos }^{\frac{3}{2}}}xdx} \hfill \\ \hfill \\ write\,\,{\sin ^3}x\,\,as\,\,{\sin ^2}xsinx \hfill \\ \hfill \\ = \int_{}^{} {{{\sin }^2}x{{\cos }^{\frac{3}{2}}}x\sin xdx} \hfill \\ \hfill \\ {\sin ^2}x = 1 - {\cos ^2}x \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {1 - {{\cos }^2}x} \right){{\cos }^{\frac{3}{2}}}x\sin xdx} \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {{{\cos }^{\frac{3}{2}}}x\sin x - {{\cos }^{\frac{7}{2}}}x\sin x} \right)} dx \hfill \\ \hfill \\ = - \int_{}^{} {{{\cos }^{\frac{3}{2}}}x\,\left( { - \sin x} \right) + \int_{}^{} {{{\cos }^{\frac{7}{2}}}} x\,\left( { - \sin x} \right)dx} \hfill \\ \hfill \\ use\,\,\int_{}^{} {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + C \hfill \\ \hfill \\ = - \frac{{{{\cos }^{\frac{5}{2}}}x}}{{\frac{5}{2}}} + \frac{{{{\cos }^{\frac{9}{2}}}x}}{{\frac{9}{2}}} + C \hfill \\ \hfill \\ solution \hfill \\ \hfill \\ = - \frac{2}{5}{\cos ^{\frac{5}{2}}}x + \frac{2}{9}{\cos ^{\frac{9}{2}}}x + C \hfill \\ \end{gathered}$