Question

∫xarcsinxdx

详解过程！谢谢

Answer 1

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

单独求∫√(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

Answer 2

原式=1/2∫arcsinxdx
2;
=1/2x
2;*arcsinx-1/2∫x
2;darcsinx
=用分布积分法∫xarcsinxdx=（1/2）∫arcsinx(dx^2)=（1/2）[x^2arcs

Answer 3

用分布积分法
∫xarcsinxdx=（1/2）∫arcsinx(dx^2)=（1/2）[x^2arcsinx-∫x^2*(1/x)dx])=（1/2）[x^2arcsinx-∫xdx]=（1/2）[x^2arcsinx-(1/2)x^2]+C=（1/2）x^2arcsinx-(1/4)x^2+C