Question

定积分22题不明白最后一步怎么直接等于Π/2了？





Answer 1

(22)
(π/2)∫(0->π) √[(cosx)^2-(cosx)^4] dx
=(π/2)∫(0->π) |cosx|sinx dx
=(π/2)[∫(0->π/2) cosx.sinx dx -∫(π/2->π) cosx.sinx dx ]
=(π/2){ [(1/2)(sinx)^2]|(0->π/2) -[(1/2)(sinx)^2]|(π/2->π) }
=(π/2) ( 1/2+1/2)
=π/2

追问

不太明白这个上限下线为什么这样变换可以说说吗

追答

∫(0->π) |cosx|sinx dx

把定积分分开成2段

=∫(0->π/2) |cosx|sinx dx +∫(π/2->π) |cosx|sinx dx

x∈(0,π/2) ， cosx>0

x∈(π/2,π) ， cosx<0

=∫(0->π/2) cosx.sinx dx -∫(π/2->π) cosx.sinx dx

= [(1/2)(sinx)^2]|(0->π/2) -[(1/2)(sinx)^2]|(π/2->π)

=1/2 +1/2

=1