Republic of the Philippines

NUEVA VIZCAYA STATE UNIVERSITY Bambang,

Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.1: MATH 2A-2S-2020-2021

College: Engineering
Campus: Bambang

BSME/BSEE/BSECE/
DEGREE PROGRAM COURSE NO. Math 2A
BSCPE
COURSE
SPECIALIZATION Mechanical Integral Calculus
TITLE
YEAR LEVEL 1st Year TIME FRAME 4 hrs/wk WK NO. 1-3 IM NO. 1

I. UNIT TITLE/CHAPTER TITLE

INTEGRATION CONCEPTS/FORMULAS

II. LESSON TITLE

1.1 Antidifferentiation/Indefinite Integral

1.2 Simple Power Formula
1.3 Simple Trigonometric Function
1.4 Logarithmic Function
1.5 Exponential Function
1.6 Inverse Trigonometric Functions
1.7 Hyperbolic Function
1.8 General Power Formula
1.9 Constant of Integration
1.10 Definite Integral

III. LESSON OVERVIEW

This lesson provides the students an introduction to indefinite integrals and how to operate
integrals of different functions, and later the definite integrals.

IV. DESIRED LEARNING OUTCOMES

At the end of the lesson, the students should be able to:

1. Know the basic principles of antiderivatives.
2. Familiarize with the formulas of integrals.
3. Use basic integration rules to find anti-derivatives.
4. Solve problems involving definite integrals.

V. COURSE CONTENT

INTRODUCTION
Often, we know the relationship involving the rate of change of two variables, but we may need to
know the direct relationship between the two variables. For example, we may know the velocity of an
object at a particular time, but we may want to know the position of the object at that time.
To find this direct relationship, we need to use the process which is opposite to differentiation.
This is called integration (or antidifferentiation).

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NUEVA VIZCAYA STATE UNIVERSITY Bambang,
Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.1: MATH 2A-2S-2020-2021
1.1 ANTI-DIFFERENTIATION / INDEFINITE INTEGRAL

The indefinite integral represents a family of functions whose derivatives are 𝑓 . The difference
between any two functions in the family is a constant. The integral key, which is used to find definite
integrals, can also be used to find indefinite integrals by simply omitting the limits of integration.

Anti-differentiation or indefinite integral perform the opposite process to differentiation. For example, if
we know that

𝑑𝑦
= 3𝑥 2 .
𝑑𝑥

We wish to perform the opposite process to differentiation. This is called "antidifferentiation" and later,
we will call it "integration".
𝑦 = 𝑥 3 is one antiderivative of

𝑑𝑦
= 3𝑥 2
𝑑𝑥

There are infinitely many other antiderivatives which would also work, for example:

𝑦 = 𝑥3 + 9 or 𝑦 = 𝑥 3 + 6 .

In general, 𝑦 = 𝑥 3 + 𝐶 is the indefinite integral of 3𝑥 2 . The number 𝐶 is called the constant of

integration.

We write: ,: and read as “the integral of 3𝑥 2 with respect to 𝑥 equals 𝑥 3 + 𝐶 ”.

Example 1:
Find the antiderivative of the function 𝑓(𝑥) = 1/𝑥.

Solution:
From our previous lesson on differential calculus,
𝑑 1
(ln|𝑥 |) = .
𝑑𝑥 𝑥

Thus, 𝐹 (𝑥 ) = ln|𝑥 | is an antiderivative of 1/𝑥. Therefore, every antiderivative of 1/𝑥 is the form
ln|𝑥 | + 𝐶 for some constant 𝐶 and every function of the form ln|𝑥 | + 𝐶 is an antiderivative of 1/𝑥.

Example 2:
Find the antiderivative of the function 𝑓(𝑥) = cos 𝑥.

Solution:
Recall that
𝑑
(sin 𝑥) = cos 𝑥.
𝑑𝑥
So 𝐹(𝑥) = sin 𝑥 is an antiderivative of cos 𝑥. Therefore, every antiderivative of cos 𝑥 is of the form
sin 𝑥 + 𝐶 for some constant 𝐶 and every function of the form sin 𝑥 + 𝐶 is an antiderivative of cos 𝑥.

1.1A. Notations of the Integral

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𝐹(𝑥) = ∫ 𝑑𝑥 = 𝑥 + 𝐶

The sign is an elongated "S", standing for "sum". Later we will see that the integral is the sum of the

areas of infinitesimally thin rectangles. This elongated “S" notation was introduced by Leibniz when he
developed the concepts of integration.

Sometimes we write a capital letter to signify integration. For example, we write 𝐹(𝑥) to mean the
integral of 𝑓(𝑥).

The function we want to find an anti-derivative, 𝑓(𝑥), is called the integrand. It contains the differential of
the variable we are integrating with respect to.

The differential 𝑑𝑥 serves to identify 𝑥 as the variable of integration.

1.1B Basic Integral

The anti-derivative of 𝑑𝑥 is 𝑥 plus a constant.

∫ 𝑑𝑥 = 𝑥 + 𝐶

1.1C Integral of a Sum and Difference Rule

The integral of the sum of two functions is equal to the sum of the integrals of each function.

