## Calculus: Early Transcendentals (2nd Edition)

$= {\sin ^{ - 1}}\,\left( {\frac{{x + 1}}{2}} \right) + C$
$\begin{gathered} \int_{}^{} {\frac{{dx}}{{\sqrt {3 - 2x - {x^2}} }}} \hfill \\ \hfill \\ completing\,the\,square \hfill \\ \hfill \\ 3 - 2x - {x^2} = - \,\left( {{x^2} + 2x - 3} \right) \hfill \\ = - \,\left( {{x^2} + 2x + 1 - 4} \right) \hfill \\ = 4 - \,\left( {{x^2} + 2x + 1} \right) \hfill \\ = 4 - \,{\left( {x + 1} \right)^2} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{\sqrt {3 - 2x - {x^2}} }}} = \int_{}^{} {\frac{{dx}}{{\sqrt {4 - \,{{\left( {x + 1} \right)}^2}} }}} \hfill \\ \hfill \\ the\,\,\,integral\,contains\,\,the\,\,form\,\,\,{a^2} - {u^2}\, \hfill \\ \hfill \\ set{\text{ }}\sin \theta = \frac{{x + 1}}{2}{\text{ and }}x = 2\sin \theta - 1 \hfill \\ \hfill \\ then \hfill \\ \hfill \\ dx = 2\cos \theta d\theta \hfill \\ \hfill \\ \sqrt {3 - 2x - {x^2}} = 2\cos \theta \hfill \\ \hfill \\ Substituting{\text{ }}these{\text{ }}factors{\text{ }} \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{\sqrt {4 - \,{{\left( {x + 1} \right)}^2}} }}} = \int_{}^{} {\frac{{2\cos \theta d\theta }}{{2\,\cos \theta }}} = \int_{}^{} {d\theta } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \theta + C \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = {\sin ^{ - 1}}\,\left( {\frac{{x + 1}}{2}} \right) + C \hfill \\ \end{gathered}$