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MathsCalculus

Methods of Integration

Examples showing how various functions can be integrated
+ View other versions (4)

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}
\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