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Maths ›Calculus ›

Methods of Integration

Examples showing how various functions can be integrated

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Contents

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

Workings

Solution

Irrational Algebraic Fraction Of The Following Kind

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

Example:

Example - Hiperbolic functions

Problem

Workings

Solution

Other forms

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

Workings

let U =

The integral can now be written as :-

Solution

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

Workings

Substitute

i.e.

Therefore

Todo

Review the following workings

thus the integral can be written as:-

Solution

Therefore

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

Workings

But

from which it can be shown that

Solution

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

Workings

Solution

Therefore

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

Workings

Solution

Integration Of Irrational Equations Of The Following Type Using Trigonometrical Substitution

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

Example:

Example -

Problem

Workings

Solution

Integration By Parts

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

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

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

this can also be written as:-

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

Example:

Example -

Problem

Workings

Solution

The Integration By Parts Twice To Regain The Original Integral

Example:

Example -

Problem

Workings

Solution

Therefore

Last Modified: 25 Jan 11 @ 19:29 Page Rendered: 2023-01-22 22:34:05

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