Question

求定积分下限∫-π/2到上限π/2sinx/(2+cosx)dx

Answer 1

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

Answer 2

被积函数sinx/(2+cosx)是奇函数

积分区间[-π/2,π/2]关于原点对称，定积分为0
不定积分为：
∫ sinx/(2+cosx)dx
= -∫1/(2+cosx)d(cosx+2)
= - ln(2+cosx) + C