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求定积分下限∫-π/2到上限π/2sinx/(2+cosx)dx
人气:428 ℃ 时间:2020-02-02 23:29:59
解答
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
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