定积分应用y=sinx (0

网友回答

S=∫[1/2 ,√3/2] { (π-arcsiny) - (arcsiny)}dy
=∫[1/2 ,√3/2] (π-2arcsiny) dy
= πy|[1/2 ,√3/2] - 2*∫[1/2 ,√3/2] arcsiny dy (arcsiny = x)
= πy|[1/2 ,√3/2] - 2*∫[π/6 ,π/3] x dsinx
= [√3-1]π/2 - 2*{xsinx|[π/6 ,π/3] - ∫[π/6 ,π/3] sinxdx }
= [√3-1]π/2 - 2*{√3π/6 - π/12] +cosx|[π/6 ,π/3] }
= [√3-1]π/2 - 2*{√3π/6 - π/12] +[1-√3]/2 }