The Missing Constant In du

Why a Constant in \(du\) Matters

Substitution is a way to recognize the chain rule in reverse. When an integrand contains a function of an inner expression \(g(x)\), it can be useful to set \(u=g(x)\). But the substitution changes more than the expression inside the function: it also changes the differential.

The detail that is easy to lose is a constant multiplier. If \(u=5x-2\), then \(du=5\,dx\), not \(du=dx\). Replacing \(dx\) by \(du\) without accounting for the \(5\) changes the integral and produces an antiderivative with the wrong scale.

Key idea: If \(u=g(x)\), then \(du=g'(x)\,dx\). Keep the full derivative factor. If a constant is dropped, correct the resulting antiderivative by that factor and verify the result by differentiating.

The Differential Relation and the Correction

Suppose \(G\) is an antiderivative of \(F\), so \(G'(u)=F(u)\). For an affine inner function \(u=ax+b\), where \(a\ne 0\), the differential relation is \(du=a\,dx\), and therefore \(dx=du/a\). Thus the factor \(1/a\) must remain in the integral.

\int F(ax+b)\,dx = \frac{1}{a}\int F(u)\,du = \frac{1}{a}G(u)+C = \frac{1}{a}G(ax+b)+C.

This formula depends on \(a\ne 0\) and on \(G'=F\). It is not a rule to divide by a number that is not actually the derivative of the inner function. More generally, if \(u=g(x)\), then \(du=g'(x)\,dx\). When the integrand includes \(g'(x)\), the differential can be replaced using the entire factor. If only a constant multiple of \(g'(x)\) appears, that constant must also be accounted for.

The same principle applies when the inner function is a polynomial. Its derivative may be a constant multiple of a factor already present in the integrand. In that case, the substitution works after keeping the corresponding constant. But a polynomial derivative that varies with \(x\) cannot be replaced by a fixed constant.

Worked Examples

Worked Example: An Affine Inner Function

Evaluate \(\int \cos(5x-2)\,dx\). The inner function is \(u=5x-2\), so its derivative is \(5\), and the full differential relation is \(du=5\,dx\). Solving for \(dx\) gives \(dx=du/5\).

Substitute both the inner expression and the differential:

\begin{aligned} \int \cos(5x-2)\,dx &=\int \cos(u)\,\frac{du}{5}\\ &=\frac{1}{5}\int \cos(u)\,du\\ &=\frac{1}{5}\sin(u)+C\\ &=\frac{1}{5}\sin(5x-2)+C. \end{aligned}

Check by differentiating the proposed antiderivative:

\frac{d}{dx}\left(\frac{1}{5}\sin(5x-2)\right) = \frac{1}{5}\cos(5x-2)\cdot 5 = \cos(5x-2).

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

Worked Example: A Polynomial Inner Function

Evaluate \(\int x(x^2+1)^4\,dx\). Set \(u=x^2+1\). Then \(du=2x\,dx\), so \(x\,dx=du/2\). The factor \(1/2\) is required even though the rest of the integrand becomes a simple power of \(u\).

\begin{aligned} \int x(x^2+1)^4\,dx &=\int u^4\,\frac{du}{2}\\ &=\frac{1}{2}\int u^4\,du\\ &=\frac{1}{2}\cdot\frac{u^5}{5}+C\\ &=\frac{(x^2+1)^5}{10}+C. \end{aligned}

Differentiate the result to check both the power rule and the factor:

\begin{aligned} \frac{d}{dx}\left(\frac{(x^2+1)^5}{10}\right) &=\frac{1}{10}\cdot 5(x^2+1)^4\cdot 2x\\ &=x(x^2+1)^4. \end{aligned}

This matches the original integrand.

Worked Example: A Cubic Inner Function

Evaluate \(\int x^2\sqrt{2x^3-1}\,dx\), on an interval where the square root is defined. Set \(u=2x^3-1\). Its derivative is \(6x^2\), so \(du=6x^2\,dx\), or \(x^2\,dx=du/6\). Omitting the \(6\) would make the proposed antiderivative six times too large.

\begin{aligned} \int x^2\sqrt{2x^3-1}\,dx &=\frac{1}{6}\int u^{1/2}\,du\\ &=\frac{1}{6}\cdot\frac{u^{3/2}}{3/2}+C\\ &=\frac{1}{6}\cdot\frac{2}{3}u^{3/2}+C\\ &=\frac{1}{9}(2x^3-1)^{3/2}+C. \end{aligned}

Check by differentiation on an interval where \(2x^3-1>0\):

\begin{aligned} \frac{d}{dx}\left(\frac{1}{9}(2x^3-1)^{3/2}\right) &=\frac{1}{9}\cdot\frac{3}{2}(2x^3-1)^{1/2}\cdot 6x^2\\ &=x^2\sqrt{2x^3-1}. \end{aligned}

The derivative equals the integrand. At a boundary point where the square root is zero, the same antiderivative formula remains valid wherever the original integrand is defined.

