Question

求∫1/1+(tanx)^adx

Answer 1

这原函数不初等，所以应该是计算0到π/2的定积分吗？
答案是π/4

设A = ∫(0,π/2) 1/[ 1 + (tanx)^a ] dx，a∈R
令x = π/2 - t，dx = - dt，tan(π/2 - t) = cot(t)
A = ∫(0,π/2) 1/[ 1 + (cotx)^a ] dx，同乘tanx
= ∫(0,π/2) (tanx)^a/[ (tanx)^a + 1 ] dx
将两个式加起来，
A + A = 2A = ∫(0,π/2) [ 1 + (tanx)^a ]/[ 1 + (tanx)^a ] dx
2A = ∫(0,π/2) dx = π/2
即A = π/4