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Integration & Antiderivatives

Emily Davis

Emily Davis

7 min read

Next Topic - Indefinite Integral Rules
Study Guide Overview

This study guide covers derivatives and antiderivatives (indefinite integrals). It explores the indefinite integrals of common functions including powers of x, exponentials, logarithmic functions, trigonometric functions, reciprocal trigonometric functions, and inverse trigonometric functions. A table of common indefinite integrals is provided for easy reference, along with practice questions and a glossary of key terms.

#Derivatives & Antiderivatives

#Table of Contents

  1. Introduction
  2. Indefinite Integrals of Common Functions
    • Powers of xxx
    • Exponentials and 1x\frac{1}{x}x1​
    • Trigonometric Functions
    • Reciprocal Trigonometric Functions
    • Inverse Trigonometric Functions
  3. Table of Common Indefinite Integrals
  4. Practice Questions
  5. Glossary
  6. Summary and Key Takeaways

#Introduction

Understanding derivatives and antiderivatives (indefinite integrals) is crucial in calculus. Differentiation and integration are inverse operations, meaning you can often reverse the process of differentiation to find the indefinite integrals of functions.

Exam Tip

Link to IB Curriculum: This section covers the fundamental concepts of calculus, focusing on differentiation and integration, which are essential for understanding the rate of change and the area under curves.

#Indefinite Integrals of Common Functions

#Indefinite Integrals of Powers of xxx

If you know the derivative of a function, you can find its indefinite integral. For any function f(x)f(x)f(x):

f′(x)=g(x)  ⟹  ∫g(x),dx=f(x)+Cf'(x) = g(x) \implies \int g(x) , dx = f(x) + Cf′(x)=g(x)⟹∫g(x),dx=f(x)+C

Key Concept

Key Concept: If ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}dxd​(xn)=nxn−1, then ∫xn,dx=1n+1xn+1+C\int x^n , dx = \frac{1}{n+1} x^{n+1} + C∫xn,dx=n+11​xn+1+C for n≠−1n \neq -1n=−1.

Special Cases:

  • ∫k,dx=kx+C\int k , dx = kx + C∫k,dx=kx+C, where kkk is a constant
  • ∫0,dx=C\int 0 , dx = C∫0,dx=C
**Worked Example:**

Find the indefinite integral ∫x10,dx\int x^{10} , dx∫x10,dx.

Solution:

Using ∫xn,dx=1n+1xn+1+C\int x^n , dx = \frac{1}{n+1} x^{n+1} + C∫xn,dx=n+11​xn+1+C,

∫x10,dx=110+1x10+1+C=111x11+C\int x^{10} , dx = \frac{1}{10+1} x^{10+1} + C = \frac{1}{11} x^{11} + C∫x10,dx=10+11​x10+1+C=111​x11+C

#Indefinite Integrals of Exponentials and 1x\frac{1}{x}x1​

For exponential functions:

ddx(ex)=ex  ⟹  ∫ex,dx=ex+C\frac{d}{dx}(e^x) = e^x \implies \int e^x , dx = e^x + Cdxd​(ex)=ex⟹∫ex,dx=ex+C

ddx(ekx)=kekx  ⟹  ∫ekx,dx=1kekx+C\frac{d}{dx}(e^{kx}) = ke^{kx} \implies \int e^{kx} , dx = \frac{1}{k} e^{kx} + Cdxd​(ekx)=kekx⟹∫ekx,dx=k1​ekx+C

For logarithmic functions:

ddx(ln⁡x)=1x  ⟹  ∫1x,dx=ln⁡∣x∣+C\frac{d}{dx}(\ln x) = \frac{1}{x} \implies \int \frac{1}{x} , dx = \ln |x| + Cdxd​(lnx)=x1​⟹∫x1​,dx=ln∣x∣+C

Note: The absolute value in ln⁡∣x∣\ln |x|ln∣x∣ allows the integral to be valid for both positive and negative values of xxx.

**Worked Example:**

Find the indefinite integral ∫e7x,dx\int e^{7x} , dx∫e7x,dx.

