# Science:Math Exam Resources/Courses/MATH105/April 2017/Question 01 (h)/Solution 1

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