W E E K 1 : B A S I C I N T E G R AT I O N R U L E S

LEARNING OUTCOMES

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

• Differentiate the indefinite and definite integrals;

• Evaluate simple integrals by reversing the process of differentiation

• Explain the need for a constant of integration when finding indefinite

integrals, and

• Apply the basic integration formulas in evaluating the indefinite and definite
integrals.
 INTEGRAL CALCULUS

A branch of mathematics concerned with the theory of and applications (the

determination of lengths, areas and volumes and in the solution of
differential equations) of integrals and integration (Merriam – Webster
dictionary)

Antiderivatives are also known as the Indefinite Integrals. They are the
reverse operation of differentiation wherein the original function being
solved for, given the derivative. A function F(x) is referred to as
antiderivative of f(x) on an interval I when F ′ x = f(x) on all x of the
interval.
 INTEGRAL CALCULUS

If F’(x) = f(x) or d(F(x)) = f(x)dx , then the anti-derivative of f(x) is defined as the
indefinite integral of f(x) and is mathematically expressed as
න𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝑐

where: f(x) is the integrand

is the integral symbol
x is the variable of integration
dx is the differential of the variable
F(x) + c is the antiderivative
c is the constant of integration
 INTEGRAL CALCULUS

• Integration is the process or method of getting the antiderivative of a given function.

• Differentiation (the process of taking the derivative) and integration or anti – differentiation (the
process of finding the integral) are inverse processes. That means that one basically “undoes” the other.

Suppose you start with the function f x = x 3 .

Differentiating f(x) gives,

d 3
x = 3x 2
dx

Integrating this derivative, we get

x3
න 3x 2 = 3 + c = x3 + c
3

As a result, we get the original function back with the addition of the arbitrary constant of integration.
 T H E F U N DA M E N TA L T H E O R E M O F C A L C U L U S

If F(x) is the antiderivative of the function f(x) i.e. F ′ x = f(x), then

‫ ׬‬f x dx = ‫ ׬‬F"(x)dx = ‫ ׬‬dF = F x + C.

Thus, the integral of the differential of a function F is equal to the function itself plus an arbitrary
constant C.

Here are some examples

𝑥4 𝑥4
1. ‫ 𝑥 ׬‬3 𝑑𝑥 = ‫𝑑 ׬‬ 4
=
4
+𝐶

1 2𝑥 1
2. ‫ 𝑒 ׬‬2𝑥 𝑑𝑥 = ‫𝑑 ׬‬ 2
𝑒 = 𝑒 2𝑥 + 𝐶
2

1 1
3. ‫ ׬‬cos(5𝑥)𝑑𝑥 = ‫𝑑 ׬‬ 5
sin(5𝑥) = sin(5𝑥) + 𝐶
5
 2 TYPES OF INTEGRAL

Definite Integral - consists of the upper and lower limit, i.e. x = a and x = b
𝑏
න 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎)
𝑎

Indefinite Integral - does not have upper and lower limit

‫ )𝑥(𝐹 = 𝑥𝑑 )𝑥(𝑓 ׬‬+ 𝑐

 B A S I C I N T E G R AT I O N R U L E S

1. Identity Rule ‫ ׬‬du = u + c

2. Constant Rule න adu = a න du

n+1
u
3. Power Rule න un du = + c,n ≠ 1
n+1
4. Sum & Difference න f(x) ± g(x) dx = න f(x)dx ± න g(x)dx
du
5. Logarithmic Rule ‫׬‬ u
= Ln u + c , n ≠ 1 n ≥ 0
 EXAMPLES

Find the indicated integral.

1
Problem 1: ‫ ׬‬2𝑥 𝑑𝑥 3

Apply the following rules:

Power Rule , and

Constant Rule

Simplify the expression

Final Answer
 EXAMPLES

1
4
Problem 2 :‫(׬‬2x +
4 − 3x 4 )dx
x
Apply the following rules:

Power Rule , and

Constant Rule

Simplify the expression

Final Answer
 EXAMPLES

𝑦 3 −64
Problem 3 : ‫𝑦 ׬‬2 +4𝑦+16 𝑑𝑦
The first step is to rewrite the function and simplify it so we can apply the one of the basic
integration formulas.
Determine factors of the numerator in the integrand

(Recall: Difference of two perfect cubes)

Cancel out common factor

Integrate applying power rule and constant rule

Final Answer
 EXAMPLES

y3 −2y2 +y−1
Problem 4 :
‫׬‬ y2
dy

Write as separate fractions

Integrate each term separately applying the following rules:

Power Rule , and

Constant Rule

Logarithmic Rule

Simplify the expression

Final Answer
 EXAMPLES

Problem 5 :‫ ׬‬3𝑥 2 − 5𝑥 + 3 𝑑𝑥

Express as

Express the radicals into fractional exponent

Integrate each term separately and simplify

Final Answer

Note: To rewrite a radical using a fractional exponent, the power to which the
radicand is raised becomes the numerator and the root becomes the denominator.
 EXAMPLES

Problem 6 :‫ ׬‬x+4
x
dx

Write as separate fractions

Express the radicals into fractional exponents

Integrate each term separately

Simplify the expression

Final Answer
 EXAMPLES

2x2 +4x−3
Problem 7 :‫׬‬ x2
dx
Write as separate fractions

Integrate each term separately

Simplify the expression

Final Answer
 EXAMPLES

2
Problem 8 : ‫׬‬0 6𝑥 2 − 4𝑥 + 5 𝑑𝑥
Integrate each term separately

Recall:

Evaluate by replacing x by the given

values of the upper limit and lower limit

Simplify

Final Answer
 EXAMPLES
1
Problem 9 : ‫׬‬0 4 + 3 𝑥 − 2𝑥 𝑥 𝑑𝑥
Express radicals to fractional exponents

Integrate each term separately

Recall:

Evaluate by replacing x by the given values of the

upper limit and lower limit

Simplify the expression

Final Answer
 EXAMPLES
2
Problem 10 : ‫׬‬−2 2x − 1 3x + 4 dx
Multiply using FOIL method

Integrate each term separately

Simplify the expression

Recall:

Evaluate by replacing x by the given values of

the upper limit and lower limit

Simplify the expression

Final Answer
 EXAMPLES
4
Problem 11: ‫׬‬1 x x − 1 dx
Distribute x then rewrite radicals to fractional exponents

Integrate each term separately

Simplify the expression

Recall:

Evaluate by replacing x by the given values of the upper limit and

lower limit
 EXAMPLES

Problem 11:
2 1 2 1
= (32) − (16) − (1) − (1) Simplify
5 2 5 2

64 2 1
= −8 − −
5 5 2

49
= Final Answer
10
 REFERENCES
Printed References
1. Stewart, J. (2019). Calculus concepts and contexts 4th edition enhanced edition. USA :
Cengage Learning, Inc.
2. Canlapan, R.B.(2017). Diwa senior high school series: basic calculus. DIWA Learning
Systems Inc.
3. Hass, J., Weir, M. & Thomas, G. (2016). University calculus early transcendentals,global
edition. Pearson Education, Inc.Larson,
4. Bacani, J.B.,et.al. (2016). Basic calculus for senior high school. Book AtBP. Publishing
Corp.
5. Balmaceda, J.M. (2016). Teaching guide in basic calculus. Commission on Higher
Education.
6. Pelias, J.G. (2016). Basic calculus. Rex Bookstore.
7. Molina, M.F. (2016). Integral calculus. Unlimited Books Library Services & Publishing
Inc.
8. Larson, R & Edwards, B. (2012). Calculus. Pasig City, Philippines. Cengage Learning
Asia Pte Ltd

Electronic Resources
1. Pulham , John. Differential Calculus. Retrieved August 7, 2020 from http://
www.maths.abdn.ac.uk/igc/tch/math1002/dif
2. Weistein, Eric. Differential Calculus. Retrieved August 7, 2020 from
http://mathworld.wolfram.com/differential calculus
3. Thomas , Christopher . Introduction to Differential Calculus. Retrieved August 7, 2020
from http:// www.usyd.edu.aav/stuserv/document/maths_learning-center
4. Purplemath. Retrieved August 7, 2020 from http://www.purplemath.com/modules/
 WEEK 2: INTEGRATION BY
SUBSTITUTION

LEARNING OUTCOME

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

• Evaluate integrals using method of substitution.

 REVIEW OF BASIC INTEGRATION
FORMULAS

Each of the basic integration rules you studied in the first week was
derived from a corresponding differentiation rule. Although you now have
all the necessary tools for differentiating algebraic, exponential, and
logarithmic functions, your set of tools for integrating these functions is
by no means complete. The main objective of this lesson is to develop
several techniques that greatly expand the set of integrals to which the
basic integration formulas can be applied.
 B A S I C I N T E G R AT I O N R U L E S

1. Identity Rule du = u + c

2. Constant Rule adu = a du

n+1
u
3. Power Rule un du = + c,n ≠ 1
n+1
4. Sum & Difference f(x) ± g(x) dx = f(x)dx ± g(x)dx
du
5. Logarithmic Rule
u
= Ln u + c , n ≠ 1 n ≥ 0
 I N T E G R AT I O N B Y S U B S T I T U T I O N

There are several techniques for rewriting an integral so that it fits one or
more of the basic formulas. One of the most powerful techniques is
integration by substitution. With this technique, identify the appropriate u
in the given integrand, then the differential du should be the remaining
integrand; otherwise, change u.

Substitute u and du in order to apply the basic rules of integration.

 G U I D E L I N E S F O R I N T E G R AT I O N B Y S U B S T I T U T I O N

Many students find it difficult to figure out the substitution since for different functions
the substitutions are also different. However, there is a general rule in substitution,
namely to change the composite function into a simple, elementary function.
Guidelines for Integration by Substitution

1. Let u be a function of x (usually part of the integrand).

2. Solve for x and dx in terms of u and du.

3. Convert the entire integral to u – variable form and try to fit it to one or more of
the basic integration formulas. If none fits, try a different substitution.

4. After integrating, rewrite the antiderivative as a function of x

 EXAMPLES

Find the indicated integral.

