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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

$\int \frac{2\,x + 3}{x^2 + 3x+7}\;dx = \ln(x^2+3x+7) + C$ $\int \frac{a}{x^2+a^2}\:dx = tan^{-1}\left (\frac{x}{a}\right ) + C$

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 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 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

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

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 $(\sin\times \cos x) dx$

$put\;\:\;\;t = tan\:\frac{x}{2}$ $but\:\;\:\;tan\:x\:=\:\frac{2\:tan\,\frac{x}{2}}{(1\:+\:tan^2\:\frac{x}{2})}$ $= \frac{2t}{1\:-\:t^2}$

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

$\int \sinh\theta \cos\theta \:d\theta$ $put\;\:\;\;\;\;U = e^\phi$ 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 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

$put\;\;\;\;x\:=\:\sin\:\theta\;\;\:and\;\therefore\;\;\;dx\:=\:\cos\:\theta\:d\theta$ $thus\;\;\;integral\:=\:\int \frac{1}{sin^2\theta\:\cos\,\theta}\:\cos\theta\:d\theta$ $=\:\int cosec^2\:\theta\:d\theta$