Derivatives & Antiderivatives (College Board AP® Calculus AB)

Study Guide

Written by: Roger B

Reviewed by: Dan Finlay

Derivatives & antiderivatives

How can I find the indefinite integrals of common functions?

Because differentiation and integration are inverse operations
- you can 'reverse' what you know about derivatives to find indefinite integrals
I.e. if $f^{'} (x) = g (x)$
- then $\int g (x) d x = f (x) + C$
This means that all your derivative results for common functions
- have indefinite integral equivalents

Indefinite integrals of powers of x

$\frac{d}{d x} (x^{n}) = n x^{n - 1}$ therefore
- $\int x^{n} d x = \frac{1}{n + 1} x^{n + 1} + C, n \neq - 1$
  - Note that you can't integrate $x^{- 1} = \frac{1}{x}$ using this rule
    - The denominator in the fraction would become zero
    - $\frac{1}{x}$ must be integrated using logarithms
Also note these two special cases
- $\int k d x = k x + C$ , where $k$ is a constant
- $\int 0 d x = C$

Worked Example

Find the indefinite integral $\int x^{10} d x$ .

Answer:

Use $\int x^{n} d x = \frac{1}{n + 1} x^{n + 1} + C, n \neq - 1$

$\int x^{10} d x = \frac{1}{10 + 1} x^{10 + 1} + C$

$\int x^{10} d x = \frac{1}{11} x^{11} + C$

Indefinite integrals of exponentials and 1/x

$\frac{d}{d x} (e^{x}) = e^{x}$ and $\frac{d}{d x} (e^{k x}) = k e^{k x}$ therefore
- $\int e^{x} d x = e^{x} + C$
- $\int e^{k x} d x = \frac{1}{k} e^{k x} + C$
$\frac{d}{d x} (a^{x}) = a^{x} \ln a$ and $\frac{d}{d x} (a^{k x}) = a^{k x} k \ln a$ therefore
- $\int a^{x} d x = \frac{1}{\ln a} a^{x} + C$
- $\int a^{k x} d x = \frac{1}{k \ln a} a^{k x} + C$
$\frac{d}{d x} (\ln x) = \frac{1}{x}$ therefore
- $\int \frac{1}{x} d x = \ln |x| + C$
  - Don't forget the modulus (absolute value) sign around the $x$
    - This allows the integral to be valid for negative values of $x$ as well as for positive values
In the above formulae, $k$ is a real number constant and $a$ is a positive real number constant

Worked Example

Find the indefinite integral $\int e^{7 x} d x$ .

Answer:

Use $\int e^{k x} d x = \frac{1}{k} e^{k x} + C$

$\int e^{7 x} d x = \frac{1}{7} e^{7 x} + C$

Indefinite integrals of trigonometric functions

$\frac{d}{d x} (\sin x) = \cos x$ and $\frac{d}{d x} (\sin k x) = k \cos k x$ therefore
- $\int \cos x d x = \sin x + C$
- $\int \cos k x d x = \frac{1}{k} \sin k x + C$
$\frac{d}{d x} (\cos x) = - \sin x$ and $\frac{d}{d x} (\cos k x) = - k \sin k x$ therefore
- $\int \sin x d x = - \cos x + C$
- $\int \sin k x d x = - \frac{1}{k} \cos k x + C$
$\frac{d}{d x} (\tan x) = \sec^{2} x$ and $\frac{d}{d x} (\tan k x) = k \sec^{2} k x$ therefore
- $\int \sec^{2} x d x = \tan x + C$
- $\int \sec^{2} k x d x = \frac{1}{k} \tan k x + C$
In the above formulae, $k$ is a real number constant

Worked Example

Find the following indefinite integrals:

(a) $\int \sin 4 x d x$

Answer:

Use $\int \sin k x d x = - \frac{1}{k} \cos k x + C$

$\int \sin 4 x d x = - \frac{1}{4} \cos 4 x + C$

(b) $\int {(\frac{1}{\cos 3 x})}^{2} d x$

Answer:

Remember $\sec x = \frac{1}{\cos x}$

${(\frac{1}{\cos 3 x})}^{2} = {(\sec 3 x)}^{2} = \sec^{2} 3 x$

Use $\int \sec^{2} k x d x = \frac{1}{k} \tan k x + C$

$\int {(\frac{1}{\cos 3 x})}^{2} d x = \int \sec^{2} 3 x d x = \frac{1}{3} \tan 3 x + C$

