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Substitution and Reversing the Chain Rule · Tutorial 1158 of 1495

A Mixed Substitution Problem Workshop

Work through varied substitution integrals by matching the inner function with its differential, transforming definite-integral bounds, and verifying the answers.

Intermediate 9 min read

What You'll Learn

  • Identify an inner function that makes a composite integrand easier to integrate
  • Match its derivative to the differential, including any necessary constant factor
  • Apply substitution to indefinite integrals and include the constant of integration
  • Transform both bounds when evaluating a definite integral in the new variable
  • Check an antiderivative by differentiating and a definite result by evaluating consistently

A Consistent Method for Mixed Substitution Problems

Substitution is most useful when an integrand contains a composite expression and a factor related to the derivative of that expression. In a mixed set of problems, the inner function may be a polynomial, an exponent, a trigonometric expression, or a quantity under a radical. The details vary, but the central task stays the same: choose a new variable that simplifies the integrand and account for its differential correctly.

For an indefinite integral, substitution produces an antiderivative in the new variable; before finishing, write the result in the original variable. For a definite integral, you may instead change both bounds to the new variable and evaluate there. Mixing these routes—for example, using the new integrand with the old bounds—does not preserve the integral.

Substitution principle: If \(u=g(x)\), then \(du=g'(x)\,dx\). In an indefinite integral, replace the inner expression and its matching differential, integrate with respect to \(u\), and substitute back. In a definite integral from \(x=a\) to \(x=b\), the corresponding bounds in \(u\) are \(g(a)\) and \(g(b)\).

What to Look For

A useful substitution often has the form \(u=g(x)\), where \(g(x)\) is a visible inner expression. Its derivative, \(g'(x)\), should appear elsewhere in the integrand, perhaps multiplied by a constant. If the integrand contains \(g'(x)\) exactly, then \(g'(x)\,dx=du\). If it contains only a constant multiple of \(g'(x)\), adjust by that constant.

The basic reason this works is the chain rule: the derivative of a composite function \(F(g(x))\) is \(F'(g(x))g'(x)\). Substitution reverses this pattern by treating \(g(x)\) as the integration variable. In each example, check that the proposed answer really has the required derivative.

1
Choose the inner expression.
Look for a composite expression whose derivative is present, or is present up to a constant factor.
2
Write the differential accurately.
Differentiate \(u=g(x)\) and solve for the factor \(dx\) that appears in the integrand.
3
Integrate in one variable.
After substitution, express the whole integral using \(u\) and \(du\), then integrate.
4
Finish and check.
For an indefinite integral, substitute back and differentiate. For a definite integral in \(u\), change both bounds and evaluate.

Worked Examples

Worked Example: A Polynomial Inner Function

Evaluate \(\int 6x(3x^2+1)^4\,dx\).

The composite expression is \(3x^2+1\). Set \(u=3x^2+1\). Then

$$ du=6x\,dx. $$

Both the inner expression and its derivative are present, so the integral becomes

$$ \int 6x(3x^2+1)^4\,dx =\int u^4\,du =\frac{u^5}{5}+C. $$

Substituting back gives \(\frac{(3x^2+1)^5}{5}+C\). To check, differentiate:

$$ \frac{d}{dx}\left(\frac{(3x^2+1)^5}{5}\right) =\frac{1}{5}\cdot 5(3x^2+1)^4\cdot 6x =6x(3x^2+1)^4. $$

This is the original integrand, so the antiderivative is correct.

Worked Example: A Missing Constant Factor

Evaluate \(\int x e^{2x^2-3}\,dx\).

Choose the exponent as the inner function: \(u=2x^2-3\). Its differential is

$$ du=4x\,dx, \qquad x\,dx=\frac{1}{4}\,du. $$

The factor \(1/4\) is necessary because the integrand contains \(x\,dx\), not \(4x\,dx\). Therefore

$$ \int x e^{2x^2-3}\,dx =\frac{1}{4}\int e^u\,du =\frac{1}{4}e^u+C =\frac{1}{4}e^{2x^2-3}+C. $$

Differentiate the proposed result:

$$ \frac{d}{dx}\left(\frac{1}{4}e^{2x^2-3}\right) =\frac{1}{4}e^{2x^2-3}\cdot 4x =xe^{2x^2-3}. $$

The derivative matches the integrand, including its coefficient.

Worked Example: A Definite Integral with Changed Bounds

Evaluate \(\int_0^{\pi/2}\sin x(2+\cos x)^2\,dx\).

Set \(u=2+\cos x\), so \(du=-\sin x\,dx\). Change each bound using the same substitution. At \(x=0\), \(u=2+\cos 0=3\); at \(x=\pi/2\), \(u=2+\cos(\pi/2)=2\). Thus

$$ \int_0^{\pi/2}\sin x(2+\cos x)^2\,dx =-\int_3^2 u^2\,du. $$

An equivalent way to handle the negative sign is to reverse the bounds:

$$ -\int_3^2 u^2\,du =\int_2^3 u^2\,du =\left[\frac{u^3}{3}\right]_2^3 =\frac{27}{3}-\frac{8}{3} =\frac{19}{3}. $$

Check the transformed evaluation by using an antiderivative in \(u\) and the mapped bounds: \(u^3/3\) evaluated from \(2\) to \(3\) gives \(19/3\), as above. The original integrand is nonnegative on the interval, and the positive result is consistent with that sign.

