## Calculus: Early Transcendentals (2nd Edition)

$= \frac{1}{8}x - \frac{1}{{32}}\sin 4x + C$
$\begin{gathered} \int_{}^{} {{{\sin }^2}x{{\cos }^2}x} \,dx \hfill \\ \hfill \\ {\text{Using}}\,\,{\text{the }}\,{\text{identities}} \hfill \\ \hfill \\ {\sin ^2}\theta = \frac{{1 - \cos 2\theta }}{2}\,\,and\,\,\,{\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2} \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {\frac{{1 - \cos \,2x}}{2}} \right)\,\left( {\frac{{1 + \cos 2x}}{2}} \right)dx} \hfill \\ \hfill \\ multiply\,\,and\,distribute \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {\frac{{1 - {{\cos }^2}2x}}{4}} \right)dx} \hfill \\ \hfill \\ = \int_{}^{} {\,\frac{1}{4}dx} - \frac{1}{4}\int_{}^{} {{{\cos }^2}2xdx} \hfill \\ \hfill \\ use\,the\,half\, - \,angle\,\,\,identity\,\,{\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2}\,\,,\, \hfill \\ \hfill \\ = \int_{}^{} {\,\frac{1}{4}dx} - \frac{1}{4}\int_{}^{} {\frac{{1 + \cos 4x}}{2}} dx \hfill \\ \hfill \\ or \hfill \\ \hfill \\ = \int_{}^{} {\,\frac{1}{4}dx} - \frac{1}{4}\int_{}^{} {\frac{1}{2}} dx - \frac{1}{4}\int_{}^{} {\frac{{\cos 4x}}{2}} dx \hfill \\ \hfill \\ \hfill \\ = \int_{}^{} {\,\frac{1}{4}dx} - \int {\frac{1}{8}} dx - \frac{1}{{32}}\int {\cos 4x} \left( 4 \right)dx \hfill \\ \hfill \\ integrate\,\,and\,simplify \hfill \\ \hfill \\ \frac{1}{4}x - \frac{1}{8}x - \frac{1}{{32}}\sin 4x + C \hfill \\ \hfill \\ = \frac{1}{8}x - \frac{1}{{32}}\sin 4x + C \hfill \\ \end{gathered}$