Lectures Notes of Mathematics I

MATHEMATICS-I Page 1

Modern University

For Technology and Information

Department of Physics and

Engineering Mathematics

Lectures Notes of

Mathematics I

MATH 101

Prepared By

Dr: Mona Samir Mehanna

Dr.Said Anwar Gouda

(First Edition 2021)


Vision

The vision of the Faculty of Engineering at MTI university is to

be a center of excellence in engineering education and scientific

research in national and global regions. The Faculty of

Engineering aims to prepare graduates meet the needs of society

and contribute to sustainable development.

Mission

The Faculty of Engineering MTI university aims to develop

distinguished graduates that can enhance in the scientific and

professional status, through the various programs which fulfill

the needs of local and regional markets. The Faculty of

Engineering hopes to provide the graduates a highly academic

level to keep up the global developments.


Contents

Chapter -1 Functions 4

1-1 Classification of Function 4

1-2 Transcendental Functions 11

Chapter - 2 Differentiation 11

2-1 Introduction 23

2-2 Rules of Differentiation 23

2-2-1 Differentiation of polynomials 25

2-2-2 Differentiation of the Logarithmic Function 27

2-2-3 Differentiation of Exponential Functions

Exercise-1

28

31

2-2-4 Differentiation of Trigonometric functions 33

2-2-5

2-2-6

Differentiation of inverse Trigonometric functions

Differentiation of hyperbolic Functions

36

39

2-2-7 Differential of Inverse Hyperbolic Functions 40

2-2-8 Implicit differentiation 43

2-2-9 Logarithmic differentiation 44

2-3 Parametric Equations 45

2-3-1 Parametric Differentiation 46

2-4 Higher Order Derivatives

Exercise - 2

49

54

General Exercise on Differentiation 57

Chapter- 3 Applications of the differential Calculus 58

3-1 L'Hopital Theorem 58

Exercise - 3 63

3-2 Power series Representation of a Function 65

3-2-1 Maclaurin series for f(x) 68

3-2-2 Taylor series for f(x) 74


Exercise - 4 78

Chapter - 4 Indefinite Integral 80

4-1 Rules of integration 80

4-1-1 Rules of Definite integral 81

4-1-2 Rules of standard integrals 81

4-2 Some properties of Indefinite integral 83

Exercise - 5 94

4-3 Powers of Trigonometric Integrals 97

Exercise - 6 106

4-4 Integration of Rational Function using Partial Fractions 108

Exercise - 7 114

4-5 Integration by Parts 116

4-5-1 Basic Techniques 116

4-5-2 Tabular integration 121

4-5-3 Reduction Formulas 123

Exercise - 8 125

4-6 Trigonometric Substitutions 126

Chapter-5

Exercise - 9

Application of the Definite Integral

131

132

5-1 Area under the curve 132

5-2 The Arc Length of a curve 135

5-3 Volume of Solid Revolution 138

Appendix

Exercise -10

Previous Exams

141

142

144

Appendix Rules Table of differentiation and Integration 160


X Y

F

1-F

x y

Chapter -1

Functions

1-1 Classification of Function

1- The function is a rule that determines to each element 𝑥 in the set 𝑋

one corresponding 𝑦 in a set 𝑌 𝑦 = 𝑓(𝑥).

Example 1: 𝑦 = 2𝑥3 + 3𝑥2 − 5

𝑥 is called the independent variable

𝑦 is the dependent variable

𝑋 is the Domain of definition of the function

𝑌 is the Range.

2- The inverse function 𝐼𝑓 𝑓: 𝑋 → 𝑌 𝑦 = 𝑓(𝑥)

𝑓−1 ∶ 𝑌 → 𝑋 𝑥 = 𝑓−1(𝑦)

𝑓−1(𝑓(𝑥)) = 𝑥, 𝑓(𝑓−1(𝑦)) = 𝑦.

Example 2:

Find the inverse function of 𝑓(𝑥) = 2𝑥2 − 1 0 ≤ 𝑥 < ∞ in

the form:

𝑦 = 𝑓−1(𝑥).

Solution:

𝑦 = 2𝑥2 − 1 → 𝑥2 =1

2(𝑦 + 1) → 𝑥 = √

𝑦+1

2


Change 𝑥 𝑤𝑖𝑡ℎ 𝑦 → 𝑦 = √𝑥+1

2 = 𝑓−1(𝑥) 𝑥 ≥ −1.

Example 3:

Find the inverse function of 𝑓(𝑥) = (𝑥5 − 1)4 in the form

𝑦 = 𝑓−1(𝑥).

Solution:

Let 𝑦 = (𝑥5 − 1)4 solving the equation for 𝑥

𝑥5 − 1 = 𝑦14 → 𝑥 = (√𝑦4 + 1)

15 = 𝑓−1(𝑦)

Interchanging x and y we get

𝑦 = (√𝑥4

+ 1)15 = 𝑓−1(𝑥).

3- Even and odd functions:

Even function 𝑓(−𝑥) = 𝑓(𝑥) symmetry about y-axis

Odd function 𝑓(−𝑥) = −𝑓(𝑥) symmetry about origin

Examples on even functions:

𝑓(𝑥) = cos 𝑥, 𝑓(𝑥) = 𝑠𝑒𝑐𝑥, 𝑓(𝑥) = |𝑥| , 𝑓(𝑥) = 𝑥0, 𝑥2. 𝑥4, …

𝑓(𝑥) = 1

𝑥2 ,

1

𝑥4, … ..

Examples on odd functions

𝑓(𝑥) = sin 𝑥, 𝑓(𝑥) = 𝑐𝑜𝑠𝑒𝑐 𝑥, 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠 𝑓(𝑥)

= csc 𝑥, 𝑓(𝑥) = tan 𝑥, 𝑓(𝑥) = 𝑐𝑜𝑡𝑥 , 𝑓(𝑥)

= 𝑥 , 𝑓(𝑥) = 𝑥3, 𝑥5. 𝑥7, …

𝑓(𝑥) =1

𝑥,

1

𝑥3 ,

1

𝑥5, … ..


Some facts about odd and even functions are:

1. The algebraic sum of even (odd) functions is again even (odd).

2. Multiplication or division of two functions of the same (or different) kind

is an even (odd) function.

3. Every function may be expressed as the sum of even and odd.

Example 4:

Say whether the following functions are even, odd or neither:

𝑎) 𝑦 = 𝑥2, 𝑏) 𝑦 = 𝑥3, 𝑐) 𝑦 = 𝑥5 + 1.

Solution:

𝑎) 𝑦 = 𝑥2

𝑓(𝑥) = 𝑥2 𝑓(−𝑥) = (−𝑥 )2 = 𝑥2 = 𝑓(𝑥)

Then 𝑦 = 𝑥2 is an even function.

𝑏) 𝑦 = 𝑥3

𝑓(𝑥) = 𝑥3 𝑓(−𝑥) = (−𝑥)3 = −𝑥3 = −𝑓(𝑥)

Then 𝑦 = 𝑥3 is an odd function.

𝑐) 𝑦 = 𝑥5 + 1

𝑓(𝑥) = 𝑥7 − 1 𝑓(−𝑥) = (−𝑥)7 − 1 = −𝑥7 − 1

≠ 𝑓(𝑥) ≠ −𝑓(𝑥)

Then 𝑦 = 𝑥7 + 1 is neither an even nor odd function.

Example 5:

Say whether the following functions are even, odd or neither:

𝑎) 𝑦 = 𝑐𝑜𝑠2𝑥 , 𝑏) 𝑦 = 𝑡𝑎𝑛5𝑥 , 𝑐) 𝑦 = sin 𝑥 + 𝑐𝑜𝑠 𝑥5.


Solution:

𝑎) 𝑦 = 𝑠𝑖𝑛2𝑥

𝑓(𝑥) = 𝑐𝑜𝑠2𝑥 = (cos 𝑥)2 𝑓(−𝑥) = (cos(−𝑥) )2

= (cos 𝑥)2 = 𝑐𝑜𝑠2𝑥 = 𝑓(𝑥)

Then 𝑦 = 𝑐𝑜𝑠2𝑥 is an even function.

𝑏) 𝑦 = 𝑡𝑎𝑛5𝑥

𝑓(𝑥) = 𝑡𝑎𝑛5𝑥 = (tan 𝑥)5 𝑓(−𝑥) = (tan(−𝑥))5

= (− tan 𝑥)5 = −𝑡𝑎𝑛5𝑥 = −𝑓(𝑥)

Then 𝑦 = 𝑡𝑎𝑛5𝑥 is an odd function.

𝑐) 𝑦 = sin 𝑥 + 𝑐𝑜𝑠 𝑥5

𝑓(𝑥) = sin 𝑥 + 𝑐𝑜𝑠 𝑥5

𝑓(−𝑥) = sin(−𝑥) + cos( −𝑥)5

= − sin( 𝑥) + cos( −𝑥5) ≠ sin( 𝑥) + cos 𝑥5 ≠ 𝑓(𝑥)

≠ −𝑓(𝑥)

Then 𝑦 = sin 𝑥 + 𝑐𝑜𝑠 𝑥5 is neither an even nor odd function.

4- Algebraic functions:

An algebraic function is a function that can be formed by a finite number of algebraic operations (addition, subtraction, multiplication, division and rising of a power) is classified as follows:

a- The polynomial functions

𝑦 = 𝑎0𝑥𝑛 + 𝑎1𝑥𝑛−1 + ⋯ + 𝑎𝑛−1𝑥 + 𝑎𝑛


y=3

X

Y

Where 𝑛 is a nonnegative integer called degree of the polynomial and

𝑎0, 𝑎1 … . . 𝑎𝑛 are real constant.

For example 𝑦 = 4𝑥3 +3

2𝑥 + 1

Is a polynomial function of the third degree.

b- The Constant Function

The equation 𝑦 = 𝑐 where c is a constant

Represents a constant function.

Example: 𝑦 = 3

It is a horizontal line at a distance 2 unit from

the 𝑥 −axis.

c- The Linear Function

𝑦 = 𝑎𝑥 + 𝑏 𝑎 ≠ 0

The graph is a straight line with slope 𝑎 and y-intercept 𝑏. The straight line can also be determined by knowing the two points of intersections with the coordinate axes. The two special cases 𝑦 =𝑥 𝑎𝑛𝑑 𝑦 = −𝑥.

Example:

𝑦 = 2𝑥 − 3,

𝑦 = |𝑥|.

d- The Quadratic Function

𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 𝑎 ≠ 0

y=x y=-x

X

Y

y=2x-3

X

Y


X

Y

X

Y

a<0

y=-x2

y=x2

y

x

3X-y=

3y= X

represents a quadratic function, its graph is a parabola symmetric

about the vertical line 𝑥 = −𝑏

2𝑎.

The parabola is open upwards if 𝑎 > 0 and downward if 𝑎 < 0

The special cases 𝑦 = 𝑥2 𝑎𝑛𝑑 𝑦 = −𝑥2.

e- The Cubic Function

𝑦 = 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑 𝑎 ≠ 0

The special case 𝑦 = − 𝑥3 𝑎𝑛𝑑 𝑦 = 𝑥3.

f- The Fractional Rational Function:

Is defined as the ratio of two polynomial

𝑦 = 𝑎0𝑥𝑛 + 𝑎1𝑥𝑛−1 + ⋯ + 𝑎𝑛−1𝑥 + 𝑎𝑛

𝑏0𝑥𝑚 + 𝑏1𝑥𝑚−1+. . . +𝑏𝑚−1𝑥 + 𝑎𝑚

Where the degree of numerator is 𝑛 and the degree of denominator is

𝑚. These functions are defined for all real values of 𝑥 with the

X

Y

a>0


exception to the values for which the denominators is zero. The

special cases 𝑦 =1

𝑥 , 𝑎𝑛𝑑 𝑦 = −

1

𝑥.

Example 6:

If 𝑓(𝑥) =𝑥−2

𝑥2+1 , find 𝑓(0), 𝑓(−1), 𝑓 (

1

𝑥).

Solution:

𝑓(0) =0−2

0+1= −2.

𝑓(−1) =−1−2

1+1= −

−3

2.

𝑓 (1

𝑥) =

1𝑥

− 2

(1𝑥)

2

+ 1

=𝑥 − 2𝑥2

1 + 𝑥2.

Example 7:

X

1/X-y=

y

X

y= 1/X y


If 𝑓(𝑥) = 3𝑥 , 𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡

i) 𝑓(𝑥 + 2) − 𝑓(𝑥 − 1) =26

3𝑓(𝑥)

ii) 𝑓(𝑥+3)

𝑓(𝑥−1)= 𝑓(4).

Solution:

i) 𝑓(𝑥 + 2) − 𝑓(𝑥 − 1) = 2𝑥 (32 −1

3) =

26

3𝑓(𝑥)

ii)

𝑓(𝑥 + 3)

𝑓(𝑥 − 1)=

3𝑥+3

3𝑥−1= 34 = 𝑓(4).

1-2 Transcendental Functions

Function which is not algebraic is called transcendental

functions are : trigonometric and inverse trigonometric functions

, the logarithmic and exponential functions and the hyperbolic

and inverse hyperbolic functions.

a- Trigonometric Functions:

𝑐𝑜𝑠2𝜃 + 𝑠𝑖𝑛2𝜃 = 1 Equation of a circle 𝑥2 + 𝑦2 = 1

sin 𝜃 =𝑂

ℎ , csc 𝜃 =

ℎ

𝑂=

1

sin 𝜃

cos 𝜃 =𝐴

ℎ , sec 𝜃 =

ℎ

𝐴=

1

cos 𝜃

tan 𝜃 =𝑂

𝐴 , cot 𝜃 =

𝐴

𝑂=

1

tan 𝜃

sin(𝑥 ± 𝑦) = sin 𝑥 cos 𝑦 ± cos 𝑥 sin 𝑦

𝜃

ℎ 𝑂

𝐴


cos(𝑥 ± 𝑦) = cos 𝑥 cos 𝑦 ∓ sin 𝑥 sin 𝑦

x

x

xxxxx

xxxxx

bababa

bababaxx

bababaxx

2

2

222

2

22

22

sin21

1cos2

]2cos1[2

1sinsincos2cos

]2cos1[2

1coscossin22sin

])(sin)(sin[2

1cossin

])(cos)(cos[2

1coscos1csccot

])(cos)(cos[2

1sinsin1sectan

−=

−=

−=−=

+==

−++=

++−=−=

+−−=−=

b- The Inverse Trigonometric Functions:

The inverse sine function is the function that assigns to each number

𝑥 in [−1,1] the unique number 𝑦 in [−𝜋

2,

𝜋

2]

Such that 𝑥 = sin 𝑦 → 𝑤𝑒 𝑐𝑎𝑛 𝑤𝑟𝑖𝑡𝑒 𝑦 = 𝑎𝑟𝑐 𝑠𝑖𝑛𝑥 = sin−1 𝑥

The domain of sin−1 𝑥 is [−1,1] and the range is [−𝜋

2,

𝜋

2].

Note: sin−1 𝑥 ≠1

sin 𝑥

The inverse cosine function is the function that assigns to each

number 𝑥 in [−1,1] the unique number 𝑦 in [0, 𝜋]

Such that 𝑥 = cos 𝑦 → 𝑤𝑒 𝑐𝑎𝑛 𝑤𝑟𝑖𝑡𝑒 𝑦 = 𝑎𝑟𝑐 𝑐𝑜𝑠𝑥 = cos−1 𝑥

The domain of cos−1 𝑥 is [−1,1] and the range is [0, 𝜋].


y=ax

1

X

y

1

y

√1 + 𝑥2

x

The inverse tangent function is the function that assigns to each

number 𝑥 the unique number 𝑦 in [−𝜋

2,

𝜋

2]

Such that 𝑥 = tan 𝑦 → 𝑤𝑒 𝑐𝑎𝑛 𝑤𝑟𝑖𝑡𝑒 𝑦 = 𝑎𝑟𝑐 𝑡𝑎𝑛𝑥 = tan−1 𝑥

The domain of tan−1 𝑥 is [−∞, ∞] and the range is [−𝜋

2,

𝜋

2].

sec−1 𝑥 = cos−1 1

𝑥 , csc−1 𝑥 = sin−1 1

𝑥.

Example 8:

Show that

sin−1𝑥

√1 + 𝑥2= tan−1 𝑥

Solution:

Let sin−1 𝑥

√1+𝑥2= 𝑦 Then

𝑥

√1+𝑥2= sin 𝑦

From figure we obtain tan 𝑦 =𝑥

1

Then tan−1 𝑥 = 𝑦 = sin−1 𝑥

√1+𝑥2.

c- The Exponential Function 𝒂𝒙:

If 𝑎 is a positive number, then an exponential function of the form

𝑦 = 𝑓(𝑥) = 𝑎𝑥

Where 𝑥 can be any real number.

The function 𝑓(𝑥) is a single – valued function

For all values of 𝑥.


y=logax

1

y

x

Properties of Exponential Functions 𝒂𝒙

1- 𝑎𝑥 > 0

2- 𝑎−𝑥 =1

𝑎𝑥

3- 𝑎𝑥+𝑦 = 𝑎𝑥𝑎𝑦

4- 𝑎𝑥−𝑦 =𝑎𝑥

𝑎𝑦

5- 𝑎0 = 𝑎𝑥−𝑥 =𝑎𝑥

𝑎𝑥 = 1

6- (𝑎𝑥)𝑦 = 𝑎𝑥𝑦

7- If 𝑎 > 1, then 𝑎𝑥 is an increasing function in any interval

8- If 0 < 𝑎 < 1 , then 𝑎𝑥 is a decreasing function in any interval

9- 𝑎𝑥 is continuous for every 𝑥.

d- Logarithmic Function to base 𝒂:

The function 𝑎𝑥 has an inverse if 𝑎 > 0 and 𝑎 ≠ 1.

The inverse of 𝑎𝑥 is called the logarithmic function to the base 𝑎

𝐼𝑓 𝑥 = 𝑎𝑦 → 𝑡ℎ𝑒𝑛 𝑦 = log𝑎 𝑥.

Properties of the Logarithmic Function

Let > 0, 𝑎 ≠ 1 , and let 𝑥, 𝑦 be positive, then the logarithmic

function has the following properties:

1- log𝑎(𝑥𝑦) = log𝑎 𝑥 + log𝑎 𝑦


2- log𝑎 (𝑥

𝑦) = log𝑎(𝑥) − log𝑎(𝑦)

3- log𝑎(1) = 0

4- log𝑎 (1

𝑥) = − log𝑎(𝑥)

5- log𝑎(𝑥 )𝑦 = 𝑦 log𝑎(𝑥) here 𝑦 can be any real number

6- log𝑎(𝑏) =1

log𝑏(𝑎) , 𝑏 > 0, 𝑏 ≠ 1

7- If 𝑎 > 1 , log𝑎(𝑥) is an increasing function of 𝑥 𝑓𝑜𝑟 𝑥 > 0

8- If 0 < 𝑎 < 1, log𝑎(𝑥) is a decreasing function of 𝑥 𝑓𝑜𝑟 𝑥 > 0

9- 𝑓(𝑥) = log𝑎 𝑥 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑥 > 0

The graph of 𝑦, 𝑦 = log𝑎(𝑥) is the reflection of the graph

𝑦 = 𝑎𝑥 about the line 𝑦 = 𝑥.

Note: Four Basic properties of Logarithms

1- 𝑦 = log𝑎 𝑥 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑎𝑡 𝑥 = 𝑎𝑦

2- 𝑥 = 𝑎𝑦 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑎𝑡 𝑦 = log𝑎 𝑥

3- log𝑎 𝑎𝑥 = 𝑥 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑥

4- 𝑎log𝑎 𝑥 = 𝑥 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑥

We discussed logarithms to any positive base, there are two bases

that are commonly used. Logarithms to base 10 are called common

logarithms, and the other base is called the natural logarithms. These

logarithms defined by using the number " 𝑒 ".


Definition :

The number 𝑒 is defined by

𝑒 = lim𝑛→∞

(1 +1

𝑛)

𝑛

= lim𝑘→0

(1 + 𝑘)1𝑘

= 1 + 1 +1

2!+

1

3!+ ⋯ +

1

𝑛!+ ⋯ = 2.71828.

Logarithmic function to the base 𝒆

Logarithms function to the base 𝑒 are called natural logarithms and

have a special notation

ln(𝑥) = log𝑒 𝑥.

Properties of the function 𝒇(𝒙) = 𝒍𝒏 𝒙

1- ln(𝑥𝑦) = ln 𝑥 + ln 𝑦

2- ln𝑥

𝑦= ln 𝑥 − ln 𝑦

3- ln 1 = 0

4- ln 𝑒 = 1

5- ln1

𝑥= − ln 𝑥

6- ln 𝑥𝑦 = 𝑦 ln 𝑥 here 𝑦 can be any real number

7- ln 𝑒𝑥 = 𝑥

8- 𝑒ln 𝑥 = 𝑥


9- ln 𝑥 = ln 𝑦 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥 = 𝑦

10- ln 𝑥𝑛 = 𝑛 ln 𝑥.

Relation between 𝐥𝐧 𝒙 𝒂𝒏𝒅 𝐥𝐨𝐠𝒂 𝒙

There exists a linear relation between ln 𝑥 𝑎𝑛𝑑 log𝑎 𝑥

log𝑎 𝑥 =ln 𝑥

ln 𝑎.

e- The exponential function 𝒆𝒙:

It is defined to be the inverse function of 𝑙𝑛 (𝑥)

𝑦 = 𝑒𝑥 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥 = ln 𝑦

Since the range of ln 𝑥 is R, then the domain of 𝑒𝑥 is 𝑅.

Properties of Exponential Functions 𝒆𝒙

1 − 𝑒𝑥 > 0

2 − 𝑒−𝑥 =1

𝑒𝑥

3 − 𝑒𝑥+𝑦 = 𝑒𝑥𝑒𝑦

4 − 𝑒𝑥−𝑦 =𝑒𝑥

𝑒𝑦

5 − 𝑒0 = 𝑒𝑥−𝑥 =𝑒𝑥

𝑒𝑥 = 1

6 − (𝑒𝑥)𝑦 = 𝑒𝑥𝑦

7- lim𝑥→∞

𝑒𝑥 = ∞ 𝑎𝑛𝑑 lim𝑥→−∞

𝑒𝑥 = 0


8- ln 𝑒𝑥 = 𝑥 , 𝑒ln 𝑥 = 𝑥 𝑖𝑓 𝑥 > 0

9- 𝑒𝑥 = 1 + 𝑥 +𝑥2

2!+

𝑥3

3!+ ⋯ … … …. is the expansion of 𝑒𝑥

about origin. This is the expansion of the function 𝑓(𝑥) = 𝑒𝑥 about

the origin 𝑥 = 0.

Example 9:

Find the inverse function for the following:

i) 𝑦 = log𝑥

3

ii) 𝑦 = tan−1 4𝑥.

Solution:

𝑖) 𝑦 = log𝑥

3→

𝑥

3= 10𝑦

𝑦 = 3. 10𝑥 , − ∞ < 𝑥 < ∞

𝑖𝑖) 𝑦 = tan−1 4𝑥 → 4𝑥 = tan 𝑦 𝑦 =1

4tan 𝑥 , −

𝜋

2< 𝑥 <

𝜋

2.

Example 10:

Which of the following functions is an odd function

i) 𝑓(𝑥) = log3+𝑥

3−𝑥

ii) 𝑓(𝑥) = log(𝑥 + √1 + 𝑥2).

Solution:


(𝑖)𝑓(𝑥) = log3+𝑥

3−𝑥, replace 𝑥 𝑏𝑦 – 𝑥 , we get

𝑓(−𝑥) = log3 − 𝑥

3 + 𝑥= − log

3 + 𝑥

3 − 𝑥= −𝑓(𝑥)

Then, the function is odd

𝑖𝑖) 𝑓(𝑥) = log(𝑥 + √1 + 𝑥2) ,replace 𝑥 𝑏𝑦 – 𝑥,

we get

𝑓(−𝑥) = [log( (−𝑥 + √1 + 𝑥2) .𝑥 + √1 + 𝑥2

𝑥 + √1 + 𝑥2)]

= log (1+𝑥2)−𝑥2

𝑥+√1+𝑥2

= log

1

x + √1 + x2

= − log (𝑥 + √1 + 𝑥2) = −𝑓(𝑥)

Therefore the function is odd function.

f- The Hyperbolic Functions:

The hyperbolic 𝑠𝑖𝑛𝑒 of 𝑥 (written sinh 𝑥 ) and the hyperbolic

𝑐𝑜𝑠𝑖𝑛𝑒 of 𝑥 (written cosh 𝑥 ) are defined by

sinh 𝑥 =𝑒𝑥−𝑒−𝑥

2 , cosh 𝑥 =

𝑒𝑥+𝑒−𝑥

2.


Properties of Hyperbolic Function

𝑐𝑜𝑠ℎ𝑥 − 𝑠𝑖𝑛ℎ𝑥 = 𝑒−𝑥 > 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ,

𝑠𝑜 𝑐𝑜𝑠ℎ𝑥 > 𝑠𝑖𝑛ℎ𝑥

𝑠𝑖𝑛ℎ(−𝑥) = −𝑠ℎ𝑖𝑛(𝑥) 𝑎𝑛𝑑 𝑐𝑜𝑠ℎ(−𝑥) = 𝑐𝑜𝑠ℎ(𝑥)

tanh x =sinh 𝑥

cosh 𝑥 =

𝑒𝑥 − 𝑒−𝑥

𝑒𝑥 + 𝑒−𝑥

coth x =cosh 𝑥

sinh 𝑥 =



sech 𝑥 =1

cosh 𝑥=

2

𝑒𝑥 + 𝑒−𝑥 ,

cosch 𝑥 =1

sinh 𝑥=

2


𝑐𝑜𝑠ℎ𝑥 + 𝑠𝑖𝑛ℎ𝑥 = 𝑒𝑥

𝑐𝑜𝑠ℎ2𝑥 − 𝑠𝑖𝑛ℎ2𝑥 = 1

1 − 𝑡𝑎𝑛ℎ2𝑥 = 𝑠𝑒𝑐ℎ2𝑥

𝑐𝑜𝑡ℎ2𝑥 − 1 = 𝑐𝑜𝑠𝑒𝑐ℎ2𝑥

cosh(𝑥 ± 𝑦) = 𝑐𝑜𝑠ℎ𝑥 𝑐𝑜𝑠ℎ𝑦 ± 𝑠𝑖𝑛ℎ𝑥 𝑠𝑖𝑛ℎ𝑦

sinh( 𝑥 ± 𝑦) = sinh 𝑥 cosh 𝑦 ± cosh 𝑥 sinh 𝑦

𝑐𝑜𝑠ℎ2𝑥 + 𝑠𝑖𝑛ℎ2𝑥 = cosh 2𝑥

sinh 2𝑥 = 2 sinh 𝑥 cosh 𝑥

cosh 𝑥 ≥ 1 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 𝑎𝑛𝑑 cosh 0 = 1 𝑎𝑛𝑑 sinh 0 = 0.


l- Inverse Hyperbolic Function:

The inverse Hyperbolic Sine: sinh−1 𝑥

𝑦 = sinh 𝑥 → 𝑥 = sinh−1 𝑦

𝑦 = cosh 𝑥 → 𝑥 = cosh−1 𝑦 , 𝑦 = tanh−1 𝑥 → 𝑦 = tanh 𝑥 .

Example 11:

Solve the equation

𝐬𝐡𝐢𝐧 𝒙 − 𝟐 𝐜𝐨𝐬𝐡 𝒙 + 𝟐 = 𝟎

Solution:

From the definition of 𝑐𝑜𝑠ℎ𝑥 𝑎𝑛𝑑 𝑠𝑖𝑛ℎ𝑥, we have

(𝑒𝑥 − 𝑒−𝑥) − 2(𝑒𝑥 + 𝑒−𝑥) + 4 = 0 𝑜𝑟 𝑒𝑥 + 3𝑒−𝑥 − 4 = 0

Multiplying by 𝑒𝑥 , we get

𝑒2𝑥 − 4𝑒 𝑥 + 3 = 0

This is a quadratic equation in 𝑒𝑥 , the solution is 𝑒𝑥 = 1, 3

Therefore the solution of the equation is 𝑥 = ln 3.

Example 12:

Solve the equation cosh(ln 𝑥) = 2 sinh(ln 𝑥) − 1.

Solution:

Using basic definitions, we have

1

2(𝑒ln 𝑥 + 𝑒− ln 𝑥) = 𝑒ln 𝑥 − 𝑒− ln 𝑥 − 1


But 𝑒ln 𝑥 = 𝑥, 𝑎𝑛𝑑 𝑒− ln 𝑥 =1

𝑥

Then 1

2(𝑥 +

1

𝑥) = 𝑥 −

1

𝑥− 1

Rearranging, we get the quadratic equation

𝑥2 − 2𝑥 − 3 = 0

The solution is 𝑥 = −1 𝑟𝑒𝑓𝑢𝑠𝑒, 𝑥 = 3.

Then the solution is 𝑥 = 3.


Chapter -2

Differentiation

2-1 Introduction

Let 𝑦 = 𝑓(𝑥) be a function defined in a certain interval. The

function 𝑦 = 𝑓(𝑥) has a definite value of the independent variable 𝑥

in this interval. If 𝑥 is incremented by ∆𝑥 (this ∆𝑥 can be positive or

negative, it doesn't matter), then the function 𝑦 will be incremented

corresponding by ∆𝑦 , then we may write 𝑦 + ∆𝑦 = 𝑓(𝑥 + ∆𝑥). Now,

the increment in y will be ∆𝑦 = 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)

We say that a change of ∆𝑥 in the independent variable 𝑥 will produce

a change of ∆𝑦 in the dependent variable 𝑦.

The ratio of Δ𝑦

Δ𝑥 is called the rate of change of 𝑦 with respect to 𝑥, and

is given by: Δ𝑦

Δ𝑥=

𝑓(𝑥+∆𝑥)−𝑓(𝑥)

∆𝑥.

