Question

What is the integral of `(1/sqrt(4-x^2))`?

Answer 1

`int1/sqrt(4-x^2)dx =1/2int 1/(sqrt(1-1/4x^2)`

Let's `u =1/2x` so `u^2=1/4x^2`

and `du = 1/2dx`

The integral become

`int1/sqrt(1-u^2)`

This is the derivative of `arcsin(u)`

`[arcsin(u)]+C` substitute back for `u = 1/2x`

`[arcsin(1/2x)]+C`