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Methods of Integration

Michael Green

Michael Green

7 min read

Study Guide Overview

This guide covers integrating composite functions using the reverse chain rule. It explains how to identify the main function, adjust and compensate for coefficients, and integrate functions raised to a power. It also details the special case of integrating f'(x)/f(x), provides practice questions and a glossary of key terms.

Integrating Composite Functions

Table of Contents

  1. Introduction to Integrating Composite Functions
  2. Steps for Integrating Composite Functions
  3. Special Case: Function Raised to a Power
  4. Integrating f(x)/f(x)f'(x)/f(x)
  5. Practice Questions
  6. Glossary
  7. Summary and Key Takeaways

Introduction to Integrating Composite Functions

What is Meant by Integrating Composite Functions?

Integrating composite functions refers to the process of integrating by inspection, often using the reverse chain rule. This method involves recognizing that the chain rule used in differentiation can be reversed for integration.

Key Concept

This method is particularly useful for integrating the product of a composite function and the derivative of its inside function.

In function notation, this method is used to integrate expressions of the form:

f(g(x))g(x),dx\int f'(g(x)) \cdot g'(x) , dx

By the chain rule:

ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

Since differentiation and integration are inverse operations:

f(g(x))g(x),dx=f(g(x))+C\int f'(g(x)) \cdot g'(x) , dx = f(g(x)) + C

Exam Tip

If coefficients do not match exactly, you can use the method of 'adjust and compensate'.

Example of Adjust and Compensate

Consider the integral:

5x2,dx5x^2 , dx

Here, 5x25x^2 is not quite the derivative of g(x)=4x3g(x) = 4x^3 because:

g(x)=12x2g'(x) = 12x^2

To adjust and compensate, we rewrite:

5x2=512(12x2)=512g(x)5x^2 = \frac{5}{12} (12x^2) = \frac{5}{12} g'(x)


Steps for Integrating Composite Functions

Step 1: Identify the Main Function

Spot the main function in the integral. For example, in:

x(5x22)6,dx\int x(5x^2 - 2)^6 , dx

The main function is (5x22)6(5x^2 - 2)^6 which would come from (5x22)7(5x^2 - 2)^7.

Step 2: Adjust and Compensate

Adjust and compensate any coefficients required in the integral. For example:

(5x22)7(5x^2 - 2)^7

Would differentiate to:

7(5x22)67 \cdot (5x^2 - 2)^6

By the chain rule, multiply by the derivative of 5x225x^2 - 2, which is 10x10x. Adjust and compensate:

x(5x22)6,dx=17×110×7×10×x(5x22)6,dx\int x(5x^2 - 2)^6 , dx = \frac{1}{7} \times \frac{1}{10} \times \int 7 \times 10 \times x(5x^2 - 2)^6 , dx

Step 3: Integrate and Simplify

Integrate and simplify the expression. For example:

17070x(5x22)6,dx\frac{1}{70} \int 70x(5x^2 - 2)^6 , dx

Is the exact derivative of:

(5x22)7(5x^2 - 2)^7

So:

x(5x22)6,dx=170(5x22)7+C\int x(5x^2 - 2)^6 , dx = \frac{1}{70} (5x^2 - 2)^7 + C

Exam Tip

After some practice, you may find Step 2 is not needed because you can do it in your head.

Common Mistake

Do use the adjust and compensate method on more awkward questions involving negatives and fractions!


Special Case: Function Raised to a Power

Integrating a Function Raised to a Power

One common example is integrating a function raised to a power. The general pattern is:

f(x)[f(x)]n,dx=1n+1[f(x)]n+1+C\int f'(x) \cdot [f(x)]^n , dx = \frac{1}{n+1} [f(x)]^{n+1} + C

For instance:

x(5x22)6,dx\int x(5x^2 - 2)^6 , dx

Step-by-step:

  1. Identify the main function: (5x22)6(5x^2 - 2)^6
  2. Adjust and compensate: 17070x(5x22)6,dx\frac{1}{70} \int 70x(5x^2 - 2)^6 , dx
  3. Integrate and simplify: 170(5x22)7+C\frac{1}{70} (5x^2 - 2)^7 + C

Integrating f(x)/f(x)f'(x)/f(x)

How to Integrate f(x)/f(x)f'(x)/f(x)?

A particularly useful special case of integrating composite functions is:

f(x)f(x),dx=lnf(x)+C\int \frac{f'(x)}{f(x)} , dx = \ln |f(x)| + C

This pattern occurs when the numerator of a fraction being integrated is the derivative of the denominator.

Exam Tip

Recognize this pattern to speed up and simplify integrals of this sort.

Example of Adjust and Compensate

Consider the integral:

x2+1x3+3x,dx\int \frac{x^2 + 1}{x^3 + 3x} , dx

Adjust and compensate:

133x2+1x3+3x,dx=133x2+3x3+3x,dx=13lnx3+3x+C\frac{1}{3} \int 3 \frac{x^2 + 1}{x^3 + 3x} , dx = \frac{1}{3} \int \frac{3x^2 + 3}{x^3 + 3x} , dx = \frac{1}{3} \ln |x^3 + 3x| + C

Common Mistake

Don't forget the modulus sign in the answer when finding integrals of this form.


