# Integration by substitution (u-sub)

> AP Calculus AB · Integration and Accumulation of Change
> Source: https://www.owlsprep.com/study/ap-calculus-ab-u6-integration-by-substitution/

This guide aligns to AP Calculus AB requirements, covering u-substitution for indefinite/definite integrals, reversing the chain rule, changing bounds, u-choice strategy, and common exam pitfalls.

**Prerequisites:** Chain rule for derivatives of composite functions; Antiderivatives of basic power, exponential, and trigonometric functions; Fundamental Theorem of Calculus for definite integrals

## Learning objectives

- Reverse the chain rule using u-substitution for indefinite integrals of composite functions
- Apply u-substitution with changing bounds to evaluate definite integrals
- Use a consistent strategy to correctly choose $u$ for common composite functions
- Avoid common u-substitution errors that lose points on the AP exam

## What Is Integration by Substitution (u-sub)?

Integration by substitution (commonly shortened to u-sub) is the core advanced integration technique for AP Calculus AB, designed explicitly to reverse the chain rule from differentiation. Per the AP CED, this topic accounts for 10-15% of Unit 6 weight, and you can expect 2-3 MCQ questions and at least one FRQ part requiring u-sub on every full AP exam.

The method works by rewriting a complicated integral of a composite function in terms of a new variable $u$, which is chosen to be the inner function of the composite. This turns an unfamiliar integral into a basic integral you already know how to solve.

**Integration by substitution (u-sub)** — A change-of-variable integration technique that reverses the chain rule, rewriting integrals of composite functions into simpler solvable integrals. Also called reverse chain rule integration.

*Notation:* $u = h(x)$ where $h(x)$ is the inner composite function

> **tip**
>
> All integration techniques on the AP Calculus AB exam build on u-sub, so mastering it is non-negotiable for earning a 4 or 5.

## U-Substitution for Indefinite Integrals

U-sub reverses the chain rule relationship. For a composite function $f(x) = g(h(x))$, the chain rule gives:

$$f'(x) = g'(h(x)) \cdot h'(x)$$

For an integral of the form $\int g'(h(x)) \cdot h'(x) dx$, set $u = h(x)$, so $du = h'(x) dx$. Substituting gives:

$$\int g'(u) du = g(u) + C = g(h(x)) + C$$

If you are only missing a constant coefficient, you can adjust by rearranging the differential and factoring the reciprocal constant out of the integral. Non-constant adjustments are never required on AP Calculus AB.

**Worked example:** Find the indefinite integral $\int 6x (2x^2 - 5)^4 dx$

1. Identify the inner function of the composite:
2. $$u = 2x^2 - 5$$
3. Compute the differential and rearrange to match the integrand:
4. $$\frac{du}{dx} = 4x \implies du = 4x dx \implies 6x dx = \frac{3}{2} du$$
5. Rewrite the integral entirely in terms of $u$:
6. $$\int (2x^2 -5)^4 \cdot 6x dx = \frac{3}{2} \int u^4 du$$
7. Integrate with respect to $u$:
8. $$\frac{3}{2} \cdot \frac{u^5}{5} + C = \frac{3u^5}{10} + C$$
9. Substitute back to $x$ for the final answer:
10. $$\frac{3(2x^2 - 5)^5}{10} + C$$

> **Exam tip:** Always substitute back to the original variable $x$ for indefinite integrals. AP exam graders will deduct full points for a correct antiderivative left in terms of $u$.

## U-Substitution for Definite Integrals (Changing Bounds)

For definite integrals, you can either substitute back to $x$ after integrating, or change the bounds of integration to match $u$, which eliminates back-substitution entirely. The bounds-changing method is faster and less error-prone on the AP exam, so it is the recommended approach.

When changing bounds for $\int_{x=a}^{x=b} f(h(x)) h'(x) dx$, after setting $u = h(x)$, calculate the lower $u$-bound as $u = h(a)$ and the upper $u$-bound as $u = h(b)$. The integral becomes:

$$\int_{h(a)}^{h(b)} f(u) du$$

You integrate directly with respect to $u$ and evaluate, with no back-substitution needed. This method is especially common for AP MCQ where you only need the final numerical value.

