Question

y=根号下（1-x的平方）的定积分求原函数

Answer 1

F(x)=∫ydx＝∫√(1-x^2)dx

令x=sint,则√(1-x^2)=cost,dx=costdt,

从而∫√(1-x^2)dx=∫cost^2dt=∫[(1+cos2t)/2]dt=∫(1/2)dt+∫[(cos2t)/2]dt

=t/2+(sin2t)/4+c=t/2+sint*cost/2+c=(arcsinx)/2+[x*√(1-x^2)]/2+c