Question
Evaluate the integral.
∫dx / (√4x^2 - 9), x > 3/2
∫dx / (√4x^2 - 9), x > 3/2
Answers
oobleck
let x = 3/2 secθ
secθ = 2/3 x
sec^2θ = 4/9 x^2
tan^2θ = sec^2θ - 1 = 4/9 x^2 - 1 = 1/9 (4x^2-9)
so √(4x^2 - 9) = 3tanθ
Now,
secθ tanθ dθ = 2/3 dx
dx = 3/2 secθ tanθ dθ
So now you have
∫ 3/2 secθ tanθ dθ / 3tanθ = 1/2 ∫secθ dθ
which is one of your standard forms
secθ = 2/3 x
sec^2θ = 4/9 x^2
tan^2θ = sec^2θ - 1 = 4/9 x^2 - 1 = 1/9 (4x^2-9)
so √(4x^2 - 9) = 3tanθ
Now,
secθ tanθ dθ = 2/3 dx
dx = 3/2 secθ tanθ dθ
So now you have
∫ 3/2 secθ tanθ dθ / 3tanθ = 1/2 ∫secθ dθ
which is one of your standard forms