Problem 1: 5 − 3x dx
Let

Replace x and dx with the appropriate u – variable forms

Express radical to fractional exponent i.e change to

Integrate applying basic integration rule

Simplify the expression

Substitute back

Final Answer
 EXAMPLES

Problem 2 : 5x − 4 7 dx
Let

Replace x and dx with the appropriate u – variable forms

Integrate applying basic integration rule

Simplify the expression

Substitute back

Final Answer
 EXAMPLES

3
2
Problem 3 : 5z − 7 zdz
2

Let

Replace x and dx with the appropriate u – variable forms

Integrate applying basic integration rule

2
Simplify the expression

1
25 Substitute back

1
25 Final Answer
 EXAMPLES

dy
Problem 4 : 5y−1

Let

Replace x and dx with the appropriate u – variable forms

Integrate applying basic integration rule

Substitute back

Final Answer
 EXAMPLES

Problem 5 : x 3 + 2x 5
6x 2 + 4 dx

Factor out 2 in the integrand

Let

Replace x and dx with the appropriate u – variable forms

Integrate applying basic integration rule

Substitute back , then simplify

Final Answer
 EXAMPLES
(2x−5)dx
Problem 6 : 3
2x2 −10x+3
Let

Replace x and dx with the appropriate u – variable forms

Integrate applying basic integration rule

Simplify the expression

Substitute back

Final Answer
 EXAMPLES

Problem 7 : x 2 3x 3 − 1 4 dx
Let

Replace x and dx with the appropriate u – variable forms

Integrate applying the basic integration rule

Simplify the expression

Substitute back

Final Answer
 EXAMPLES

Let
z+1
3 dz 2
du
z 2 + 2z + 1 u= z+1 du = 2 z + 1 dz = (z + 1)dz
2
z+1 Rewrite z 2 + 2z + 1 as z + 1 2
= dz
3
z+1 2 Replace z and dz with the appropriate u – variable forms

1 du 1 1
−3
= 3 ∗ Express 3 as u
u 2 u
1 −
1
= u 3 du Integrate applying the basic integration rule
2
2
1 u3
= ∗ +c Simplify the expression
2 2
3
1 3 2 2
= ∗ (u)3 + c Substitute back u = z + 1
2 2
3 2
= z+1 3 +c Final Answer
4
 EXAMPLES
4
1 5
v −7
3
Problem 9 : 2 dv
v3
Let

Replace x and dx with the appropriate u – variable forms

Integrate applying the basic integration rule

Simplify the expression

Substitute back and simplify

Final Answer
 EXAMPLES
dx
Problem 10 :
x Lnx 2
Let

Replace x and dx with the appropriate u – variable forms

Integrate applying the basic integration rule

Simplify the expression

Substitute back

Final Answer
 EXAMPLES

𝑒 2𝑥
Problem 11 : 𝑑𝑥
1+𝑒 2𝑥
Let

Rewrite the function in the integrand

Replace x and dx with the appropriate u – variable forms

Integrate applying the basic integration rule

Simplify the expression

Substitute back

Final Answer
 EXAMPLES
dx
Problem 12 : x+2
Let

Replace x and dx with the appropriate u – variable forms

Integrate applying the basic integration rule

Substitute back

Simplify the expression

= Final Answer
 REFERENCES
Printed References
1. Stewart, J. (2019). Calculus concepts and contexts 4th edition enhanced edition. USA :
Cengage Learning, Inc.
2. Canlapan, R.B.(2017). Diwa senior high school series: basic calculus. DIWA Learning
Systems Inc.
3. Hass, J., Weir, M. & Thomas, G. (2016). University calculus early transcendentals,global
edition. Pearson Education, Inc.Larson,
4. Bacani, J.B.,et.al. (2016). Basic calculus for senior high school. Book AtBP. Publishing
Corp.
5. Balmaceda, J.M. (2016). Teaching guide in basic calculus. Commission on Higher
Education.
6. Pelias, J.G. (2016). Basic calculus. Rex Bookstore.
7. Molina, M.F. (2016). Integral calculus. Unlimited Books Library Services & Publishing
Inc.
8. Larson, R & Edwards, B. (2012). Calculus. Pasig City, Philippines. Cengage Learning
Asia Pte Ltd

Electronic Resources
1. Pulham , John. Differential Calculus. Retrieved August 7, 2020 from http://
www.maths.abdn.ac.uk/igc/tch/math1002/dif
2. Weistein, Eric. Differential Calculus. Retrieved August 7, 2020 from
http://mathworld.wolfram.com/differential calculus
3. Thomas , Christopher . Introduction to Differential Calculus. Retrieved August 7, 2020
from http:// www.usyd.edu.aav/stuserv/document/maths_learning-center
4. Purplemath. Retrieved August 7, 2020 from http://www.purplemath.com/modules/
Learning Outcomes

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

1. Use trigonometric identities to write integrands in alternative forms;

and

2. Evaluate integral of trigonometric functions

Calculus II: Integral Calculus

Introduction
For functions other than the algebraic expressions, the integration
process is quite different from the integration rules applied to
Transcendental Functions.

Trigonometric, exponential and logarithmic functions are collectively

called Transcendental Functions. They are functions that cannot be
written using the algebraical operations of addition, subtraction,
multiplication or division. Thus, their integrals are evaluated using a
set of rules beyond those of algebraic functions.

Calculus II: Integral Calculus

Introduction

This week’s lesson focuses mainly on the Integration of

Trigonometric Functions. The Basic Integration Formulas and
Integration by Substitution will still be used and applied as we
present and discuss examples on the integral of trigonometric
functions. But before that, may we first recall the following
differentiation of trigonometric functions:

Calculus II: Integral Calculus

Differentiation of Trigonometric Functions
1. d du 4. d 2
du
sin u = cosu cot u = −csc u
dx dx dx dx
5. d du
2. d du sec u = secutanu
cos u = −sinu dx dx
dx dx

3. d du 6. d du
2
tan u = sec u csc u = −cscucotu
dx dx dx dx

Calculus II: Integral Calculus

Integration of Trigonometric Functions

නsin u du = − cos u + c නcsc u cot u du = − csc u + c

නcos u du = sin u + c නtan u du = −Ln cos u + c = Ln sec u + c

නsec 2 u du = tan u + c නcot u du = Ln sin u + c

නcsc 2 u du = − cot u + c නsec u du = Ln sec u + tan u + c

නsec u tan u du = sec u + c නcsc u du = −Ln csc u + cot u + c

Calculus II: Integral Calculus

 Example 1
The first step is to rewrite the function and simplify it so
we can apply Rule No. 6 of integral of trigonometric
න 𝒄𝒔𝒄 𝟓 𝐱 𝒄𝒐𝒕 𝟓 𝒙 𝒅𝒙 functions.
du
u = 5x du = 5dx = dx
5
du
= න csc u cot u Replace x and dx with the appropriate u – variable forms
5
1
= න csc u cot u du Integrate using Rule No. 6
5
1
= − csc u + c Substitute back u = 5x
5
𝟏
= − 𝒄𝒔𝒄 𝟓 𝐱 + 𝐜 Final Answer
𝟓
Calculus II: Integral Calculus
 Example 2
The first step is to rewrite the function and simplify it so
න𝒔𝒊𝒏 𝟗 𝒙𝒅𝒙 we can apply Rule No. 1 of integral of trigonometric
functions.
du
u = 9x du = 9dx = dx
9
du Replace x and dx with the appropriate u – variable
= නsin u ∗ forms
9
1
= නsinudu Integrate using Rule No. 1
9
1
= − cos u + c Substitute back u = 9x
9
𝟏
= − 𝒄𝒐𝒔 𝟗𝐱 + 𝐜 Final Answer
𝟗
Calculus II: Integral Calculus
 Example 3

න 𝟓𝒙𝟑 − 𝟐 𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙

= 5 න x 3 − 2 නsec 2 x dx Integrate each term separately and apply Rule No. 3

𝟓𝐱 𝟒 Final Answer
= − 𝟐 𝐭𝐚𝐧 𝐱 + 𝐜
𝟒

Calculus II: Integral Calculus

 Example 4
𝒄𝒐𝒔 𝟕 𝒙
න 𝒅𝒙 Rewrite the function and apply integration by substitution.
𝟓 + 𝒔𝒊𝒏 𝟕 𝒙 u = 5 + sin 7 x du = 7cos7xdx
du
= cos7x dx
7

1 du
=න ∗ Replace x and dx with the appropriate u – variable forms
u 7
1 du
= න Apply Basic Integration Formula
7 u
1
= Ln u + c Substitute back u = 5 + sin7x
7
𝟏
= 𝐋𝐧 𝟓 + 𝐬𝐢𝐧 𝟕 𝐱 + 𝐜 Final Answer
𝟕
Calculus II: Integral Calculus
 Example 5
𝒔𝒊𝒏 𝟓𝒙 − 𝟏 Rewrite the function and recall that
න 𝒅𝒙 sin u
𝒄𝒐𝒔 𝟓𝒙 − 𝟏 tan u =
cos u
du
= නtan 5x − 1 dx Let u = 5x − 1 du = 5dx = dx
5
du
= න tan u Replace x and dx with the appropriate u – variable forms
5
1
= නtan u du Integrate using Rule No. 7
5
1
= − Ln cos u + c Substitute back u = 5 + sin7x
5
𝟏
= − 𝐋𝐧 𝐜𝐨𝐬 𝟓𝒙 − 𝟏 + 𝐜 Final Answer
𝟓
Calculus II: Integral Calculus
 Example 6

𝒄𝒐𝒔𝟐 𝒙 Rewrite the function and recall that

න 𝒅𝒙 cos2 x = 1 − sin2 x
𝟏 + 𝒔𝒊𝒏 𝒙

1 − sin2 x Factor the numerator

=න dx a2 − b2 = (a − b)(a + b)
1 + sin x
1 − sin x 1 + sin x
=න dx Cancel common factors
1 + sin x
Integrate each term separately and apply Rule No. 1 and
= ‫(׬‬1 − sin x) dx Basic Integration Formula

= 𝐱 + 𝐜𝐨𝐬 𝐱 + 𝐜 Final Answer

Calculus II: Integral Calculus

 Example 7
Rewrite the function and recall that
න𝒕𝒂𝒏𝟐 𝟑 𝒚𝒅𝒚 tan2 x = sec 2 x − 1

= න sec 2 3 y − 1 dy Integrate each term separately

du
= න sec 2 3 ydy − නdy Let u = 3y du = 3dy = dy
3
du Integrate using Rule No. 3 and Basic Integration
= න sec 2 u ∗ − නdy Formula
3
1
= tan u − y + c Substitute back u = 3y
3
𝟏
= 𝐭𝐚𝐧( 𝟑𝐲) − 𝐲 + 𝐜 Final Answer
𝟑
Calculus II: Integral Calculus
 Example 8
𝟏
න 𝐝𝐱
𝒔𝒊𝒏 𝟑 𝐱 𝒕𝒂𝒏 𝟑 𝒙
1 1 Rewrite the function and apply the following identities
=න ∗ dx 1 1
sin 3 x tan 3 x csc 3x =
sin 3 x
and cot 3x =
tan 3 x
du
= නcsc 3 x cot 3 xdx Let u = 3x du = 3dx = dx
3
du
= නcsc u cot u ∗ Integrate applying Rule No. 6
3
1
= − csc u + c Substitute back u = 3x
3
𝟏
= − 𝐜𝐬𝐜 𝟑𝒙 + 𝐜 Final Answer
𝟑
Calculus II: Integral Calculus
 Example 9
𝟐
න 𝒔𝒊𝒏 𝟓 𝜽 − 𝒄𝒐𝒔 𝟓 𝜽 𝐝𝛉

= න sin2 5 θ − 2 sin 5 θ cos 5 θ + cos 2 5 θ dθ Expand the binomial and rearrange the terms

