Difficulty: Introductory-Intermediate | Prerequisites: Basic integration rules, chain rule for derivatives.

Big Picture

U-substitution is the integration counterpart of the chain rule for differentiation. It is the single most-used technique for evaluating integrals that are not immediately recognisable from a table of antiderivatives, and it underpins every other integration method you will meet in this course. If you are comfortable differentiating composite functions, u-substitution is the natural next step: you are simply running the chain rule in reverse. Almost every Calculus II exam opens with two or three integrals that require it.

TL;DR

When an integral contains a composite function and something that looks like the derivative of the inner function, you rename the inner function u, rewrite the whole integral in terms of u, integrate, and substitute back. For definite integrals you can convert the limits too, so you never need to return to x.

Key Terms

U-substitution (substitution method, change of variables)

A technique that simplifies an integral by replacing a complicated inner expression with a single variable u, and rewriting dx in terms of du. Think of it as "undoing" the chain rule: you spot the composite structure, name the inside piece u, and the integral collapses into something you already know how to handle.

Integrand

The function being integrated, i.e. the expression that sits between the integral sign and dx. In simple terms, it is the thing you are trying to find the antiderivative of.

Differential (du)

The derivative of your chosen substitution multiplied by dx. When you set u = g(x), the differential is du = g'(x) dx. In simple terms, du is the "price of admission" for switching from x-world to u-world: if the leftover x-terms in your integral do not line up with du, the substitution will not work cleanly.

Antiderivative (indefinite integral)

A function F(x) whose derivative is the integrand f(x). In simple terms, it is the reverse of differentiation. The "+C" at the end accounts for the fact that any constant disappears when you differentiate.

Constant of integration (C)

An arbitrary constant added to every indefinite integral, because differentiation destroys constant terms. Think of it as a reminder that infinitely many functions share the same derivative, each shifted vertically by a different amount.

Core Content: The U-Substitution Method

The Central Idea

The chain rule says that if F'(x) = f(x), then the derivative of F(g(x)) is f(g(x)) · g'(x). Running that statement backwards gives you u-substitution:

\int f(g(x)) \cdot g'(x)\, dx = F(g(x)) + C

You set u = g(x), so du = g'(x) dx, and the integral becomes:

\int f(u)\, du = F(u) + C

Once you integrate in u, substitute g(x) back in for u.

Step-by-Step Procedure

1. Look at the integrand and identify an inner function g(x) whose derivative g'(x) also appears (possibly off by a constant factor).

1. Set u = g(x).

1. Differentiate: du = g'(x) dx. Solve for dx if needed, or match the remaining x-terms to du.

1. Rewrite the entire integral in terms of u and du. No x should remain.

1. Integrate with respect to u.

1. Substitute back: replace u with g(x).

1. Add +C (indefinite integrals).

For Definite Integrals

When limits of integration are given, you have two options:

• Convert the limits. If x goes from a to b, then u goes from g(a) to g(b). Evaluate entirely in u; no back-substitution needed.

• Back-substitute first. Find the antiderivative in terms of x, then apply the original limits.

Converting the limits is usually faster and less error-prone.

Worked Examples From the Lecture

Example 1: Radical with a Polynomial Inside

\int x\sqrt{1+3x^2}\, dx

Set u = 3x² + 1, so du = 6x dx, which means x dx = du/6.

The integral becomes:

\frac{1}{6}\int \sqrt{u}\, du = \frac{1}{6} \cdot \frac{2u^{3/2}}{3} = \frac{(3x^2+1)^{3/2}}{9} + C

The key recognition here is that x dx is sitting right next to the square root of a function of x², and the derivative of 3x² + 1 produces the 6x you need.

Example 2: Trigonometric Powers

\int \sec^2(x)\tan^3(x)\, dx

Set u = tan x, so du = sec²x dx.

The integral becomes:

\int u^3\, du = \frac{u^4}{4} = \frac{\tan^4 x}{4} + C

Whenever you see sec²x next to powers of tan x, u = tan x is the natural substitution, because sec²x dx becomes du.

Example 3: Splitting a Fraction

\int \frac{1+x}{1+x^2}\, dx

Split into two separate integrals:

\int \frac{1}{1+x^2}\, dx + \int \frac{x}{1+x^2}\, dx

The first integral is arctan(x) by direct recognition.

For the second, set u = x² + 1, so du = 2x dx, meaning x dx = du/2:

\frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln|u| = \frac{\ln(x^2+1)}{2}

Final answer:

\arctan(x) + \frac{\ln(x^2+1)}{2} + C

This example shows that you sometimes need to break an integral apart before any substitution is possible.

Example 4: Definite Integral with Trig Composition

\int_0^{\pi/2} \cos(x)\sin(\sin x)\, dx

Set u = sin x, so du = cos x dx. Convert the limits: when x = 0, u = 0; when x = π/2, u = 1.

