∫(0,π/4) ln(1+tanx) dx Let y = π/4 - x then dy = -dxWhen x = 0,y = π/4,when x = π/4,y = 0J = ∫(0,π/4) ln(1+tanx) dx= ∫(π/4,0) ln[1+tan(π/4-y)] -dy= ∫(0,π/4) ln[1 + (tan(π/4)-tany)/(1+tan(π/4)tany)] dy= ∫(0,π/4) ln[1 + (1-tany)/(1
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∫(0~π/4) ln(1+tanx)dx 怎么算
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