题目

证明∫(上π,下0)xf(sinx)dx=π/2∫(上π,下0)f(sinx)dx
f(x)在区间[0,1]连续

提问时间：2020-11-03

答案

∫(上π,下π/2)xf(sinx)dx=(令t=x-π/2)=∫(上π/2,下0)(t+π/2)f(sint)dt=∫(上π/2,下0)tf(sint)dt+π/2∫(上π/2,下0)f(sint)dtπ/2∫(上π,下π/2)f(sinx)dx=(令t=x-π/2)=π/2∫(上π/2,下0)f(sint)dt看清楚了...