Answer
$$y{\sin ^{ - 1}}\sqrt y - \frac{1}{2}{\sin ^{ - 1}}\sqrt y + \frac{1}{2}\sqrt y \sqrt {1 - {y^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^{ - 1}}} \sqrt y dy \cr
& \cr
& {\text{Let }}t = \sqrt y ,\,\,\,\,\,\,\,dt = \frac{1}{{2\sqrt y }}dy,\,\,\,\,\,\,2\sqrt y dt = dy \to 2tdt = dy \cr
& {\text{Write the integral in terms of }}t \cr
& \int {{{\sin }^{ - 1}}} \sqrt y dy = \int {{{\sin }^{ - 1}}} t\left( {2tdt} \right) \cr
& \int {2t{{\sin }^{ - 1}}t} dt \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}\,\,\,\,\,u = {\sin ^{ - 1}}t,\,\,\,\,du = \frac{1}{{\sqrt {1 - {t^2}} }}dt\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = 2tdt,\,\,\,\,v = {t^2} \cr
& \cr
& {\text{Integration by parts formula }}\int {udv} = uv - \int {vdu.{\text{ T}}} {\text{hen}}{\text{,}} \cr
& \int {2t{{\sin }^{ - 1}}t} dt = {t^2}{\sin ^{ - 1}}t - \int {{t^2}} \left( {\frac{1}{{\sqrt {1 - {t^2}} }}} \right)dt \cr
& \int {2t{{\sin }^{ - 1}}t} dt = {t^2}{\sin ^{ - 1}}t - \int {\frac{{{t^2}}}{{\sqrt {1 - {t^2}} }}dt} \cr
& \cr
& {\text{Integrating }}\int {\frac{{{t^2}}}{{\sqrt {1 - {t^2}} }}dt} \cr
& {\text{let }}t = \sin \theta ,\,\,\,\,\,dt = \cos \theta d\theta \cr
& \,\,\,\int {\frac{{{t^2}}}{{\sqrt {1 - {t^2}} }}dt} = \int {\frac{{{{\sin }^2}\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}\left( {\cos \theta d\theta } \right)} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {\frac{{{{\sin }^2}\theta }}{{\cos \theta }}\left( {\cos \theta d\theta } \right)} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {{{\sin }^2}\theta } d\theta \cr
& {\text{Use the identity co}}{{\text{s}}^2}\theta = \frac{1}{2} - \frac{1}{2}\cos 2\theta \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\int {\left( {1 - \cos 2\theta } \right)} d\theta \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\theta - \frac{1}{2}\left( {\frac{1}{2}\sin 2\theta } \right) + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\theta - \frac{1}{4}\sin 2\theta + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\theta - \frac{1}{2}\sin \theta \cos \theta + C \cr
& \cr
& {\text{Replace }}\theta = {\sin ^{ - 1}}t \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}{\sin ^{ - 1}}t - \frac{1}{2}\sin \left( {{{\sin }^{ - 1}}t} \right)\cos \left( {{{\sin }^{ - 1}}t} \right) + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}{\sin ^{ - 1}}t - \frac{1}{2}t\sqrt {1 - {t^2}} + C \cr
& {\text{Then}}{\text{,}} \cr
& {t^2}{\sin ^{ - 1}}t - \int {\frac{{{t^2}}}{{\sqrt {1 - {t^2}} }}dt} \cr
& = {t^2}{\sin ^{ - 1}}t - \frac{1}{2}{\sin ^{ - 1}}t + \frac{1}{2}t\sqrt {1 - {t^2}} + C \cr
& \cr
& {\text{Substitute }}t = \sqrt y \cr
& = {\left( {\sqrt y } \right)^2}{\sin ^{ - 1}}\sqrt y - \frac{1}{2}{\sin ^{ - 1}}\sqrt y + \frac{1}{2}\sqrt y \sqrt {1 - {{\left( {\sqrt y } \right)}^2}} + C \cr
& = y{\sin ^{ - 1}}\sqrt y - \frac{1}{2}{\sin ^{ - 1}}\sqrt y + \frac{1}{2}\sqrt y \sqrt {1 - {y^2}} + C \cr} $$
