Answer
\[\begin{gathered}
= 2{x^2}{\cos ^{ - 1}}\,\left( {10x} \right) - \frac{x}{{10}}\sqrt {1 - 100{x^2}} + \frac{1}{{100}}{\sin ^{ - 1}}\,\left( {10x} \right) + C \hfill \\
\hfill \\
\end{gathered} \]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {4x{{\cos }^{ - 1}}\,\left( {10x} \right)dx} \hfill \\
\hfill \\
Integrate\,\,by\,\,tables,\,\,use\,\,the\,\,formula \hfill \\
\hfill \\
\int_{}^{} {{x^n}{{\cos }^{ - 1}}xdx} = \frac{{{x^{n + 1}}{{\cos }^{ - 1}}x}}{{n + 1}} + \frac{1}{{n + 1}}\int_{}^{} {\frac{{{x^{n + 1}}}}{{\sqrt {1 - {x^2}} }}dx} \hfill \\
\hfill \\
\int_{}^{} {\frac{{{x^2}}}{{\sqrt {{a^2} - {x^2}} }}\,dx} = - \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\,\left( {\frac{x}{a}} \right) + C \hfill \\
\hfill \\
set\,\,,\,\,u = 10x\,\,\,\,\,\,then\,\,\,\frac{{du}}{{10}} = dx \hfill \\
\hfill \\
\int_{}^{} {4x{{\cos }^{ - 1}}\,\left( {10x} \right)dx} = \frac{4}{{100}}\int_{}^{} {u{{\cos }^{ - 1}}\,udu} \hfill \\
\hfill \\
= \frac{4}{{100}}\,\,\left[ {\frac{{{u^2}{{\cos }^{ - 1}}u}}{2} + \frac{1}{2}\int_{}^{} {\frac{{{u^2}}}{{\sqrt {1 - {u^2}} }}\,dx} } \right] \hfill \\
\hfill \\
= \frac{2}{{100}}{u^2}{\cos ^{ - 1}}u + \frac{1}{{50}}\,\,\left[ { - \frac{u}{2}\sqrt {1 - {u^2}} + \frac{1}{2}{{\sin }^{ - 1}}u} \right] \hfill \\
\hfill \\
substitute\,\,back\,\,u = 10x \hfill \\
\hfill \\
\int_{}^{} {4x{{\cos }^{ - 1}}\,\left( {10x} \right)dx} \, = 2{x^2}{\cos ^{ - 1}}\,\left( {10x} \right) - \frac{x}{{10}}\sqrt {1 - 100{x^2}} + \frac{1}{{100}}{\sin ^{ - 1}}\,\left( {10x} \right) + C \hfill \\
\hfill \\
\end{gathered} \]
