微积分∫π/2 -π/2 SINXCOX DX

Question

怎么求解，知道的带个步骤说下

Answer 1

因为∫sinx*cosx dx =(1/4)∫sin2x d2x= (-1/4)cos2x+c

所以∫π/2 -π/2 sinx*cosxdx=∫π/2dx-(π/2)∫sinx*cosx dx
=(π/2)x+(π/8)cos2x+c

Answer 2

标准答案 注：sin2x＝2sinx*cosx
∫π/2dx -π/2∫sinx*cosxdx
=∫π/2dx-π/4∫2sinx*cosx dx
=∫π/2dx-π/4∫sin2xdx
=πx/2+π/8cos2x+C

Answer 3

∫sinx*cosx dx =1/4∫sin2x d2x=-1/4cos2x+c
∫π/2 -π/2 sinx*cosx=0本回答被提问者采纳