Transforing improper integral into normal integral

Question

I had some basic integral:
$$
I=\int _0^2 \frac{dx}{\sqrt{2x-x^2}}
$$
I have subsituted $u=\sqrt{\frac{x}{2}}$ and got that
$$
I=2[\arcsin(u)]_0^1=\pi
$$
At the beginning $I$ was improper integral and the function was undefined for both $0$ and $2$, but $\arcsin$ is defined in both $0$ and $1$, so I don't need to calculate any limits? The final value is correct, but I don't know whether I can transform an improper integral into normal integral in such way.

Answer 1

There always exists a substitution to turn an improper integral into a nonimproper integral for absolutely integrable singularities, which means that they weren't true singularities to begin with.

Take for example

$$\int_0^1 \frac{1}{\sqrt{x}}\:dx$$

and let $x = u^2 \implies dx = 2u du$

$$\int_0^1 \frac{2u}{u}\:du = \int_0^1 2\:du = 2$$