**Worked example:** Evaluate the definite integral $\int_{x=1}^{x=3} 4x e^{x^2} dx$

1. Choose the inner function as the exponent of the composite exponential:
2. $$u = x^2$$
3. Compute the differential and adjust for the constant coefficient:
4. $$du = 2x dx \implies 4x dx = 2 du$$
5. Change the bounds of integration to match $u$:
6. $$x=1 \implies u = 1^2 = 1; \quad x=3 \implies u = 3^2 = 9$$
7. Rewrite and integrate in terms of $u$:
8. $$2 \int_{1}^{9} e^u du = 2 \left[ e^u \right]_1^9$$
9. Evaluate using the Fundamental Theorem of Calculus:
10. $$2(e^9 - e^1) = 2e^9 - 2e$$

> **Exam tip:** Write down your new u-bounds immediately after setting $u$, before you rewrite the integral. This eliminates the common mistake of accidentally using the original x-bounds when integrating with respect to $u$.

## U-Choice Strategy for Non-Linear Inner Functions

Most u-sub problems on the AP exam use non-linear inner functions, so having a consistent strategy for choosing $u$ is critical. The number one rule of thumb for AP AB: if you see a function and its derivative (up to a constant multiple) in the integrand, the function is your $u$.

Common non-linear inner functions tested on AP AB include powers of trigonometric functions, logarithms, polynomials under roots, and exponential functions. If you end up needing a non-constant term of $x$ to complete $du$, you have almost certainly chosen the wrong $u$.

**Worked example:** Find the indefinite integral $\int \frac{\tan^2(2x) \sec^2(2x)}{1/2} dx$

1. Identify that $\tan(2x)$ raised to a power has derivative $\sec^2(2x)$ (up to a constant), so set:
2. $$u = \tan(2x)$$
3. Compute the differential and adjust for the constant:
4. $$\frac{du}{dx} = 2 \sec^2(2x) \implies \frac{1}{2} du = \sec^2(2x) dx$$
5. Simplify the original integrand and substitute:
6. $$2 \int \tan^2(2x) \sec^2(2x) dx = 2 \int u^2 \cdot \frac{1}{2} du = \int u^2 du$$
7. Integrate and substitute back to $x$:
8. $$\frac{u^3}{3} + C = \frac{\tan^3(2x)}{3} + C$$

**Worked example:** The marginal profit of a small bakery selling $q$ loaves of bread is $MP(q) = \frac{2q}{q^2 + 100}$, measured in hundreds of dollars per loaf. What is the total change in profit when increasing production from 10 loaves to 20 loaves? Round to the nearest whole dollar.

1. Total change in profit is the integral of marginal profit over the interval:
2. $$\text{Total Change} = \int_{10}^{20} \frac{2q}{q^2 + 100} dq$$
3. The denominator is a function whose derivative is in the numerator, so set:
4. $$u = q^2 + 100 \implies du = 2q dq$$
5. Change bounds to match $u$:
6. $$q=10 \implies u=200; \quad q=20 \implies u=500$$
7. Integrate and evaluate:
8. $$\int_{200}^{500} \frac{1}{u} du = \left[ \ln u \right]_{200}^{500} = \ln\left(\frac{500}{200}\right) = \ln(2.5) \approx 0.9163$$
9. Convert units (hundreds of dollars) for the final answer:
10. $0.9163 \times 100 \approx 92$, so total profit increases by approximately \$92.

**Check your understanding**

Test your understanding of definite u-sub:

1. Evaluate $\int_{1}^{e^2} \frac{(\ln x)^2}{x} dx$. Which is the correct value?

   - $\frac{8}{3}$
   - $\frac{(e^2)^3}{9} - \frac{1}{9}$
   - $\frac{2}{3}$
   - $e^6 - 1$

   *Why:* Correct. Set $u = \ln x$, which gives bounds from 0 to 2, leading to $\int_0^2 u^2 du = \frac{8}{3}$. Distractors come from incorrect bounds or back-substitution errors.

> **Exam tip:** Never change your $u$ to adjust for a missing constant factor. Just rearrange the differential to get the correct multiple of $du$, and factor the constant out of the integral. Changing $u$ for a constant will always introduce unnecessary errors.

