求定积分∫(0到π/2)[cos(x/2)-sin(x/2)]^2dx
问题描述：
求定积分∫(0到π/2)[cos(x/2)-sin(x/2)]^2dx
运用到什么三角变换?
答
[cos(x/2)-sin(x/2)]²=[cos²(X/2)+sin²(x/2)]+2sin(x/2)cos(x/2)=1+sinx∫(π/2,0)[cos(x/2)-sin(x/2)]^2dx=∫(π/2,0) 1+sinx dx=x|(π/2,0)-cosx|(π/2,0)=(π/2)+1