I am reading "A Course in Analysis vol.2" by Kazuo Matsuzaka.
In this book, the author wrote as follows:
1) $f(x)$ is continuous on an interval $I$.
2) $\phi(t)$ is continuous and differentiable on an interval $J$, and $\phi(t)$ is continuous on $J$.
3) the range of $\phi(t)$ is $I$.Let $\int f(x) dx = F(x)$.
Then, $\frac{d}{dt} F(\phi(t)) = f(\phi(t)) \phi'(t)$.
So, $$\int f(\phi(t)) \phi'(t) dt = F(\phi(t)).$$
So, $$\int f(x) dx = \int f(\phi(t)) \phi'(t) dt.$$
If I want to get $\int f(x) dx$ and I apply this theorem, I will get $F(\phi(t))$ instead of $F(x)$.
If $\phi$ has the inverse function, then I can get $F(x) = F(\phi(\phi^{-1}(x)))$.
The author didn't assume that $\phi$ has the inverse function.
Why?
Note that from condition (1), we have that $x\in I.$ Also, from (3), we have that $\phi(t)\in I.$ Thus, we may compose $f(x)$ with $\phi(t).$ That is, we set $x=\phi(t).$ So getting $F(\phi(t))$ is the same as getting $F(x).$