Methods of Integration

Examples showing how various functions can be integrated

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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.

Simple Algebraic Equations

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

Except when n = -1 Then

$\int \frac{dx}{x} = Ln\,x + C$

Rational Algebraic Functions Whose Denominator Factorizes

Here is a worked example

$\int \frac{x}{(x\:-\:1)(x\:-\:2)}\:dx$

$= \int \left( \frac{-\,1}{x\:-\:1}\:+\:\frac{2}{x\:-\:2} \right)\:dx$

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

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

Rational Algebraic Functions Whose Denominators Do Not Factorize

$\int\frac{f^{'}(x)}{f(x)} = \Ln\:f(x) + C$

Here are some examples

$\int \frac{2\,x\:+\:3}{x^2\:+\:3\,x\:+\:7}\;dx = \Ln(x^2\:+\:3x\:+\:7) + C$

$\int \frac{a}{x^2\:+\:a^2}\:dx = tan^{-1}\left (\frac{x}{a}\right ) + C$

Example:

Example - Simple example

Problem

$\int \frac{1}{x^2\:+\:8x\:+\:25}\,dx$

Workings

$= \frac{1}{3}\int \frac{3}{(x+4)^2+3^2}\:dx$

Solution

$= \frac{1}{3}\:tan^{-1}\left (\frac{x\:+\:4}{3}\right ) + C$

Irrational Algebraic Fraction Of The Following Kind

$\frac{ax\:+\:b}{\sqrt[]{px^2\:+\:qx\:+\:s}}\;\:\:\;where\;p\:\neq\:0$

Example:

Example - Hiperbolic functions

Problem

$\int \frac{1}{\sqrt{x^2+2x-3}}\; dx$

Workings

$= \int \frac{1}{\sqrt[]{(x+1)^2-4}}$

$=cosh^{-1}(\frac{x+1}{2}) + C$

Solution

Other forms

$\int \frac{b}{(x + a)^2-d^2}\:dx = b \cosh^{-1} \frac{x + a}{d}$

$\int \frac{b}{\sqrt[]{(x+a)^2+d^2}} = b \sinh^{-1} \frac{x + a}{d}$

$\int \frac{b}{\sqrt[]{d^2-(x+a)^2}} = b \sin^{-1} \frac{x + c}{d}$

An Irrational Function Of The Following Type

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

$\text{let}\;\;\;U = Ln\:x$

$\therefore\;\:\;\frac{dU}{dx} = \frac{1}{x}$

$\text{and}\;\;\;dU = \frac{1}{x}\:dx$

thus the original equation can now be rewritten as :-

$\int \frac{Ln\:x}{x}\:dx = \int \frac{U}{x}\:.\:x\:dU$

$\text{And}\;\;\;\int U\:dU = \frac{1}{2}U^2 + C$

$\therefore\:\;\;\int \frac{Ln\:x}{x}\:dx = \frac{1}{2}(Ln\;x)^2 + C$

Example:

Example -

Problem

Find the integral of

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

Workings

let U =

$\sqrt[]{x^3\:+\:a^2}$

$\therefore\:\:\;U^2\:=\:x^3\:+\:a^2\;\;\;and\:\;\;2U\:\frac{dU}{dx}\:=\:3x^2$

$\therefore\;\;\;\frac{2}{3}\,U\:dU = x^2\:dx$

The integral can now be written as :-

$\int \frac{U}{U^2\:-\:a^2}\:dU$

$= \frac{2}{3}\int\frac{U^2}{U^2\:-\:a^2}\;dU$

$= \frac{2}{3}\int \left(1\:+\:\frac{a^2}{(U\:-\:a)(U\:+\:a)} \right)dU$

$= \frac{2}{3}\int \left(1\:+\:\frac{\frac{1}{2}a}{U\:-\:a}\:-\:\frac{\frac{1}{2}a}{U\:+\:a} \right)\:dU$

$= \frac{2}{3}\left[U\:+\:\frac{1}{2}\: a\:Ln\:\frac{U\:-\:a}{U\:+\:a}\right]\;+\:C$

Solution

$=\:\frac{2}{3}\left[\sqrt[]{x^3\:+\:a^2}\:+\:\frac{1}{2}\:a\:Ln\:\frac{\sqrt[]{x^3\,+\:a^2}\:-\:a}{\sqrt[]{x^3\:+\:a^2}\:+\:a} \right]$

An Irrational Function Containing

$\sqrt[n]{ax\:+\:b}$

substitute

$U = \sqrt[n]{ax\:+\:b}\;\;\;i.e.\;\;\;U^n\:=\:ax\:+\:b$

$\therefore\;\;\;\frac{n}{a}\:\;U^{(n\:-\:1)}\:dU = dx$

So the integral is now rational in $\inline U\:dU$

Example:

Example -

Problem

Find the integral of

$\int \frac{1}{x+\sqrt{2x - 1}}\;dx$

Workings

Substitute

$U=\sqrt{2x - 1}$

i.e.