Recall:
= න sin2 5 θ + cos 2 5 − 𝟐 𝐬𝐢𝐧 𝟓 𝛉 𝐜𝐨𝐬 𝟓 𝛉 dθ sin2 u + cos2 u = 1
sin 2 u = 2 sin u cos u

Integrate separately
= ‫ ׬‬1 − sin 1 0θ dθ du
Let u = 10θ du = 10dθ = dθ
10

Calculus II: Integral Calculus

 Example 9

du
= ‫ ׬‬dθ − ‫ ׬‬sin u Apply Rule No.1 and Basic Integration Formula
10
1
= θ + cos u + c Substitute back u = 10θ
10
𝟏
=𝛉+ 𝐜𝐨𝐬 𝟏𝟎𝛉 + 𝐜 Final Answer
𝟏𝟎

Calculus II: Integral Calculus

 Example 10
𝒄𝒐𝒔 𝟖 𝒙
න 𝟐
𝒅𝒙
𝒄𝒐𝒔 𝟒 𝒙
cos 2 (4x) Apply the Double angle identity
=න 2
dx cos2x = cos 2 x − sin2 x
cos 4 x
cos 2 4 x − sin2 4 x
=න 2
dx Separate the terms
cos 4 x
cos2 4 x sin2 4 x Recall that
=න − dx sin2 4 x
2
cos 4 x cos 4 x 2
2
= tan2 4 x
cos 4 x
Recall that
= න 1 − tan2 4x dx tan2 4 x = sec 2 4 x − 1

Calculus II: Integral Calculus

 Example 10

= න 1 − (sec 2 4 x − 1) dx Simplify the expression

Integrate each term separately, then apply BIF and

= න 2 − sec 2 4 x dx Rule No. 3

𝟏
= 𝟐𝒙 − 𝒕𝒂𝒏 𝟒 𝐱 + 𝐜 Final Answer
𝟒

Calculus II: Integral Calculus

 Example 11

𝒅𝒙
න
𝟏 + 𝒔𝒊𝒏 𝒙
1 1 − sin x Multiply both the numerator and denominator of the
=න ∗ dx integrand by 1 – sinx
1 + sin x 1 − sin x
1 − sin x Recall that
=න 2
dx 1 − sin2 u = cos 2 u
1 − sin x
1 − sin x
=න 2
dx Separate the terms
cos x

Calculus II: Integral Calculus

 Example 11

1 1 sin x Recall:
=න 2
− ∗ dx 1
= sec x and
sin x
= tanx
cos x cos x cos x cos x cos x

Integrate each term separately applying

= න sec 2 x − sec x tan x dx Rule No. 3 and Rule No. 5

= 𝐭𝐚𝐧 𝐱 − 𝐬𝐞𝐜 𝐱 + 𝐜 Final Answer

Calculus II: Integral Calculus

 Example 12
𝒔𝒊𝒏𝟑 𝒙 Split sin3 x as sinx ∗ sin2 x
න 𝒅𝒙
𝒄𝒐𝒔𝒙 − 𝟏 Rewrite cosx − 1 as − 1 − cos x

sin2 x sin x Apply the Trigonometric Identity

=න dx sin2 x = 1 − cos 2 x
− 1 − cos x
(1 − cos2 x) sin x Determine factors of 1 − cos2 x
= −න dx 1 − cos2 x = (1 − cosx)(1 + cosx)
(1 − cos x)
(1 − cos x)(1 + cos x) sin x
= −න dx Cancel out common factors
(1 − cos x)

= − න(1 + cos x) sin x dx Distribute sin x

= − න(sin x + cos x sin x)dx Apply the identity sin2x = 2sinxcosx

Calculus II: Integral Calculus

 Example 12

sin 2 x Integrate each term separately applying Rule No. 3

= − න(sin x + )dx du
2 Let u = 2x du = 2dx
2
= dx

sin u d u
= − නsin x dx − න Simplify the expression
2 2
1
= − නsin x dx − නsin u du Apply Rule No. 3
4
1
= cos x + cos u + c Substitute back u = 2x
4
𝟏
= 𝒄𝒐𝒔 𝒙 + 𝒄𝒐𝒔 𝟐 𝐱 + 𝐜 Final Answer
𝟒

Calculus II: Integral Calculus

 Example 13

𝟒 𝒔𝒊𝒏𝟐 𝒙 𝒄𝒐𝒔𝟐 𝒙
න 𝐝𝐱
𝒔𝒊𝒏 𝟐 𝐱 𝒄𝒐𝒔 𝟐 𝒙

2 sin x cos x 2 Rewrite

=න dx 4 sin2 x cos 2 x as 2sinxcosx 2
sin 2 x cos 2 x
sin 2 x 2 Apply the identity:
=න dx sin2x = 2sinx cosx
sin 2 x cos 2 x
sin 2 x 2
=න dx Cancel out the common term
sin 2 x cos 2 x
Calculus II: Integral Calculus
 Example 13
sin 2 x Recall
=න dx sin 2 x
= tan2x
cos 2 x cos 2 x
du
= නtan 2 x dx Let u = 2x du = 2dx = dx
2

du
= නtan u Integrate using Rule No. 7
2
1
= Ln cos u + c Substitute back u = 2x
2
𝟏
= − 𝐋𝐧 𝐜𝐨𝐬 𝟐 𝐱 + 𝐜 Final Answer
𝟐
Calculus II: Integral Calculus
Note:

The presence of the absolute value symbol is

needed to ensure that the natural logarithm
will have a defined value, because logarithms
are not defined for negative n umbers.

Calculus II: Integral Calculus

Printed References

1. Stewart, J. (2019). Calculus concepts and contexts 4th edition enhanced edition. USA : Cengage
Learning, Inc.
2. Canlapan, R.B.(2017). Diwa senior high school series: basic calculus. DIWA Learning Systems Inc.
3. Hass, J., Weir, M. & Thomas, G. (2016). University calculus early transcendentals,global edition.
Pearson Education, Inc.Larson,
4. Bacani, J.B.,et.al. (2016). Basic calculus for senior high school. Book AtBP. Publishing Corp.
5. Balmaceda, J.M. (2016). Teaching guide in basic calculus. Commission on Higher Education.
6. Pelias, J.G. (2016). Basic calculus. Rex Bookstore.
7. Molina, M.F. (2016). Integral calculus. Unlimited Books Library Services & Publishing Inc.
8. Larson, R & Edwards, B. (2012). Calculus. Pasig City, Philippines. Cengage Learning Asia Pte Ltd

Calculus II: Integral Calculus

Electronic Resources

1. Pulham , John. Differential Calculus. Retrieved August 7, 2020 from http://

www.maths.abdn.ac.uk/igc/tch/math1002/dif
2. Weistein, Eric. Differential Calculus. Retrieved August 7, 2020 from
http://mathworld.wolfram.com/differential calculus
3. Thomas , Christopher . Introduction to Differential Calculus. Retrieved August 7,
2020 from http:// www.usyd.edu.aav/stuserv/document/maths_learning-center
4. Purplemath. Retrieved August 7, 2020 from http://www.purplemath.com/modules/

Calculus II: Integral Calculus

Learning Outcomes

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

evaluate indefinite and definite integrals of
trigonometric functions by method of substitution.

Calculus II: Integral Calculus

Introduction

The topic for this week is a continuation of the previous

lesson, Integration of Trigonometric Functions but
examples to be presented and discussed will be more
complicated than the examples in the previous lesson.

Calculus II: Integral Calculus

Introduction

Again, we recall the following rules on the integration of

trigonometric functions:
Integration of Trigonometric Functions
Let u be the angle of any trigonometric function

Calculus II: Integral Calculus

Integration of Trigonometric Functions

නsin u du = − cos u + c නcsc u cot u du = − csc u + c

නcos u du = sin u + c නtan u du = −Ln cos u + c = Ln sec u + c

නsec 2 u du = tan u + c නcot u du = Ln sin u + c

නcsc 2 u du = − cot u + c නsec u du = Ln sec u + tan u + c

නsec u tan u du = sec u + c නcsc u du = −Ln csc u + cot u + c

Calculus II: Integral Calculus

Example 1

du
න 𝟏 − 𝒔𝒊𝒏 𝟓 𝒙 𝟏𝟎 𝒄𝒐𝒔 𝟓 𝒙 𝐝𝐱 Let u = 1 − sin 5 x du = −5cos5xdx − 5 = cos5xdx
Replace x and dx with the appropriate u – variable forms

du
= න u10 Simplify the expression
−5
1 10
= − නu du Integrate using the basic integration rule
5
1 u11
=− 5 11 +c Substitute back u = 1 − sin 5 x
𝟏𝟏
𝟏 − 𝒔𝒊𝒏 𝟓 𝒙
=− +𝐜 Final Answer
𝟓𝟓

Calculus II: Integral Calculus

Example 2

Let
𝒔𝒆𝒄 𝟐 𝐱 𝒕𝒂𝒏 𝟐 𝒙 du
න 𝒅𝒙 u = sec 2 x − 1 du = 2 sec 2 x tan 2 xdx 2
=
𝟑
𝒔𝒆𝒄 𝟐 𝒙 − 𝟏 sec 2 x tan 2 xdx
Replace x and dx with the appropriate u – variable forms
1 du du −3
1
= න 1 Rewrite 1 as u
2
u3 u3
1 −3
1
= න u du Integrate using the basic integration rule
2

Calculus II: Integral Calculus

Example 2

2
1 u3
= +c Simplify the expression
2 2
3
3 2
= u3 + c Substitute back u = 1 − sin 5 x
4
𝟑 𝟐
= 𝒔𝒆𝒄 𝟐 𝒙 − 𝟏 𝟑 +𝐜 Final Answer
𝟒

Calculus II: Integral Calculus

Example 3
Let
𝒔𝒆𝒄𝟐 𝒙 du
න 𝒅𝒙 u = a + b tan x du = b sec 2 b xdx = sec 2 x dx
𝐚 + 𝐛 𝒕𝒂𝒏 𝒙 b
Replace x and dx with the appropriate u – variable forms
1 du
=න ∗
u b
1 du
= න Integrate using the basic integration rule
b u
1
= Ln u + c Substitute back u = a + b tan x
b
𝟏
= 𝐋𝐧 𝐚 + 𝐛 𝒕𝒂𝒏 𝒙 + 𝐜 Final Answer
𝒃
Calculus II: Integral Calculus
Example 4
Let

𝝅
u = cos x + sin x
𝟒 𝒄𝒐𝒔 𝒙 − 𝒔𝒊𝒏 𝒙
න 𝒅𝒙 du = − sin x + cos x dx
𝟎 𝒄𝒐𝒔 𝒙 + 𝒔𝒊𝒏 𝒙
du = cos x − sin x dx
Replace x and dx with the appropriate u – variable forms
π
4 du
=න Integrate using the basic Integration rule
0 u
Recall:
b
π
4 න f(x) dx = F(b) − F(a)
= Ln u 0 a

