O Level (Sec 3 & 4)Additional Mathematics
Techniques of Integration for O Level A Math
Master integration for O Level & IGCSE Additional Mathematics (4047/0606). Learn power rule, exponential, trigonometric integration, and reverse differentiation. Includes 6 practice questions with step-by-step solutions by Timothy Gan.
Timothy Gan
December 10, 2021
Understanding Integration as the Reverse of Differentiation
We have already learnt how to differentiate a function and how to take the derivative from the previous article. Is there a way for us to get back the original equation? Yes, there is. In this article, we will show you how to get back the original equation from its derivative using integration – a topic that studies the relationship between a whole and its parts.
Many students find the idea of integration to be a bit intimidating. Here's a slightly more sophisticated way to think about integration: as the inverse of differentiation. It's a useful trick, and it's easy enough to remember. Integration is an operation we use all the time, to find areas under graphs, the volume of a solid, and many others.
Integration is a fundamental topic that can be found in many mathematics syllabi such as the IGCSE Additional Mathematics (0606) and Singapore SEAB Additional Mathematics (4047).
Table of Contents
Jump to any section to learn at your own pace
Basics of Integration
Many students find the idea of integration to be a bit intimidating. Here's a slightly more sophisticated way to think about integration: as the inverse of differentiation. It's a useful trick, and it's easy enough to remember. Integration is an operation we use all the time, to find areas under graphs, the volume of a solid, and many others.
Key Concept: Integration and differentiation are inverse operations. Differentiation is a way of finding the rate at which something changes. Integration is a way of finding the total amount of change.
The Constant of Integration: When we integrate, we always add a constant $c$ at the end. This is because when we differentiate a constant, it becomes zero. So when we reverse the process (integrate), we need to account for any constant that might have been there originally.
Notation: The integral sign $\int$ represents integration, and $dx$ indicates we're integrating with respect to $x$.
### Quick Reference: Essential Integration Formulas
Below is a comprehensive table of all the integration formulas you'll need for O Level Additional Mathematics. Bookmark this section for quick reference!
Standard Integration Formulas
Power Functions
| Basic Form | Linear Form (ax + b) |
|---|---|
$\int x^n \, dx = \frac{x^{n+1}}{n+1} + c, \quad n \neq -1$ | $\int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c$ |
Integration of $\frac{1}{ax+b}$
| Basic Form | Linear Form (ax + b) |
|---|---|
$\int \frac{1}{x} \, dx = \ln|x| + c$ | $\int \frac{1}{ax+b} \, dx = \frac{1}{a}\ln|ax+b| + c$ |
Integration of Exponential Functions
| Basic Form | Linear Form (ax + b) |
|---|---|
$\int e^x \, dx = e^x + c$ | $\int e^{ax+b} \, dx = \frac{1}{a}e^{ax+b} + c$ |
Integration of Trigonometric Functions
| Basic Form | Linear Form (ax + b) |
|---|---|
$\int \cos x \, dx = \sin x + c$ | $\int \cos(ax+b) \, dx = \frac{\sin(ax+b)}{a} + c$ |
$\int \sin x \, dx = -\cos x + c$ | $\int \sin(ax+b) \, dx = -\frac{\cos(ax+b)}{a} + c$ |
$\int \sec^2 x \, dx = \tan x + c$ | $\int \sec^2(ax+b) \, dx = \frac{\tan(ax+b)}{a} + c$ |
Integration of Quadratic Trigonometric Functions
| Basic Form | Linear Form (ax + b) |
|---|---|
$\int \cos^2 x \, dx = \int \frac{\cos 2x + 1}{2} \, dx$ | [Use $\cos 2A = 2\cos^2 A - 1$] |
$\int \sin^2 x \, dx = \int \frac{1 - \cos 2x}{2} \, dx$ | [Use $\cos 2A = 1 - 2\sin^2 A$] |
$\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx$ | [Use $\sec^2 A = 1 + \tan^2 A$] |
Integration of Power Functions
The power rule is the only integration rule that you'll use when you're first learning how to integrate. The power rule for integration is a way to find the integral of powers of $x$. The power rule for integration is applicable only to monomials and polynomials without any brackets.
