Integration Using Substitution (College Board AP® Calculus AB)

Study Guide

Written by: Roger B

Reviewed by: Dan Finlay

Determining indefinite integrals using u-substitutions

What is integration by substitution?

Substitution simplifies an integral by defining an alternative variable (usually $u$ ) in terms of the original variable (usually $x$ )
- The integral in $u$ is much easier to solve than the original integral in $x$
- The substitution can be reversed at the end to get the answer in terms of $x$

How do I integrate simple functions using u-substitution?

In a simple integral involving substitution, you will usually be integrating a composite function (i.e., 'function of a function')
- These can also be solved 'by inspection'
  - See the 'Integrals of Composite Functions' study guide
- Substitution can be a safer method when 'by inspection' is awkward or difficult to spot
STEP 1
Identify the substitution to be used – it will be the secondary (or 'inside') function in a composite function
- I.e. if the integral involves $f (g (x))$ , let $u = g (x)$
- E.g. $\int 3 x \cos (x^{2} - 5) d x$
  - Let $u = x^{2} - 5$
STEP 2
Differentiate the substitution and rearrange
- $\frac{d u}{d x}$ can be treated like a fraction (i.e. “multiply by $d x$ " to get rid of fractions)
- E.g. $u = x^{2} - 5 \Rightarrow \frac{d u}{d x} = 2 x$
  - Then $d u = 2 x d x \Rightarrow x d x = \frac{1}{2} d u$
STEP 3
Replace all parts of the integral
- All $x$ terms should be replaced with equivalent $u$ terms, including $d x$
- E.g. $\int 3 x \cos (x^{2} - 5) d x = 3 \int \cos (x^{2} - 5) \cdot x d x$
  - So $\int 3 x \cos (x^{2} - 5) d x = 3 \int \cos u \cdot \frac{1}{2} d u = \frac{3}{2} \int \cos u d u$
STEP 4
Integrate
- E.g. $\frac{3}{2} \int \cos u d u = \frac{3}{2} \sin u + C$
  - Don't forget the constant of integration
STEP 5
Substitute $x$ back in
- Replace $u$ everywhere with the equivalent expression for $x$
- E.g. $\frac{3}{2} \sin u + C = \frac{3}{2} \sin (x^{2} - 5) + C$
  - So $\int 3 x \cos (x^{2} - 5) d x = \frac{3}{2} \sin (x^{2} - 5) + C$

Worked Example

Find the indefinite integral $\int \frac{x^{2} - 1}{2 x^{3} - 6 x + 7} d x$ .

Answer:

Here the 'main' function in the composite function is $\frac{1}{(. . .)}$ , and the 'inside function' is $2 x^{3} - 6 x + 7$

This is an integral that could also be solved by using $\int \frac{f^{'} (x)}{f (x)} d x = \ln |f (x)| + C$ , but here we'll use substitution

Let $u = 2 x^{3} - 6 x + 7$

Differentiate the substitution and rearrange

$\frac{d u}{d x} = 6 x^{2} - 6 d u = 6 (x^{2} - 1) d x \Rightarrow (x^{2} - 1) d x = \frac{1}{6} d u$

Replace all parts of the integral

$\int \frac{x^{2} - 1}{2 x^{3} - 6 x + 7} d x = \int \frac{1}{u} \cdot \frac{1}{6} d u = \frac{1}{6} \int \frac{1}{u} d u$

Integrate

$\frac{1}{6} \int \frac{1}{u} d u = \frac{1}{6} \ln |u| + C$

Substitute $x$ back in

$\int \frac{x^{2} - 1}{2 x^{3} - 6 x + 7} d x = \frac{1}{6} \ln |2 x^{3} - 6 x + 7| + C$

How do I integrate more complicated functions using u-substitution?

