(根号下a^2-x^2)dx/x^2(a>0)不定积分····求具体解答过程···我解到cos^2t/sin^2t就不会了

Question

(根号下a^2-x^2)dx/x^2(a>0)不定积分····求具体解答过程···我解到cos^2t/sin^2t就不会了

Answer 1

OK,解决.
令x = a • sinθ,dx = a • cosθ dθ
√(a² - x²) = √(a² - a² • sin²θ) = a • cosθ
∫ √(a² - x²)/x² dx
= ∫ (a² • cos²θ)/(a² • sin²θ) dθ
= ∫ cot²θ dθ
= ∫ (csc²θ - 1) dθ
= - cotθ - θ + C
= -√(a² - x²)/x - arcsin(x/a) + C
_______________________________
由于sinθ = x/a
所以cosθ = √(1 - sin²θ) = √(1 - x²/a²) = √(a² - x²)/a
cosθ = cosθ/sinθ = √(a² - x²)/x