原式=1/2∫arcsinxdx²

=1/2x²*arcsinx-1/2∫x²darcsinx

=1/2x²*arcsinx-1/2∫x²/√(1-x²)dx

=1/2x²*arcsinx+1/2∫-x²/√(1-x²)dx

=1/2x²*arcsinx+1/2∫(1-x²-1)/√(1-x²)dx

=1/2x²*arcsinx+1/2∫[(1-x²)/√(1-x²)-1/√(1-x²)]dx

=1/2x²*arcsinx+1/2∫[√(1-x²)-1/√(1-x²)]dx

=1/2x²*arcsinx+1/2∫√(1-x²)dx-arcsinx

單獨求∫√(1-x²)dx

令x=sina

√(1-x²)=cosa

sin2a=2sinacosa=2x√(1-x²)

dx=cosada

∫√(1-x²)dx

=∫cosa*cosada

=∫(1+cos2a)/2 da

=1/2∫da+1/4∫cos2ad2a

=a/2+sin2a/4

=arcsinx/2+2x√(1-x²)/4

=arcsinx/2+x√(1-x²)/2

所以原式=1/2x²*arcsinx+(arcsinx)/4+x√(1-x²)/4-arcsinx+C