# Integration by substitution (u-substitution)

> AP Calculus BC · Unit 6: Integration and Accumulation of Change
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u6-integration-by-substitution/

This guide covers u-substitution for AP Calculus BC, including indefinite/definite integrals, adjusting for missing constants, common exam pitfalls, and worked AP-style examples. Mastery of this foundational technique is required for all advanced integration methods on the exam.

**Prerequisites:** Chain rule for derivatives; Antiderivatives of basic functions; Definite integral notation and properties

## Learning objectives

- Recognize when to use u-substitution for integrating composite functions
- Execute u-substitution correctly for indefinite integrals
- Perform definite u-substitution with changing limits of integration
- Adjust substitution for missing constant factors
- Avoid common u-substitution pitfalls on the AP exam

## What is U-Substitution?

Integration by substitution (commonly called u-substitution) is the core technique for integrating composite functions, and it is the inverse of the chain rule for differentiation. It rewrites complex integrands into a form that matches the standard library of basic antiderivatives for power, trigonometric, exponential, and logarithmic functions.

This topic makes up 17–20% of your total AP Calculus BC exam score as part of Unit 6, and it is a prerequisite for all advanced integration techniques covered in BC, including integration by parts, trigonometric substitution, and partial fractions.

**U-Substitution** — A change of variables technique that reverses the chain rule to simplify integration of composite functions, also called the reverse chain rule.

*Notation:* $u = g(x)$

> **tip**
>
> U-substitution appears in both MCQ and FRQ sections every year, so mastering this technique is non-negotiable for a high score.

## Indefinite U-Substitution for Composite Functions

Indefinite u-substitution is used to find the general antiderivative (with constant of integration) of a composite function. It follows directly from reversing the chain rule:

$$\frac{d}{dx}\left[f(g(x))\right] = f'(g(x)) \cdot g'(x)$$

Integrating both sides gives the core identity:

$$\int f'(g(x)) g'(x) dx = f(g(x)) + C$$

To apply substitution, set $u = g(x)$, the inner function of the composite. By the chain rule, $\frac{du}{dx} = g'(x)$, which rearranges to $du = g'(x) dx$. Substituting simplifies the integral to $\int f'(u) du = f(u) + C$, which you can integrate with basic rules, then substitute back $u = g(x)$ to get the final result in terms of $x$. The key requirement is that all $x$-terms must be replaced with $u$-terms after substitution.

**Worked example:** Evaluate $\int 2x \cos(x^2 + 3) dx$

1. Identify the inner composite function: the argument of cosine is $x^2 + 3$, so set $u = x^2 + 3$.
2. Calculate $du$:

   $$\frac{du}{dx} = 2x \implies du = 2x dx$$
3. This $du$ exactly matches the remaining terms in the integrand. Substitute into the original integral:

   $$\int \cos(x^2 + 3) \cdot 2x dx = \int \cos(u) du$$
4. Integrate with respect to $u$:

   $$\int \cos(u) du = \sin(u) + C$$
5. Substitute back $u = x^2 + 3$ to get the final antiderivative in terms of $x$:

   $$\sin(x^2 + 3) + C$$
6. Differentiating the result confirms it matches the original integrand.

> **tip**
>
> Always substitute back to the original variable for indefinite integrals; leaving your answer in terms of $u$ will cost you points on FRQ and is incorrect on MCQ.

## Definite U-Substitution and Changing Limits

For definite integrals, the most efficient approach (and the one expected on the AP exam) is to change the limits of integration to match the new variable $u$, eliminating the need to substitute back to $x$ after integration. For an integral over $x = a$ to $x = b$, if $u = g(x)$, the lower limit for $u$ is $g(a)$ and the upper limit for $u$ is $g(b)$. The formula becomes:

$$\int_{x=a}^{x=b} f'(g(x)) g'(x) dx = \int_{u=g(a)}^{u=g(b)} f'(u) du$$

This method reduces arithmetic error by removing the extra substitution step required if you first find the indefinite antiderivative in terms of $x$. While you can technically solve for the indefinite antiderivative, substitute back, then evaluate at the original $x$ limits, this adds unnecessary work and room for error.

