2个回答
展开全部
arctanx +arctan(1/x) = π/2
u=-x
du =-dx
I
=∫(-π/4->π/4) 5sinx. [e^x/(1+e^(2x)] dx
=∫(π/4->-π/4) -5sinu. [e^(-u)/(1+e^(-2u)] ( -du)
=∫(-π/4->π/4) -5sinx. [e^(-x)/(1+e^(-2x)] dx
=∫(-π/4->π/4) -5sinx. [e^(x)/(1+e^(2x)] dx
2I=∫(-π/4->π/4) 5sinx. [e^x/(1+e^(2x)] dx+∫(-π/4->π/4) -5sinx. [e^(x)/(1+e^(2x)] dx
=0
I=0
u=-x
du =-dx
I
=∫(-π/4->π/4) 5sinx. [e^x/(1+e^(2x)] dx
=∫(π/4->-π/4) -5sinu. [e^(-u)/(1+e^(-2u)] ( -du)
=∫(-π/4->π/4) -5sinx. [e^(-x)/(1+e^(-2x)] dx
=∫(-π/4->π/4) -5sinx. [e^(x)/(1+e^(2x)] dx
2I=∫(-π/4->π/4) 5sinx. [e^x/(1+e^(2x)] dx+∫(-π/4->π/4) -5sinx. [e^(x)/(1+e^(2x)] dx
=0
I=0
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
展开全部
如下图所示，换元法，变换被积函数，然后用三角函数和指数函数的基本性质来做:
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
