Question

定积分 请问哪里错了

正确答案是1/2

Answer 1

∫(0->π) √[(sinx)^2. (1-(sinx)^2) ]
=∫(0->π/2) sinx.cosx dx - ∫(π/2->π) sinx.cosx dx
=(1/2)∫(0->π/2) sin2x dx -(1/2) ∫(π/2->π) sin2x dx
=-(1/4)[cos2x]|(0->π/2) +(1/4)[cos2x]|(π/2->π)
=1/2 +1/2
=1

追问

请问为什么要从π/2分段呢？

追答

x∈(0,π/2) => cosx >0 , √(cosx)^2 = cosx
x∈(π/2,π) => cosx <0, √(cosx)^2 = -cosx