## Calculus: Early Transcendentals (2nd Edition)

$= \frac{1}{2}{x^2}{\sin ^{ - 1}}\,\left( {2x} \right) + \frac{x}{8}\sqrt {1 - 4{x^2}} - \frac{1}{{16}}{\sin ^{ - 1}}\,\left( {2x} \right) + C$
$\begin{gathered} \int_{}^{} {x{{\sin }^{ - 1}}2xdx} \hfill \\ \hfill \\ Integrate\,\,by\,\,tables,\,\,use\,\,the\,\,formula \hfill \\ \hfill \\ \int_{}^{} {{x^n}{{\sin }^{ - 1}}xdx} = \frac{{{x^{n + 1}}{{\sin }^{ - 1}}x}}{{n + 1}} - \frac{1}{{n + 1}}\int_{}^{} {\frac{{{x^{n + 1}}}}{{\sqrt {1 - {x^2}} }}dx} \hfill \\ \hfill \\ \int_{}^{} {\frac{{{x^2}}}{{\sqrt {{a^2} - {x^2}} }}dx} = - \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\,\left( {\frac{x}{a}} \right) + C \hfill \\ \hfill \\ Let\,\,,\,\,u = 2x\,\,\,\,\,\,then\,\,\,\,\frac{{du}}{2} = dx \hfill \\ Therefore, \hfill \\ \hfill \\ \int_{}^{} {x{{\sin }^{ - 1}}\,\left( {2x} \right)dx} = \frac{1}{4}\int_{}^{} {usi{n^{ - 1}}} \,\,udu \hfill \\ \hfill \\ = \frac{1}{4}\,\,\left[ {\frac{{{u^2}{{\sin }^{ - 1}}u}}{2} - \frac{1}{2}\int_{}^{} {\frac{{{u^2}}}{{\sqrt {1 - {u^2}} }}dx} } \right] \hfill \\ \hfill \\ = \frac{1}{4}{u^2}{\sin ^{ - 1}}u - \frac{1}{8}\,\,\left[ { - \frac{u}{2}\sqrt {1 - {u^2}} + \frac{1}{2}{{\sin }^{ - 1}}u} \right] \hfill \\ \hfill \\ substitute\,\,back\,\,u = 2x \hfill \\ \hfill \\ = \frac{1}{2}{x^2}{\sin ^{ - 1}}\,\left( {2x} \right) + \frac{x}{8}\sqrt {1 - 4{x^2}} - \frac{1}{{16}}{\sin ^{ - 1}}\,\left( {2x} \right) + C \hfill \\ \hfill \\ \,\,using\,\,a\,\,\,CAS\,\,we\,\,get\,\,the\,\,same\,\,result \hfill \\ \hfill \\ \int_{}^{} {x{{\sin }^{ - 1}}2xdx} = \frac{1}{2}{x^2}{\sin ^{ - 1}}\,\left( {2x} \right) + \frac{x}{8}\sqrt {1 - 4{x^2}} - \frac{1}{{16}}{\sin ^{ - 1}}\,\left( {2x} \right) + C \hfill \\ \end{gathered}$