> 数学 >
求定积分∫(上限π/4,下限0)ln(1+tanx)dx,
人气:284 ℃ 时间:2019-09-22 07:51:32
解答
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...
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版