Question

证明：若函数f(x)在[0,1]上连续,则∫xf(sinx)dx=π/2∫f(sinx)dx (上限 π,下限 0)

Answer 1

令u=π-x,du=-dx,u：π--->0,则
∫[0--->π] xf(sinx)dx
=-∫[π--->0] (π-u)f(sin(π-u))du
=∫[0--->π] (π-u)f(sinu)du
=π∫[0--->π] f(sinu)du-∫[0--->π] uf(sinu)du
积分变量可随便换字母
=π∫[0--->π] f(sinx)dx-∫[0--->π] xf(sinx)dx
将 -∫[0--->π] xf(sinx)dx 移到等式左边与左边合并,然后除去系数
∫[0--->π] xf(sinx)dx=π/2∫[0--->π] f(sinx)dx