I want to solve the integral $$\int \frac{1}{x\sqrt{1-x^2}}\, \mathrm dx$$ using the trigonometric substitution $x=\sin{t}$ with $t\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ since the function sine is invertible in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].$
Now my question is: I know that I have to write that $\int \frac{1}{x\sqrt{1-x^2}}\, \mathrm dx$ equals $\displaystyle\int \frac{1}{\sin{t}\sqrt{1-\sin^2{t}}}\cos{t}\, \mathrm dt$ instead of $\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\sin{t}\sqrt{1-\sin^2{t}}}\cos{t}\, \mathrm dt.$ But why is the integral still indefinite? When I tranform $x$ to $t\in [-\frac{\pi}{2}, \frac{\pi}{2}],$ I get an integrand whose variable is defined in this particular interval, so I would say that the integral becomes definite and it is no longer indefinite.
Hint
For real domain, the original integral could be $$\int_a^b\dfrac{dx}{x\sqrt{1-x^2}}$$ where $-1\le a,b\le1$
Also, if trigonometric substitution is not mandatory, for $$\int\dfrac{dx}{x^{2n+1}\sqrt{1-x^2}},$$ we can choose $\sqrt{1-x^2}=y$ to eliminate the radical