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# Methods of Integration

Examples showing how various functions can be integrated

## 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.

## Simple Algebraic Equations

$\int&space;x^n\:dx&space;=&space;\frac{1}{n\:+\:1}\:x^{n\:+\:1}&space;+&space;C$

Except when n = -1 Then
$\int&space;\frac{dx}{x}&space;=&space;Ln\,x&space;+&space;C$

## Rational Algebraic Functions Whose Denominator Factorizes

Here is a worked example
$\int&space;\frac{x}{(x\:-\:1)(x\:-\:2)}\:dx$
$=&space;\int&space;\left(&space;\frac{-\,1}{x\:-\:1}\:+\:\frac{2}{x\:-\:2}&space;\right)\:dx$

$=\:-\:\Ln(x\:-\:1)\:+\:2\,\Ln(x\:-\:2)&space;+&space;C$

$=&space;\Ln\left[K\frac{(x\:-\:2)^2}{(x\:-\:1)}&space;\right]&space;+&space;C$

## Rational Algebraic Functions Whose Denominators Do Not Factorize

$\int\frac{f^{'}(x)}{f(x)}&space;=&space;\Ln\:f(x)&space;+&space;C$
Here are some examples
$\int&space;\frac{2\,x\:+\:3}{x^2\:+\:3\,x\:+\:7}\;dx&space;=&space;\Ln(x^2\:+\:3x\:+\:7)&space;+&space;C$
$\int&space;\frac{a}{x^2\:+\:a^2}\:dx&space;=&space;tan^{-1}\left&space;(\frac{x}{a}\right&space;)&space;+&space;C$

Example:
##### Example - Simple example
Problem
$\int&space;\frac{1}{x^2\:+\:8x\:+\:25}\,dx$
Workings
$=&space;\frac{1}{3}\int&space;\frac{3}{(x+4)^2+3^2}\:dx$
Solution
$=&space;\frac{1}{3}\:tan^{-1}\left&space;(\frac{x\:+\:4}{3}\right&space;)&space;+&space;C$

## Irrational Algebraic Fraction Of The Following Kind

$\frac{ax\:+\:b}{\sqrt[]{px^2\:+\:qx\:+\:s}}\;\:\:\;where\;p\:\neq\:0$
Example:
##### Example - Hiperbolic functions
Problem
$\int&space;\frac{1}{\sqrt{x^2+2x-3}}\;&space;dx$
Workings
$=&space;\int&space;\frac{1}{\sqrt[]{(x+1)^2-4}}$
$=cosh^{-1}(\frac{x+1}{2})&space;+&space;C$
Solution
Other forms
$\int&space;\frac{b}{(x&space;+&space;a)^2-d^2}\:dx&space;=&space;b&space;\cosh^{-1}&space;\frac{x&space;+&space;a}{d}$
$\int&space;\frac{b}{\sqrt[]{(x+a)^2+d^2}}&space;=&space;b&space;\sinh^{-1}&space;\frac{x&space;+&space;a}{d}$
$\int&space;\frac{b}{\sqrt[]{d^2-(x+a)^2}}&space;=&space;b&space;\sin^{-1}&space;\frac{x&space;+&space;c}{d}$

## An Irrational Function Of The Following Type

$\int&space;\frac{Ln\:x}{x}\;dx$

$\text{let}\;\;\;U&space;=&space;Ln\:x$
$\therefore\;\:\;\frac{dU}{dx}&space;=&space;\frac{1}{x}$
$\text{and}\;\;\;dU&space;=&space;\frac{1}{x}\:dx$
thus the original equation can now be rewritten as :-
$\int&space;\frac{Ln\:x}{x}\:dx&space;=&space;\int&space;\frac{U}{x}\:.\:x\:dU$

$\text{And}\;\;\;\int&space;U\:dU&space;=&space;\frac{1}{2}U^2&space;+&space;C$
$\therefore\:\;\;\int&space;\frac{Ln\:x}{x}\:dx&space;=&space;\frac{1}{2}(Ln\;x)^2&space;+&space;C$
Example:
##### Example -
Problem

Find the integral of
$\frac{\sqrt[]{x^3\:+\:a^2}}{x}\:dx$

Workings
let U =
$\sqrt[]{x^3\:+\:a^2}$
$\therefore\:\:\;U^2\:=\:x^3\:+\:a^2\;\;\;and\:\;\;2U\:\frac{dU}{dx}\:=\:3x^2$
$\therefore\;\;\;\frac{2}{3}\,U\:dU&space;=&space;x^2\:dx$

