Answer
$$\frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2} - 1$$
Work Step by Step
$$\eqalign{
& \int_0^{1/\sqrt 2 } {2x{{\sin }^{ - 1}}\left( {{x^2}} \right)} dx \cr
& {\text{Using the integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}\,\,\,\,\,u = {\sin ^{ - 1}}\left( {{x^2}} \right),\,\,\,\,du = \frac{{2x}}{{\sqrt {1 - {{\left( {{x^2}} \right)}^2}} }}\,\,\,dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = 2xdx,\,\,\,\,v = {x^2} \cr
& \cr
& {\text{Integration by parts formula }}\int {udv} = uv - \int {vdu.{\text{ T}}} {\text{hen}}{\text{,}} \cr
& \int {2x{{\sin }^{ - 1}}\left( {{x^2}} \right)} dx = {x^2}{\sin ^{ - 1}}\left( {{x^2}} \right) - \int {{x^2}} \frac{{2x}}{{\sqrt {1 - {{\left( {{x^2}} \right)}^2}} }}\,\,\,dx \cr
& \int {2x{{\sin }^{ - 1}}\left( {{x^2}} \right)} dx = {x^2}{\sin ^{ - 1}}\left( {{x^2}} \right) - \int {\frac{{2{x^3}}}{{\sqrt {1 - {x^4}} }}} \,dx \cr
& \int {2x{{\sin }^{ - 1}}\left( {{x^2}} \right)} dx = {x^2}{\sin ^{ - 1}}\left( {{x^2}} \right) + \frac{1}{2}\int {{{\left( {1 - {x^4}} \right)}^{ - 1/2}}\left( { - 4{x^3}} \right)} \,dx \cr
& {\text{using the power rule for integration}} \cr
& = {x^2}{\sin ^{ - 1}}\left( {{x^2}} \right) + \frac{1}{2}\left( {\frac{{{{\left( {1 - {x^4}} \right)}^{1/2}}}}{{1/2}}} \right) + C \cr
& = {x^2}{\sin ^{ - 1}}\left( {{x^2}} \right) + \sqrt {1 - {x^4}} + C \cr
& \cr
& \int_0^{1/\sqrt 2 } {2x{{\sin }^{ - 1}}\left( {{x^2}} \right)} dx = \left( {{x^2}{{\sin }^{ - 1}}\left( {{x^2}} \right) + \sqrt {1 - {x^4}} } \right)_0^{1/\sqrt 2 } \cr
& {\text{Evaluating}} \cr
& = \left( {{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}{{\sin }^{ - 1}}\left( {{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right) + \sqrt {1 - {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^4}} } \right) - \left( {{{\left( 0 \right)}^2}{{\sin }^{ - 1}}\left( {{{\left( 0 \right)}^2}} \right) + \sqrt {1 - {{\left( 0 \right)}^4}} } \right) \cr
& = \left( {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right) + \sqrt {1 - \frac{1}{4}} } \right) - \left( 1 \right) \cr
& = \frac{1}{2}\left( {\frac{\pi }{6}} \right) + \sqrt {\frac{3}{4}} - 1 \cr
& = \frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2} - 1 \cr} $$
