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)
MATHEMATICS-I Page 1
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.
MATHEMATICS-I Page 2
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
MATHEMATICS-I Page 3
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
MATHEMATICS-I Page 4
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
MATHEMATICS-I Page 5
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, … ..
MATHEMATICS-I Page 6
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.
MATHEMATICS-I Page 7
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
MATHEMATICS-I Page 8
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
MATHEMATICS-I Page 9
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
MATHEMATICS-I Page 10
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
MATHEMATICS-I Page 11
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
MATHEMATICS-I Page 12
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].
MATHEMATICS-I Page 13
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.
MATHEMATICS-I Page 14
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
MATHEMATICS-I Page 15
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 ".
MATHEMATICS-I Page 16
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
MATHEMATICS-I Page 17
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
MATHEMATICS-I Page 18
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:
MATHEMATICS-I Page 19
(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.
MATHEMATICS-I Page 20
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 =
𝑒𝑥 + 𝑒−𝑥
𝑒𝑥 − 𝑒−𝑥
sech 𝑥 =1
cosh 𝑥=
2
𝑒𝑥 + 𝑒−𝑥 ,
cosch 𝑥 =1
sinh 𝑥=
2
𝑒𝑥 − 𝑒−𝑥
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.
MATHEMATICS-I Page 21
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
MATHEMATICS-I Page 22
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.
MATHEMATICS-I Page 23
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
MATHEMATICS-I Page 24
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
𝑓′(𝑥) =𝑑𝑦
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
MATHEMATICS-I Page 25
𝑓′(𝑥) =𝑑𝑦
𝑑𝑥= lim
∆𝑥→0
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
∆𝑥
𝑓′(𝑥) =𝑑𝑦
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)
MATHEMATICS-I Page 26
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)
MATHEMATICS-I Page 27
(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.
MATHEMATICS-I Page 28
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)
MATHEMATICS-I Page 29
(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
MATHEMATICS-I Page 30
(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 ).
MATHEMATICS-I Page 31
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
MATHEMATICS-I Page 32
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 )
MATHEMATICS-I Page 33
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.
MATHEMATICS-I Page 34
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.
MATHEMATICS-I Page 35
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.
MATHEMATICS-I Page 36
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
MATHEMATICS-I Page 37
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.
MATHEMATICS-I Page 38
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 .
MATHEMATICS-I Page 39
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
MATHEMATICS-I Page 40
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 .
MATHEMATICS-I Page 41
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
MATHEMATICS-I Page 42
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.
MATHEMATICS-I Page 43
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)
MATHEMATICS-I Page 44
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
MATHEMATICS-I Page 45
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)
MATHEMATICS-I Page 46
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
MATHEMATICS-I Page 47
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
MATHEMATICS-I Page 48
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.
MATHEMATICS-I Page 49
Example 16:
If 𝑥 = sin 𝑡 𝑎𝑛𝑑 𝑦 = sin 𝑛𝑡
Show that (1 − 𝑥2)𝑦′′ − 𝑥𝑦′ + 𝑛2𝑦 = 0.
Solution:
𝑦′ =
𝑑𝑦𝑑𝑡𝑑𝑥𝑑𝑡
= 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:
MATHEMATICS-I Page 50
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
MATHEMATICS-I Page 51
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=
𝑑𝑦𝑑𝑡𝑑𝑥𝑑𝑡
=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
MATHEMATICS-I Page 52
= 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.
MATHEMATICS-I Page 53
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.
MATHEMATICS-I Page 54
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 )
MATHEMATICS-I Page 55
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
MATHEMATICS-I Page 56
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.
MATHEMATICS-I Page 57
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 .
MATHEMATICS-I Page 58
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=
𝑣𝑎𝑙𝑢𝑒
𝑣𝑎𝑙𝑢𝑒
1- ∞
∞ ,
𝟎
𝟎
Example 1:
Find the following limits:
a) lim 𝑥→0
𝑠𝑖𝑛2𝑥
𝑥=
0
0
MATHEMATICS-I Page 59
= 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
𝑒𝑥 − 𝑒−𝑥
2 sin 2𝑥=
0
0
= lim𝑥→0
𝑒𝑥 + 𝑒−𝑥
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
MATHEMATICS-I Page 60
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 .
MATHEMATICS-I Page 61
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
MATHEMATICS-I Page 62
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
∞
∞
𝐥𝐢𝐦𝒙→𝟎
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
MATHEMATICS-I Page 63
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
MATHEMATICS-I Page 64
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
MATHEMATICS-I Page 65
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
MATHEMATICS-I Page 66
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 + ⋯
MATHEMATICS-I Page 67
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!