Power Rule for Integration:
$\int x^n \, dx = \frac{x^{n+1}}{n+1} + c, \quad n \neq -1$
$\int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c$
The rule is simple:
1. Take the expression you're integrating ($x$ or $ax+b$) and add one to the exponent: $n+1$
2. Divide the whole thing by that same value $n+1$
3. For $ax+b$, remember to divide by the coefficient of $x$ as well (which is $a$)
4. Finally, don't forget to add the constant $c$
Important Note: The power rule doesn't work when $n = -1$. For $\int x^{-1} \, dx = \int \frac{1}{x} \, dx$, we use the logarithm rule instead (covered in the next section).
Examples:
- $\int x^3 \, dx = \frac{x^4}{4} + c$
- $\int x^{-2} \, dx = \frac{x^{-1}}{-1} + c = -\frac{1}{x} + c$
- $\int (2x+3)^5 \, dx = \frac{(2x+3)^6}{2 \times 6} + c = \frac{(2x+3)^6}{12} + c$
Integration of Power Functions
Integration of Reciprocal Functions
The reciprocal function is a function whose value is the multiplicative inverse of another value. Integration of reciprocal functions is a powerful extension of the power rule of integration. However, this integral rule does not follow the usual power rule for integrals.
Reciprocal Integration Formulas:
$\int \frac{1}{x} \, dx = \ln|x| + c$
$\int \frac{1}{ax+b} \, dx = \frac{1}{a} \ln|ax+b| + c$
Why Natural Logarithm?
We obtain a natural logarithm as the answer, instead of a power function. This is because the derivative of $\ln x$ is $\frac{1}{x}$, so when we reverse the process (integrate), we get back to the logarithm.
Important Notes:
- The absolute value $|x|$ ensures the logarithm is defined for both positive and negative values of $x$
- For $ax+b$, we must divide by the coefficient $a$
- This formula explains why $n \neq -1$ in the power rule – because $x^{-1}$ has its own special integration formula
Examples:
- $\int \frac{5}{x} \, dx = 5\ln|x| + c$
- $\int \frac{1}{3x+2} \, dx = \frac{1}{3}\ln|3x+2| + c$
Integration of Reciprocal Functions
Integration of Exponential Functions
In general, we cannot apply the power rule to the exponent on $e$. But fret not, exponential functions can be integrated using special formulas.
Exponential Integration Formulas:
$\int e^x \, dx = e^x + c$
$\int e^{ax+b} \, dx = \frac{1}{a} e^{ax+b} + c$
The Remarkable Property of $e^x$:
The function $e^x$ is its own derivative, which means it's also its own integral (plus a constant)! This unique property makes exponential functions particularly important in calculus and applications involving growth and decay.
Key Points:
- The integral of $e^x$ is simply $e^x + c$ – the function stays the same
- For $e^{ax+b}$, we must divide by the coefficient of $x$ (which is $a$)
- The integral of exponential functions is easy to calculate compared to many other functions
- This function may also be solved using integration by substitution or integration by parts (advanced techniques)
Examples:
- $\int e^{2x} \, dx = \frac{1}{2}e^{2x} + c$
- $\int 3e^{5x-1} \, dx = 3 \cdot \frac{1}{5}e^{5x-1} + c = \frac{3}{5}e^{5x-1} + c$
- $\int e^{-x} \, dx = -e^{-x} + c$ (represents exponential decay)
Integration of Exponential Functions
Integration of Trigonometric Functions
Previously, we have learnt the basics of trigonometric derivatives from the topic Techniques and Applications of Differentiation. Now, let's look at the integration of trigonometric functions.