If ∆𝑥 becomes smaller and smaller and approaches to zero ( we write

∆𝑥 → 0) 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 will called the derivative of

function 𝑓(𝑥) with respect to the independent variable 𝑥, and is

denoted by 𝑓′(𝑥) and we have:

𝑓′(𝑥) = lim∆𝑥→0

Δ𝑦

Δ𝑥

Or 𝑓′(𝑥) = lim∆𝑥→0

𝑓(𝑥+∆𝑥)−𝑓(𝑥)

∆𝑥


This notation for the derivative is not the only one used. Alternative

symbols 𝑑𝑦

𝑑𝑥 ,

𝑑

𝑑𝑥𝑓(𝑥), 𝐷𝑓(𝑥), 𝑦′.

Example 1:

Using definition find the derivative of the function

𝑦 = 4𝑥2 − 5.

Solution:

𝑓(𝑥) = 𝑦 = 4𝑥2 − 5

𝑓(𝑥 + ∆𝑥) = 4(𝑥 + ∆𝑥)2 − 5

𝑓′(𝑥) =𝑑𝑦

𝑑𝑥= lim

∆𝑥→0

𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)

∆𝑥


𝑑𝑥= lim

∆𝑥→0

[4(𝑥 + ∆𝑥)2 − 5] − [2𝑥2 − 5]

∆𝑥

= lim∆𝑥→0

4𝑥2 + 8𝑥∆𝑥 + 4(∆𝑥)2 − 5 − 4𝑥2 + 5

∆𝑥

= lim∆𝑥→0

[8𝑥 + 4∆𝑥] = 8𝑥.

Example 2:

Using definition find the derivative of the function

𝑦 =−5

𝑥.

Solution:

𝑓(𝑥) = 𝑦 =−5

𝑥

𝑓(𝑥 + ∆𝑥) =−5

𝑥 + ∆𝑥



𝑑𝑥= lim

∆𝑥→0

𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)

∆𝑥


𝑑𝑥= lim

∆𝑥→0

−5𝑥 + ∆𝑥

−−5𝑥

∆𝑥

= lim∆𝑥→0

−5𝑥 + 5(𝑥 + ∆𝑥)𝑥(𝑥 + ∆𝑥)

∆𝑥

= lim∆𝑥→0

5∆𝑥𝑥(𝑥 + ∆𝑥)

∆𝑥

= lim∆𝑥→0

5

𝑥(𝑥 + ∆𝑥)=

5

𝑥2.

2-2 -1 General Derivative Rules

(power Rule)

1) 𝑑

𝑑𝑥 𝑐 = 0 , 𝑐 is a constant.

2) 𝑑

𝑑𝑥𝑥𝑛 = 𝑛𝑥𝑛−1 , for any real number 𝑛.

𝑰𝒇 𝒇(𝒙)𝒂𝒏𝒅 𝒈(𝒙) 𝒂𝒓𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒃𝒍𝒆 𝒂𝒕 𝒙 𝒂𝒏𝒅 𝒄 𝒊𝒔 𝒂𝒏𝒚 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕, 𝒕𝒉𝒆𝒏

3)𝑑

𝑑𝑥[𝑓(𝑥) ± 𝑔(𝑥)] = 𝑓′(𝑥) ± 𝑔′(𝑥),

4)𝑑

𝑑𝑥[𝑐 𝑓(𝑥)] = 𝑐 𝑓′(𝑥),

5)𝑑

𝑑𝑥[𝑓(𝑥). 𝑔(𝑥)] = 𝑓′(𝑥). 𝑔(𝑥) + 𝑔′(𝑥). 𝑓(𝑥), (𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑅𝑢𝑙𝑒)

6)𝑑

𝑑𝑥[𝑓(𝑥)

𝑔(𝑥)] =

𝑓′(𝑥). 𝑔(𝑥) − 𝑔′(𝑥). 𝑓(𝑥)

[𝑔(𝑥)]2, 𝑔(𝑥) ≠ 0(𝑄𝑢𝑜𝑡𝑖𝑒𝑛𝑡 𝑅𝑢𝑙𝑒)


7)𝑑

𝑑𝑥[(𝑓(𝑥))

𝑛] = 𝑛 (𝑓(𝑥))

𝑛−1. 𝑓′(𝑥), for any real number 𝑛.

8)𝑇ℎ𝑒 𝑐ℎ𝑎𝑖𝑛 𝑟𝑢𝑙𝑒 𝑒𝑛𝑎𝑏𝑙𝑒𝑠 𝑢𝑠 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 𝑜𝑓 𝑚𝑎𝑛𝑦 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒

𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑓(𝑔(𝑥)).

𝐼𝑓 𝑦 = 𝑓(𝑢) 𝑎𝑛𝑑 𝑢 = 𝑔(𝑥), 𝑑𝑒𝑓𝑖𝑛𝑒 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

𝑦 = 𝑚(𝑥) = 𝑓[𝑔(𝑥)], 𝑡ℎ𝑒𝑛 𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑢.𝑑𝑢

𝑑𝑥, 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑

𝑑𝑦

𝑑𝑢 𝑎𝑛𝑑

𝑑𝑢

𝑑𝑥 𝑒𝑥𝑖𝑠𝑡

𝑜𝑟 𝑚′(𝑥) = 𝑓′[𝑔(𝑥)]. 𝑔′(𝑥), 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑡ℎ𝑎𝑡 𝑓′[𝑔(𝑥)]𝑎𝑛𝑑 𝑔′(𝑥)𝑒𝑥𝑖𝑠𝑡.

Example 3:

Find the derivative of the following functions:

(a)𝑦 = (3√𝑥 − 𝑥5 + 6𝑥3)15

(𝑏)𝑦 = √2 − 4𝑥 + 6𝑥54

(𝑐)𝑦 =1

(5 − 4𝑥2)20

Solution:

(𝑎)𝑦′ = 15(3√𝑥 − 𝑥5 + 6𝑥3)14

(3

2 𝑥

−12 − 5𝑥4 + 18𝑥2)

(b)𝑦′ =1

4(2 − 4𝑥 + 6𝑥5)

−34 (−4 + 30𝑥4)


(𝑐)𝑦′ = −20(5 − 4𝑥2)−21(−8𝑥)

2-2-2 Differentiation of the Logarithmic function

𝒅

𝒅𝒙𝐥𝐨𝐠𝒂 𝒙 =

𝟏

𝒙𝐥𝐨𝐠𝒂 𝒆 =

𝟏

𝒙.

𝟏

𝐥𝐧 𝒂

Proof:

Let

𝑦 = log𝑎 𝑥 , then from first principles, we have

∆𝑦 = log𝑎(𝑥 + ∆𝑥) − log𝑎 𝑥 = log𝑎 𝑥+∆𝑥

𝑥= log𝑎(1 +

∆𝑥

𝑥)

Hence

Δ𝑦

Δ𝑥=

1

∆𝑥log𝑎 (1 +

∆𝑥

𝑥) =

1

𝑥

𝑥

∆𝑥log𝑎 (1 +

∆𝑥

𝑥)

=1

𝑥log𝑎 (1 +

∆𝑥

𝑥)

𝑥∆𝑥

Let ∆𝑥

𝑥= 𝛼 , then as ∆𝑥 → 0 , 𝑡ℎ𝑒𝑛 𝛼 → 0 and

𝑑𝑦

𝑑𝑥= lim

∆𝑥→0

Δ𝑦

Δ𝑥= lim

𝛼→0

1

𝑥log𝑎(1 + 𝛼)

1𝛼 =

1

𝑥log𝑎 𝑒 =

1

𝑥 ln 𝑎

And if 𝒖 is a function of 𝑥 , we have

𝒅

𝒅𝒙𝐥𝐨𝐠𝒂 𝒖 =

𝟏

𝒖 𝒅𝒖

𝒅𝒙𝐥𝐨𝐠𝒂 𝒆 =

𝟏

𝒖 𝐥𝐧 𝒂.𝒅𝒖

𝒅𝒙

If 𝑎 replaced by 𝑒, then we have the special case

𝒅

𝒅𝒙𝐥𝐧 𝒙 =

𝟏

𝒙. 𝒂𝒏𝒅

𝒅

𝒅𝒙𝐥𝐧 𝒖 =

𝟏

𝒖 .𝒅𝒖

𝒅𝒙.


2-2-3 Differentiation of Exponential functions

𝒅

𝒅𝒙𝒂𝒙 = 𝒂𝒙𝐥𝐧 𝐚.

Proof:

Let 𝑦 = 𝑎𝑥 , then 𝑥 = log𝑎 𝑦

Then 𝑑𝑥

𝑑𝑦=

1

𝑦log𝑎 𝑒 then

dy

dx= 𝑦 log𝑒 𝑎 = 𝑎𝑥 ln 𝑎.

And if 𝒖 is a function of 𝑥 , we have

𝑑

𝑑𝑥𝑎𝒖 = 𝑎𝒖

d𝒖

dx ln a.

Now If 𝑎 is replaced by 𝑒, then we have the special case

𝒅

𝒅𝒙𝒆𝒙 = 𝒆𝒙.

Examples 4:

Find 𝒅𝒚

𝒅𝒙 :

(a)y = 5𝑥3

(𝑏)𝑦 = 𝑒𝑥2

(𝑐)𝑦 = 𝑙𝑜𝑔2(3𝑥4)

(𝑑)𝑦 = 𝑙𝑛𝑥5

(e)y = ln√3 − 2𝑥2

(𝑓)𝑦 = 3𝑥2. √2𝑥 − 𝑥63

. 𝑙𝑜𝑔4(5𝑥3)


(g)y =𝑥4. 𝑒2𝑥8

√2𝑥2 + 9𝑥6

(ℎ)𝑦 = 𝑙𝑛5𝑡 𝑥 = √𝑡2 + 5

(𝑖)𝑦 = 𝑢4 − 2𝑢 𝑢 = 𝑒2𝑥5

Solution:

(a)𝑦′ = 5𝑥3. 3𝑥2. 𝑙𝑛5

(𝑏)𝑦′ = 𝑒𝑥2. 2𝑥

(𝑐)𝑦′ =12𝑥3

𝑙𝑛2. 3𝑥4

(d)𝑦′ =5𝑥4

𝑥5

(𝑒)𝑦′ =

12

(3 − 2𝑥2)−12 (−4𝑥)

√3 − 2𝑥2

(𝑓)𝑦′ = 3𝑥2. 2𝑥. 𝑙𝑛3. √2𝑥 − 𝑥63

. 𝑙𝑜𝑔4(5𝑥3)

+ 3𝑥2.1

3. (2𝑥 − 𝑥6)

−23 . (2 − 6𝑥5). 𝑙𝑜𝑔4(5𝑥3)

+ 3𝑥2. √2𝑥 − 𝑥63

.15𝑥2

𝑙𝑛4. (5𝑥3).

(𝑔)𝑦′ =

√2𝑥2 + 9𝑥6

(4𝑥3. 𝑒2𝑥8+ 𝑥4. 𝑒2𝑥8

. 16𝑥7) − (𝑥4. 𝑒2𝑥8).

16

(2𝑥2 + 9𝑥)−56 . (4𝑥 + 9)

(√2𝑥2 + 9𝑥6

)2


(ℎ)𝑦 = 𝑙𝑛5𝑡 ⇒𝑑𝑦

𝑑𝑡=

5

5𝑡, 𝑥 = √𝑡2 + 5 ⇒

𝑑𝑥

𝑑𝑡=

1

2(𝑡2 + 5)

−1

2 . 2𝑡

dy

dx=

𝑑𝑦

𝑑𝑡.

𝑑𝑡

𝑑𝑥=

1

𝑡. (𝑡2 + 5)

12.

1

𝑡=

√𝑡2 + 5

𝑡2=

𝑥

𝑥2 − 5 .

(𝑖)𝑦 = 𝑢4 − 2𝑢 ⇒𝑑𝑦

𝑑𝑢= 4𝑢3 − 2,

𝑢 = 𝑒2𝑥5⇒

𝑑𝑢

𝑑𝑥= 𝑒2𝑥5

. 10𝑥4

dy

dx=

𝑑𝑦

𝑑𝑢.𝑑𝑢

𝑑𝑥= (4𝑢3 − 2). (𝑒2𝑥5

. 10𝑥4) = 𝑥4. (40𝑢4 − 20𝑢 ).


Exercise 1

Find 𝒅𝒚

𝒅𝒙 𝒊𝒇:

𝟏) 𝒚 = 𝟒𝒙𝟏𝟎 −𝟐

√𝒙𝟓𝟑 − 𝟏𝟒𝒙 − 𝟎. 𝟎𝟎𝟓

𝟐)𝒚 = 𝟐𝟎𝒆𝟐𝒙𝟕

𝟑)𝒚 = 𝒍𝒐𝒈 (𝒙𝟓. √𝟐𝒙 − 𝟓𝒙𝟗𝟔)

𝟒)𝒚 =𝟐𝟓𝒙 + 𝟑

√𝒍𝒏𝟖𝒙𝟕 +𝟐

𝒙𝟐𝟒

𝟓) 𝒚 =(𝟑𝒙𝟑 − 𝟒𝒙)𝟔

𝒙𝟗 √𝒙 + 𝒍𝒏𝒙

𝟔) 𝒚 = (𝟐𝒙𝟐 −𝟑

𝒙√𝑿)

𝟓𝟒⁄

. 𝒍𝒐𝒈𝟑(𝟖𝒙𝟐). 𝒆−𝟒𝒙𝟑


𝟕)𝒚 = 𝒍𝒐𝒈𝟖𝟗𝒙 +√𝟐𝟏𝒙𝟓

+ 𝒍𝒏 𝟖𝒙

𝒍𝒐𝒈𝟒𝒙𝟒

𝟖)𝑰𝒇 𝒚𝟐 = 𝒙𝟐(𝟑 + 𝒙), 𝒑𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕: 𝒚 (𝒅𝟐𝒚

𝒅𝒙𝟐) + (

𝒅𝒚

𝒅𝒙)𝟐 − 𝟑𝒙 = 𝟑

𝟗)𝒚 = √𝟐𝜽𝟐 + 𝟕 , 𝒙 = (𝟏 − 𝟐𝜽)𝟐, 𝒑𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕:

(𝒅𝒙

𝒅𝜽) + 𝟒 [𝟏 − 𝒚 (

𝒅𝒚

𝒅𝜽)] = 𝟎.

𝟏𝟎)𝒚 = √√𝟒𝒙𝟓 + 𝟑𝒙 − 𝒙𝟓. 𝒆√𝒙𝟐−𝟐

𝟏𝟏)𝒚 = 𝒍𝒐𝒈 (𝒆−𝟗𝒙𝟐. (𝟒−𝟑𝒙𝟒)𝟏𝟏

(𝟓𝒙+𝒙𝟒)).

𝟏𝟐)𝒚 = 𝐥𝐨𝐠(𝟖−𝟐𝒙 − √𝒙𝟐 + 𝟑)

𝟏𝟑)𝒚 = 𝒍𝒏 (𝒆𝟏𝟎𝟐𝒙. √𝟑−𝟒𝒙𝟗𝟑

𝒙𝟓 )


2-2-4 Differentiation of Trigonometric functions

1) 𝒅

𝒅𝒙𝐬𝐢𝐧 𝒙 = 𝐜𝐨𝐬 𝒙.

Proof :

Let 𝑦 = 𝑓(𝑥) = sin 𝑥 , 𝑡ℎ𝑒𝑛

∆𝑦 = 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) = sin(𝑥 + ∆𝑥) − sin 𝑥

= 2 cos(𝑥 + ∆𝑥 + 𝑥)

2sin

(𝑥 + ∆𝑥 − 𝑥)

2

= 2 cos (𝑥 +∆𝑥

2) sin

∆𝑥

2

∆𝑦

∆𝑥=

2 cos (𝑥 +∆𝑥2 ) sin

∆𝑥2

∆𝑥

𝑑𝑦

𝑑𝑥= lim

∆𝑥→0

∆𝑦

∆𝑥= lim

∆𝑥→0 cos (𝑥 +

∆𝑥

2) lim

∆𝑥2 →0

sin

∆x2

∆x2

But lim∆𝑥

2→0

sin

∆x

2∆x

2

= 1

Then

𝑑𝑦

𝑑𝑥=

𝑑

𝑑𝑥sin 𝑥 = cos 𝑥.

Moreover, if 𝑢 is a function of 𝑥, then according to the chain rule , we

have

𝒅

𝒅𝒙𝐬𝐢𝐧 𝒖 = 𝐜𝐨𝐬 𝒖

𝒅𝒖

𝒅𝒙.


Example 4:

a) 𝑑

𝑑𝑥cos(2𝑥2 + 5𝑥 + 1) = −(4𝑥 + 5) sin(2𝑥2 + 5𝑥 + 1).

b) 𝑑

𝑑𝑥 𝑠𝑖𝑛3(2𝑥2 + 5𝑥 + 1) =

𝑑

𝑑𝑥[sin(2𝑥2 + 5𝑥 + 1)]3

= 3[sin(2𝑥2 + 5𝑥 + 1) ]2[cos(2𝑥2 + 5𝑥 + 1)](4𝑥 + 5).

2) 𝒅

𝒅𝒙𝐜𝐨𝐬 𝒙 = − 𝐬𝐢𝐧 𝒙.

Moreover, if 𝒖 is a function of 𝑥, then according to the chain rule

, we have

𝒅

𝒅𝒙𝐜𝐨𝐬 𝒖 = − 𝐬𝐢𝐧 𝒖

𝒅𝒖

𝒅𝒙.

3) 𝒅

𝒅𝒙𝐭𝐚𝐧 𝒙 = 𝒔𝒆𝒄𝟐𝒙.

Proof:

𝑑

𝑑𝑥tan 𝑥 =

𝑑

𝑑𝑥(

𝑠𝑖𝑛 𝑥

𝑐𝑜𝑠 𝑥) =

𝑐𝑜𝑠 𝑥(𝑐𝑜𝑠 𝑥) − 𝑠𝑖𝑛 𝑥(−𝑠𝑖𝑛 𝑥)

𝑐𝑜𝑠2𝑥=

1

𝑐𝑜𝑠2𝑥

= 𝑠𝑒𝑐2𝑥.

Moreover, if 𝑢 is a function of 𝑥, then according to the chain rule,

we have

𝒅

𝒅𝒙𝐭𝐚𝐧 𝒖 = 𝒔𝒆𝒄𝟐𝒖

𝒅𝒖

𝒅𝒙.


4) 𝐝

𝐝𝐱𝐜𝐨𝐭 𝒖 = −𝐜𝐨𝐬𝐜𝟐𝒖

𝐝𝒖

𝐝𝐱.

5) 𝐝

𝐝𝐱𝐬𝐞𝐜 𝒖 = 𝐬𝐞𝐜 𝒖. 𝐭𝐚𝐧 𝒖

𝐝𝒖

𝐝𝐱.

Proof:

d

dxsec x =

𝑑

𝑑𝑥(

1

𝑐𝑜𝑠 𝑥) =

−1(− sin 𝑥)

𝑐𝑜𝑠2𝑥=

1

cos 𝑥 .

sin 𝑥

cos 𝑥

d

dx(sec 𝑥) = sec 𝑥 . tan 𝑥

Moreover , if 𝒖 is a function of 𝑥, then according to the chain

rule , we have

𝐝

𝐝𝐱𝐬𝐞𝐜 𝒖 = 𝐬𝐞𝐜 𝒖 . 𝐭𝐚𝐧 𝐮

𝐝𝒖

𝐝𝐱.

6) 𝐝

𝐝𝐱𝐜𝐨𝐬𝐞𝐜 𝒖 = −𝐜𝐨𝐬𝐞𝐜 𝒖 . 𝐜𝐨𝐭 𝒖

𝐝𝒖

𝐝𝐱.

Example 5:

a) 𝑑

𝑑𝑥𝑡𝑎𝑛4(2𝑥3) =

𝑑

𝑑𝑥[tan(2𝑥3)]4 =

4[tan ( 2𝑥3)]3[𝑠𝑒𝑐2(2𝑥3)]. 6𝑥2 = 24𝑥2[𝑡𝑎𝑛3(2𝑥3). 𝑠𝑒𝑐2(2𝑥3)].

b) 𝑑

𝑑𝑥𝑠𝑒𝑐3(√𝑥) = 3 sec2(√𝑥). sec(√𝑥) tan( √𝑥). (

1

2√𝑥) =

3𝑠𝑒𝑐3(√𝑥).tan( √𝑥 )

2√𝑥.


2-2-5 Differentiation of inverse Trigonometric functions

For inverse functions we have the following important theorem

Theorem:

If the function 𝑦 = 𝑓(𝑥) has an inverse of the form

𝑥 = 𝜗(𝑦)

Then 𝑑𝑦

𝑑𝑥=

1𝑑𝑥

𝑑𝑦

.

This theorem allows us to find the derivative of a function if we know

the derivative of its inverse and vice versa.

For the inverse trigonometric function, we have:

1) 𝑑

𝑑𝑥sin−1 𝑥 =

1

√1−𝑥2

2) 𝑑

𝑑𝑥cos−1 𝑥 =

−1

√1−𝑥2

3) 𝑑

𝑑𝑥tan−1 𝑥 =

1

1+𝑥2

4) 𝑑

𝑑𝑥cot−1 𝑥 =

−1

1+𝑥2

5) 𝑑

𝑑𝑥sec−1 𝑥 =

1

𝑥√𝑥2−1

6) 𝑑

𝑑𝑥cosec−1 𝑥 =

−1

𝑥√𝑥2−1.

We will prove (1) and all rules can prove by the same way

1) Let 𝑦 = sin−1 𝑥 , → 𝑥 = sin 𝑦 , 𝑑𝑥

𝑑𝑦= cos 𝑦


Hence 𝑑𝑦

𝑑𝑥=

1

cos 𝑦=

1

√1−𝑠𝑖𝑛2𝑦=

1

√1−𝑥2.

And if 𝒖 is a function of 𝑥 , then

𝒅

𝒅𝒙𝐬𝐢𝐧−𝟏 𝒖 =

𝟏

√𝟏−𝒖𝟐 𝒅𝒖

𝒅𝒙 .

Example 6:

a) 𝑑

𝑑𝑥sin−1 𝑥3 =

1

√1−𝑥6. 3𝑥2.

b) 𝑑

𝑑𝑥cos−1( 𝑠𝑖𝑛𝑥2) = −

1

√1−𝑠𝑖𝑛2𝑥2. (cos 𝑥2)(2𝑥)

= −2𝑥𝑐𝑜𝑠𝑥2

𝑐𝑜𝑠𝑥2= −2𝑥.

c) 𝑑

𝑑𝑥cot−1(𝑠𝑖𝑛2𝑥) =

−1

1+𝑠𝑖𝑛4𝑥. 2 sin 𝑥 cos 𝑥 =

− sin 2𝑥

1+𝑠𝑖𝑛4𝑥.

d) 𝑑

𝑑𝑥sec−1(𝑥2) =

1

𝑥2√𝑥4−1. 2𝑥 =

2

𝑥√𝑥4−1.

Examples 7:

a) 𝑑

𝑑𝑥log6( 2 sin 𝑥 + tan−1 𝑥) =

1

2sin 𝑥+tan−1 𝑥[2 cos 𝑥 +

1

1+𝑥2] log6 𝑒 =[2cos 𝑥+

1

1+𝑥2]

(2 sin 𝑥+tan−1 𝑥) ln 6.


b) 𝑑

𝑑𝑥ln 𝑠𝑒𝑐3 𝑥 =

𝑑

𝑑𝑥3 ln sec 𝑥 = 3

1

sec 𝑥sec 𝑥 tan 𝑥 = 3 tan 𝑥.

c) 𝑑

𝑑𝑥ln(𝑥cos−1 𝑥) =

𝑑

𝑑𝑥(cos−1 𝑥 ln 𝑥) =

cos−1 𝑥

𝑥−

ln 𝑥

√1−𝑥2.

d) 𝑑

𝑑𝑥sec−1(ln 𝑥) =

1

ln 𝑥√𝑙𝑛2𝑥−1.

1

𝑥.

(𝑒)𝑑

𝑑𝑥logtan 𝑥(2𝑥 + 5) =

𝑑

𝑑𝑥(ln (2𝑥 + 5)

ln(𝑡𝑎𝑛𝑥))

𝑑

𝑑𝑥logtan 𝑥(2𝑥 + 5)

= ln( tan 𝑥)

22𝑥 + 5

− [ln( 2𝑥 + 5)]sec2 𝑥tan 𝑥

ln2( tan 𝑥).

Examples 8:

a) 𝑑

𝑑𝑥2𝑥 sin−1(𝑥) = 2𝑥 sin−1( 𝑥)(x.

1

√1−𝑥2+ sin−1(𝑥)) ln 2.

b) 𝑑

𝑑𝑥𝑒cos−1( 𝑥2) = 𝑒cos−1( 𝑥2).

−1

√1−𝑥4. 2𝑥.

c) 𝑑

𝑑𝑥tan−1(5 − 𝑒3𝑥) =

−3

1+(5−𝑒3𝑥)2 . 𝑒3𝑥 .


2-2-6 Differentiation of hyperbolic Functions

1) 𝑑

𝑑𝑥sinh 𝑥 = cosh 𝑥

2) 𝑑

𝑑𝑥cosh 𝑥 = sinh 𝑥

3) 𝑑

𝑑𝑥tanh 𝑥 = sech2 𝑥

4) 𝑑

𝑑𝑥coth 𝑥 = −cosech2 𝑥

5) 𝑑

𝑑𝑥sech 𝑥 = − sech 𝑥 tanh 𝑥

6) 𝑑

𝑑𝑥cosech 𝑥 = − cosech 𝑥 coth 𝑥.

We prove (1) and the rest of the rules left as exercise.

𝒅

𝒅𝒙𝐬𝐢𝐧𝐡 𝒙 = 𝐜𝐨𝐬𝐡 𝒙.

Proof:

From the definition of hyperbolic function in terms of the exponential function, we have

𝑑

𝑑𝑥sinh 𝑥 =

d

𝑑𝑥(1

2(𝑒𝑥 − 𝑒−𝑥)) =

1

2(𝑒𝑥 + 𝑒−𝑥) = cosh 𝑥.

Moreover if 𝑣 is a function of , we have

1) 𝑑

𝑑𝑥sinh 𝑢 = cosh 𝑢 .

𝑑𝑢

𝑑𝑥

2) 𝑑

𝑑𝑥cosh 𝑢 = sinh 𝑢 .

𝑑𝑢

𝑑𝑥

3) 𝑑

𝑑𝑥tanh 𝑢 = sech2 𝑢 .

𝑑𝑢

𝑑𝑥


4) 𝑑

𝑑𝑥coth 𝑢 = −cosech2 𝑢 .

𝑑𝑢

𝑑𝑥

5) 𝑑

𝑑𝑥sech 𝑢 = − sech 𝑢 tanh 𝑢 .

𝑑𝑢

𝑑𝑥

6) 𝑑

𝑑𝑥cosech 𝑢 = − cosech 𝑢 coth 𝑢 .

𝑑𝑢

𝑑𝑥.

Examples 9:

a) 𝑑

𝑑𝑥sinh(𝑡𝑎𝑛3𝑥) = cosh(𝑡𝑎𝑛3𝑥). 3 tan2 𝑥. 𝑠𝑒𝑐2𝑥.

b) 𝑑

𝑑𝑥sech(ln sec 𝑥) =

− sech(ln sec 𝑥). tanh (ln sec 𝑥).sec 𝑥 tan 𝑥

sec 𝑥

= − tan 𝑥 . sech(ln sec 𝑥). tanh(ln sec 𝑥) .

2-2-7 Differential of Inverse Hyperbolic Functions

1) 𝑑

𝑑𝑥sinh−1 𝑥 =

1

√𝑥2+1

2) 𝑑

𝑑𝑥cosh−1 𝑥 =

1

√𝑥2−1, 𝑥 > 1

3) 𝑑

𝑑𝑥tanh−1 𝑥 =

1

1−𝑥2 , |𝑥| < 1

4) 𝑑

𝑑𝑥coth−1 𝑥 =

1

1−𝑥2 , |𝑥| > 1

5) 𝑑

𝑑𝑥sech−1 𝑥 = −

1

𝑥√1−𝑥2 , 0 < 𝑥 < 1

6) 𝑑

𝑑𝑥csch−1 𝑥 = −

1

𝑥√1+𝑥2 .


Proof case (1) 𝒅

𝒅𝒙𝐬𝐢𝐧𝐡−𝟏 𝒙 =

𝟏

√𝒙𝟐+𝟏.

Let 𝑦 = sinh−1 𝑥 ,

then 𝑥 = sinh 𝑦 and 𝑑𝑥

𝑑𝑦= cosh 𝑦

Then

𝑑𝑦

𝑑𝑥=

1

cosh 𝑦=

1

√𝑠𝑖𝑛ℎ2𝑥 +1=

1

√𝑥2+1.

By the same way all rules

Moreover, if 𝑢 is a function of 𝑥 , we have

1) 𝑑

𝑑𝑥sinh−1 𝑢 =

1

√𝑣2+1 𝑑𝑢

𝑑𝑥

2) 𝑑

𝑑𝑥cosh−1 𝑢 =

1

√𝑢2−1 𝑑𝑢

𝑑𝑥

3) 𝑑

𝑑𝑥tanh−1 𝑢 =

1

1−𝑢2 𝑑𝑢

𝑑𝑥

4) 𝑑

𝑑𝑥coth−1 𝑢 =

1

1−𝑢2 𝑑𝑢

𝑑𝑥

5) 𝑑

𝑑𝑥sech−1 𝑢 = −

1

𝑢√1−𝑢2 𝑑𝑢

𝑑𝑥

6) 𝑑

𝑑𝑥csch−1 𝑢 = −

1

𝑢√1+𝑢2 𝑑𝑢

𝑑𝑥.

