求定积分,[-π/2,π/2],((cosx)^(1/2))乘以（sinx）的绝对值 dx

考虑曲线y=√(cosx) * sinx在[-π/2,0],曲线在x轴下,在[0,π/2],曲线在x轴上∴∫[-π/2,π/2] √(cosx) * |sinx| dx= -∫[-π/2,0] √(cosx) * sinx dx + ∫[0,π/2] √(cosx) * sinx dx,曲线在x轴下,需加上负号= ∫[-π/2,0] √(cosx) d(cosx) - ∫[0,π/2] √(cosx) d(cosx)= (2/3)[cosx]^(3/2)[-π/2,0] - (2/3)[cosx]^(3/2)[0,π/2]= (2/3)(1-0) - (2/3)(0-1)= 2/3 + 2/3= 4/3