=4
如图
∫(-π->π) √[1-(sinx)^2] dx=-∫(-π->-π/2) cosx dx + ∫(-π/2->π/2) cosx dx - ∫(π/2->π) cosx dx=-[sinx](-π->-π/2) +[sinx]|(-π/2->π/2) - [sinx]|(π/2->π)=1 +2 +1 =4
