Methods of Integration

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Contents
  1. Introduction
  2. Simple Algebraic equations
  3. Rational Algebraic Functions whose Denominator Factorizes
  4. Rational Algebraic Functions whose Denominators do not Factorize
  5. Irrational Algebraic Fraction of the following kind
  6. An Irrational Function
  7. An Irrational Function Containing
  8. Simple Trigonometrical Functions
  9. Using Trigonometrical formula
  10. Any Trigonometrical formula
  11. Any Hyperbolic Function
  12. Simple equations
  13. Any hyperbolic equation
  14. Integration of Irrational Equations of the Following Type using Trigonometrical Substitution
  15. Integration by Parts
  16. The Integration by Parts twice to regain the Original Integral

Examples showing how various functions can be integrated

Calculus Maths

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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.

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

$\int\frac{f^{'}(x)}{f(x)}\:dx = \ln\:f(x) + C$ 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 1
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 1
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

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

$\text{let}\;\;\;U = \ln\:x$ $\therefore\;\:\;\frac{dU}{dx} = \frac{1}{x}$ and $dx = x\:dU$ 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 1
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 1
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

$\int \cos\:x\:dx = sin\:x\:+\:C$ $\int \sin\:x\:dx\:= -\:\cos\:x + C$ $\int \tan\:x\:dx = \ln\:\sec\:x + C$ $\int \sec^2\:x\,dx\:= \tan\:x + C$ $\int \sin^4\,x\:\cos\:x\:dx = \frac{1}{5}\:\sin^5\,x + C$

Using Trigonometrical formula

Example 1
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 1
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

$\int (\sinh\,\theta\:\cos\,\theta)\:d \theta$ $put\;\:\;\;\;\;U = e^\theta$ Then $\sinh\:\theta = \frac{U\:-\:\frac{1}{U}}{2}\:=\:\frac{U^2\:-\:1}{2U}$ $\cosh\:\theta\:=\:\frac{U\:+\:\frac{1}{U}}{2}\:=\:\frac{U^2\:+\:1}{2U}$

Example 1
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 1
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

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

$$\therefore\;\;\;uv = \int u\:\frac{dv}{dx}\:dx\:+\:\int v\:\frac{du}{dx}\:dx$$
(3)

$\int u\:\frac{dv}{dx}\:dx\:=\:uv\:-\:\int v\:\frac{du}{dx}\:dx$ this can also be written as:- $\int \:u (v)^{'}\:dx = u\,v\:-\:\int v\:(u)'\:dx$

Example 1
Problem

$\int x \cdot cos x\:dx$

Workings

$=\int x \cdot (\sin x)'\:dx = x \sin x-\int \sin x\:dx$

Solution

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

The Integration by Parts twice to regain the Original Integral

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