Evaluate by replacing x by the given values of the upper

limit and lower limit
Calculus II: Integral Calculus
Example 4

π
= Ln cos x + sin x4 Integrate using the basic integration rule
0
π π
= Ln cos + sin
4 4 Simplify the expression
− Ln cos 0 + sin 0
2 2 Simplify the expression
= Ln + − Ln1
2 2
= 𝐋𝐧 𝟐 Final Answer

Calculus II: Integral Calculus

Example 5
1 1
Let u = x du = dx 2du = dx
𝟏 2 x x
න 𝒔𝒊𝒏 𝒙 𝒅𝒙
𝒙 Replace x and dx with the appropriate u – variable
forms

= නsin u (2du)

= 2 නsin u du Apply integration of trigonometric functions

= −2 cos u + c Substitute back u = x

= −𝟐 𝒄𝒐𝒔 𝒙 + 𝐜 Final Answer

Calculus II: Integral Calculus

Example 6
3 2
du
න𝒙𝟐 𝒔𝒆𝒄𝟐 𝒙𝟑 𝒅𝒙 Let u = x du = 3x dx = x 2 dx
3
Replace x and dx with the appropriate u – variable forms
2
du
= න sec u ∗
3
1
= නsec 2 u du
3 Apply integration of trigonometric functions

1
= tan u + c Substitute back u = x 3
3
𝟏
= 𝒕𝒂𝒏 𝒙𝟑 + 𝐜 Final Answer
𝟑

Calculus II: Integral Calculus

Example 7
𝟑
𝒄𝒐𝒔 𝒙 3 3 du 1
න 𝒅𝒙 Let u = du = − 2 dx − = 2 dx
𝒙𝟐 x x 3 x
Replace x and dx with the appropriate u – variable forms

1
= න cos u ∗ − du
3

1
= − නcos u du Apply integration of trigonometric functions
3
1 3
= − sin u + c Substitute back u =
3 x
𝟏 𝟑
= − 𝒔𝒊𝒏 +𝐜 Final Answer
𝟑 𝒙

Calculus II: Integral Calculus

Example 8
Let u = cos x du = − sin x dx − du = sin x dx
න𝒄𝒐𝒔 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝒙 𝒅𝒙 Replace x and dx with the appropriate u – variable
forms

= නcos u −du Simplify

= − නcos u du Apply integration of trigonometric functions

= − sin u + c 3
Substitute back u =
x
= − 𝒔𝒊𝒏 𝒄𝒐𝒔 𝒙 + 𝒄 Final Answer

Calculus II: Integral Calculus

Example 9
𝒄𝒔𝒄𝟑𝒙𝒄𝒐𝒕𝟑𝒙
න 𝒅𝒙 1 1
−2
𝒄𝒔𝒄 𝟑 𝒙 𝑅𝑒𝑤𝑟𝑖𝑡𝑒 𝑎𝑠 csc 3 x
csc 3 x

Let
1 du
−2
= න csc 3 x csc 3 x cot 3 xdx u = csc 3 x du = −3 csc 3 x cot 3 xdx − = csc 3 x cot 3 xdx
3
Replace x and dx with the appropriate u – variable forms
1 1
−2
=− න u du Integrate using the basic Integration rule
3
1
1 u2
=− ∗ +c Simplify the expression
3 1
2
2 1
= − u2 + c Simplify the expression
3
𝟐 𝟏
= − 𝒄𝒔𝒄 𝟑 𝒙 𝟐 +𝐜 Final Answer
𝟑

Calculus II: Integral Calculus

Example 10
𝟐
𝒅𝒚
න 𝟑 Factor out y in the denominator of the integrand
𝟏 𝒚 +𝒚

2
dy Express the numerator of the integrand as
=න
1 y y2 + 1 y 2 + 1 − y 2 which is just equal to 1
2
y2 + 1 − y2
=න dy Separate the terms
1 y y2 + 1
2
y2 + 1 y2
=න 2 + 1 − y y 2 + 1 dy Simplify the expression
1 y y
2
1 y
=න − 2 dy Integrate separately
1 y y + 1
2
1 2
y du
= න dy − න Let u = y 2 + 1 du = 2dy = dy
y y 2 + 1 dy 2
1 1 Replace y and dy with the appropriate u – variable forms
Calculus II: Integral Calculus
Example 10
2 2
1 du
= න dy − න Integrate separately
1 y 1 2u
2
1 Substitute back u = y 2 + 1
= Ln y − Ln u
2 1

2 b
1 Recall: ‫׬‬a f(x) dx = F(b) − F(a)
= Ln y − Ln y 2 + 1
2 1 Evaluate by replacing y by the given values of the upper limit and
lower limit

1 1
= Ln(2) − Ln(5) − Ln(1) − Ln(2
2 2 Simplify the expression

𝟑 𝟏
= 𝐋𝐧(𝟐) − 𝐋𝐧(𝟓) Final Answer
𝟐 𝟐

Calculus II: Integral Calculus

Example 11
Recall that
u 1 + cos u
cos =
𝝅 2 2
න 𝟏 + 𝒄𝒐𝒔 𝒙 𝐝𝐱
𝟎
u 1 + cos u
cos =
2 2
u
2 cos = 1 + cos u
2
π x 1
x Let u = 2 du = 2 dx 2du = dx
=න 2 cos dx
0 2 Replace x and dx with the appropriate u – variable forms
π
= 2 න cos u (2du)
0
π
= 2 2 න cos u du Apply Integration of trigonometric functions
0

Calculus II: Integral Calculus

Example 11

π x
= 2 ∗ 2 sin u Substitute back u =
0 2
Recall:
x π b
= 2 ∗ 2 sin න f(x) dx = F(b) − F(a)
2 0 a
Evaluate by replacing x by the given values of the upper
limit and lower limit
π 0
= 2 ∗ 2 sin − 2 ∗ 2 sin Simplify the expression
2 2
=2 2−0
=𝟐 𝟐 Final Answer

Calculus II: Integral Calculus

Example 12
Recall that
𝒄𝒐𝒕 𝒙
න 𝒅𝒙 cos x
𝟏 − 𝒄𝒐𝒕𝟐 𝒙 cot x =
sin x
cos x
= න sin x 2 dx Simplify the expression
cos x
1−
sin2 x
cos x
= න 2 sin x 2 dx Multiply the numerator of the integrand by the reciprocal of its
sin x − cos x denominator
sin2 x
cos x sin2 x
න ∗ dx Simplify the expression
sinx sin2 x − cos 2 x

cos x sin x
=න 2 dx Express sin2 x − cos 2 x as − (cos 2 x − sin2 x)
sin x − cos2 x

Calculus II: Integral Calculus

Example 12
cos x sin x Recall that
= −න dx
cos2 x − sin2 x sin2x = 2sinxcosx & cos2x = cos2 x − sin2 x
1 sin 2 x Recall that
=− න dx sin x
2 cos 2 x tan x =
cos x
1 du
= − නtan 2 x dx Let u = 2x du = 2dx = dx
2 2
Replace x and dx with the appropriate u – variable forms
1 du
= − නtanu
2 2 Apply integration of trigonometric functions

1
= − Ln u + c Substitute back u = 2x
4
1
= − Ln cos 2 x + c Final Answer
4

Calculus II: Integral Calculus

Example 13
න𝒔𝒊𝒏 𝟑 𝒙 (𝒄𝒐𝒔 𝟔 𝐱 + 𝟏)𝟒 𝐝𝐱 Express cos6x as cos2(3x)

= නsin 3 x cos 2 (3x) + 1 4 dx Recall that

cos 2x = 2 cos2 x − 1

= නsin 3 x (2 cos2 3 x − 1) + 1 4 dx Simplify the expression

= නsin 3 x 2 cos2 ( 3x) 4 dx Simplify the expression

= නsin 3 x (16 cos8 3 x)dx

Calculus II: Integral Calculus

Example 13
du
= 16 නsin 3 x (cos8 3 x)dx Let u = cos 3x du = −3 sin 3x dx − = sin3xdx
3
Replace x and dx with the appropriate u – variable forms
du
8
= −16 න u Simplify
3
16
= − නu8 du Apply basic integration rule
3

16 u9
=− ∗ +c Substitute back u = cos 3x then simplify
3 9
𝟏𝟔
=− (𝒄𝒐𝒔 𝟑 𝐱)𝟗 + 𝒄 Final Answer
𝟐𝟕

Calculus II: Integral Calculus

Example 14
Recall that
න𝒔𝒊𝒏 𝟐 𝐱 𝒄𝒐𝒔 𝟐 𝒙𝒅𝒙
sin 2x = 2sinx cosx

sin 4 x du
=න dx Let u = 4x du = 4 dx = dx
2 4
Replace x and dx with the appropriate u – variable forms
1 du
= න sin u Apply integration of trigonometric functions
2 4

1 1
= − cos u + c Substitute back u = 4x
2 4

1 1
= − cos 4 x + c Simplify the expression
2 4

𝟏
= − 𝒄𝒐𝒔 𝟒 𝐱 + 𝐜 Final Answer
𝟖

Calculus II: Integral Calculus

Example 15

න𝒕𝒂𝒏𝟒 𝒙 𝒅 𝐱 Split tan4 x as tan2 x ∗ tan2 x

= නtan2 x tan2 x d x Recall ∶ tan2 x = sec 2 x − 1

= නtan2 x sec 2 x − 1 d x Distribute tan2 x

= න tan2 x sec 2 x − tan2 x d x Integrate each term separately

Calculus II: Integral Calculus

Example 15

Let u = tan x du = sec 2 x dx

= නtan2 x sec 2 x d x − නtan2 x d x Replace x and dx with the appropriate u – variable
forms

Apply basic integration rule & integration of

= නu2 du − න sec 2 x −1 dx trigonometric functions
u3 Substitute back u = tan x
= − tan x + x + c
3
𝒕𝒂𝒏𝟑 𝒙
= − 𝒕𝒂𝒏 𝒙 + 𝐱 + 𝐜 Final Answer
𝟑

Calculus II: Integral Calculus

Printed References

1. Stewart, J. (2019). Calculus concepts and contexts 4th edition enhanced edition. USA : Cengage Learning, Inc.
2. Canlapan, R.B.(2017). Diwa senior high school series: basic calculus. DIWA Learning Systems Inc.
3. Hass, J., Weir, M. & Thomas, G. (2016). University calculus early transcendentals,global edition. Pearson
Education, Inc.Larson,
4. Bacani, J.B.,et.al. (2016). Basic calculus for senior high school. Book AtBP. Publishing Corp.
5. Balmaceda, J.M. (2016). Teaching guide in basic calculus. Commission on Higher Education.
6. Pelias, J.G. (2016). Basic calculus. Rex Bookstore.
7. Molina, M.F. (2016). Integral calculus. Unlimited Books Library Services & Publishing Inc.
8. Larson, R & Edwards, B. (2012). Calculus. Pasig City, Philippines. Cengage Learning Asia Pte Ltd

