## Calculus: Early Transcendentals (2nd Edition)

$= {\sin ^{ - 1}}\,\left( {\frac{{\theta + 3}}{6}} \right) + C$
$\begin{gathered} \int_{}^{} {\frac{{d\theta }}{{\sqrt {27 - 6\theta - {\theta ^2}} }}} \hfill \\ \hfill \\ complete\,\,\,the\,\,square\,\,and\,\,factor \hfill \\ \hfill \\ = \int_{}^{} {\frac{{d\theta }}{{\sqrt {36 - \,\left( {9 + 6\theta + {\theta ^2}} \right)} }}} \hfill \\ \hfill \\ = \int_{}^{} {\frac{{\,\,d\theta }}{{\sqrt {{6^2} - \,{{\left( {\theta + 3} \right)}^2}} }}} \hfill \\ \hfill \\ set\,\,\,\left( {\theta + 3} \right) = t\, \to \,\,\,d\theta = dt \hfill \\ \hfill \\ \int_{}^{} {\frac{{d\theta }}{{\sqrt {27 - 6\theta - {\theta ^2}} }}} = \int_{}^{} {\frac{{dt}}{{\sqrt {{6^2} - {t^2}} }}} \hfill \\ \hfill \\ {\text{Using}}\,\,\int_{}^{} {\frac{{dx}}{{{a^2} - {x^2}}}} = \sin {\,^{ - 1}}\frac{x}{a} + C \hfill \\ \hfill \\ = {\sin ^{ - 1}}\frac{t}{6} + C \hfill \\ \hfill \\ substituting\,back\,\,t = \theta + 3 \hfill \\ \hfill \\ = {\sin ^{ - 1}}\,\left( {\frac{{\theta + 3}}{6}} \right) + C \hfill \\ \end{gathered}$