Substitution: Indefinite Integrals
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
The basic integration table handles $\int x^5\,dx$ and $\int \cos x\,dx$, but what about $\int x^5(x^6 + 1)^{10}\,dx$ or $\int \cos(5x)\,dx$? These integrands involve compositions of functions -- and the chain rule is the derivative rule for compositions. The Substitution Rule is the chain rule run backward: it replaces a composed expression with a simpler one, turning an unfamiliar integral into a familiar one.
Quick Reference
Substitution Rule. Let $u = g(x)$ where $g$ is differentiable and $f$ is continuous on the range of $g$. Then $$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du.$$
Procedure:
- Identify $u = g(x)$ (an inner function).
- Compute $du = g'(x)\,dx$.
- Rewrite the integral entirely in terms of $u$.
- Evaluate the simpler $\int f(u)\,du$.
- Substitute back: replace $u$ with $g(x)$.
Key Concepts
1. Why Substitution Works
By the chain rule: $\dfrac{d}{dx}[F(g(x))] = F'(g(x))\cdot g'(x)$. Therefore: $$\int F'(g(x))\cdot g'(x)\,dx = F(g(x)) + C.$$
With $u = g(x)$, $du = g'(x)\,dx$, and $F' = f$, this reads: $\int f(u)\,du = F(u) + C$. The substitution is just a notational repackaging of this identity.
2. Recognizing the Pattern
The substitution pattern $\int f(g(x))\cdot g'(x)\,dx$ requires:
- An inner function $g(x)$ whose derivative $g'(x)$ (or a constant multiple of it) appears in the integrand.
- An outer function $f$ applied to $g(x)$.
Look for a factor in the integrand that is the derivative (up to a constant) of another factor or of an expression inside a more complex factor.
3. Handling the $du$ -- Constant Multiples
If $du$ matches the integrand up to a constant factor, compensate:
Example. $\displaystyle\int \sin(5x)\,dx$.
Let $u = 5x$, $du = 5\,dx$, so $dx = \dfrac{du}{5}$.
$\displaystyle\int \sin(5x)\,dx = \int \sin u\cdot\frac{du}{5} = \frac{1}{5}(-\cos u) + C = -\frac{1}{5}\cos(5x) + C$.
“Any substitution will simplify the integral.” Substitution only helps when $g'(x)$ (or a constant multiple) is already present in the integrand. If you choose $u$ and the remaining integral still contains $x$ in a complicated way that cannot be eliminated, a different $u$ is needed.
Worked Example
Evaluate $\displaystyle\int 2x\sqrt{x^2 + 1}\,dx$.
Step 1 -- Choose $u$. The expression inside the square root is $x^2 + 1$, and its derivative $2x$ appears in the integrand. Let $u = x^2 + 1$.
Step 2 -- Compute $du$. $du = 2x\,dx$, so $2x\,dx = du$.
Step 3 -- Rewrite. $\displaystyle\int 2x\sqrt{x^2+1}\,dx = \int \sqrt{u}\,du = \int u^{1/2}\,du$.
Step 4 -- Integrate. $\displaystyle\int u^{1/2}\,du = \dfrac{u^{3/2}}{3/2} + C = \dfrac{2}{3}u^{3/2} + C$.
Step 5 -- Back-substitute. $= \dfrac{2}{3}(x^2 + 1)^{3/2} + C$.
Verification. $\dfrac{d}{dx}\left[\dfrac{2}{3}(x^2+1)^{3/2}\right] = \dfrac{2}{3}\cdot\dfrac{3}{2}(x^2+1)^{1/2}\cdot 2x = 2x\sqrt{x^2+1}$. Confirmed.
Evaluate $\displaystyle\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx$.
Let $u = \sqrt{x} = x^{1/2}$, $du = \dfrac{1}{2\sqrt{x}}\,dx$, so $\dfrac{dx}{\sqrt{x}} = 2\,du$.
$\displaystyle\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx = \int e^u\cdot 2\,du = 2e^u + C = 2e^{\sqrt{x}} + C$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Leaving $x$ in the integral after substitution | $\int 2x(x^2+1)^5\,dx = \int 2x\cdot u^5\,du$ | All $x$ must be replaced: $2x\,dx = du$, so the integral is $\int u^5\,du$ |
| Forgetting to back-substitute | Leaving the answer as $\frac{u^6}{6} + C$ | Replace $u$ with the original expression in $x$: $\frac{(x^2+1)^6}{6} + C$ |
| Choosing $u$ without $g'(x)$ in the integrand | Setting $u = x^3 + 1$ in $\int x^5(x^3+1)^4\,dx$ when $3x^2 \neq x^5$ (off by more than a constant) | Check that $du$ matches (up to a constant) what remains in the integrand |
Leveled Practice
Level 1 -- Direct Substitution
Problem 1. Evaluate $\displaystyle\int (3x+1)^4\,dx$.
Show answer
$u = 3x+1$, $du = 3\,dx$, so $dx = \frac{du}{3}$.
$\displaystyle\int u^4\,\frac{du}{3} = \frac{1}{3}\cdot\frac{u^5}{5} + C = \frac{(3x+1)^5}{15} + C$.
Level 2 -- Derivative Factor Present
Problem 2. Evaluate $\displaystyle\int x^2 e^{x^3}\,dx$.
Show answer
$u = x^3$, $du = 3x^2\,dx$, so $x^2\,dx = \frac{du}{3}$.
$\displaystyle\int e^u\,\frac{du}{3} = \frac{1}{3}e^u + C = \frac{1}{3}e^{x^3} + C$.
Level 3 -- Completing the Substitution
Problem 3. Evaluate $\displaystyle\int \frac{\ln x}{x}\,dx$.
Show answer
$u = \ln x$, $du = \frac{1}{x}\,dx$.
$\displaystyle\int u\,du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C$.
Mastery Checklist
Mental Model
Substitution compresses the “action” in an integrand. The integral $\int 2x(x^2+1)^{10}\,dx$ contains a complicated inner expression $(x^2+1)$ raised to a high power, but the factor $2x$ is exactly the derivative of that inner expression. Substituting $u = x^2+1$ “unzips” the chain rule: the complexity is absorbed into $u$, and what remains is the simple $\int u^{10}\,du$.
After integrating and back-substituting, the chain rule is “re-zipped.” Verification by differentiation always checks that the zip was done correctly.
Connections
Looking back
- Chain rule (Section 2.5): Substitution reverses the chain rule.
- Basic integration formulas (Section 4.4): The integral in $u$ must match one of these standard forms.
Looking ahead
- Substitution for definite integrals (Section 4.5): The same method, but the bounds of integration change when $u$ changes.
- Integration by parts (Chapter 7): The next main integration technique, reversing the product rule.
Back to Integration Foundations | Next: Substitution -- Definite Integrals