Question

定积分计算中的三角函数问题？

请问图中第一行的cosudu/(sinu+cosu) 是如何转化为sinudu/(cosu+sinu)

Answer 1

分析与证明过程如图所示

Answer 2

let
x=π/2-u
dx=-du
u=0, x=π/2
u=π/2, x=0
∫(0->π/2) cosu /(sinu +cosu) du
=∫(π/2->0) [sinx /(sinx +cosx) ] (-dx)
=∫(0->π/2) [sinx /(sinx +cosx) ] dx
=∫(0->π/2) [sinu /(sinu +cosu) ] du