设f(x)为连续函数,证明:∫(0,π)f(丨cosx丨)dx=2∫(0,π/2)f(sinx)dx

Question

如题所述

Answer 1

:∫(0,π)f(丨cosx丨)dx=∫(0,π/2)f(cosx)dx+:∫(π/2,π)f(-cosx)dx
对1个积分，x=π/2-t
∫(0,π/2)f(cosx)dx==∫(π/2,0)f(sinx)d(-t)=∫(0,π/2)f(sint)dt=∫(0,π/2)f(sinx)dx
对2个积分，x=π-t
∫(π/2,π)f(-cosx)dx=∫(π/2,0)f(sint)d(-t)=∫(0,π/2)f(sint)dt=∫(0,π/2)f(sinx)dx
所以：:∫(0,π)f(丨cosx丨)dx=2∫(0,π/2)f(sinx)dx

Answer 2

设t=x-π/2
左边=∫(-π/2,π/2)f(丨cos(t+π/2)丨)dt
=∫(-π/2,π/2)f(丨sint丨)dt
因为f(丨sint丨)是偶函数
所以=2∫(0,π/2)f(丨sint丨)dt
又因为0<=t<=π/2时，sint>=0
所以=2∫(0,π/2)f(sinx)dx=右边本回答被提问者采纳