求定积分∫π/20sinxcosx/(1+cos2x)
∫π/20[sinxcox/(1+cos2x)]dx=-∫π/20[cosx/(1+cos2x)]dcosx=-(1/2)∫π/20d(1+cos2x)/(1+cos2x)=-(1/2)ln(1+cos2x)∣π/20=(1/2)ln2
求定积分∫π/20sinxcosx/(1+cos2x)
∫π/20[sinxcox/(1+cos2x)]dx=-∫π/20[cosx/(1+cos2x)]dcosx=-(1/2)∫π/20d(1+cos2x)/(1+cos2x)=-(1/2)ln(1+cos2x)∣π/20=(1/2)ln2