∫tan (sin−1x)1−x2 dx =
tan2(sin−1x)2+C
log|(sin−1x)|+C
−12log (1−x2)+C
log|1+x2|+C
sin−1x=t⇒11−x2dx=dt
∫tant dt=log|sect|+C
=−12log(1−x2)+C