Derivatives & Antiderivatives (College Board AP® Calculus AB)

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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

Worked Example

Indefinite integrals of exponentials and 1/x

Worked Example

Indefinite integrals of trigonometric functions

Worked Example

Indefinite integrals of reciprocal trigonometric functions

Worked Example

Indefinite integrals using inverse trigonometric functions

Table of common indefinite integrals

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Author: Roger B

Author: Dan Finlay