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