Let I=∫(sin−1x)2dx
∫(sin−1x)2dx=∫1×(sin−1x)2dxApplying integration by parts, we have
I=x(sin−1x)2−∫x(2sin−1x√1−x2)dx
I=x(sin−1x)2−2∫sin−1x(x√1−x2)dx
Now solving for ∫sin−1x(x√1−x2)dx,
Let
u=sin−1x
v=x√1−x2⇒∫vdx=−√1−x2
∴∫sin−1x(x√1−x2)dx=sin−1x(−√1−x2)−∫−√1−x2√1−x2dx=−√1−x2sin−1x+x+C
∴I=x(sin−1x)2−2(−√1−x2sin−1x+x)+C
⇒I=x(sin−1x)2+2√1−x2sin−1x−2x+C
Hence ∫(sin−1x)2dx=x(sin−1x)2+2√1−x2sin−1x−2x+C.