$x=\frac{U^2+1}{2}$

Therefore

$U\:dU = dx$

Todo

Review the following workings

thus the integral can be written as:-

$\int \frac{1}{\frac{U^2+1}{2}+U}\:dU$

$= 2\,\int \frac{U}{(U^2+1)}dU$

$= 2\int\left(\frac{1}{U+1}-\frac{1}{(U+1)^2} \right)$

$= 2\ln (U+1)+\frac{2}{U+1}+C$

Solution

Therefore

$\int \frac{1}{x+\sqrt{2x-1}}\:dx = 2 \ln \left\{ \sqrt{2x-1} +1 \right\} +\frac{2}{\sqrt{2x-1} + 1}+ C$

Simple Trigonometrical Functions

$\int \cos\:x\:dx = sin\:x\:+\:C$

$\int \sin\:x\:dx\:= -\:\cos\:x + C$

$\int \tan\:x\:dx = Ln\:\sec\:x + C$

$\int \sec^2\:x\,dx\:= \tan\:x + C$

$\int \sin^4\:\cos\:x\:dx = \frac{1}{5}\:\sin^5\,x + C$

Using Trigonometrical Formula

Example:

Example -

Problem

To find the integral of

$(\cos \:5x\:\cos\:2x)\;dx$

Workings

But

$\cos\:A\:+\:B = 2\:\cos\frac{A\:+\:B}{2}\:\cos\frac{A\,-\:B}{2}$

from which it can be shown that

$\int\cos\,5x\:\cos\,2x\:dx = \frac{1}{2}\int (\cos\,7x\:+\:\cos\,3x)\:dx$

Solution

$=\:\frac{1}{14}\,\sin\,7x\:+\:\frac{1}{6}\sin\,3x + C$

Any Trigonometrical Formula

To integrate any trigonometrical function such as $\inline (\sin\times \cos x) dx$

$put\;\:\;\;t = tan\:\frac{x}{2}$

$but\:\;\:\;tan\:x\:=\:\frac{2\:tan\,\frac{x}{2}}{(1\:+\:tan^2\:\frac{x}{2})}$

$= \frac{2t}{1\:-\:t^2}$

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Example:

Example -

Problem

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

Workings

$= \int \frac{1}{\frac{2t}{1\:+\:t^2}}\;X\;\frac{2}{1\:+\:t^2}\;dt$

$= \int \frac{1}{t}\:dt = \ln\:t + C$

Solution

Therefore

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

Any Hyperbolic Function

Simple Equations

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

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

Any Hyperbolic Equation

$\intf(sinh\,\theta\:\cos\,\theta)\:d\theta$

$put\;\:\;\;\;\;U = e^\phi$

Then

$\sinh\:\theta = \frac{U\:-\:\frac{1}{U}}{2}\:=\:\frac{U^2\:-\:1}{2U}$

$\cosh\:\phi\:=\:\frac{U\:+\:\frac{1}{U}}{2}\:=\:\frac{u^2\:+\:1}{2U}$

Example:

Example -

Problem

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

Workings

$=\:2\:\tan{_1}\:U+C = 2\:\tan^{-1}\,e^\phi+C$

Solution

$=\:2\:\tan{_1}\:U\:+\:C\;\; = \;\;2\:\tan^{-1}\,e^\phi\:+\:C$

Integration Of Irrational Equations Of The Following Type Using Trigonometrical Substitution

$\sqrt[]{ax^2\:+\:bx\:+\:C}$

Example:

Example -

Problem

$\int \frac{1}{x^2\:\sqrt[]{1\:-\:x^2}}\:dx$

Workings

$put\;\;\;\;x\:=\:\sin\:\theta\;\;\:and\;\therefore\;\;\;dx\:=\:\cos\:\theta\:d\theta$

$thus\;\;\;integral\:=\:\int \frac{1}{sin^2\theta\:\cos\,\theta}\:\cos\theta\:d\theta$

$=\:\int cosec^2\:\theta\:d\theta$

Solution

$=\:\cot\:\theta\:+\:C\;\;=\:-\:\frac{\sqrt[]{1\:-\:x^2}}{x}\:+\:C$

Integration By Parts

$\frac{d\,(uv)}{dx}\:=\:u\:\frac{dv}{dx}\:+\:v\:\frac{du}{dx}$

$\therefore\;\;\;uv\:=\:\int u\:\frac{dv}{dx}\:dx\:+\:\int v\:\frac{du}{dx}\:dx$

(2)

$\int u\:\frac{dv}{dx}\:dx\:=\:uv\:-\:\int v\:\frac{du}{dx}\:dx$

this can also be written as:-

$\int \:u\:(v)^{'}\:dx\:=\:u\,v\:-\:\int v\:(u)'\:dx$

Example:

Example -

Problem

$\int \:x\:cos\,x\:dx\:$

Workings

$=\:\int x\:(\sin\:x)^1}\:dx\:=\:x\:sin\,x\:-\:\int \:\sin\,(x)$

Solution

$=\:x\:\sin\,x\:+\:\cos\,x\:+\:C$

The Integration By Parts Twice To Regain The Original Integral

Example:

Example -

Problem

$\int e^{3x}\cos2x\;dx=\int e^{3x}\left ( \frac{1}{2}\sin2x \right )'\;dx$

Workings

$=\frac{1}{2}e^{3x}\sin x- \frac{3}{2}\int\sin2x\;e^{3x}\;dx$

$=\frac{1}{2}e^{3x}\sin x- \frac{3}{2}\int e^{3x}-\left ( \frac{1}{2}\cos2x \right )'dx$

$=\frac{1}{2}e^{3x}\sin x- \frac{3}{2}\left [ -\frac{1}{2}e^{3x}\cos2x + \frac{1}{2}\int \cos2x\;e^{3x} \right ]dx$

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

Solution

Therefore

$\int e^{3x}\;\cos2x\;dx=\frac{2}{13}e^{3x}\;\sin2x+\frac{3}{13}e^{3x}\cos2x+C$