Examples 10:

a) Find the derivative of the function

logcos 𝑥( tanh−1 𝑥).

Solution:

𝑦 = logcos 𝑥( tanh−1 𝑥) =ln(tanh−1 𝑥)

ln cos 𝑥


Then

𝑑𝑦

𝑑𝑥=

1

ln2 cos 𝑥[ln cos 𝑥 .

1

(tanh−1 𝑥).

1

1−𝑥2 + ln (tanh−1 𝑥) tan 𝑥].

(b)Find the derivative of cos−1(1−𝑥2

1+𝑥2) with respect to tan−1(2𝑥

1−𝑥2).

Solution:

Let

y = cos−1(1−𝑥2

1+𝑥2 ) , 𝑎𝑛𝑑 𝑢 = tan−1(2𝑥

1−𝑥2)

Hence it is required to find 𝑑𝑦

𝑑𝑢 From the chain rule, we have

𝑑𝑦

𝑑𝑢=

𝑑𝑦

𝑑𝑥.𝑑𝑥

𝑑𝑢

𝑑𝑦

𝑑𝑥= −

−2𝑥(1 + 𝑥2) − 2𝑥(1 − 𝑥2)

(1 + 𝑥2)2√1 − (1 − 𝑥2

1 + 𝑥2)2

= 2

1 + 𝑥2

𝑑𝑢

𝑑𝑥=

1

1+(2𝑥

1−𝑥2)2 .

2(1−𝑥2) −2𝑥(−2𝑥)

(1−𝑥2)2 =2

1+𝑥2

Thus 𝑑𝑦

𝑑𝑢= 1.


2-2-8 Implicit differentiation

If the relation between 𝑥 and 𝑦 is implicit in the form 𝑓(𝑥, 𝑦) = 0,

then we can still obtain 𝑑𝑦

𝑑𝑥 by simply differentiating the whole equation

term by term, then rearranging. In the process we treat 𝑦 as a function

of 𝑥.

Examples 11 :

a) Find 𝑑𝑦

𝑑𝑥 for the following implicit equation :

4𝑥2 + 𝑥𝑦 + 𝑦2 − 2𝑥 + 3𝑦 = 1.

Solution:

8𝑥 + 𝑥𝑑𝑦

𝑑𝑥+ 𝑦 + 2𝑦

𝑑𝑦

𝑑𝑥− 2 + 3

𝑑𝑦

𝑑𝑥= 0

Solving for 𝑑𝑦

𝑑𝑥 , we get

𝑑𝑦

𝑑𝑥= −

8𝑥 + 𝑦 − 2

𝑥 + 2𝑦 + 3.

b) Find 𝑑𝑦

𝑑𝑥 for the following implicit equation :

tan−1 (𝑥

𝑦) = 3𝑥2.

Solution:

𝑑

𝑑𝑥(tan−1 (

𝑥

𝑦) = 3𝑥2)


1

1 +𝑥2

𝑦2

.𝑦 − 𝑥𝑦′

𝑦2= 6𝑥 𝑜𝑟

𝑦 − 𝑥𝑦′

𝑦2 + 𝑥2= 6𝑥

→ 𝑦 − 𝑥𝑦′ = 6𝑥𝑦2 + 6𝑥3

Then solving for 𝑦′ . we finally get

𝑦′ =𝑦

𝑥− 6𝑥2 − 6𝑦2.

2-2-9 Logarithmic differentiation

Sometimes, taking logarithms first and then differentiating the

resultant equation help in the process of finding the derivative,

especially when the variables occur also in the indices, i.e., the so

called exponential-power functions of the form [𝑢(𝑥)]𝑣(𝑥).

This technique, called Logarithmic Differentiation, depends on finding

the derivative of logarithm of the given function; it is illustrated in the

following examples:

Example 12:

Differentiate the following function with respect to 𝑥:

𝑦 =(𝑥 + 1)3 √𝑥 − 1

(𝑥2 + 4)4 𝑒4𝑥.

Solution:

Taking the logarithm of the given function, we get

ln 𝑦 = 3 ln(𝑥 + 1) +1

2ln(𝑥 − 1) − 4 ln(𝑥2 + 4) − 4𝑥


Now, we find 𝑦′ by implicit differentiation

1

𝑦𝑦′ =

3

𝑥 + 1+

1

2(𝑥 − 1)−

8𝑥

𝑥2 + 4− 4

Substituting for 𝑦 and rearranging, we obtain

𝑦′ =(𝑥 + 1)3 √𝑥 − 1

(𝑥2 + 4)4 𝑒4𝑥[

3

𝑥 + 1+

1

2(𝑥 − 1)−

8𝑥

𝑥2 + 4− 4].

Example 13:

Find 𝑑𝑦

𝑑𝑥 for the following function

𝑦 = (tan 𝑥)𝑥2.

Solution

Taking the logarithm of the given function, we get

ln 𝑦 = 𝑥2(ln tan 𝑥 )

Differentiate both sides w.r.t. 𝑥, then

1

𝑦𝑦′ = 𝑥2

1

tan 𝑥sec2 𝑥 + 2𝑥(ln tan 𝑥)

By substitution

𝑦′ = 𝑥2(𝑡𝑎𝑛)𝑥2−1 sec2 𝑥 + (tan 𝑥)𝑥2(2𝑥 ln sin 𝑥).

2-3 Parametric Equations

Usually the equation of a curve is given in Cartesian form as 𝑦 =

𝑓(𝑥). this is only one of many representations for a curve.

The equations:

𝑥 = 𝜑(𝑡)

𝑦 = 𝜏(𝑡) (1)


Where 𝑡 assumes values that lie in some interval [𝑡1, 𝑡2], are called

the parametric equations of a curve, 𝑡 is the parameter and parametric

is the way the curve is represented by equations (1). To each value of

𝑡 corresponds values of 𝑥 𝑎𝑛𝑑 𝑦 (the functions 𝜑 𝑎𝑛𝑑 𝜏 are assumed

to be single-valued).

If one regards the value of 𝑥 𝑎𝑛𝑑 𝑦 as coordinates of a point in the

coordinate 𝒙𝒚 − 𝒑𝒍𝒂𝒏𝒆, then to each value of 𝑡 there will correspond

will describe a certain curve.

The explicit expression of the dependence of 𝑦 𝑜𝑛 𝑥 is obtained by

eliminating the parameter 𝑡 from equations (1).

A simple example of this parametric representation is the equations of

a circle with radius 𝑟 and center 𝑎 the origin

𝑥 = 𝑟 cos 𝜃 𝑎𝑛𝑑 𝑦 = 𝑟 sin 𝜃

The parameter is now 𝜃. To obtain the Cartesian form, we eliminate 𝜃

from these two equations as follows.

Dividing each equation by 𝑟 , squaring then adding , we obtain

𝑐𝑜𝑠2𝜃 + 𝑠𝑖𝑛2𝜃 =𝑥2

𝑟2+

𝑦2

𝑟2

𝑥2 + 𝑦2 = 𝑟2.

2-3-1 Parametric Differentiation

If a function is represented parametrically by

𝑥 = 𝑥(𝑡) 𝑎𝑛𝑑 𝑦 = 𝑦(𝑡)

Then


𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑡.

𝑑𝑡

𝑑𝑥 (chain rule)

Or 𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑡𝑑𝑥

𝑑𝑡

=𝑑𝑦

𝑑𝑡

𝑑𝑥

𝑑𝑡⁄

This formula permits us to find the derivative 𝑑𝑦

𝑑𝑥 without having to find

the explicit expression of 𝑦 as a function of 𝑥.

Note: the expression obtained for the first derivative is a function of

parameter 𝑡, for higher derivatives, we have

𝑑2𝑦

𝑑𝑥2=

𝑑

𝑑𝑥(

𝑑𝑦

𝑑𝑥) =

𝑑

𝑑𝑡(

𝑑𝑦

𝑑𝑥) .

𝑑𝑡

𝑑𝑥

Or

𝑑2𝑦

𝑑𝑥2=

𝑑𝑑𝑡

(𝑑𝑦𝑑𝑥

)

𝑑𝑥𝑑𝑡

Similarly 𝑑3𝑦

𝑑𝑥3 =

𝑑

𝑑𝑡(

𝑑2𝑦

𝑑𝑥2)

𝑑𝑥

𝑑𝑡

.

Example 14:

Find 𝑑𝑦

𝑑𝑥 and

𝑑2𝑦

𝑑𝑥2 for the equation given parametrically as

𝑦 = a sin 𝑡𝑥 = a cos 𝑡

} 0 ≤ 𝑡 ≤ 𝜋, 𝑎 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

Solution


𝑑𝑦

𝑑𝑥=

𝑑𝑦𝑑𝑡𝑑𝑥𝑑𝑡

=𝑎 𝑐𝑜𝑠 𝑡

−𝑎 sin 𝑡= − cot 𝑡

𝑑2𝑦

𝑑𝑥2=

𝑑𝑑𝑡

(𝑑𝑦𝑑𝑥

)

𝑑𝑥𝑑𝑡

=𝑐𝑜𝑠𝑒𝑐2𝑡

−𝑎 sin 𝑡= −

1

𝑎𝑐𝑜𝑠𝑒𝑐3𝑡.

Example 15:

Show that 𝑑𝑦

𝑑𝑥= tan 𝜃 𝑤ℎ𝑒𝑛

𝑥 = 𝑎 (cos 𝜃 + ln tan𝜃

2) 𝑎𝑛𝑑 𝑦 = 𝑎 sin 𝜃.

Solution:

𝑑𝑦

𝑑𝜃= 𝑎 cos 𝜃 𝑎𝑛𝑑

𝑑𝑥

𝑑𝜃= 𝑎 (− sin 𝜃 +

1

tan𝜃

2

𝑠𝑒𝑐2 𝜃

2.

1

2)

= 𝑎 (− sin 𝜃 +1 + 𝑡𝑎𝑛2 𝜃

2

2 tan𝜃2

)

But we know that

sin 𝜃 =2 tan

𝜃2

1 + 𝑡𝑎𝑛2 𝜃2

Then , 𝑑𝑥

𝑑𝜃= 𝑎 (− sin 𝜃 +

1

sin 𝜃) = 𝑎

𝑐𝑜𝑠2𝜃

sin 𝜃

Thus 𝑑𝑦

𝑑𝑥=

acos 𝜃

𝑎𝑐𝑜𝑠2𝜃/ sin 𝜃= tan 𝜃.


Example 16:

If 𝑥 = sin 𝑡 𝑎𝑛𝑑 𝑦 = sin 𝑛𝑡

Show that (1 − 𝑥2)𝑦′′ − 𝑥𝑦′ + 𝑛2𝑦 = 0.

Solution:

𝑦′ =


= 𝑛 cos 𝑛𝑡

cos 𝑡

𝑦′′ =

𝑑𝑑𝑡

𝑦′

𝑑𝑥𝑑𝑡

= 𝑛−𝑛 cos 𝑡 sin 𝑛𝑡 + cos 𝑛𝑡 sin 𝑡

𝑐𝑜𝑠2𝑡.

1

cos 𝑡

=−𝑛2 sin 𝑛𝑡 + sin 𝑡

𝑛 cos 𝑛𝑡cos 𝑡

𝑐𝑜𝑠2𝑡

But 𝑥 = sin 𝑡 , 𝑎𝑛𝑑 𝑦 = sin 𝑛𝑡

Then 𝑦′′ =−𝑛2𝑦+𝑥𝑦′

1−𝑠𝑖𝑛2𝑡=

−𝑛2𝑦+𝑥𝑦′

1−𝑥2

Rearranging, we get

(1 − 𝑥2)𝑦′′ − 𝑥𝑦′ + 𝑛2𝑦 = 0.

2-4 Higher Order Derivatives

The derivative 𝑓′(𝑥) is a function derived from a function 𝑦 = 𝑓(𝑥).

By differentiating the first derivative 𝑓′(𝑥), we obtain yet another

function called the second derivative. This second derivative is

denoted by 𝑓′′(𝑥). The second derivative is commonly denoted by:


𝑓′′(𝑥), 𝑦′′,𝑑2𝑦

𝑑𝑥2 , 𝐷𝑥

2(𝑦) 𝑜𝑟 𝑦2

The derivative of the second derivative is called the third derivative

and is commonly denoted by

𝑦′′′ , 𝑓′′′(𝑥) ,𝑑

𝑑𝑥(

𝑑2𝑦

𝑑𝑥2) ,

𝑑3𝑦

𝑑𝑥3 , 𝐷𝑥

3(𝑦) 𝑜𝑟 𝑦3

In general, a derivative of the nth order is denoted by

𝑦(𝑛) , 𝑓(𝑛)(𝑥) ,𝑑

𝑑𝑥(

𝑑𝑛−1𝑦

𝑑𝑥𝑛−1) ,𝑑𝑛𝑦

𝑑𝑥𝑛 , 𝐷𝑥𝑛(𝑦) 𝑜𝑟 𝑦𝑛 .

Solved Examples

Find 𝑑𝑦

𝑑𝑥 for the following:

1- 𝑦 = 3𝑥4

3 + 4√𝑥 + sec 𝑥 tan 𝑥 → 𝑦′ = 4𝑥1

3 +2

√𝑥+

sec 𝑥. 𝑠𝑒𝑐2𝑥 +tan 𝑥 . sec 𝑥. tan 𝑥

𝑦′ = 4𝑥13 +

2

√𝑥+ 𝑠𝑒𝑐3𝑥 + sec 𝑥 tan2 𝑥.

2- 𝑦 = cot(cos 𝑥2) → 𝑦′ = −𝑐𝑜𝑠𝑒𝑐2(cos 𝑥2). (−2𝑥 sin 𝑥2).

3- 4𝑦 = 𝑥3 + sin 𝑦 → 4𝑦′ = 3𝑥2 + cos 𝑦 𝑦′

𝑦′(4 − cos 𝑦) = 3𝑥2 → 𝑦′ =3𝑥2

(4 − cos 𝑦).

4- Find the slope of the tangent to the curve defined by

𝑥 = 𝑎(𝑡 − sin 𝑡) , 𝑦 = 𝑎(1 − cos 𝑡) , 0 ≤ 𝑡 ≤ 2𝜋


𝑎𝑡 𝑡 =𝜋

3.

Solution:

𝑥 = 𝑎(𝑡 − sin 𝑡) → 𝑑𝑥

𝑑𝑡= 𝑎(1 − cos 𝑡)

𝑦 = 𝑎(1 − cos 𝑡) → 𝑑𝑦

𝑑𝑡= 𝑎 sin 𝑡

𝑑𝑦

𝑑𝑥=


=𝑎 sin 𝑡

𝑎(1 − cos 𝑡)=

sin 𝑡

1 − cos 𝑡

𝑑𝑦

𝑑𝑥|

𝑡=𝜋3

=sin 𝑡

1 − cos 𝑡|

𝑡=𝜋3

= √𝟑.

5- If 𝑦 = sec 𝑥 prove that:

𝑦 (𝑑2𝑦

𝑑𝑥2) + (

𝑑𝑦

𝑑𝑥)

2

= 𝑦2(3𝑦2 − 2).

Solution

𝑦 = sec 𝑥 →𝑑𝑦

𝑑𝑥= sec 𝑥 tan 𝑥

𝑑2𝑦

𝑑𝑥2= sec 𝑥 𝑠𝑒𝑐2𝑥 + tan 𝑥 sec 𝑥 tan 𝑥

= sec 𝑥( 𝑠𝑒𝑐2𝑥 + 𝑡𝑎𝑛2𝑥)

Left hand side

𝑦 (𝑑2𝑦

𝑑𝑥2) + (

𝑑𝑦

𝑑𝑥)

2

= sec 𝑥 [sec 𝑥( 𝑠𝑒𝑐2𝑥 + 𝑡𝑎𝑛2𝑥)] + [sec 𝑥 tan 𝑥]2


= 𝑠𝑒𝑐2𝑥 ( 𝑠𝑒𝑐2𝑥 + 𝑡𝑎𝑛2𝑥) + 𝑠𝑒𝑐2𝑥 𝑡𝑎𝑛2𝑥

= 𝑠𝑒𝑐2𝑥(𝑠𝑒𝑐2𝑥 + 2𝑡𝑎𝑛2𝑥)

= 𝑠𝑒𝑐2𝑥[𝑠𝑒𝑐2𝑥 + 2(𝑠𝑒𝑐2𝑥 − 1)]

= 𝑠𝑒𝑐2𝑥(3𝑠𝑒𝑐2𝑥 − 2) = 𝑦2(3𝑦2 − 2) = 𝑅𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑 𝑠𝑖𝑑𝑒.

6 − 𝑦 = sin−1(𝑡𝑎𝑛√𝑥)

𝑦′ =1

√1 − (tan √𝑥)2

. 𝑠𝑒𝑐2√𝑥.1

2√𝑥.

7 − y = sec−1(cos 𝑥3)

𝑦′ =1

(cos 𝑥3)√(cos 𝑥3)2 − 1 . (−3𝑥2𝑠𝑖𝑛𝑥3)

𝑦′ =−3𝑥2𝑡𝑎𝑛𝑥3

√(cos2 𝑥3) − 1.


8-If 𝑦 = sin−1 𝑥 prove that

(1 − 𝑥2)𝑦′′ − 𝑥𝑦′ = 0.

Solution

𝑦 = sin−1 𝑥 → 𝑦′ =1

√1−𝑥2

𝑦′√1 − 𝑥2 = 1

𝑦′′√1 − 𝑥2 + 𝑦′ −2𝑥

2√1 − 𝑥2= 0

𝑦′′(1 − 𝑥2) − 𝑥𝑦′ = 0.


Exercise 2

Find 𝒅𝒚

𝒅𝒙 𝒊𝒇:

𝟏)𝒚 = 𝒄𝒐𝒔√𝒍𝒐𝒈𝟏𝟒𝒙 − 𝟓

𝟐)𝒚 = 𝒕𝒂𝒏𝒆𝒙𝟐(𝒄𝒐𝒔−𝟏(𝒍𝒐𝒈𝟖𝟖𝒙))

𝟑)𝒚 = 𝒄𝒐𝒔𝒆𝒄−𝟏 (𝒍𝒏𝒙𝟑. √𝟏𝟏𝒙𝟐 − 𝟐𝒙)

𝟒)𝒚 = 𝒕𝒂𝒏−𝟏(𝒍𝒐𝒈𝟓 √𝟏 − 𝒙𝟑

)

𝟓)𝒚 = 𝒔𝒆𝒄𝟑(𝒙 − 𝐬𝐢𝐧𝟓 𝒙)

𝟔) 𝒚 = 𝐜𝐨𝐭−𝟑(𝐬𝐢𝐧 𝟐𝒙𝟐)

𝟕)𝒚 = 𝒄𝒐𝒔−𝟏(𝒄𝒐𝒕𝒙𝟒)

𝟖)𝒚 = 𝒕𝒂𝒏𝟖 (𝒄𝒐𝒕𝟐𝒙

√𝟑𝟒𝒙 + 𝒙𝟑𝟓 )


𝟗)𝒚 = 𝒍𝒐𝒈𝒔𝒊𝒏𝒉𝟒𝒙(𝒔𝒆𝒄−𝟏𝟒𝒙𝟐)

𝟏𝟎)𝒚 = 𝒍𝒐𝒈√𝟏𝟖−𝟐𝒄𝒐𝒔𝒙(𝒄𝒐𝒕𝒉−𝟏𝒙𝟖)

𝟏𝟏)𝒆𝒙𝟑𝒚𝟐. 𝒄𝒐𝒔𝒉−𝟏(𝒙𝟐) + 𝟓𝒄𝒐𝒔(𝒙𝟑+𝟒𝒚𝟐)

= 𝒔𝒊𝒏(𝒍𝒏𝒙𝒚)

𝟏𝟐)𝒚 = √(𝟐𝒙 − 𝒄𝒐𝒔−𝟏𝒆𝒙𝟑

)𝟑

. (𝒔𝒊𝒏𝒉−𝟏𝟓𝒙 + 𝒍𝒏𝟑𝒙)𝟏𝟎

(𝒔𝒆𝒄𝒉−𝟏𝟑𝒙𝟒+ 𝟓)

𝟔. √𝒍𝒐𝒈𝟐𝒙 − 𝒄𝒐𝒔𝒆𝒄𝒉−𝟏𝟐𝒙

𝟗

𝟖

𝟏𝟑)𝒚 = 𝒙𝒕𝒂𝒏𝟐𝒙 + (𝒄𝒐𝒔𝒙)𝒔𝒊𝒏−𝟏𝟒𝒙

𝟏𝟒)𝒚 = √𝒙𝟑. 𝒕𝒂𝒏𝒉−𝟏𝒙𝟐 + 𝒕𝒂𝒏−𝟏𝒆𝟑𝒙

𝟏𝟓)𝒚 =𝒆(𝒄𝒐𝒔𝒉−𝟏𝒙)

√𝟏 − 𝒔𝒆𝒄𝒉−𝟏(𝒄𝒐𝒕𝒉−𝟏𝟐𝒙𝟑)𝟔

𝟏𝟔) 𝟖𝒙𝟐− √𝟑𝒙𝒚 + 𝒚𝟑 = 𝒍𝒐𝒈𝟕


𝟏𝟕) 𝒍𝒏(𝒚𝒙𝟐) = 𝒆𝒙𝒚 + 𝐭𝐚𝐧𝟐 𝒚

𝟏𝟖) 𝒕𝒂𝒏𝟑(𝒙𝒚𝟐 + 𝒚) = √𝒙𝟑

+ 𝟐𝟏

II-Find 𝒅𝟐𝒚

𝒅𝒙𝟐 if :

a) 𝒚 =𝟓

𝒙𝟑 + 𝟓√𝒙 − 𝟖𝟐𝒙. 𝐬𝐢𝐧𝐡 𝟓𝒙.

b) 𝒙𝟒 + 𝟕𝒙𝟖𝒚 − 𝒚𝟐 = 𝟑𝒙𝟐.


General Exercise on Differentiation

Find the first derivative of the following functions:

1- 𝑦 = √𝑥123+

𝑠𝑒𝑐5𝑥

𝑙𝑛2𝑥−1.

2- 𝑦 = 2𝑥. ln (𝑥 − 2𝑐𝑜𝑠ℎ−13𝑥).

3- 𝑦 = 𝑐𝑜𝑠𝑥−1 + 𝑐𝑜𝑠−1𝑥.

4- 𝑦 = cot 5𝑥4 +𝑠𝑖𝑛ℎ4𝑥

𝑥20 .

5- 𝑦 = 𝑥. cosh (2𝑥 + 8𝑦)

6- 𝑦 + 2𝑦 + 9𝑥𝑦 = 4𝑠𝑒𝑐−1𝑥.

7- 𝑦 = 𝑠𝑒𝑐3𝑡 + log(𝑐𝑜𝑠ℎ𝑡) , 𝑥 = 𝑡3 + ln (4𝑡 − 1).

8- 𝑦 = 𝑐𝑜𝑠−1𝑡 + 𝑐𝑜𝑠ℎ𝑡−1 , 𝑥 = 𝑠𝑖𝑛−12𝑡 + 4𝑡𝑎𝑛ℎ3𝑡.

9- 𝑥𝑥 + 𝑦𝑦 = 3

10-2𝑥. 𝑐𝑜𝑠𝑥 − 𝑒𝑥𝑦 = 4𝑐𝑜𝑠ℎ−1𝑦.

11- 𝑦 = 𝑐𝑜𝑡ℎ52𝑥6 . 𝑙𝑜𝑔3(4𝑥3 − 8).

12- 𝑦 = tanh−1(𝑥2 − 5) . 𝑒sin 2𝑥.

13 − 𝑦 =(√3𝑥3+𝑥)

5 . 𝑠𝑒𝑐ℎ4𝑥 .

ln (𝑐𝑜𝑡2𝑥) .𝑒cos 𝑥 .


Chapter-3

Applications of the Differential Calculus

3-1 L'Hopital Theorem

Suppose 𝑓 𝑎𝑛𝑑 𝑔 are differentiable on an open interval (𝑎, 𝑏)

containing 𝑐 , except at 𝑐 itself. If 𝑓(𝑥)/𝑔(𝑥) has the indeterminate

form 𝟎

𝟎,

∞

∞, at 𝑥 = 𝑐 and if

𝑔′(𝑥) ≠ 0 𝑓𝑜𝑟 𝑥 = 𝑐 , then

lim𝑥→𝑐

𝑓(𝑥)

𝑔(𝑥)= lim

𝑥→𝑐

𝑓′(𝑥)

𝑔′(𝑥)

Provided either

lim𝑥→𝑐

𝑓′(𝑥)

𝑔′(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑠 𝑜𝑟 lim

𝑥→𝑐

𝑓′(𝑥)

𝑔′(𝑥)= ∞

𝑖. 𝑒. , =0

𝑣𝑎𝑙𝑢𝑒=

𝑣𝑎𝑙𝑢𝑒

0=



1- ∞

∞ ,

𝟎

𝟎

Example 1:

Find the following limits:

a) lim 𝑥→0

𝑠𝑖𝑛2𝑥

𝑥=

0

0


= lim 𝑥→0

2 cos 2𝑥

1=

2(1)

1= 2.

b) lim 𝑥→0

𝑒𝑥 + 𝑒−𝑥 − 2

1 − cos 2𝑥=

0

0

= lim𝑥→0


2 sin 2𝑥=

0

0

= lim𝑥→0


4 cos 2𝑥=

2

4=

1

2.

Example 2:

lim𝑥→0

ln 𝑥

cot 𝑥=

∞

∞= lim 𝑥→0

1𝑥

−𝑐𝑜𝑠𝑒𝑐2𝑥= lim

𝑥→0

−𝑠𝑖𝑛2𝑥

𝑥=

0

0

= −2 lim𝑥→0

sin 𝑥 cos 𝑥

1= 0.

2- 𝟎. ∞ 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏𝒕𝒐 ∞

∞𝒐𝒓

𝟎

𝟎

Example 3:

lim𝑥→2

(2 − 𝑥) tan𝜋𝑥

4= 0. ∞

= lim 𝑥→2

2 − 𝑥

cot𝜋𝑥4

=0

0, = lim

𝑥→1

−1

−𝑐𝑜𝑠𝑒𝑐2 𝜋𝑥4

.𝜋4

=4

𝜋.

3- ∞ − ∞ 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏𝒕𝒐 ∞

∞ 𝒐𝒓

𝟎

𝟎


Example 4:

lim𝑥→

𝜋2

(sec 𝑥 − tan 𝑥) = ∞ − ∞

= lim𝑥→

𝜋2

(1

cos 𝑥−

sin 𝑥

cos 𝑥)

= lim𝑥→

𝜋2

1 − sin 𝑥

cos 𝑥=

0

0, lim

𝑥→𝜋2

−cos 𝑥

−sin 𝑥= 0.

= ∞ − ∞ = lim𝑥→

𝜋2

(1

cos 𝑥−

sin 𝑥

cos 𝑥)

= lim𝑥→

𝜋2

1 − sin 𝑥

cos 𝑥=

0

0, lim

𝑥→𝜋2

−cos 𝑥

−sin 𝑥= 0.

4- 𝟎𝟎 → 𝟎 𝐥𝐧 𝟎 = 𝟎. −∞

∞𝟎 → 𝟎 𝐥𝐧 ∞ = 𝟎. ∞

𝟏∞ → ∞ 𝐥𝐧 𝟏 = ∞. 𝟎

Example 5:

lim𝑥→0

𝑥𝑥 = 00 , 𝑝𝑢𝑡 𝑦 = 𝑥𝑥 ,

ln 𝑦 = 𝑥 ln 𝑥

lim𝑥→0

ln 𝑦 = lim𝑥→0

𝑥 ln 𝑥 = 0. ∞

= lim𝑥→0

ln 𝑥

1𝑥

= lim𝑥→0

1𝑥

−1𝑥2

= 0, ln 𝑦 = 0 , y = 𝑒0 = 1 .


Examples 6:

lim𝑥→

𝜋2

(tan 𝑥)cos 𝑥

Solutions:

lim𝑥→

𝜋2

(tan 𝑥)cos 𝑥 = ∞0

𝑝𝑢𝑡 𝑦 = (tan 𝑥)cos 𝑥

ln 𝑦 = ln(tan 𝑥)cos 𝑥 = (𝑐𝑜𝑠𝑥) ln(tan 𝑥) = 0. ∞

= lim𝑥→

𝜋2

𝑙𝑛(tan 𝑥)

sec 𝑥=

∞

∞

= lim𝑥→

𝜋2

𝑠𝑒𝑐2𝑥tan 𝑥

sec 𝑥 tan 𝑥= lim

𝑥→𝜋2

sec 𝑥

𝑡𝑎𝑛2𝑥=

∞

∞

= lim 𝑥→

𝜋2

1cos 𝑥

𝑠𝑖𝑛2𝑥𝑐𝑜𝑠2𝑥

= lim𝑥→

𝜋2

cos 𝑥

𝑠𝑖𝑛2𝑥=

0

1= 0

= lim 𝑥→

𝜋2

𝑦 = 𝑒0 = 1

Then lim𝑥→

𝜋

2

(tan 𝑥)cos 𝑥 = 1.

In the table below. We give a list of indeterminate forms. We stress

however, that only the forms 0

0 𝑎𝑛𝑑

∞

∞ can be evaluated directly.

In oter cases it is necessary to bring the expression into the form


0

0 𝑎𝑛𝑑

∞

∞ . This is done either by some algebraic manipulations or

taking logarithm manipulations or taking logarithms.

