# Methods of Integration

Examples showing how various functions can be integrated

**Contents**

- Introduction
- Simple Algebraic Equations
- Rational Algebraic Functions Whose Denominator Factorizes
- Rational Algebraic Functions Whose Denominators Do Not Factorize
- Irrational Algebraic Fraction Of The Following Kind
- An Irrational Function Of The Following Type
- An Irrational Function Containing
- Simple Trigonometrical Functions
- Using Trigonometrical Formula
- Any Hyperbolic Function
- Integration Of Irrational Equations Of The Following Type Using Trigonometrical Substitution
- Integration By Parts
- Page Comments

## Introduction

The following methods of Integration cover all the Normal Requirements of A.P.; A. level; The International Baccalaureate as well as Engineering Degree Courses. It does not cover approximate methods such as The Trapezoidal Rule or Simpson's Rule. These will be covered in another paper.## Rational Algebraic Functions Whose Denominators Do Not Factorize

Here are some examplesExample:

##### Example - Simple example

Problem

Workings

Solution

## Irrational Algebraic Fraction Of The Following Kind

Example:

##### Example - Hiperbolic functions

Problem

Workings

Solution

Other forms

## An Irrational Function Of The Following Type

thus the original equation can now be rewritten as :-

Example:

##### Example -

Problem

Find the integral of

Workings

let U =
The integral can now be written as :-

Solution

## An Irrational Function Containing

substitute
So the integral is now rational in

Example:

##### Example -

Problem

Find the integral of

Workings

Substitute
i.e.
Therefore

### Todo

- Review the following workings thus the integral can be written as:-

Solution

Therefore

## Using Trigonometrical Formula

Example:

##### Example -

Problem

To find the integral of

Workings

But
from which it can be shown that

Solution

### Any Trigonometrical Formula

- To integrate any trigonometrical function such as

Example:

##### Example -

Problem

\int \cosec\:x\:dx = \int \frac{1}{sin\,x}\:dx

Workings

Solution

Therefore

\int \cosec\:x\:dx = \ln\,\tan\frac{x}{2} + C

### Any Hyperbolic Equation

- Then

Example:

##### Example -

Problem

\int \sech\:\phi\:d\phi=\int\frac{2U}{1+U^2}\:.\:\frac{1}{U}\:dU

Workings

Solution