The procedure here is exactly the same as for integrating simpler functions
- However the substitution to use may not be as obvious
- Practice questions like this to improve your integration by substitution skills
E.g. $\int x \sqrt{x - 4} d x$
- Note that this is not an integral that can be solved 'by inspection' (i.e. by the 'reverse chain rule')
- Identify the substitution
  - Let $u = x - 4$
- Differentiate the substitution and rearrange
  - $\frac{d u}{d x} = 1 \Rightarrow d u = d x$
- Replace all parts of the integral
  - $u = x - 4 \Rightarrow x = u + 4$
  - So $\int x \sqrt{x - 4} d x = \int (u + 4) \sqrt{u} d u$
- Integrate
  - $\int (u + 4) \sqrt{u} d u = \int (u + 4) u^{\frac{1}{2}} d u = \int (u^{\frac{3}{2}} + 4 u^{\frac{1}{2}}) d u = \frac{2}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}} + C$
- Substitute $x$ back in
  - $\int x \sqrt{x - 4} d x = \frac{2}{5} {(x - 4)}^{\frac{5}{2}} + \frac{8}{3} {(x - 4)}^{\frac{3}{2}} + C$

Worked Example

Find the indefinite integral $\int \frac{d x}{\sqrt{16 - 9 x^{2}}}$ .

Answer:

To spot the substitution to use here it helps to recall the standard integral $\int \frac{d x}{\sqrt{1 - x^{2}}} = \arcsin x + C$

First rearrange the integral slightly

$\int \frac{d x}{\sqrt{16 - 9 x^{2}}} = \int \frac{d x}{4 \sqrt{1 - \frac{9}{16} x^{2}}} = \frac{1}{4} \int \frac{d x}{\sqrt{1 - \frac{9}{16} x^{2}}}$

Now the substitution to use is more obvious

Let $u = \frac{3}{4} x \Rightarrow u^{2} = \frac{9}{16} x^{2}$

Differentiate the substitution and rearrange

$\frac{d u}{d x} = \frac{3}{4} \Rightarrow d x = \frac{4}{3} d u$

Replace all parts of the integral

$\frac{1}{4} \int \frac{d x}{\sqrt{1 - \frac{9}{16} x^{2}}} = \frac{1}{4} \int \frac{\frac{4}{3} d u}{\sqrt{1 - u^{2}}} = \frac{1}{3} \int \frac{d u}{\sqrt{1 - u^{2}}}$

Integrate

$\frac{1}{3} \int \frac{d u}{\sqrt{1 - u^{2}}} = \frac{1}{3} \arcsin u + C$

Substitute $x$ back in

$\int \frac{d x}{\sqrt{16 - 9 x^{2}}} = \frac{1}{3} \arcsin (\frac{3}{4} x) + C$

Evaluating definite integrals using u-substitutions

How do I evaluate definite integrals using u-substitution?

Definite integrals can also be solved using u-substitution
- You just need to rewrite the integration limits in terms of u as well
E.g. $\int_{4}^{8} x \sqrt{x - 4} d x$
- We've already seen that this can be integrated using the substitution $u = x - 4$
  - with $d u = d x$ and $x = u + 4$ following from this
- We just need to change the integral limits as well
  - When $x = 4$ , $u = 4 - 4 = 0$
  - When $x = 8$ , $u = 8 - 4 = 4$
- Now when we replace all parts of the integral we get
  - $\int_{4}^{8} x \sqrt{x - 4} d x = \int_{0}^{4} (u + 4) \sqrt{u} d u$
- Integrate, then evaluate the definite integral using the $u$ values
  - $\begin{array}{rcl} \int_{0}^{4} (u + 4) \sqrt{u} d u & = & {[\frac{2}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}}]}_{0}^{4} \\ = & \frac{2}{5} \cdot 32 + \frac{8}{3} \cdot 8 - (\frac{2}{5} \cdot 0 + \frac{8}{3} \cdot 0) \\ = & \frac{512}{15} \end{array}$
- Note that there is no need to substitute $x$ back in to evaluate the definite integral!

Worked Example

Evaluate the definite integral $\int_{1}^{2} \frac{2 x + 3}{3 x^{2} + 9 x - 5} d x$ .

