## Calculus: Early Transcendentals (2nd Edition)

$= x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} \, + C$
$\begin{gathered} \int_{}^{} {{{\sin }^{ - 1}}x\,dx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ dv = dx\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,v = x \hfill \\ u = {\sin ^{ - 1}}x\,\, \to \,\,\,\,du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = x{\sin ^{ - 1}}x - \int_{}^{} {\frac{x}{{\sqrt {1 - {x^2}} }}\,dx} \hfill \\ \hfill \\ rewrite \hfill \\ \hfill \\ = x{\sin ^{ - 1}}x + \frac{1}{2}\int_{}^{} {\frac{{\,\left( { - 2x} \right)dx}}{{\sqrt {1 - {x^2}} }}} \hfill \\ \hfill \\ integrate\,\,{\text{using }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} \hfill \\ \hfill \\ x{\sin ^{ - 1}}x + \frac{1}{2}\frac{{\,\left( {\sqrt {1 - {x^2}} } \right)}}{{\frac{1}{2}}} + C \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} \, + C \hfill \\ \end{gathered}$