题目

赶快∫【-π,π】[sinx/(x^2+cosx)]dx和∫【-π/4,-π/3】(sinx+cosx)dx求解

提问时间：2020-07-27

答案

∫【-π,π】[sinx/(x^2+cosx)]dx 奇函数
所以原式=0
∫【-π/4,-π/3】(sinx+cosx)dx
=(-cosx+sinx)|(-π/4,-π/3)
=-cos(-3/π）+cos(-π/4）+sin(-π/3)-sin(π/4)
=（根号2）-1/2-（根号3）/2