Calculus II: Integral Calculus

Electronic Resources

1. Pulham , John. Differential Calculus. Retrieved August 7, 2020 from http://

www.maths.abdn.ac.uk/igc/tch/math1002/dif
2. Weistein, Eric. Differential Calculus. Retrieved August 7, 2020 from
http://mathworld.wolfram.com/differential calculus
3. Thomas , Christopher . Introduction to Differential Calculus. Retrieved August 7, 2020 from http://
www.usyd.edu.aav/stuserv/document/maths_learning-center
4. Purplemath. Retrieved August 7, 2020 from http://www.purplemath.com/modules/

Calculus II: Integral Calculus

Integration of Exponential and
Hyperbolic Functions
Objectives

At the end of the lesson, the students should be

able to:

Evaluate the integral of natural exponential and

exponential functions

Evaluate the integral of hyperbolic functions

Introduction

Because the natural logarithmic function is increasing on its

entire domain, ten by the inverse function theorem, it has an
inverse that is also an increasing function. The inverse of 𝑙𝑛 is
called the natural exponential function, denoted by 𝑒, which
we now formally define. The natural exponential function is
the inverse of the natural logarithmic function; it is, therefore,
defined by

𝑒 𝑥 = 𝑦 if and only if 𝑥 = 𝑙𝑛𝑦

Introduction

If 𝑎is any positive number and 𝑥 is any real

number, the function 𝑓 defined by 𝑓 𝑥 = 𝑎 𝑥 is
called the exponential function to the base 𝑎.
The exponential function to the base 𝑎 satisfies
the same properties as the natural exponential
function
Introduction
Certain combinations of the exponential function 𝑒 𝑥 and 𝑒 −𝑥 occur
frequently in mathematics, science, and engineering. These functions are
called hyperbolic functions. They are defined as follows:

𝑒 𝑥 − 𝑒 −𝑥 𝑐𝑜𝑠ℎ𝑥
1. 𝑠𝑖𝑛ℎ𝑥 = 4. 𝑐𝑜𝑡ℎ𝑥 =
2 𝑠𝑖𝑛ℎ𝑥
𝑒 + 𝑒 −𝑥
𝑥 1
2. 𝑐𝑜𝑠ℎ𝑥 = 5. 𝑠𝑒𝑐ℎ𝑥 =
2 𝑐𝑜𝑠ℎ𝑥
𝑠𝑖𝑛ℎ𝑥 1
3. 𝑡𝑎𝑛ℎ𝑥 = 6. 𝑐𝑠𝑐ℎ𝑥 =
𝑐𝑜𝑠ℎ𝑥 𝑠𝑖𝑛ℎ𝑥
Introduction

This week’s lesson focuses mainly on the Integration

of Natural Exponential and Hyperbolic Functions. The
Basic Integration Formulas and Integration by
Substitution will still be used and applied as we
present and discuss examples on the integral of natural
exponential and hyperbolic functions. But before that,
may we first recall the following differentiation of
natural exponential and hyperbolic functions:
Introduction

Differentiation of Exponential Functions

The following formulas are used for differentiating exponential functions. The
symbol u denotes an arbitrary differentiable function of x.

𝑑 𝑢 𝑢
𝑑𝑢
1. 𝑒 =𝑒
𝑑𝑥 𝑑𝑥
𝑑 𝑢 𝑢
𝑑𝑢
2. 𝑎 = 𝑎 𝑙𝑛𝑎 𝑎 > 0, 𝑎 ≠ 1
𝑑𝑥 𝑑𝑥
Introduction
Differentiation of Hyperbolic Functions
The following formulas are used for differentiating exponential functions. The
symbol u denotes an arbitrary differentiable function of x.

𝑑 𝑑𝑢 𝑑 2
𝑑𝑢
1. 𝑠𝑖𝑛ℎ𝑢 = 𝑐𝑜𝑠ℎ𝑢 4. 𝑐𝑜𝑡ℎ𝑢 = − 𝑐𝑠𝑐ℎ 𝑢
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑 𝑑𝑢 𝑑 𝑑𝑢
2. 𝑐𝑜𝑠ℎ𝑢 = 𝑐𝑜𝑠ℎ𝑢 5. 𝑠𝑒𝑐ℎ𝑢 = −𝑠𝑒𝑐ℎ𝑢 𝑡𝑎𝑛ℎ𝑢
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑 2
𝑑𝑢 𝑑 𝑑𝑢
3. 𝑡𝑎𝑛ℎ𝑢 = 𝑠𝑒𝑐ℎ 𝑢 6. 𝑐𝑠𝑐ℎ𝑢 = −𝑐𝑠𝑐ℎ𝑢 𝑐𝑜𝑡ℎ𝑢
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
Introduction

Integration of Exponential Functions

1. න 𝑒 𝑢 𝑑𝑢 = 𝑒 𝑢 + 𝐶

𝑢
𝑎
2. න 𝑎𝑢 𝑑𝑢 = +𝐶 𝑎 > 0, 𝑎 ≠ 1
𝑙𝑛 𝑎
Introduction

Integration of Hyperbolic Functions

Let 𝑢 be the angle of any trigonometric function

1. න 𝑐𝑜𝑠ℎ𝑢𝑑𝑢 = 𝑠𝑖𝑛ℎ𝑢 + 𝐶 5. න 𝑠𝑒𝑐ℎ𝑢𝑡𝑎𝑛ℎ𝑢𝑑𝑢 = −𝑠𝑒𝑐ℎ𝑢 + 𝐶

2. න 𝑠𝑖𝑛ℎ𝑢𝑑𝑢 = 𝑐𝑜𝑠ℎ𝑢 + 𝐶 6. න 𝑐𝑠𝑐ℎ𝑢𝑐𝑜𝑡ℎ𝑢𝑑𝑢 = −𝑐𝑠𝑐ℎ𝑢 + 𝐶

3. න 𝑠𝑒𝑐ℎ2 𝑢𝑑𝑢 = 𝑡𝑎𝑛ℎ𝑢 + 𝐶 7. න 𝑡𝑎𝑛ℎ𝑢𝑑𝑢 = 𝑙𝑛 𝑐𝑜𝑠ℎ𝑢 + 𝐶

4. න 𝑐𝑠𝑐ℎ2 𝑢𝑑𝑢 = −𝑐𝑜𝑡ℎ𝑢 + 𝐶 8. න 𝑐𝑜𝑡ℎ𝑑𝑢 = 𝑙𝑛 𝑠𝑖𝑛ℎ𝑢 + 𝐶

Example 1
*Integration of exponential function
𝟏. න 𝒆−𝟑𝒙 𝒅𝒙

Let: න 𝑒 𝑢 𝑑𝑢 = 𝑒 𝑢 + 𝐶
𝑢 = −3𝑥
𝑑𝑢 = −3𝑑𝑥
𝑑𝑢 Substitute u and du to the given
− = 𝑑𝑥
3
𝑑𝑢 Bring out the constant then integrate the function
= න 𝑒𝑢 ∙ −
3
1 Substitute back 𝑢 = −3𝑥
= − න 𝑒 𝑢 𝑑𝑢
3
1 𝑢
=− 𝑒 +𝐶
3
𝟏 −𝟑𝒙 Final answer
=− 𝒆 +𝑪
𝟑
Example 2
𝒅𝒗 *Integration of exponential function
𝟐. න 𝒗
𝒆
Rewrite the given problem
𝑑𝑣 1
න 𝑣 = න 𝑒 −𝑣 𝑑𝑣 Note: 𝑥 = 𝑒 −𝑥
𝑒 𝑒

Let: න 𝑒 𝑢 𝑑𝑢 = 𝑒 𝑢 + 𝐶
𝑢 = −𝑣
𝑑𝑢 = −𝑑𝑣
−𝑑𝑢 = 𝑑𝑣 Substitute u and du to the given

= න 𝑒 𝑢 ∙ −𝑑𝑢

= − න 𝑒 𝑢 𝑑𝑢
Substitute back 𝑢 = −𝑣
= −𝑒 𝑢 + 𝐶
𝟏 Final answer
= −𝒆−𝒗 + 𝑪 𝒐𝒓 − 𝒗
+𝑪
𝒆
Example 3
*Integration of exponential function
𝟑. න 𝒆𝒕 − 𝒆−𝒕 𝟐 𝒅𝒕 Applying law of exponent ** 𝑒 𝑡 𝑒 −𝑡 = 1
Before integrating, expand first the integrand න 𝑒 𝑢 𝑑𝑢 = 𝑒 𝑢 + 𝐶
= න(𝑒 2𝑡 − 2𝑒 𝑡 𝑒 −𝑡 + 𝑒 −2𝑡 )𝑑𝑡

= න 𝑒 2𝑡 𝑑𝑡 − 2 න 𝑑𝑡 + න 𝑒 −2𝑡 𝑑𝑡

𝑑𝑢 𝑑𝑢
= න 𝑒𝑢 ∙ 𝑢
− 2 න 𝑑𝑡 + න 𝑒 ∙ − න 𝑒 2𝑡 𝑑𝑡 න 𝑒 −2𝑡 𝑑𝑡
2 2
1 1 𝑢 = 2𝑡 𝑢 = −2𝑡
= න 𝑒 𝑑𝑢 − 2 න 𝑑𝑡 − න 𝑒 𝑢 𝑑𝑢
𝑢
2 2
𝑑𝑢 = 2𝑑𝑡 𝑑𝑢 = −2𝑑𝑡
1 𝑢 1 𝑢
= 𝑒 − 2𝑡 − 𝑒 + 𝐶 𝑑𝑢 𝑑𝑢
2 2 = 𝑑𝑡 − = 𝑑𝑡
2 2
Example 3

Do not cancel out the first and last term

because they have different ‘u’ Substitute
back

𝟏 𝟐𝒕 𝟏 −𝟐𝒕 Final answer

= 𝒆 − 𝟐𝒕 − 𝒆 +𝑪
𝟐 𝟐
Example 4
𝟏 *Integration of exponential function
𝒆𝒙
𝟒. න 𝒅𝒙
𝒙𝟐
Let න 𝑒 𝑢 𝑑𝑢 = 𝑒 𝑢 + 𝐶
1
𝑢=
𝑥
1 1 Substitute u and du to the given
𝑑𝑢 = − 2 𝑑𝑥 → −𝑑𝑢 = 2 𝑑𝑥
𝑥 𝑥
= න 𝑒 𝑢 ∙ (−𝑑𝑢)

= − න 𝑒 𝑢 𝑑𝑢
1
Substitute back 𝑢 = 𝑥
= −𝑒 𝑢 + 𝐶
𝟏 Final answer
= −𝒆𝒙 +𝑪
Example 5
*Integration of exponential function
𝟓. න 𝒆𝒔𝒊𝒏𝟒𝒙 𝒄𝒐𝒔𝟒𝒙𝒅𝒙