**Worked example:** Evaluate $\int_{x=0}^{x=2} 3x^2 e^{x^3} dx$

1. Identify the inner function: the exponent of $e$ is $x^3$, so set $u = x^3$.
2. Calculate $du$:

   $$du = 3x^2 dx$$
3. Change the limits of integration: when $x=0$, $u = 0^3 = 0$ (new lower limit); when $x=2$, $u = 2^3 = 8$ (new upper limit).
4. Substitute into the definite integral:

   $$\int_{0}^{8} e^u du$$
5. Evaluate directly at the $u$ limits, no substitution back required:

   $$\left[e^u\right]_{0}^{8} = e^8 - e^0 = e^8 - 1$$

> **tip**
>
> Always confirm the direction of limits after changing variables: if $g(x)$ is decreasing over $[a,b]$, your upper $u$-limit will be smaller than your lower $u$-limit, which is fine—do not swap them unless you are explicitly correcting a sign error.

## U-Substitution with a Missing Constant Factor

A very common scenario in u-substitution problems is when $du$ does not exactly match the remaining $x$-terms in the integrand, but only differs by a constant multiple. For example, if we have $\int x \cos(x^2) dx$, setting $u = x^2$ gives $du = 2x dx$, but the integrand only has $x dx$. In this case, we can rearrange the $du$ equation to solve for the missing term: $x dx = \frac{du}{2}$. This works only when the difference is a constant factor—you can always pull constant factors out of an integral, so this adjustment is valid.

**Worked example:** Evaluate $\int \frac{2x + 1}{2x^2 + 2x + 1} dx$

1. Notice the numerator is proportional to the derivative of the denominator, so set $u = 2x^2 + 2x + 1$.
2. Calculate $du$:

   $$\frac{du}{dx} = 4x + 2 = 2(2x + 1) \implies du = 2(2x + 1) dx$$
3. Rearrange to get the $(2x + 1) dx$ term from the integrand: $(2x + 1) dx = \frac{du}{2}$.
4. Substitute into the integral:

   $$\int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{1}{u} du$$
5. Integrate and substitute back:

   $$\frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|2x^2 + 2x + 1| + C$$

> **warning**
>
> Never adjust a non-constant leftover term by pulling it out as a constant; if you have a leftover $x$ term that cannot be rewritten in terms of $u$, your initial $u$ choice is wrong and you need a different approach.

**Check your understanding**

Test your understanding with this AP-style multiple choice question:

1. Evaluate $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx = ?$

   - A) $\frac{2 e^{\sqrt{x}}}{\sqrt{x}} + C$
   - B) $2 e^{\sqrt{x}} + C$
   - C) $\frac{1}{2} e^{\sqrt{x}} + C$
   - D) $e^{\sqrt{x}} + C$

   *Why:* Correct. Set $u = \sqrt{x}$, so $du = \frac{dx}{2\sqrt{x}}$, which rearranges to $2 du = \frac{dx}{\sqrt{x}}$. Substituting gives $\int 2 e^u du = 2 e^u + C = 2 e^{\sqrt{x}} + C$.