Basic Trigonometric Integration Formulas:
$\int \cos x \, dx = \sin x + c$
$\int \sin x \, dx = -\cos x + c$
$\int \sec^2 x \, dx = \tan x + c$
For Linear Expressions $ax + b$:
$\int \cos(ax+b) \, dx = \frac{\sin(ax+b)}{a} + c$
$\int \sin(ax+b) \, dx = -\frac{\cos(ax+b)}{a} + c$
$\int \sec^2(ax+b) \, dx = \frac{\tan(ax+b)}{a} + c$
The Special Relationship:
The integrals and the derivatives are inter-related. Yes, it is the reverse of the derivatives of trigonometric functions!
- Since $\frac{d}{dx}(\sin x) = \cos x$, we have $\int \cos x \, dx = \sin x + c$
- Since $\frac{d}{dx}(\cos x) = -\sin x$, we have $\int \sin x \, dx = -\cos x + c$
- Since $\frac{d}{dx}(\tan x) = \sec^2 x$, we have $\int \sec^2 x \, dx = \tan x + c$
Important Note: Watch out for the negative sign when integrating $\sin x$!
Examples:
- $\int \cos 3x \, dx = \frac{\sin 3x}{3} + c$
- $\int \sin 2x \, dx = -\frac{\cos 2x}{2} + c$
- $\int \sec^2 5x \, dx = \frac{\tan 5x}{5} + c$
Integration of Trigonometric Functions
Integration of Quadratic Trigonometric Functions
The trigonometric identities are special equations used to simplify complex trigonometric expressions. The best thing is that the basic trigonometric functions and their identities can also be used to evaluate integrals involving trigonometric functions. Now, let's look at the integration of quadratic trigonometric functions.
Key Strategy: Convert squared trigonometric functions to linear form using double angle identities.
Important Identities for Integration:
$\int \cos^2 x \, dx = \int \frac{\cos 2x + 1}{2} \, dx$
$\int \sin^2 x \, dx = \int \frac{1 - \cos 2x}{2} \, dx$
$\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx$
Why Use These Identities?
We use these identities because we cannot directly integrate $\cos^2 x$ or $\sin^2 x$ using the basic rules. By converting them to expressions involving $\cos 2x$ or $\sec^2 x$, we can apply our standard integration formulas.
The Double Angle Formulas Used:
- $\cos 2A = 2\cos^2 A - 1$ → $\cos^2 A = \frac{\cos 2A + 1}{2}$
- $\cos 2A = 1 - 2\sin^2 A$ → $\sin^2 A = \frac{1 - \cos 2A}{2}$
- $\sec^2 A = 1 + \tan^2 A$ → $\tan^2 A = \sec^2 A - 1$
Step-by-Step Process:
1. Identify the squared trigonometric function
2. Apply the appropriate identity to convert to linear form
3. Integrate using standard formulas
4. Simplify your answer
Examples:
$\int \cos^2 x \, dx = \int \frac{\cos 2x + 1}{2} \, dx = \frac{1}{2}\left(\frac{\sin 2x}{2} + x\right) + c = \frac{x}{2} + \frac{\sin 2x}{4} + c$
$\int \sin^2 x \, dx = \int \frac{1 - \cos 2x}{2} \, dx = \frac{1}{2}\left(x - \frac{\sin 2x}{2}\right) + c = \frac{x}{2} - \frac{\sin 2x}{4} + c$
$\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \tan x - x + c$
Integration of Quadratic Trigonometric Functions
Integration As A Reverse Process of Differentiation
From above, we know that integration and differentiation are inverse operations. Differentiation is a way of finding the rate at which something changes. Integration is a way of finding the total amount of change.
The Fundamental Relationship:
If $\frac{d}{dx}[F(x)] = f(x)$, then $\int f(x) \, dx = F(x) + c$
This relationship is the foundation of the Fundamental Theorem of Calculus.
