Question

求解：∫（0到π+ u）/4 f(sinx)

Answer 1

本题你漏了积分限，积分限应是[0--->π]
∫[0--->π] xf(sinx)dx
做变量代换，令x=π-u，则dx=-du，u：π--->0
=-∫[π--->0] (π-u)f(sin(π-u))du
=∫[0--->π] (π-u)f(sinu)du
=∫[0--->π] πf(sinu)du-∫[0--->π] uf(sinu)du
定积分可随便换积分变量
=∫[0--->π] πf(sinx)dx-∫[0--->π] xf(sinx)dx
将-∫[0--->π] xf(sinx)dx移动等式左边与左边合并得
2∫[0--->π] xf(sinx)dx=π∫[0--->π] f(sinx)dx
即：∫[0--->π] xf(sinx)dx=π/2∫[0--->π] f(sinx)dx