计算定积分S(0-π)(1+cosx)^1/2dx

1+cosx=2cos²(x/2)
所以原积分可拆分为两个积分计算

S(0-π)(1+cosx)^1/2dx
=S(0-π/2)(根号2)cos(x/2)dx+S(π/2-π)(根号2)(-cos(x/2))dx
=[2(根号2)sin(x/2)](0-π/2)-[2(根号2)sin(x/2)](π/2-π)
=(2根号2)[(sinπ/4-sin0)-(sinπ/2-sinπ/4)]
=(2根号2)[2sinπ/4-sinπ/2]
=(2根号2)[根号2-1]
=4-2(根号2)