Question

这个积分怎么求？

Answer 1

let

u=π/2 -x
du =-dx
x=0, u=π/2
x=π/2, u=0
I
=∫(0->π/2) dx/[ 1+ (tanx)^2018]
=∫(π/2->0) -du/[ 1+ (cotu)^2018]
=∫(0->π/2) dx/[ 1+ (cotx)^2018]
2I
=∫(0->π/2) dx/[ 1+ (tanx)^2018] + ∫(0->π/2) dx/[ 1+ (cotx)^2018]
=∫(0->π/2) [ 2+(tanx)^2018 + (cotx)^2018 ]/{[1+ (tanx)^2018].[ 1+ (cotx)^2018]} dx
=∫(0->π/2) [ 2+(tanx)^2018 + (cotx)^2018 ]/[2+ (tanx)^2018]+ (cotx)^2018]} dx
=∫(0->π/2) dx
=π/2
I=π/4
=>∫(0->π/2) dx/[ 1+ (tanx)^2018] =π/4

追问

好方法。接下来该怎么办呢？

追答

分母
{[1+ (tanx)^2018].[ 1+ (cotx)^2018]} =2+ (tanx)^2018]+ (cotx)^2018
=分子
整个分数=1

追问

谢谢你。