高数,求不定积分∫√(x^2 -4)dx 我该怎么思考这种求不定积分的题目呢?望高手总结.

Question

如题所述

Answer 1

假设x > 2,可令x = 2secθ
secθ = x/2、tanθ = √(x² - 4)/2
dx = 2secθtanθ dθ
∫ √(x² - 4) dx
= ∫ √(4sec²θ - 4)[2secθtanθ dθ]
= ∫ (2tanθ)(2secθtanθ) dθ
= 4∫ tan²θsecθ dθ
= 4∫ (sec²θ - 1)secθ dθ
= 4I - 4∫ secθ dθ
I = ∫ sec³θ dθ = ∫ secθ[sec²θ dθ] = ∫ secθ d[tanθ]
= secθtanθ - ∫ tanθ d[secθ]
= secθtanθ - ∫ tanθ(secθtanθ dθ)
= secθtanθ - ∫ (sec²θ - 1)secθ dθ
= secθtanθ - I + ∫ secθ dθ
2I = secθtanθ + ∫ secθ dθ
I = (1/2)secθtanθ + (1/2)∫ secθ dθ
原式 = 4I - 4∫ secθ dθ
= 4[(1/2)secθtanθ + (1/2)∫ secθ dθ] - 4∫ secθ dθ
= 2secθtanθ - 2∫ secθ dθ
= 2secθtanθ - 2ln|secθ + tanθ| + C
= 2[x/2][√(x² - 4)/2] - 2ln|x/2 + √(x² - 4)/2| + C
= (x/2)√(x² - 4) - 2ln|x + √(x² - 4)| + C