Integration by substitution (u-sub) — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Indefinite and definite integrals via u-substitution, changing bounds for definite integrals, reversing the chain rule, integrating composite functions, common substitutions for linear and nonlinear inner functions, and adjusting for constant leading coefficients.

You should already know: Chain rule for derivatives of composite functions. Antiderivatives of basic power, exponential, and trigonometric functions. The Fundamental Theorem of Calculus for definite integrals.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. 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 Calculus AB Course and Exam Description (CED), this topic accounts for approximately 10-15% of the total unit weight for Integration and Accumulation of Change, and it appears on both multiple-choice (MCQ) and free-response (FRQ) sections of the exam. 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, using existing antiderivative rules. Synonyms for the method include change-of-variables integration and reverse chain rule integration. 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 on the exam.

2. U-Substitution for Indefinite Integrals

The most basic application of u-sub is finding indefinite integrals (antiderivatives) of composite functions. Recall that the chain rule for derivatives states that for a composite function $f(x)=g(h(x))$, the derivative is: $$f^{\prime }(x)=g^{\prime }(h(x))\cdot h^{\prime }(x)$$ U-sub reverses this relationship. If you have an integral of the form $\int g^{\prime }(h(x))\cdot h^{\prime }(x)dx$, you set $u=h(x)$ (the inner function of the composite). By definition of the differential, $du=h^{\prime }(x)dx$. Substituting into the original integral gives: $$\int g^{\prime }(u)du=g(u)+C=g(h(x))+C$$

Not all integrals have $h^{\prime }(x)$ as a perfect factor. If you are only missing a constant coefficient, you can adjust by rearranging the differential and factoring the constant reciprocal out of the integral. For example, if $du=2h^{\prime }(x)dx$, then $\frac{1}{2}du=h^{\prime }(x)dx$, which you can substitute directly.

Worked Example

Find the indefinite integral $\int 6x(2x^2-5{)}^{4}dx$.

1. Identify the composite inner function: $(2x^2-5{)}^4$ is a composite with inner function $u=2x^2-5$.
2. Compute the differential: $\frac{du}{dx}=4x⟹du=4xdx$. Rearrange to match the integrand: $\frac{6xdx}{4xdx}=\frac{3}{2}⟹6xdx=\frac{3}{2}du$.
3. Rewrite the integral entirely in terms of $u$: $\int (2x^2-5{)}^4\cdot 6xdx=\int u^4\cdot \frac{3}{2}du=\frac{3}{2}\int u^{4}du$.
4. Integrate with respect to $u$: $\frac{3}{2}\cdot \frac{u^5}{5}+C=\frac{3u^5}{10}+C$.
5. Substitute back to $x$ to get the final antiderivative: $\frac{3(2x^2-5{)}^5}{10}+C$.

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 $u$.

3. U-Substitution for Definite Integrals (Changing Bounds)

For definite integrals, you can either find the full indefinite antiderivative, substitute back to $x$, then evaluate at the original bounds, or change the bounds of integration to match the new variable $u$, which eliminates the need for back-substitution. The bounds-changing method is faster and less error-prone on the AP exam, so it is the recommended approach for most problems.

When changing bounds, for a definite integral ${\int }_{x=a}^{x=b}f(h(x))h^{\prime }(x)dx$, after setting $u=h(x)$, you calculate the lower u-bound as $u=h(a)$ and the upper u-bound as $u=h(b)$. You then rewrite the integral as: $${\int }_{h(a)}^{h(b)}f(u)du$$ You integrate directly with respect to $u$ and evaluate using the Fundamental Theorem of Calculus, with no back-substitution needed. This method is especially common on AP MCQ, where you only need the final numerical value of the integral.

Worked Example

Evaluate the definite integral ${\int }_{x=1}^{x=3}4x{e}^{x^2}dx$.

1. Choose the inner function: $u=x^2$, the exponent of the composite exponential function.
2. Compute the differential: $\frac{du}{dx}=2x⟹du=2xdx$. Adjust for the constant: $4xdx=2\cdot 2xdx=2du$.
3. Change the bounds of integration: When $x=1$, $u=1^2=1$ (lower bound). When $x=3$, $u=3^2=9$ (upper bound).
4. Rewrite the integral in terms of $u$: ${\int }_1^{9}2e^{u}du=2{\int }_1^9{e}^{u}du$.
5. Integrate and evaluate: $2{\left[e^u\right]}_1^9=2(e^9-e^1)=2e^9-2e$.

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 $u$.

4. Non-Linear Inner Functions and U-Choice Strategy

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. On AP AB, you will never need to adjust for non-constant missing factors of $x$, so 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 the inner function: We have $\tan (2x)$ raised to the 2nd power, and ${\sec }^2(2x)$ is the derivative of $\tan (2x)$ (up to a constant). Set $u=\tan (2x)$.
2. Compute the differential: $\frac{du}{dx}=2{\sec }^2(2x)⟹du=2{\sec }^2(2x)dx⟹\frac{1}{2}du={\sec }^2(2x)dx$.
3. Simplify and substitute: The original integrand simplifies to $2{\tan }^2(2x){\sec }^2(2x)dx$, so substituting gives $2\int u^2\cdot \frac{1}{2}du=\int u^{2}du$.
4. Integrate and substitute back: $\frac{u^3}{3}+C=\frac{{\tan }^3(2x)}{3}+C$.