Let: න 𝑒 𝑢 𝑑𝑢 = 𝑒 𝑢 + 𝐶
𝑢 = 𝑠𝑖𝑛4𝑥
𝑑𝑢
𝑑𝑢 = 4𝑐𝑜𝑠4𝑥𝑑𝑥 → = 𝑐𝑜𝑠4𝑥𝑑𝑥 Substitute u and du to the given
4
𝑑𝑢 Bring out the constant then integrate the function
𝑢
= න𝑒 ∙
4

1
= න 𝑒 𝑢 𝑑𝑢
4 Substitute back 𝑢 = 𝑠𝑖𝑛4𝑥
1 𝑢
= 𝑒 +𝐶
4
𝟏 Final answer
= 𝒆𝒔𝒊𝒏𝟒𝒙 + 𝑪
𝟒
Example 6
*Integration of exponential function
𝟔. න 𝟓𝟑−𝟐𝒙 𝒅𝒙 𝑢
𝑎
න 𝑎𝑢 𝑑𝑢 = +𝐶
Let: 𝑙𝑛 𝑎
𝑎=5
𝑢 = 3 − 2𝑥 Substitute a, u, and du to the given
𝑑𝑢
𝑑𝑢 = −2𝑑𝑥 → − = 𝑑𝑥
2
𝑑𝑢 Bring out the constant then integrate the function
𝑢
= න𝑎 ∙ −
2
𝑢
1 1 𝑎 Substitute back 𝑎 = 5, 𝑢 = 3 − 2𝑥
= − න 𝑎𝑢 𝑑𝑢 = − +𝐶
2 2 𝑙𝑛𝑎
Apply law of logarithm
1 53−2𝑥
=− +𝐶 𝑎𝑙𝑛𝑏 = 𝑙𝑛𝑏 𝑎
2 𝑙𝑛5

𝟓𝟑−𝟐𝒙 Final answer

=− +𝑪
𝒍𝒏𝟐𝟓
Example 7
*Integration of exponential function
𝟕. න 𝟑𝒙 𝟐𝒙 𝒅𝒙 Before integrating, rewrite the given applying law of
exponent
𝑚 𝑥 𝑛 𝑥 = 𝑚𝑛 𝑥
න 3𝑥 2𝑥 𝑑𝑥 = න 6𝑥 𝑑𝑥
𝑢
𝑎
න 𝑎𝑢 𝑑𝑢 = +𝐶
Let: 𝑙𝑛 𝑎
𝑎=6
𝑢=𝑥
Substitute a, u, and du to the given
𝑑𝑢 = 𝑑𝑥

= න 𝑎𝑢 𝑑𝑢

𝑎𝑢 Substitute back 𝑎 = 6, 𝑢 = 𝑥
= +𝐶
𝑙𝑛 𝑎
𝟔𝒙 Final answer
= +𝑪
𝒍𝒏𝟔
Example 8
*Integration of exponential function
𝟑
𝟖. න 𝟒𝟐𝒙 𝒅𝒙 Before integrating, rewrite the given
problem
1 2𝑥 𝑢
න 2𝑥
4 3 𝑑𝑥 = න 4 3 𝑑𝑥 𝑎
න 𝑎𝑢 𝑑𝑢 = +𝐶
𝑙𝑛 𝑎
Let:
𝑎=4
2𝑥
𝑢=
3
2 3
𝑑𝑢 = 𝑑𝑥 → 𝑑𝑢 = 𝑑𝑥 Substitute a, u, and du to the given
3 2
Example 8
3 Bring out the constant then integrate the function
𝑢
= න𝑎 ∙ 𝑑𝑢
2

3 2𝑥
= න 𝑎𝑢 𝑑𝑢 Substitute back 𝑎 = 4, 𝑢 = 3
2

3 𝑎𝑢 Apply law of logarithm

= +𝐶
2 𝑙𝑛 𝑎 𝑎𝑙𝑛𝑏 = 𝑙𝑛𝑏 𝑎
2𝑥
343
= +𝐶
2 𝑙𝑛4
2𝑥
3∙43
= +𝐶
𝑙𝑛42
𝟐𝒙 Final answer
𝟑∙𝟒𝟑
= +𝑪
𝒍𝒏𝟏𝟔
Example 9

𝟑𝒙 *Integration of exponential function

𝟗. න 𝒙 𝒅𝒙 Before integrating, rewrite the given
𝟐
applying law of exponent
𝑥
3𝑥 3 𝑚𝑥 𝑚 𝑥
න 𝑥 𝑑𝑥 = න 𝑑𝑥 =
2 2 𝑛 𝑥 𝑛
𝑢
Let: 𝑎
න 𝑎𝑢 𝑑𝑢 = +𝐶
3 𝑙𝑛 𝑎
𝑎=
2
𝑢=𝑥
Substitute a, u, and du to the given
𝑑𝑢 = 𝑑𝑥
Example 9

= න 𝑎𝑢 𝑑𝑢
3
Substitute back 𝑎 = 2 , 𝑢 = 𝑥
𝑎𝑢
= +𝐶
𝑙𝑛 𝑎
𝑥
3
Apply law of logarithm
2
= +𝐶 𝑎
3 ln = 𝑙𝑛𝑎 − 𝑙𝑛𝑏
ln 𝑏
2

𝟑𝒙 Final answer
= 𝒙 +𝑪
𝟐 (𝒍𝒏𝟑 − 𝒍𝒏𝟐)
Example 10
*Integration of hyperbolic function
𝟏𝟎. න 𝒔𝒊𝒏𝒉 𝟑𝒙 − 𝟏 𝒅𝒙
න 𝑠𝑖𝑛ℎ𝑢𝑑𝑢 = 𝑐𝑜𝑠ℎ𝑢 + 𝐶
Let:
𝑢 = 3𝑥 − 1
𝑑𝑢 Substitute u and du to the given
𝑑𝑢 = 3𝑑𝑥 → = 𝑑𝑥
3
𝑑𝑢 Bring out the constant then integrate the function
= න 𝑠𝑖𝑛ℎ𝑢 ∙
3

1
= න 𝑠𝑖𝑛ℎ𝑢𝑑𝑢
3 Substitute back 𝑢 = 3𝑥 − 1
1
= 𝑐𝑜𝑠ℎ𝑢 + 𝐶
3
𝟏 Final answer
= 𝒄𝒐𝒔 𝒉 𝟑𝒙 − 𝟏 + 𝑪
𝟑
Example 11

𝒔𝒆𝒄𝒉𝟐(𝒍𝒏𝒙) *Integration of hyperbolic function

𝟏𝟏. න 𝒅𝒙
𝒙
Let: න 𝑠𝑒𝑐ℎ2 𝑢𝑑𝑢 = 𝑡𝑎𝑛ℎ𝑢 + 𝐶
𝑢 = 𝑙𝑛𝑥
1 Substitute u and du to the given
𝑑𝑢 = 𝑑𝑥
𝑥

= න sech2 𝑢 (𝑑𝑢)
Substitute back 𝑢 = 𝑙𝑛𝑥
= 𝑡𝑎𝑛ℎ𝑢 + 𝐶
= 𝒕𝒂𝒏𝒉 𝒍𝒏𝒙 + 𝑪 Final answer
Example 12
*Integration of hyperbolic function
𝟏𝟐. න 𝒄𝒔𝒄𝒉𝟐 𝟏 − 𝒙𝟐 𝒙𝒅𝒙

Let: න 𝑐𝑠𝑐ℎ2 𝑢𝑑𝑢 = −𝑐𝑜𝑡ℎ𝑢 + 𝐶

𝑢 = 1 − 𝑥2
𝑑𝑢
𝑑𝑢 = −2𝑥𝑑𝑥 → − = 𝑥𝑑𝑥 Substitute u and du to the given
2
𝑑𝑢 Bring out the constant then integrate the function
= න 𝑐𝑠𝑐ℎ2 𝑢 ∙ −
2

1
= − න 𝑐𝑠𝑐ℎ2 𝑢 𝑑𝑢
2
1
=− −𝑐𝑜𝑡ℎ𝑢 + 𝐶 Substitute back 𝑢 = 1 − 𝑥 2
2
1
= 𝑐𝑜𝑡ℎ𝑢 + 𝐶
2
𝟏 Final answer
= 𝒄𝒐𝒕𝒉𝒖 𝟏 − 𝒙𝟐 + 𝑪
𝟐
Example 13
𝟏 𝟏 *Integration of hyperbolic function
𝟏𝟑. න 𝒄𝒔𝒄𝒉 𝒙𝒄𝒐𝒕𝒉 𝒙𝒅𝒙
𝟐 𝟐
Let: න 𝑐𝑠𝑐ℎ𝑢𝑐𝑜𝑡ℎ𝑢𝑑𝑢 = −𝑐𝑠𝑐ℎ𝑢 + 𝐶
1
𝑢= 𝑥
2
1
𝑑𝑢 = 𝑑𝑥 → 2𝑑𝑢 = 𝑑𝑥
2 Substitute u and du to the given
Bring out the constant then integrate the function
= න 𝑐𝑠𝑐ℎ 𝑢 𝑐𝑜𝑡ℎ 𝑢 ∙ 2𝑑𝑢

= 2 න 𝑐𝑠𝑐ℎ 𝑢 𝑐𝑜𝑡ℎ 𝑢 𝑑𝑢
1
Substitute back 𝑢 = 𝑥
2
= 2 −𝑐𝑠𝑐ℎ𝑢 + 𝐶
= −2𝑐𝑠𝑐ℎ𝑢 + 𝐶
𝟏 Final answer
= −𝟐𝒄𝒔𝒄𝒉 𝒙 + 𝑪
𝟐
References

Printed References
1. Stewart, J. (2019). Calculus concepts and contexts 4th edition enhanced edition. USA :
Cengage Learning, Inc.
2. Canlapan, R.B.(2017). Diwa senior high school series: basic calculus. DIWA Learning
Systems Inc.
3. Hass, J., Weir, M. & Thomas, G. (2016). University calculus early transcendentals,global
edition. Pearson Education, Inc.Larson,
4. Bacani, J.B.,et.al. (2016). Basic calculus for senior high school. Book AtBP. Publishing Corp.
5. Balmaceda, J.M. (2016). Teaching guide in basic calculus. Commission on Higher Education.
6. Pelias, J.G. (2016). Basic calculus. Rex Bookstore.
7. Molina, M.F. (2016). Integral calculus. Unlimited Books Library Services & Publishing Inc.
8. Larson, R & Edwards, B. (2012). Calculus. Pasig City, Philippines. Cengage Learning Asia Pte
Ltd
References

Electronic Resources
1. Pulham , John. Differential Calculus. Retrieved August 7, 2020 from http://
www.maths.abdn.ac.uk/igc/tch/math1002/dif
2. Weistein, Eric. Differential Calculus. Retrieved August 7, 2020 from
http://mathworld.wolfram.com/differential calculus
3. Thomas , Christopher . Introduction to Differential Calculus. Retrieved August 7,
2020 from http:// www.usyd.edu.aav/stuserv/document/maths_learning-center
4. Purplemath. Retrieved August 7, 2020 from http://www.purplemath.com/modules/
Product of Sines and Cosines
 Objective

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

to:

Develop the methods of evaluating integrals which

cannot be evaluated directly by any of the standard
formulas.
 Introduction

Some trigonometric integrals cannot be evaluated directly from their

given forms,
න 𝑠𝑖𝑛𝑈𝑐𝑜𝑠𝑉𝑑𝑥

න 𝑐𝑜𝑠𝑈𝑐𝑜𝑠𝑉𝑑𝑥

න 𝑠𝑖𝑛𝑈𝑠𝑖𝑛𝑉𝑑𝑥

where U and V are two different angles.

