∫(π/2→0)(cosx/2-sinx/2)^2dx
问题描述：
∫(π/2→0)(cosx/2-sinx/2)^2dx
答
∫(π/2→0)(cosx/2-sinx/2)^2dx
=
∫(π/2→0)(1-sinx)dx
=(π/2→0)x+cosx=1-π/2