Indeterminate Forms Example

𝟎

𝟎

𝐥𝐢𝐦𝒙→𝟎

𝐬𝐢𝐧 𝒙

𝒙= 𝟏

∞

∞


𝐥𝐧 𝒙

𝟏

𝒙𝟐

= 𝟎

𝟎. ∞ 𝐥𝐢𝐦𝒙→𝟎

𝒙 𝐥𝐧 𝒙 = 𝟎

∞ − ∞ 𝐥𝐢𝐦𝒙→𝟎

(𝒄𝒐𝒔𝒆𝒄 𝒙 − 𝐜𝐨𝐭 𝒙) = 𝟎

𝟎𝟎 𝐥𝐢𝐦𝒙→𝟎

𝒙𝒂𝒙 = 𝟏 , 𝒂 ≠ 𝟏

∞𝟎 𝐥𝐢𝐦𝒙→∞

𝒙𝟏𝒙 = 𝟏

𝟏∞ 𝐥𝐢𝐦𝒙→𝟎

(𝟏 + 𝒙)𝟏𝒙 = 𝒆


Exercise 3

Evaluate the following limits:

𝟏)𝐥𝐢𝐦𝒙→𝟎

𝐬𝐢𝐧(𝟑 𝒙)

𝟖𝒔𝒊𝒏(𝟕𝒙)

𝟐) 𝐥𝐢𝐦𝒙→

𝝅𝟐

𝟐 𝒄𝒐𝒔𝒙

𝟐𝒙 − 𝝅

𝟑) 𝐥𝐢𝐦𝒙→𝟎

( 𝒙. 𝒄𝒐𝒕𝒙)

𝟒)𝐥𝐢𝐦𝒙→𝟎

𝒆𝟐𝒙 − 𝟏

𝒙

𝟓) 𝐥𝐢𝐦𝒙→𝟎

𝒔𝒊𝒏𝒉𝒙 − 𝒔𝒊𝒏𝒙

𝒙𝟑

𝟔) 𝐥𝐢𝐦𝒙→𝟎

𝒕𝒂𝒏𝒙 + 𝒔𝒆𝒄𝒙 − 𝟏

𝒕𝒂𝒏𝒙 − 𝒔𝒆𝒄𝒙 + 𝟏

𝟕) 𝐥𝐢𝐦𝒙→𝟎

𝒙𝒙𝟐


𝟖) 𝐥𝐢𝐦𝒙→∞

𝒆−𝒙 + 𝒆𝒙

𝒆𝒙 + 𝒙𝟐

𝟗) 𝐥𝐢𝐦𝒙→𝟎

(𝟏 + 𝟑𝒙)𝟔𝒙

𝟏𝟎) 𝐥𝐢𝐦𝒙→𝟎

𝐬𝐢𝐧 𝒙 − 𝒙

𝐭𝐚𝐧 𝒙 − 𝒙

𝟏𝟏) 𝐥𝐢𝐦𝒙→∞

𝟐𝒆𝒙

𝒙

𝟏𝟐) 𝐥𝐢𝐦𝒙→𝟎

(𝒆𝟑𝒙 − 𝟓𝒙)𝟏𝒙

𝟏𝟑)𝐥𝐢𝐦𝒙→∞

(𝟏 +𝟑

𝒙)

𝟐𝒙

𝟏𝟒)𝐥𝐢𝐦𝒙→∞

(𝐥𝐧(𝒙 + 𝟏) − 𝒍𝒏𝒙)

𝟏𝟓 ) 𝐥𝐢𝐦𝒙→𝟎

𝒙𝟐

𝟓 − 𝒆−𝒙 − 𝒆𝒙

𝟏𝟔)𝐥𝐢𝐦𝒙→𝟎

𝟐𝟎 − 𝒄𝒐𝒔𝒙

𝒙𝟐 + 𝟒𝒙

𝟏𝟕) 𝐥𝐢𝐦𝒙→

𝝅𝟐

(𝐜𝐨𝐬 𝒙)𝐜𝐨𝐬𝟐 𝒙


3-2 Power Series Representation of a Function

Numerical computations using power series provide the basis for the

design of calculators and the construction of mathematical tables.

We are going to study some of the properties of power series and

show that any function, under certain assumptions, can be expanded

into (approximated by) a power series.

We recall the binomial expansion

(1 + 𝑥)𝑛 = 1 + 𝑛𝑥 +𝑛(𝑛 − 1)

2!𝑥2 +

𝑛(𝑛 − 1)(𝑛 − 2)

3!𝑥3 + ⋯

The right hand side is called an infinite power series for 𝑛 not a positive

integer. We recall that this expansion is true only if |𝑥| < 1.

By saying the expansion is true we mean that the series on the right

hand side is convergent and that |𝑥| < 1 is the interval of

convergence.

We now say that the function on the left (in this case (1 + 𝑥)𝑛) is

expanded into an infinite power series about the origin (𝑥 = 0).

Definition: If a function 𝑓(𝑥) is defined by

𝑓(𝑥) = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + ⋯ + 𝑎𝑛𝑥𝑛 + ⋯

= ∑ 𝑎𝑛𝑥𝑛 … … … … … … … … … … … (1)

∞

𝑛=0

Where 𝑎0, 𝑎1, 𝑎2, … , 𝑎𝑛, … are constants, we say that

∑ 𝑎𝑛𝑥𝑛

∞

𝑛=0


Is a power series representation for the function 𝑓(𝑥). And the

series in 𝑥 or about 𝑥 = 0.

Note: A power series of the form

∑ 𝑎𝑛(𝑥 − 𝑏)𝑛

∞

𝑛=0

= 𝑎0 + 𝑎1(𝑥 − 𝑏) + 𝑎2(𝑥 − 𝑏)2 + ⋯ + 𝑎𝑛(𝑥 − 𝑏)𝑛

+ ⋯ … … … … … … (2)

Is called a power series in (𝑥 − 𝑏) or about 𝑥 = 𝑏.

The values of 𝑥 for which a power series converges is called its

interval of convergence. Clearly, (1) converges for 𝑥 = 0 and (2)

converges for 𝑥 = 𝑏. There may be other values of 𝑥 for which power

series (1) or (2) converges. They can converge for all values of 𝑥 or

for all values of 𝑥 on some finite interval (closed, open, and half open)

having as midpoint 𝑥 = 0 for (1) or 𝑥 = 𝑏 for (2).

Two important questions may now be asked, if a function 𝑓(𝑥) has a

power series representation of the form

𝑓(𝑥) = ∑ 𝑎𝑛𝑥𝑛

∞

𝑛=0

𝑜𝑟 𝑓(𝑥) = ∑ 𝑎𝑛(𝑥 − 𝑏)𝑛

∞

𝑛=0

What are the values of 𝑎𝑛?

Moreover, for what values of 𝑥 is this series convergent?

To answer these questions, suppose that

𝑓(𝑥) = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + ⋯ + 𝑎𝑛𝑥𝑛 + ⋯


Differentiating successively with respect to 𝑥, we get

𝑓′(𝑥) = 𝑎1 + 2𝑎2𝑥 + 3𝑎3𝑥2 + 4𝑎4𝑥3 … = ∑ 𝑛𝑎𝑛𝑥𝑛−1

∞

𝑛=1

𝑓′′ (𝑥) = 2𝑎2 + (3.2)𝑎3𝑥 + (4.3)𝑎4𝑥2 … = ∑ 𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛−2

∞

𝑛=2

𝑓′′′ (𝑥) = (3.2)𝑎3 + (4.3.2)𝑎4𝑥 + … = ∑ 𝑛(𝑛 − 1)(𝑛 − 2)𝑎𝑛𝑥𝑛−3

∞

𝑛=3

And for any positive integer 𝑘, we have

𝑓(𝑘) (𝑥) = ∑ 𝑛(𝑛 − 1)(𝑛 − 2) … (𝑛 − 𝑘 + 1)𝑎𝑛𝑥𝑛−𝑘

∞

𝑛=𝑘

Now for 𝑥 = 0, we obtain

𝑓(0) = 𝑎0, 𝑓′(0) = 𝑎1, 𝑓′′(0) = 2𝑎2, 𝑓′′′(0) = (3.2)𝑎3.

And for every positive integer 𝑘,

𝑓(𝑘) (0) = 𝑘(𝑘 − 1)(𝑘 − 2) … (1)𝑎𝑘

If we let 𝑘 = 𝑛, then

𝑓(𝑛) (0) = 𝑛! 𝑎𝑛

Solving the preceding equations for

𝑎0, 𝑎1, 𝑎2, … , 𝑎𝑛, … , we obtain

𝑎0 = 𝑓(0), 𝑎1 = 𝑓′(0), 𝑎2 = 𝑓′′(0)

2! , 𝑎3 =

𝑓′′′(0)

3! ,

And in general

𝑎𝑛 = 𝑓(𝑛)(0)

𝑛!


We have determined the constants 𝑎𝑛 𝑓𝑜𝑟 𝑛 = 0, 1, 2, …, and we

have the following:

3-2-1 Maclaurin Series for 𝒇(𝒙)

If a function 𝑓(𝑥) together with its derivatives of all orders exist at 𝑥 =

0, then it has a power series representation of the form

𝑓(𝑥) = = ∑ 𝑎𝑛𝑥𝑛

∞

𝑛=0

With 𝑎𝑛 =𝑓(𝑛)(0)

𝑛!. And

𝑓(𝑥) = 𝑓(0) + 𝑓′(0)𝑥 +𝑓′′(0)

2!𝑥2 + ⋯ +

𝑓(𝑛)(0)

𝑛!𝑥𝑛 + ⋯

For every 𝑥 in the interval of convergence of the series. And in

compact form

𝑓(𝑥) = = ∑𝑓(𝑛)(0)

𝑛!𝑥𝑛

∞

𝑛=0

.

Example 1:

Find the Maclaurin expansion for 𝑒𝑥 .

Solution: We have

𝑓(𝑥) = 𝑒𝑥 𝑓(0) = 1

𝑓′(𝑥) = 𝑒𝑥 𝑓′(0) = 1

𝑓′′(𝑥) = 𝑒𝑥 𝑓′′(0) = 1

And so on


Substituting these values in Maclaurin series. We obtain

𝑒𝑥 = 1 + 𝑥 +𝑥2

2!+

𝑥3

3!+ ⋯ +

𝑥𝑛

𝑛!+ ⋯

And in compact form

𝑒𝑥 = ∑ 𝑥𝑛

𝑛!

∞

𝑛=0

.

It can be shown that the expansion is true for all finite values of 𝑥.

Note: the statement in this example can be stated as : find the

expansion of the function 𝑒𝑥 about 𝑥 = 0 (the origin).

Example 2:

Find the Maclaurin expansion for ln(1 + 𝑥).

Solution: We have

𝑓(𝑥) = ln(1 + 𝑥) 𝑓(0) = 0

𝑓′(𝑥) =1

1 + 𝑥 𝑓′(0) = 1

𝑓′′ (𝑥) = −1

(1 + 𝑥)2 𝑓′′(0) = −1

𝑓′′′(𝑥) = 1.2

(1 + 𝑥)3 𝑓′′′(0) = 2!

And in general

𝑓(𝑛)(𝑥) = (−1)𝑛 1.2.3 … (𝑛 − 1)

(1 + 𝑥)𝑛

𝑓(𝑛)(0) = (−1)𝑛−1(𝑛 − 1)!


Substituting these values in Maclaurin series, we obtain

ln(1 + 𝑥) = 𝑥 −𝑥2

2+

𝑥3

3−

𝑥4

4+ ⋯ (−1)𝑛

𝑥𝑛+1

𝑛 + 1+ ⋯

And in compact form

ln(1 + 𝑥) = ∑(−1)𝑛 𝑥𝑛+1

𝑛 + 1

∞

𝑛=0

We state here some useful Maclaurin

𝑒𝑥 = 1 + 𝑥 +𝑥2

2!+

𝑥3

3!+ ⋯ = ∑

𝑥𝑛

𝑛!

∞

𝑛=0

𝑒−𝑥 = 1 − 𝑥 +𝑥2

2!−

𝑥3

3!+ ⋯ = ∑(−1)𝑛

𝑥𝑛

𝑛!

∞

𝑛=0

sin 𝑥 = 𝑥 −𝑥3

3!+

𝑥5

5!− ⋯ = ∑(−1)𝑛

𝑥2𝑛+1

(2𝑛 + 1)!

∞

𝑛=0

cos 𝑥 = 1 −𝑥2

2!+

𝑥4

4!− ⋯ = ∑(−1)𝑛

𝑥2𝑛

(2𝑛 )!

∞

𝑛=0

sinh 𝑥 = 𝑥 +𝑥3

3!+

𝑥5

5!+ ⋯ = ∑

𝑥2𝑛+1

(2𝑛 + 1)!

∞

𝑛=0

cosh 𝑥 = 1 +𝑥2

2!+

𝑥4

4!+ ⋯ = ∑

𝑥2𝑛

(2𝑛 )!

∞

𝑛=0

(1 + 𝑥)𝑛 = 1 + 𝑛𝑥 +𝑛(𝑛 − 1)

2!𝑥2 +

𝑛(𝑛 − 1)(𝑛 − 2)

3!𝑥3 + ⋯ , |𝑥| < 1


1

1 ± 𝑥= 1 ∓ 𝑥 + 𝑥2 ∓ 𝑥3 + 𝑥4 + ⋯ , |𝑥| < 1

ln(1 + 𝑥) = 𝑥 −𝑥2

2+

𝑥3

3−

𝑥4

4+ ⋯ (−1)𝑛

𝑥𝑛+1

𝑛 + 1+ ⋯

Note: Several other expansions can be obtained by performing

operations (differentiation, integration, addition, etc…) on one or more

of the known expansions.

Example 3:

Find the Maclaurin expansion for

𝑖) sin 𝑥 𝑖𝑖) √1 + sin 𝑥.

Solution: We have

𝑖) 𝑓(𝑥) = sin 𝑥 𝑓(0) = 0

𝑓′(𝑥) = cos 𝑥 𝑓′(0) = 1

𝑓′′ (𝑥) = − sin 𝑥 𝑓′′(0) = 0

𝑓′′′(𝑥) = − cos 𝑥 𝑓′′′(0) = −1

And so on

Substituting these values in Maclaurin series, we obtain

sin 𝑥 = 𝑥 −𝑥3

3!+

𝑥5

5!− ⋯ = ∑(−1)𝑛

𝑥2𝑛+1

(2𝑛 + 1)!.

∞

𝑛=0

It can be shown that the expansion is true for all finite values of 𝑥.


ii) If we start the process of differentiation, the derivative of this

function will be cumbersome (difficult). Instead, we know from the

trigonometric identities that

√1 + sin 𝑥 = cos𝑥

2+ sin

𝑥

2

We also know the expression of cos𝑥

2 𝑎𝑛𝑑 sin

𝑥

2.

Replacing 𝑥 𝑏𝑦 𝑥

2 𝑖𝑛 (𝑖), we get

sin𝑥

2=

𝑥

2−

𝑥3

23. 3!+

𝑥5

25. 5!− ⋯ + (−1)𝑛

𝑥2𝑛+1

22𝑛+1. (2𝑛 + 1)!+ ⋯

Differentiating, we obtain

cos𝑥

2= 1 −

𝑥2

22. 2!+

𝑥4

24. 4!− ⋯ + (−1)𝑛

𝑥2𝑛

22𝑛 . (2𝑛 )!+ ⋯

Then the required series will be the sum of these two series, hence

√1 + sin 𝑥 = cos𝑥

2+ sin

𝑥

2

= 1 +𝑥

2−

𝑥2

22. 2!−

𝑥3

23. 3!+

𝑥4

24. 4!+

𝑥5

25. 5!− ⋯

Note: some important facts are now said:

1- The series resulting from term wise differentiation of another

series has the same interval of convergence as the original

series.

2- The series resulting from term wise integration of another series


term wise has the same interval of convergence as the original

series.

3- The series resulting from adding (subtracting) two or more

series has the least interval of convergence of all the original

series.

Example 4:

Find the expression of 𝑒𝑥 cos 𝑥 𝑎𝑏𝑜𝑢𝑡 𝑥 = 0.

Solution: We know that 𝑒𝑥 = 1 + 𝑥 +𝑥2

2!+

𝑥3

3!+ ⋯

And cos 𝑥 = 1 −𝑥2

2!+

𝑥4

4!− ⋯

Then 𝑒𝑥 cos 𝑥 = (1 + 𝑥 +𝑥2

2!+

𝑥3

3!+ ⋯ ) (1 −

𝑥2

2!+

𝑥4

4!− ⋯ )

= 1 + 𝑥 −2𝑥3

3!+

22𝑥4

4!+ ⋯

Example 5: Find the expansion of ln(1 + 𝑥 + 𝑥2) 𝑎𝑏𝑜𝑢𝑡 𝑥 = 0.

Solution:

ln(1 + 𝑥 + 𝑥2) = ln1 − 𝑥3

1 − 𝑥= ln(1 − 𝑥3) − ln(1 − 𝑥)

= −𝑥3 −𝑥6

2− ⋯ − (−𝑥 −

𝑥2

2−

𝑥3

3−

𝑥4

4− ⋯ )

= 𝑥 +1

2𝑥2 −

2

3𝑥3 +

1

4𝑥4 + ⋯


3-2-2 Taylor Series for 𝒇(𝒙)

Here the expansion of 𝑓(𝑥) is found about 𝑥 = 𝑏. We also say the

resultant series is increasing power of (𝑥 − 𝑏). Assume that

𝑓(𝑥) = 𝑎0 + 𝑎1(𝑥 − 𝑏) + 𝑎2(𝑥 − 𝑏)2 + ⋯ + 𝑎𝑛(𝑥 − 𝑏)𝑛.

Differentiation successively with respect to 𝑥, we get

𝑓′(𝑥) = 𝑎1 + 2𝑎2(𝑥 − 𝑏) + 3𝑎3(𝑥 − 𝑏)2 + ⋯ = ∑ 𝑛

∞

𝑛=1

𝑎𝑛(𝑥 − 𝑏)𝑛−1

𝑓′′(𝑥) = 2𝑎2 + (3.2)𝑎3(𝑥 − 𝑏) + ⋯ = ∑ 𝑛(𝑛 − 1)

∞

𝑛=2


𝑓′′′(𝑥) = (3.2)𝑎3 + (4.3.2 )𝑎 4(𝑥 − 𝑏) + ⋯

= ∑ 𝑛(𝑛 − 1)(𝑛 − 2)

∞

𝑛=3


And for any positive integer 𝑘, we have

𝑓(𝑘)(𝑥) = ∑ 𝑛(𝑛 − 1)(𝑛 − 2) … (𝑛 − 𝑘 + 1)

∞

𝑛=2

𝑎𝑛(𝑥 − 𝑏)𝑛−𝑘

Now, for 𝑥 = 𝑏, we obtain

𝑓(𝑏) = 𝑎0, 𝑓′(𝑏) = 𝑎1, 𝑓′′(𝑏) = 2𝑎2, 𝑓′′′(𝑏) = (3.2)𝑎3.

And for every positive integer k,

𝑓(𝑘) (𝑏) = 𝑘(𝑘 − 1)(𝑘 − 2) … (1)𝑎𝑘

If we let 𝑘 = 𝑛, then

𝑓(𝑛) (𝑏) = 𝑛! 𝑎𝑛


Solving the preceding equations for

𝑎0, 𝑎1, 𝑎2, … , 𝑎𝑛, … , we obtain

𝑎0 = 𝑓(𝑏), 𝑎1 = 𝑓′(𝑏), 𝑎2 = 𝑓′′(𝑏)

2! , 𝑎3 =

𝑓′′′(𝑏)

3! ,

And in general

𝑎𝑛 = 𝑓(𝑛)(𝑏)

𝑛!

We have determined the constants 𝑎𝑛 𝑓𝑜𝑟 𝑛 = 0, 1, 2, …, and we

have the following

𝑓(𝑥) = 𝑓(𝑏) + 𝑓′(𝑏)(𝑥 − 𝑏) +𝑓′′(𝑏)

2!(𝑥 − 𝑏)2 + ⋯ +

𝑓(𝑛)(𝑏)

𝑛!(𝑥 − 𝑏)𝑛

+ ⋯

For every 𝑥 in the interval of convergence of the series. And in

compact form

𝑓(𝑥) = ∑𝑓(𝑛)(𝑏)

𝑛!(𝑥 − 𝑏)𝑛

∞

𝑛=0

This is the Taylor expansion of 𝑓(𝑥) about 𝑥 = 𝑏.

Note:

1- If 𝑏 = 0, then Taylor expansion reduces to Maclaurin

expansion.

2- If we replace 𝑥 about (𝑥 − 𝑏), we obtain another form for Taylor

expansion

𝑓(𝑥 − 𝑏) = 𝑓(𝑏) + 𝑓′(0)𝑥 +𝑓′′(𝑏)

2! 𝑥 2 + ⋯ +

𝑓(𝑛)(𝑏)

𝑛! 𝑥 𝑛 + ⋯


This form is particularly useful when we need to compute the

approximate value of a function near 𝑥 = 𝑏.

For example, to compute the approximate value of sin 310.

3-The interval of convergence is written as |𝑥 − 𝑏| < 𝑐 , where 𝑐 can

be a finite or infinite number.

Example 1:

Obtain the expansion of the following functions in powers of

(𝑥 − 3): (𝑖) 𝑒𝑥3 (𝑖𝑖) ln 𝑥.

Solution:

(i) 𝑓(𝑥) = 𝑒𝑥

2 𝑓(2) = 𝑒

𝑓′(𝑥) =1

3𝑒

𝑥3 𝑓′(3) =

1

3𝑒

𝑓′′(𝑥) =1

32𝑒

𝑥3 𝑓′′(3) =

1

32𝑒

𝑓′′′(𝑥) =1

33𝑒

𝑥3 𝑓′′′(3) =

1

33𝑒

And so on

And in general

𝑓(𝑛)(𝑥) =1

3𝑛𝑒

𝑥3 𝑓(𝑛) (3) =

1

3𝑛𝑒

Substituting these values in Taylor series. We obtain

𝑒𝑥3 = 𝑒 (1 +

1

3

(𝑥 − 3)

1!+

1

32

(𝑥 − 3)2

2!+ … +

1

3𝑛

(𝑥 − 3)𝑛

𝑛!+ ⋯ ).


(𝑖𝑖) 𝑓(𝑥) = ln 𝑥 𝑓(3) = ln 3

𝑓′(𝑥) =1

𝑥 𝑓′(3) =

1

3

𝑓′′ (𝑥) = −1

𝑥2 𝑓′′(3) = −

1

32

𝑓′′′(𝑥) = 2

𝑥3 𝑓′′′(3) =

2

33

And so on, Substituting these values in Taylor series, we obtain

ln 𝑥 = ln 3 + 1

3 (𝑥 − 3)

1!−

1

32

(𝑥 − 3)2

2!+

2

33

(𝑥 − 3)3

3!− ⋯

Example 2:

Represent 𝑓(𝑥) = 𝑠𝑖𝑛 𝑥 in a Taylor series at 𝑥 = 𝜋

3

Solution: we have

𝑓(𝑥) = 𝑠𝑖𝑛𝑥 𝑓 (𝜋

3) =

√3

2

𝑓′(𝑥) = 𝑐𝑜𝑠𝑥 𝑓` (𝜋

3) =

1

2

𝑓"(𝑥) = −𝑠𝑖𝑛𝑥 𝑓"(𝜋

3) = −

√3

2

𝑓′′′(𝑥) = −𝑐𝑜𝑠𝑥 𝑓′′′ (𝜋

3) = −

1

2

And so on, hence the Taylor series expansion about 𝜋

3

corresponding to 𝑠𝑖𝑛𝑥 is

𝑠𝑖𝑛𝑥 = √3

2 +

1

2.1 ! ( 𝑥 −

𝜋

3 ) −

√3

2.2 !(𝑥 −

𝜋

3)

2−

1

2.3!(𝑥 −

𝜋

3)

3+ ⋯.


Exercise 4

I-Find the Maclaurin series for 𝒇(𝒙) (four non zero terms)

1)𝑦 = 𝑒2𝑥. 𝑐𝑜𝑠ℎ𝑥

2)y = cosh4x

3)𝑓(𝑥) = x2 𝑒−𝑥

4)y = 𝑐𝑜𝑠22𝑥

5)𝑦 = 𝑒5𝑥

6) 𝑓(𝑥) = l n (1 + 2𝑥

1 − 2𝑥)

7)𝑓(𝑥) = 𝑥 sin 𝑥

8)𝑓(𝑥) =𝑠𝑖𝑛2𝑥

𝑥

9)𝑦 = 𝑠𝑖𝑛2𝑥 + 𝑠𝑖𝑛ℎ3𝑥


II-Find the Taylor series for 𝑓(𝑥), expanded about x=a :

1)𝑓(𝑥) = ln (1 + 𝑥) . 𝑎 = 1.

2)𝑓(𝑥) = 2𝑥 − 𝑥3 . 𝑎 = −1.

3)𝑓(𝑥) = 𝑠𝑖𝑛ℎ3𝑥. 𝑎 = 5.

4)𝑓(𝑥) = cos2 𝑥. 𝑎 = 𝜋.

5)𝑓(𝑥) = √1 − 3𝑥3

. 𝑎 =1

3.

6)𝑓(𝑥) = cos 𝑥. 𝑎 =𝜋

3.

7)𝑓(𝑥) = sin 𝑥. 𝑎 =𝜋

2.


Chapter - 4

Indefinite Integral

Integration is the opposite of Differentiation

Since 𝑑

𝑑𝑥(𝑔(𝑥)) = 𝑓(𝑥)

Antiderivative for 𝑓(𝑥) is the indefinite integral

∫ 𝑓(𝑥) 𝑑𝑥 = 𝑔(𝑥) + 𝑐.

Where 𝑐 is the constant of integration.

4-1 Rules of integration

∫ 𝑓(𝑥) 𝑑𝑥 = 𝑔(𝑥) + 𝑐.

C is the constant of integration

∫ 0 𝑑𝑥 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

∫ 𝑎𝑓(𝑥) 𝑑𝑥 = 𝑎 ∫ 𝑓(𝑥) 𝑑𝑥.

a is a constant

∫(𝑓(𝑥) ± 𝑔(𝑥)) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 ± ∫ 𝑔(𝑥) 𝑑𝑥.

∫(𝑓(𝑥). 𝑔(𝑥)) 𝑑𝑥 ≠ ∫ 𝑓(𝑥) 𝑑𝑥. ∫ 𝑔(𝑥) 𝑑𝑥.


∫𝑓(𝑥)

𝑔(𝑥)𝑑𝑥 ≠

∫ 𝑓(𝑥) 𝑑𝑥

∫ 𝑔(𝑥) 𝑑𝑥 .

4-1-1 Rules on definite integral

∫ 𝑓(𝑥)𝑏

𝑎

𝑑𝑥 =

𝑔(𝑥)

|

𝑎

𝑏

= 𝑔(𝑏) − 𝑔(𝑎).

∫ 0𝑏

𝑎

𝑑𝑥 = 0.

∫ 𝑓(𝑥)𝑏

𝑎

𝑑𝑥 = − ∫ 𝑓(𝑥)𝑎

𝑏

𝑑𝑥.

∫ 𝑓(𝑥)𝑎

−𝑎𝑑𝑥 = 2 ∫ 𝑓(𝑥)

𝑎

0𝑑𝑥 If 𝑓(𝑥) is even.

∫ 𝑓(𝑥)𝑎

−𝑎𝑑𝑥 = 0 If 𝑓(𝑥) is odd.

4-1-2 Rules of standard integrals:

1) ∫ 𝑥𝑛 𝑑𝑥 =𝑥𝑛+1

𝑛+1+ 𝑐 , 𝑛 ≠ −1.

2) ∫(𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 =(𝑎𝑥+𝑏)𝑛+1

𝑎(𝑛+1)+ 𝑐 , 𝑛 ≠ −1.

3) ∫1

𝑥 𝑑𝑥 = ln|𝑥| + 𝑐.

4) ∫1

𝑎𝑥+𝑏𝑑𝑥 =

1

𝑎ln|𝑎𝑥 + 𝑏| + 𝑐.

5) ∫ 𝑒𝑥 𝑑𝑥 = 𝑒𝑥 + 𝑐.

6) ∫ 𝑎𝑥 𝑑𝑥 =𝑎𝑥

ln 𝑎+ 𝑐.


7) ∫ 𝑒(𝑎𝑥+𝑏) 𝑑𝑥 =1

𝑎 𝑒(𝑎𝑥+𝑏) + 𝑐.

8) ∫ sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝑐.

9) ∫ sin( 𝑎𝑥 + 𝑏) 𝑑𝑥 = −1

𝑎cos(ax +𝑏) + 𝑐.

10) ∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐.

11) ∫ cos(a 𝑥 + 𝑏) 𝑑𝑥 =1

𝑎 sin(a 𝑥 + 𝑏) + 𝑐.

12) ∫ sec2 𝑥 𝑑𝑥 = tan 𝑥 + 𝑐.

13) ∫ cosec2 𝑥 𝑑𝑥 = − cot 𝑥 + 𝑐.

14) ∫ sec 𝑥. tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝑐.

15) ∫ 𝑐𝑜𝑠𝑒𝑐 𝑥. cot 𝑥 𝑑𝑥 = − cosec 𝑥 + 𝑐.

16) ∫ sec 𝑥 𝑑𝑥 = ∫sec 𝑥(sec 𝑥+tan 𝑥)

(sec 𝑥+tan 𝑥)𝑑𝑥

= 𝑙𝑛|sec 𝑥 + tan 𝑥| + 𝑐.

17) ∫ cosec 𝑥 𝑑𝑥 = ∫cosec 𝑥(cosec 𝑥−cotan 𝑥)

(cosec 𝑥−cotan 𝑥)𝑑𝑥

= 𝑙𝑛|cosec 𝑥 − cotan 𝑥| + 𝑐.

18) ∫ tan 𝑥 𝑑𝑥 = ∫𝑠𝑖𝑛𝑥

cos 𝑥𝑑𝑥 = − ln(cos 𝑥) =

= ln(sec 𝑥) + 𝑐.

19) ∫ cot 𝑥 𝑑𝑥 = ln(𝑠𝑖𝑛𝑥) + 𝑐.

