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MathsCalculus

Methods of Integration

Examples showing how various functions can be integrated
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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}}
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 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
thus the integral can be written as:-
\(\int \frac{1}{\frac{U^2\:+\:1}{2}\:+\:a}\: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\,(\sqrt[]{2x\,-\:1}\:+\:1\:+\:\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 (\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}
22109/integ.jpg
+
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}
\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