 Introduction

But those integrals can be evaluated with the aid of

the following trigonometric formulas:

2𝑠𝑖𝑛𝑈𝑐𝑜𝑠𝑉 = 𝑠𝑖𝑛 𝑈 + 𝑉 + 𝑠𝑖𝑛 𝑈 − 𝑉

2𝑐𝑜𝑠𝑈𝑐𝑜𝑠𝑉 = 𝑐𝑜𝑠 𝑈 + 𝑉 + 𝑐𝑜𝑠 𝑈 − 𝑉
2𝑠𝑖𝑛𝑈𝑠𝑖𝑛𝑉 = 𝑐𝑜𝑠 𝑈 − 𝑉 − 𝑐𝑜𝑠(𝑈 + 𝑉)
 Introduction

Integrating both sides we have,

න 2𝑠𝑖𝑛𝑈𝑐𝑜𝑠𝑉𝑑𝑥 = න [𝑠𝑖𝑛 𝑈 + 𝑉 + 𝑠𝑖𝑛 𝑈 − 𝑉 ]𝑑𝑥

න 2𝑐𝑜𝑠𝑈𝑐𝑜𝑠𝑉𝑑𝑥 = න [𝑐𝑜𝑠 𝑈 + 𝑉 + 𝑐𝑜𝑠 𝑈 − 𝑉 ] 𝑑𝑥

න 2𝑠𝑖𝑛𝑈𝑠𝑖𝑛𝑉𝑑𝑥 = න [𝑐𝑜𝑠 𝑈 − 𝑉 − 𝑐𝑜𝑠(𝑈 + 𝑉)]𝑑𝑥

 Example 1

𝟏. න 𝐜𝐨𝐬 𝟔𝒙 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙

1
= න 2𝑐𝑜𝑠6𝑥 𝑐𝑜𝑠2𝑥 𝑑𝑥
2
2
1 Multiply to the given expression to apply
2
= න(cos 6𝑥 + 2𝑥 + cos(6𝑥 − 2𝑥))𝑑𝑥 the corresponding trigonometric formula.
2
let 𝑢 = 6𝑥 𝑎𝑛𝑑 𝑣 = 2𝑥
1
= න(cos 8𝑥 + cos(4𝑥))𝑑𝑥
2
1 sin 8𝑥 sin 4𝑥 Apply the integral of trigonometric
= + +𝑐
2 8 4 functions then simplify.
𝒔𝒊𝒏𝟖𝒙 𝒔𝒊𝒏𝟒𝒙
= + +𝒄 Final answer
𝟏𝟔 𝟖
 Example 2

𝟐. න 𝟑 𝐬𝐢𝐧 𝟓𝒙 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙

3
= න 2 sin 5𝑥 𝑐𝑜𝑠4𝑥 𝑑𝑥 2
2 Multiply 2 to the given
3 expression to apply the
= න(sin(5𝑥 + 4𝑥) + sin(5𝑥 − 4𝑥))𝑑𝑥 corresponding trigonometric
2
formula.
3 let 𝑢 = 5𝑥 𝑎𝑛𝑑 𝑣 = 4𝑥
= න(sin 9𝑥 + sin(𝑥))𝑑𝑥
2

3 − cos 9𝑥 Apply the integral of

= − 𝑐𝑜𝑠𝑥 + 𝑐 trigonometric functions then
2 9
simplify.
𝒄𝒐𝒔𝟗𝒙 𝟑𝒄𝒐𝒔𝒙
=− − +𝒄 Final answer
𝟔 𝟐
 Example 3
𝟑. න 𝐬𝐢𝐧 𝟓𝒙 𝒔𝒊𝒏𝒙 𝒅𝒙

1
= න 2 sin 5𝑥 𝑠𝑖𝑛𝑥 𝑑𝑥 2
2 Multiply 2 to the given
1 expression to apply the
= න(cos(5𝑥 − 𝑥) − cos(5𝑥 + 𝑥))𝑑𝑥 corresponding trigonometric
2
formula.
1 let 𝑢 = 5𝑥 𝑎𝑛𝑑 𝑣 = 𝑥
= න(cos 4𝑥 − cos(6𝑥))𝑑𝑥
2

1 sin 4𝑥 sin 6𝑥 Apply the integral of

= − +𝑐 trigonometric functions then
2 4 6
simplify.
𝒔𝒊𝒏𝟒𝒙 𝒔𝒊𝒏𝟔𝒙
= − +𝒄 Final answer
𝟖 𝟏𝟐
 Example 4
𝟒. න 𝐬𝐢𝐧(𝟒𝒙 − 𝟑) 𝐜𝐨𝐬(𝒙 + 𝟓) 𝒅𝒙

1
= න 2 sin(4𝑥 − 3) cos(𝑥 + 5) 𝑑𝑥
2
2
1 Multiply 2 to the given expression to
= න(sin(4𝑥 − 3 + 𝑥 + 5) apply the corresponding
2
trigonometric formula.
+ sin(4𝑥 − 3 − 𝑥 − 5))𝑑𝑥 let 𝑢 = 4𝑥 − 3 𝑎𝑛𝑑 𝑣 = 𝑥 + 5
1
= න(sin(5𝑥 + 2) + sin(3𝑥 − 8))𝑑𝑥
2

1 −cos(5𝑥 + 2) cos(3𝑥 − 8) Apply the integral of trigonometric

= − +𝑐
2 5 3 functions and simplify.
−𝐜𝐨𝐬(𝟓𝒙 + 𝟐) 𝐜𝐨𝐬(𝟑𝒙 − 𝟖)
= − +𝒄 Final answer
𝟏𝟎 𝟔
 Example 5

𝟓. න 𝐜𝐨𝐬(𝟑𝒙 − 𝟐𝝅) 𝐜𝐨𝐬(𝒙 + 𝝅) 𝒅𝒙

1 2
= න 2 cos(3𝑥 − 2𝜋) cos(𝑥 + 𝜋) 𝑑𝑥 Multiply 2 to the given
2
expression to apply the
1
= න(cos(3𝑥 − 2𝜋 + 𝑥 + 𝜋) corresponding trigonometric
2 formula.
+ cos(3𝑥 − 2𝜋 − 𝑥 − 𝜋))𝑑𝑥
1 let 𝑢 = 3𝑥 − 2𝜋 𝑎𝑛𝑑 𝑣 =
= න(cos(4𝑥 − 𝜋) + cos(2𝑥 − 3𝜋))𝑑𝑥 𝑥+𝜋
2
 Example 5

1 sin(4𝑥 − 𝜋) sin(2𝑥 − 3𝜋) Apply the integral of

= + +𝑐
2 4 2 trigonometric functions.
1
= sin 4𝑥 cos 𝜋 − 𝑐𝑜𝑠4𝑥 𝑠𝑖𝑛𝜋 Recall: 𝑐𝑜𝑠𝜋 = −1, 𝑠𝑖𝑛𝜋 =
8
1 0
+ (sin 2𝑥 cos 3𝜋 − 𝑐𝑜𝑠2𝑥 𝑠𝑖𝑛3𝜋)
4
𝐬𝐢𝐧 𝟒𝒙 𝐬𝐢𝐧 𝟐𝒙
=− − +𝒄 Final answer
𝟖 𝟒
 Example 6

𝟔. න 𝐬𝐢𝐧 𝟗𝒙 𝐜𝐨𝐬 𝟒𝒙 𝒅𝒙 Using the trigonometric formula

Apply integration
= න sin 9𝑥 cos 4𝑥 𝑑𝑥
simplify
1
= න sin(9𝑥 + 4𝑥) + sin(9𝑥 − 4𝑥)
2
1
= න( sin 13𝑥 + sin 5𝑥) 𝑑𝑥
2
1 cos 13𝑥 cos 5𝑥
= − − +𝑐
2 13 5
𝐜𝐨𝐬 𝟏𝟑𝒙 𝐜𝐨𝐬 𝟓𝒙 Final answer
=− − +𝒄
𝟐𝟔 𝟏𝟎
 Example 7

𝟕. න 𝐜𝐨𝐬 𝟗𝒙 𝐜𝐨𝐬 𝟒𝒙 𝒅𝒙 Using the trigonometric formula

Apply integration
= න cos 9𝑥 cos 4𝑥 𝑑𝑥
simplify
1
= න cos(9𝑥 + 4𝑥) + cos(9𝑥 − 4𝑥)
2
1
= න( cos 13𝑥 + cos 5𝑥) 𝑑𝑥
2
1 sin 13𝑥 sin 5𝑥
= + +𝑐
2 13 5
𝐬𝐢𝐧 𝟏𝟑𝒙 𝐬𝐢𝐧 𝟓𝒙 Final answer
= + +𝒄
𝟐𝟔 𝟏𝟎
 Example 8

𝟖. න 𝐬𝐢𝐧 𝟗𝒙 𝐬𝐢𝐧 𝟒𝒙 𝒅𝒙 Using the trigonometric

formula
= න sin 9𝑥 sin 4𝑥 𝑑𝑥 Apply integration

1 simplify
= න cos(9𝑥 − 4𝑥) − cos(9𝑥 + 4𝑥)
2
1
= න( cos 5𝑥 − cos 13𝑥) 𝑑𝑥
2
1 sin 5𝑥 sin 13𝑥
= − +𝑐
2 5 13
𝐬𝐢𝐧 𝟓𝒙 𝐬𝐢𝐧 𝟏𝟑𝒙 Final answer
= − +𝒄
𝟏𝟎 𝟐𝟔
 Example 9
𝟗. න 𝐬𝐢𝐧 𝟐𝒙 𝐜𝐨𝐬 𝟑𝒙 𝒅𝒙 Using the trigonometric formula
sin −𝑥 = − sin 𝑥
= න sin 2𝑥 cos 3𝑥 𝑑𝑥 Apply integration

1 simplify
= න sin(2𝑥 + 3𝑥) + sin(2𝑥 − 3𝑥)
2
1
= න( sin 5𝑥 + sin(−𝑥)) 𝑑𝑥
2
1
= න( sin 5𝑥 − sin 𝑥) 𝑑𝑥
2
1 cos 5𝑥
= − + cos 𝑥 + 𝑐
2 5
𝐜𝐨𝐬 𝟓𝒙 𝐜𝐨𝐬 𝒙 Final answer
=− + +𝒄
𝟏𝟎 𝟐
 Example 10
𝟏𝟎. න 𝐬𝐢𝐧 𝟒𝒙 𝐬𝐢𝐧 𝟑𝒙 𝒅𝒙 Using the trigonometric
formula
= න sin 4𝑥 sin 3𝑥 𝑑𝑥 Apply integration

1 simplify
= න cos(4𝑥 − 3𝑥) − cos(4𝑥 + 3𝑥)
2
1
= න( cos 𝑥 − cos 7𝑥) 𝑑𝑥
2
1 sin 7𝑥
= sin 𝑥 − +𝑐
2 7
𝐬𝐢𝐧 𝒙 𝐬𝐢𝐧 𝟕𝒙 Final answer
= − +𝒄
𝟐 𝟏𝟒
Powers of Sines and Cosines
Introduction

In general, we shall integrate a trigonometric integral of the form,

න 𝑠𝑖𝑛𝑚 𝑣𝑐𝑜𝑠 𝑛 𝑣 𝑑𝑥

Where “v” is a differentiable function of x and m, n are real numbers.

