高数题求学霸解答

Question

如题所述

Answer 1

从积分限可以看出0≤z≤y≤x≤1,变换积分顺序
原式=∫[0→1] dz∫[z→1] dx∫[z→x] sinz/(1-z)²dy
=∫[0->1] dz∫[z->1] (x-z)sinz/(1-z)² dx
=∫[0->1] dz (x²/2 - xz)sinz/(1-z)² |(z→1)
=∫[0->1] sinz /2 dz
= -cosz /2 |(0→1)
= (-cos1 + 1)/2