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

The integral of the difference of two functions is equal to the difference of the integrals of each function.

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

1.1D Integral of a Constant Multiple Rule

Move the constant outside the integral before integrating.

∫ 𝑐𝑓(𝑥)𝑑𝑥 = 𝑐 ∫ 𝑓(𝑥)𝑑𝑥

Example 1:
Integrate the function 𝑓(𝑥) = 𝑥 + 4.

Solution:

∫(𝑥 + 4) 𝑑𝑥 = ∫ 𝑥 𝑑𝑥 + ∫ 4𝑑𝑥

= ∫ 𝑥 𝑑𝑥 + 4 ∫ 𝑑𝑥

𝑥2
= + 4𝑥 + 𝐶
2
Example 2:
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Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.1: MATH 2A-2S-2020-2021
Find the anti-derivative of the function 𝑓(𝑥) = 5.
Solution:
∫ 5 𝑑𝑥 = 5 ∫ 𝑑𝑥 = 5𝑥 + 𝐶

1.2 SIMPLE POWER FORMULA

𝒙𝒏+𝟏
∫ 𝒙𝒏 𝒅𝒙 = +𝑪 𝒏 ≠ −𝟏
𝒏+𝟏

The Simple Power Formula is a useful tool to find antiderivatives of functions expressed as powers of 𝑥
alone.

Example 1:
Integrate the function 𝑓(𝑥) = 𝑥 6 .
Solution:
6
𝑥 6+1
∫ 𝑥 𝑑𝑥 = +𝐶
6+1

𝑥7
= +𝐶
7

Example 2:
Evaluate
5
∫ (3𝑥 2 + √𝑥 − ) 𝑑𝑥.
𝑥3
Solution:
Our first step is to re-write the exponents so it is easier to integrate:

5
∫ (3𝑥 2 + √𝑥 − ) 𝑑𝑥 = ∫(3𝑥 2 + 𝑥 1/2 − 5𝑥 −3 )𝑑𝑥
𝑥3

Then apply theorems 1.1C and 1.2;

5 3𝑥 3 𝑥 3/2 5𝑥 −2
∫ (3𝑥 2 + √𝑥 − 3 ) 𝑑𝑥 = + − +𝐶
𝑥 3 3 −2
2
2𝑥 3/2 5𝑥 −2
= 𝑥3 + + +𝐶
3 2

PRACTICE PROBLEMS 1.2

Evaluate the following integrals.
3
1. ∫ √𝑥 2 𝑑𝑥 6. ∫(27𝑒 9𝑥 + 𝑒 12𝑥 )1/3 𝑑𝑥
𝑥+1
2. ∫(3𝑥 + 5) 𝑑𝑥 7. ∫ 𝑑𝑥
√𝑥
1 (𝑥 + 1)3
3. ∫ √𝑥 (𝑥 + ) 𝑑𝑥 8. ∫
𝑥 √𝑥 + 1
2
5𝑥 + 7 √𝑥
4. ∫ 𝑑𝑥 9. ∫ 3 𝑑𝑥
𝑥 4/3 √𝑥
5. ∫(5𝑥 4 − 8𝑥 3 + 9𝑥 2 − 2𝑥 + 7) 𝑑𝑥 10. ∫ sin1/3 𝑥 cos 𝑥 𝑑𝑥

1.3 SIMPLE TRIGONOMETRIC FUNCTIONS

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INSTRUCTIONAL MODULE
IM No.1: MATH 2A-2S-2020-2021

We obtain the following integral formulas by reversing the formulas for differentiation of trigonometric
functions:
∫ 𝐬𝐢𝐧 𝒙 𝒅𝒙 = − 𝐜𝐨𝐬 𝒙 + 𝑪 ∫ 𝐜𝐨𝐬 𝒙 𝒅𝒙 = 𝐬𝐢𝐧 𝒙 + 𝑪

∫ 𝐬𝐞𝐜 𝟐 𝒙 𝒅𝒙 = 𝐭𝐚𝐧 𝒙 + 𝑪 ∫ 𝐜𝐬𝐜 𝟐 𝒙 𝒅𝒙 = −𝐜𝐨𝐭 𝒙 𝒅𝒙 + 𝑪

∫ 𝐬𝐞𝐜 𝒙 𝐭𝐚𝐧 𝒙 𝒅𝒙 = 𝐬𝐞𝐜 𝒙 + 𝑪 ∫ 𝐜𝐬𝐜 𝒙 𝐜𝐨t 𝒙 𝒅𝒙 = 𝐬𝐞𝐜 𝒙 + 𝑪

We will limit our discussion for simple trigonometric functions for those functions with known derivatives
for the reason that some trigonometric functions, even though it is in its basic form, when we integrate
them yields to a complicated process and they will be discussed in different subject lessons.

The integrals of tangent, cotangent, secant, and cosecant will be discussed in the Logarithmic
Functions.

Below is the list of some useful trigonometric identities.

