求详细过程，谢谢

1+sinx =sin2(x/2)+cos2(x/2)+2sin(x/2)cos(x/2) =[sin(x/2)+cos(x/2)]2 =[√2sin(x/2 + π/4)]2 原式 =∫(0→2π) √[√2sin(x/2 + π/4)]2 sinx dx =∫(0→3π/2) [√2sin(x/2 + π/4)] sinx dx + ∫(3π/2→2π) -[√2sin(x/2 + π/4)] sinx dx =(-√2/2)∫(0→3π/2) [cos(3x/2 + π/4) - cos(x/2 -π/4)] +(√2/2)∫(3π/2→2π) [cos(3x/2 + π/4) - cos(x/2 -π/4)] =(-√2/2) [(2/3)sin(3x/2 + π/4) - 2sin(x/2 -π/4)] |(0→3π/2) + (√2/2) [(2/3)sin(3x/2 + π/4) - 2sin(x/2 -π/4)] |(3π/2→2π) =(-√2/2) [(2/3)sinπ -2sinπ/2 -(2/3)sinπ/4 -2sinπ/4]+(√2/2) [(2/3)sin13π/4 -2sin3π/4 -(2/3)sinπ +2sinπ/2] =自己算下吧