We first write ∫ arcsin ( x ) d x = ∫ 1 ⋅ arcsin ( x ) d x {\displaystyle \int \arcsin(x)\,dx=\int 1\cdot \arcsin(x)\,dx} and use integration by parts.
Thus let u = arcsin ( x ) {\displaystyle u=\arcsin(x)} and d v = 1 ⋅ d x . {\displaystyle dv=1\cdot dx.} We then have d u = 1 1 − x 2 d x {\displaystyle du={\frac {1}{\sqrt {1-x^{2}}}}\,dx} and v = x . {\displaystyle v=x.}
Applying integration by parts then gives
In order to evaluate the resulting integral, we use the substitution u = 1 − x 2 , d u = − 2 x d x , {\displaystyle u=1-x^{2},\,du=-2x\,dx,} obtaining
Hence ∫ arcsin ( x ) d x = x arcsin ( x ) + 1 − x 2 + C . {\displaystyle \int \arcsin(x)\,dx=\color {blue}x\arcsin(x)+{\sqrt {1-x^{2}}}+C.}
and use integration by parts.
and 
We then have 
and 
obtaining