## Common pitfalls

- **Wrong:** For definite integrals, you change variables to $u$, then substitute back to $x$ and plug in the $u$-limits to evaluate.
  - Why it fails: Students mix the two methods for evaluating definite u-sub, confusing the changed-limits approach with the substitute-back approach.
  - Correct: If you change the limits to $u$-values, evaluate the antiderivative directly at the $u$-limits, do not substitute back to $x$. If you do not change limits, keep the original $x$-limits and substitute back to $x$ before evaluating.
- **Wrong:** When $du$ differs by a constant factor, you forget the reciprocal constant, e.g. writing $\int x \cos(x^2) dx = \sin(x^2) + C$ instead of $\frac{1}{2}\sin(x^2) + C$.
  - Why it fails: Students rush through substitution and forget to account for the extra constant factor.
  - Correct: Always explicitly solve for the $x$-term in the $du$ equation, so the constant factor is included before you integrate.
- **Wrong:** Choosing $u$ as the outer function instead of the inner function of the composite, for example setting $u = \cos(x^2 + 3)$ instead of $u = x^2 + 3$.
  - Why it fails: Students do not remember the reverse chain rule structure, which targets the inner function for substitution.
  - Correct: Always set $u$ equal to the inside function of the composite, the $g(x)$ in the $f(g(x))$ chain rule structure.
- **Wrong:** Forgetting the absolute value when integrating $\frac{1}{u}$, writing $\ln u + C$ instead of $\ln|u| + C$.
  - Why it fails: Students remember the antiderivative of $\frac{1}{x}$ but forget the rule carries over to substituted variables.
  - Correct: Whenever you integrate $\int \frac{1}{u} du$, immediately write $\ln|u| + C$ before substituting back to $x$.
- **Wrong:** After substituting $u$ and $du$, you leave a leftover $x$ term in the integrand and treat it as a constant when integrating with respect to $u$.
  - Why it fails: Students rush to integrate and do not check that all $x$-terms are rewritten.
  - Correct: After substitution, always check for leftover $x$-terms; if any remain, rewrite them using $u = g(x)$ or pick a new $u$.

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Indefinite U-Sub Rule | $\int f(g(x)) g'(x) dx = \int f(u) du = f(g(x)) + C$ | $u = g(x)$, $du = g'(x) dx$; always substitute back to $x$ |
| Definite U-Sub (Change Limits) | $\int_a^b f(g(x)) g'(x) dx = \int_{g(a)}^{g(b)} f(u) du$ | No need to substitute back to $x$; evaluate directly at $u$ limits |
| Missing Constant Adjustment | If $du = k \cdot (\text{remaining x term})$, then $(\text{remaining x term}) = \frac{du}{k}$ | Only valid for constant $k$; never use for non-constant $x$ terms |
| Rational Integrand U-Sub | Set $u = $ denominator | Use when numerator = $k \cdot \frac{d}{dx}(\text{denominator})$; results in $\ln\|u\| + C$ |
| Exponential Integrand U-Sub | Set $u = $ exponent of $e^{f(x)}$ | Simplifies to $\int e^u du = e^u + C$ |
| Trigonometric Integrand U-Sub | Set $u = $ inner argument of the trig function | Matches reverse chain rule structure |
| Antiderivative of $1/u$ | $\int \frac{1}{u} du = \ln\|u\| + C$ | Absolute value is required for all $u \neq 0$ |
| Indefinite Integral Rule | Always add $+C$ | Forgetting $+C$ costs points on all indefinite integral problems |

## What's next

Integration by u-substitution is the foundational integration technique that every advanced integration method builds on for AP Calculus BC. Immediately after mastering u-sub, you will apply it to integration using algebraic rearrangement for rational functions, where u-sub is almost always used as a final step after manipulation. More importantly, u-substitution is required for all other key BC integration topics: integration by parts, trigonometric substitution, and partial fraction decomposition all rely on recognizing substitution structures and correctly executing substitution steps to get the final result. Without mastering u-sub, you will struggle to complete complex integration problems on the AP exam, even if you understand the more advanced techniques.

- [Integration by parts](https://www.owlsprep.com/study/ap-calculus-bc-u6-integration-by-parts/)
- [Integration with long division and completing the square](https://www.owlsprep.com/study/ap-calculus-bc-u6-integration-with-long-division-and/)
- [Integration using partial fractions (BC only)](https://www.owlsprep.com/study/ap-calculus-bc-u6-integration-using-partial-fractions/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ap-calculus-bc-u6-integration-by-substitution/