Question

What is the integral of (sqrt(1 - x^2) / x)?
Thanks

Answer 1

let x = sinθ, then:
sqrt(1 - x^2) / x = sqrt(1 - (sinθ)^2)/sinθ = sqrt((cosθ)^2)/sinθ = cosθ/sinθ = cotθ
then take the integral of cotθ

Answer 2

you will also have to replace \(dx\) with \(\cos\theta d\theta\)

Answer 3

So when you let \(x =\sin\theta\) then \(dx = \cos\theta d\theta\). Then
\[\int \frac{\sqrt{1-x^2}}{x} dx = \int \frac{\cos^2\theta }{\sin\theta} d\theta\]
Rewrite \(\cos^2\theta\) as \(1- \sin^2\theta\) and break up the fraction,
\[\int \frac{\cos^2\theta }{\sin\theta} d\theta = \int \frac{1 - \sin^2\theta }{\sin\theta} d\theta = \int (csc\theta - \sin\theta) d\theta\]

Answer 4

Thanks!