∫sin(x+π/3) dx=∫sin(x+π/3) d(x+π/3)
=-cos(x+π/3)+C
所以原式=-cos(π/3+π/3)+cos(π+π/3)
=-cos(2π/3)-cos(π/3)
=1/2-1/2
=0
π到π/3 sin（x+π/3）dx 定积分
∫sin(x+π/3) dx=∫sin(x+π/3) d(x+π/3)
=-cos(x+π/3)+C
所以原式=-cos(π/3+π/3)+cos(π+π/3)
=-cos(2π/3)-cos(π/3)
=1/2-1/2
=0