Question

∫(arcsinx)^2 dx (上限1,下限0) (π^2)/4 - 2

Answer 1

∵∫arcsinxdx/√(1-x²)=[(arcsinx)²]│-∫arcsinxdx/√(1-x²) (应用分部积分法)

==>2∫arcsinxdx/√(1-x²)=[(arcsinx)²]│ (把∫arcsinxdx/√(1-x²)移项)

∴∫arcsinxdx/√(1-x²)=(1/2)[(arcsinx)²]│

=(1/2)((π/2)²-0²)

=π²/8