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Substitution and Reversing the Chain Rule · Tutorial 1155 of 1495

The Missing Differential Factor

A substitution changes the variable and its differential together; learn to track the derivative factor so the resulting integral is equivalent to the original.

Intermediate 9 min read

What You'll Learn

  • Explain why replacing an inner expression by a new variable also requires rewriting the differential.
  • Identify the derivative factor that must accompany a substitution.
  • Apply substitutions when the integrand contains the required factor, including its sign.
  • Diagnose an invalid substitution when the differential factor is absent.
  • Verify each proposed antiderivative by differentiating it.

Why the Differential Must Change

A substitution can make an integral look simpler by naming a complicated inner expression with a new variable. But changing the name of that expression alone does not transform the integral. The differential must change too, and that change introduces a derivative factor.

For example, if \(u=x^2+3\), then an expression such as \((x^2+3)^4\) becomes \(u^4\). But \(dx\) does not become \(du\): differentiating the substitution gives \(du=2x\,dx\). The factor \(2x\) is part of the change of variable. Ignoring it generally replaces the original integral by a different one.

Key principle: When \(u=g(x)\), the differential changes according to \(du=g'(x)\,dx\). A substitution is valid only when the integrand and differential are both rewritten consistently.

The Chain Rule Behind Substitution

The substitution rule for antiderivatives is the chain rule read in reverse. Suppose \(H'(u)=F(u)\), and let \(g\) be differentiable. Applying the chain rule to the composite function \(H(g(x))\) gives

$$ \frac{d}{dx}H(g(x)) = H'(g(x))g'(x) = F(g(x))g'(x). $$

Therefore an integrand of the form \(F(g(x))g'(x)\) has antiderivative \(H(g(x))\). In differential notation, setting \(u=g(x)\) gives \(du=g'(x)\,dx\), so the factor \(g'(x)\,dx\) can be replaced by \(du\). This reasoning justifies the substitution; it is not a license to replace only the inner expression.

Theorem: If \(H'=F\) and \(g\) is differentiable on an interval, then \(\frac{d}{dx}H(g(x))=F(g(x))g'(x)\) there. Consequently, an integral whose integrand is \(F(g(x))g'(x)\) can be evaluated by finding an antiderivative \(H\) of \(F\) and composing it with \(g\).

The essential check is whether the integrand contains the derivative of the chosen inner function, including its sign and any numerical factor. If it does not, then the substitution may still be possible, but the remaining factor must also be transformed. The new integral is not automatically just the original-looking expression with \(g(x)\) renamed \(u\).

A Correct Substitution, Step by Step

Worked Example: Matching the Derivative Factor

Evaluate \(\int x(x^2+3)^4\,dx\). The inner expression \(x^2+3\) suggests \(u=x^2+3\). Differentiate this definition:

$$ u=x^2+3, \qquad du=2x\,dx. $$

The integrand contains \(x\,dx\), which is half of \(du\). Rearranging the differential equation gives \(x\,dx=\frac12\,du\). Thus the complete integral transforms as follows:

$$ \int x(x^2+3)^4\,dx = \int u^4\left(\frac12\,du\right) = \frac12\int u^4\,du = \frac12\cdot\frac{u^5}{5}+C = \frac{u^5}{10}+C. $$

Substitute \(u=x^2+3\) back:

$$ \frac{(x^2+3)^5}{10}+C. $$

Check by differentiating:

$$ \frac{d}{dx}\left(\frac{(x^2+3)^5}{10}\right) = \frac{5}{10}(x^2+3)^4(2x) = x(x^2+3)^4. $$

The derivative recovers the original integrand, so the antiderivative is correct.

Notice that the substitution did not mean “replace \(x^2+3\) by \(u\) and leave everything else untouched.” It meant replacing \(x^2+3\) by \(u\) and replacing the matching part \(x\,dx\) by \(\frac12\,du\).

The Sign of the Differential Factor Matters

The derivative factor can be negative. Its sign must be included when converting the differential. Omitting that sign changes the answer.

Worked Example: Keeping a Negative Factor

Evaluate \(\int \sin x\,\cos^5 x\,dx\). Choose \(u=\cos x\). Differentiation gives

$$ u=\cos x, \qquad du=-\sin x\,dx, \qquad \sin x\,dx=-du. $$

Since the original integrand contains \(\sin x\,dx\), it becomes \(-du\), not \(du\):

$$ \int \sin x\,\cos^5 x\,dx = -\int u^5\,du = -\frac{u^6}{6}+C = -\frac{\cos^6 x}{6}+C. $$

Check the result:

$$ \frac{d}{dx}\left(-\frac{\cos^6 x}{6}\right) = -\frac{1}{6}\cdot 6\cos^5 x(-\sin x) = \sin x\,\cos^5 x. $$

The two negative signs, one from the antiderivative and one from differentiating cosine, produce the positive integrand required.