Solution:

Using ∫ekx,dx=1kekx+C\int e^{kx} , dx = \frac{1}{k} e^{kx} + C∫ekx,dx=k1​ekx+C,

∫e7x,dx=17e7x+C\int e^{7x} , dx = \frac{1}{7} e^{7x} + C∫e7x,dx=71​e7x+C

#Indefinite Integrals of Trigonometric Functions

For trigonometric functions:

ddx(sin⁡x)=cos⁡x  ⟹  ∫cos⁡x,dx=sin⁡x+C\frac{d}{dx}(\sin x) = \cos x \implies \int \cos x , dx = \sin x + Cdxd​(sinx)=cosx⟹∫cosx,dx=sinx+C

ddx(cos⁡x)=−sin⁡x  ⟹  ∫sin⁡x,dx=−cos⁡x+C\frac{d}{dx}(\cos x) = -\sin x \implies \int \sin x , dx = -\cos x + Cdxd​(cosx)=−sinx⟹∫sinx,dx=−cosx+C

ddx(tan⁡x)=sec⁡2x  ⟹  ∫sec⁡2x,dx=tan⁡x+C\frac{d}{dx}(\tan x) = \sec^2 x \implies \int \sec^2 x , dx = \tan x + Cdxd​(tanx)=sec2x⟹∫sec2x,dx=tanx+C

**Worked Example:**

Find the following indefinite integrals:

(a) ∫sin⁡4x,dx\int \sin 4x , dx∫sin4x,dx

Solution:

Using ∫sin⁡kx,dx=−1kcos⁡kx+C\int \sin kx , dx = -\frac{1}{k} \cos kx + C∫sinkx,dx=−k1​coskx+C,

∫sin⁡4x,dx=−14cos⁡4x+C\int \sin 4x , dx = -\frac{1}{4} \cos 4x + C∫sin4x,dx=−41​cos4x+C

(b) ∫(1cos⁡3x)2,dx\int \left(\frac{1}{\cos 3x}\right)^2 , dx∫(cos3x1​)2,dx

Solution:

Remember sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}secx=cosx1​,

(1cos⁡3x)2=(sec⁡3x)2=sec⁡23x\left(\frac{1}{\cos 3x}\right)^2 = (\sec 3x)^2 = \sec^2 3x(cos3x1​)2=(sec3x)2=sec23x

Using ∫sec⁡2kx,dx=1ktan⁡kx+C\int \sec^2 kx , dx = \frac{1}{k} \tan kx + C∫sec2kx,dx=k1​tankx+C,

∫(1cos⁡3x)2,dx=∫sec⁡23x,dx=13tan⁡3x+C\int \left(\frac{1}{\cos 3x}\right)^2 , dx = \int \sec^2 3x , dx = \frac{1}{3} \tan 3x + C∫(cos3x1​)2,dx=∫sec23x,dx=31​tan3x+C

#Indefinite Integrals of Reciprocal Trigonometric Functions

For reciprocal trigonometric functions:

ddx(sec⁡x)=tan⁡xsec⁡x  ⟹  ∫tan⁡xsec⁡x,dx=sec⁡x+C\frac{d}{dx}(\sec x) = \tan x \sec x \implies \int \tan x \sec x , dx = \sec x + Cdxd​(secx)=tanxsecx⟹∫tanxsecx,dx=secx+C

ddx(csc⁡x)=−cot⁡xcsc⁡x  ⟹  ∫cot⁡xcsc⁡x,dx=−csc⁡x+C\frac{d}{dx}(\csc x) = -\cot x \csc x \implies \int \cot x \csc x , dx = -\csc x + Cdxd​(cscx)=−cotxcscx⟹∫cotxcscx,dx=−cscx+C

ddx(cot⁡x)=−csc⁡2x  ⟹  ∫csc⁡2x,dx=−cot⁡x+C\frac{d}{dx}(\cot x) = -\csc^2 x \implies \int \csc^2 x , dx = -\cot x + Cdxd​(cotx)=−csc2x⟹∫csc2x,dx=−cotx+C

**Worked Example:**

Find the indefinite integral ∫cot⁡5xcsc⁡5x,dx\int \cot 5x \csc 5x , dx∫cot5xcsc5x,dx.

Solution:

Using ∫cot⁡kxcsc⁡kx,dx=−1kcsc⁡kx+C\int \cot kx \csc kx , dx = -\frac{1}{k} \csc kx + C∫cotkxcsckx,dx=−k1​csckx+C,

∫cot⁡5xcsc⁡5x,dx=−15csc⁡5x+C\int \cot 5x \csc 5x , dx = -\frac{1}{5} \csc 5x + C∫cot5xcsc5x,dx=−51​csc5x+C

