Question

求学霸解答，要详细过程喔

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
若有用，望采纳，谢谢。

Answer 2

找邓超

追问

😒😒真的很急好不好