When the Differential Factor Is Missing

Consider an integral whose integrand contains a power of \(x^2+1\), but no factor \(2x\). If we set \(u=x^2+1\), then \(du=2x\,dx\). There is no \(x\,dx\) in the integral to replace by \(\frac12\,du\). Treating \(dx\) as if it were \(du\) would silently discard the derivative factor.

Worked Example: Correcting a Missing-Factor Error

Evaluate \(\int (x^2+1)^2\,dx\). A tempting but incorrect calculation is to set \(u=x^2+1\) and write

$$ \int (x^2+1)^2\,dx \ \not=\ \int u^2\,du = \frac{u^3}{3} = \frac{(x^2+1)^3}{3}. $$

The error occurs at the first equality: \(du=2x\,dx\), so \(dx\ne du\). In fact, on the interval \(x>0\), the substitution gives \(x=\sqrt{u-1}\) and

$$ dx=\frac{du}{2\sqrt{u-1}}. $$

Thus this substitution transforms the integral into

$$ \int (x^2+1)^2\,dx = \int \frac{u^2}{2\sqrt{u-1}}\,du, $$

not into \(\int u^2\,du\). The remaining factor shows that this substitution has not simplified the problem. Instead, expand the original integrand and integrate each term:

$$ (x^2+1)^2=x^4+2x^2+1, $$
$$ \int (x^2+1)^2\,dx = \int (x^4+2x^2+1)\,dx = \frac{x^5}{5}+\frac{2x^3}{3}+x+C. $$

Check by differentiating:

$$ \frac{d}{dx}\left(\frac{x^5}{5}+\frac{2x^3}{3}+x\right) = x^4+2x^2+1 = (x^2+1)^2. $$

By contrast, differentiating the tempting but incorrect answer gives

$$ \frac{d}{dx}\left(\frac{(x^2+1)^3}{3}\right) = \frac{1}{3}\cdot 3(x^2+1)^2(2x) = 2x(x^2+1)^2, $$

which has an extra factor \(2x\). It is not the original integrand.

A Reliable Check Before Integrating

Before carrying out a substitution, write down the proposed \(u\) and differentiate it. Then compare the resulting \(du\) with the differential factors actually present in the integral. The comparison may show that a factor is already present, is present up to a numerical multiplier or sign, or is missing.

1
Name the inner expression.
Set \(u=g(x)\) when that expression is the one you intend to simplify.
2
Differentiate the substitution.
Compute \(du=g'(x)\,dx\), retaining the full derivative factor.
3
Match the differential.
Rewrite the part of the integrand involving \(g'(x)\,dx\) in terms of \(du\). Do not replace \(dx\) by \(du\) alone.
4
Carry any leftover factor.
If the integrand does not contain the full derivative factor, keep and transform whatever remains; the substitution may not simplify the integral.
5
Check the result.
Differentiate the proposed antiderivative and compare it with the original integrand.

Why This Error Is So Easy to Make

The inner expression often attracts attention because replacing it by \(u\) visibly shortens the integrand. The differential factor can look like minor notation, even though it carries essential information about how the variable changes. The chain rule explains why that information cannot be omitted: differentiating a composite function always multiplies by the derivative of its inner function.

A useful way to diagnose a proposed answer is to differentiate it before accepting it. If the result contains an extra factor such as \(2x\), or is missing a negative sign, the substitution likely mishandled the differential. This check is direct and does not depend on whether the intermediate substitution looked plausible.

Takeaway: Changing \(g(x)\) to \(u\) requires changing \(g'(x)\,dx\) to \(du\). If the integrand does not supply that derivative factor, do not treat the substitution as though it did.

Check Your Understanding

For each question, pay particular attention to the relationship between the chosen inner expression and its differential.

  1. If \(u=x^2+5\), what is \(du\), and what expression involving \(x\,dx\) can be replaced by a multiple of \(du\)?
  2. Why is it incorrect to transform \(\int (x^2+1)^3\,dx\) directly into \(\int u^3\,du\) using \(u=x^2+1\)?
  3. When \(u=\sin x\), what is \(du\), and what sign must be used to rewrite \(\cos x\,dx\)?
  4. What does differentiating a proposed antiderivative reveal about a missing derivative factor?
  5. In \(\int 3x^2(x^3+2)^4\,dx\), identify the inner expression and its differential before integrating.

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