计算∫(-π/3),π/)][x/(1+cosx)]dx

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/11/17 03:10:37

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/6)=2tan(π/6)=2(√3)/3

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