定积分 求1+cos^2x开根号0到π的积分

let
u=π-x
du=-dx
x=π/2, u=π/2
x=π, u=0
I
=∫(0->π) √[1+(cosx)^2] dx
=∫(0->π/2) √[1+(cosx)^2] dx + ∫(π/2->π) √[1+(cosx)^2] dx
=∫(0->π/2) √[1+(cosx)^2] dx + ∫(π->0) √[1+(cosu)^2] (-du)
=∫(0->π/2) √[1+(cosx)^2] dx + ∫(0->π) √[1+(cosx)^2] dx
=2∫(0->π/2) √[1+(cosx)^2] dx本回答被网友采纳

x=π, u=0 I =∫(0->π) √[1+(cosx)^2] dx =∫(0->π/2) √[1+(cosx)^2] dx + ∫(π/2->π) √[1+(cosx)^2] dx =∫(0->π/2) √[1+(cosx)^2] dx + ∫(π->0) √[1+(cosu...