Defining Identities:
sin 𝑥 cos 𝑥
tan 𝑥 = cot 𝑥 =
cos 𝑥 sin 𝑥
1 1
sec 𝑥 = csc 𝑥 =
cos 𝑥 sin 𝑥

Pythagorean Identities:
sin2 𝑥 + cos 2 𝑥 = 1
tan2 𝑥 + 1 = sec 2 𝑥
cot 2 𝑥 + 1 = csc 2 𝑥

Double Angle Formula:

sin 2𝑥 = 2 sin 𝑥 cos 𝑥
cos 2𝑥 = cos 2 𝑥 − sin2 𝑥
= 2cos 2 𝑥 − 1
= 1 − 2 sin2 𝑥

Example 1:
Evaluate
∫(3 sin 𝑥 − 4 sec 2 𝑥)𝑑𝑥.
Solution:
∫(3 sin 𝑥 − 4 sec 2 𝑥)𝑑𝑥 = 3 ∫ sin 𝑥 𝑑𝑥 − 4 ∫ sec 2 𝑥 𝑑𝑥
= −3 cos 𝑥 − 4 tan 𝑥 + 𝐶

Example 2:
Find the integral of the function 𝑓(𝑥) = cos(2𝑥 − 6).
Solution:
∫ cos(2𝑥 − 6) 𝑑𝑥
𝑑(2𝑥 − 6) = 2𝑑𝑥
1
∫ cos(2𝑥 − 6) 𝑑𝑥 = ∫ 2 cos(2𝑥 − 6)𝑑𝑥
2
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Nueva Vizcaya
INSTRUCTIONAL MODULE
IM No.1: MATH 2A-2S-2020-2021
1
= sin(2𝑥 − 6) + 𝐶
2

PRACTICE PROBLEMS 1.3

Evaluate the following integrals.
1. ∫(sin 𝑥 − cos 𝑥) 𝑑𝑥

2. ∫(1 − sin 𝑥) cos 𝑥 𝑑𝑥

3. ∫ sec(2𝑥 − 3) tan(2𝑥 − 3) 𝑑𝑥

4. ∫(cos 𝑥 − 1) sin 𝑥 𝑑𝑥

5. ∫(𝑥 + 1) cos(𝑥 2 + 2𝑥) 𝑑𝑥

1.4 LOGARITHMIC FUNCTION

The simple power formula is valid for all values of n except n = −1.
If n = −1, we need to take the opposite of the derivative of the logarithmic function to solve such cases:

𝒅𝒙
∫ = 𝐥𝐧 |𝒙| + 𝑪
𝒙

The ∣ ∣ (absolute value) signs around the 𝑥 are necessary since the log of a negative number is not
defined.

In words, this means that if we have the derivative of a function in the numerator (top) of a fraction, and
the function in the denominator (bottom) of the fraction, then the integral of the fraction will be the natural
logarithm of the function.

Example 1:
Evaluate
2𝑥 3
∫ 𝑑𝑥.
𝑥4 + 1

Solution:
Since the derivative of the denominator is 4𝑥 3 𝑑𝑥, we can rewrite the problem into the form
2𝑥 3 1 4𝑥 3 𝑑𝑥
∫ 4 𝑑𝑥 = ∫ 4
𝑥 +1 2 𝑥 +1

1
= ln|𝑥 4 + 1| + 𝐶
2

Example 2:
Evaluate

𝑑𝑥
∫
𝑥 ln 𝑥

Solution:
𝑑𝑥
The derivative of ln 𝑥 is .
𝑥

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NUEVA VIZCAYA STATE UNIVERSITY Bambang,
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INSTRUCTIONAL MODULE
IM No.1: MATH 2A-2S-2020-2021
𝑑𝑥 1 𝑑𝑥
∫ =∫
𝑥 ln 𝑥 ln 𝑥 𝑥

= ln|ln 𝑥 | + 𝐶

Example 3:
Find the integral of tan 𝑥.
Solution:
By trigonometric identity,

sin 𝑥
∫ tan 𝑥 𝑑𝑥 = ∫ 𝑑𝑥
cos 𝑥
𝑑(cos 𝑥) = − sin 𝑥 𝑑𝑥

−sin 𝑥
∫ tan 𝑥 𝑑𝑥 = − ∫ 𝑑𝑥
cos 𝑥

= − ln|cos 𝑥 | + 𝐶

Example 3:
Find the integral of cot 𝑥.
Solution:
cos 𝑥
∫ cot 𝑥 𝑑𝑥 = ∫ 𝑑𝑥
sin 𝑥

= ln|sin 𝑥 | + 𝐶

Example 4:
Evaluate
∫ sec 𝑥 𝑑𝑥.
Solution:
Expand the fraction by multiplying tan 𝑥 + sec 𝑥,
tan 𝑥 + sec 𝑥
∫ sec 𝑥 𝑑𝑥 = ∫ sec 𝑥 𝑑𝑥 ( )
tan 𝑥 + sec 𝑥
sec 𝑥 tan 𝑥 + sec 2 𝑥
=∫ 𝑑𝑥
tan 𝑥 + sec 𝑥

