and f(x)=[x-a(1)][x-a(2)]...[x-a(n)], evaluate
(1) ∫_[0~1] [f(x)]²/[√(1-x²)]dx
(2) ∫_[0~1] x^n f(x)/[√(1-x²)] dx
更新1:
Sorry! a(k)=cos[(2k-1)π/(2n)]
