导函数是y=(1-x^2)^(1/2) 求原函数?????

∫y
=∫√(1-x^2) dx
令x=cost,则dx=sint dt.

∫√(1-x^2) dx
=∫√(1-cos^2 t) sint dt
=∫sin^2 t dt
=(1/4)∫(1-cos2t) d 2t
=t/2 -(1/4) sin2t +C
=(1/2)arccos x -(1/4)sin2arccos x +C
=(1/2)arccos x -(1/2)x·√(1-x^2) +C

1/2*x*(1-x^2)^(1/2)+1/2*asin(x)+C

你是不是要求定积分啊?那是不用求原函数的!要用数形结合.及x^2+y^2=1,是一个圆