Indefinite integrals of reciprocal trigonometric functions

$\frac{d}{d x} (\sec x) = \tan x \sec x$ and $\frac{d}{d x} (\sec k x) = k \tan k x \sec k x$ therefore
- $\int \tan x \sec x d x = \sec x + C$
- $\int \tan k x \sec k x d x = \frac{1}{k} \sec k x + C$
$\frac{d}{d x} (\csc x) = - \cot x \csc x$ and $\frac{d}{d x} (\csc k x) = - k \cot k x \csc k x$ therefore
- $\int \cot x \csc x d x = - \csc x + C$
- $\int \cot k x \csc k x d x = - \frac{1}{k} \csc k x + C$
$\frac{d}{d x} (\cot x) = - \csc^{2} x$ and $\frac{d}{d x} (co t k x) = - k \csc^{2} k x$ therefore
- $\int \csc^{2} x d x = - \cot x + C$
- $\int \csc^{2} k x d x = - \frac{1}{k} \cot k x + C$
In the above formulae, $k$ is a real number constant

Worked Example

Find the indefinite integral $\int \cot 5 x \csc 5 x d x$

Use $\int \cot k x \csc k x d x = - \frac{1}{k} \csc k x + C$

$\int \cot 5 x \csc 5 x d x = - \frac{1}{5} \csc 5 x + C$

Indefinite integrals using inverse trigonometric functions

$\frac{d}{d x} (\arcsin x) = \frac{1}{\sqrt{1 - x^{2}}}, - 1 < x < 1,$ therefore
- $\int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin x + C, - 1 < x < 1$
$\frac{d}{d x} (arc \cos x) = - \frac{1}{\sqrt{1 - x^{2}}}, - 1 < x < 1,$ therefore
- $\int \frac{1}{\sqrt{1 - x^{2}}} d x = - arc \cos x + C, - 1 < x < 1$
You can see that $\int \frac{1}{\sqrt{1 - x^{2}}} d x$ is either $\arcsin x + C$ or $- \arccos x + C$
- Usually $\arcsin$ is used when finding indefinite integrals of this form
$\frac{d}{d x} (arc \tan x) = \frac{1}{1 + x^{2}}$ therefore
- $\int \frac{1}{1 + x^{2}} d x = arc \tan x + C$

Table of common indefinite integrals

In the table below, $k$ is a real number constant and $a$ is a positive real number constant

Standard derivative	Corresponding indefinite integral
$\frac{d}{d x} (x^{n}) = n x^{n - 1}$	$\int x^{n} d x = \frac{1}{n + 1} x^{n + 1} + C, n \neq - 1$
$\frac{d}{d x} (k x) = k$	$\int k d x = k x + C$
derivative of a constant is zero	$\int 0 d x = C$
$\frac{d}{d x} (e^{k x}) = k e^{x}$	$\int e^{k x} d x = \frac{1}{k} e^{k x} + C$
$\frac{d}{d x} (a^{k x}) = a^{k x} k \ln a$	$\int a^{k x} d x = \frac{1}{k \ln a} a^{k x} + C$
$\frac{d}{d x} (\ln x) = \frac{1}{x}$	$\int \frac{1}{x} d x = \ln \|x\| + C$
$\frac{d}{d x} (\sin k x) = k \cos k x$	$\int \cos k x d x = \frac{1}{k} \sin k x + C$
$\frac{d}{d x} (\cos k x) = - k \sin k x$	$\int \sin k x d x = - \frac{1}{k} \cos k x + C$
$\frac{d}{d x} (\tan k x) = k \sec^{2} k x$	$\int \sec^{2} k x d x = \frac{1}{k} \tan k x + C$
$\frac{d}{d x} (\sec k x) = k \tan k x \sec k x$	$\int \tan k x \sec k x d x = \frac{1}{k} \sec k x + C$
$\frac{d}{d x} (\csc k x) = - k \cot k x \csc k x$	$\int \cot k x \csc k x d x = - \frac{1}{k} \csc k x + C$
$\frac{d}{d x} (co t k x) = - k \csc^{2} k x$	$\int \csc^{2} k x d x = - \frac{1}{k} \cot k x + C$
$\frac{d}{d x} (\arcsin x) = \frac{1}{\sqrt{1 - x^{2}}}, - 1 < x < 1$	$\int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin x + C, - 1 < x < 1$
$\frac{d}{d x} (arc \tan x) = \frac{1}{1 + x^{2}}$	$\int \frac{1}{1 + x^{2}} d x = arc \tan x + C$

Last updated: 8 August 2024

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Derivatives & Antiderivatives (College Board AP® Calculus AB)

Study Guide

Derivatives & antiderivatives

How can I find the indefinite integrals of common functions?

Indefinite integrals of powers of x

Indefinite integrals of exponentials and 1/x

Indefinite integrals of trigonometric functions

Indefinite integrals of reciprocal trigonometric functions

Indefinite integrals using inverse trigonometric functions

Table of common indefinite integrals

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