The integral can now be written as :-
$\int&space;\frac{U}{U^2\:-\:a^2}\:dU$
$=&space;\frac{2}{3}\int\frac{U^2}{U^2\:-\:a^2}\;dU$
$=&space;\frac{2}{3}\int&space;\left(1\:+\:\frac{a^2}{(U\:-\:a)(U\:+\:a)}&space;\right)dU$
$=&space;\frac{2}{3}\int&space;\left(1\:+\:\frac{\frac{1}{2}a}{U\:-\:a}\:-\:\frac{\frac{1}{2}a}{U\:+\:a}&space;\right)\:dU$
$=&space;\frac{2}{3}\left[U\:+\:\frac{1}{2}\:&space;a\:Ln\:\frac{U\:-\:a}{U\:+\:a}\right]\;+\:C$
Solution
$=\:\frac{2}{3}\left[\sqrt[]{x^3\:+\:a^2}\:+\:\frac{1}{2}\:a\:Ln\:\frac{\sqrt[]{x^3\,+\:a^2}\:-\:a}{\sqrt[]{x^3\:+\:a^2}\:+\:a}&space;\right]$

## An Irrational Function Containing

$\sqrt[n]{ax\:+\:b}$
substitute
$U&space;=&space;\sqrt[n]{ax\:+\:b}\;\;\;i.e.\;\;\;U^n\:=\:ax\:+\:b$
$\therefore\;\;\;\frac{n}{a}\:\;U^{(n\:-\:1)}\:dU&space;=&space;dx$
So the integral is now rational in $\inline&space;U\:dU$
Example:
##### Example -
Problem
Find the integral of
$\int&space;\frac{1}{x+\sqrt{2x&space;-&space;1}}\;dx$
Workings
Substitute
$U=\sqrt{2x&space;-&space;1}$
i.e.
$x=\frac{U^2+1}{2}$

Therefore
$U\:dU&space;=&space;dx$

### Todo

Review the following workings

thus the integral can be written as:-
$\int&space;\frac{1}{\frac{U^2+1}{2}+U}\:dU$
$=&space;2\,\int&space;\frac{U}{(U^2+1)}dU$
$=&space;2\int\left(\frac{1}{U+1}-\frac{1}{(U+1)^2}&space;\right)$
$=&space;2\ln&space;(U+1)+\frac{2}{U+1}+C$
Solution
Therefore
$\int&space;\frac{1}{x+\sqrt{2x-1}}\:dx&space;=&space;2&space;\ln&space;\left\{&space;&space;\sqrt{2x-1}&space;+1&space;\right\}&space;&space;+\frac{2}{\sqrt{2x-1}&space;+&space;1}+&space;C$

## Simple Trigonometrical Functions

$\int&space;\cos\:x\:dx&space;=&space;sin\:x\:+\:C$
$\int&space;\sin\:x\:dx\:=&space;-\:\cos\:x&space;+&space;C$
$\int&space;\tan\:x\:dx&space;=&space;Ln\:\sec\:x&space;+&space;C$
$\int&space;\sec^2\:x\,dx\:=&space;\tan\:x&space;+&space;C$
$\int&space;\sin^4\:\cos\:x\:dx&space;=&space;\frac{1}{5}\:\sin^5\,x&space;+&space;C$

## Using Trigonometrical Formula

Example:
##### Example -
Problem

To find the integral of
$(\cos&space;\:5x\:\cos\:2x)\;dx$

Workings
But
$\cos\:A\:+\:B&space;=&space;2\:\cos\frac{A\:+\:B}{2}\:\cos\frac{A\,-\:B}{2}$
from which it can be shown that
$\int\cos\,5x\:\cos\,2x\:dx&space;=&space;\frac{1}{2}\int&space;(\cos\,7x\:+\:\cos\,3x)\:dx$
Solution
$=\:\frac{1}{14}\,\sin\,7x\:+\:\frac{1}{6}\sin\,3x&space;+&space;C$

### Any Trigonometrical Formula

To integrate any trigonometrical function such as $\inline&space;(\sin\times&space;&space;\cos&space;x)&space;dx$
$put\;\:\;\;t&space;=&space;tan\:\frac{x}{2}$
$but\:\;\:\;tan\:x\:=\:\frac{2\:tan\,\frac{x}{2}}{(1\:+\:tan^2\:\frac{x}{2})}$
$=&space;\frac{2t}{1\:-\:t^2}$
Example:
##### Example -
Problem
$\int&space;\cosec\:x\:dx&space;=&space;\int&space;\frac{1}{sin\,x}\:dx$
Workings
$=&space;\int&space;\frac{1}{\frac{2t}{1\:+\:t^2}}\;X\;\frac{2}{1\:+\:t^2}\;dt$
$=&space;\int&space;\frac{1}{t}\:dt&space;=&space;\ln\:t&space;+&space;C$
Solution
Therefore
$\int&space;\cosec\:x\:dx&space;=&space;\ln\,\tan\frac{x}{2}&space;+&space;C$

