![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
求定积分∫[1/x√(x²-1)]dx x属于(-2,-1)
展开全部
∫ 1/[x√(x² - 1)] dx,x∈[-2,-1]
√(sec²θ - 1) = √(tan²θ) = ±tanθ
令x = secθ,dx = secθtanθ dθ,θ∈[2π/3,π]
∫ (secθtanθ)/(- secθtanθ) dθ
= - θ
= - [(π) - (2π/3)]
= - π/3
√(sec²θ - 1) = √(tan²θ) = ±tanθ
令x = secθ,dx = secθtanθ dθ,θ∈[2π/3,π]
∫ (secθtanθ)/(- secθtanθ) dθ
= - θ
= - [(π) - (2π/3)]
= - π/3
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