Using Given Derivatives to Find Integrals:
Is it possible to solve an integral question if you are only given a function and its derivative? We can! The answer is surprisingly simple.
If we know that $\frac{d}{dx}[F(x)] = f(x)$, then we can immediately write:
$\int f(x) \, dx = F(x) + c$
Strategy for "Hence" Questions:
Many exam questions follow this pattern:
1. Show that the derivative of some function equals another expression
2. Hence, find the integral of that expression
The key word "hence" tells you to use the result from part (1) to solve part (2). You don't need to integrate from scratch – you already know the answer from the differentiation!
Important Technique:
When the integral looks similar but not exactly the same as the derivative you showed, you may need to:
- Factor out constants
- Multiply/divide by constants to match the form
- Adjust for missing or extra terms
Example Pattern:
If we showed that $\frac{d}{dx}[(2x-1)\sqrt{x+3}] = \frac{6x+11}{2\sqrt{x+3}}$
Then we can find: $\int \frac{6x+11}{\sqrt{x+3}} \, dx$
Notice the integral has $\sqrt{x+3}$ in the denominator instead of $2\sqrt{x+3}$. So we need to account for the factor of 2:
$\int \frac{6x+11}{\sqrt{x+3}} \, dx = 2(2x-1)\sqrt{x+3} + c$
Practice Questions with Detailed Solutions
Work through these problems to master integration techniques
Question 1
Integration of Power Functions
Question
Integrate the following expressions with respect to $x$:
(a) $10x^4 - 5x^2 + x + 3$
(b) $\frac{6}{x^3} - \frac{2}{\sqrt{x}} + 8x^3$
(c) $\frac{4x^5 + 3x^4 + x}{6x^3}$
Step-by-Step Solution
- 1(a) Apply power rule to each term:
- 2$\int (10x^4 - 5x^2 + x + 3) \, dx$
- 3$= \frac{10x^5}{5} - \frac{5x^3}{3} + \frac{x^2}{2} + 3x + c$
- 4$= 2x^5 - \frac{5}{3}x^3 + \frac{1}{2}x^2 + 3x + c$
- 5(b) Rewrite with negative/fractional powers first:
- 6$\int \left(6x^{-3} - 2x^{-1/2} + 8x^3\right) dx$
- 7$= \frac{6x^{-2}}{-2} - \frac{2x^{1/2}}{1/2} + \frac{8x^4}{4} + c$
- 8$= -3x^{-2} - 4x^{1/2} + 2x^4 + c = -\frac{3}{x^2} - 4\sqrt{x} + 2x^4 + c$
- 9(c) Simplify first by dividing each term:
- 10$\int \frac{4x^5 + 3x^4 + x}{6x^3} \, dx = \int \left(\frac{2}{3}x^2 + \frac{1}{2}x + \frac{1}{6}x^{-2}\right) dx$
- 11$= \frac{2}{3} \cdot \frac{x^3}{3} + \frac{1}{2} \cdot \frac{x^2}{2} + \frac{1}{6} \cdot \frac{x^{-1}}{-1} + c$
- 12$= \frac{2}{9}x^3 + \frac{1}{4}x^2 - \frac{1}{6x} + c$
Answer:
(a) $2x^5 - \frac{5}{3}x^3 + \frac{1}{2}x^2 + 3x + c$, (b) $-\frac{3}{x^2} - 4\sqrt{x} + 2x^4 + c$, (c) $\frac{2}{9}x^3 + \frac{1}{4}x^2 - \frac{1}{6x} + c$
