如题所述
如图所示：
一个是换元，一个是凑微分，原理都是一样的
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第1个回答 2018-08-26
∫(0->π/2) dx/(1+cosx)^2
=∫(0->π/2) dx/( [2cos(x/2)]^2 )^2
=(1/4) ∫(0->π/2) [sec(x/2)]^4 dx
=(1/2) ∫(0->π/2) [sec(x/2)]^2 dtan(x/2)
=(1/2) ∫(0->π/2) ( 1+ [tan(x/2)]^2 ) dtan(x/2)
=(1/2) [ tan(x/2) +(1/3)[tan(x/2)]^3 ]|(0->π/2)
=(1/2) [ 1 +1/3 ]
= 2/3
=∫(0->π/2) dx/( [2cos(x/2)]^2 )^2
=(1/4) ∫(0->π/2) [sec(x/2)]^4 dx
=(1/2) ∫(0->π/2) [sec(x/2)]^2 dtan(x/2)
=(1/2) ∫(0->π/2) ( 1+ [tan(x/2)]^2 ) dtan(x/2)
=(1/2) [ tan(x/2) +(1/3)[tan(x/2)]^3 ]|(0->π/2)
=(1/2) [ 1 +1/3 ]
= 2/3