MATHEMATICS-I Page 68
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
MATHEMATICS-I Page 69
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)!
MATHEMATICS-I Page 70
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
MATHEMATICS-I Page 71
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.
MATHEMATICS-I Page 72
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
MATHEMATICS-I Page 73
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 + ⋯
MATHEMATICS-I Page 74
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
𝑎𝑛(𝑥 − 𝑏)𝑛−2
𝑓′′′(𝑥) = (3.2)𝑎3 + (4.3.2 )𝑎 4(𝑥 − 𝑏) + ⋯
= ∑ 𝑛(𝑛 − 1)(𝑛 − 2)
∞
𝑛=3
𝑎𝑛(𝑥 − 𝑏)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
MATHEMATICS-I Page 75
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 + ⋯
MATHEMATICS-I Page 76
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!+ ⋯ ).
MATHEMATICS-I Page 77
(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+ ⋯.
MATHEMATICS-I Page 78
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
MATHEMATICS-I Page 79
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.
MATHEMATICS-I Page 80
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.
MATHEMATICS-I Page 81
∫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.
MATHEMATICS-I Page 82
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.
MATHEMATICS-I Page 83
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.
MATHEMATICS-I Page 84
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:
MATHEMATICS-I Page 85
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:
MATHEMATICS-I Page 86
∫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.
MATHEMATICS-I Page 87
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.
MATHEMATICS-I Page 88
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
MATHEMATICS-I Page 89
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
MATHEMATICS-I Page 90
(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
MATHEMATICS-I Page 91
(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
MATHEMATICS-I Page 92
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
MATHEMATICS-I Page 93
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
)(
)(
MATHEMATICS-I Page 94
Exercise 5
Evaluate the following integrals:
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
MATHEMATICS-I Page 95
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
MATHEMATICS-I Page 96
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
MATHEMATICS-I Page 97
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
Evaluate the following integrals:
𝐄𝐱𝐚𝐦𝐩𝐥𝐞𝟏: ∫𝒄𝒐𝒔𝟓𝒙
𝒔𝒊𝒏𝟐𝒙𝒅𝒙
𝑃𝑢𝑡 𝑢 = 𝑠𝑖𝑛𝑥 ⟹ 𝑑𝑢 = 𝑐𝑜𝑠𝑥 𝑑𝑥
𝐼 = ∫𝑐𝑜𝑠4𝑥
𝑠𝑖𝑛2𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥 = ∫
(𝑐𝑜𝑠2𝑥)2
𝑠𝑖𝑛2𝑥𝑐𝑜𝑠𝑥 𝑑𝑥
MATHEMATICS-I Page 98
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
MATHEMATICS-I Page 99
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
MATHEMATICS-I Page 100
= − [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.
MATHEMATICS-I Page 101
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
MATHEMATICS-I Page 102
∫ 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.
MATHEMATICS-I Page 103
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)
MATHEMATICS-I Page 104
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
MATHEMATICS-I Page 105
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
𝑰 =𝒕𝒂𝒏𝟒𝒙
𝟒− ∫(𝒔𝒆𝒄𝟐𝒙. 𝒕𝒂𝒏𝒙 − 𝒕𝒂𝒏𝒙) 𝒅𝒙
𝑰 =𝒕𝒂𝒏𝟒𝒙
𝟒−
𝒕𝒂𝒏𝟐𝒙
𝟐+ 𝒍𝒏|𝒔𝒆𝒄𝒙| + 𝒄.
MATHEMATICS-I Page 106
Exercise 6
Evaluate the following integrals:
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
MATHEMATICS-I Page 107
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
MATHEMATICS-I Page 108
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).
MATHEMATICS-I Page 109
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
Evaluate the following integrals:
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 )
MATHEMATICS-I Page 110
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)
Using zeros from equation(1), and substitute in (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
MATHEMATICS-I Page 111
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)
Using zeros from equation(1), and substitute in (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.
MATHEMATICS-I Page 112
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)
Using zeros from equation(1), and substitute in (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.
MATHEMATICS-I Page 113
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)
Using zeros from equation(1), and substitute in (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.
MATHEMATICS-I Page 114
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
MATHEMATICS-I Page 115
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
MATHEMATICS-I Page 116
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
MATHEMATICS-I Page 117
. 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 .
