Question
Evaluate the integral using the indicated trigonometric substitution.
∫8/(x^2(sqrt(x^2−9))dx and x=3sec(θ)
∫8/(x^2(sqrt(x^2−9))dx and x=3sec(θ)
Answers
so what's the trouble? Make the sub!
x = 3secθ
x^2-9 = 9sec^2θ - 9 = 9(sec^1θ - 1) = 9tan^2θ
sinθ = √(x^2−9)/x
dx = 3secθ tanθ dθ
Now just plug and chug
∫8/(x^2√(x^2−9)) dx
= ∫8/(9sec^2θ * 3tanθ) * 3secθ tanθ dθ
= ∫8/(9secθ) dθ
= 8/9 ∫cosθ dθ
= 8/9 sinθ
= 8/9 * √(x^2−9)/x + C
x = 3secθ
x^2-9 = 9sec^2θ - 9 = 9(sec^1θ - 1) = 9tan^2θ
sinθ = √(x^2−9)/x
dx = 3secθ tanθ dθ
Now just plug and chug
∫8/(x^2√(x^2−9)) dx
= ∫8/(9sec^2θ * 3tanθ) * 3secθ tanθ dθ
= ∫8/(9secθ) dθ
= 8/9 ∫cosθ dθ
= 8/9 sinθ
= 8/9 * √(x^2−9)/x + C