Why Matching the Differential Matters
A substitution begins with a chosen inner function, but choosing that function is only part of the work. You must also check what happens to the rest of the integrand. If \(u=g(x)\), then the derivative of \(g\) determines \(du\). The substitution is especially direct when the integrand contains the factor \(g'(x)\,dx\), because that entire factor can be replaced by \(du\).
If the integrand contains a constant multiple of \(g'(x)\,dx\), the substitution still has a simple form, but the constant must be carried through the calculation. If the remaining factor is not proportional to \(g'(x)\), it cannot be replaced by a constant multiple of \(du\). These distinctions prevent missing factors and unsupported substitutions.
What the Differential \(du\) Represents
Suppose \(g\) is differentiable and set \(u=g(x)\). The chain rule gives \(du=g'(x)\,dx\), the differential form of the relationship between the variables. In a substitution in an indefinite integral, this tells us how to express the integral using \(u\): an occurrence of \(g'(x)\,dx\) becomes \(du\). The other part of the integrand must also be expressible in terms of \(u\) for the result to be an integral in \(u\) alone.
There are three useful possibilities when comparing a factor \(h(x)\,dx\) in the integrand with \(du\):
- Exact match: \(h(x)=g'(x)\), so \(h(x)\,dx=du\).
- Constant-multiple match: \(h(x)=c\,g'(x)\) for a constant \(c\), so \(h(x)\,dx=c\,du\).
- No constant-multiple match: there is no constant \(c\) for which \(h(x)=c\,g'(x)\) throughout the relevant interval.
The third case means that this factor cannot be replaced by a constant multiple of \(du\) as it stands. It does not prove that the integral has no antiderivative or that no other method or substitution could work. It simply identifies a limit of the proposed match.
A Justification from the Chain Rule
The substitution method works because differentiation reverses the change of variable. Let \(F\) be an antiderivative of a function \(f\), so \(F'(u)=f(u)\), and let \(u=g(x)\). By the chain rule, differentiating \(F(g(x))\) produces the factor \(g'(x)\). Thus an integrand containing \(f(g(x))g'(x)\) is the derivative of \(F(g(x))\).
Let \(F'(u)=f(u)\).
The resulting composite function is \(F(g(x))\).
The chain rule gives \(\frac{d}{dx}F(g(x))=F'(g(x))g'(x)=f(g(x))g'(x)\).
Therefore, wherever the functions are defined and differentiable, \(\int f(g(x))g'(x)\,dx=F(g(x))+C\).
This is a complete justification for an exact differential match. If the integrand instead contains \(c\,g'(x)\), with \(c\) constant, then the chain-rule expression must be scaled accordingly; the constant cannot simply be ignored.
Worked Example: An Exact Match
Worked Example: An Exact Match
Evaluate \(\int 2x(x^2+4)^3\,dx\) using \(u=x^2+4\).
First differentiate the chosen inner function:
The integrand contains \(2x\,dx\) exactly, so replace that factor with \(du\). Also, \(x^2+4=u\):
Using the power rule for antiderivatives, \(\int u^3\,du=\frac{u^4}{4}+C\). Substitute back \(u=x^2+4\):
Check by differentiating the proposed antiderivative:
The derivative equals the original integrand, so the result is correct.
Worked Example: A Constant-Multiple Match
Worked Example: A Constant-Multiple Match
Evaluate \(\int 5x\cos(5x^2)\,dx\) using \(u=5x^2\).
Differentiate \(u\):
The integrand contains \(5x\,dx\), which is half of \(10x\,dx\):
The factor \(\cos(5x^2)\) becomes \(\cos u\). Keeping the factor \(\frac{1}{2}\) gives
Check by differentiation:
This confirms the answer. The inner derivative was not present exactly: the integrand contained half of it, and that factor had to remain in the integral.
Worked Example: Testing for a Match
Worked Example: Testing for a Match
For the integral \(\int (x+1)e^{x^2+3x}\,dx\), test whether \(u=x^2+3x\) makes the factor \((x+1)\,dx\) equal to a constant multiple of \(du\).
Differentiate the proposed inner function:
A constant-multiple match would require a constant \(c\) such that \(x+1=c(2x+3)\) for every \(x\) in the interval. Comparing the coefficients of \(x\) requires \(1=2c\), so \(c=\frac{1}{2}\). Comparing the constant terms requires \(1=3c\), so \(c=\frac{1}{3}\). These requirements conflict.
Therefore, \((x+1)\,dx\) is neither exactly \(du\) nor a constant multiple of \(du\) for this choice of \(u\). The test does not evaluate the integral; it shows that this proposed differential match does not convert the factor into a constant multiple of \(du\). No antiderivative is asserted here.
A Reliable Matching Test
When using a chosen inner function \(u=g(x)\), do not decide by appearance alone that the differential matches. Calculate \(g'(x)\), write \(du=g'(x)\,dx\), and compare the complete factor multiplying \(dx\) in the integrand with \(g'(x)\). If needed, test whether their ratio is a constant, rather than a function of \(x\).
For \(u=g(x)\), calculate \(g'(x)\).
Record \(du=g'(x)\,dx\), including every numerical factor.
Check whether the integrand contains \(g'(x)\,dx\), \(c\,g'(x)\,dx\) for a constant \(c\), or neither.
Replace the matched differential factor by \(du\), retaining any constant multiple.
Differentiate the proposed antiderivative and compare it with the original integrand.
Common Pitfall: Losing a Constant
A frequent error is to write \(du\) correctly but then replace a different factor in the integrand as if the match were exact. In the second worked example, \(du=10x\,dx\), not \(5x\,dx\). The integrand's factor \(5x\,dx\) is \(\frac{1}{2}du\).
This answer fails the derivative check:
The correction is to keep the factor \(\frac{1}{2}\) from \(5x\,dx=\frac{1}{2}du\). Then \(\frac{1}{2}\sin(5x^2)+C\) differentiates to \(5x\cos(5x^2)\), as required. The check exposes the error because the incorrect result produces twice the original integrand.
Why the Match Is Useful
Substitution changes the variable used to describe the integral, but it does not make the derivative factor optional. The differential records precisely how the inner quantity changes as \(x\) changes. Matching it is what lets the chain rule run in reverse. An exact match makes the conversion direct; a constant-multiple match remains usable if the constant is preserved; a nonproportional factor is a warning that the proposed substitution does not produce the desired differential form on its own.
In practice, writing \(du\) on a separate line is a small step that prevents a common source of errors. It also makes the reasoning visible: the substitution is justified by an equality of differential factors, not by the mere presence of a familiar-looking inner expression.
Check Your Understanding
For each question, pay attention to the derivative of the chosen inner function and to the entire factor multiplying \(dx\).
- If \(u=x^2+7\), what is \(du\), and does \(2x\,dx\) match it exactly?
- If \(u=4x^2\), express \(3x\,dx\) as a constant multiple of \(du\).
- For \(u=x^2+2x\), can \((x+1)\,dx\) be replaced by \(du\) exactly? Show the differential.
- What condition must hold for a factor \(h(x)\,dx\) to equal a constant multiple of \(du\), when \(u=g(x)\)?
- Why does differentiating a proposed antiderivative help detect a missing constant factor?
