Difference between the two substitution methods in integration 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?
 A: 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.
A: 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 $F(x) = G(t(x))$.
