如何证明∫[0,π]xf(sinx)dx=π∫[0,π/2]f(sinx)dx

问题描述:

如何证明∫[0,π]xf(sinx)dx=π∫[0,π/2]f(sinx)dx

http://zhidao.baidu.com/question/341650003.html?oldq=1
满意请采纳

-2sinx*(1-(sinx)^2)^0.5在[pi/2,pi] 已不满足(积分区间为0到π)∫xf(sinx)dx=(π/2)∫f(sinx)dx的分条件,

如图

计算∫[π/2,π]xf(sinx)dx
令x=π-t 得
∫[π/2,π]xf(sinx)dx=∫[π/2,0] (π-t)f(sin(π-t))d(π-t)
=∫[0,π/2] (π-t)f(sint)dt=π∫[0,π/2] f(sint)dt-∫[0,π/2]t f(sint)dt
∫[0,π]xf(sinx)dx=∫[0,π/2]t f(sint)dt+∫[π/2,π]xf(sinx)dx=π∫[0,π/2]f(sint)dt