𝑑(tan 𝑥 + sec 𝑥) = (sec 2 𝑥 + sec 𝑥 tan 𝑥) 𝑑𝑥

∫ sec 𝑥 𝑑𝑥 = ln|tan 𝑥 + sec 𝑥 | + 𝐶

Example 5:
Evaluate
∫ csc 𝑥 𝑑𝑥.
Solution:
Expand the fraction by multiplying csc 𝑥 + cot 𝑥,
csc 𝑥 + cot 𝑥
∫ csc 𝑥 𝑑𝑥 = ∫ csc 𝑥 𝑑𝑥 ( )
csc 𝑥 + cot 𝑥
csc 2 𝑥 + csc 𝑥 cot 𝑥
=∫ 𝑑𝑥
csc 𝑥 + cot 𝑥

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𝑑(csc 𝑥 + cot 𝑥) = − (cot 𝑥 csc 𝑥 + csc 2 𝑥 𝑑𝑥

(csc 2 𝑥 + csc 𝑥 cot 𝑥)

∫ 𝑐𝑠 c 𝑥 𝑑𝑥 = − ∫ − 𝑑𝑥
csc 𝑥 + cot 𝑥

= − ln|csc 𝑥 + cot 𝑥 | + 𝐶

PRACTICE PROBLEMS 1.4

Evaluate the following integrals:
𝑑𝑥
1. ∫ 6. ∫ cot 𝑥 𝑑𝑥
20 − 𝑥

𝑑𝑥 𝑥2 + 𝑥 + 1
2. ∫ 7. ∫ 𝑑𝑥
𝑥(1 + 2 ln 𝑥) 𝑥2 + 1

𝑥2 + 1 𝑥 2 + 2𝑥 + 3
3. ∫ 𝑑𝑥 8. ∫ 𝑑𝑥
𝑥 3 + 3𝑥 𝑥 3 + 3𝑥 2 + 9𝑥 + 1

𝑥2
4. ∫ tan 𝑥 𝑑𝑥 9. ∫ 𝑑𝑥
3 − 𝑥3

sin 𝑥 1
5. ∫ 𝑑𝑥 10. ∫ 𝑑𝑥
1 + cos 𝑥 6𝑥 − 5

1.5 EXPONENTIAL FUNCTION

By reversing the process in obtaining the derivative of the exponential function, we obtain the
remarkable result:

∫ 𝒆𝒙 𝒅𝒙 = 𝐞𝒙 + 𝑪

It is remarkable because the integral is the same as the expression we started with. That is, ex.

Example 1:
Evaluate

∫ 3𝑒 4𝑥 𝑑𝑥.
Solution:
The derivative of the exponent of 𝑒 is 4𝑑𝑥.
4𝑑𝑥
∫ 3𝑒 4𝑥 𝑑𝑥 = ∫ 3𝑒 4𝑥
4
3
= ∫ 𝑒 4𝑥 4𝑑𝑥
4
3
= 𝑒 4𝑥 + 𝐶
4

Example 2:
Evaluate
4
∫ 4𝑒 𝑥 𝑥 3 𝑑𝑥.
Solution:
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INSTRUCTIONAL MODULE
IM No.1: MATH 2A-2S-2020-2021
The derivative of the exponent of 𝑒 is 4𝑥 3 𝑑𝑥. Therefore, a direct integration is possible.

4 4
∫ 𝑒 𝑥 4𝑥 3 𝑑𝑥 = 𝑒 𝑥 + 𝐶

Example 3:
Evaluate
2
∫ 𝑥𝑒 −𝑥 𝑑𝑥.

Solution:
The derivative of the exponent of 𝑒 is -2𝑥 𝑑𝑥. Rewrite the problem into its new form.
2 2 −2𝑥 𝑑𝑥
∫ 𝑥𝑒 −𝑥 𝑑𝑥 = ∫ 𝑒 −𝑥
2
1 2
= − ∫ 𝑒 −𝑥 − 2𝑥 𝑑𝑥
2
1 2
= − 𝑒 −𝑥 + 𝐶
2

PRACTICE PROBLEMS 1.5

Evaluate the following integrals.

𝑑𝑥
1. ∫ 6. ∫(𝑒 4𝑥 − 𝑒 −4𝑥 ) 𝑑𝑥
𝑒 2−3𝑥

4
2. ∫ 2𝑥 3 𝑒 𝑥 𝑑𝑥 7. ∫ 𝑒 2𝑥 𝑑𝑥

3. ∫ 𝑒 1−𝑥 𝑑𝑥 8. ∫(𝑒 𝑥 + 𝑥)𝑑𝑥

2
4. ∫(2 − 3𝑒 𝑥 )𝑑𝑥 9. ∫ 𝑥𝑒 2𝑥 𝑒 (1−𝑥) 𝑑𝑥

𝑒 2𝑥 3 2 +𝑥
5. ∫ 𝑒 5𝑥 ( + 3𝑥 ) 𝑑𝑥 10. ∫(2𝑥 + 1)𝑒 𝑥 𝑑𝑥
7 𝑒

1.6 INVERSE TRIGONOMETRIC FUNCTIONS

By using the derivatives of inverse trigonometric identities and by reversing those differentiation
processes, we can obtain the following integrals.

