## Calculus: Early Transcendentals (2nd Edition)

$= - \,\left( {\frac{{\sqrt {9 - {x^2}} }}{x}} \right) - {\sin ^{ - 1}}\,\left( {\frac{x}{3}} \right) + C$
$\begin{gathered} \int_{}^{} {\frac{{\sqrt {9 - {x^2}} }}{{{x^2}}}dx} \hfill \\ \hfill \\ the\,\,\,integral\,contains\,\,the\,\,form\,\,{x^2} - {a^2}\,\,so. \hfill \\ \hfill \\ x = \sin \theta \,\,\,then\,\,\,dx = 3\cos \theta d\theta \hfill \\ and\,\,\sqrt {9 - {x^2}} = 3\cos \theta \hfill \\ \hfill \\ Substituting{\text{ }}these{\text{ }}factors{\text{ }} \hfill \\ \hfill \\ = \int_{}^{} {\frac{{3\cos \theta }}{{9{{\sin }^2}\theta }}} \,\left( {3\cos \theta } \right)d\theta \,\, = \int_{}^{} {\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}d\theta } \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_{}^{} {{{\cot }^2}\theta d\theta } = \int_{}^{} {\,\left( {{{\csc }^2}\theta - 1} \right)d\theta } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = - \cot \theta - \theta + C \hfill \\ \hfill \\ substitute\,\,for\,\,\cot \theta {\text{ and }}\theta \hfill \\ \hfill \\ = - \,\left( {\frac{{\sqrt {9 - {x^2}} }}{x}} \right) - {\sin ^{ - 1}}\,\left( {\frac{x}{3}} \right) + C \hfill \\ \end{gathered}$