2. 考研
  3. ∫0π/2sin2x/(1+esinx)dx=∫0π/2d(sin2x)/(1+esinx)=∫01dx/(1+ex)=∫01exdx/(e-x+1)=-∫01d(e-x+1)/(e-x+1)=-ln(e-x+1)｜01=ln2e/(e+1)．

∫0π/2sin2x/(1+esinx)dx=∫0π/2d(sin2x)/(1+esinx)=∫01dx/(1+ex)=∫01exdx/(e-x+1)=-∫01d(e-x+1)/(e-x+1)=-ln(e-x+1)｜01=ln2e/(e+1)．

admin2021-10-18  24
问题
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答案∫0π/2sin2x/(1+esinx)dx=∫0π/2d(sin2x)/(1+esinx)=∫01dx/(1+ex)=∫01exdx/(e-x+1)=-∫01d(e-x+1)/(e-x+1)=-ln(e-x+1)｜01=ln2e/(e+1)．
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考研数学二

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