𝑑𝑥 𝑥
∫ = arcsin (
𝑎
)+ 𝐶 Form ➀
√𝑎2 − 𝑥 2

𝑑𝑥 1 𝑥
∫ = arctan ( ) + 𝐶 Form ➁
𝑎2 +𝑥 2 𝑎 𝑎

𝑑𝑥 1 |𝑥 |
∫ = arcsec ( ) + 𝐶 Form ➂
𝑥√𝑥 2 − 𝑎2 𝑎 𝑎
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IM No.1: MATH 2A-2S-2020-2021

Example 1:
Evaluate
𝑑𝑥
∫ .
√1 − 𝑥 2

Solution:
This is similar to the form of the first formula so we can directly integrate it.
𝑑𝑥
∫ = arcsin 𝑥 + 𝐶
√1 − 𝑥 2

Example 2:
Evaluate
𝑑𝑥
∫ .
9 + 𝑥2

This is in the form of the second formula with a = 3.

𝑑𝑥 1 𝑥
∫ 2
= arctan ( ) + 𝐶
9+𝑥 3 3

Example 3:
Evaluate
𝑑𝑥
∫ .
𝑥√𝑥 2 − 16

This is in the form of the third formula with a = 4.

𝑑𝑥 1 |𝑥 |
∫ = arcsec ( ) + 𝐶
𝑥√𝑥 2 − 16 4 4

PRACTICE PROBLEMS 1.6

𝑒𝑥
1. ∫ 𝑑𝑥
1 + 𝑒 2𝑥

𝑑𝑥
2. ∫
𝑥√1 − ln2 𝑥

𝑑𝑥
3. ∫
√16 − 𝑥 2

7
4. ∫ 𝑑𝑥
𝑥2 − 2𝑥 + 5

5
5. ∫ 𝑑𝑥
𝑥2 + 10𝑥 + 34

1.7 HYPERBOLIC FUNCTIONS

Hyperbolic functions are analogues of the ordinary trigonometric functions, but defined using the
hyperbola rather than the circle.
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Hyperbolic functions occur in the calculations of angles and distances in hyperbolic geometry.
They also occur in the solutions of many linear differential equations (such as the equation
defining a catenary), cubic equations, and Laplace's equation in Cartesian coordinates. Laplace's
equations are important in many areas of physics, including electromagnetic theory, heat
transfer, fluid dynamics, and special relativity.

Definition of the hyperbolic functions in exponential form;

𝑒 𝑥 − 𝑒 −𝑥 𝑒 𝑥 + 𝑒 −𝑥
sinh 𝑥 = cosh 𝑥 =
2 2

Note: sinh 𝑥 is read as “the hyperbolic sine of 𝑥”, and so on.

Hyperbolic Identities:

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

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

−1 + cosh 2𝑥 1 + cosh 2𝑥
sinh2 𝑥 = cosh2 𝑥 =
2 2

sinh 2𝑥 = 2 sinh 𝑥 cosh 𝑥 cosh 2𝑥 = sinh2 𝑥 + cosh2 𝑥

Integrals of the Hyperbolic Functions

∫ 𝐜𝐨𝐬𝐡 𝒙 𝒅𝒙 = 𝐬𝐢𝐧𝐡 𝒖 + 𝑪 ∫ 𝐬𝐢𝐧𝐡 𝒙 𝒅𝒙 = 𝐜𝐨𝐬𝐡 𝒙 + 𝑪

Example 1:
Calculate
𝑑𝑥
∫ .
1 + cosh 𝑥
Solution:
𝑒 𝑥 + 𝑒 −𝑥
cosh 𝑥 =
2

𝑒 𝑥 + 𝑒 −𝑥 2 + 𝑒 𝑥 + 𝑒 −𝑥
1 + cosh 𝑥 = 1 + =
2 2
𝑥 + 𝑒 2𝑥 + 1
1 2𝑒
2 + 𝑒𝑥 + 𝑥
= 𝑒 = 𝑒𝑥
2 2

2𝑒 𝑥 + 𝑒 2𝑥 + 1 (𝑒 𝑥 + 1)2
= =
2𝑒 𝑥 2𝑒 𝑥

𝑑𝑥 2𝑒 𝑥 𝑑𝑥
∫ =∫ 𝑥
1 + cosh 𝑥 (𝑒 + 1)2

= 2 ∫ 𝑒 𝑥 (𝑒 𝑥 + 1)−2 𝑑𝑥

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We can proceed integrating by using the power formula,

𝑑𝑥 2(𝑒 𝑥 + 1)−1
∫ = +𝐶
1 + cosh 𝑥 −1
2
=− 𝑥 +𝐶
𝑒 +1

Example 2:
Evaluate
∫ 𝑒 −𝑥 sinh 2𝑥 𝑑𝑥 .

