What is the function y if dy/dx=square root(1-x^2)/x^2?

We'll have to integrate dy to get the primitive funtion
Y.

dy =
sqrt(1-x^2)dx/x^2

We'll integrate both
sides:

`int` dy = `int`
sqrt(1-x^2)dx/x^2

Let x = sin t => dx = cos t
dt

t = arcsin x

`int`
sqrt(1-x^2)dx/x^2 = `int` sqrt[1-(sin t)^2]*cos t dt/(sin
t)^2

We'll apply Pythagorean identity to
numerator:

`int` [sqrt (cos t)^2]*cos t dt/(sin t)^2 =
`int` (cos t)^2dt/(sin t)^2

`int` (cos t)^2dt/(sin t)^2 =
`int` [1-(sin t)^2]dt/(sin t)^2

`int` [1-(sin t)^2]dt/(sin
t)^2 =`int` ` ` dt/(sin t)^2 - `int` dt

`int` [1-(sin
t)^2]dt/(sin t)^2 = -cot t - t + C

`int` sqrt(1-x^2)dx/x^2
= - cot (arcsin x) - arcsin x + C

The
requested primitive function Y is: Y = - cot (arcsin x) - arcsin x +
C.