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  3. ∫-π/2π/2(x+sin2x)/(1+cosx)2dx．

∫-π/2π/2(x+sin2x)/(1+cosx)2dx．

admin2022-10-25  63
问题 ∫-π/2π/2(x+sin2x)/(1+cosx)2dx．
选项
答案∫-π/2π/2(x+sin2x)/(1+cosx)2dx=∫-π/2π/2sin2x/(1+cosx)2dx=2∫0π/2sin2x/(1+cosx)2dx=2∫0π/2(1-cos2x)/(1+cosx)2dx=2∫0π/2(1-cosx)/(1+cosx)dx=2∫0π/2(-1+2/(1+cosx))dx=-π+4∫0π/2dx/(1+cisx)=-π+4tanx/2｜0π/2=4-π．
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考研数学三

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