Why the Correction Works

The correction is not an arithmetic trick. It follows directly from the chain rule. If \(G'(u)=F(u)\), then differentiating a scaled composite function gives

\frac{d}{dx}\left(\frac{1}{a}G(ax+b)\right) = \frac{1}{a}G'(ax+b)\cdot a = F(ax+b), \qquad a\ne 0.

The factor \(a\) from differentiating the inner function cancels the factor \(1/a\) outside. That cancellation is precisely what is lost if the constant in \(du=a\,dx\) is ignored.

The same verification works for a general differentiable inner function \(g\) wherever the relevant functions are defined. If \(G'=F\), then

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

This shows why an integrand of the form \(F(g(x))g'(x)\) has antiderivative \(G(g(x))+C\). If the integrand instead contains \(c\,F(g(x))g'(x)\) for a nonzero constant \(c\), an antiderivative is \(c\,G(g(x))+C\). The derivative must match the complete integrand, including all constant factors.

A Constant Factor Is Not Always Enough

For \(u=ax+b\), the derivative of the inner function is the constant \(a\). For a polynomial inner function, however, its derivative may depend on \(x\). A fixed multiplier can repair a missing constant, but it cannot replace a missing variable factor.

For example, consider \(\int (x^2+1)^4\,dx\) and try \(u=x^2+1\). Here \(du=2x\,dx\), but the integrand contains no factor \(x\). There is no constant \(c\) for which \(dx=c\,du\) holds for all \(x\), because \(du=2x\,dx\) includes the variable factor \(2x\). This substitution does not turn the integral into a constant multiple of \(\int u^4\,du\).

A derivative check exposes the mistake in claiming \((x^2+1)^5/5\) as an antiderivative:

\frac{d}{dx}\left(\frac{(x^2+1)^5}{5}\right) = \frac{1}{5}\cdot 5(x^2+1)^4\cdot 2x = 2x(x^2+1)^4,

which is not \((x^2+1)^4\). The extra \(2x\) shows that a nonconstant factor is missing. This check does not rule out other methods of evaluating the integral; it shows only that the proposed substitution and antiderivative do not match the integrand.

A Reliable Check for a Missing Constant

When a substitution produces an answer that looks plausible but may have lost a constant, use a derivative check rather than relying on how familiar the formula appears. The following sequence keeps the bookkeeping explicit.

Identify the inner function.
Write \(u=g(x)\) and calculate its full derivative \(g'(x)\).

Write the complete differential.
State \(du=g'(x)\,dx\), then solve for the part of \(dx\) used in the substitution.

Keep any constant multiplier.
If replacing a factor in the original integral introduces \(1/c\) or \(c\), carry it through the integration.

Differentiate the proposed answer.
Use the chain rule and check that every factor, including the inner derivative, reproduces the original integrand.

Common Mistake: Treating \(du\) as \(dx\)

A common incorrect calculation sets \(u=5x-2\), notes that the integrand becomes \(\cos(u)\), and writes \(\int\cos(u)\,du=\sin(u)+C\). Substituting back gives \(\sin(5x-2)+C\), but this has silently replaced \(dx\) by \(du\) even though \(du=5\,dx\).

The correction is \(dx=du/5\), so the integral is \(\frac{1}{5}\sin(5x-2)+C\). Differentiating the incorrect result confirms the error:

\frac{d}{dx}\sin(5x-2) = 5\cos(5x-2),

which is five times the required integrand. A derivative check is especially useful here because it identifies both the size and the source of the error: the inner derivative \(5\) was not balanced by a factor \(1/5\).

What to Remember

The essential habit is to record the whole relation \(du=g'(x)\,dx\), not just the new inner variable. With an affine inner function, the derivative is a constant, and omitting it changes the scale of the answer. With a polynomial inner function, calculate its derivative before deciding whether the missing factor is a constant or a variable expression. In either case, differentiating the proposed antiderivative gives a direct and decisive check.

Check Your Understanding

For each question, pay attention to the full derivative of the inner function and the factor that remains in the integral.

For \(u=7x+3\), what is \(du\), and what expression for \(dx\) should be used in a substitution?
Evaluate \(\int \sin(4x)\,dx\), keeping the constant from \(du\), and verify the answer by differentiation.
In \(\int x(x^2+3)^2\,dx\), what is \(du\) when \(u=x^2+3\), and what constant factor must be included?
For \(u=x^3+1\), can \(dx\) be replaced by a constant multiple of \(du\) for all \(x\)? Explain using \(du\).
Differentiate \(\frac{1}{6}(3x+1)^4\). What factor would be missing if the coefficient \(1/3\) were omitted from an antiderivative of \((3x+1)^3\)?

What You'll Learn

Why a Constant in \(du\) Matters

The Differential Relation and the Correction