Substitution Strategies
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 1.5: “Substitution” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/1-5-substitution |
| Textbook used in class | Stewart, Calculus, Section 4.5: “The Substitution Rule” |
Opening Scenario
Knowing the mechanics of $u$-substitution is not enough to handle unfamiliar integrals. The hard part is choosing $u$. Several strategies make that choice, and a few signs show when substitution is the right tool at all.
Quick Reference
When to try substitution: The integrand contains a composed function $f(g(x))$ where $g'(x)$ (or a constant multiple) also appears.
Heuristics for choosing $u$:
- Try $u = $ the expression inside a parenthesis, radical, exponent, or denominator.
- Try $u = $ the argument of a trigonometric or exponential function.
- If the integrand is a product $h(x)\cdot k(x)$ where $h = g'$ and $k = f(g)$, set $u = g(x)$.
When substitution does not apply directly: The derivative of your candidate $u$ does not appear and cannot be forced in (up to a constant). Consider algebraic manipulation, a trigonometric identity, or a different technique (integration by parts, partial fractions).
Key Concepts
1. Reading the Integrand for Structure
Substitution unzips the chain rule. The chain rule produces products of the form $f'(g(x))\cdot g'(x)$. To find this structure in an integrand:
- Look for a function and its derivative. In $\int x e^{x^2}\,dx$, the factor $x$ is (up to a constant) the derivative of $x^2$, which is the exponent. Set $u = x^2$.
- Look inside composite expressions. In $\int \cos^3 x\,\sin x\,dx$, think of $\cos x$ as the inner function. Its derivative is $-\sin x$, and $\sin x$ is present. Set $u = \cos x$.
- Check the exponent or denominator. In $\int \dfrac{x}{(x^2+1)^5}\,dx$, the denominator contains $x^2 + 1$ and $x$ is (up to a constant) its derivative. Set $u = x^2 + 1$.
2. Adjusting for Missing Constants
If $g'(x)$ is missing from the integrand but differs only by a constant factor, multiply and divide by that constant:
$\displaystyle\int \cos(3x)\,dx$: the derivative of $3x$ is $3$, but $3$ is absent. Write $\cos(3x)\,dx = \dfrac{1}{3}[3\,\cos(3x)\,dx]$; now $3\,dx = du$ and the integral becomes $\dfrac{1}{3}\int\cos u\,du$.
3. Rearranging Before Substituting
Sometimes a small algebraic manipulation reveals the substitution structure.
Example. $\displaystyle\int \frac{x+1}{\sqrt{x}}\,dx$. No obvious composite function -- rewrite first: $\dfrac{x+1}{\sqrt{x}} = x^{1/2} + x^{-1/2}$, then integrate term by term using the power rule. No substitution is needed.
Example. $\displaystyle\int x\sqrt{x+1}\,dx$. Set $u = x+1$, so $x = u-1$ and $dx = du$:
$\displaystyle\int (u-1)\sqrt{u}\,du = \int (u^{3/2} - u^{1/2})\,du = \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} + C$.
Back-substitute: $= \dfrac{2}{5}(x+1)^{5/2} - \dfrac{2}{3}(x+1)^{3/2} + C$.
4. When Substitution Fails
Substitution is not universal. It does not work for products where neither factor is (a constant multiple of) the derivative of the other -- for example, $\int x\sin x\,dx$. Such integrals require integration by parts. Substitution also cannot directly handle rational functions like $\int \dfrac{1}{x^2-1}\,dx$; those require partial fractions.
Recognizing when substitution is unlikely to work saves time. If you have tried two or three natural candidates for $u$ and none eliminate all the $x$’s from the integral, substitute does not apply -- look for another technique.
“If I try hard enough, I can always find a substitution.” Not every integral yields to substitution. The techniques in a first calculus course (substitution, integration by parts, partial fractions, trig substitution) together cover a wide range of integrals, but many integrals have no closed form at all. The skill is knowing which technique to try first, and when to stop and switch.
Worked Examples
Example 1. $\displaystyle\int \frac{\ln x}{x}\,dx$.
