求定积分1/x^2(1+x^2)^1/2 上限根号3,下限1
1个回答
展开全部
令x = tanθ,dx = sec²θdθ,x∈[1,√3]→θ∈[π/4,π/3]
∫(1~√3) 1/[x²√(1 + x²)] dx
= ∫(π/4~π/3) sec²θ/(tan²θsecθ) dθ
= ∫(π/4~π/3) 1/cosθ • cos²θ/sin²θ dθ
= ∫(π/4~π/3) cscθcotθ dθ
= - cscθ |(π/4~π/3)
= - 1/sin(π/3) + 1/sin(π/4)
= √2 - 2/√3
∫(1~√3) 1/[x²√(1 + x²)] dx
= ∫(π/4~π/3) sec²θ/(tan²θsecθ) dθ
= ∫(π/4~π/3) 1/cosθ • cos²θ/sin²θ dθ
= ∫(π/4~π/3) cscθcotθ dθ
= - cscθ |(π/4~π/3)
= - 1/sin(π/3) + 1/sin(π/4)
= √2 - 2/√3
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