Video Solution
Watch the step-by-step video explanation for this question
Question 2
Integration of Reciprocal Functions
Question
Integrate the following expressions with respect to $x$:
(a) $\int \frac{5}{6x} \, dx$
(b) $\int \frac{1}{4x-3} \, dx$
(c) $\int \frac{8 + (4x-3)^5}{(4x-3)^6} \, dx$
(d) $\int \frac{5}{2-5x} \, dx$
Step-by-Step Solution
- 1(a) Factor out the constant:
- 2$\int \frac{5}{6x} \, dx = \frac{5}{6} \int \frac{1}{x} \, dx = \frac{5}{6} \ln|x| + c$
- 3(b) Use reciprocal formula with $ax+b$:
- 4$\int \frac{1}{4x-3} \, dx = \frac{1}{4} \ln|4x-3| + c$
- 5(c) Separate the fraction first:
- 6$\int \frac{8 + (4x-3)^5}{(4x-3)^6} \, dx = \int \left(8(4x-3)^{-6} + (4x-3)^{-1}\right) dx$
- 7$= 8 \cdot \frac{(4x-3)^{-5}}{-5 \cdot 4} + \frac{1}{4}\ln|4x-3| + c$
- 8$= -\frac{2}{5(4x-3)^5} + \frac{1}{4}\ln|4x-3| + c$
- 9(d) Factor out the constant:
- 10$\int \frac{5}{2-5x} \, dx = 5 \int \frac{1}{2-5x} \, dx = 5 \cdot \frac{1}{-5} \ln|2-5x| + c$
- 11$= -\ln|2-5x| + c$
Answer:
(a) $\frac{5}{6}\ln|x| + c$, (b) $\frac{1}{4}\ln|4x-3| + c$, (c) $-\frac{2}{5(4x-3)^5} + \frac{1}{4}\ln|4x-3| + c$, (d) $-\ln|2-5x| + c$
Video Solution
Watch the step-by-step video explanation for this question
Question 3
Integration of Exponential Functions
Question
Integrate the following expressions with respect to $x$:
(a) $\int e^{2x-1} \, dx$
(b) $\int \frac{e^{4x} + 5}{e^{4x}} \, dx$
(c) $\int \frac{(e^{2x} - e^x)^2}{e^x} \, dx$
(d) $\int \frac{1 - e^{4-3x}}{e^{4-3x}} \, dx$
Step-by-Step Solution
- 1(a) Apply exponential formula:
- 2$\int e^{2x-1} \, dx = \frac{e^{2x-1}}{2} + c$
- 3(b) Simplify first:
- 4$\int \frac{e^{4x} + 5}{e^{4x}} \, dx = \int \left(1 + 5e^{-4x}\right) dx$
- 5$= x + 5 \cdot \frac{e^{-4x}}{-4} + c = x - \frac{5}{4}e^{-4x} + c$
- 6(c) Expand the numerator first:
- 7$\int \frac{(e^{2x} - e^x)^2}{e^x} \, dx = \int \frac{e^{4x} - 2e^{3x} + e^{2x}}{e^x} \, dx$
- 8$= \int (e^{3x} - 2e^{2x} + e^x) \, dx$
- 9$= \frac{e^{3x}}{3} - 2 \cdot \frac{e^{2x}}{2} + e^x + c = \frac{1}{3}e^{3x} - e^{2x} + e^x + c$
- 10(d) Simplify the fraction:
- 11$\int \frac{1 - e^{4-3x}}{e^{4-3x}} \, dx = \int (e^{-(4-3x)} - 1) \, dx$
- 12$= \int (e^{3x-4} - 1) \, dx = \frac{e^{3x-4}}{3} - x + c$
Answer:
(a) $\frac{e^{2x-1}}{2} + c$, (b) $x - \frac{5}{4}e^{-4x} + c$, (c) $\frac{1}{3}e^{3x} - e^{2x} + e^x + c$, (d) $\frac{e^{3x-4}}{3} - x + c$
Video Solution
Watch the step-by-step video explanation for this question
Question 4
Integration of Trigonometric Functions