Worked Example

Example: Finding an Expression for f(x)f(x)

Let ff be a function whose derivative, ff', is given by:

f(x)=5x2sin(2x3)f'(x) = 5x^2 \sin(2x^3)

Given that the graph of ff passes through the point (0,1)(0, 1), find an expression for f(x)f(x).

Solution:

Use:

f(x)=f(x),dxf(x) = \int f'(x) , dx

  1. Identify the main function: sin()\sin(\dots)
  2. Recognize that sin\sin comes from cos\cos, so use the chain rule:

ddx[cos(2x3)]=sin(2x3)6x2\frac{d}{dx}[\cos(2x^3)] = -\sin(2x^3) \cdot 6x^2

  1. Adjust and compensate:

f(x)=5x2sin(2x3),dx=56(sin(2x3)6x2),dxf(x) = \int 5x^2 \sin(2x^3) , dx = -\frac{5}{6} \int (-\sin(2x^3) \cdot 6x^2) , dx

  1. Integrate:

f(x)=56cos(2x3)+Cf(x) = -\frac{5}{6} \cos(2x^3) + C

Since the graph of ff goes through (0,1)(0, 1):

f(0)=1f(0) = 1

56cos(0)+C=1-\frac{5}{6} \cos(0) + C = 1

56+C=1-\frac{5}{6} + C = 1

C=116C = \frac{11}{6}

Therefore:

f(x)=56cos(2x3)+116f(x) = -\frac{5}{6} \cos(2x^3) + \frac{11}{6}


Practice Questions

Practice Question

Question 1: Evaluate the integral:

x(3x2+2)4,dx\int x(3x^2 + 2)^4 , dx

Solution:

Identify the main function: (3x2+2)4(3x^2 + 2)^4

Adjust and compensate:

151656x(3x2+2)4,dx\frac{1}{5} \cdot \frac{1}{6} \int 5 \cdot 6 \cdot x(3x^2 + 2)^4 , dx

Integrate and simplify:

130(3x2+2)5+C\frac{1}{30} \left( 3x^2 + 2 \right)^5 + C

Practice Question

Question 2: Find the indefinite integral:

3x2x3+1,dx\int \frac{3x^2}{x^3 + 1} , dx

Solution:

Recognize the pattern:

f(x)f(x),dx=lnf(x)+C\int \frac{f'(x)}{f(x)} , dx = \ln |f(x)| + C

Identify f(x)=x3+1f(x) = x^3 + 1, so f(x)=3x2f'(x) = 3x^2.

Therefore:

3x2x3+1,dx=lnx3+1+C\int \frac{3x^2}{x^3 + 1} , dx = \ln |x^3 + 1| + C


Glossary

  • Composite Function: A function composed of two or more functions, such as f(g(x))f(g(x)).
  • Chain Rule: A fundamental rule in calculus used to differentiate composite functions.
  • Adjust and Compensate: A technique used to handle coefficients that do not match exactly in integration.
  • Main Function: The primary function being integrated, often identified by its structure.
  • Inverse Operations: Operations that undo each other, such as differentiation and integration.

Summary and Key Takeaways

Summary

  • Integrating composite functions involves using the reverse chain rule.
  • Identify the main function and adjust and compensate for coefficients.
  • Special cases include functions raised to a power and integrals of the form f(x)/f(x)f'(x)/f(x).
  • Always check your work by differentiating if possible.

Key Takeaways

  • Recognize patterns in integrals to simplify the process.
  • Use the adjust and compensate method for coefficients.
  • Practice spotting the main function and applying the reverse chain rule.

Exam Strategy

  1. Identify Patterns: Quickly recognize patterns like f(x)/f(x)f'(x)/f(x) to speed up integration.
  2. Check Work: If time allows, differentiate your result to ensure it matches the original function.
  3. Adjust and Compensate: Don't forget to adjust and compensate for coefficients.

Real-World Applications

Integrating composite functions is useful in physics for calculating areas, volumes, and solving differential equations that model real-world phenomena.


Encouragement: Keep practicing! The more you work with these integrals, the more intuitive they will become. Good luck!

Question 1 of 9

Which of the following represents the general form for integrating a composite function using reverse chain rule? 🧐

f(g(x))g(x),dx=f(g(x))+C\int f(g(x)) \cdot g'(x) , dx = f(g(x)) + C

f(g(x))g(x),dx=f(g(x))+C\int f'(g(x)) \cdot g'(x) , dx = f(g(x)) + C

f(x)g(x),dx=f(g(x))+C\int f'(x) \cdot g'(x) , dx = f(g(x)) + C

f(g(x)),dx=f(g(x))+C\int f(g(x)) , dx = f(g(x)) + C