Does trigonometric substitution make an indefinite integral definite? 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.
 A: 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
A: 
$$\int \frac{1}{x\sqrt{1-x^2}}\, \mathrm dx\tag1$$
using the substitution $x=\sin{t}$ with $t\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ $$\int \frac{\cos{t}}{\sin{t}\sqrt{1-\sin^2{t}}} \mathrm dt\tag2$$

Restricting $(2)$'s domain to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (actually, this can be changed to any interval where $\sin$ is invertible and has range $[-1,1]$) is imposed purely so that if $(1)$ is given integration limits, we will choose a valid corresponding pair of integration limits for $(2);$ for example, $$\int_{-\pi/2}^{\pi/2} \frac{\cos{t}}{\sin{t}\sqrt{1-\sin^2{t}}} \mathrm dt\ne \int_{0.25}^{0.5} \frac{1}{x\sqrt{1-x^2}}\, \mathrm dx\ne \int_{3\pi-\arcsin(0.25)}^{\arcsin(0.5)} \frac{\cos{t}}{\sin{t}\sqrt{1-\sin^2{t}}} \mathrm dt$$ even though $$\sin\big(3\pi-\arcsin(0.25)\big)=0.25.$$
On the other hand, if $(2)$ is considered without regard to $(1),$ then its domain is $\mathbb R{\setminus}\left\{\frac{k\pi}2\mid k\in\mathbb Z\right\}.$
The domain of $(1)$ is $(-1,0)\cup(0,1).$
Having a proper subset of $\mathbb R$ as domain does not make any of the above indefinite integrals definite though. An indefinite integral is just a general antiderivative, which has the same domain as its integrand (for example, the domain of $\int\frac1x\,\mathrm dx$ excludes $0,$ so it is illegal to plug in $-3$ and $5$ as its integration limits). In contrast, a definite integral denotes an area associated with its integration limits.