## Common pitfalls

- **Wrong:** Leaving indefinite integrals in terms of $u$ instead of substituting back to $x$
  - Why it fails: Students get used to the bounds-changing method for definite integrals and forget that indefinite integrals require an answer in the original variable
  - Correct: Always replace $u$ with its original expression in $x$ before writing your final answer for an indefinite integral
- **Wrong:** Keeping the original $x$-bounds when integrating a definite integral in terms of $u$
  - Why it fails: Students rush and skip the step of calculating new bounds, or forget that the variable of integration changed
  - Correct: Immediately after setting $u$, write down the new lower and upper bounds for $u$ next to your work, before you rewrite the integral
- **Wrong:** Choosing $u$ as the outer function instead of the inner function of the composite
  - Why it fails: Students memorize "pick the complicated part" but misidentify which part is the inner composite
  - Correct: For any composite $f(g(x))$, $u$ is always $g(x)$, the inner function
- **Wrong:** When $du = k h'(x) dx$, writing $k du = h'(x) dx$ instead of $\frac{1}{k} du = h'(x) dx$
  - Why it fails: Students mix up algebra when rearranging the differential equation
  - Correct: Always write $\frac{du}{dx} = h'(x)$ first, then rearrange term by term to get $h'(x) dx$ in terms of $du$
- **Wrong:** Adding the constant of integration $C$ before integrating, resulting in an extra factor of $C$
  - Why it fails: Students rush and add $C$ too early, incorrectly treating it as a variable
  - Correct: Add the single constant $C$ once, immediately after integrating with respect to $u$, before substituting back to $x$

## Cheatsheet

| Category | Rule / Formula | Notes |
| --- | --- | --- |
| Core Reverse Chain Rule | $\int g'(h(x)) h'(x) dx = g(h(x)) + C$ | Applies to any composite function $g(h(x))$ |
| U-Substitution Definition | $u = h(x), \; du = h'(x) dx$ | $u$ is always the inner function of the composite |
| Indefinite U-Sub Step | $\int f(h(x)) h'(x) dx = \int f(u) du = F(u) + C = F(h(x)) + C$ | Always substitute back to $x$ for final answer |
| Definite U-Sub (Change Bounds) | $\int_{x=a}^{x=b} f(h(x)) h'(x) dx = \int_{h(a)}^{h(b)} f(u) du$ | No back-substitution needed after integration |
| Constant Adjustment | If $du = k h'(x) dx$, then $\frac{1}{k} du = h'(x) dx$ | Only constant factors need adjustment on AP AB |
| Power Composite U-Choice | $\int x^n (a x^{n+1} + b)^k dx \implies u = a x^{n+1} + b$ | Works for all constant $a,b,k,n$ |
| Log/Exp U-Choice | $\int \frac{f'(\ln x)}{x} dx \implies u = \ln x$; $\int f'(e^x) e^x dx \implies u = e^x$ | Derivatives simplify perfectly for substitution |
| Trig Composite U-Choice | $\int f'(\sin x) \cos x dx \implies u = \sin x$; $\int f'(\cos x) \sin x dx \implies u = \cos x$ | Watch for the negative sign when adjusting $du$ for cosine |

## What's next

U-substitution is the foundational integration technique for all more advanced integration concepts on the AP Calculus AB syllabus. Next, you will apply u-substitution to find net area, the area between curves, and volumes of revolution, where you must integrate composite functions correctly to earn full points. Without mastering u-sub steps like changing bounds and adjusting for constants, you will not be able to solve these application problems correctly. U-sub also underpins understanding of the Fundamental Theorem of Calculus with variable bounds, a common heavily tested topic on the AP exam, and is required for almost all application of integration problems in Unit 6 and Unit 8 of the AP CED.

- [Integration with long division and completing the square](https://www.owlsprep.com/study/ap-calculus-ab-u6-integration-with-long-division-and/)
- [Selecting techniques for antidifferentiation](https://www.owlsprep.com/study/ap-calculus-ab-u6-selecting-techniques-for-antidifferentiation/)
- [Differential Equations Overview](https://www.owlsprep.com/study/ap-calculus-ab-u7-overview/)

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