The Core Concept
What is U-Substitution?
U-substitution is the integration counterpart of the chain rule for differentiation. When an integrand contains a composite function, we replace the inner function with $u$ to simplify the integral into a basic form we can evaluate directly.
The Intuition
Spot the Pattern
You need a function and its derivative (or a constant multiple of it) both present in the integrand.
Substitute
Replace the inner function with $u$, and rewrite $dx$ in terms of $du$. This transforms the messy integral into a clean one.
Integrate
Evaluate the resulting integral in $u$ using basic antiderivative rules.
Back-Substitute
Replace $u$ with the original expression in $x$ to get your final answer.
The 5-Step Method
Follow these steps every time — no exceptions:
Choose $u$
Pick $u$ = the inner function (usually what's inside a power, root, trig function, or exponent). Look for a function whose derivative is also present.
Find $\dfrac{du}{dx}$, solve for $dx$
Differentiate $u$ with respect to $x$. Rearrange to express $dx$ in terms of $du$: $\;dx = \dfrac{du}{u'}$.
Substitute everything
Replace the inner expression with $u$ AND replace $dx$ with the $du$ expression. The result should have no remaining $x$'s.
Integrate in $u$
Evaluate $\int f(u)\,du$ using the basic rules (power rule, trig integrals, exponential rules).
Back-substitute (indefinite integrals only)
Replace $u$ with your original expression in $x$. Add $+ C$.
How to Choose $u$ — Decision Guide
Inside a Power
$\int (3x+1)^5\,dx$ → let $u = 3x+1$
Inside a Root
$\int \sqrt{2x-1}\,dx$ → let $u = 2x-1$
Exponent of $e$
$\int e^{3x}\,dx$ → let $u = 3x$
Inside Trig
$\int \cos(2x)\,dx$ → let $u = 2x$
Pro tip: If your derivative brings an unwanted constant (like $\frac{1}{3}$), just multiply outside the integral to compensate — you can always divide it out.
Essential Formulas
Basic Antiderivatives (after substitution, $u$ is the variable)
Definite Integral with U-Sub — Limits Rule
When evaluating definite integrals, change the limits of integration to $u$-values:
This means you never need to back-substitute — just evaluate the $u$-integral at the new limits directly.
Worked Examples
These are based on the problems from your study guide. Every step is shown.
Example 1 — Definite Integral with Trig
Full Solution
Note: $\sin(\pi) = 0$ and $\sin(0) = 0$, so the integral evaluates to zero.
Example 2 — Sum of Two Integrals (Splitting)
Full Solution
Key technique: when the integrand is a sum, split it and apply u-sub independently to each part.
Example 3 — Definite Integral, Composite Polynomial
Full Solution
Key technique: recognize when a factor outside cancels with the derivative of $u$, leaving only a constant multiplier.
Example 4 — Double Substitution (Nested)
Full Solution
Key technique: apply u-sub twice in sequence. Start with the most deeply nested expression ($\sqrt{x}$), then substitute again for the remaining composite.
Definite Integrals — Special Rules
Two Valid Methods
| Method | Steps | When to Use |
|---|---|---|
| Change Limits | Convert $a, b$ to $u$-values → integrate in $u$ → evaluate directly | Preferred — faster, no back-substitution needed |
| Keep Original Limits | Integrate in $u$ → back-substitute → evaluate at $x = a, b$ | When you forget to change limits (still valid) |
Limit Conversion Practice
| $u$ substitution | $x = a$ | New lower $u$ | $x = b$ | New upper $u$ |
|---|---|---|---|---|
| $u = 2x$ | $0$ | $0$ | $\pi/2$ | $\pi$ |
| $u = x^2+2x-1$ | $-1$ | $-2$ | $0$ | $-1$ |
| $u = 3x+1$ | $0$ | $1$ | $2$ | $7$ |
| $u = x^2$ | $1$ | $1$ | $3$ | $9$ |
Practice Problems
Click a problem to expand it. Use the hint before revealing the full solution.
P1 — Basic U-Sub (Indefinite)
Solution
P2 — Exponential U-Sub
Solution
P3 — Trigonometric U-Sub
Solution
P4 — Definite Integral with Changed Limits
Solution
P5 — Rational / Log Form
Solution
P6 — Tricky: Adjust for Missing Factor
Solution
P7 — Challenge: Trig with Definite Bounds
Solution
Common Errors to Avoid
These are the mistakes that cost students points. Know them cold.
After substituting $u$, you must also convert the limits $a$ and $b$ to $u$-values using $u = g(a)$ and $u = g(b)$. Leaving old $x$-limits on a $u$-integral gives the wrong answer.
An answer with $u$ in it is incomplete for an indefinite integral. Always replace $u$ with the original expression in $x$ before writing your final answer.
Every indefinite integral has a constant of integration. Omitting it on a test is a guaranteed point deduction.
When $du$ gives a coefficient you don't want (e.g., $du = 5\,dx$), you must divide: $dx = \frac{du}{5}$, bringing $\frac{1}{5}$ outside the integral. Students often multiply instead of divide, or drop the factor entirely.
Choosing a $u$ whose derivative doesn't appear in the integrand means you'll have leftover $x$-terms that can't be cancelled. If you get stuck with $x$'s remaining, reconsider your choice of $u$.
$\int u^{1/2}\,du = \frac{u^{3/2}}{3/2} = \frac{2}{3}u^{3/2}$, not $\frac{1}{2}u^{-1/2}$ (that's the derivative). Raise the power, divide by it.
$\int \sin(u)\,du = -\cos(u) + C$ (negative!). Students often write $+\cos(u)$. When back-substituting trig, carefully track the sign.
Quick Quiz
Test yourself. Each question is multiple choice. Instant feedback after each answer.
For $\displaystyle\int (6x)(3x^2+1)^4\,dx$, the best choice of $u$ is:
Evaluate $\displaystyle\int_1^2 2x\,e^{x^2}\,dx$.
$\displaystyle\int \cos(5x)\,dx = $?
$\displaystyle\int \frac{4x^3}{x^4+7}\,dx = $?
Which expression equals $\displaystyle\int \sqrt{2x-1}\,dx$?
Score: —
Rapid-Reference Cheat Sheet
U-Sub Antiderivatives at a Glance
Quick Checklist Before You Submit
U-Choice Quick Reference
| Integrand Structure | Choose $u =$ | Example |
|---|---|---|
| $(f(x))^n \cdot f'(x)$ | inner function | $(x^2+1)^3 \cdot 2x$ |
| $e^{f(x)} \cdot f'(x)$ | exponent | $e^{3x} \cdot 3$ |
| $\dfrac{f'(x)}{f(x)}$ | denominator | $\dfrac{2x}{x^2+1}$ |
| $\sin(f(x)) \cdot f'(x)$ | trig argument | $\sin(x^2) \cdot 2x$ |
| $\dfrac{1}{\sqrt{x}} \cdot g(\sqrt{x})$ | $\sqrt{x}$ | $\dfrac{\cos\sqrt{x}}{\sqrt{x}}$ |