Solution:
𝑒 𝑥 − 𝑒 −𝑥
sinh 𝑥 =
2
𝑒 2𝑥 − 𝑒 −2𝑥
sinh 2𝑥 =
2
𝑒 2𝑥 − 𝑒 −2𝑥
∫ 𝑒 −𝑥 sinh 2𝑥 𝑑𝑥 = ∫ 𝑒 −𝑥 ( ) 𝑑𝑥
2

1
= ∫(𝑒 𝑥 − 𝑒 −3𝑥 )𝑑𝑥
2
1 1
= ∫ 𝑒 𝑥 𝑑𝑥 − ∫ 𝑒 −3𝑥 𝑑𝑥
2 2
1 1
∫ 𝑒 −𝑥 sinh 2𝑥 𝑑𝑥 = ∫ 𝑒 𝑥 𝑑𝑥 + ∫ −3𝑒 −3𝑥 𝑑𝑥
2 6
1 𝑥 1 −3𝑥
= 𝑒 + 𝑒 +𝐶
2 6

PRACTICE PROBLEMS 1.7

1. ∫ sinh2 𝑥 𝑑𝑥

2. ∫ cosh2 𝑥 𝑑𝑥

3. ∫ 𝑒 2𝑥 cosh 𝑥 𝑑𝑥

𝑑𝑥
4. ∫
sinh 𝑥 + 2 cosh 𝑥

5. ∫ sinh 2𝑥 cosh 3𝑥 𝑑𝑥

1.8 GENERAL POWER FORMULA

In this section, we apply the following formula to trigonometric, logarithmic and exponential functions.

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1.8A U-SUBSTITUTION
The method of u-substitution is a method for algebraically simplifying the form of a function so
that its antiderivative can be easily recognized.

EXAMPLE:
Evaluate
∫ 𝑥 3 √1 − 𝑥 2 𝑑𝑥.

Use the variable 𝑢 to substitute the value 1 − 𝑥 2 . Then, by further algebraic operations;
𝑑𝑢 = −2𝑥 𝑑𝑥
𝑥2 = 1 − 𝑢

We can rewrite the problem into the form

−2𝑥 𝑑𝑥
∫ 𝑥 3 √1 − 𝑥 2 𝑑𝑥 = ∫ 𝑥 2 √1 − 𝑥 2 ( )
−2
1
∫ 𝑥 3 √1 − 𝑥 2 𝑑𝑥 = − ∫ 𝑥 2 √1 − 𝑥 2 (−2𝑥 𝑑𝑥)
2

Then we substitute the 𝑢 counterparts;

1
∫ 𝑥 3 √1 − 𝑥 2 𝑑𝑥 = − ∫(1 − 𝑢) √𝑢 𝑑𝑢
2
1
∫ 𝑥 3 √1 − 𝑥 2 𝑑𝑥 = − ∫(𝑢1/2 − 𝑢3/2 ) 𝑑𝑢
2

Note that the new form of the problem can be easily integrated by using the integral of difference rule
together with the simple power formula.

1 𝑢3/2 𝑢5/2
∫ 𝑥 3 √1 − 𝑥 2 𝑑𝑥 = − [ − ]+𝐶
2 3 5
2 2

1 2𝑢3/2 2𝑢5/2
∫ 𝑥 3 √1 − 𝑥 2 𝑑𝑥 = − [ − ]+𝐶
2 3 5

𝑢5/2 𝑢3/2
∫ 𝑥 3 √1 − 𝑥 2 𝑑𝑥 = − +𝐶
5 3

1.8B THE GENERAL POWER FORMULA

𝑢𝑛+1
∫ 𝑢𝑛 𝑑𝑢 = +𝐶 𝑛 ≠ −1
𝑛+1

When using the General Power Rule, you must first identify a factor 𝑢 of the integrand that is raised to a
power. Then, you must show that its derivative is also a factor of the integrand.
Many times, part of the derivative is missing from the integrand, and in some cases, you can make the
necessary adjustments to apply the General Power Rule.

Example 1:
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Evaluate
∫ sin1/3 𝑥 cos 𝑥 𝑑𝑥 .
Solution:
Our options are to either choose 𝑢 = sin 𝑥, 𝑢 = sin1/3 𝑥 or𝑢 = cos 𝑥. However, only the first one of these
works in this problem.
𝑢 = sin 𝑥, 𝑑𝑢 = cos 𝑥 𝑑𝑥.

∫ sin1/3 𝑥 cos 𝑥 𝑑𝑥 = ∫ 𝑢1/3 𝑑𝑢

𝑢4/3
= +𝐶
4
3
3𝑢4/3
= +𝐶
4

3 sin4/3 𝑥
= +𝐶
4

Example 2:
Evaluate
∫ 2√1 − 𝑒 −𝑥 𝑒 −𝑥 𝑑𝑥.
Solution:
Let 𝑢 = 1 − 𝑒 −𝑥 , 𝑑𝑢 = 𝑒 −𝑥 𝑑𝑥.

∫ 2√1 − 𝑒 −𝑥 𝑒 −𝑥 𝑑𝑥 = ∫ 2√𝑢 𝑑𝑢

= 2 ∫ 𝑢1/2 𝑑𝑢

𝑢3/2
= 2( )+𝐶
3
2

4𝑢3/2
= +𝐶
3

4
= (1 − 𝑒 −𝑥 )3/2 + 𝐶
3

Example 3:
Evaluate

(1 − 2 ln 𝑥)
∫ 𝑑𝑥.
𝑥

Solution:
2
Let 𝑢 = 1 − 2 ln 𝑥 , 𝑑𝑢 = − 𝑑𝑥.
𝑥

(1 − 2 ln 𝑥) 1
∫ 𝑑𝑥 = − ∫ 𝑢 𝑑𝑢
𝑥 2
1 𝑢2
=− ( )+𝐶
2 2

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𝑢2
=− +𝐶
4
(1 − 2 ln 𝑥)2
=− +𝐶
4