Worked Example: A Definite Integral with a Radical

Evaluate \(\int_0^3 2x\sqrt{x^2+1}\,dx\).

Take \(u=x^2+1\). Then \(du=2x\,dx\). The bounds become \(u=1\) when \(x=0\), and \(u=10\) when \(x=3\). Substituting the integrand and both bounds gives

$$ \int_0^3 2x\sqrt{x^2+1}\,dx =\int_1^{10}\sqrt{u}\,du =\int_1^{10}u^{1/2}\,du. $$

Integrate and evaluate:

$$ \left[\frac{2}{3}u^{3/2}\right]_1^{10} =\frac{2}{3}\left(10^{3/2}-1^{3/2}\right) =\frac{2}{3}(10\sqrt{10}-1). $$

For a check, the antiderivative in the substituted variable is \(2u^{3/2}/3\), whose derivative is \(u^{1/2}\). At the mapped endpoints it gives \(2(10\sqrt{10})/3-2/3\), which is the stated value. Both bounds are essential to this evaluation.

Why the Substitution Works

The substitution method is justified by the chain rule and the Fundamental Theorem of Calculus. Here is the definite-integral argument. Suppose \(g\) is differentiable on \([a,b]\), and \(f\) is continuous on the interval containing the values \(g(x)\) for \(a\leq x\leq b\). Let \(F\) be an antiderivative of \(f\), so \(F'(u)=f(u)\). By the chain rule,

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

Thus \(F(g(x))\) is an antiderivative of the integrand \(f(g(x))g'(x)\). The Fundamental Theorem of Calculus then gives

$$ \int_a^b f(g(x))g'(x)\,dx =F(g(b))-F(g(a)) =\int_{g(a)}^{g(b)} f(u)\,du. $$

This also explains the bound rule: the lower \(x\)-endpoint \(a\) maps to \(g(a)\), and the upper endpoint \(b\) maps to \(g(b)\). If \(g\) decreases, the mapped upper bound may be smaller than the mapped lower bound. That is valid; the resulting orientation is accounted for by the integral itself.

When this definite-integral form applies: The integrand must be expressible as \(f(g(x))g'(x)\), with the stated continuity and differentiability conditions on the interval. If the derivative factor is present only up to a constant, first account for that constant. When using \(u\) as the variable of integration, use the corresponding \(u\)-bounds.

A Common Bounds Error

A frequent mistake is to substitute the integrand but keep the original bounds. Consider \(\int_0^1 2x(x^2+1)^2\,dx\). With \(u=x^2+1\), we have \(du=2x\,dx\), and the correct new bounds are \(u=1\) and \(u=2\). An incorrect calculation keeps \(0\) and \(1\):

$$ \text{Incorrect:}\qquad \int_0^1 2x(x^2+1)^2\,dx \overset{\text{wrong bounds}}{=} \int_0^1 u^2\,du =\frac{1}{3}. $$

The correction is to map both endpoints before evaluating:

$$ \int_0^1 2x(x^2+1)^2\,dx =\int_1^2 u^2\,du =\left[\frac{u^3}{3}\right]_1^2 =\frac{8}{3}-\frac{1}{3} =\frac{7}{3}. $$

This can also be checked using the original variable: an antiderivative is \((x^2+1)^3/3\), and its values at \(1\) and \(0\) differ by \(8/3-1/3=7/3\). The incorrect answer came from combining an integrand in \(u\) with limits in \(x\), which describe different variables.

Checking Your Work

A short check often catches a missing factor or an incomplete substitution. For an indefinite integral, differentiate the final answer after substituting back. For a definite integral evaluated in \(u\), check that the endpoints are the images of the original endpoints and that the antiderivative is evaluated in that same variable. In either case, keep track of a negative sign if \(du\) has the opposite sign from the differential factor in the original integral.

Takeaway: A substitution is complete only when the integrand, differential, and—if the integral is definite—both bounds are expressed consistently in the new variable. Check the final result in the form in which it is reported.

Check Your Understanding

For each problem, identify the substitution and show the differential or bounds needed to complete it.

  1. Evaluate \(\int 8x(4x^2+5)^3\,dx\), then differentiate your result to check it.
  2. For \(\int x e^{x^2+2}\,dx\), what constant factor is needed when replacing \(x\,dx\) with a multiple of \(du\)?
  3. Evaluate \(\int_0^1 2x(x^2+3)^2\,dx\) using \(u=x^2+3\). What are the new bounds?
  4. For \(\int_0^{\pi/2}\sin x(1+\cos x)^2\,dx\), what are the transformed bounds and what sign appears in \(du\)?
  5. State one check that can reveal an error in an indefinite substitution and one that can reveal an error in a definite substitution.

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