Question
Integrate the following expressions with respect to $x$:
(a) $7\cos 3x + \sec^2 x$
(b) $\frac{1}{x^3} - \sin(5x+2)$
(c) $\sin 3x - \sec^2 6x$
Step-by-Step Solution
- 1(a) Apply trigonometric formulas:
- 2$\int (7\cos 3x + \sec^2 x) \, dx = 7 \cdot \frac{\sin 3x}{3} + \tan x + c$
- 3$= \frac{7}{3}\sin 3x + \tan x + c$
- 4(b) Integrate term by term:
- 5$\int \left(x^{-3} - \sin(5x+2)\right) dx = \frac{x^{-2}}{-2} - \frac{-\cos(5x+2)}{5} + c$
- 6$= -\frac{1}{2x^2} + \frac{1}{5}\cos(5x+2) + c$
- 7(c) Apply trigonometric formulas:
- 8$\int (\sin 3x - \sec^2 6x) \, dx = -\frac{\cos 3x}{3} - \frac{\tan 6x}{6} + c$
Answer:
(a) $\frac{7}{3}\sin 3x + \tan x + c$, (b) $-\frac{1}{2x^2} + \frac{1}{5}\cos(5x+2) + c$, (c) $-\frac{\cos 3x}{3} - \frac{\tan 6x}{6} + c$
Video Solution
Watch the step-by-step video explanation for this question
Question 5
Integration of Quadratic Trigonometric Functions
Question
Integrate the following expressions with respect to $x$:
(a) $3\sec^2 4x + \sin 3x$
(b) $6\cos^2 x - \sec^2 2x + x$
(c) $\sin^2 2x + \cos^2 x$
Step-by-Step Solution
- 1(a) Direct integration (no identity needed for $\sec^2$):
- 2$\int (3\sec^2 4x + \sin 3x) \, dx = 3 \cdot \frac{\tan 4x}{4} - \frac{\cos 3x}{3} + c$
- 3$= \frac{3}{4}\tan 4x - \frac{1}{3}\cos 3x + c$
- 4(b) Use identity for $\cos^2 x$:
- 5$\int (6\cos^2 x - \sec^2 2x + x) \, dx = \int \left(6 \cdot \frac{\cos 2x + 1}{2} - \sec^2 2x + x\right) dx$
- 6$= \int (3\cos 2x + 3 - \sec^2 2x + x) \, dx$
- 7$= 3 \cdot \frac{\sin 2x}{2} + 3x - \frac{\tan 2x}{2} + \frac{x^2}{2} + c$
- 8$= \frac{3}{2}\sin 2x - \frac{1}{2}\tan 2x + 3x + \frac{x^2}{2} + c$
- 9(c) Use identities for both squared terms:
- 10$\int (\sin^2 2x + \cos^2 x) \, dx = \int \left(\frac{1 - \cos 4x}{2} + \frac{\cos 2x + 1}{2}\right) dx$
- 11$= \int \left(\frac{1}{2} - \frac{1}{2}\cos 4x + \frac{1}{2}\cos 2x + \frac{1}{2}\right) dx$
- 12$= \int \left(1 - \frac{1}{2}\cos 4x + \frac{1}{2}\cos 2x\right) dx$
- 13$= x - \frac{1}{2} \cdot \frac{\sin 4x}{4} + \frac{1}{2} \cdot \frac{\sin 2x}{2} + c$
- 14$= x - \frac{1}{8}\sin 4x + \frac{1}{4}\sin 2x + c$
Answer:
(a) $\frac{3}{4}\tan 4x - \frac{1}{3}\cos 3x + c$, (b) $\frac{3}{2}\sin 2x - \frac{1}{2}\tan 2x + 3x + \frac{x^2}{2} + c$, (c) $x - \frac{1}{8}\sin 4x + \frac{1}{4}\sin 2x + c$
Video Solution
Watch the step-by-step video explanation for this question
Question 6
Integration Using Given Derivative
Question
Show that $\frac{d}{dx}[(2x-1)\sqrt{x+3}] = \frac{6x+11}{2\sqrt{x+3}}$