\int_0^1 \sin(u)\, du = [-\cos(u)]_0^1 = -\cos(1) + \cos(0) = 1 - \cos(1) \approx 0.4597

Example 5: Definite Integral with ln(x)

\int_e^{e^4} \frac{dx}{x\sqrt{\ln x}}

Set u = ln x, so du = (1/x) dx. Convert the limits: when x = e, u = 1; when x = e⁴, u = 4.

\int_1^4 u^{-1/2}\, du = 2\sqrt{u}\Big|_1^4 = 2(2) - 2(1) = 2

Formulas to Memorise

The substitution rule (indefinite):

\int f(g(x))\, g'(x)\, dx = \int f(u)\, du \quad \text{where } u = g(x)

The substitution rule (definite):

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

Common pairings worth knowing by sight:

• If u = xⁿ + c, then du = nxⁿ⁻¹ dx

• If u = sin x, then du = cos x dx

• If u = cos x, then du = -sin x dx

• If u = tan x, then du = sec²x dx

• If u = eˣ, then du = eˣ dx

• If u = ln x, then du = (1/x) dx

Common Misconceptions

• Students often forget to convert dx to du. Setting u = 3x² + 1 without also replacing x dx with du/6 leaves you with a meaningless hybrid of two variables.

• Students sometimes change the variable but leave x-terms in the integral. After substitution, every part of the integrand, including leftover x's and the dx, must be expressed purely in u and du.

• When working definite integrals, students frequently convert to u but forget to change the limits of integration, then evaluate using the original x-limits on a u-expression. This produces a wrong answer.

• A common error is choosing the wrong expression for u. A good u is one whose derivative du already appears (or nearly appears) elsewhere in the integrand. If you have to introduce new x-terms to make the substitution work, you have probably picked the wrong u.

Why It Matters / Exam Flags

⚠️ U-substitution appears on virtually every Calculus II exam. Expect two to four integrals that require it, ranging from straightforward polynomial-inside-radical problems to trig compositions.

⚠️ Definite integral problems will often test whether you remembered to convert the limits. Markers look for this specifically.

⚠️ Some exam questions combine u-substitution with algebra (splitting fractions, factoring, long division) before the substitution step. If the integral does not look substitutable at first glance, simplify the integrand first.

⚠️ You may be asked to evaluate an integral that requires recognising a standard form (arctan, arcsin) after a substitution. Know the derivatives of inverse trig functions.

Quick Self-Test

1. True or False: After substituting u = g(x), it is acceptable to have both x and u in the integral at the same time.
   Answer: False. The integral must be entirely in terms of u and du.

1. Fill in the blank: If u = sin x, then du = ______.
   Answer: cos x dx

1. True or False: When evaluating a definite integral by substitution, you must always substitute back to x before applying the limits.
   Answer: False. You can convert the limits to u-values and never return to x.

1. Fill in the blank: For the integral of x·√(1 + 3x²) dx, a good choice of u is ______.
   Answer: u = 1 + 3x² (because its derivative, 6x dx, matches the leftover x dx up to a constant).

1. True or False: If you set u = x³ and the integrand has no x² term, the substitution will not work.
   Answer: True. du = 3x² dx, and without x² in the integrand you cannot form du.

Practice Q&A

Q: Evaluate ∫ 2x · cos(x²) dx.

A: Set u = x², du = 2x dx. The integral becomes ∫ cos(u) du = sin(u) + C = sin(x²) + C.

Q: Evaluate ∫₀¹ xe^(x²) dx.

A: Set u = x², du = 2x dx, so x dx = du/2. Limits become u = 0 to u = 1. The integral is (1/2)∫₀¹ eᵘ du = (1/2)(e - 1).

Q: Evaluate ∫ (3x²)/(1 + x³) dx.

A: Set u = 1 + x³, du = 3x² dx. The integral becomes ∫ du/u = ln|u| + C = ln|1 + x³| + C.

Q: Evaluate ∫ sin(x) · cos³(x) dx.

A: Set u = cos x, du = -sin x dx. The integral becomes -∫ u³ du = -u⁴/4 + C = -cos⁴(x)/4 + C.

Q: What is the first thing to check when deciding whether u-substitution applies?

A: Check whether the integrand contains a composite function f(g(x)) and whether g'(x) (or a constant multiple of it) appears as a factor alongside it.

Connections to Other Topics

U-substitution is the foundation for integration by parts, which handles products where substitution alone is not enough. Many integration-by-parts problems still require a u-substitution inside one of their sub-integrals.

Trigonometric integrals (powers of sin and cos, sec and tan) rely heavily on substitution with u = sin x or u = cos x combined with Pythagorean identities.

Partial fractions, the next major technique in most Calculus II courses, often produces simpler rational integrals that still need a final u-substitution step to finish.

Related Terms / Search Tags

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