反常函数(0,1)x눀arcsinxdx⼀根号下1-x눀

2025-06-23 09:58:51
推荐回答(2个)
回答1:

回答2:

原式=-∫1/2*√(1-x²)arcsinxd(1-x²)
=-∫1/2*2/3arcsinxd(1-x²)^(3/2)
=-1/3*(1-x²)^(3/2)*arcsinx+1/3∫(1-x²)^(3/2)darcsinx
=-1/3*(1-x²)^(3/2)*arcsinx+1/3∫(1-x²)^(3/2)*1/√(1-x²)dx
=-1/3*(1-x²)^(3/2)*arcsinx+1/3∫(1-x²)dx
=-1/3*(1-x²)*√(1-x²)*arcsinx+x/3-x³/9+C