#Indefinite Integrals Using Inverse Trigonometric Functions

For inverse trigonometric functions:

ddx(arcsin⁡x)=11−x2,,−1<x<1  ⟹  ∫11−x2,dx=arcsin⁡x+C\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}, , -1 < x < 1 \implies \int \frac{1}{\sqrt{1-x^2}} , dx = \arcsin x + Cdxd​(arcsinx)=1−x2​1​,,−1<x<1⟹∫1−x2​1​,dx=arcsinx+C

ddx(arccos⁡x)=−11−x2,,−1<x<1  ⟹  ∫11−x2,dx=−arccos⁡x+C\frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1-x^2}}, , -1 < x < 1 \implies \int \frac{1}{\sqrt{1-x^2}} , dx = -\arccos x + Cdxd​(arccosx)=−1−x2​1​,,−1<x<1⟹∫1−x2​1​,dx=−arccosx+C

Note: Usually, arcsin⁡\arcsinarcsin is used when finding indefinite integrals of this form.

ddx(arctan⁡x)=11+x2  ⟹  ∫11+x2,dx=arctan⁡x+C\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2} \implies \int \frac{1}{1+x^2} , dx = \arctan x + Cdxd​(arctanx)=1+x21​⟹∫1+x21​,dx=arctanx+C

#Table of Common Indefinite Integrals

In the table below, kkk is a real number constant and aaa is a positive real number constant:

Standard DerivativeCorresponding Indefinite Integral
ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}dxd​(xn)=nxn−1∫xn,dx=1n+1xn+1+C,,n≠−1\int x^n , dx = \frac{1}{n+1} x^{n+1} + C, , n \neq -1∫xn,dx=n+11​xn+1+C,,n=−1
ddx(kx)=k\frac{d}{dx}(kx) = kdxd​(kx)=k∫k,dx=kx+C\int k , dx = kx + C∫k,dx=kx+C
Derivative of a constant is zero∫0,dx=C\int 0 , dx = C∫0,dx=C
ddx(ekx)=kekx\frac{d}{dx}(e^{kx}) = ke^{kx}dxd​(ekx)=kekx∫ekx,dx=1kekx+C\int e^{kx} , dx = \frac{1}{k} e^{kx} + C∫ekx,dx=k1​ekx+C
ddx(akx)=akxkln⁡a\frac{d}{dx}(a^{kx}) = a^{kx} k \ln adxd​(akx)=akxklna∫akx,dx=1kln⁡aakx+C\int a^{kx} , dx = \frac{1}{k \ln a} a^{kx} + C∫akx,dx=klna1​akx+C
ddx(ln⁡x)=1x\frac{d}{dx}(\ln x) = \frac{1}{x}dxd​(lnx)=x1​∫1x,dx=ln⁡\int \frac{1}{x} , dx = \ln∫x1​,dx=ln
ddx(sin⁡kx)=kcos⁡kx\frac{d}{dx}(\sin kx) = k \cos kxdxd​(sinkx)=kcoskx∫cos⁡kx,dx=1ksin⁡kx+C\int \cos kx , dx = \frac{1}{k} \sin kx + C∫coskx,dx=k1​sinkx+C
ddx(cos⁡kx)=−ksin⁡kx\frac{d}{dx}(\cos kx) = -k \sin kxdxd​(coskx)=−ksinkx∫sin⁡kx,dx=−1kcos⁡kx+C\int \sin kx , dx = -\frac{1}{k} \cos kx + C∫sinkx,dx=−k1​coskx+C
ddx(tan⁡kx)=ksec⁡2kx\frac{d}{dx}(\tan kx) = k \sec^2 kxdxd​(tankx)=ksec2kx∫sec⁡2kx,dx=1ktan⁡kx+C\int \sec^2 kx , dx = \frac{1}{k} \tan kx + C∫sec2kx,dx=k1​tankx+C
ddx(sec⁡kx)=ktan⁡kxsec⁡kx\frac{d}{dx}(\sec kx) = k \tan kx \sec kxdxd​(seckx)=ktankxseckx∫tan⁡kxsec⁡kx,dx=1ksec⁡kx+C\int \tan kx \sec kx , dx = \frac{1}{k} \sec kx + C∫tankxseckx,dx=k1​seckx+C
ddx(csc⁡kx)=−kcot⁡kxcsc⁡kx\frac{d}{dx}(\csc kx) = -k \cot kx \csc kxdxd​(csckx)=−kcotkxcsckx∫cot⁡kxcsc⁡kx,dx=−1kcsc⁡kx+C\int \cot kx \csc kx , dx = -\frac{1}{k} \csc kx + C∫cotkxcsckx,dx=−k1​csckx+C
ddx(cot⁡kx)=−kcsc⁡2kx\frac{d}{dx}(\cot kx) = -k \csc^2 kxdxd​(cotkx)=−kcsc2kx∫csc⁡2kx,dx=−1kcot⁡kx+C\int \csc^2 kx , dx = -\frac{1}{k} \cot kx + C∫csc2kx,dx=−k1​cotkx+C
ddx(arcsin⁡x)=11−x2,,−1<x<1\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}, , -1 < x < 1dxd​(arcsinx)=1−x2​1​,,−1<x<1∫11−x2,dx=arcsin⁡x+C,,−1<x<1\int \frac{1}{\sqrt{1-x^2}} , dx = \arcsin x + C, , -1 < x < 1∫1−x2​1​,dx=arcsinx+C,,−1<x<1
ddx(arctan⁡x)=11+x2\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}dxd​(arctanx)=1+x21​∫11+x2,dx=arctan⁡x+C\int \frac{1}{1+x^2} , dx = \arctan x + C∫1+x21​,dx=arctanx+C

