定积分计算 ∫[0→π／2]（cosx）^2dx

Answer 1

这道题先给cos^2降次 cos^2=(cos2x+1)/2
∫[0→π／2]（cos2x+1)/2dx=1/2∫[0→π／2]（cos2x+1)dx=1/2∫[0→π／2]cos2xdx+1/2∫[0→π／2]1dx=1/2(1/2sin2x)|[0→π／2]+1/2x|[0→π／2]
=1/4sinπ-1/4sin0+1/2(π/2)-0=π/4