PRACTICE PROBLEMS 1.8

Evaluate the following functions:

1. ∫(3𝑥 + 2)5 𝑑𝑥 6. ∫ 𝑒 𝑥 (1 − 𝑒 𝑥 )(1 + 𝑒 𝑥 )10 𝑑𝑥

sin 𝑥 𝑥2
2. ∫ 𝑑𝑥 7. ∫ 𝑑𝑥
√1 + cos 𝑥 (1 + 𝑥 3 )2
3 2
3. ∫(1 + tan 2𝑥)2 sec 2 2𝑥 𝑑𝑥 8. ∫ √𝑥 (4 − 𝑥 2 ) 𝑑𝑥
2 −1
4. ∫ 30𝑒 −3𝑥 (1 + 3𝑒 −𝑥 )5 𝑑𝑥 9. ∫(4 − √𝑥) ( ) 𝑑𝑥
2√𝑥
8𝑒 𝑥 (3 + 𝑒 𝑥 ) 2𝑑𝑥
5. ∫ 𝑑𝑥 10. ∫
√𝑒 2𝑥 + 6𝑒 𝑥 + 1 (1 + 2𝑥)2

1.9 CONSTANT OF INTEGRATION

In calculus, the constant of integration, often denoted by 𝐶 , is a constant added to the end of an
antiderivative of a function 𝑓(𝑥) to indicate that the indefinite integral of 𝑓(𝑥) (i.e., the set of all
antiderivatives of 𝑓(𝑥) ) on a connected domain, is only defined up to an additive constant. This constant
expresses an ambiguity inherent in the construction of antiderivatives.

More specifically, if a function 𝑓(𝑥) is defined on an interval, and 𝐹(𝑥) is an antiderivative of 𝑓(𝑥), then the
set of all antiderivatives of 𝑓(𝑥) is given by the function 𝐹(𝑥) + 𝐶 , where 𝐶 is an arbitrary constant
(meaning that any value of 𝐶 would make 𝐹(𝑥) + 𝐶 a valid antiderivative).

Example 1:
𝑑𝑦
An equation = 3𝑥 2 + 4, has a point P (1, 3) on its curve. Find the constant of integration.
𝑑𝑥

Solution:
𝑦 = ∫(3𝑥 2 + 4)𝑑𝑥
𝑥3
= 3( ) + 4𝑥 + 𝐶
3
= 𝑥 3 + 4𝑥 + 𝐶

Since the curve passes at point P(1, 3);

3 = 13 + 4(1) + 𝐶
𝐶 = −2
Example 2:
What is 𝑓(𝑥) = ∫(𝑥 3 − 𝑥)𝑑𝑥 if 𝑓(2) = 4?

Solution:
𝑥4 𝑥2
𝑓(𝑥) = ∫(𝑥 3 − 𝑥)𝑑𝑥 = − +𝐶
4 2
24 22
𝑓(2) = 4 = − +𝐶
4 2
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16 4
4= − +𝐶
4 2
4 =4−2+𝐶
𝐶=2

𝑥4 𝑥2
𝑓(𝑥) = − +2
4 2

𝑥 4 − 2𝑥 2 + 8
𝑓(𝑥) =
4

PRACTICE PROBLEMS 1.9

1. If 𝑓(2) = 1, find the constant of integration for
𝑑𝑥
𝑓(𝑥) = ∫ .
𝑥+3
2. What is 𝑓(𝑥) = ∫ √𝑥 + 3𝑑𝑥 if 𝑓(1) = 7?
3. What is 𝑓(𝑥) = ∫(𝑥 2 + 𝑥 + 3)𝑑𝑥 if 𝑓(2) = 3.
4. Find the equation of the curve that passes through (1, 13) and having a slope of -4x at that point.

1.10 DEFINITE INTEGRAL

When integration was introduced as the reverse of differentiation, the integrals were indefinite integrals.
The result of finding an indefinite integral is usually a function plus a constant of integration. Definite
integrals, so called because the result will be a definite answer, usually a number, with no constant of
integration. Definite integrals have many applications, for example in finding areas bounded by curves,
and finding volumes of solids.

Definite integrals can be recognized by numbers written to the upper and lower right of the integral sign.
The quantity

𝑥=𝑏
∫ 𝑓 (𝑥 )𝑑𝑥
𝑥=𝑎

is called the definite integral of 𝑓(𝑥) from a to b. The numbers a and b are known as the lower and upper
limits of the integral. This integral is commonly written as

𝑏
∫ 𝑓 (𝑥 )𝑑𝑥
𝑎

First of all, the integration of is performed in the normal way. However, to show we are dealing with a
definite integral, the result is usually enclosed in square brackets and the limits of integration are written
on the right bracket:

𝑏 𝑏
∫ 𝑓 (𝑥 )𝑑𝑥 = 𝐹(𝑥) |
𝑎 𝑎
= 𝐹 (𝑏) − 𝐹(𝑎)

where; 𝐹(𝑥) is the integral of 𝑓(𝑥);

𝐹(𝑏) is the value of the integral at the upper limit 𝑥 = 𝑏; and
𝐹(𝑎) is the value of the integral at the lower limit 𝑥 = 𝑎.

