Why Give a Name to an Inner Expression?
When an integrand contains a composite expression, it can be useful to name the inner expression. For example, in \(\cos(x^2)\), we might set \(u=x^2\). But naming \(x^2\) alone does not explain what happens to the rest of the integral. The notation \(du\) records how the new variable changes with \(x\), and the Chain Rule tells us exactly what that notation means.
The key relationship is \(du=g'(x)\,dx\) when \(u=g(x)\). It is a compact way to keep track of the derivative required by the chain rule. It does not mean that \(dx\) and \(du\) are independent numerical factors that can be manipulated without justification.
What Does \(du\) Mean?
Suppose \(u=g(x)\), where \(g\) is differentiable. If \(x\) changes by a small amount \(\Delta x\), the corresponding change in \(u\) is \(\Delta u=g(x+\Delta x)-g(x)\). Differentiability means that this change is approximated, to first order, by \(g'(x)\Delta x\): the difference between \(\Delta u\) and \(g'(x)\Delta x\), divided by \(\Delta x\), tends to zero as \(\Delta x\) tends to zero.
The differential records this first-order, linear part of the change. In the \(x\) coordinate, the differential of \(x\) is written \(dx\). The differential of \(u=g(x)\) is therefore defined by
This equation says that the differential of the new variable is the derivative of \(g\) multiplied by the differential of the original variable. It is not a separate rule about multiplying ordinary numbers. It is notation grounded in the linear approximation of \(g\), and the Chain Rule makes it useful in integration.
For instance, if \(u=x^2\), then \(g'(x)=2x\), so the differential relationship is \(du=2x\,dx\). This does not say that \(du\) is a number equal to \(2x\), or that \(dx\) can always be cancelled like a numerical factor. It records how the variable change and its derivative fit together.
Why the Substitution Notation Works
Let \(f\) be continuous on an interval containing the values of \(g\), and let \(A\) be an antiderivative of \(f\) there, so \(A'(u)=f(u)\). Assume \(g\) is differentiable. The Chain Rule gives the derivative of the composite function \(A(g(x))\). Writing \(u=g(x)\), we obtain
Thus \(A(g(x))\) is an antiderivative of \(f(g(x))g'(x)\). This is why the matching factor \(g'(x)\) matters: it is the factor supplied by differentiating the composite antiderivative. In differential notation, the same chain of reasoning is written as
Consequently, an indefinite integral of this form can be rewritten as
Here \(C\) is an arbitrary constant. The equality is justified by the chain rule: differentiating the proposed result \(A(g(x))+C\) gives the original integrand. The notation makes the change of variable visible, but the derivative identity is the reason the substitution is valid.
Worked Examples
Worked Example: Rewriting a Composite Integrand
Evaluate \(\int 2x\cos(x^2)\,dx\). Set \(u=x^2\). Then \(g'(x)=2x\), so
The integrand contains exactly the factor \(2x\,dx\) represented by \(du\). Therefore the integral becomes
Check by differentiating the proposed antiderivative:
This is the original integrand, so the result is correct.
Worked Example: Using a Polynomial Inner Function
Evaluate \(\int (2x+1)\bigl(3(x^2+x)^2-2\bigr)\,dx\). Choose \(u=x^2+x\). Its derivative and differential are
After this change, \(x^2+x\) is replaced by \(u\), and the matching factor \((2x+1)\,dx\) is represented by \(du\):
To check, differentiate the result. Using the Chain Rule gives
This agrees with the integrand, verifying the antiderivative.
Worked Example: A Change of Variable with \(du=dx\)
Evaluate \(\int (x+3)^2\,dx\). Let \(u=x+3\). Since \(g'(x)=1\), the differential relationship is \(du=1\,dx=dx\). Also, \(x+3=u\). Thus
Differentiate the result to check it:
The derivative is the original integrand. In this example, \(du=dx\) because the chosen inner function has derivative \(1\); that relationship follows from its derivative, not from a general rule that different variables have identical differentials.
How to Read the Notation Carefully
In \(\int f(u)\,du\), the symbol \(du\) identifies the variable of integration. When \(u=g(x)\), the equation \(du=g'(x)\,dx\) records the chain-rule relationship between that variable and \(x\). After integrating with respect to \(u\), substitute \(g(x)\) back if the answer is to be written as a function of \(x\).
A useful way to read the change is: “the factor \(g'(x)\,dx\) becomes \(du\), while \(g(x)\) becomes \(u\).” Both changes belong together. Replacing \(g(x)\) by \(u\) but leaving a factor that depends on \(x\) in the integral generally does not produce an integral purely in \(u\). Nor is it justified to erase a factor simply because a differential symbol appears nearby.
The notation may look like a cancellation of \(g'(x)\) and \(dx\), but the justification is different. We first identify a composite function and apply the Chain Rule. That establishes that the transformed integral has the desired antiderivative. Differential notation packages this reasoning into a compact form; it does not replace the reasoning.
A Common Mistake: Ignoring the Derivative Factor
Consider \(\int \cos(x^2)\,dx\). A tempting but incorrect calculation is to set \(u=x^2\), write \(du=2x\,dx\), and then claim
The proposed answer fails a direct check:
which is not the integrand \(\cos(x^2)\). The error is that the original integral does not contain the factor \(2x\) needed for \(du=2x\,dx\). The symbol \(du\) represents \(2x\,dx\) in this change of variable; it cannot replace \(dx\) alone while leaving the factor unmatched.
This correction diagnoses why that particular substitution does not directly give the claimed antiderivative. It does not assert that every integrand lacking a visible matching factor is impossible to integrate by any method. The specific rule used here applies when the derivative of the inner function is present in the integrand.
Summary
If \(u=g(x)\), the differential relationship is \(du=g'(x)\,dx\). This notation follows from the Chain Rule and records the first-order change in the new variable. When the integrand has the form \(f(g(x))g'(x)\), substitution rewrites it as \(f(u)\,du\), and an antiderivative in \(u\) becomes an antiderivative in \(x\) after substituting \(g(x)\) back.
The essential discipline is to treat the differential notation as a guide to a chain-rule argument, not as permission to manipulate \(dx\) and \(du\) as independent numerical factors. A derivative check confirms whether the resulting expression is an antiderivative of the original integrand.
Check Your Understanding
Use the relationship between differentials and the Chain Rule to answer these questions.
- If \(u=x^3\), what is \(du\) in terms of \(x\) and \(dx\)?
- For what reason does the integrand \(f(g(x))g'(x)\) become \(f(u)\,du\) when \(u=g(x)\)?
- In the integral \(\int \cos(x^2)\,dx\), which factor needed for the direct substitution \(u=x^2\) is absent?
- Why is \(du=g'(x)\,dx\) not a statement that \(du\) and \(dx\) are independent ordinary numbers?
- What is one way to verify an indefinite integral obtained by substitution?
