Question

计算定积分I=∫(0→π)f(sinx)/[f(sinx)+f(cosx)]*dx,其中f(x)为连续函数,且f(sin？

Answer 1

令u=π/2 -x
则x=π/2 -u
原积分=∫(π/2→0) f( sin(π/2 -u) ) / [f( sin(π/2 -u) ) + f( cos(π/2 -u) )] d(π/2 -u)
= - ∫(π/2→0) f(cosu) / [f(cosu) + f(sinu)] du
= ∫(0→π/2) f(cosu) / [f(cosu) + f(sinu)] du
积分变量的字符对积分的结果没有影响,因此
∫(0→π/2)f(sinx)/[f(sinx)+f(cosx)]*dx = ∫(0→π/2)f(cosx)/[f(sinx)+f(cosx)]*dx
则
2 ∫(0→π/2)f(sinx)/[f(sinx)+f(cosx)]*dx
=∫(0→π/2)f(sinx)/[f(sinx)+f(cosx)]*dx + ∫(0→π/2)f(cosx)/[f(sinx)+f(cosx)]*dx
=∫(0→π/2) [f(sinx)+f(cosx)] / [f(sinx)+f(cosx)] dx
=∫(0→π/2) 1 dx
= x|(0→π/2)
= π/2 - 0
= π/2
则原积分 = π/4,3,令x = π/2 - u，dx = - du
当x = 0，u = π/2
当x = π/2，u = 0
I = ∫(0-->π/2) f(sinx)/[f(sinx) + f(cosx)] dx
= ∫(π/2-->0) f(sin(π/2 - u))/[f(sin(π/2 - u)) + f(cos(π/2 - u))] (- du)
= ∫(0-->π...,1,计算定积分I=∫(0→π)f(sinx)/[f(sinx)+f(cosx)]*dx,其中f(x)为连续函数,且f(sinx)+f(cosx)不等于0
修改一下，上下限是：（0→π/2）