## Any Hyperbolic Function

### Simple Equations

$\int&space;\sech^2\:\theta\;d\theta&space;=&space;\tanh\:\theta&space;+&space;C$

$\int\:\cosh^2\:\theta\:d\theta=\:\frac{1}{2}\theta\:+\:\frac{1}{4}\:\sinh\:2\,\theta&space;+&space;C$

### Any Hyperbolic Equation

$\intf(sinh\,\theta\:\cos\,\theta)\:d\theta$
$put\;\:\;\;\;\;U&space;=&space;e^\phi$
Then
$\sinh\:\theta&space;=&space;\frac{U\:-\:\frac{1}{U}}{2}\:=\:\frac{U^2\:-\:1}{2U}$
$\cosh\:\phi\:=\:\frac{U\:+\:\frac{1}{U}}{2}\:=\:\frac{u^2\:+\:1}{2U}$
Example:
##### Example -
Problem
$\int&space;\sech\:\phi\:d\phi=\int\frac{2U}{1+U^2}\:.\:\frac{1}{U}\:dU$
Workings
$=\:2\:\tan{_1}\:U+C&space;=&space;2\:\tan^{-1}\,e^\phi+C$
Solution
$=\:2\:\tan{_1}\:U\:+\:C\;\;&space;=&space;\;\;2\:\tan^{-1}\,e^\phi\:+\:C$

## Integration Of Irrational Equations Of The Following Type Using Trigonometrical Substitution

$\sqrt[]{ax^2\:+\:bx\:+\:C}$
Example:
##### Example -
Problem
$\int&space;\frac{1}{x^2\:\sqrt[]{1\:-\:x^2}}\:dx$
Workings
$put\;\;\;\;x\:=\:\sin\:\theta\;\;\:and\;\therefore\;\;\;dx\:=\:\cos\:\theta\:d\theta$
$thus\;\;\;integral\:=\:\int&space;\frac{1}{sin^2\theta\:\cos\,\theta}\:\cos\theta\:d\theta$
$=\:\int&space;cosec^2\:\theta\:d\theta$
Solution
$=\:\cot\:\theta\:+\:C\;\;=\:-\:\frac{\sqrt[]{1\:-\:x^2}}{x}\:+\:C$

## Integration By Parts

$\frac{d\,(uv)}{dx}\:=\:u\:\frac{dv}{dx}\:+\:v\:\frac{du}{dx}$
$\therefore\;\;\;uv\:=\:\int&space;u\:\frac{dv}{dx}\:dx\:+\:\int&space;&space;&space;v\:\frac{du}{dx}\:dx$
$\int&space;u\:\frac{dv}{dx}\:dx\:=\:uv\:-\:\int&space;v\:\frac{du}{dx}\:dx$
this can also be written as:-
$\int&space;\:u\:(v)^{'}\:dx\:=\:u\,v\:-\:\int&space;v\:(u)'\:dx$
Example:
##### Example -
Problem
$\int&space;\:x\:cos\,x\:dx\:$
Workings
$=\:\int&space;x\:(\sin\:x)^1}\:dx\:=\:x\:sin\,x\:-\:\int&space;\:\sin\,(x)$
Solution
$=\:x\:\sin\,x\:+\:\cos\,x\:+\:C$

### The Integration By Parts Twice To Regain The Original Integral

Example:
##### Example -
Problem

$\int&space;e^{3x}\cos2x\;dx=\int&space;e^{3x}\left&space;(&space;\frac{1}{2}\sin2x&space;\right&space;)'\;dx$
Workings
$=\frac{1}{2}e^{3x}\sin&space;x-&space;\frac{3}{2}\int\sin2x\;e^{3x}\;dx$
$=\frac{1}{2}e^{3x}\sin&space;x-&space;\frac{3}{2}\int&space;e^{3x}-\left&space;(&space;\frac{1}{2}\cos2x&space;\right&space;)'dx$
$=\frac{1}{2}e^{3x}\sin&space;x-&space;\frac{3}{2}\left&space;[&space;-\frac{1}{2}e^{3x}\cos2x&space;+&space;\frac{1}{2}\int&space;\cos2x\;e^{3x}&space;\right&space;]dx$

$\frac{13}{4}\int&space;e^{3x}\;\cos2x\;dx=\frac{1}{2}e^{3x}\;\sin2x+\frac{3}{4}e^{3x}\cos2x$
Solution
Therefore
$\int&space;e^{3x}\;\cos2x\;dx=\frac{2}{13}e^{3x}\;\sin2x+\frac{3}{13}e^{3x}\cos2x+C$