20) ∫ sinh 𝑥 𝑑𝑥 = cosh 𝑥 + 𝑐.

21) ∫ sinh(a 𝑥 + 𝑏) 𝑑𝑥 =1

𝑎cosh(𝑎𝑥 + 𝑏) + 𝑐.

22) ∫ cosh 𝑥 𝑑𝑥 = sinh 𝑥 + 𝑐.


23) ∫ cosh(a 𝑥 + 𝑏) 𝑑𝑥 =1

𝑎sinh(𝑎𝑥 + 𝑏) + 𝑐.

24) ∫ sech2 𝑥 𝑑𝑥 = tanh 𝑥 + 𝑐.

25) ∫ cosech2 𝑥 𝑑𝑥 = − coth 𝑥 + 𝑐.

26) ∫ sech 𝑥 . tanh 𝑥 𝑑𝑥 = − sech 𝑥 +𝑐.

27) ∫ 𝑐𝑜𝑠𝑒𝑐ℎ 𝑥. 𝑐𝑜𝑡ℎ 𝑥 𝑑𝑥 = −𝑐𝑜𝑠𝑒𝑐ℎ 𝑥 + 𝑐.

28) ∫ tanh 𝑥 𝑑𝑥 = ln|𝑐𝑜𝑠ℎ𝑥| + 𝑐.

29) ∫ coth 𝑥 𝑑𝑥 = ln|(sinh 𝑥)| + 𝑐.

4-2 Some properties of Indefinite integral

1- (∫ 𝑓(𝑥) 𝑑𝑥)′ = (𝑔(𝑥) + 𝑐)′ = 𝑓(𝑥).

2- 𝑑(∫ 𝑓(𝑥) 𝑑𝑥) = 𝑓(𝑥) + 𝑐.

3- ∫ 𝑑𝑓(𝑥) = 𝑓(𝑥) + 𝑐.

4- ∫ 𝑓(𝑎𝑥) 𝑑𝑥 =1

𝑎𝑔(𝑎𝑥) + 𝑐.

5- ∫ 𝑓( 𝑥 + 𝑏) 𝑑𝑥 = 𝑔(𝑥 + 𝑏) + 𝑐.

6- ∫ 𝑓(𝑎𝑥 + 𝑏) 𝑑𝑥 =1

𝑎𝑔(𝑎𝑥 + 𝑏) + 𝑐.

7- ∫𝑓(𝑥)′

𝑓(𝑥)𝑑𝑥 = ∫

𝑑𝑓(𝑥)

𝑓(𝑥)= ln 𝑓(𝑥) + 𝑐.

8- ∫ 𝑓𝑛(𝑥). 𝑓′(𝑥) 𝑑𝑥 = ∫(𝑓(𝑥))𝑛

𝑑𝑓(𝑥) =(𝑓(𝑥))

𝑛+1

𝑛+1+ 𝑐 𝑛 ≠ −1.

9- ∫𝑓(𝑥)′

√𝑓(𝑥)𝑑𝑥 = 2√𝑓(𝑥) + 𝑐.


Examples 1:

Evaluate the following integrals:

a) ∫ 𝑥−4 𝑑𝑥 =𝑥−3

−3+ 𝑐.

b) ∫ √𝑥53𝑑𝑥 = ∫ 𝑥

5

3 𝑑𝑥 =3𝑥

83

8+ 𝑐.

c) ∫𝑥4+2𝑥−5

𝑥4 𝑑𝑥 = ∫(1 + 2𝑥−3 − 5𝑥−4) 𝑑𝑥

= 𝑥 + 2𝑥−2

−2+

5

3𝑥−3 + 𝑐

= 𝑥 −1

𝑥2+

5

3𝑥3+ 𝑐.

d) ∫𝑑𝑥

√𝑥−2= 2√𝑥 − 2 + 𝑐.

e) ∫(3𝑥 + 7)5 𝑑𝑥 = (3𝑥+7)6

6.

1

3+ 𝑐 =

(3𝑥+7)6

18+ 𝑐.

Examples 2:

a) ∫ 3𝑥 . 𝑒𝑥𝑑𝑥 = ∫(3𝑒)𝑥𝑑𝑥 = (3𝑒)𝑥

ln 3𝑒=

3𝑥𝑒𝑥

ln 3 +ln 𝑒=

3𝑥𝑒𝑥

ln 3 +1.

b) ∫ (√𝑎

𝑥+ √

𝑥

𝑎)

2

𝑑𝑥 = ∫ (𝑎

𝑥+

𝑥

𝑎+ 2) 𝑑𝑥 = 𝑎 ln 𝑥 +

𝑥2

2𝑎+ 2𝑥 + 𝑐.

Example 3:

Evaluate ∫1+𝑥+𝑥2+𝑥3

1−𝑥5 𝑑𝑥.

Solution:


𝐼 = ∫1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 − 𝑥4

1 − 𝑥5𝑑𝑥

= ∫1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4

1 − 𝑥5𝑑𝑥 − ∫

𝑥4

1 − 𝑥5𝑑𝑥

= ∫1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4

(1 − 𝑥)(1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4)𝑑𝑥 −

1

5∫

5 𝑥4

1 − 𝑥5𝑑𝑥

= ∫ 1

(1 − 𝑥)𝑑𝑥 −

1

5∫

5 𝑥4

1 − 𝑥5𝑑𝑥

= − ln(1 − 𝑥) +1

5ln(1 − 𝑥5) + 𝑐.

Example 4:

Evaluate the integral: ∫sin 𝑥

𝑐𝑜𝑠2𝑥𝑑𝑥.

Solution:

∫sin 𝑥

𝑐𝑜𝑠2𝑥𝑑𝑥 = ∫

1

𝑐𝑜𝑠2𝑥𝑑𝑐𝑜𝑠𝑥 = ∫ 𝑐𝑜𝑠−2𝑥 𝑑𝑐𝑜𝑠𝑥

=𝑐𝑜𝑠−1𝑥

−1+ 𝑐 =

1

𝑐𝑜𝑠 𝑥+ 𝑐 = sec 𝑥 + 𝑐.

Or

∫sin 𝑥

𝑐𝑜𝑠2𝑥𝑑𝑥 = ∫ tan 𝑥 sec 𝑥 𝑑𝑥 = sec 𝑥 + 𝑐.

Example 5:

Evaluate the integral:


∫1

𝑥(1 + 𝑙𝑛2𝑥)𝑑𝑥.

Solution:

𝐿𝑒𝑡: 𝑢 = 𝑙𝑛𝑥

∫1

𝑥(1+𝑙𝑛2𝑥)𝑑𝑥 = ∫

1

(1+𝑙𝑛2𝑥)𝑑(ln 𝑥) =

∫1

1+𝑢2 𝑑𝑢 = tan−1 𝑢 + 𝑐 = tan−1(ln 𝑥) + 𝑐.

Example 6:

Evaluate the integral

∫ 𝑥3(𝑥4 + 5)3𝑑𝑥.

Solution:

∫ 𝑥3(𝑥4 + 5)3𝑑𝑥 =1

4∫(𝑥4 + 5)3𝑑𝑥4 =

(𝑥4 + 5)4

16+ 𝑐.

Example 7:

Evaluate:

∫ 𝑥2

√4 + 𝑥3𝑑𝑥.

Solution:

∫ 𝑥2

√4 + 𝑥3𝑑𝑥 =

1

3∫

1

√4 + 𝑥3𝑑𝑥3 =

2

3√4 + 𝑥3 + 𝑐.


Example 8:

Evaluate:

∫ 1

𝑒𝑥 + 1𝑑𝑥.

Solution:

∫ 1

𝑒𝑥 + 1𝑑𝑥 = ∫

𝑒−𝑥

1 + 𝑒−𝑥𝑑𝑥 = − ln|1 + 𝑒−𝑥| + 𝑐.

By multiplying to 𝑒−𝑥 to the denominator and numerator.

Example 9:

Evaluate:

∫ 𝑥 − 1

𝑥2 − 2𝑥 + 4𝑑𝑥.

Solution:

∫ 𝑥 − 1

𝑥2 − 2𝑥 + 4𝑑𝑥 =

1

2∫

2( 𝑥 − 1)

𝑥2 − 2𝑥 + 4𝑑𝑥 =

1

2ln|𝑥2 − 2𝑥 + 4| + 𝑐.


Solved Examples

1) −3 12x dx

2)( )

−32 1x

xdx

3) 132 + xx dx

4) 123 + xx dx

CxCuduudxduxu +−=+==−= 34

34

31

)12(8

38

32

1212

( ) CxCuduuxdxduxu +−−=+−==−=−−−

22232 1

41

41

2121

( ) CxCuduudxxduxu ++=+==+= 23

323

23 19

29

23

1312

1

-1n , 1

))(()())((

1

++

=+

cn

xfdxxfxf

n

n


Let xdxxxxdxduxu 2.12

121 222 +=+= =

( ) uu −12

1 du = duuu

− 21

23

21

= .

5) )sin( x dx

Let == udxduxu sin1

du =

CxCu +−=+− )cos(1

cos1

.

6)( )

x

xcos dx

Let == udxx

duxu cos22

1du

= 2 sin u + C = ( ) Cx +sin2 .

7) xxcossin dx

(𝑖) 𝐿𝑒𝑡 𝑢 = 𝑠𝑖𝑛 𝑥

( ) ( ) CxxCuu ++−+=+− 23

225

223

25

13

115

13

15

1

CxCuuduxdxdu +=+== 22 sin

2

1

2

1cos


(𝑖𝑖) 𝐿𝑒𝑡 𝑢 = 𝑐𝑜𝑠 𝑥

(iii) Use the identity sin 2x = 2 sin x cos x and the

x 2u = substitution

and

8) )5(sec2 x dx

Let u = 5x

CxCuududxdu +=+ == )5tan(5

1tan5

1sec5

15 2

9) xtan sec2x dx

Let u = tan x

CxCuduuxdxdu +=+== 23

23

21

2 tan3

23

2sec

10) xx tansec2

dx

CxCuuduxdxdu +−=+−=−−= 22 cos

2

1

2

1sin

dxdu 2=

CxCuuduxdxxdxx +−=+−=== 2cos4

1cos

4

1sin

4

12sin

2

1cossin


(i) Let u = tan x

(ii) Let u = sec x

11) xx tansec3

dx

Let u = sec x

12)( ) ( )

2

1tan1sec

x

xx dx

13) x

x

2cos

sin dx

Let u = cos x

.

14) dxxx )4cos()4(sin3

CxCuuduxdxdu +=+== 222 tan

21

21sec

CxCuuduxdxxdu +=+== 22 sec

21

21tansec

CxCuduuxdxxdu +=+== 332 sec

31

31tansec

( ) Cx

Cuuduudxx

dux

u +−=+−=−−== 1secsectansec112

Cx

Cu

Cuduuxdxdu +=+=+=−−= −−

cos

11sin 12


Let u = sin(4x)

.

15) +1x

x dx = Let u = x + 1 and x = u – 1

du=

.

1) = dxxx )(tansecsec 22 Let u = tan x

CxCuuduxdxdu +=+== )tan(tantansecsec 22.

2) = xdx3cos Use the identity and the

substitutions u = sin x And du = cos x dx

= dx to get

.

CxCuduudxxdu +=+== )4(sin16

1

16

1

4

1)4cos(4 443

dxdu =

−=

−

−2

12

11uudu

u

u

CxxCuu ++−+=+− 21

23

21

23

)1(2)1(3

223

2

1cossin 22 =+ xx

== xdxxxdx 23 coscoscos( ) − xx 2sin1cos

−xdxcos CxxCuxduuxxdxx +−=+−=−= 3322 sin

3

1sin

3

1sinsincossin


Solved Examples-2

1) cxxdxxx

x++=

+

+35ln

35

310 2

2.

2) cxedxxe

e x

x

x

++=+

+4ln

4

43 3

3

3

3) cxdxx

x++=

+3ln

21

3

2

2 .

4) cxdxx

xdxxx

+== lnlnln

)1(

ln

1.

cxxdxxx

xx++=

+

−sincosln

sincos

sincos5)

cxfdxxf

xf+=

)(ln

)(

)(


Exercise 5


𝟏) ∫ 𝒄𝒐𝒔𝒆𝒄𝟐𝟖𝒙 𝒅𝒙

𝟐) ∫ 𝒔𝒊𝒏𝟐𝟐𝒙 𝒅𝒙

𝟑) ∫ 𝒄𝒐𝒕𝟐𝒙 𝒅𝒙

𝟒) ∫ 𝒄𝒐𝒕𝒉𝟐𝒙 𝒅𝒙

𝟓) ∫ 𝒙𝟐. 𝒔𝒊𝒏𝒉𝟓𝒙𝟑 𝒅𝒙

𝟔) ∫√𝟐−𝒄𝒐𝒕𝜽

𝒔𝒊𝒏𝟐𝜽𝒅𝜽

𝟕) ∫𝐜𝐨𝐬𝛉

(𝟐𝟓+𝒔𝒊𝒏𝜽)𝟑 𝐝𝛉

𝟖) ∫𝟏

𝐜𝐨𝐬𝟐𝟔𝐱 𝐝𝐱

𝟗) ∫ 𝒙𝟐 𝒆𝒙𝟑 𝒅𝒙


𝟏𝟎) ∫ 𝟖𝒙𝟓 𝒙𝟒 𝒅𝒙

𝟏𝟏) ∫ 𝒄𝒐𝒔𝒆𝒄𝟓𝒙 . 𝐝𝐱

𝟏𝟐) ∫ 𝐬𝐢𝐧 𝟒𝐱 𝐜𝐨𝐬𝟑 𝟒𝐱 𝐝𝐱

𝟏𝟑) ∫𝒔𝒆𝒄𝟏𝟓𝒙

𝒄𝒐𝒕𝟔𝒙 𝐝𝐱

𝟏𝟒) ∫𝒔𝒆𝒄𝟐𝜽. 𝒅𝜽

√𝟏 − 𝐭𝐚𝐧𝟐 𝐱

𝟏𝟓) ∫ 𝐭𝐚𝐧𝟐𝟓 𝐱 . 𝐝𝐱

𝟏𝟔) ∫𝟐 + 𝟑𝐱

𝟐𝟓 + 𝐱𝟐. 𝐝𝐱

𝟏𝟕) ∫𝟒𝐱

√𝐱𝟐 + 𝟒𝒙 + 𝟕𝐝𝐱

𝟏𝟖) ∫ (𝐬𝐞𝐜 𝒙 + 𝐭𝐚𝐧 𝒙)(𝟏 − 𝐬𝐢𝐧 𝒙) 𝐝𝐱


𝟏𝟗) ∫ 𝟏

𝟐√𝒙√√𝒙 + 𝟏 𝐝𝐱

𝟐𝟎) ∫ −𝟓

𝟏 − 𝟒𝒙𝟐 𝐝𝐱

𝟐𝟏) ∫ 𝟏

𝒆𝒙 + 𝟏 𝐝𝐱

𝟐𝟐) ∫ 𝒙 + 𝟖

𝒙 + 𝟑 𝐝𝐱

𝟐𝟑) ∫ 𝒄𝒐𝒔𝒉√𝒙

√𝒙 𝐝𝐱

𝟐𝟒) ∫ 𝟏

(𝒙 − 𝟐)𝟕 𝐝𝐱

𝟐𝟓) ∫ (𝒙 + √𝒙)𝟑

𝐝𝐱

𝟐𝟔) ∫ 𝒄𝒐𝒔𝒙 − 𝒔𝒊𝒏𝒙

𝒄𝒐𝒔𝒙 + 𝒔𝒊𝒏𝒙 𝐝𝐱


Trigonometric Integrals Powers of3 -4

Integrals of type ∫ 𝒔𝒊𝒏𝒎𝒙 . 𝒄𝒐𝒔𝒏𝒙. 𝒅𝒙

where m and n are nonnegative integers.

METHOD OF INTEGRATION:

(i) If m is odd, then 𝑢 = 𝑐𝑜𝑠 𝑥 ⟹ 𝑑𝑢 = −𝑠𝑖𝑛𝑥 𝑑𝑥.

Use the identity: 𝑠𝑖𝑛2𝑥 = 1 − 𝑐𝑜𝑠2𝑥 , then your integration will take

the form:∫(𝑐𝑜𝑠𝑥)𝑟(−𝑠𝑖𝑛𝑥)𝑑𝑥.

(ii) If n is odd, then 𝑢 = 𝑠𝑖𝑛𝑥 ⟹ 𝑑𝑢 = 𝑐𝑜𝑠𝑥 𝑑𝑥 .

Use the identity: 𝑐𝑜𝑠2𝑥 = 1 − 𝑠𝑖𝑛2𝑥 , then your integration will take

the form :∫(𝑠𝑖𝑛𝑥)𝑟(𝑐𝑜𝑠𝑥)𝑑𝑥.

(iii) If both m and n are even, then use the identities for the

double angle:

𝑐𝑜𝑠2𝑥 =1

2(1 + 𝑐𝑜𝑠2𝑥) 𝑠𝑖𝑛2𝑥 =

1

2(1 − 𝑐𝑜𝑠2𝑥)

𝑠𝑖𝑛𝑥. 𝑐𝑜𝑠𝑥 =1

2 𝑠𝑖𝑛2𝑥


𝐄𝐱𝐚𝐦𝐩𝐥𝐞𝟏: ∫𝒄𝒐𝒔𝟓𝒙

𝒔𝒊𝒏𝟐𝒙𝒅𝒙

𝑃𝑢𝑡 𝑢 = 𝑠𝑖𝑛𝑥 ⟹ 𝑑𝑢 = 𝑐𝑜𝑠𝑥 𝑑𝑥

𝐼 = ∫𝑐𝑜𝑠4𝑥

𝑠𝑖𝑛2𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥 = ∫

(𝑐𝑜𝑠2𝑥)2

𝑠𝑖𝑛2𝑥𝑐𝑜𝑠𝑥 𝑑𝑥


𝐼 = ∫(1 − 𝑠𝑖𝑛2𝑥 )2

𝑠𝑖𝑛2𝑥𝑐𝑜𝑠𝑥 𝑑𝑥 = ∫

(1 − 𝑢2)2

𝑢2𝑑𝑢

𝐼 = ∫ (1

𝑢2− 2 + 𝑢2) 𝑑𝑢 = 𝑢−1 − 2𝑢 +

𝑢3

3+ 𝑐

𝐼 = (𝑠𝑖𝑛𝑥)−1 − 2𝑠𝑖𝑛𝑥 +(𝑠𝑖𝑛𝑥)3

3+ 𝑐.

𝑬𝒙𝒂𝒎𝒑𝒍𝒆𝟐: ∫ 𝒔𝒊𝒏𝟑𝒙 𝒔𝒆𝒄𝒙 𝒅𝒙

𝑃𝑢𝑡 𝑢 = 𝑐𝑜𝑠𝑥 ⟹ 𝑑𝑢 = −𝑠𝑖𝑛𝑥 𝑑𝑥

𝐼 = ∫𝑠𝑖𝑛3𝑥

𝑐𝑜𝑠𝑥 𝑑𝑥 = ∫

𝑠𝑖𝑛2𝑥

𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 𝑑𝑥 = ∫

1 − 𝑐𝑜𝑠2𝑥

𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 𝑑𝑥

𝐼 = ∫1 − 𝑢2

𝑢 (−𝑑𝑢) = ∫ (𝑢 −

1

𝑢) 𝑑𝑢 =

𝑢2

2− 𝑙𝑛|𝑢| + 𝑐

𝐼 =1

2𝑐𝑜𝑠2𝑥 − 𝑙𝑛|𝑐𝑜𝑠𝑥| + 𝑐.

𝐄𝐱𝐚𝐦𝐩𝐥𝐞𝟑: ∫ 𝒔𝒊𝒏𝟓𝒙 𝒄𝒐𝒔𝟑𝒙 𝒅𝒙

𝑃𝑢𝑡 𝑢 = 𝑠𝑖𝑛𝑥 ⟹ 𝑑𝑢 = 𝑐𝑜𝑠𝑥 𝑑𝑥

𝐼 = ∫ 𝑠𝑖𝑛5𝑥 𝑐𝑜𝑠2𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥 = ∫ 𝑠𝑖𝑛5𝑥 (1 − 𝑠𝑖𝑛2𝑥 ) 𝑐𝑜𝑠𝑥 𝑑𝑥

𝐼 = ∫ 𝑢5 (1 − 𝑢2)𝑑𝑢 =𝑢6

6−

𝑢8

8+ 𝑐 =

𝑠𝑖𝑛6𝑥

6−

𝑠𝑖𝑛8𝑥

8+ 𝑐.

𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟒: ∫𝒔𝒊𝒏𝟐𝒙

𝒄𝒐𝒔𝟔𝒙 𝒅𝒙


𝑃𝑢𝑡 𝑢 = 𝑡𝑎𝑛𝑥 ⟹ 𝑑𝑢 = 𝑠𝑒𝑐2𝑥 𝑑𝑥

𝐼 = ∫𝑠𝑖𝑛2𝑥

𝑐𝑜𝑠2𝑥

1

𝑐𝑜𝑠2𝑥

1

𝑐𝑜𝑠2𝑥𝑑𝑥 = ∫ 𝑡𝑎𝑛2𝑥 (1 + 𝑡𝑎𝑛2𝑥) 𝑠𝑒𝑐2𝑥 𝑑𝑥

𝐼 = ∫ 𝑢2(1 + 𝑢2)𝑑𝑢 =𝑢3

3+

𝑢5

5+ 𝑐 =

𝑡𝑎𝑛3𝑥

3+

𝑡𝑎𝑛5𝑥

5+ 𝑐.

𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟓: ∫ 𝒔𝒊𝒏𝟐𝒙 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙

𝐼 = ∫(𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥)2 𝑐𝑜𝑠2𝑥 𝑑𝑥

𝐼 = ∫ (1

2 𝑠𝑖𝑛2𝑥)

2

(1

2(𝑐𝑜𝑠2𝑥 + 1)) 𝑑𝑥

𝐼 =1

8∫[( 𝑠𝑖𝑛2𝑥)2𝑐𝑜𝑠2𝑥 + (𝑠𝑖𝑛2𝑥)2]𝑑𝑥

𝐼 = 1

16

(𝑠𝑖𝑛2𝑥)3

3+

1

16∫(1 − 𝑐𝑜𝑠4𝑥)𝑑𝑥

𝐼 =1

16[(𝑠𝑖𝑛2𝑥)3

3+ 𝑥 −

𝑠𝑖𝑛4𝑥

4] + 𝑐.

Example 6: ∫ 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙

𝐼 = ∫ 𝑐𝑜𝑠2𝑥 𝑑𝑥 =1

2∫(1 + cos 2𝑥)𝑑𝑥 =

1

2[𝑥 +

1

2sin 2𝑥] + 𝑐.

Example 7: ∫ 𝒔𝒊𝒏𝟑𝒙 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙

𝑰 = ∫ 𝒔𝒊𝒏𝟑𝒙 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙 = − ∫(𝟏 − 𝒄𝒐𝒔𝟐𝒙)𝒄𝒐𝒔𝟒𝒙 𝒔𝒊𝒏𝒙 𝒅𝒙


= − [𝟏

𝟓𝒄𝒐𝒔𝟓𝒙 −

𝟏

𝟕𝒄𝒐𝒔𝟕𝒙] + 𝒄.

Example 8: ∫ 𝒔𝒊𝒏𝟒𝒙 𝒅𝒙

𝑰 = ∫ 𝑠𝑖𝑛4𝒙 𝒅𝒙 = ∫(𝒔𝒊𝒏𝟐𝒙 )2 𝒅𝒙 = ∫(1

2 (𝟏 − 𝒄𝒐𝒔𝟐𝒙))2 𝒅𝒙

=𝟏

𝟒∫(1 − 2𝑐𝑜𝑠2𝑥 + 𝑐𝑜𝑠22𝑥)𝑑𝑥

=𝟏

𝟒(𝑥 − 𝑠𝑖𝑛2𝑥 +

1

2(𝑥 +

1

4𝑠𝑖𝑛4𝑥))

=𝟏

𝟒𝑥 −

𝟏

𝟒𝑠𝑖𝑛2𝑥 +

1

8𝑥 +

1

32𝑠𝑖𝑛4𝑥.

Example 9: ∫𝒔𝒊𝒏𝟑(√𝒙 ) 𝒄𝒐𝒔𝟓(√𝒙)

√𝒙 𝒅𝒙

𝐼 = ∫𝑠𝑖𝑛3(√𝑥 ) 𝑐𝑜𝑠5(√𝑥)

√𝑥 𝑑𝑥 = 2 ∫ 𝑠𝑖𝑛3(√𝑥 ) 𝑐𝑜𝑠5(√𝑥 )𝑑√𝑥

𝑰 = −𝟐 ∫ 𝒔𝒊𝒏𝟐(√𝒙 ) 𝒄𝒐𝒔𝟓(√𝒙) 𝒅𝒄𝒐𝒔√𝒙

= −𝟐 ∫(𝟏 − 𝒄𝒐𝒔𝟐√𝒙)𝒄𝒐𝒔𝟓 (√𝒙) 𝒅𝒄𝒐𝒔√𝒙

= −𝟐 ∫(𝒄𝒐𝒔𝟓√𝒙 − 𝒄𝒐𝒔𝟕√𝒙) 𝒅𝒄𝒐𝒔√𝒙

= −𝟏

𝟑𝒄𝒐𝒔𝟔(√𝒙) +

𝟏

𝟒𝒄𝒐𝒔𝟖(√𝒙) + 𝒄.


Example 10: ∫ 𝒄𝒐𝒔𝟓𝒙 𝒅𝒙

𝑰 = ∫ 𝒄𝒐𝒔𝟓𝒙 𝒅𝒙 = ∫(𝟏 − 𝒔𝒊𝒏𝟐𝒙)𝟐 𝒄𝒐𝒔𝒙 𝒅𝒙

= ∫(𝟏 − 𝟐𝒔𝒊𝒏𝟐𝒙 + 𝒔𝒊𝒏𝟒𝒙) 𝒄𝒐𝒔𝒙 𝒅𝒙

= 𝐬𝐢𝐧 𝒙 −𝟐

𝟑𝒔𝒊𝒏𝟑𝒙 +

𝟏

𝟓𝒔𝒊𝒏𝟓𝒙 + 𝒄.

Example 11: ∫ 𝒔𝒊𝒏𝟐𝒙 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙

Solution:

∫ 𝒔𝒊𝒏𝟐𝒙 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙

=𝟏

𝟒∫ 𝒔𝒊𝒏𝟐𝟐𝒙 𝒅𝒙

=𝟏

𝟖∫(𝟏 − 𝐜𝐨𝐬 𝟒𝒙) 𝒅𝒙 =

𝟏

𝟖[𝒙 −

𝟏

𝟒𝒔𝒊𝒏𝟒𝒙] + 𝒄.

Example 12: ∫ 𝒔𝒊𝒏𝟏

𝟑𝒙 𝒄𝒐𝒔𝟑𝒙 𝒅𝒙

∫ 𝐬𝐢𝐧𝟏𝟑𝐱 𝐜𝐨𝐬𝟑𝐱 𝐝𝐱 = ∫ 𝐬𝐢𝐧

𝟏𝟑𝐱 𝐜𝐨𝐬𝟐𝐱 . 𝐜𝐨𝐬𝐱 𝐝𝐱

∫ 𝐬𝐢𝐧𝟏𝟑𝐱 (𝟏 − 𝐬𝐢𝐧𝟐

𝐱) . 𝐜𝐨𝐬𝐱 𝐝𝐱 =𝟑

𝟒𝒔𝒊𝒏

𝟒𝟑𝒙 −

𝟑

𝟏𝟎𝒔𝒊𝒏

𝟏𝟎𝟑 𝒙 + 𝒄

Note:

∫ 𝐬𝐢𝐧 𝒎𝒙 𝐜𝐨𝐬 𝒏𝒙 𝒅𝒙 =𝟏

𝟐∫[𝐬𝐢𝐧(𝒎 − 𝒏)𝒙 + 𝐬𝐢𝐧(𝒎 + 𝒏)𝒙] 𝒅𝒙

∫ 𝐜𝐨𝐬 𝒎𝒙 𝐜𝐨𝐬 𝒏𝒙 𝒅𝒙 =𝟏

𝟐∫[𝐜𝐨𝐬(𝒎 − 𝒏)𝒙 + 𝐜𝐨𝐬(𝒎 + 𝒏)𝒙] 𝒅𝒙


∫ 𝐬𝐢𝐧 𝒎𝒙 𝐬𝐢𝐧 𝒏𝒙 𝒅𝒙 =𝟏

𝟐∫[𝐜𝐨𝐬(𝒎 − 𝒏)𝒙 − 𝐜𝐨𝐬(𝒎 + 𝒏)𝒙] 𝒅𝒙

Example 13: ∫ 𝐬𝐢𝐧 𝟑𝒙 𝐜𝐨𝐬 𝟓𝒙 𝒅𝒙

∫ 𝐬𝐢𝐧 𝟑𝒙 𝐜𝐨𝐬 𝟓𝒙 𝒅𝒙 =𝟏

𝟐∫[𝐬𝐢𝐧(−𝟐𝒙) + 𝐬𝐢𝐧 𝟖𝒙] 𝒅𝒙

=𝟏

𝟐∫[𝐬𝐢𝐧 𝟖𝒙 − 𝐬𝐢𝐧 𝟐𝒙] 𝒅𝒙 = −

𝟏

𝟏𝟔𝒄𝒐𝒔 𝟖𝒙 +

𝟏

𝟒𝒄𝒐𝒔𝟐𝒙 + 𝒄.