Answer:

Choose the substitution

Let $u = 3 x^{2} + 9 x - 5$

Differentiate the substitution and rearrange

$\frac{d u}{d x} = 6 x + 9 d u = 3 (2 x + 3) d x \Rightarrow (2 x + 3) d x = \frac{1}{3} d u$

Find the integration limits in terms of $u$

When $x = 1$ , $u = 3 {(1)}^{2} + 9 (1) - 5 = 7$

When $x = 2$ , $u = 3 {(2)}^{2} + 9 (2) - 5 = 25$

Replace all parts of the integral, including the integration limits

$\int_{1}^{2} \frac{2 x + 3}{3 x^{2} + 9 x - 5} d x = \int_{7}^{25} \frac{1}{u} \cdot \frac{1}{3} d u = \frac{1}{3} \int_{7}^{25} \frac{1}{u} d u$

Integrate and evaluate the definite integral

$\frac{1}{3} \int_{7}^{25} \frac{1}{u} d u = \frac{1}{3} {[\ln u]}_{7}^{25} = \frac{1}{3} (\ln 25 - \ln 7)$

Use laws of logarithms to simplify the final answer

$\int_{1}^{2} \frac{2 x + 3}{3 x^{2} + 9 x - 5} d x = \frac{1}{3} \ln (\frac{25}{7})$

Last updated: 14 August 2024

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Previous:Integrals of Composite FunctionsNext:Integration Using Completing the Square

Integration Using Substitution (College Board AP® Calculus AB)

Study Guide

Determining indefinite integrals using u-substitutions

What is integration by substitution?

How do I integrate simple functions using u-substitution?

How do I integrate more complicated functions using u-substitution?

Evaluating definite integrals using u-substitutions

How do I evaluate definite integrals using u-substitution?

You've read 0 of your 5 free study guides this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

Unit 1: Limits & Continuity

Limits

Definition of a Limit

Infinite Limits & Limits at Infinity

Properties of Limits

Evaluating Limits Analytically

Evaluating Limits Numerically & Graphically

Squeeze Theorem & Trigonometric Limits

Summary of Limits

Selecting Procedures for Determining Limits

Continuity

Continuity

Removable Discontinuities

Non-removable Discontinuities

Continuous Functions

Intermediate Value Theorem

Unit 2: Differentiation Definition & Fundamental Properties

Definition of Differentiation

Average Rate of Change

Instantaneous Rate of Change

Derivatives & Tangents

Estimating the Derivative at a Point

Differentiability & Continuity

Fundamental Properties of Differentiation

Derivative Rules

Derivatives of Exponentials and Logarithms

Derivatives of Sine and Cosine Functions

The Product Rule

The Quotient Rule

Derivatives of Tangent and Reciprocal Trig Functions

Unit 3: Differentiation of Composite, Implicit & Inverse Functions

Differentiation of Composite & Inverse Functions

The Chain Rule

The Inverse Function Theorem

Derivatives of Inverse Trigonometric Functions

Higher-Order Derivatives

Implicit Differentiation

Implicit Differentiation

Summary of Differentiation

Selecting Procedures for Calculating Derivatives

Unit 4: Contextual Applications of Differentiation

Rates of Change & Related Rates

Meaning of a Derivative in Context

Motion in a Straight Line

Related Rates

Linearization

Approximating Values of a Function

L'Hospital's Rule

L'Hospital's Rule

Unit 5: Analytical Applications of Differentiation

Graphs of Functions & Their Derivatives

Mean Value Theorem

Extreme Value Theorem

Critical Points

Increasing & Decreasing Functions

First Derivative Test for Local Extrema

Candidates Test for Global Extrema

Concavity of Functions

Second Derivative Test for Local Extrema

Graphs of f, f' & f''

Optimization Problems

Behaviors of Implicit Relations

Second Derivatives of Implicit Functions

Critical Points of Implicit Relations

Unit 6: Integration & Accumulation of Change

Integration & Antiderivatives

Indefinite Integrals

Derivatives & Antiderivatives

Indefinite Integral Rules