∫-π/2π/2(x+sin2x)/(1+cosx)2dx.

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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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