Example 14: ∫ 𝐬𝐢𝐧 𝟑𝒙 𝐜𝐨𝐬𝟐 𝒙 𝒅𝒙

𝑰 = ∫ 𝐬𝐢𝐧 𝟑𝒙 𝐜𝐨𝐬𝟐 𝒙 𝒅𝒙

= ∫ 𝐜𝐨𝐬 𝐱 [𝐬𝐢𝐧 𝟑𝒙 𝐜𝐨𝐬 𝒙] 𝒅𝒙

=𝟏

𝟐∫ 𝒄𝒐𝒔 𝒙 [𝒔𝒊𝒏𝟒𝒙 + 𝒔𝒊𝒏𝟐𝒙] 𝒅𝒙

=𝟏

𝟐∫(𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟒𝒙 + 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟐𝒙)𝒅𝒙

=𝟏

𝟒∫( 𝒔𝒊𝒏 𝟓𝒙 + 𝒔𝒊𝒏 𝟑𝒙 + 𝒔𝒊𝒏 𝟑𝒙 + 𝒔𝒊𝒏 𝒙)𝒅𝒙

=𝟏

𝟒∫( 𝒔𝒊𝒏 𝟓𝒙 + 𝟐 𝒔𝒊𝒏 𝟑𝒙 + 𝒔𝒊𝒏 𝒙)𝒅𝒙

= −𝟏

𝟒[𝟏

𝟓𝒄𝒐𝒔 𝟓𝒙 +

𝟐

𝟑𝒄𝒐𝒔 𝟑𝒙 + 𝒄𝒐𝒔 𝒙] + 𝒄.


Trigonometric integrals depends on identity of trigonometric.

1 + 𝑡𝑎𝑛2𝑥 = 𝑠𝑒𝑐2𝑥

∫ 𝒕𝒂𝒏𝒏𝒙. 𝒔𝒆𝒄𝒎𝒙 𝒅𝒙

We have three cases:

𝟏 − ∫ 𝒕𝒂𝒏𝒐𝒅𝒅 𝒔𝒆𝒄𝒂𝒏𝒚 𝒏𝒖𝒎𝒃𝒆𝒓𝒅𝒙

• ∫ ∎∎ (𝒔𝒆𝒄𝒙 𝒕𝒂𝒏𝒙 𝒅𝒙)

• 𝒕𝒂𝒏𝟐𝒙 = 𝒔𝒆𝒄𝟐𝒙 − 𝟏

𝟐 − ∫ 𝒔𝒆𝒄𝒆𝒗𝒆𝒏 𝒕𝒂𝒏𝒂𝒏𝒚 𝒏𝒖𝒎𝒃𝒆𝒓𝒅𝒙

• ∫ ∎∎ (𝒔𝒆𝒄𝟐𝒙 𝒅𝒙)

• 𝒔𝒆𝒄𝟐𝒙 = 𝒕𝒂𝒏𝟐𝒙 + 𝟏

𝟑 − ∫ 𝒔𝒆𝒄𝒐𝒅𝒅 𝒕𝒂𝒏𝒆𝒗𝒆𝒏𝒅𝒙

• Integration by parts

Example 1:

∫ 𝒕𝒂𝒏𝟑𝒙 𝒔𝒆𝒄𝟑𝒙 𝒅𝒙

Solution

𝑰 = ∫(𝒔𝒆𝒄𝟐𝒙 − 𝟏) 𝒔𝒆𝒄𝟐𝒙 (𝒔𝒆𝒄𝒙 𝒕𝒂𝒏𝒙 𝒅𝒙)


𝑰 = ∫(𝒔𝒆𝒄𝟒𝒙 − 𝒔𝒆𝒄𝟐𝒙)(𝒔𝒆𝒄𝒙 𝒕𝒂𝒏𝒙 𝒅𝒙)

𝑰 = ∫ (𝒔𝒆𝒄𝟒𝒙(𝒔𝒆𝒄𝒙 𝒕𝒂𝒏𝒙 𝒅𝒙) − 𝒔𝒆𝒄𝟐𝒙(𝒔𝒆𝒄𝒙 𝒕𝒂𝒏𝒙 𝒅𝒙))

𝑰 =𝒔𝒆𝒄𝟓𝒙

𝟓+

𝒔𝒆𝒄𝟑𝒙

𝟑+ 𝒄.

Example 2:

𝑰 = ∫ 𝒕𝒂𝒏𝟒𝒙 𝒔𝒆𝒄𝟒𝒙 𝒅𝒙

𝑰 = ∫ 𝒕𝒂𝒏𝟒𝒙 . 𝒔𝒆𝒄𝟐𝒙 (𝒔𝒆𝒄𝟐𝒙 𝒅𝒙)

𝑰 = ∫ 𝒕𝒂𝒏𝟒𝒙 (𝟏 + 𝒕𝒂𝒏𝟐𝒙) 𝒔𝒆𝒄𝟐𝒙 𝒅𝒙 = ∫(𝒕𝒂𝒏𝟒𝒙 + 𝒕𝒂𝒏𝟔𝒙)𝒔𝒆𝒄𝟐𝒙 𝒅𝒙

𝑰 = ∫(𝒕𝒂𝒏𝟒𝒙. 𝒔𝒆𝒄𝟐𝒙 𝒅𝒙 + 𝒕𝒂𝒏𝟔𝒙. 𝒔𝒆𝒄𝟐𝒙 𝒅𝒙) =𝒕𝒂𝒏𝟓𝒙

𝟓+

𝒕𝒂𝒏𝟕𝒙

𝟕+ 𝒄

Example 3:

𝑰 = ∫ 𝑠𝑒𝑐4𝑥 𝑑𝑥.

Solution:

𝐼 = ∫ 𝑠𝑒𝑐2𝑥 . (𝑠𝑒𝑐2𝑥 𝑑𝑥) = ∫(1 + 𝑡𝑎𝑛2𝑥) 𝑠𝑒𝑐2𝑥 𝑑𝑥 =

𝐼 = ∫(𝑠𝑒𝑐2𝑥 𝑑𝑥 + 𝑡𝑎𝑛2𝑥. 𝑠𝑒𝑐2𝑥 𝑑𝑥) = 𝑡𝑎𝑛𝑥 +𝑡𝑎𝑛3𝑥

3+ 𝑐


Example 4:

𝑰 = ∫ 𝑡𝑎𝑛4𝑥 𝑑𝑥 .

Solution:

𝐼 = ∫(𝑠𝑒𝑐2𝑥 − 1) . (𝑡𝑎𝑛2𝑥 𝑑𝑥)

𝐼 = ∫(𝑠𝑒𝑐2𝑥. 𝑡𝑎𝑛2𝑥 − 𝑡𝑎𝑛2𝑥) . 𝑑𝑥 =𝑡𝑎𝑛3𝑥

3− ∫ 𝑡𝑎𝑛2𝑥 𝑑𝑥

𝐼 =𝑡𝑎𝑛3𝑥

3− ∫(𝑠𝑒𝑐2𝑥 − 1)𝑑𝑥 =

𝑡𝑎𝑛3𝑥

3− 𝑡𝑎𝑛𝑥 + 𝑥 + 𝑐.

Example 5:

𝑰 = ∫ 𝒕𝒂𝒏𝟓𝒙 𝒅𝒙

Solution:

𝑰 = ∫ 𝒕𝒂𝒏𝟐𝒙 . 𝒕𝒂𝒏𝟑𝒙 𝒅𝒙 = ∫(𝒔𝒆𝒄𝟐𝒙 − 𝟏) . (𝒕𝒂𝒏𝟑𝒙 𝒅𝒙)

𝑰 = ∫(𝒔𝒆𝒄𝟐𝒙. 𝒕𝒂𝒏𝟑𝒙 − 𝒕𝒂𝒏𝟑𝒙) . 𝒅𝒙 =𝒕𝒂𝒏𝟒𝒙

𝟒− ∫ 𝒕𝒂𝒏𝟑𝒙 𝒅𝒙

𝑰 =𝒕𝒂𝒏𝟒𝒙

𝟒− ∫ 𝒕𝒂𝒏𝟐𝒙. 𝒕𝒂𝒏𝒙. 𝒅𝒙 =

𝒕𝒂𝒏𝟒𝒙

𝟒− ∫(𝒔𝒆𝒄𝟐𝒙 − 𝟏)𝒕𝒂𝒏𝒙 𝒅𝒙


𝟒− ∫(𝒔𝒆𝒄𝟐𝒙. 𝒕𝒂𝒏𝒙 − 𝒕𝒂𝒏𝒙) 𝒅𝒙


𝟒−

𝒕𝒂𝒏𝟐𝒙

𝟐+ 𝒍𝒏|𝒔𝒆𝒄𝒙| + 𝒄.


Exercise 6


𝟏) ∫ 𝒔𝒊𝒏𝟑𝒙 . 𝒔𝒊𝒏𝟕𝒙. 𝒅𝒙

𝟐) ∫ 𝒄𝒐𝒔𝟏𝟒𝒙. 𝒔𝒊𝒏𝟏𝟎𝒙. 𝒅𝒙

𝟑) ∫ 𝒄𝒐𝒔𝟓𝒙. 𝒅𝒙

𝟒) ∫ 𝒄𝒐𝒔𝟓𝒙. 𝒔𝒊𝒏𝟐𝒙. 𝒅𝒙

𝟓) ∫ 𝒔𝒊𝒏𝟒𝒙. 𝒅𝒙

𝟔) ∫𝐬𝐢𝐧𝟒 𝒙

𝒄𝒐𝒔𝟔𝒙𝒅𝒙

𝟕) ∫ 𝒔𝒆𝒄𝟔𝒙. 𝐝𝐱

𝟖) ∫ 𝒕𝒂𝒏𝟓𝒙. 𝒔𝒆𝒄𝟒𝒙. 𝐝𝐱


𝟗) ∫ 𝒕𝒂𝒏𝟔𝒙. 𝒅𝒙

𝟏𝟎) ∫𝐜𝐨𝐬 𝐱 𝐝𝐱

√𝐬𝐢𝐧𝟐 𝐱𝟑

𝟏𝟏) ∫ 𝐬𝐢𝐧 𝟐𝐱 𝐜𝐨𝐬𝟔 𝟐𝐱 𝐝𝐱

𝟏𝟐) ∫ 𝒕𝒂𝒏𝟕𝒙. 𝐝𝐱

𝟏𝟑) ∫ 𝒄𝒐𝒕𝟓𝒙. 𝐝𝐱

𝟏𝟒) ∫ 𝒔𝒊𝒏𝟐𝟑𝒙. 𝒄𝒐𝒔𝟐𝟒𝐱. 𝐝𝐱

𝟏𝟓) ∫ 𝒄𝒐𝒔𝟕𝒙. 𝐝𝐱

𝟏𝟔) ∫ 𝒔𝒊𝒏𝟓𝟐𝒙. 𝐝𝐱

𝟏𝟕) ∫ 𝐜𝐨𝐬 𝟖𝐱 𝐜𝐨𝐬𝟐𝐱 𝐝𝐱


4-4 Integration of Rational Function using Partial Fractions

If the integrand (the expression after the integral sign) is in the form of

an algebraic fraction and the integral cannot be evaluated by simple

methods, the fraction needs to be expressed in partial fractions before

integration takes place.

We decompose fractions into partial fractions because:

• It makes certain integrals much easier to do, and • It is used in the Laplace transform, which we meet later.

So, if we needed to integrate the fraction, we could simplify our integral

in the following way:

∫6𝑥 + 13

𝑥2 + 5𝑥 + 6𝑑𝑥 = ∫

1

𝑥 + 2𝑑𝑥 + ∫

5

𝑥 + 3𝑑𝑥

We integrate the two fractions using what we learned in Basic

Logarithmic Form:

∫6𝑥 + 13

𝑥2 + 5𝑥 + 6𝑑𝑥 = ∫

1

𝑥 + 2𝑑𝑥 + ∫

5

𝑥 + 3𝑑𝑥

= ln(𝑥 + 2) + 5 ln(𝑥 + 3) + 𝑐

𝑨𝒍𝒈𝒆𝒃𝒓𝒂𝒊𝒄 𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏 =𝒏𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓

𝒅𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓

Definition of proper Fraction

A proper fraction is a fraction where the degree of numerator (the top

equation) is less than the degree of denominator (the bottom

equation).

http://www.intmath.com/laplace-transformation/7-inverse-laplace-transform.php

http://www.intmath.com/methods-integration/2-integration-logarithmic-form.php

http://www.intmath.com/methods-integration/2-integration-logarithmic-form.php


Definition of improper Fraction

A proper fraction is a fraction where the degree of numerator (the top

equation) is greater than or equal to the degree of denominator (the

bottom equation).

Partial Fraction can be applied only for proper Fraction. :Remark

Denominator containing…

Expression Form of partial Fractions

a-Linear factor 𝑓(𝑥)

(𝑥 + 𝑎)(𝑥 + 𝑏)

𝐴

𝑥 + 𝑎+

𝐵

𝑥 + 𝑏

b-Repeated linear factors

𝑓(𝑥)

(𝑥 + 𝑎)3

𝐴

𝑥 + 𝑎+

𝐵

(𝑥 + 𝑎)2

+𝐶

(𝑥 + 𝑎)3

c-Quadratic term which cannot be factored

𝑓(𝑥)

(𝑎𝑥2 + 𝑏𝑥 + 𝑐)(𝑔𝑥 + ℎ)

𝐴𝑥 + 𝐵

(𝑎𝑥2 + 𝑏𝑥 + 𝑐)

+𝐶

(𝑔𝑥 + ℎ)

Examples


𝐄𝐱𝐚𝐦𝐩𝐥𝐞𝟏: ∫𝒙 + 𝟏

𝒙𝟑 + 𝒙𝟐 − 𝟔𝒙𝒅𝒙

𝑥 + 1

𝑥3 + 𝑥2 − 6𝑥=

𝑥 + 1

𝑥(𝑥2 + 𝑥 − 6)=

𝑥 + 1

𝑥(𝑥 − 2)(𝑥 + 3)

𝑥 + 1

𝑥3 + 𝑥2 − 6𝑥 =

𝐴

𝑥+

𝐵

𝑥 − 2+

𝐷

𝑥 + 3 → (1 )


𝑥 + 1 = 𝐴(𝑥 − 2)(𝑥 + 3) + 𝐵𝑥(𝑥 + 3) + 𝐷𝑥(𝑥 − 2) → (2)

Using zeros from equation(1), and substitute in (2)

𝑃𝑢𝑡 𝑥 = 0 𝑖𝑛 𝑒𝑞. (2) ⟹ 1 = 𝐴(−2)(3) ⟹ 𝐴 =−1

6

𝑃𝑢𝑡 𝑥 = 2 𝑖𝑛 𝑒𝑞. (2) ⟹ 3 = 𝐵(2)(5) ⟹ 𝐵 =3

10

𝑃𝑢𝑡 𝑥 = −3 𝑖𝑛 𝑒𝑞. (2) ⟹ −2 = 𝐷(−3)(−5) ⟹ 𝐷 =−2

15

𝐼 =−1

6∫

𝑑𝑥

𝑥+

3

10∫

𝑑𝑥

𝑥 − 2−

2

15∫

𝑑𝑥

𝑥 + 3

𝐼 =−1

6 𝑙𝑛|𝑥| +

3

10 𝑙𝑛|𝑥 − 2| −

2

15 𝑙𝑛|𝑥 + 3| + 𝑐

𝐄𝐱𝐚𝐦𝐩𝐥𝐞𝟐: ∫𝑥2

(𝑥 + 1)(𝑥 − 1)2 𝑑𝑥

𝑥2

(𝑥 + 1)(𝑥 − 1)2=

𝐴

𝑥 + 1+

𝐵

𝑥 − 1+

𝐶

(𝑥 − 1)2 → (𝟏)

𝑥2 = 𝐴(𝑥 − 1)2 + 𝐵(𝑥 + 1)(𝑥 − 1) + 𝐶(𝑥 + 1) → (𝟐)


𝑃𝑢𝑡 𝑥 = 1 𝑖𝑛 𝑒𝑞. (2) ⟹ 1 = 𝐶(2) ⟹ 𝐶 =1

2

𝑃𝑢𝑡 𝑥 = −1 𝑖𝑛 𝑒𝑞. (2) ⟹ 1 = 𝐴(4) ⟹ 𝐴 =1

4

Comparing coefficients of 𝑥2: 1 = 𝐴 + 𝐵 ⟹ 𝐵 =3

4


𝐼 =1

4∫

𝑑𝑥

𝑥 + 1+

3

4∫

𝑑𝑥

𝑥 − 1+

1

2∫

𝑑𝑥

(𝑥 − 1)2

𝐼 =1

4 𝑙𝑛|𝑥 + 1| +

3

4 𝑙𝑛|𝑥 − 1| −

1

2(𝑥 − 1)−1 + 𝑐.

𝐄𝐱𝐚𝐦𝐩𝐥𝐞𝟑: ∫2𝑥3 − 4𝑥 − 8

(𝑥2 − 𝑥)(𝑥2 + 4) 𝑑𝑥

𝐼 = ∫2𝑥3 − 4𝑥 − 8

(𝑥2 − 𝑥)(𝑥2 + 4) 𝑑𝑥 = ∫

2𝑥3 − 4𝑥 − 8

𝑥 (𝑥 − 1)(𝑥2 + 4)𝑑𝑥

2𝑥3 − 4𝑥 − 8

𝑥 (𝑥 − 1)(𝑥2 + 4)=

𝐴

𝑥+

𝐵

𝑥 − 1+

𝐶𝑥 + 𝐷

𝑥2 + 4 → (𝟏)

2𝑥3 − 4𝑥 − 8 = 𝐴(𝑥 − 1)(𝑥2 + 4) + 𝐵 𝑥 (𝑥2 + 4) + (𝐶𝑥 + 𝐷)𝑥 (𝑥 − 1)

→ (𝟐)


Put x = 0 in eq. (2) ⟹ −8 = A(−1)(4) ⟹ A = 2

Put x = 1 in eq. (2) ⟹ −10 = B(1)(5) ⟹ B = −2

Comparing coefficients of 𝑥3: 2 = 𝐴 + 𝐵 + 𝐶 ⟹ 𝐶 = 2

Comparing coefficients of 𝑥2: 0 = −𝐴 − 𝐶 + 𝐷 ⟹ 𝐷 = 4

𝐼 = ∫( 2

𝑥−

2

𝑥 − 1+

2𝑥 + 4

𝑥2 + 4) 𝑑𝑥 = ∫(

2

𝑥−

2

𝑥 − 1+

2𝑥

𝑥2 + 4+

4

𝑥2 + 4) 𝑑𝑥

𝐼 = 2 𝑙𝑛|𝑥| − 2 𝑙𝑛|𝑥 − 1| + 𝑙𝑛|(𝑥2 + 4)| + 2 𝑡𝑎𝑛−1 (𝑥

2) + 𝑐.

Remark:

For improper fraction we use division to obtain proper fraction before

we use the rules of partial fractions.


𝐄𝐱𝐚𝐦𝐩𝐥𝐞𝟒: ∫𝒙𝟓 + 𝟑

𝒙𝟑 − 𝒙 𝒅𝒙

𝑥2 + 1

𝑥3 − 𝑥 𝑥5 + 3

𝑥5 − 𝑥3

𝑥3 + 3 𝑥3 − 𝑥

𝑥 + 3

𝐼 = ∫𝑥5 + 3

𝑥3 − 𝑥 𝑑𝑥 = ∫ (𝑥2 + 1 +

𝑥 + 3

𝑥3 − 𝑥) 𝑑𝑥

=𝑥3

3+ 𝑥 + ∫

𝑥 + 3

𝑥(𝑥 − 1)(𝑥 + 1)𝑑𝑥

𝑥 + 3

𝑥(𝑥 − 1)(𝑥 + 1)=

𝐴

𝑥+

𝐵

𝑥 − 1+

𝐶

𝑥 + 1 → (1)

𝑥 + 3 = 𝐴(𝑥 − 1)(𝑥 + 1) + 𝐵𝑥(𝑥 + 1) + 𝐶𝑥(𝑥 − 1) → (2)


𝑃𝑢𝑡 𝑥 = 0 𝑖𝑛 𝑒𝑞. (2) ⟹ 3 = 𝐴(−1)(1) ⟹ 𝐴 = −3

𝑃𝑢𝑡 𝑥 = 1 𝑖𝑛 𝑒𝑞. (2) ⟹ 4 = 𝐵(2)(1) ⟹ 𝐵 = 2

𝑃𝑢𝑡 𝑥 = −1 𝑖𝑛 𝑒𝑞. (2) ⟹ 2 = 𝐶(−1)(−2) ⟹ 𝐶 = 1

𝐼 =𝑥3

3+ 𝑥 + ∫ (

−3

𝑥+

2

𝑥 − 1+

1

𝑥 + 1) 𝑑𝑥

𝐼 =𝑥3

3+ 𝑥 − 3𝑙𝑛|𝑥| + 2𝑙𝑛|𝑥 − 1| + 𝑙𝑛|𝑥 + 1| + 𝑐.


𝐄𝐱𝐚𝐦𝐩𝐥𝐞𝟓: ∫𝟐𝒙𝟐 − 𝟏

(𝒙 − 𝟏)(𝒙 + 𝟐)𝒅𝒙

2

𝒙𝟐 + 𝒙 − 𝟐 𝟐𝒙𝟐 − 𝟏

2𝑥2 + 2𝑥 − 4

−2𝑥 + 3

𝐼 = ∫𝟐𝒙𝟐 − 𝟏

(𝒙 − 𝟏)(𝒙 + 𝟐) 𝒅𝒙 = ∫ 2 + ∫

−2𝑥 + 3

(𝑥 − 1)(𝑥 + 2)𝑑𝑥

−2𝑥 + 3

(𝑥 − 1)(𝑥 + 2)=

𝐴

𝑥 − 1+

𝐵

𝑥 + 2 → (1)

−2𝑥 + 3 = 𝐴 (𝑥 + 2) + 𝐵(𝑥 − 1) → (2)


𝑃𝑢𝑡 𝑥 = 1 𝑖𝑛 𝑒𝑞. (2) ⟹ 1 = 𝐴(3) ⟹ 𝐴 =1

3

𝑃𝑢𝑡 𝑥 = −2 𝑖𝑛 𝑒𝑞. (2) ⟹ 7 = 𝐵(−3) ⟹ 𝐵 = −7

3

𝐼 = 2𝑥 + ∫ (

13

𝑥 − 1−

73

𝑥 + 2) 𝑑𝑥

𝐼 = 2𝑥 +1

3𝑙𝑛|𝑥 − 1| −

7

3𝑙𝑛|𝑥 + 2| + 𝑐.


Exercise 7

Using Partial Fraction Evaluate the following integrals:

𝟏) ∫𝟐𝒙𝟐 − 𝟑𝒙 − 𝟏

𝒙𝟑 + 𝟐𝒙𝟐 − 𝟑𝒙𝒅𝒙

𝟐) ∫𝟐𝒙 + 𝟏

(𝒙𝟐 + 𝟐𝒙 + 𝟒)(𝒙 − 𝟐) 𝒅𝒙

𝟑) ∫𝒙𝟑 + 𝟒𝒙𝟐

𝒙𝟐 + 𝟓𝒙 + 𝟒 𝒅𝒙

𝟒) ∫𝒙𝟐 + 𝟏

(𝒙 − 𝟓)(𝒙 + 𝟏)𝟐𝒅𝒙

𝟓) ∫𝐬𝐢𝐧 𝒙

𝐜𝐨𝐬 𝒙 (𝟏 + 𝒄𝒐𝒔𝟐𝒙)𝒅𝒙

𝟔) ∫𝟏

𝒙𝟑 + 𝒙𝟐𝒅𝒙


𝟕) ∫𝒙𝟒

𝟏 + 𝒙𝟐𝒅𝒙

𝟖) ∫𝒙 + 𝟐

𝒙𝟑 − 𝟒𝒙𝟐 − 𝟓𝒙𝒅𝒙

𝟗) ∫𝒙 + 𝟐

𝒙𝟑 − 𝟔𝒙𝟐 + 𝟗𝒙𝒅𝒙

𝟏𝟎) ∫𝟑𝒙𝟑 − 𝟓𝒙𝟐 − 𝟏𝟏𝒙 + 𝟗

𝒙𝟐 − 𝟐𝒙 − 𝟑𝒅𝒙

𝟏𝟏) ∫𝟐 − 𝒙

𝒙𝟐 + 𝟓𝒙𝒅𝒙

𝟏𝟐) ∫𝟑𝒙 + 𝟏𝟏

𝒙𝟐 − 𝒙 − 𝟔𝒅𝒙


4-5 INTEGRATION BY PARTS

4-5-I Basic Technique

By the Product Rule for Derivatives,

. )()()()()()( xfxgxgxfxgxfdx

d+=

Thus,

=+=+ dxxfxgdxxgxfxgxfdxxfxgxgxf )()()()()()()()()()(

. −= dxxfxgxgxfdxxgxfxgxf )()()()()()()()(

This formula for integration by parts often makes it possible to reduce

A simpler integral.

By a complicated integral involving a product, we have:

dxxfduxfu )()( ==

𝑑𝑣 = 𝑔′(𝑥)𝑑𝑥 ⇒ 𝑣 = 𝑔(𝑥).

we get the more common formula for integration by parts:

∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑢𝑑𝑣.

: Example 1

. xdxxlnFind

Solution


. Thus, 2

2

1xdxxv ==and dx

xduxdxdv

1==and xu ln=Let

−

=−===

2

2

1)(ln))((lnln xxvduuvudvdxxxxdxx

=+

−=−=

Cxxxdxxxxdxx

x 2222

2

1

2

1ln

2

1

2

1ln

2

11

2

1

Cxxx +− 22

4

1ln

2

1

It is possible that when you set up an integral using integration by

parts, the resulting integral will be more complicated than the original

integral. In this case, change your substitutions for u and dv.

. dxxxsinFind Example 2:

Let xu sin= and dxxdudxxdv cos== and

2

2

1xdxxv == . Thus, ∫ 𝑥𝑠𝑖𝑛𝑥𝑑𝑥 = ∫(𝑠𝑖𝑛𝑥)(𝑥𝑑𝑥) = ∫ 𝑢𝑑𝑣 = 𝑢𝑣 −

∫ 𝑣𝑑𝑢 = (𝑠𝑖𝑛𝑥)(1

2𝑥2) −

1

2∫ 𝑥2𝑐𝑜𝑠𝑥𝑑𝑥.

is more dxxx cos2

1 2

Notice that this resulting integral

complicated than the original one.

dxduxdxdv == sinand xu =Therefore, let

= dxxxsin. Thus, −== xdxxv cossinand

=+−=−−−=−= xdxxxdxxxxvduuvudv coscos)cos()cos)((

Cxxx ++−= sincos .


. dxxln: Find Example 3

. == xdxv 1and dxx

dudxdv1

1 ==and xu ln=Let

Thus,

=

−=−=== )(

1))((ln)1)((lnln xdx

xxxvduuvudvdxxdxx

. +−=−= Cxxxdxxx ln1ln

. dxxarctan: Find Example 4

and xu arctan=As in the previous example, let

dxx

dudxdv21

1

+==

. Thus, == xdxv 1and

−xxarctan =−== vduuvudvdxxarctan

=+

−=+

−=

+ dx

x

xxxdx

x

xxxdx

xx

222 1

2

2

1arctan

1arctan

1

1

.Cxxx ++− )1ln(2

1arctan 2

.dxx

x 2

ln: Find Example 5

anddxx

dudxxdxx

dv11 2

2=== −

and xu ln=Let

.1

1 12

xxdxxv −=−== −−


−

−=−==

=

xxvduuvudvdx

xxdx

x

x 1)(ln

1)(ln

ln22

Thus,

=+−−

=+−

=+−

=

−

− Cxx

xdxx

x

xdx

xx

x

xdx

x

1lnln1ln11 2

2

.Cx

x+

−− 1ln

Sometimes, it is necessary to use integration by parts more than

once.

.dxxx cos2

: Find Example 6

and xdxdudxxdv 2cos ==and 2xu =Let

. Thus, == xdxxv sincos

−=− =−== xdxxxxxxdxxxvduuvudvdxxx sin2sin))(sin2())(sin(cos 222

Notice that

is less complicated than the original dxxxsinthe resulting integral

one, but integration by parts is needed to evaluate it.

Let xu = and dxdudxxdv == sin and −== xdxxv cossin .

Thus,

xxxdxxxxvduuvudvdxxx sincos)cos()cos(sin +−=−−−=−==

.sin2cos2sincos 22 cxxxxdxxx +−+= Finally, we get


Note:

We able to integrate dxxx cos2

more easily with tabular

integration; this technique will be described later.

The next illustration of repeated integration by parts deserves

special attention.

.dxxe x sin: Find Example 7

Let xeu = and dxedudxxdv x== sin and

xxdxv cossin −== . Thus,

=−−− =−== ))(cos()cos)((sin dxexxevduuvudvdxxe xxx

xdxexe xx coscos +−

Notice that integration by parts is now needed to evaluate xdxe x cos .

Let xeu = and dxeduxdxdv x== cos and

== xxdxv sincosThus,

− =−== ))((sinsincos dxexxevduuvudvxdxe xxx

xdxexedxex xxx sinsin))((sin −=

−+−=+− = dxxexexexdxexexdxe xxxxxx sinsincoscoscossin

= dxxe x sin +−= xexedxxe xxx sincossin2Thus,

. Cxe x +sin2

1+− xe x cos

2

1

The next example illustrates an interesting type of integral that

surprisingly require integration by parts.


Example 8: ( )dxxsin .

Let === dxuduxuxu 22

( ) == uduuuduudxx sin2)2)((sinsin

In example 2, we got the following using integration by parts:

= duuu sinor Cxxxxdxx ++− = sincossin

.Cuuu ++− sincos

Thus, ( ) dxxsin = duuu sin2 =

( ) .)sin(2cos2sin2cos2 cxxxCuuu ++−=++−

and this does check.

In general, to evaluate ( ) dxxf n , let

=== − dxdunuxuxu nnn 1

( ) dxxf n =

− duufun n )(1.

