∫(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 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 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
∫(0,π/4) ln(1+tanx) dx
Let y = π/4 - x then dy = -dx
When x = 0,y = π/4,when x = π/4,y = 0
J = ∫(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+tany)] dy
= ∫(0,π/4) ln[(1+tany+1-tany)/(1+tany)] dy
= ∫(0,π/4) [ln(2) - ln(1+tany)] dy /*这一行到
= ln(2) * ∫(0,π/4) dy - J 这一行的转换是为什么!*/
2J = ln(2) * (π/4-0)
J = (π*ln2)/8
难道 ln(1+tanx)dx=ln(1+tany)dx
是的话 为什么相等?

∫(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
不是说ln(1+tanx)dx=ln(1+tany)dx这两个一样,这两者不能化等号
而是∫(0,π/4) ln(1+tanx) dx 和对于∫(0,π/4) ln(1+tany) dy
当积分形式一样 而被积函数和对应积分变量一样,对应的积分变量取值一样,那么做出来结果是一样的,因为定积分其实质上是一个数
正如∫(0,1) xdx=1/2 ∫(0,1) ydy=1/2 这两个定积分的结果是一样的

∫(0~π/4) ln(1+tanx)dx 怎么算 求定积分∫(上限π/4,下限0)ln(1+tanx)dx, 就是关于那个定积分“求定积分:∫ln(tanx)dx (o≤x≤π/2),积分是限是π/2,下限是0的一些问题∫ln(tanx)dx=∫[0,π/2] ln(tanx)dx=∫[0,π/4]ln(tanx)dx+∫[π/4,π/2]ln(tanx)dx=∫[0,π/4]ln(tanx)dx+∫[π/4,π/2]lncot(π/2-x 求定积分:∫ ln(1+tanx)dx (o≤x≤π/4)求定积分:∫ln(1+tanx)dx (o≤x≤π/4) 求不定积分:∫ln(1+tanx)dx (o≤x≤π/4)求不定积分:∫ln(1+tanx)dx (o≤x≤π/4) 计算定积分：∫(0,π/4)ln(1+tanx)dx=其中0是下限,π/4是上限 求定积分∫ln(1 tanx)dx(o≤x≤π/4) ∫ln（1+tanx）dx= ∫(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 澄哥ln(1+tanX)在0到π?上的定积分怎么算 x→0,lim(1-cosx)[x-ln(1+tanx)]/sinx^4的极限 lim（x→0）（1-cosx）[x-ln（1+tanx）]/（sinx）^4 ∫[0→π/4] secx dx =ln|secx + tanx| |[0→π/4]这个积分怎么求出来的? ∫secxdx=ln|secx+tanx| ∫secxdx=ln|secx+tanx|+C lim1/(ln(tanx+1)-1/sinx) ln（1/tanx）的导数,求细解 x趋向于0+,lim(ln(tan4x)/ln(tanx))