integration - Question about u-substitution method.

How can we really use u-substitution method for finding integrals if "dx" in the integral is just a notation, not a number. At first we say that it is just a part of notation, but then, we use it as a number to find "du" (although it is also just a part a notation and we can't really find it for a new integral which we write after substitution). Maybe I misunderstood something. Thanks.

Answer

Let's do the proof for indefinite integrals:

$$\int f(u(x))u'(x)dx=\int f(u)du$$

Let $g$ be an antiderivative of $f$ on some interval $D$ (so $g'(x)=f(x)$ on $D$), and let $X$ be another interval and $u:X\to D$ a differentiable function that we will use for substitution.

Now, by solving the RHS, we obtain $\int f(u)du = g(u) + C$, and by substituting $u$, we obtain $g(u(x))+C$. Now, take the derivative on the domain $X$:

$$(g(u(x)+C)'=g'(u(x))u'(x)=f(u(x))u'(x)$$

(the chain rule), so the result we got by solving the RHS satisfies the LHS too: $g(u(x))+C$ is an antiderivative of $f(u(x))u'(x)$.

The reverse is true because all the other antiderivatives in LHS differ from this one by a constant (because $X$ is connected), and can be obtained by varying the constant on the RHS.

The condition that $X$ is an interval cannot be weakened to include sets that are not connected, because the LHS could have different constants applied to the multiple connected components of $X$, while the RHS has a single constant.