The factor $\frac{1}{x}$ is the derivative of $\ln x$. Set $u = \ln x$, $du = \frac{1}{x}\,dx$.
$\displaystyle\int u\,du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C$.
Example 2. $\displaystyle\int \sin^4 x\,\cos x\,dx$.
The factor $\cos x$ is the derivative of $\sin x$. Set $u = \sin x$, $du = \cos x\,dx$.
$\displaystyle\int u^4\,du = \frac{u^5}{5} + C = \frac{\sin^5 x}{5} + C$.
Example 3. $\displaystyle\int \tan x\,dx$.
Write $\tan x = \dfrac{\sin x}{\cos x}$. The numerator $\sin x$ is (up to sign) the derivative of $\cos x$. Set $u = \cos x$, $du = -\sin x\,dx$:
$\displaystyle\int \frac{\sin x}{\cos x}\,dx = \int \frac{-du}{u} = -\ln|u| + C = -\ln|\cos x| + C$.
This can also be written $\ln|\sec x| + C$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Choosing $u$ that does not simplify the integrand | Setting $u = x^2$ in $\int x^3 e^x\,dx$ (no $x^2$ in the remaining part) | The derivative $2x$ must match the remaining factor; $\int x^3 e^x\,dx$ requires integration by parts |
| Forgetting to rewrite $x$ in terms of $u$ when $x$ remains after substituting $du$ | Partial substitution in $\int x\sqrt{x+1}\,dx$ without replacing $x$ with $u-1$ | Express $x = u - 1$ and substitute fully |
| Applying substitution when the integrand has no composite structure | Trying $u = x+1$ in $\int \frac{1}{x+1}\,dx$ | This actually works: $u = x+1$ gives $\int \frac{du}{u} = \ln|u| + C = \ln|x+1| + C$. Check carefully before giving up |
Leveled Practice
Level 1 -- Identify $u$
Problem 1. For each integral, state the most natural choice of $u$ and find $du$. Do not integrate.
(a) $\displaystyle\int e^{x^2}\cdot 2x\,dx$ (b) $\displaystyle\int \frac{\cos(\ln x)}{x}\,dx$ (c) $\displaystyle\int (5x^4 + 1)(x^5 + x)^3\,dx$
Show answer
(a) $u = x^2$, $du = 2x\,dx$.
(b) $u = \ln x$, $du = \frac{1}{x}\,dx$.
(c) $u = x^5 + x$, $du = (5x^4 + 1)\,dx$.
Level 2 -- Full Computation
Problem 2. Evaluate $\displaystyle\int \cos^5 x\,\sin x\,dx$.
Show answer
$u = \cos x$, $du = -\sin x\,dx$.
$\displaystyle\int u^5(-du) = -\frac{u^6}{6} + C = -\frac{\cos^6 x}{6} + C$.
Level 3 -- Rewrite First
Problem 3. Evaluate $\displaystyle\int \frac{e^x}{e^x + 1}\,dx$.
Show answer
$u = e^x + 1$, $du = e^x\,dx$.
$\displaystyle\int \frac{du}{u} = \ln|u| + C = \ln(e^x + 1) + C$.
(No absolute value needed since $e^x + 1 > 0$.)
Mastery Checklist
Mental Model
Choosing $u$ is pattern-matching. Ask: “Is there a function inside another function here?” If yes, set $u$ equal to the inner function, compute $du$, and check whether $du$ (or a constant multiple) is already present. If it is, the substitution will work. If not -- if eliminating $x$ entirely requires more than multiplying by a constant -- then either rewrite the integrand first, or try a different technique altogether.
The goal is always the same: reduce the integral to one of the standard forms in the basic table.
Connections
Looking back
- Substitution mechanics (Section 4.5): This lesson extends the mechanical procedure to the strategic question of when and how to choose $u$.
- Algebraic manipulation: Rewriting the integrand (distributing, combining fractions, factoring) often reveals the substitution.
Looking ahead
- Integration by parts (Chapter 7): The technique for products where neither factor is the derivative of the other.
- Trigonometric substitution (Chapter 7): For integrals involving $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$.
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