求定积分下限∫-π/2到上限π/2sinx/(2+cosx)dx

分类: 作业答案 • 2021-12-31 09:37:45

问题描述：

答

原式=-∫-π/2到上限π/2dcosx/(2+cosx)
=-∫-π/2到上限π/2d(2+cosx)/(2+cosx)
=-ln(2+cosx)-π/2到上限π/2
=-[ln(2+0)-ln(2-0)]
=0