evaluate the indefinite integral sqrt(4-x^2)/x^2 using trigonometric substitution and the triangle thing idk

Question

get stuck after getting the integral of cot^2theta dtheta.

Answer 1

∫[√(4 - x²) / x²]dx Let x = 2sinθ. Then dx = 2cosθdθ

So, the given integral is equivalent to ∫[√(4 - 4sin²θ) / (4sin²θ)]2cosθdθ = ∫cot²θdθ =

∫[[csc²θ - 1]dθ = -cotθ - θ + C

Since x = 2sinθ, sinθ = x/2 (so θ = Sin^-1(x/2)) and using right triangle trigonometry, cotθ = √(4 - x²) / x

So, the answer is : - √(4 - x²) / x - Sin^-1(x/2) + C

Answer 2

Why?

it is 4/x^2 - 1 = 4x^(-2) - 1

integral is (-4)x^(-1) - x = -4/x - x = (-4-x^2)/x