MATHEMATICS-I Page 118
. 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 −=−== −−
MATHEMATICS-I Page 119
−
−=−==
=
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
MATHEMATICS-I Page 120
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.
MATHEMATICS-I Page 121
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
MATHEMATICS-I Page 122
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=
+
-
-
+
-
MATHEMATICS-I Page 123
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
MATHEMATICS-I Page 124
+−=−+− −−− 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 .
MATHEMATICS-I Page 125
Exercise 8
Evaluate the following integrals:
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
MATHEMATICS-I Page 126
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:
Evaluate the following integrals:
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
MATHEMATICS-I Page 127
= 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
MATHEMATICS-I Page 128
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
MATHEMATICS-I Page 129
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
MATHEMATICS-I Page 130
= 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.
MATHEMATICS-I Page 131
Exercise 9
Evaluate the following integrals:
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
MATHEMATICS-I Page 132
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
MATHEMATICS-I Page 133
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
MATHEMATICS-I Page 134
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
MATHEMATICS-I Page 135
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
MATHEMATICS-I Page 136
(∆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
MATHEMATICS-I Page 137
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.
MATHEMATICS-I Page 138
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
MATHEMATICS-I Page 139
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
MATHEMATICS-I Page 140
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
MATHEMATICS-I Page 141
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.
MATHEMATICS-I Page 142
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.
MATHEMATICS-I Page 143
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.
MATHEMATICS-I Page 144
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.
MATHEMATICS-I Page 145
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
MATHEMATICS-I Page 146
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
ln 𝑦 = lim𝑥→0
4 ln sec2 𝑥
sin 𝑥=
0
0 using L'hopital rule
ln 𝑦 = lim𝑥→0
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
MATHEMATICS-I Page 147
(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
MATHEMATICS-I Page 148
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
MATHEMATICS-I Page 149
Attempt all Questions Number of Questions: 6 Number of Pages: 2
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
MATHEMATICS-I Page 150
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
MATHEMATICS-I Page 151
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
Attempt all Questions Number of Questions: 6 Number of Pages: 2
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)
MATHEMATICS-I Page 152
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
MATHEMATICS-I Page 153
Engineering Mathematics and Physics
Department
Math1 Code: Math 101
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
MATHEMATICS-I Page 154
Engineering Mathematics and Physics
Department
Math1 Code: Math 101
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
MATHEMATICS-I Page 155
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
Question 1, (10 Marks)
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
MATHEMATICS-I Page 156
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)
MATHEMATICS-I Page 157
𝒚′
𝒚=
−𝟐
(𝟐𝒙)√𝟏 − (𝟐𝒙)𝟐+
𝟏𝟓𝒙𝟒𝒔𝒆𝒄𝟐𝒙𝟓
𝒕𝒂𝒏𝒙𝟓+
𝟏
𝟏 − 𝒙
𝒚′ = 𝒚−𝟐
(𝟐𝒙)√𝟏 − (𝟐𝒙)𝟐+
𝟏𝟓𝒙𝟒𝒔𝒆𝒄𝟐𝒙𝟓
𝒕𝒂𝒏𝒙𝟓+
𝟏
𝟏 − 𝒙
Question 2, (4 Marks)
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 + ⋯
MATHEMATICS-I Page 158
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:
(𝒊) 𝐥𝐢𝐦𝒙→𝟎
(𝟏 − 𝟔𝒙)𝟏𝟐𝒙
, (𝒊𝒊) 𝐥𝐢𝐦 𝒙→𝟎
𝒄𝒐𝒔𝟒𝒙 − 𝒄𝒐𝒔𝟓𝒙
𝟏 − 𝒄𝒐𝒔𝟑𝒙.
Solution
(𝒊) 𝐥𝐢𝐦𝒙→𝟎
(𝟏 − 𝟔𝒙)𝟏𝟐
𝒙
= (1)∞ undetermined value
𝒍𝒆𝒕 𝒚 = (𝟏 − 𝟔𝒙)𝟏𝟐𝒙 𝒕𝒂𝒌𝒆 𝒍𝒏 𝒐𝒇 𝒃𝒐𝒕𝒉 𝒔𝒊𝒅𝒆𝒔
𝒍𝒏𝒚 = 𝒍𝒏(𝟏 − 𝟔𝒙)𝟏𝟐𝒙
𝐥𝐢𝐦𝒙→𝟎
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
MATHEMATICS-I Page 159
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
MATHEMATICS-I Page 160
MATHEMATICS-I Page 161
MATHEMATICS-I Page 162
MATHEMATICS-I Page 163
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.