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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Leaving indefinite integrals in terms of $u$ instead of substituting back to $x$. Why: Students get used to the bounds-changing method for definite integrals and forget that indefinite integrals require an answer in the original variable. Correct move: Always end your indefinite integral calculation by replacing $u$ with its original expression in $x$ before writing your final answer.
• Wrong move: Keeping the original $x$-bounds when integrating a definite integral in terms of $u$. Why: Students rush and skip the step of calculating new bounds, or forget that the variable of integration changed. Correct move: Immediately after setting $u$, write down the new lower and upper bounds for $u$ next to your work, before you rewrite the integral.
• Wrong move: Choosing $u$ as the outer function instead of the inner function of the composite. For example, choosing $u=(x^2+3{)}^5$ instead of $u=x^2+3$ in $\int 4x(x^2+3{)}^{5}dx$. Why: Students memorize "pick the complicated part" but misidentify which part is the inner composite. Correct move: When writing the integrand, label the outer and inner functions of any composite: for $f(g(x))$, $u$ is always $g(x)$, the inner function.
• Wrong move: When $du=k{h}^{\prime }(x)dx$, writing $kdu=h^{\prime }(x)dx$ instead of $\frac{1}{k}du=h^{\prime }(x)dx$. Why: Students mix up algebra when rearranging the differential equation. Correct move: Always write $\frac{du}{dx}=h^{\prime }(x)$ first, then rearrange term by term to get $h^{\prime }(x)dx$ in terms of $du$.
• Wrong move: Adding the constant of integration $C$ before integrating, resulting in an extra factor of $C$ multiplied by $u$ after substitution. Why: Students rush and add $C$ before integrating, then incorrectly treat it as a variable. Correct move: Add the single constant $C$ once, immediately after integrating with respect to $u$, before substituting back to $x$.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Evaluate ${\int }_1^{e^2}\frac{(\ln x{)}^2}{x}dx$. Which of the following is the correct value? A) $\frac{8}{3}$ B) $\frac{(e^2{)}^3}{9}-\frac{1}{9}$ C) $\frac{2}{3}$ D) $e^6-1$

Worked Solution: We use u-sub for definite integrals with changing bounds. Let $u=\ln x$, so $du=\frac{1}{x}dx$, which matches the remaining term in the integrand. Change bounds: when $x=1$, $u=\ln 1=0$; when $x=e^2$, $u=\ln (e^2)=2$. Rewrite the integral as ${\int }_0^2{u}^{2}du$. Integrate to get ${\left[\frac{u^3}{3}\right]}_0^2=\frac{8}{3}-0=\frac{8}{3}$. The distractors B and D result from incorrect bounds or back-substitution error, and C results from miscalculating the upper bound. The correct answer is A.

Question 2 (Free Response)

Let $f(x)=4x\sin (x^2+2)$. (a) Find the general antiderivative of $f(x)$. (b) Evaluate ${\int }_0^{\sqrt{\pi }}4x\sin (x^2+2)dx$. (c) Given that $F(x)$ is an antiderivative of $f(x)$ satisfying $F(0)=1$, find $F(1)$.

Worked Solution: (a) Use u-sub for indefinite integral: Let $u=x^2+2$, so $du=2xdx⟹2du=4xdx$. The integral becomes $\int 2\sin udu=-2\cos u+C=-2\cos (x^2+2)+C$. (b) Change bounds: When $x=0$, $u=0+2=2$; when $x=\sqrt{\pi }$, $u=\pi +2$. Evaluate: ${\int }_2^{\pi +2}2\sin udu=-2{\left[\cos u\right]}_2^{\pi +2}=-2(\cos (\pi +2)-\cos 2)=-2(-\cos 2-\cos 2)=4\cos 2$. (c) The general antiderivative is $F(x)=-2\cos (x^2+2)+C$. Use $F(0)=1$: $-2\cos (2)+C=1⟹C=1+2\cos 2$. Substitute $x=1$: $F(1)=-2\cos (3)+1+2\cos 2=1+2(\cos 2-\cos 3)$.

Question 3 (Application / Real-World Style)

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

Worked Solution: Total change in profit is the integral of marginal profit from $q=10$ to $q=20$, so $\text{Total change}={\int }_{10}^{20}\frac{2q}{q^2+100}dq$. Use u-sub: Let $u=q^2+100$, so $du=2qdq$. Change bounds: $q=10⟹u=100+100=200$; $q=20⟹u=400+100=500$. Rewrite: ${\int }_{200}^{500}\frac{1}{u}du={\left[\ln u\right]}_{200}^{500}=\ln 500-\ln 200=\ln \left(\frac{500}{200}\right)=\ln (2.5)\approx 0.9163$. Since the units are hundreds of dollars, this is $0.9163\times 100\approx 92$ dollars. In context, increasing production from 10 to 20 loaves increases total profit by approximately $92.

7. Quick Reference Cheatsheet

8. 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 with the disk and washer methods, where you will need to integrate composite functions to get correct results. Without mastering the steps of u-sub, including changing bounds and adjusting for constants, any problem involving integrating composite functions in these applications will be impossible to solve correctly. U-sub also underpins understanding of the Fundamental Theorem of Calculus with variable bounds, a common AP exam topic. This topic feeds directly into all application of integration topics in Unit 6 and Unit 8 of the AP Calculus AB CED.

Net area from definite integrals Area between curves Volumes of revolution Fundamental Theorem of Calculus