#Practice Questions

Practice Question

Question 1:

Evaluate the indefinite integral ∫5x4,dx\int 5x^4 , dx∫5x4,dx.

Answer:

Using ∫xn,dx=1n+1xn+1+C\int x^n , dx = \frac{1}{n+1} x^{n+1} + C∫xn,dx=n+11​xn+1+C,

∫5x4,dx=5⋅14+1x4+1+C=x5+C\int 5x^4 , dx = 5 \cdot \frac{1}{4+1} x^{4+1} + C = x^5 + C∫5x4,dx=5⋅4+11​x4+1+C=x5+C

Practice Question

Question 2:

Evaluate the indefinite integral ∫12x,dx\int \frac{1}{2x} , dx∫2x1​,dx.

Answer:

Using ∫1x,dx=ln⁡∣x∣+C\int \frac{1}{x} , dx = \ln |x| + C∫x1​,dx=ln∣x∣+C,

∫12x,dx=12∫1x,dx=12ln⁡∣x∣+C\int \frac{1}{2x} , dx = \frac{1}{2} \int \frac{1}{x} , dx = \frac{1}{2} \ln |x| + C∫2x1​,dx=21​∫x1​,dx=21​ln∣x∣+C

Practice Question

Question 3:

Evaluate the indefinite integral ∫cos⁡5x,dx\int \cos 5x , dx∫cos5x,dx.

Answer:

Using ∫cos⁡kx,dx=1ksin⁡kx+C\int \cos kx , dx = \frac{1}{k} \sin kx + C∫coskx,dx=k1​sinkx+C,

∫cos⁡5x,dx=15sin⁡5x+C\int \cos 5x , dx = \frac{1}{5} \sin 5x + C∫cos5x,dx=51​sin5x+C

Practice Question

Question 4:

Evaluate the indefinite integral ∫e−3x,dx\int e^{-3x} , dx∫e−3x,dx.

Answer:

Using ∫ekx,dx=1kekx+C\int e^{kx} , dx = \frac{1}{k} e^{kx} + C∫ekx,dx=k1​ekx+C,

∫e−3x,dx=1−3e−3x+C=−13e−3x+C\int e^{-3x} , dx = \frac{1}{-3} e^{-3x} + C = -\frac{1}{3} e^{-3x} + C∫e−3x,dx=−31​e−3x+C=−31​e−3x+C

#Glossary

  • Indefinite Integral: An integral without specific limits, representing a family of functions.
  • Constant of Integration (CCC): An arbitrary constant added to the result of an indefinite integral.
  • Inverse Operations: Operations that reverse the effect of each other, like differentiation and integration.
  • Logarithm (ln⁡x\ln xlnx): The natural logarithm, the inverse function of the exponential function exe^xex.
  • Absolute Value (∣x∣|x|∣x∣): The non-negative value of xxx, regardless of its sign.

#Summary and Key Takeaways

  • Differentiation and integration are inverse operations.
  • To find the indefinite integral, you reverse the differentiation process.
  • Common integrals include powers of xxx, exponential functions, trigonometric functions, and their reciprocals.
  • Always include the constant of integration (CCC) in your indefinite integrals.
  • Use the provided table of common indefinite integrals as a quick reference.
Exam Tip

Exam Tip: Remember to check the conditions for the variables when integrating, especially for logarithmic and inverse trigonometric functions.

#Key Takeaways

  • Differentiation and integration are closely related.
  • Memorize the basic indefinite integrals for quick reference.
  • Practice problems to strengthen your understanding and application of integration rules.

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Question 1 of 12

What is the indefinite integral of x2x^2x2?

x33\frac{x^3}{3}3x3​

x32+C\frac{x^3}{2} + C2x3​+C

x33+C\frac{x^3}{3} + C3x3​+C

2x+C2x + C2x+C