Integration by substitution (u-sub)

AP Calculus AB· AP Calculus AB CED — Integration and Accumulation of Change· 14 min read

1. What Is Integration by Substitution (u-sub)?★☆☆☆☆⏱ 3 min

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.

📘 Definition

Integration by substitution (u-sub)

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

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.

2. U-Substitution for Indefinite Integrals★★☆☆☆⏱ 4 min

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 $\int g^{'} (h (x)) \cdot h^{'} (x) d x$ , set $u = h (x)$ , so $d u = h^{'} (x) d x$ . 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 $\int 6 x (2 x^{2} - 5)^{4} d x$

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 $u$ .

3. U-Substitution for Definite Integrals (Changing Bounds)★★☆☆☆⏱ 4 min

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 $\int_{x = a}^{x = b} f (h (x)) h^{'} (x) d x$ , 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 $\int_{x = 1}^{x = 3} 4 x e^{x^{2}} d x$

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 $u$ .

4. U-Choice Strategy for Non-Linear Inner Functions★★★☆☆⏱ 3 min

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 $d u$ , you have almost certainly chosen the wrong $u$ .

📐 Worked Example

Find the indefinite integral $\int \frac{t a n ^{2} ( 2 x ) s e c ^{2} ( 2 x )}{1/2} d x$

1
Identify that $tan (2 x)$ raised to a power has derivative $sec^{2} (2 x)$ (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 $M P (q) = \frac{2 q}{q ^{2} + 100}$ , 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.

✓ Quick check

Test your understanding of definite u-sub:

Evaluate $\int_{1}^{e^{2}} \frac{( l n x ) ^{2}}{x} d x$ . Which is the correct value?
- $\frac{8}{3}$
- $\frac{( e ^{2} ) ^{3}}{9} - \frac{1}{9}$
- $\frac{2}{3}$
- $e^{6} - 1$
Reveal answer
A —
Correct. Set $u = ln x$ , which gives bounds from 0 to 2, leading to $\int_{0}^{2} u^{2} d u = \frac{8}{3}$ . 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 $d u$ , and factor the constant out of the integral. Changing $u$ for a constant will always introduce unnecessary errors.

5. Common Pitfalls

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 replace $u$ with its original expression in $x$ before writing your final answer for an indefinite integral

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

Why:

Students memorize "pick the complicated part" but misidentify which part is the inner composite

Correct move:

For any composite $f (g (x))$ , $u$ is always $g (x)$ , the inner function

Wrong move:

When $d u = k h^{'} (x) d x$ , writing $k d u = h^{'} (x) d x$ instead of $\frac{1}{k} d u = h^{'} (x) d x$

Why:

Students mix up algebra when rearranging the differential equation

Correct move:

Always write $\frac{d u}{d x} = h^{'} (x)$ first, then rearrange term by term to get $h^{'} (x) d x$ in terms of $d u$

Wrong move:

Adding the constant of integration $C$ before integrating, resulting in an extra factor of $C$

Why:

Students rush and add $C$ too early, incorrectly treating 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. Quick Reference Cheatsheet

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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · AP Calculus AB
3 MCQ, 1 FRQ part
2024 · AP Calculus AB
Marginal profit application question

Going deeper

parent_unitUnit 6 Integration and Accumulation of Change OverviewFull unit curriculum overview

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.