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,円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円}{x\:-\:1}\:+\:\frac{2}{x\:-\:2} \right)\:dx
=\:-\:\Ln(x\:-\:1)\:+\:2,円\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,円x\:+\:3}{x^2\:+\:3,円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\:dUExample:
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,円dx\:= \tan\:x + C
\int \sin^4\:\cos\:x\:dx = \frac{1}{5}\:\sin^5,円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,円\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,円\theta + C
Any Hyperbolic Equation
- \intf(sinh,円\theta\:\cos,円\theta)\:d\thetaput\;\:\;\;\;\;U = e^\phiThen\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,円(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,円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