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FACULTY OF ENGINEERING
DEPT. OF ENGINEERING MATHEMATICS
PROGRAM: BACHELOR OF “CIVIL/PETROLEUM/ELECTRICAL/MOBILE &
SATELLITE” ENGINEERING
COURSE UNIT:
EMT1123: ENGINEERING MATHEMATICS 1
Lecturer: Dr.Dr.Eng. Robert Setekera
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 1
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Lecture Eight
In this lecture we will cover:
Indefinite integrals
Integration of Logarithmic and Exponential functions
Integration of Trigonometric functions
Introduction
Indefinite Integrals
When the limits of the integral are not given, we can write
𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑥) + 𝑐, if and only if 𝐹 (𝑥) = 𝑓(𝑥). (1)
Where 𝑐 is an arbitrary constant, called a constant of integration.
The expression on the left in (1) is called an indefinite integral of 𝑓.
Example
1
𝑥 𝑑𝑥 = 𝑥 + 𝑐.
3
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 2
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1 1
8𝑡 − 5 + 𝑑𝑡 = 2𝑡 − 5𝑡 − + 𝑐.
𝑡 𝑡
Note:
(1) 𝐷 ∫ 𝑓(𝑥)𝑑𝑥 = 𝐷 [𝐹(𝑥) + 𝑐] + 𝑐.
(2) ∫ 𝑓(𝑥)𝑑𝑥 = [∫ 𝑓(𝑥) 𝑑𝑥]
(3) ∫ 𝑥 𝑑𝑟 = 𝑥 + 𝑐, 𝑟 ≠ −1.
(4) ∫[𝑓(𝑥) + 𝑔(𝑥)] 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑔(𝑥) 𝑑𝑥
(5) ∫ 𝑐 𝑓(𝑥) 𝑑𝑥 𝑐 ∫ 𝑓(𝑥) 𝑑𝑥, 𝑐 is areal number.
Example
/
𝑥
𝐷 (𝑥 + 5) = + 𝑐.
√𝑥 + 5
∫ 𝑑𝑥 = √𝑥 + 5 + 𝑐
√
(6) 𝐷 𝐹 𝑔(𝑥) = 𝐹 𝑔(𝑥) 𝑔 (𝑥) = 𝑓 𝑔(𝑥) 𝑔 (𝑥).
∫ 𝑓 𝑔(𝑥) 𝑔 (𝑥)𝑑𝑥 = 𝐹 𝑔(𝑥) + 𝑐, where 𝐹 = 𝑓.
Now if we let 𝑢 = 𝑔(𝑥), then
𝑑𝑢
= 𝑔 (𝑥) or 𝑑𝑢 = 𝑔 (𝑥)𝑑𝑥
𝑑𝑥
This approach is known as change of variable or a method of substitution.
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 3
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(7) Using formula above, we can write
1
𝑢 𝑑𝑢 = 𝑢 +𝑐
𝑟+1
where 𝑢 = 𝑔(𝑥), 𝑑𝑢 = 𝑔 (𝑥)𝑑𝑥 and 𝑟 ≠ −1.
Example
Find ∫(2𝑥 + 1) 𝑥 𝑑𝑥
Solution
Let 𝑢 = 2𝑥 + 1
= 6𝑥 𝑜𝑟 𝑑𝑢 = 6𝑥 𝑑𝑥.
∫(2𝑥 + 1) 𝑥 𝑑𝑥 = ∫(2𝑥 + 1) 6𝑥 𝑑𝑥
= ∫(𝑢) 𝑑𝑢 = 𝑢 + 𝑐
/
Since 𝑢 = 2𝑥 + 1 ==> 𝑥 =
∫(2𝑥 + 1) 𝑥 𝑑𝑥 = (2𝑥 + 1) + 𝑘
= [(2𝑥 + 1) ] + 𝑘
= [(2𝑥 + 1) ] + 𝐶
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 4
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OR
𝑢 = 2𝑥 + 1, 𝑑𝑢 = 6𝑥 𝑑𝑥.
