Question about u-substitution method.

Question

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 1

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.

Answer 2

Conceptually, the reason has to do with the fact that you are inducing a change in your coordinate system. You are correct that the dx is just a dummy variable, but a u-substitution also changes the FUNCTION ITSELF from $f(x)$ to $g = (f \circ x)(u)$. You can show using the chain-rule that even though the function you are integrating has now changed, that the VALUE of the integral will be the same as long as you make the appropriate modifications to the limits of integration and account for the relationship between your old coordinate system and the new one (this is the dx = .. du part). Really it is the same function "in disguise", but in a new coordinate system that is frequently easier to deal with.