Question

两道关于积分的证明题1、证明∫0到π(sin x)^ndx=2倍∫0到π/2 (sin x)^ndx2、证明：连续函数f

Answer 1

∫0到π(sin x)^ndx=∫0到π/2 (sin x)^ndx+∫π/2到π (sin x)^ndx

根据定理,f(x)在[0,1]连续,∫0到π/2f(sinx)dx=∫0到π/2f(cos)dx

∫π/2到π (sin x)^ndx=∫0到π/2 (sin( x+π/2))^ndx=∫0到π/2(cos x)^ndx=∫0到π/2 (sin x)^ndx

因此∫0到π(sin x)^ndx=∫0到π/2 (sin x)^ndx+∫π/2到π (sin x)^ndx=2*∫0到π/2 (sin x)^ndx

2、

续函数f(x)是偶函数f(t)=f(-t)

g(x)=∫0到xf(t)dt

g(-x)=∫0到(-x)f(t)dt=∫0到xf(-t)d(-t)=-∫0到xf(t)d(t)=-g(x)

所以g（x）为奇函数