calculus - Justification of substitution in finding indefinite integrals

I know that if we want to compute a definite integral by substitution then

$$\int_{a}^b f(\phi(t))\phi '(t)~dt = \int_{\phi(a)}^{\phi(b)}f(x)~dx$$
I know that this method can go from left to right or right to left.

If we want to compute indefinite integral of a function using substitution, we express the integral as a composition of two function times the derivative of the function i.e.

$$\int f(t)~dt = \int g(\phi(t))\phi '(t)~dt = G(\phi(t)) + C \tag{1}$$
where $G$ is an antiderivative of $g$ and $C$ is the constant of integration. However, how can we compute the indefinite integral of a function that can't be expressed in aforementioned form? For example for evaluating functions like $\dfrac{1}{\sqrt{4 - x^2}}$ we can make the substitution $x = 2\sin \theta$ and find $dx$ in terms of $d\theta$. Substituting its value in the integral, it becomes solvable. However, pretending that $\dfrac{dx}{d\theta}$ is a fraction and substituting the value of $dx$ in the integral just because it "works" is not very rigorous. Is there a rigorous statement for solving indefinite integrals which can't be expressed in aforementioned form using substitution just like we have for definite integrals or can they be computed using $(1)$ in a way that I can't figure out?

Answer

You ask a good question. In the example you mentioned, the first step is simply writing down an antiderivative in abstract form, which is

$$\tag 1 \int_0^x \frac{1}{\sqrt {4-t^2}}\,dt,\,\,|x|<2.$$

We know that $(1)$ is an antiderivative for the integrand by the FTC. Now consider the function $t=2\sin s.$ This function maps $(-\pi/2,\pi/2)$ to $(-2,2)$ nicely. By the first equation you wrote,

$$\tag 2 \int_0^x \frac{1}{\sqrt {4-t^2}}\,dt = \int_0^{\arcsin x/2} \frac{2\cos s}{\sqrt {4-(2\sin s)^2}}\,ds.$$

After we realize that the denominator on the right equals $2\cos s,$ we get $(1) = \arcsin x/2,$ and we're done.

This is the justification behind the scenes, although after a while you'll be cancelling $dx$'s with the best (worst) of them.