CodeCogs

Examples showing how various functions can be integrated

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

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

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

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

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