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

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Indefinite u-substitution for composite functions, definite integral u-substitution with changing limits of integration, reverse chain rule application, and u-sub for trigonometric, exponential, and rational integrands.

You should already know: Chain rule for derivatives, antiderivatives of basic functions, definite integral notation and net change interpretation.

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 BC 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 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. According to the AP Calculus BC Course and Exam Description (CED), this topic is part of Unit 6: Integration and Accumulation of Change, which accounts for 17–20% of the total AP exam score. U-substitution appears in both multiple-choice (MCQ) and free-response (FRQ) sections every year, and it is a prerequisite for every advanced integration technique covered in BC, from integration by parts to partial fraction decomposition. Synonyms for this method include the reverse chain rule and change of variables integration.

2. 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: if we have a derivative of the form $\frac{d}{dx}\left[f(g(x))\right]=f^{\prime }(g(x))\cdot g^{\prime }(x)$, integrating both sides gives: $$\int f^{\prime }(g(x))g^{\prime }(x)dx=f(g(x))+C$$ To apply substitution, we set $u=g(x)$, the inner function of the composite. By the chain rule, $\frac{du}{dx}=g^{\prime }(x)$, which rearranges to $du=g^{\prime }(x)dx$. Substituting into the original integral simplifies it to $\int f^{\prime }(u)du=f(u)+C$, which we can integrate using basic antiderivative 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; if a non-constant $x$-term remains, your initial $u$ choice is incorrect.

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 $du$: $\frac{du}{dx}=2x$, so $du=2xdx$, which exactly matches the remaining terms in the integrand.
3. Substitute into the original integral: $\int \cos (x^2+3)\cdot 2xdx=\int \cos (u)du$.
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 $x$: $\sin (x^2+3)+C$. (Differentiating the result confirms it matches the original integrand.)

Exam 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.

3. Definite U-Substitution and Changing Limits of Integration

For definite integrals, the most efficient approach (and the one most commonly 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 a definite 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: $${\int }_{x=a}^{x=b}{f}^{\prime }(g(x))g^{\prime }(x)dx={\int }_{u=g(a)}^{u=g(b)}{f}^{\prime }(u)du$$ 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 $du$: $du=3x^{2}dx$, which matches the remaining terms in the integrand.
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: ${\left[e^u\right]}_0^8=e^8-e^0=e^8-1$, no substitution back required.

Exam 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.

4. 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 $du=2xdx$, but the integrand only has $xdx$, not $2xdx$. In this case, we can rearrange the $du$ equation to solve for the missing term: $xdx=\frac{du}{2}$. 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. If the leftover term is a non-constant function of $x$ that cannot be rewritten in terms of $u$, u-substitution will not work for that integrand, and you need another technique like integration by parts.

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 $du$: $\frac{du}{dx}=4x+2=2(2x+1)$, so $du=2(2x+1)dx$.
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: $\frac{1}{2}\ln |u|+C=\frac{1}{2}\ln |2x^2+2x+1|+C$.

Exam tip: 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.

5. Common Pitfalls (and how to avoid them)

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

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Evaluate $\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=?$ A) $\frac{2e^{\sqrt{x}}}{\sqrt{x}}+C$ B) $2e^{\sqrt{x}}+C$ C) $\frac{1}{2}{e}^{\sqrt{x}}+C$ D) $e^{\sqrt{x}}+C$

Worked Solution: First, identify the inner function of the composite: the exponent of $e$ is $\sqrt{x}$, so set $u=\sqrt{x}=x^{1/2}$. Calculate $du$: $\frac{du}{dx}=\frac{1}{2\sqrt{x}}$, so $du=\frac{dx}{2\sqrt{x}}$. Rearranging gives $2du=\frac{dx}{\sqrt{x}}$, which matches the remaining terms in the integrand. Substitute to get $\int e^u\cdot 2du=2e^u+C$, then substitute back $u=\sqrt{x}$ to get $2e^{\sqrt{x}}+C$. Correct answer: $\boxed{B}$

Question 2 (Free Response)

Let $f(x)=x(x^2-4{)}^3$. (a) Find the general indefinite integral of $f(x)dx$. (b) Given that the antiderivative $F(x)$ passes through $(1,5)$, find the constant of integration $C$. (c) Evaluate ${\int }_0^{2}x(x^2-4{)}^{3}dx$.

Worked Solution: (a) Let $u=x^2-4$, so $du=2xdx$, which rearranges to $xdx=\frac{du}{2}$. Substitute into the integral: $$\int u^3\cdot \frac{du}{2}=\frac{1}{2}\cdot \frac{u^4}{4}+C=\frac{(x^2-4{)}^4}{8}+C$$

(b) Substitute $F(1)=5$ to solve for $C$: $$5=\frac{(1-4{)}^4}{8}+C=\frac{81}{8}+C⟹C=5-\frac{81}{8}=-\frac{41}{8}$$

(c) Change limits of integration: when $x=0$, $u=0^2-4=-4$; when $x=2$, $u=2^2-4=0$. Evaluate: $${\int }_{-4}^0\frac{u^3}{2}du=\frac{1}{8}{\left[u^4\right]}_{-4}^0=\frac{1}{8}\left(0-256\right)=-32$$

Question 3 (Application / Real-World Style)

The rate of change of the number of bacteria in a petri dish $t$ hours after the start of an experiment is given by $r(t)=2t{e}^{0.5t^2}$ bacteria per hour. What is the net change in the number of bacteria from $t=0$ to $t=3$ hours? Round to the nearest whole number and include units.

Worked Solution: Net change is the integral of the rate function over the interval, so we calculate ${\int }_0^{3}2t{e}^{0.5t^2}dt$. Let $u=0.5t^2$, so $du=tdt$, which means $2tdt=2du$. Change limits: $t=0⟹u=0$; $t=3⟹u=0.5(9)=4.5$. The integral becomes: $${\int }_0^{4.5}2e^{u}du=2{\left[e^u\right]}_0^{4.5}=2\left(e^{4.5}-1\right)\approx 2(90.02-1)\approx 178$$ Interpretation: The bacteria population increases by approximately 178 bacteria over the 3-hour period.

7. Quick Reference Cheatsheet

8. What's Next

Integration by u-substitution is the foundational integration technique that every advanced integration method builds on. Immediately after mastering u-sub, you will apply it to integration using algebraic rearrangement (long division, completing the square) for rational functions, where u-sub is almost always used as a final step after algebraic 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.

Follow-up topics: Integration by parts Trigonometric substitution Partial fraction decomposition Net change from rate functions