解题过程如下: 原式=-∫sinx dcos =-∫√(1-cos2x) dcosx =(1/2)[-cosx (1-(cosx)^2)^(1/2)+arccos(cosx))] (x=0, π/2) =x/2-sin2x/4 (x=0, π/2) = ∫ dx(1-cos2x)/2
解题过程如下: 原式=-∫sinx dcos =-∫√(1-cos2x) dcosx =(1/2)[-cosx (1-(cosx)^2)^(1/2)+arccos(cosx))] (x=0, π/2) =x/2-sin2x/4 (x=0, π/2) = ∫ dx(1-cos2x)/2
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