计算∫(-π/3),π/)][x/(1+cosx)]dx

计算[-π/3,π/3]∫[x/(1+cosx)]dx(原题的上限π可能有错,若是π,则此积分的值为无穷大)原式=[-π/3,π/3]∫[x/2cos²(x/2)]dx=[-π/3,π/3]∫d(x/2)/cos²(x/2)=tan(x/2)︱[-π/3,π/3]=tan(π/6)-tan(-π/...