∫sin−1⁡x1−x23/2dx is equal to

Let I=∫sin−1⁡x1−x21−x2dxPut⁡sin−1⁡x=t⇒11−x2=dtdx⇒dx=1−x2dt∴I=∫t1−sin2⁡t1−x2⋅1−x2dt=∫t1−sin2⁡tdt=∫  tsec2⁡dt ∵1−sin2⁡t=cos2⁡t=t∫sec2⁡tdt−∫ddtt∫sec2⁡tdtdt                                                               [integration by parts] =ttan⁡t−∫tan⁡tdt=ttan⁡t+log⁡cos⁡t+C=sin−1⁡xx1−x2+log⁡1−x2+C [fromfigure tan⁡t=x1−x2 and cos⁡t=1−x2 ]