𝑑𝑢 = 𝑥 𝑑𝑥.
Using change of variables. Now substitute directly for 𝑥 𝑑𝑥, we have
1
∫(2𝑥 + 1) 𝑥 𝑑𝑥 = ∫ 𝑢 𝑑𝑢
6
1
= 𝑢 𝑑𝑢
6
1 8
= 𝑢 +𝑐
48
1 8
= (2𝑥3 + 1) + 𝑐
48
Example 2:
Find ∫ 𝑥 √7 − 6𝑥 𝑑𝑥
Solution
Let 𝑢 = 7 − 6𝑥 ==> 𝑑𝑢 = −12𝑥 𝑑𝑥
−1
==> 𝑥 𝑑𝑥 = 𝑑𝑢
12
Using change of variables,
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 5
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−1
𝑥 7 − 6𝑥 𝑑𝑥 = √𝑢 𝑑𝑢
12
−1 /
= 𝑢 𝑑𝑢
12
−1 𝑢 /
= + 𝑐
12 4/3
−1 3 /
= (7 − 6𝑥 ) + 𝑐
12 4
−1 /
= (7 − 6𝑥 ) + 𝑐.
16
Theorem
( )
( )
we change the limit of integration from 𝑎 to 𝑏 to functional limits of
𝑔(𝑎) 𝑡𝑜 𝑔(𝑏).
Example
Evaluate ∫ 𝑑𝑥
√
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 6
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Solution
3 1
𝑑𝑥 = 3 𝑑𝑥
√5𝑥 − 1 √5𝑥 − 1
Now let 𝑢 = 5𝑥 − 1, ==> 𝑑𝑢 = 5𝑑𝑥.
We change the limits of integration.
For
𝑥 = 2, 𝑢=9
𝑥 = 10, 𝑢 = 49
Using the above theorem, we have
1 3 1
3 𝑑𝑥 = 5𝑑𝑥
√5𝑥 − 1 5 √5𝑥 − 1
3 1
= 𝑑𝑢
5 √𝑢
3 /
= 𝑢 𝑑𝑢
5
3 /
= 2𝑢
5
24
= .
5
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 7
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Personal Exercise
Evaluate
(a) ∫(3𝑥 + 1) 𝑑𝑥
(b) ∫(2𝑥 − 3) 𝑥𝑑𝑥
(c) ∫ √𝑡 − 1𝑡 𝑑𝑡
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 8
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EMT1123: Engineering Mathematics 1 Lecturer Notes Page 9
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Integration of Logarithmic
and Exponential Functions
In part 1 of this lecture, we looked at the definitions of logarithmic and
exponential functions. We will not repeat these definitions here, we are going
to just use them. In case one needs to know more about them, I refer you to
the notes on differentiation in part 1 of this course.
Integration of Logarithmic Functions
The Integral of an logarithmic function is defined as:
Example
Find
Solution
Let 𝑢 = 3𝑥 − 5 ==> 𝑑𝑢 = 6𝑥𝑑𝑥 or 𝑑𝑢 = 𝑥 𝑑𝑥
Then
𝑥 1/6
𝑑𝑥 = 𝑑𝑢
3𝑥 − 5 𝑢
1 1
= 𝑑𝑢
6 𝑢
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 10
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1
= ln|𝑢| + 𝑐
6
Example 2
Evaluate
Solution
Let 𝑢 = 9 − 2𝑥 ==> 𝑑𝑢 = −2𝑑𝑥 or − 𝑑𝑢 = 𝑑𝑥
Then
1 1
= 2 𝑢
𝑑𝑢
−1
= ln|𝑢| + 𝑐
2
−1
= ln|9 − 2𝑥 | + 𝑐
2
Using Fundamental Theorem of Calculus
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 11
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−1 4
ln 9 − 2𝑥 2
2
−1 4
ln 1 − ln 5 2
2
1
ln 5 .