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Properties of the Definite Integrals
A. Reversing the Interval
We can interchange the limits on any definite integral, all that we need to do is tack a minus sign
onto the integral when we do.

𝑏 𝑎
∫ 𝑓 (𝑥 )𝑑𝑥 = − ∫ 𝑓(𝑥)𝑑𝑥
𝑎 𝑏

B. Adding/Subtracting Functions

𝑏 𝑎 𝑏
∫ 𝑓 (𝑥 )𝑑𝑥 ± 𝑔(𝑥 )𝑑𝑥 = ∫ 𝑓 (𝑥 )𝑑𝑥 ± ∫ 𝑔(𝑥 )𝑑𝑥
𝑎 𝑏 𝑎

C. Interval of Zero Length

𝑎
∫ 𝑓 (𝑥 )𝑑𝑥 = 0
𝑎

D. Adding of Intervals

𝑏 𝑐 𝑏
∫ 𝑓 (𝑥 )𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑓(𝑥)𝑑𝑥
𝑎 𝑎 𝑐

Example 1:
Evaluate
5
∫ (3𝑥 2 + 4𝑥 + 1).
1
Solution:
1) Treat first the problem as a definite integral but omit the arbitrary constant C.
5
3𝑥 3 4𝑥 2
∫ (3𝑥 2 + 4𝑥 + 1) = + +𝑥
1 3 2
= 𝑥 3 + 2𝑥 2 + 𝑥

2) Next, substitute the upper limit into the integral.

[(5)3 + 2(5)2 + 5] = 125 + 50 + 5 = 180

3) Then substitute the lower limit into the integral.

[(1)3 + 2(1)2 + 1] = 1 + 2 + 1 = 4

4) Subtract the result of (3) from the result of (2).

5 5
∫ (3𝑥 2 + 4𝑥 + 1) = 𝑥 3 + 2𝑥 + 𝑥 |
1 1

= [(5)3 + 2(5)2 + 5] − [(1)3 + 2(1)2 + 1]

= 180 − 4
= 176
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Example 2: Using substitution

Evaluate
0
∫ 𝑥 3 (1 − 2𝑥 4 )3 𝑑𝑥 .
−1

Solution:
Let 𝑢 = 1 − 2𝑥 4 , 𝑑𝑢 = −8𝑥 3 𝑑𝑥
0
1 0
∫ 𝑥 3 (1 − 2𝑥 4 )3 𝑑𝑥 = − ∫ −8𝑥 3 (1 − 2𝑥 4 )3 𝑑𝑥
−1 8 −1
\
1 0
= − ∫ 𝑢3 𝑑𝑢
8 −1

1 𝑢4
=− [ ]
8 4

𝑢4
=−
32

a) First approach, using the variable 𝑥.

0 (1 − 2𝑥 4 )4 0
∫ 𝑥 3 (1 − 2𝑥 4 )3 𝑑𝑥 = − |
−1 32
−1

{1 − 2(0)4 } − {1 − 2(−1)4 }
= −[ ]
32

=0

b) Second approach, with variable changed.

If we use the transformation of 𝑥iinto 𝑢, then we must change also the upper and
lower limits.

𝑥2 =0
∫ 𝑥 3 (1 − 2𝑥 4 )3 𝑑𝑥
𝑥1= −1

𝑢 = 1 − 2𝑥 4
𝑢2 = 1 − 2(0)4 = 1
𝑢1 = 1 − 2(−1)4 = −1

0
𝑢4 1
∫ 𝑥 3 (1 − 2𝑥 4 )3 𝑑𝑥 = − |
−1 32
−1

14 (−1)4
= −[ − ]
32 32

1 1
= −[ − ]
32 32

=0
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PRACTICE PROBLEMS 1.10

Evaluate the following definite integrals.

9
1. ∫ (2𝑥 + 3√𝑥)𝑑𝑥
4

−2
2. ∫ 𝑒 −𝑥 𝑑𝑥
−4

1.6
6
3. ∫ (5 + ) 𝑑𝑥
1.2 𝑥4

4
4. ∫ ln 𝑥 𝑑𝑥
2

5. If it is known that

8 5 8
∫ 𝑓(𝑥)𝑑𝑥 = 10, and ∫ 𝑓(𝑥)𝑑𝑥 = 5, find the value of ∫ 𝑓(𝑥)𝑑𝑥.
0 0 5

VI. REFERENCES:

Books/Printed Resources:

Larson, Ron et. al. Calculus Early Transcendentals 4th Ed., Houghton
Mifflin Company, New York, 2007

Hughes-Hallet, D. et. al. Calculus Multivariable 7th Ed., John Wiley and Sons,
Inc, New Jersey, 2017

Prepared by:

ENGR. JAY CON T. ACOSTA

Emergency Instructor

Recommending approval:

ENGR. LARRY P. REMOLAZO, M.Eng.

Program Chair, BSME

Approved by:

ENGR. MARY B. PASION, D.Eng.

Dean, College of Engineering

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