4-5-2 Tabular Integration

Integrals of the form dxxgxf )()( , in which f can be differentiated

repeatedly to become zero and g can be integrated repeatedly

without difficulty, can be evaluated using tabular integration

Example1: Find dxex x23

.

f (x) and its derivatives g (x) and its antiderivatives


𝑒2𝑥

𝑥3

1

2𝑒2𝑥

3𝑥2

1

4𝑒2𝑥

6𝑥

1

8𝑒2𝑥

6

1

16𝑒2𝑥

0

Cexeexex xxxx +−+−+ 222223

16

6

8

6

4

3

2

1= dxex x23

.Cexeexex xxxx +−+− 222223

8

3

4

3

4

3

2

1=

+

-

-

+

-


4-5-3 Reduction Formulas

Integration by parts can be used to derive reduction formulas for

integrals. These are formulas that express an integral involving a

power of a function in terms of an integral that involves a lower

power of that function.

Example 1:

Prove the reduction formula dxxnxxdxx nnn

− = −1)(ln)(ln)(ln

and use the result to find dxx2)(ln .

Solution:

Let nxu )(ln= and

== − dx

xxndudxdv n 1)(ln 1

and == xdxv 1 .

Thus,

−=

− =−== −− dxxnxxdx

xxnxxxvduuvudvdxx nnnnn 11 )(ln)(ln

1)(ln)()(ln)(ln

Thus,

.2ln2)(ln]1)(ln[2)(lnln2)(ln)(ln 2222 Cxxxxxdxxxxxdxxxxdxx ++− =−−=−=

Example 2: Prove the reduction formula

+−=− xx

ndxx nn cos)(sin

1)(sin 1 dxx

n

n n

−− 2)(sin

1.

Solution:

Let 1)(sin −= nxu and

−==−== − xxdxvdxxxndudxxdv n cossin and cos))(sin1(sin 2

Thus, ===− udvdxxxdxx nn )(sin)(sin)(sin 1

=−−−−=− −− dxxxnxxxvduuv nn cos))(sin1)(cos()cos()(sin 21


+−=−+− −−− xxdxxxnxx nnn cos)(sincos)(sin)1(cos)(sin 1221

−−+−=−− −−− dxxnxxdxxxn nnn 2122 )(sin)1(cos)(sin)sin1()(sin)1(

−+−=− −− dxxnxxdxxndxxn nnnn 21 )(sin)1(cos)(sin)(sin)(sin)1(

−

+−=−− dxx

n

nxx

ndxx nnn 21 )(sin

1cos)(sin

1)(sin .


Exercise 8


𝟏) ∫ 𝒙𝟖. 𝒍𝒏𝒙𝟑. 𝒅𝒙

𝟐) ∫ 𝒙𝟐. 𝒔𝒆𝒄𝒉𝟐𝒙. 𝒅𝒙

𝟑) ∫ 𝒄𝒐𝒔−𝟏𝒙. 𝒅𝒙

𝟒) ∫ 𝒙𝟒. 𝒆𝟑𝒙. 𝒅𝒙

𝟓) ∫ 𝒔𝒊𝒏(𝒍𝒏𝒙). 𝒅𝒙

𝟔) ∫ 𝒄𝒐𝒔(√𝒙). 𝒅𝒙

𝟕) ∫ 𝒔𝒆𝒄𝟓𝒙. 𝒅𝒙

𝟖) ∫ 𝒄𝒐𝒔𝒏𝒙. 𝒅𝒙


4- 6 Trigonometric Substitutions Integration

The idea behind the trigonometric substitution is quite simple: to

replace expressions involving square roots with expressions that

involve standard trigonometric functions, but no square roots.

Integrals involving trigonometric functions are often easier to solve

than integrals involving square roots.

In order to evaluate integrals containing radicals we use the

trigonometric substitutions

Expression Substitution Identity

1 √𝒂𝟐 − 𝒙𝟐 𝒙 = 𝒂 𝒔𝒊𝒏𝜽 𝒐𝒓 𝒙 = 𝒂𝒄𝒐𝒔𝜽 𝒔𝒊𝒏𝟐𝜽 + 𝒄𝒐𝒔𝟐𝜽 = 𝟏

2 √𝒂𝟐 + 𝒙𝟐 𝒙 = 𝒂 𝒕𝒂𝒏𝜽 𝒐𝒓 𝒙 = 𝒂 𝒔𝒊𝒏𝒉 𝒕 𝟏 + 𝒕𝒂𝒏𝟐𝜽 = 𝒔𝒆𝒄𝟐𝜽

3 √𝒙𝟐 − 𝒂𝟐 𝒙 = 𝒂 𝒔𝒆𝒄𝜽 𝒐𝒓 𝒙 = 𝒂 𝒄𝒐𝒔𝒉 𝒕 𝒄𝒐𝒔𝒉𝟐𝒕 − 𝒔𝒊𝒏𝒉𝟐𝒕 = 𝟏

Examples:


Example 1:

∫𝒙𝟐

√𝟗 − 𝒙𝟐𝒅𝒙

Solution

x = 3 sinθ ⟹ dx = 3cosθ dθ

I = ∫9 sin2θ

√9 − 9 sin2θ 3cosθ dθ = 9 ∫

sin2θ cosθ

cosθ dθ


= 9 ∫ sin2θ dθ =9

2 ∫(1 − cos2θ) dθ

=9

2(θ −

1

2 sin2θ) + c =

9

2θ −

9

4sin2θ + c

I =9

2 sin−1 (

x

3) −

x

2 √9 − x2 + c.

Example 2:

∫𝒙𝟐

(𝟒 + 𝒙𝟐)𝟑

𝟐⁄𝒅𝒙

Solution

𝑥 = 2 𝑡𝑎𝑛𝜃 ⟹ 𝑑𝑥 = 2 𝑠𝑒𝑐2𝜃 𝑑𝜃

𝐼 = ∫4 𝑡𝑎𝑛2𝜃

432 (1 + 𝑡𝑎𝑛2𝜃)

32

2 𝑠𝑒𝑐2𝜃 𝑑𝜃 = ∫𝑡𝑎𝑛2𝜃

𝑠𝑒𝑐𝜃 𝑑𝜃 = ∫

𝑠𝑖𝑛2𝜃

𝑐𝑜𝑠𝜃𝑑𝜃

𝐼 = ∫1 − 𝑐𝑜𝑠2𝜃

𝑐𝑜𝑠𝜃𝑑𝜃 = ∫(𝑠𝑒𝑐𝜃 − 𝑐𝑜𝑠𝜃)𝑑𝜃

𝐼 = 𝑙𝑛|𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃| − 𝑠𝑖𝑛𝜃 + 𝑐 = 𝑙𝑛 |√𝑥2 + 4

2+

𝑥

2| −

𝑥

√𝑥2 + 4+ 𝑐.

Example 3:

∫𝒙𝟑

(𝟗 + 𝒙𝟐)𝟑

𝟐⁄𝒅𝒙

Solution

𝑥 = 3 𝑡𝑎𝑛𝜃 ⟹ 𝑑𝑥 = 3 𝑠𝑒𝑐2𝜃 𝑑𝜃

√𝟗 − 𝒙𝟐

x

3


𝐼 = ∫27 𝑡𝑎𝑛3𝜃

(9 + 9𝑡𝑎𝑛2𝜃)32

3 𝑠𝑒𝑐2𝜃 𝑑𝜃 = 3 ∫𝑡𝑎𝑛3𝜃

𝑠𝑒𝑐𝜃 𝑑𝜃

𝐼 = 3 ∫𝑡𝑎𝑛2𝜃

𝑠𝑒𝑐2𝜃 𝑡𝑎𝑛𝜃 𝑠𝑒𝑐𝜃 𝑑𝜃 = 3 ∫

𝑠𝑒𝑐2𝜃 − 1

𝑠𝑒𝑐2𝜃 𝑡𝑎𝑛𝜃 𝑠𝑒𝑐𝜃 𝑑𝜃

𝐼 = 3 ∫(1 − 𝑠𝑒𝑐−2𝜃) 𝑡𝑎𝑛𝜃 𝑠𝑒𝑐𝜃 𝑑𝜃 = 3𝑠𝑒𝑐𝜃 +3

𝑠𝑒𝑐𝜃+ 𝑐

𝐼 =9

√𝑥2+9+ √𝑥2 + 9 + 𝑐.

Example 4:

∫√𝒙𝟐 − 𝟏𝟔

𝒙 𝒅𝒙

Solution

𝑥 = 4𝑠𝑒𝑐𝜃 ⟹ 𝑑𝑥 = 4𝑠𝑒𝑐𝜃 𝑡𝑎𝑛𝜃 𝑑𝜃

𝐼 = ∫√16 𝑠𝑒𝑐2𝜃 − 16

4𝑠𝑒𝑐𝜃4𝑠𝑒𝑐𝜃 𝑡𝑎𝑛𝜃 𝑑𝜃 = 4 ∫ 𝑡𝑎𝑛2𝜃 𝑑𝜃

𝐼 = 4 ∫(𝑠𝑒𝑐2𝜃 − 1)𝑑𝜃 = 4(𝑡𝑎𝑛𝜃 − 𝜃) + 𝑐

𝐼 = 4 (√𝑥2 − 16 − 4𝑠𝑒𝑐−1 (𝑥

4)) + 𝑐.

Example 5:

∫𝒅𝒙

𝒙𝟒√𝒙𝟐 − 𝟏

Solution

𝑥 = 𝑐𝑜𝑠ℎ𝑢 ⟹ 𝑑𝑥 = 𝑠𝑖𝑛ℎ𝑢 𝑑𝑢


𝐼 = ∫𝑠𝑖𝑛ℎ𝑢 𝑑𝑢

𝑐𝑜𝑠ℎ4𝑢√𝑐𝑜𝑠ℎ2𝑢 − 1= ∫

𝑠𝑖𝑛ℎ𝑢 𝑑𝑢

𝑐𝑜𝑠ℎ4𝑢 . 𝑠𝑖𝑛ℎ𝑢

𝐼 = ∫ 𝑠𝑒𝑐ℎ4𝑢 𝑑𝑢 = ∫(1 − 𝑡𝑎𝑛ℎ2𝑢)𝑠𝑒𝑐ℎ2𝑢 𝑑𝑢

𝐼 = ∫ 𝑠𝑒𝑐ℎ2𝑢 𝑑𝑢 − ∫(𝑡𝑎𝑛ℎ2𝑢)𝑠𝑒𝑐ℎ2𝑢 𝑑𝑢

𝐼 = 𝑡𝑎𝑛ℎ𝑢 −𝑡𝑎𝑛ℎ3𝑢

3+ 𝑐 =

√𝑥2 − 1

𝑥−

1

3(

√𝑥2 − 1

𝑥)

3

+ 𝑐.

Example 6:

𝑥2𝑑𝑥

√4𝑥 − 𝑥2

Solution

∫𝑥2𝑑𝑥

√4𝑥 − 𝑥2 = ∫

𝑥2

√4 − (𝑥 − 2)2𝑑𝑥

𝑥 − 2 = 2𝑠𝑖𝑛 𝜃 → 𝑑𝑥 = 2𝑐𝑜𝑠𝜃𝑑𝜃

√4𝑥 − 𝑥2 = 2𝑐𝑜𝑠𝜃

∫𝑥2𝑑𝑥

√4𝑥 − 𝑥2 = ∫

(2 + 2𝑠𝑖𝑛𝜃)2. 2𝑐𝑜𝑠𝜃

2𝑐𝑜𝑠𝜃𝑑𝜃

= 4 ∫[1 + 2𝑠𝑖𝑛𝜃 + 𝑠𝑖𝑛2𝜃]𝑑𝜃

= 4 ∫[1 + 2𝑠𝑖𝑛𝜃 +1

2(1 − 𝑐𝑜𝑠2𝜃)]𝑑𝜃

= 4 [𝜃 − 2𝑐𝑜𝑠𝜃 +1

2𝜃 −

1

4𝑠𝑖𝑛2𝜃] + 𝑐

√𝟒 − (𝒙 − 𝟐)𝟐

x-2 2


= 4 [3

2𝜃 − 2𝑐𝑜𝑠𝜃 −

1

4𝑠𝑖𝑛2𝜃] + 𝑐

= 6𝜃 − 8𝑐𝑜𝑠𝜃 − 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 + 𝑐

= 6 sin−1𝑥 − 2

2− 4√4𝑥 − 𝑥2 −

1

2(𝑥 − 2)√4𝑥 − 𝑥2 + 𝑐.

Example 8:

∫𝒅𝒙

√𝟒𝒙𝟐 − 𝟏

Solution

4𝑥2 = 𝑠𝑒𝑐2𝜃 ⟹ 2𝑥 = 𝑠𝑒𝑐𝜃 ⟹ 2𝑑𝑥 = 𝑠𝑒𝑐𝜃 𝑡𝑎𝑛𝜃 𝑑𝜃

𝐼 = ∫

12

𝑠𝑒𝑐𝜃 𝑡𝑎𝑛𝜃 𝑑𝜃

√𝑡𝑎𝑛2𝜃=

1

2∫ 𝑠𝑒𝑐𝜃 𝑑𝜃 =

1

2𝑙𝑛|𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃| + 𝑐.

𝐼 =1

2𝑙𝑛 |2𝑥 + √4𝑥2 − 1| + 𝑐.


Exercise 9


𝟏) ∫𝟖𝒙𝟐

(𝟏𝟒𝟒 + 𝒙𝟐)𝟐𝒅𝒙

𝟐) ∫𝒙𝟐

(𝟔𝟒 + 𝒙𝟐)𝟑𝟐

𝒅𝒙

𝟑) ∫√𝒙𝟐 − 𝟕

𝒙 𝒅𝒙

𝟒) ∫√𝟏𝟔 − 𝒙𝟐

𝟐𝒙𝟐𝒅𝒙

𝟓) ∫𝒙𝟑

√𝒙𝟐 + 𝟏𝒅𝒙

𝟔) ∫ √𝒙𝟐 − 𝟓 𝒅𝒙


a dx b

y

y

x

c

d

y

d

x

y

x

Chapter -5

Application of the Definite Integral

5-1 Area under the curve

If 𝑓(𝑥) > 0 and continuous in the interval [𝑎, 𝑏] the area between

the curve 𝑓(𝑥) and 𝑥 − 𝑎𝑥𝑖𝑠 and the straight lines 𝑥 = 𝑎 , 𝑥 = 𝑏

𝐴 = ∫ 𝑓(𝑥)𝑑𝑥 =𝑏

𝑎 ∫ 𝑦 𝑑𝑥.𝑏

𝑎

If 𝑓(𝑥) > 0 and continuous in the interval [𝑐, 𝑑] the area between

the curve 𝑔(𝑦) and 𝑦 − 𝑎𝑥𝑖𝑠 and the straight lines 𝑦 = 𝑐, 𝑦 = 𝑑

𝐴 = ∫ 𝑔(𝑦)𝑑𝑦 =𝑏

𝑎 ∫ 𝑥 𝑑𝑦.𝑑

𝑐


X

bdx a

y

x

(x)2=f2y

(x)1=f1y

The Area between two Curves

If 𝑦1 = 𝑓1(𝑥) , 𝑦2 = 𝑓2(𝑥)

𝐴 = ∫ |𝑓1(𝑥) − 𝑓2(𝑥)|𝑑𝑥 =𝑏

𝑎

∫ |𝑦1 − 𝑦2| 𝑑𝑥 𝑏

𝑎

Example 1:

Find the area under the curve 𝑦 = sin 𝑥 ,

𝑓𝑟𝑜𝑚 𝑥 = 0 𝑡𝑜 𝑥 = 𝜋

Solution:

since 𝐴 = ∫ 𝑓(𝑥)𝑑𝑥 =𝑏

𝑎 ∫ 𝑦 𝑑𝑥 𝑏

𝑎

𝑦 = 𝑠𝑖𝑛𝑥

𝑎 = 0 , 𝑏 = 𝜋


𝐴 = ∫ 𝑓(𝑥)𝑑𝑥 =𝑏

𝑎

∫ 𝑦 𝑑𝑥 𝑏

𝑎

= ∫ sin 𝑥 𝑑𝑥 𝜋

0

= −cos 𝑥|0𝜋 = − cos 𝜋 + cos 0 = +1 + 1 = 2 .

Example 2:

Find the area between the curve

𝑥 = 3𝑦 − 𝑦2 , 𝑓𝑟𝑜𝑚 𝑦 = 1 𝑡𝑜 𝑦 = 2

Solution:

since 𝐴 = ∫ 𝑔(𝑦)𝑑𝑦 =𝑑

𝑐 ∫ 𝑥 𝑑𝑦 𝑏

𝑎

𝑥 = 3𝑦 − 𝑦2

𝑐 = 1 , 𝑑 = 2

𝐴 = ∫ 𝑔(𝑥)𝑑𝑥 =𝑑

𝑐

∫ 𝑥 𝑑𝑦 𝑑

𝑐

= ∫ (3𝑦 − 𝑦2) 𝑑𝑦 2

1

=3𝑦2

2−

𝑦3

3|1

2

= (6 −8

3) − (

3

2−

1

3)

=36 − 16 − 9 + 2

6=

13

6

.

Example 3:

Find the area under the curves

𝑦 + 𝑥 = 1 𝑎𝑛𝑑 𝑦 = 𝑥2 − 1

Solution:

since


𝐴 = ∫ |𝑓1(𝑥) − 𝑓2(𝑥)|𝑑𝑥 =𝑏

𝑎

∫ |𝑦1 − 𝑦2| 𝑑𝑥 𝑏

𝑎

𝑦1 = 1 − 𝑥 , 𝑦2 = 𝑥2 − 1

To get a and b we get the point of intersection between the two curves

𝑦1 = 1 − 𝑥 = 𝑦2 = 𝑥2 − 1

1 − 𝑥 = 𝑥2 − 1 → 𝑥2 + 𝑥 − 2 = 0 → (𝑥 − 1)(𝑥 + 2) = 0

𝑎 = −2 , 𝑏 = 1

𝐴 = ∫ |𝑓1(𝑥) − 𝑓2(𝑥)|𝑑𝑥 =𝑏

𝑎

∫ |𝑦1 − 𝑦2| 𝑑𝑥 𝑏

𝑎

= ∫ (𝑥2 + 𝑥 − 2)𝑑𝑥 =𝑥3

3+

𝑥2

2− 2𝑥|

−2

11

−2

= (1

3+

1

2− 2) − (

−8

3+

4

2+ 4)

=2 + 3 − 12 + 16 − 12 − 24

6= |

−27

6| =

27

6.

5-2 The arc length of a curve

Suppose we have 𝑦 = 𝑓(𝑥) in the Cartesian form and we want to

compute the arc length 𝐴𝐵 from the curve bounded by = 𝑎 , 𝑥 = 𝑏 as

shown in the curve

∆𝐿

∆𝑥

x

y

x

𝑥 + ∆𝑥 b a

∆𝑦


(∆𝑙)2 = (∆𝑥)2 + (∆𝑦)2

∆𝑙

∆𝑥= √1 + (

∆𝑦

∆𝑥)

2

𝑎𝑡 ∆𝑥 → 0

𝑑𝑙

𝑑𝑥= √1 + (

𝑑𝑦

𝑑𝑥)

2

𝑑𝑙 = √1 + (𝑑𝑦

𝑑𝑥)

2

𝑑𝑥

𝑙 = ∫ 𝑑𝑙

𝐴𝐵

= ∫ √1 + (𝑑𝑦

𝑑𝑥)

2𝑏

𝑎

𝑑𝑥

Or

𝑙 = ∫ √1 + (𝑑𝑥

𝑑𝑦)

2𝑑

𝑐

𝑑𝑦

Example 4:

Find the arc length to the curve 𝑦 = ln(𝑠𝑒𝑐𝑥)

𝑓𝑟𝑜𝑚 𝑥 = 0 𝑡𝑜 𝑥 =𝜋

3

Solution:

𝑦′ =𝑑𝑦

𝑑𝑥=

1

𝑠𝑒𝑐𝑥. sec 𝑥 tan 𝑥 = tan 𝑥

√1 + (𝑑𝑦

𝑑𝑥)

2

= √1 + 𝑡𝑎𝑛2𝑥 = sec 𝑥


𝑙 = ∫ √1 + (𝑑𝑦

𝑑𝑥)

2𝑏

𝑎

𝑑𝑥 = ∫ sec 𝑥

𝜋3

0

𝑑𝑥 = [ln(sec 𝑥 + tan 𝑥)]0

𝜋3

= ln(2 + √3).

Example 5:

Find the arc length to the curve

𝑥 =1

4𝑦4 +

1

8𝑦2 𝑓𝑟𝑜𝑚 𝑦 = 1 𝑡𝑜 𝑦 = 2

Solution:

𝑙 = ∫ √1 + (𝑑𝑥

𝑑𝑦)

2𝑏

𝑎

𝑑𝑦

𝑑𝑥

𝑑𝑦= 𝑦3 −

1

4𝑦−3

1 + (𝑑𝑥

𝑑𝑦)

2

= 1 + (𝑦3 −1

4𝑦−3)2 = 1 + 𝑦6 −

1

2+

1

16𝑦−6

= 𝑦6 +1

2+

1

16𝑦−6 = (𝑦3 +

1

4𝑦−3)

2

𝑙 = ∫ (𝑦3 +1

4𝑦−3) 𝑑𝑦 = [

1

4𝑦4 −

1

8𝑦−2]

1

22

1

= (4 −1

32) − (

1

4−

1

8) =

128 − 1 − 8 + 4

32=

123

32.


5-3 Volume of Solid of Revolution

a- Revolution about x- axis

If the area of the curve 𝑦 = 𝑓(𝑥) bounded by 𝑥 = 𝑎 , 𝑥 = 𝑏

revolved about 𝑥-axis (axis of rotation) we find a circle its

radius is 𝑦 and its area 𝜋𝑦2 and the resultant is a vertical

circular disk its axis of rotation is x-axis and its base is a

circle with area 𝜋𝑦2 and height 𝑑𝑥 then its volume

𝑑𝑣 = 𝜋𝑦2𝑑𝑥

Then 𝑉 = ∫ 𝑑𝑣 = 𝜋 ∫ 𝑦2𝑑𝑥𝑏

𝑎

b-Rotation about y- axis

Similarly, if the area 𝑥 = 𝑔(𝑦) bounded by 𝑦 = 𝑐, 𝑦 = 𝑑 rotate about

y-axis then its volume of rotation

𝑉 = 𝜋 ∫ 𝑥2𝑑𝑦𝑑

𝑐


Example 6:

Find the volume of revolution of the area bounded by the curves

𝑥 = 0 , 𝑦 = 0, 𝑥 + 𝑦 = 1

About a) x-axis b) y-axis

Solution:

a) About x-axis

The slide is 𝑦𝑑𝑥

Then

𝑉 = ∫ 𝜋𝑦2𝑑𝑥1

0

= 𝜋 ∫(1 − 𝑥)2𝑑𝑥

1

0

= 𝜋 ∫(1 − 2𝑥 + 𝑥2) 𝑑𝑥 = 𝜋 [𝑥 − 𝑥2 +𝑥3

3]

0

11

0

= 𝜋 (1

3− 0) =

𝜋

3.

b) About y-axis

The slide is 𝑥𝑑𝑦

Then 𝑉 = ∫ 𝜋𝑥2𝑑𝑦1

0= 𝜋 ∫ (1 − 𝑦)2𝑑𝑦

1

0

= 𝜋 ∫(1 − 2𝑦 + 𝑦2) 𝑑𝑦 = 𝜋 [𝑦 − 𝑦2 +𝑦3

3]

0

11

0

= 𝜋 (1

3− 0) =

𝜋

3.

Example 7:

Find the volume of revolution of the area bounded by the curves

𝑦 = 𝑥2, 𝑥 = 4, 𝑥 = 2, 𝑦 = 0 about 𝑥 − 𝑎𝑥𝑖𝑠

X

dx

y

y+x=1

Y


Solution

The slide is 𝑦𝑑𝑥

Then

𝑉 = ∫ 𝜋𝑦2𝑑𝑥4

2

= 𝜋 ∫ 𝑥4𝑑𝑥 = 𝜋 [𝑥5

5]

2

44

2

= 𝜋 [45 − 25

5] =

(32). 31

5=

992

5.

X

Y

x=2 dx

2y=x

X=4


Exercise 10

Application of the definite Integration

Find the area between the following curves:

1 − 𝑦 = 𝑥2 , 𝑦 = 𝑥.

2 − 𝑦 = 2𝑥 − 𝑥2 , 𝑦 = −3.

3 − 𝑦 = 𝑥4 − 2𝑥2 , 𝑦 = 2𝑥2.

4 − 𝑦 = 𝑥2 , 𝑦 = −𝑥, 𝑥 = −2, 𝑥 = 4.

5 − 𝑦 = 𝑒𝑥 , 𝑦 = 𝑒−𝑥 , 𝑥 = 0 , 𝑥 = 2.

Calculate the arc length for the following curves in the

indicated interval:

6 − 𝑦 = ln(sec 𝑥) , [0,𝜋

3].

7 − 𝑦 = 𝑥2 , [0, 1].

8 − 𝑦 = ln(x + √𝑥2 − 1) , [1, 2].

9 − 𝑦 = ln(1 − 𝑥2) , [1

4,

3

4].

Find the volume of revolution the area bounded by the

following curves about x-axis:

10 − 𝑦 = 𝑥4 , 𝑥 = 2 , 𝑦 = 0.

11 − 𝑦 = 𝑥2 + 1 , 𝑥 = 1 , 𝑥 = −1 , 𝑦 = 0.


Appendix

Graphs, symmetries and periodicities of sin, cos and tan

The graphs of the three major functions are very important and you need to learn the characteristics of each.

The sine function

1. This graph is continuous (there are no breaks). 2. The range is -1 ≤ sin θ ≤ +1. 3. The shape of the graph from θ = 0 to θ = 2π is repeated every

2π radians. 4. This is called a periodic or cyclic function and the width of the

repeating pattern that is measured on the horizontal axis, is called the period. The sine wave has a period of 2π, a maximum value of +1, and a minimum value of -1.

5. The greatest value of the sine wave is called the amplitude.

The cosine function

1. This graph is continuous. 2. The range is -1 ≤ cos θ ≤ +1. 3. It has a period of 2π. 4. The shape is the same as the sine wave but displaced a

distance of π ⁄ 2 to the left on the horizontal axis. This is called a phase shift.


The tan function

The tan function is found using:

It therefore follows that tan θ = 0, when sin θ = 0, and tan θ is undefined when cos θ = 0.

1. This graph is continuous, but is undefined when

2. The range of values for tan θ is unlimited.

3. It has a period of π.


Department : Basic Science

Subject : MATH.I

Code : MATH 101

Academic year : 2016-2017

Sem.(fall) stSemester : 1

Time Allowed : 90 Min.

Total mark : 30

TERM EXAM MATHEMATICS -MID

Attempt all Questions Number of Questions: 6 Number of Pages: 1

1- Find 𝑑𝑦

𝑑𝑥 if (15 marks)

a) y = cos3 (1

2x2 − 5x + 1).

b) 𝑦 = logcos x (sinhx).

c) y =sec x

1−tan x− 4 csch x.

d) x4y4 − cos−1(4x) = tan(3x2).

e) y = tanh−1(x2 − 5) . esin 2x _____________________________________________________

2- Prove that d

dx(sin−1x) =

1

√1−x2. (2 marks)

____________________________________________________

3- If y2 = x2(3 + x), Prove that:y (d2y

dx2) + (dy

dx)

2= 3 + 3x. (2 marks)

______________________________________________________

4- Find d2y

dx2 when x = a sinht , y = b cosht (4 marks)

Where 𝑎, 𝑏 are constants.

______________________________________________________

5- Find the Taylor expansion series for the function:

(1 + x)5 centered at a = 1. (3 marks)

______________________________________________________

6- Find the following limits: (4 marks)

a) lim 𝑥→0

(𝑥)𝑥2. 𝑏) lim

𝑥→1(1 − 𝑥) tan

𝜋𝑥

2.


Modern University For Information and Technology Academic year202/2021

Department: Engineering Mathematics

and Physics Semester : Fall

Specialization: Prepartory Exam Date: 11/3/2021

Final Written Examination Subject : Math I Code: MATH 101

Examiner: Dr.Said Gouda

Dr.Mona Mehanna Time Allowed: 2 hours

Question 1, (10Marks)

Find 𝑑𝑦

𝑑𝑥 if

𝑦 = 𝑡𝑎𝑛ℎ−1(3𝑥2 − 1) . 3ln √3𝑥3

Solution:

𝑦′ = 𝑡𝑎𝑛ℎ−1(3𝑥2 − 1) . 3ln √3𝑥3

(ln 3) 1

3𝑥

+ 3ln √3𝑥3

.6𝑥

1 − (3𝑥2 − 1)2

a) 𝑥3𝑦2 = 𝑠𝑖𝑛2(𝑥𝑦) − 𝑐𝑜𝑠𝑒𝑐−1 (

1

𝑥𝑦)

Solution: 3 𝑥2𝑦2 + 2𝑥3𝑦𝑦′

= 2𝑠𝑖𝑛 (𝑥𝑦) cos(𝑥𝑦) (𝑥𝑦′ + 𝑦) + 1

(1

𝑥𝑦) √(1

𝑥𝑦)2

− 1

(1

𝑥𝑦)

2

(𝑥𝑦′ + 𝑦)

c) If 3𝑦3 = 2𝑥2 , Prove that:

𝑦 (𝑑2𝑦

𝑑𝑥2) + 2 (𝑑𝑦

𝑑𝑥)

2=

4

9𝑦

Solution:

9𝑦2𝑦′ = 4𝑥 → 18𝑦(𝑦′)2 + 9𝑦2𝑦′′ = 4 → 𝑦 (𝑑2𝑦

𝑑𝑥2) + 2 (𝑑𝑦

𝑑𝑥)

2=

4

9𝑦

Question 2, (10 Marks)

a) Find 𝑑𝑦

𝑑𝑥 for y = logtan 𝑥(2𝑥−3 + 10)

Number of Pages:1 Number of Questions: 3 Attempt all Questions


Solution: y = logtan 𝑥(2𝑥−3 + 10) =ln(2𝑥−3+10)

ln tan 𝑥

y′ = ln(tan 𝑥).

