There exist two methods of integration by substitution.
The first is:
While the second is:
Unfortunately, I'm having some trouble in understanding clearly the difference between the two; they appear to me as the same thing. Can someone explain to me (possibly by words, I'm not asking mathematical rigorous demonstrations) where is intuitively the actual difference?
On
The differing usage of the symbols in the two theorems may cause some confusion.
I will stick to the same symbols across the two theorems.
Let's start with the function $f(x)$ and it's antiderivative $F(x) = \int f(x)\; dx (+C)$.
Besides this, let's write the $\varphi$ from the theorem in the more convenient form $x=x(t)$.
Case 1:
Theorem 1.18 refers to the case that you know $F(x)$. Then, you also know the antiderivative of $f(x(t))\frac{dx}{dt}$
$$\int f(x(t))\frac{dx}{dt}\; dt = F(x(t)) (+C)$$
This is just a direct consequence of the chain rule of differentiation.
Case 2:
Theorem 1.27 refers to the case that you don't know $F(x)$, but you know the antiderivative
$$\int f(x(t))\frac{dx}{dt}\; dt = G(t) (+C)$$
Now, the question is, how does $G(t)$ relate to $F(x)$?
If you can invert $x=x(t)$, then you you can write $t= t(x)$ (this plays the role of $\varphi^{-1}$ from the theorem). Now, having in mind that $x(t(x)) = x$ and $\frac{dx}{dt}\cdot \frac{dt}{dx} = 1$, the chain rule gives
$$\frac{dG(t(x))}{dx} = \frac{dG(t(x))}{dt}\frac{dt}{dx} = f(x)\frac{dx}{dt}\cdot \frac{dt}{dx}=f(x)$$
This means, if you can invert $x=x(t)$, the searched for antiderivative of $f(x)$ is $F(x) = G(t(x))$.
One way of looking at these is that Theorem 1.18 discusses "straightforward substitution", while an example of Theorem 1.27 is trigonometric substitution.
For Theorem 1.18, consider for example $f(u) = \frac{1}{2}\cos u$, which has antiderivative $F(u) = \frac{1}{2}\sin u$. Let $\varphi(x)=x^2$. Then to integrate $x\cos x^2$, we get $$\int x\cos x^2\,dx = \int f(\varphi(x))\varphi'(x)\,dx = F(\varphi(x)) = \frac{1}{2}\sin x^2,$$ which is correct.
For Theorem 1.27, consider instead trigonometric substitution. For example, let $f(x) = \frac{1}{\sqrt{1-x^2}}$, and let $\varphi(t) = \sin t$. Then $$\int f(x)\,dx = \int f(\varphi(t))\varphi'(t)\,dt = \int\frac{\cos t}{\cos t}\,dt = t+C\,dt.$$ Substituting $\varphi^{-1}(t) = \arcsin x$ for $t$ gives $$\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C,$$ which is correct.