2
OR
Use the change of limits; Let 𝑢 = 9 − 2𝑥 ==> 𝑑𝑢 = −2𝑑𝑥
For
𝑥 = 2, 𝑢 = 5
𝑥 = 4, 𝑢 = 1
Then
1
= 2
1
= 2
−1
= [ln|𝑢|]
2
−1
= [ln(1) − ln(5)]
2
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 12
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1
= ln(5).
2
Example 3
Find
√ln 𝑥
𝑑𝑥
𝑥
Solution
Let 𝑢 = ln 𝑥, ==> 𝑑𝑢 = 𝑑𝑥
√ln 𝑥 1
𝑑𝑥 = √ln 𝑥 . 𝑑𝑥
𝑥 𝑥
= √𝑢 𝑑𝑢
2 /
= 𝑢 + 𝑐
3
2 /
= (ln 𝑥 ) + 𝑐
3
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 13
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Integration of Exponential Functions
Since 𝐷 𝑒 = 𝑒 , we can write the integral of exponential functions as:
𝑒 𝑑𝑥 = 𝑒 + 𝑐.
In general form
𝑒 𝑑𝑢 = 𝑒 + 𝑐, 𝑢 = 𝑔(𝑥).
Example
Find
𝑒 /
𝑑𝑥
𝑥
Solution
Let 𝑢 = , ==> 𝑑𝑢 = (−3/𝑥 )𝑑𝑥
−1 1
==> 𝑑𝑢 = 𝑑𝑥
3 𝑥
Then
𝑒 / −1
𝑑𝑥 = 𝑒 𝑑𝑢
𝑥 3
−1
= 𝑒 𝑑𝑢
3
−1
= (𝑒 ) + 𝑐
3
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 14
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−1 /
= 𝑒 + 𝑐.
3
Example 2
Compute the following integral
𝑒 /
𝑑𝑥
𝑥
Solution
Using the Fundamental Theorem of Calculus, we have
𝑒 / −1 /
𝑑𝑥 = 𝑒
𝑥 3
−1 /
= 𝑒 − 𝑒 .
3
We can also use substitution method plus change of limits of integration.
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 15
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Assignment 2
Evaluate
(a) ∫ 𝑑𝑥
(b) ∫ 𝑑𝑥
(c) ∫ 𝑑𝑥
𝑒𝑥
(d) ∫ 𝑥 𝑑𝑥
𝑒 +1
(e) ∫ 𝑑𝑥
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 16
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Integration of Trigonometric Functions
As seen in Part I of this lecture, trigonometric functions are:
sin 𝑥 , cos 𝑥 , tan 𝑥 , sec 𝑥 , cot 𝑥 , csc 𝑥 and there inverses. Here we will look at
how to integrate such functions.
Cosine function: 𝐜𝐨𝐬 𝒙
We know that
(sin 𝑥) = cos 𝑥,
Then, if we introduce integration with respect to variable 𝑥 on both sides , we
have
∫ (sin 𝑥) 𝑑𝑥 = ∫ cos 𝑥 𝑑𝑥,
∫ 𝑑(sin 𝑥) = ∫ cos 𝑥 𝑑𝑥,
sin 𝑥 + 𝑐 = ∫ cos 𝑥 𝑑𝑥,
Thus,
cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐, where 𝑐 is a constant of integration.
In general form:
cos 𝑢 𝑑𝑢 = sin 𝑢 + 𝑐, where 𝑢 = 𝑔(𝑥).
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 17
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Example
Find ∫ 𝑥 cos(𝑥 ) 𝑑𝑥
Solution
Let 𝑢 = 𝑥 , ==> 𝑑𝑢 = 2𝑥 𝑑𝑥, ==> 𝑑𝑢 = 𝑥 𝑑𝑥
1
𝑥 cos(𝑥 ) 𝑑𝑥 = 𝑑𝑢 cos(𝑢)
2
1
= cos(𝑢)𝑑𝑢
2
1
= sin(𝑢) + 𝑐
2
1
= sin(𝑥 ) + 𝑐 .
