求定积分∫[1/x√(x²-1)]dx x属于（-2,-1）
问题描述：
求定积分∫[1/x√(x²-1)]dx x属于（-2,-1）
答
∫ 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