−6𝑥−4

(2𝑥−3 + 10)− ln(2𝑥−3 + 10).

𝑠𝑒𝑐2 𝑥tan 𝑥

(ln( tan 𝑥))2

b) lim𝑥→0 (sec 2𝑥)4 csc 𝑥

Solution: lim𝑥→0

(sec 2𝑥)4 csc 𝑥 = (1)∞

𝑦 = (sec 2𝑥)4 csc 𝑥 → ln 𝑦 = 4 csc 𝑥 ln sec 2𝑥 =4 ln sec2 𝑥

sin 𝑥

lim𝑥→0


4 ln sec2 𝑥

sin 𝑥=

0

0 using L'hopital rule


8 tan 2𝑥

cos 𝑥= 0

𝑦 = 𝑒0 = 1 c) Find the Maclaurin expansion ( only three non-zeroes terms) for

𝑓(𝑥) = 𝑥𝑐𝑜𝑠 (1

2𝑥2)

Solution:

Maclaurin expansion cos 𝑥 = 1 − 𝑥2

2!+

𝒙𝟒

𝟒!+ ⋯

cos (𝑥2

2) = 1 −

(𝑥2

2)

2

2!+

(𝑥2

2)

𝟒

𝟒!+ ⋯

x cos (𝑥2

2) = 𝑥 −

𝑥 (𝑥2

2)

2

2!+

𝑥 (𝑥2

2)

𝟒

𝟒!+ ⋯

x cos (𝑥2

2) = 𝑥 −

𝑥5

8+

𝑥𝟗

𝟑𝟖𝟒 + ⋯

Question 3, (20 Marks) Evaluate (i) then use it in (ii):

(1) (𝑖) ∫ 𝑠𝑒𝑐2𝑥 𝑑𝑥 , (𝑖𝑖) ∫ 𝑡𝑎𝑛6𝑥 𝑑𝑥

Solution:

(𝑖) ∫ 𝑠𝑒𝑐2𝑥 𝑑𝑥 = tan 𝑥 + 𝐶


(𝑖𝑖) ∫ 𝑡𝑎𝑛6𝑥 𝑑𝑥 = ∫ 𝑡𝑎𝑛2 𝑥 . 𝑡𝑎𝑛4𝑥 𝑑𝑥 = ∫(𝑠𝑒𝑐2 𝑥 − 1)𝑡𝑎𝑛4 𝑥 𝑑𝑥

=𝑡𝑎𝑛5𝑥

5− ∫(𝑠𝑒𝑐2 𝑥 − 1)𝑡𝑎𝑛2 𝑥 𝑑𝑥

=𝑡𝑎𝑛5𝑥

5−

𝑡𝑎𝑛3𝑥

3+ ∫(𝑠𝑒𝑐2 𝑥 − 1) 𝑑𝑥

=𝑡𝑎𝑛5𝑥

5−

𝑡𝑎𝑛3𝑥

3+ tan 𝑥 − 𝑥 + 𝐶

(2)(𝑖) ∫ 𝑐𝑜𝑠−13 𝑥. 𝑠𝑖𝑛𝑥. 𝑑𝑥, (𝑖𝑖) ∫

𝑠𝑖𝑛3𝑥

√𝑐𝑜𝑠𝑥3 𝑑𝑥

Solution:

(𝑖) ∫ 𝑐𝑜𝑠−13 𝑥. 𝑠𝑖𝑛𝑥. 𝑑𝑥 = −

3

2𝑐𝑜𝑠

23𝑥 + 𝐶

(𝑖𝑖) ∫𝑠𝑖𝑛3𝑥

√𝑐𝑜𝑠𝑥3 𝑑𝑥 = ∫(1 − 𝑐𝑜𝑠2𝑥) 𝑐𝑜𝑠

−13 𝑥 sin 𝑥 𝑑𝑥

= ∫ 𝑐𝑜𝑠−13 𝑥 sin 𝑥 𝑑𝑥 − ∫ 𝑐𝑜𝑠

53𝑥 sin 𝑥 𝑑𝑥

= −3

2𝑐𝑜𝑠

23𝑥 +

3

8 𝑐𝑜𝑠

83𝑥 + 𝐶

(3)(𝑖) ∫ 𝑒−5𝑥 𝑑𝑥, (𝑖𝑖) ∫ 𝑥2 𝑒−5𝑥 𝑑𝑥

Solution:

(𝑖) ∫ 𝑒−5𝑥 𝑑𝑥 =𝑒−5𝑥

−5 + 𝐶

(𝑖𝑖) ∫ 𝑥2 𝑒−5𝑥 𝑑𝑥 = 𝑥2𝑒−5𝑥

−5− 2𝑥

𝑒−5𝑥

25+ 2

𝑒−5𝑥

−125+ 𝐶

(4)(𝑖) ∫1

𝑥 − 1𝑑𝑥, (𝑖𝑖) ∫

3𝑥2 + 𝑥 + 1

(𝑥 − 1)(𝑥2 + 4)𝑑𝑥

Solution:

(𝑖) ∫1

𝑥 − 1𝑑𝑥 = ln(𝑥 − 1) + 𝐶

(𝑖𝑖) ∫3𝑥2 + 𝑥 + 1

(𝑥 − 1)(𝑥2 + 4)𝑑𝑥 = ∫

𝐴

𝑥 − 1𝑑𝑥 + ∫

𝐵𝑥 + 𝐶

𝑥2 + 4𝑑𝑥

= 𝐴 ln(𝑥 − 1) +𝐵

2ln(𝑥2 + 4) +

𝐶

2tan−1

𝑥

2+ 𝐾


3𝑥2 + 𝑥 + 1

(𝑥 − 1)(𝑥2 + 4)=

𝐴

𝑥 − 1+

𝐵𝑥 + 𝐶

𝑥2 + 4

𝐴 = 1 3𝑥2 + 𝑥 + 1 = 𝑥2 + 4 + (𝑥 − 1)(𝐵𝑥 + 𝐶) , 3 = 1 + 𝐵 → 𝐵 = 2 , 1 = 4 − 𝐶 → 𝐶 = 3

∫3𝑥2 + 𝑥 + 1

(𝑥 − 1)(𝑥2 + 4)𝑑𝑥 = ln(𝑥 − 1) + ln(𝑥2 + 4) +

3

2tan−1

𝑥

2+ 𝐾

(5)(𝑖) ∫ 𝑠𝑖𝑛2𝑥 𝑑𝑥, (𝑖𝑖) ∫𝑥2

√121 − 𝑥2𝑑𝑥

Solution:

(𝑖) ∫ 𝑠𝑖𝑛2𝑥 𝑑𝑥 =1

2∫(1 − 𝑐𝑜𝑠2𝑥)𝑑𝑥 =

1

2(𝑥 −

sin 2𝑥

2) + 𝐾

(𝑖𝑖) ∫𝑥2

√121 − 𝑥2𝑑𝑥 → 𝑥

= 11 sin 𝑡 , 𝑑𝑥 = 11 cos 𝑡 𝑑𝑡, √121 − 𝑥2 = 11 cos 𝑡

∫𝑥2

√121 − 𝑥2𝑑𝑥 = ∫

121 𝑠𝑖𝑛2𝑡

11 𝑐𝑜𝑠𝑡 11 𝑐𝑜𝑠𝑡 𝑑𝑡 =

121

2(𝑡 −

sin 2𝑡

2) + 𝐾

𝑥 = 11 sin 𝑡 → 𝑡 = sin−1𝑥

11→ sin 2𝑡 = 2 sin 𝑡 cos 𝑡 = 2

𝑥

11

√121 − 𝑥2

11

∫𝑥2

√121 − 𝑥2𝑑𝑥 =

121

2(sin−1

𝑥

11−

𝑥

11

√121 − 𝑥2

11) + 𝐾

∫𝑥2

√121 − 𝑥2𝑑𝑥 = (

121

2sin−1

𝑥

11−

𝑥

2

√121 − 𝑥2

1) + 𝐾

Best wishes, Dr.Said Gouda, Dr.Mona Mehanna



1- Find 𝑑𝑦

𝑑𝑥 if (12 marks) each (3 marks)

a) 𝑦 = cos4(𝑐𝑜𝑡√𝑥).

b) 𝑦 = (𝑡𝑎𝑛2𝑥)sin 3𝑥 .

c) 𝑦 = 𝑡𝑎𝑛−1(𝑥3 − 1) . 3ln 2𝑥

d) Find 𝑑𝑦

𝑑𝑥 for the following implicit equation:

𝑥3𝑦2 = 𝑠𝑖𝑛ℎ−1(𝑥𝑦) + 4.

______________________________________________________

2- Evaluate the following: (6 marks) each (2 marks) a) If 4𝑦3 = 3𝑥4 , Prove that:

𝑦(𝑦′′) + 2(𝑦′)2 =3𝑥2

𝑦

b) Evaluate the limits: lim

𝑥→0[𝑥. cot (2𝑥)]

c) Find the Maclurin series for 𝒆𝒙 𝒂𝒏𝒅 𝒄𝒐𝒔𝒙 . Then find the first 3 non

zero terms of Maclurin series (expansion) for 𝒆𝒙 . 𝒄𝒐𝒔𝒙 ______________________________________________________

3- Evaluate the following integrals: (8 marks) each (2 marks)

𝑎) ∫3𝑥6 + 2𝑥2 − 5

𝑥4𝑑𝑥.

𝑏) ∫5𝑥2

√4 + 2𝑥3𝑑𝑥.

Engineering Mathematics and Physics Department

Math1 Code: Math 101

Final Exam: 9-01-2016

Time Allowed: 2 Hours

Faculty of Engineering

Academic year: 2016-2017

Semester: Fall

Examiners: Dr. Said Gouda

Dr.Mona Mehanna

Answer All questions No. Of questions: 4 Total Mark: 40


𝑐) ∫ sec 𝑥 𝑑𝑥.

𝑑) ∫ cos(4𝑥). cos(6𝑥) 𝑑𝑥.

______________________________________________________

4- Evaluate the integral: (6 marks) each ( 2 marks)

a) ∫1

𝑥(1+𝑙𝑛2𝑥)𝑑𝑥.

b) ∫ 𝑠𝑖𝑛1

3𝑥. cos3 𝑥 𝑑𝑥

c) ∫ 𝑠𝑒𝑐3𝑥. tan 𝑥 𝑑𝑥

______________________________________________________

5- Evaluate the integral: (6 marks) each ( 2 marks)

a) ∫ln 𝑥

𝑥2𝑑𝑥

b) ∫sin 𝑥

cos 𝑥(1+𝑐𝑜𝑠2𝑥)𝑑𝑥

c) ∫𝑥2

√9−𝑥2𝑑𝑥 .

6- Find the area between the following curves: (2 marks)

𝑦 + 𝑥 = 1 𝑎𝑛𝑑 𝑦 = 𝑥2 − 1

WITH MY BEST WISHES,

Dr. Said Gouda – Dr. Mona Samir


Department: Basic Science

Subject : MATH.I

Code : MATH 101

Date :27-05-2017

Academic year: 2016-2017

Semester :Spring 2017

Time Allowed:120 Min.

Total mark : 40

FINALEXAM MATHEMATICS.I


1- Find 𝑑𝑦

𝑑𝑥 if (12 marks) each (3 marks)

a) 𝑦 = (√𝑥−1

𝑥 ) (𝑥3 +

4

𝑥3 ) .

b) 𝑦 = (𝑐𝑜𝑠2𝑥)1

𝑥2 .

c) 𝑦 = 𝑡𝑎𝑛ℎ−1(3𝑥−2 − 5) . 5ln 2𝑥

d) Find 𝑑𝑦

𝑑𝑥 for the following implicit equation:

1

𝑦+ 𝑥3𝑦2 = 𝑐𝑜𝑠ℎ−1(𝑥𝑦) .

2- Evaluate the following: (8 marks) each (2 marks)

a- Show that (1 − 𝑥2)𝑦′′ − 𝑥𝑦′ = 0.

If If 𝑦 = cos−1 𝑥 .

b- lim𝑥→0

(sec 𝑥)2 csc 𝑥

c- Find the Maclaurin expansion ( only three non-zeroes terms) for

𝑓(𝑥) = 𝑥𝑐𝑜𝑠 (1

2𝑥2).

d- Find 𝑑𝑦

𝑑𝑥 for y = logtan 𝑥(2𝑥−3 + 10)

3- Evaluate the following integrals: (8 marks) each (2 marks)


𝑎) ∫1

√𝑥(𝑥 +

1

𝑥)

2

𝑑𝑥.

𝑏) ∫1

𝑥𝑙𝑛3𝑥𝑑𝑥.

𝑐) ∫𝑠𝑒𝑐2 𝑥 𝑑𝑥

√𝑡𝑎𝑛2 𝑥3

.

d) ∫ sin(4𝑥). sin(6𝑥) 𝑑𝑥

4- Evaluate the following integrals: (4 marks) each ( 2 marks)

a) ∫(3𝑠𝑖𝑛2 𝑥 + 8 𝑐𝑜𝑠2 𝑥)−1 𝑑𝑥

b) ∫ 𝑡𝑎𝑛5 𝑥 𝑠𝑒𝑐3 𝑥 𝑑𝑥

______________________________________________________

5- Evaluate the integral: (6 marks) each( 2 marks)

a) ∫ 𝑐𝑜𝑠√𝑥 𝑑𝑥

b) ∫cos x

sin x(1+sin2x)dx

c) ∫1

√ 16−𝑥2𝑑𝑥

_____________________________________________________

6- Find the area between the following curves: (2 marks)

𝑦 + 2𝑥 − 1 = 0 𝑎𝑛𝑑 𝑦 − 𝑥2 + 2 = 0

WITH MY BEST WISHES Dr. Said Gouda


Engineering Mathematics and Physics

Department


Final Exam: 30 / 5 / 2019

Time Allowed: 2 Hours Faculty of Engineering

Academic year: 2018 / 2019

Semester : Spring

Examiners: Dr. Mona Mehanna

Dr. Marwa Hani Maneea

Answer All questions No. Of questions: 4 Total Mark: 40

Question 1

(a)Find 𝒅𝒚

𝒅𝒙

(𝒊)𝒕𝒂𝒏 (𝟐𝒙

𝟑𝒚) − 𝒔𝒊𝒏𝒉−𝟏(𝒆𝒚𝟔

) = 𝒄𝒐𝒕𝒉(𝟖𝒙)

(𝒊𝒊)𝒚 = (√𝒄𝒐𝒔𝟒(𝟑𝟐𝒙𝟐) + 𝟓

𝟕) + (𝒙𝟖. 𝒍𝒐𝒈𝟐𝟓𝒙)

(b) Evaluate:

(𝒊) 𝐥𝐢𝐦𝒙→𝟎

(𝒄𝒐𝒔𝒙)𝒄𝒐𝒔𝒆𝒄𝒙 (𝒊𝒊) 𝐥𝐢𝐦𝒙→

𝝅𝟒

𝒔𝒊𝒏𝒙 − 𝒄𝒐𝒔𝒙

(𝒙 −𝝅𝟒)

Question 2

(i)A region in the xy-plane is bounded by the following curves:

𝒚 = 𝟐 − 𝒙𝟐 , 𝒚 = −𝒙

(a) (a)find the area of this region.

(b) (b)Compute the volume 𝑽𝒙 of the solid generated by rotating this region one complete

revolution about the x-axis.

(c) (ii)Find the first three non zero terms of Taylor series for the function:

(d) 𝒚 = √𝟏 + 𝒔𝒊𝒏𝟐𝒙 , 𝒂 =𝝅

𝟒.

(e) Question3

(𝒊) ∫ 𝒔𝒆𝒄𝒉𝟐𝒙 𝒅𝒙 (𝒊𝒊) ∫(𝟓𝒙𝟐 + 𝟐)𝟔. 𝒙 𝒅𝒙 (𝒊𝒊𝒊) ∫𝟏

𝒙√𝟒𝒙𝟐−𝟏 𝒅𝒙 (𝒊𝒗) ∫ 𝒙. 𝒆𝒙𝟐

𝒅𝒙

Question4

(𝒊) ∫ 𝒙𝟔 𝒍𝒏𝒙𝟒 𝒅𝒙 (𝐢𝐢) ∫ 𝒄𝒐𝒔𝟑𝒙 𝒔𝒊𝒏𝟒𝒙 𝒅𝒙 (𝒊𝒊𝒊) ∫𝒙𝟓+𝟑

𝒙𝟑−𝒙 𝒅𝒙 (𝒊𝒗) ∫

𝒅𝒙

√𝟐𝟓+𝒙𝟐 Best Wishes

12

8

8

12


Engineering Mathematics and Physics

Department


Mid Term Exam: 2 / 4 / 2019

Time Allowed: 1 Hour Faculty of Engineering

Academic year: 2018 / 2019

Semester: Spring

Examiners: Dr. Mona Mehanna

Dr.MarwaHani Maneea

Answer All questions No. Of questions:3 Total Mark: 25

Question 1

Find 𝒅𝒚

𝒅𝒙

(𝒊)𝒚 = √𝒔𝒆𝒄−𝟏(𝒆−𝟒𝒙𝟐) + 𝒄𝒐𝒔𝒉𝟓𝒙

𝟑 (𝒊𝒊)𝒚 = 𝒍𝒐𝒈𝒔𝒊𝒏𝒙(𝒙𝟑)

(𝒊𝒊)𝒚 = (𝒕𝒂𝒏𝒙)𝒔𝒊𝒏𝒉−𝟏(𝟐𝒙) (𝒊𝒊𝒊)𝒚 =𝟖𝒔𝒆𝒄𝒙

𝒍𝒏𝒙𝟓

Question 2

Find the first three non zero terms of Maclaurin series for the function:

(f)

𝑭(𝒙) =𝒆𝟑𝒙

√𝟏 + 𝟐𝒙

(g)

Question3

Evaluate:

(𝒊) 𝐥𝐢𝐦𝒙→𝟏

(𝒙)𝟓

(𝒙−𝟏)⁄ (𝒊𝒊) 𝐥𝐢𝐦

𝒙→𝟎

𝒕𝒂𝒏𝟓𝒙

𝒕𝒂𝒏𝟐𝟎𝒙

Best Wishes


Modern University For Information and Technology Academic year 2020/2021

Department: Engineering Mathematics and

Physics Department Semester Fall

Specialization: Preparatory Exam Date 29/11/2020

Solution Mid Term Examination

Subject : MathI Code Math101

Examiner: Dr.Said Gouda

Dr.Mona Mehanna TimeAllowed: 60 minutes


Find 𝒅𝒚

𝒅𝒙

(𝒊) 𝒔𝒆𝒄𝒉−𝟏(𝒙𝟐𝒚) = 𝒍𝒏(𝒔𝒊𝒏𝒉𝒙) + 𝒄𝒐𝒔𝒉(𝟐𝒚)

Solution

𝟐𝒙𝒚 + 𝒙𝟐𝒚′

(𝒙𝟐𝒚)√𝟏 − (𝒙𝟐𝒚)𝟐

=𝒄𝒐𝒔𝒉𝒙

𝒔𝒊𝒏𝒉𝒙+ 𝒔𝒊𝒏𝒉(𝟐𝒚)𝟐𝒚 𝒍𝒏𝟐 𝒚′

𝒚′[

𝒙𝟐

(𝒙𝟐𝒚)√𝟏−(𝒙𝟐𝒚)𝟐

− 𝟐𝒚 𝒍𝒏𝟐 𝒔𝒊𝒏𝒉(𝟐𝒚)

]= 𝒕𝒂𝒏𝒉 𝒙 − 𝟐𝒙𝒚

(𝒙𝟐𝒚)√𝟏−(𝒙𝟐𝒚)𝟐

𝒚′ =(𝒙𝟐𝒚)√𝟏 − (𝒙𝟐𝒚)

𝟐 𝒕𝒂𝒏𝒉𝒙 − 𝟐𝒙𝒚

𝒙𝟐 − (𝒙𝟐𝒚)√𝟏 − (𝒙𝟐𝒚)𝟐

𝟐𝒚 𝒍𝒏𝟐 𝒔𝒊𝒏𝒉(𝟐𝒚)

(𝒊𝒊) 𝒚 = 𝒍𝒐𝒈𝒙(𝟕𝒙𝟓)

Number of Pages: 1 Number of Questions: 3 Attempt all Questions


Solution

𝒚 = 𝒍𝒐𝒈𝒙(𝟕𝒙𝟓) =

𝒍𝒏(𝟕𝒙𝟓)

𝒍𝒏 𝒙

𝒚′ =

(𝒍𝒏 𝒙) 𝟑𝟓𝒙𝟒

(𝟕𝒙𝟓)

−𝒍𝒏(𝟕𝒙𝟓

)𝒙

(𝒍𝒏 𝒙)𝟐=

𝟑𝟓𝒙𝟒 𝒍𝒏 𝒙 − 𝟕𝒙𝟒𝒍𝒏(𝟕𝒙𝟓)

𝟕𝒙𝟓(𝒍𝒏 𝒙)𝟐

𝒚′ =𝟓

𝒙(𝒍𝒏 𝒙) −

𝒍𝒏(𝟕𝒙𝟓)

𝒙(𝒍𝒏 𝒙)𝟐

(𝒊𝒊𝒊) 𝒚 =𝒆𝒄𝒐𝒔𝒆𝒄−𝟏(𝟐𝒙). 𝒕𝒂𝒏𝟑(𝒙𝟓)).

√𝟏−𝒙

Solution

Take ln of both sides

𝒍𝒏𝒚 = 𝒍𝒏(𝒆𝒄𝒐𝒔𝒆𝒄−𝟏(𝟐𝒙). 𝒕𝒂𝒏𝟑(𝒙𝟓))

√𝟏 − 𝒙)

𝒍𝒏𝒚 = 𝒍𝒏 (𝒆𝒄𝒐𝒔𝒆𝒄−𝟏(𝟐𝒙)) + 𝟑𝒍𝒏(𝒕𝒂𝒏𝒙𝟓) −𝟏

𝟐𝒍𝒏(𝟏 − 𝒙)

Differentiate with respect to x and 𝒍𝒏 (𝒆𝒄𝒐𝒔𝒆𝒄−𝟏(𝟐𝒙)) =

𝒄𝒐𝒔𝒆𝒄−𝟏(𝟐𝒙)


𝒚′

𝒚=

−𝟐

(𝟐𝒙)√𝟏 − (𝟐𝒙)𝟐+

𝟏𝟓𝒙𝟒𝒔𝒆𝒄𝟐𝒙𝟓

𝒕𝒂𝒏𝒙𝟓+

𝟏

𝟏 − 𝒙

𝒚′ = 𝒚−𝟐

(𝟐𝒙)√𝟏 − (𝟐𝒙)𝟐+

𝟏𝟓𝒙𝟒𝒔𝒆𝒄𝟐𝒙𝟓

𝒕𝒂𝒏𝒙𝟓+

𝟏

𝟏 − 𝒙


Find the first four non zero terms of Taylor series for the function:

𝑭(𝒙) = √𝟐 + 𝟐𝒙𝟑

𝒂 = 𝟐.

Solution

𝒇(𝒙) = (𝟐 + 𝟐𝒙)𝟏𝟑 𝒇(𝟐) = (𝟐 + 𝟐(𝟐))

𝟏𝟑 (𝟔)

𝟏𝟑

𝒇′(𝒙) =𝟐

𝟑(𝟐

+ 𝟐𝒙)−𝟐𝟑

𝒇′(𝒙)

=𝟐

𝟑(𝟐 + 𝟐(𝟐))−

𝟐𝟑

𝟐

𝟑(𝟔)

−𝟐𝟑

𝒇′′(𝒙) =−𝟖

𝟗(𝟐

+ 𝟐𝒙)−𝟓𝟑

𝒇′′(𝒙)

=−𝟖

𝟗(𝟐 + 𝟐(𝟐))−

𝟓𝟑

−𝟖

𝟗(𝟔)

−𝟓𝟑

𝒇′′′(𝒙) = 𝟖𝟎

𝟐𝟕(𝟐

+ 𝟐𝒙)−𝟖𝟑

𝒇′′′(𝒙)

=𝟖𝟎

𝟐𝟕(𝟐 + 𝟐(𝟐))−

𝟖𝟑

𝟖𝟎

𝟐𝟕(𝟔)

−𝟖𝟑

By Taylor expansion about a=2

𝒇(𝒙) = 𝒇(𝒂) + 𝒇′ (𝒂)(𝒙 − 𝒂) +𝒇′′(𝒂)

𝟐!(𝒙 − 𝒂)𝟐 +

𝒇′′′(𝒂)

𝟑!(𝒙 − 𝒂)𝟑 + ⋯

𝒇(𝒙) = 𝒇(𝟐) + 𝒇′ (𝟐)(𝒙 − 𝟐) +𝒇′′(𝟐)

𝟐!(𝒙 − 𝟐)𝟐 +

𝒇′′′(𝟐)

𝟑!(𝒙 − 𝟐)𝟑 + ⋯


𝒇(𝒙) = (𝟔)𝟏𝟑 +

𝟐

𝟑(𝟔)

−𝟐𝟑

(𝒙 − 𝟐) +−𝟖

𝟏𝟖(𝟔)

−𝟓𝟑

(𝒙 − 𝟐)𝟐

+𝟖𝟎

𝟐𝟕(𝟔)(𝟔)

−𝟖𝟑

(𝒙 − 𝟐)𝟑 + ⋯

Question 3, (6Marks)

Evaluate:


(𝟏 − 𝟔𝒙)𝟏𝟐𝒙

, (𝒊𝒊) 𝐥𝐢𝐦 𝒙→𝟎

𝒄𝒐𝒔𝟒𝒙 − 𝒄𝒐𝒔𝟓𝒙

𝟏 − 𝒄𝒐𝒔𝟑𝒙.

Solution


(𝟏 − 𝟔𝒙)𝟏𝟐

𝒙

= (1)∞ undetermined value

𝒍𝒆𝒕 𝒚 = (𝟏 − 𝟔𝒙)𝟏𝟐𝒙 𝒕𝒂𝒌𝒆 𝒍𝒏 𝒐𝒇 𝒃𝒐𝒕𝒉 𝒔𝒊𝒅𝒆𝒔

𝒍𝒏𝒚 = 𝒍𝒏(𝟏 − 𝟔𝒙)𝟏𝟐𝒙


𝒍𝒏𝒚 = 𝐥𝐢𝐦𝒙→𝟎

𝟏𝟐

𝒙𝒍𝒏(𝟏 − 𝟔𝒙) = 𝐥𝐢𝐦

𝒙→𝟎

𝟏𝟐𝒍𝒏(𝟏 − 𝟔𝒙)

𝒙

𝒍𝒏𝒚 =0

0 𝑏𝑦 𝐿′ℎ𝑜𝑝𝑖𝑡𝑎𝑙𝑟𝑢𝑙𝑒

𝒍𝒏𝒚 = 𝐥𝐢𝐦𝒙→𝟎

(𝟏𝟐𝒍𝒏(𝟏 − 𝟔𝒙))′

𝑥′= 𝐥𝐢𝐦

𝒙→𝟎

(𝟏𝟐 ( −𝟔 ))

(𝟏 − 𝟔𝒙)= −72


𝒍𝒏𝒚 = −72 𝑡ℎ𝑒𝑛 𝑦 = 𝑒−72 𝐥𝐢𝐦𝒙→𝟎

(𝟏 − 𝟔𝒙)𝟏𝟐𝒙

= = 𝑒−72

(𝒊𝒊) 𝐥𝐢𝐦 𝒙→𝟎

𝒄𝒐𝒔𝟒𝒙 − 𝒄𝒐𝒔𝟓𝒙

𝟏 − 𝒄𝒐𝒔𝟑𝒙

Solution

𝐥𝐢𝐦 𝒙→𝟎

𝒄𝒐𝒔𝟒𝒙−𝒄𝒐𝒔𝟓𝒙

𝟏−𝒄𝒐𝒔𝟑𝒙=

0

0 by L'hopital rules

𝐥𝐢𝐦 𝒙→𝟎

(𝒄𝒐𝒔𝟒𝒙 − 𝒄𝒐𝒔𝟓𝒙)′

(𝟏 − 𝒄𝒐𝒔𝟑𝒙)′= 𝐥𝐢𝐦

𝒙→𝟎

−𝟒𝒔𝒊𝒏𝟒𝒙 + 𝟓𝒔𝒊𝒏𝟓𝒙

𝟑𝒔𝒊𝒏𝟑𝒙

=0

0

by L'hopital rules

= 𝐥𝐢𝐦 𝒙→𝟎

(−𝟒𝒔𝒊𝒏𝟒𝒙 + 𝟓𝒔𝒊𝒏𝟓𝒙)′

(𝟑𝒔𝒊𝒏𝟑𝒙)′

= 𝐥𝐢𝐦 𝒙→𝟎

−𝟏𝟔𝒄𝒐𝒔𝟒𝒙 + 𝟐𝟓𝒄𝒐𝒔𝟓𝒙

𝟗𝒄𝒐𝒔𝟑𝒙=

9

9= 1

GOOD LUCK





References

1. Calculus Early Transcendental Functions Robert T.Smith -

Roland B. Minton, 4th Edition.

2. Mathematics for Calculus, by J. Stewart, L. Redlin and S.

Watson. Published by Brooks/Cole. ISBN: 9781305701618

(7th Edition 2015).

3. Salas and Hille’s, “Calculus”, Sixth Edition.

4.Calculus, Swokowski, classic edition.

Documents

Lectures Notes of Mathematics I