We use the substitution u = arcsin y , v = y {\displaystyle u=\arcsin y,v=y} , then u ′ = 1 1 − y 2 , v ′ = 1. {\displaystyle u^{\prime }={\frac {1}{\sqrt {1-y^{2}}}},v^{\prime }=1.}
Then using integration by parts we have ∫ arcsin y d y = y arcsin y − ∫ y 1 − y 2 d y {\displaystyle \int \arcsin ydy=y\arcsin y-\int {\frac {y}{\sqrt {1-y^{2}}}}dy} .
The second integral on the right can be evaluated as ∫ y 1 − y 2 d y = − 1 − y 2 + C {\displaystyle \int {\frac {y}{\sqrt {1-y^{2}}}}dy=-{\sqrt {1-y^{2}}}+C} for some constant C {\displaystyle C} .
Thus the answer is y arcsin y + 1 − y 2 + C , where C is a constant. {\displaystyle \color {blue}y\arcsin y+{\sqrt {1-y^{2}}}+C,{\text{ where }}C{\text{ is a constant.}}}
, then 
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for some constant 
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