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求定积分∫[1/x√(x²-1)]dx x属于(-2,-1)
人气:325 ℃ 时间:2020-05-26 14:27:45
解答
∫ 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
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