2
Sine function: 𝐬𝐢𝐧 𝒙
We know that
(cos 𝑥) = − sin 𝑥,
Then, if we introduce integration with respect to variable 𝑥 on both sides , we
have
∫ (cos 𝑥) 𝑑𝑥 = − ∫ sin 𝑥 𝑑𝑥,
∫ 𝑑(cos 𝑥) = − ∫ sin 𝑥 𝑑𝑥,
−cos 𝑥 + 𝑐 = ∫ sin 𝑥 𝑑𝑥,
Thus,
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 18
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sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝑐, where 𝑐 is a constant of integration.
In general form:
sin 𝑢 𝑑𝑢 = −cos 𝑢 + 𝑐, where 𝑢 = 𝑔(𝑥).
Example
Evaluate ∫ sin(5𝑥) 𝑑𝑥
Solution
Let 𝑢 = 5𝑥, ==> 𝑑𝑢 = 5 𝑑𝑥, ==> 𝑑𝑢 = 𝑑𝑥
1
sin(5𝑥) 𝑑𝑥 = sin(𝑢) 𝑑𝑢
5
1
= sin(𝑢)𝑑𝑢
5
−1
= cos(𝑢) + 𝑐
5
−1
= cos(5𝑥) + 𝑐 .
5
Example
Show that ∫ sin(𝑥) cos(𝑥) 𝑑𝑥 = sin (𝑥) + 𝑐
Solution
(Personal exercise)
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 19
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Tangent function:
tan 𝑥 𝑑𝑥 = −ln|cos 𝑥| + 𝑐, where 𝑐 is a constant of integration.
In general form:
tan 𝑢 𝑑𝑢 = −ln|cos 𝑢| + 𝑐, where 𝑢 = 𝑔(𝑥).
Additionally,
sec 𝑥 𝑑𝑥 = tan 𝑥 + 𝑐, where 𝑐 is a constant of integration.
In general form:
sec 𝑢 𝑑𝑢 = tan 𝑢 + 𝑐, where 𝑢 = 𝑔(𝑥).
Cotangent function:
cot 𝑥 𝑑𝑥 = ln|sin 𝑥| + 𝑐, where 𝑐 is a constant of integration.
In general form:
cot 𝑢 𝑑𝑢 = ln|sin 𝑢| + 𝑐, where 𝑢 = 𝑔(𝑥).
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 20
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Cosecant function:
cosec 𝑥 𝑑𝑥 = ln|cosec 𝑥 − cot 𝑥| + 𝑐,
where 𝑐 is a constant of integration.
In general form:
cosec 𝑢 𝑑𝑢 = ln|cosec 𝑢 − cot 𝑢| + 𝑐, where 𝑢 = 𝑔(𝑥).
Other important integrals:
1 1
sin 𝑥 𝑑𝑥 = 𝑥 − sin 2𝑥 + 𝑐,
2 4
1 1
cos 𝑥 𝑑𝑥 = 𝑥 + sin 2𝑥 + 𝑐,
2 4
tan 𝑥 𝑑𝑥 = tan 𝑥 − 𝑥 + 𝑐,
cot 𝑥 𝑑𝑥 = −cot 𝑥 − 𝑥 + 𝑐,
𝑒
𝑒 sin 𝑏𝑥 𝑑𝑥 = (a sin 𝑏𝑥 − b cos 𝑏𝑥) + 𝑐,
a +b
𝑒
𝑒 cos 𝑏𝑥 𝑑𝑥 = (a cos 𝑏𝑥 + b sin 𝑏𝑥) + 𝑐,
a +b
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 21
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Personal exercise:
Prove these integrals.
{End of lEcturE}
EMT1123: Engineering Mathematics 1 Lecturer Notes Page 22