Hence, find $\int \frac{6x+11}{\sqrt{x+3}} \, dx$
Step-by-Step Solution
- 1Part 1: Show the derivative
- 2Let $y = (2x-1)\sqrt{x+3} = (2x-1)(x+3)^{1/2}$
- 3Using product rule: $\frac{dy}{dx} = (2x-1) \cdot \frac{d}{dx}[(x+3)^{1/2}] + (x+3)^{1/2} \cdot \frac{d}{dx}[2x-1]$
- 4$= (2x-1) \cdot \frac{1}{2}(x+3)^{-1/2} + (x+3)^{1/2} \cdot 2$
- 5$= \frac{2x-1}{2\sqrt{x+3}} + 2\sqrt{x+3}$
- 6Combine fractions: $= \frac{2x-1}{2\sqrt{x+3}} + \frac{2\sqrt{x+3} \cdot 2\sqrt{x+3}}{2\sqrt{x+3}}$
- 7$= \frac{2x-1 + 4(x+3)}{2\sqrt{x+3}} = \frac{2x-1+4x+12}{2\sqrt{x+3}} = \frac{6x+11}{2\sqrt{x+3}}$ ✓
- 8Part 2: Find the integral
- 9From part 1, we know: $\frac{d}{dx}[(2x-1)\sqrt{x+3}] = \frac{6x+11}{2\sqrt{x+3}}$
- 10Notice the integral has $\sqrt{x+3}$ (not $2\sqrt{x+3}$) in denominator
- 11Multiply derivative by 2: $\frac{d}{dx}[2(2x-1)\sqrt{x+3}] = \frac{6x+11}{\sqrt{x+3}}$
- 12Therefore: $\int \frac{6x+11}{\sqrt{x+3}} \, dx = 2(2x-1)\sqrt{x+3} + c$
Answer:
$\int \frac{6x+11}{\sqrt{x+3}} \, dx = 2(2x-1)\sqrt{x+3} + c$
Video Solution
Watch the step-by-step video explanation for this question
Key Formulas to Remember
Power Rule
$\int x^n \, dx = \frac{x^{n+1}}{n+1} + c, \quad n \neq -1$
Note: Add 1 to exponent, divide by new exponent
Power Rule (Linear)
$\int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c$
Note: Remember to divide by coefficient of x
Reciprocal Function
$\int \frac{1}{x} \, dx = \ln|x| + c$
Note: Special case when n = -1
Reciprocal (Linear)
$\int \frac{1}{ax+b} \, dx = \frac{1}{a}\ln|ax+b| + c$
Note: Natural logarithm with coefficient adjustment
Exponential Function
$\int e^x \, dx = e^x + c$
Note: The function is its own integral
Exponential (Linear)
$\int e^{ax+b} \, dx = \frac{1}{a}e^{ax+b} + c$
Note: Divide by coefficient of x
Cosine
$\int \cos x \, dx = \sin x + c$
Note: Reverse of sine differentiation
Sine
$\int \sin x \, dx = -\cos x + c$
Note: Note the negative sign
Secant Squared
$\int \sec^2 x \, dx = \tan x + c$
Note: Reverse of tangent differentiation
Special Identities to Memorize
- $\int \cos^2 x \, dx = \int \frac{\cos 2x + 1}{2} \, dx$ (use $\cos 2x = 2\cos^2 x - 1$)
- $\int \sin^2 x \, dx = \int \frac{1 - \cos 2x}{2} \, dx$ (use $\cos 2x = 1 - 2\sin^2 x$)
- $\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx$ (use $\sec^2 x = 1 + \tan^2 x$)
- $\int \cos(ax+b) \, dx = \frac{\sin(ax+b)}{a} + c$
- $\int \sin(ax+b) \, dx = -\frac{\cos(ax+b)}{a} + c$
- $\int \sec^2(ax+b) \, dx = \frac{\tan(ax+b)}{a} + c$
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