求不定积分∫[√(1-x ²)] arcsinx dx.
设arcsinx=u,则x=sinu,dx=cosudu,代入原式得：
原式=∫[√(1-sin²u)]ucosudu=∫ucos²udu=(1/2)∫u(1+cos2u)du=(1/2)[∫udu+∫ucos2udu]
=(1/2)[u²/2+(1/2)∫udsin(2u)]=u²/4+(1/4)[usin2u-∫sin2udu]=u²/4+(1/4)usin2u-(1/8)∫sin2ud(2u)
=u²/4+(1/4)u(2sinucosu)+(1/8)cos2u+C=(1/4)(arcsinx)²+(1/2)(arcsinx)[x√(1-x²)]+(1/8)(1-2x²)+C
其中sinucosu=sinu√(1-sin²u)=x√(1-x²)；cos2u=1-2sin²u=1-2x².