If 𝑚 = 𝑛 = 1 or 𝑚 = 1, 𝑛 ≠ 1 or 𝑚 ≠ 1, 𝑛 = 1, the integral can
easily be evaluated by the method of substitution as discussed.

We shall deal with the integrals of the general form given above but
with 𝑚, 𝑛 ≠ 1.
Case I

CASE 1: When “m” is a positive odd integer and “n” is any number, may we write

𝑠𝑖𝑛𝑚 𝑣𝑐𝑜𝑠 𝑛 𝑣 = 𝑠𝑖𝑛𝑚−1 𝑣 𝑐𝑜𝑠 𝑛 𝑣 sin 𝑣

Since m is odd, then 𝑚 − 1 is even and therefore we may use the trigonometric
identity
𝑠𝑖𝑛2 𝑣 = 1 − 𝑐𝑜𝑠 2 𝑣
To express 𝑠𝑖𝑛𝑚−1 𝑣 in terms of the powers of cos 𝑣. Then the given integral is
reduced to the form,

න(𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 cos 𝑣) sin 𝑣 𝑑𝑥

Which can now be evaluated by letting 𝑢 = cos 𝑣.

Example
𝟏. න 𝒔𝒊𝒏𝟑 𝟒𝒙 𝒄𝒐𝒔𝟐 𝟒𝒙 𝒅𝒙

න 𝑠𝑖𝑛3 4𝑥 𝑐𝑜𝑠 2 4𝑥 𝑑𝑥 = න 𝑠𝑖𝑛2 4𝑥 𝑐𝑜𝑠 2 4𝑥 sin 4𝑥 𝑑𝑥

= න( 1 − 𝑐𝑜𝑠 2 4𝑥) 𝑐𝑜𝑠 2 4𝑥 sin 4𝑥 𝑑𝑥

Let 𝑢 = cos 4𝑥 ,
𝑑𝑢 = −4 sin 4𝑥 𝑑𝑥
= න( 𝑐𝑜𝑠 2 4𝑥 − 𝑐𝑜𝑠 4 4𝑥) sin 4𝑥 𝑑𝑥 𝑑𝑢
− = sin 4𝑥 𝑑𝑥
4
1 𝑐𝑜𝑠 3 4𝑥 𝑐𝑜𝑠 5 4𝑥
=− − +𝑐
4 3 5 Simplify
𝒄𝒐𝒔𝟑 𝟒𝒙 𝒄𝒐𝒔𝟓 𝟒𝒙
= − + +𝒄
𝟏𝟐 𝟐𝟎 Final answer
 Example
𝟐. න 𝒔𝒊𝒏𝟓 𝒙 𝒄𝒐𝒔𝟐 𝒙 𝒅𝒙

Manipulate the integrand before applying

න 𝑠𝑖𝑛5 𝑥 𝑐𝑜𝑠 2 𝑥 𝑑𝑥 = න 𝑠𝑖𝑛4 𝑥 𝑐𝑜𝑠 2 𝑥 sin 𝑥 𝑑𝑥 integration.

= න(𝑠𝑖𝑛2 𝑥)2 𝑐𝑜𝑠 2 𝑥 sin 𝑥 𝑑𝑥

= න(1 − 𝑐𝑜𝑠 2 𝑥)2 𝑐𝑜𝑠 2 𝑥 sin 𝑥 𝑑𝑥 Let 𝑢 = cos 𝑥 ,

𝑑𝑢 = − sin 𝑥 𝑑𝑥
= න( 1 − 2𝑐𝑜𝑠 2 𝑥 + 𝑐𝑜𝑠 4 𝑥)𝑐𝑜𝑠 2 𝑥 sin 𝑥 𝑑𝑥

= න( 𝑐𝑜𝑠 2 𝑥 − 2𝑐𝑜𝑠 4 𝑥 + 𝑐𝑜𝑠 6 𝑥) sin 𝑥 𝑑𝑥

𝑐𝑜𝑠 3 𝑥 2𝑐𝑜𝑠 5 𝑥 𝑐𝑜𝑠 7 𝑥

=− − + +𝑐
3 5 7 Simplify
𝒄𝒐𝒔𝟑 𝒙 𝟐𝒄𝒐𝒔𝟓 𝒙 𝒄𝒐𝒔𝟕 𝒙
= − + − +𝒄
𝟑 𝟓 𝟕 Final answer
Case II

CASE 2: When “m” is any number and “n” is a positive odd integer, we may write

𝑠𝑖𝑛𝑚 𝑣 𝑐𝑜𝑠 𝑛 𝑣 = 𝑠𝑖𝑛𝑚 𝑣 𝑐𝑜𝑠 𝑛−1 𝑣 cos 𝑣

And then use the identity 𝑐𝑜𝑠 2 𝑣 = 1 − 𝑠𝑖𝑛2 𝑣 to reduced the integral to the form,

න 𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 sin 𝑣 cos 𝑣 𝑑𝑥

Which can now be evaluated letting 𝑢 = sin 𝑣

Example

𝟏. න 𝒔𝒊𝒏𝟐 𝒙 𝒄𝒐𝒔𝟑 𝒙 𝒅𝒙

න 𝑠𝑖𝑛2 𝑥 𝑐𝑜𝑠 3 𝑥 𝑑𝑥 = න 𝑠𝑖𝑛2 𝑥 𝑐𝑜𝑠 2 𝑥 cos 𝑥 𝑑𝑥

= න 𝑠𝑖𝑛2 𝑥 (1 − 𝑠𝑖𝑛2 𝑥) cos 𝑥 𝑑𝑥 Let 𝑢 = sin 𝑥

𝑑𝑢 = cos 𝑥 𝑑𝑥
= න(𝑠𝑖𝑛2 𝑥 − 𝑠𝑖𝑛4 𝑥) cos 𝑥 𝑑𝑥

𝒔𝒊𝒏𝟑 𝒙 𝒔𝒊𝒏𝟓 𝒙
= − +𝒄 Final answer
𝟑 𝟓
 Example
𝟐. න 𝒄𝒐𝒔𝟓 𝒙 𝒅𝒙

න 𝑐𝑜𝑠 5 𝑥 𝑑𝑥 = න 𝑐𝑜𝑠 4 𝑥 cos 𝑥 𝑑𝑥

= න(𝑐𝑜𝑠 2 𝑥)2 cos 𝑥 𝑑𝑥 Manipulate the integrand before

applying integration

= න(1 − 𝑠𝑖𝑛2 𝑥)2 cos 𝑥 𝑑𝑥

= න(1 − 2𝑠𝑖𝑛2 𝑥 + 𝑠𝑖𝑛4 𝑥) cos𝑥 𝑑𝑥

= න cos 𝑥 𝑑𝑥 − 2 න 𝑠𝑖𝑛2 𝑥 cos 𝑥 𝑑𝑥 + න 𝑠𝑖𝑛4 𝑥 cos𝑥 𝑑𝑥 Apply integration by substitution on

the 2nd and 3rd term letting 𝑢 = sin 𝑥
𝟐𝒔𝒊𝒏𝟑 𝒙 𝒔𝒊𝒏𝟓 𝒙
= 𝐬𝐢𝐧 𝒙 − + +𝒄 Final answer
𝟑 𝟓
Case III

CASE 3: When “m” and “n” are both even integers (either both positive or one
positive and one zero), we may write
𝑚 𝑛
𝑚 𝑛 2 2
𝑠𝑖𝑛 𝑣 𝑐𝑜𝑠 𝑣 = (𝑠𝑖𝑛 𝑣) 2 (𝑐𝑜𝑠 𝑣) 2

then we may use one or both of the following identities:

1−𝑐𝑜𝑠2𝑣 1+𝑐𝑜𝑠2𝑣
𝑠𝑖𝑛2 𝑣 = ; 𝑐𝑜𝑠 2 𝑣 =
2 2

To reduce the given integral into an integrable form. Then identities above are
used repeatedly when m and n or both are greater than 2.
 Example
𝟏. න 𝒔𝒊𝒏𝟒 𝒙 𝒅𝒙 1 − 𝑐𝑜𝑠2𝑥
𝑠𝑖𝑛2 𝑥 =
2
න 𝑠𝑖𝑛4 𝑥 𝑑𝑥 = න(𝑠𝑖𝑛2 𝑥)2 𝑑𝑥

2 1 + 𝑐𝑜𝑠4𝑥
1 − cos 2𝑥 𝑐𝑜𝑠 2 2𝑥 =
=න 𝑑𝑥 2
2
1
= න(1 − 2𝑐𝑜𝑠2𝑥 + 𝑐𝑜𝑠 2 2𝑥) 𝑑𝑥
4
1 1 + cos 4𝑥
= න(1 − 2𝑐𝑜𝑠2𝑥 + ) 𝑑𝑥
4 2
1 3 1
= න( − 2𝑐𝑜𝑠2𝑥 + cos 4𝑥) 𝑑𝑥
4 2 2
1 3𝑥 1 simplify
= − sin 2𝑥 + sin 4𝑥 + 𝑐
4 2 8
𝟑𝒙 𝐬𝐢𝐧 𝟐𝒙 𝐬𝐢𝐧 𝟒𝒙 Final answer
= − + +𝒄
𝟖 𝟒 𝟑𝟐
 Example
𝟐. න 𝒔𝒊𝒏𝟒 𝒙 𝒄𝒐𝒔𝟐 𝒙 𝒅𝒙

න 𝑠𝑖𝑛4 𝑥 𝑐𝑜𝑠 2 𝑥 𝑑𝑥 = න(𝑠𝑖𝑛2 𝑥)2 𝑐𝑜𝑠 2 𝑥 𝑑𝑥

2 Expand
1 − cos 2𝑥 1 + cos 2𝑥
=න 𝑑𝑥
2 2 Combine like terms
1
= න( 1 − 2 cos 2𝑥 + 𝑐𝑜𝑠 2 2𝑥)(1 + cos 2𝑥) 𝑑𝑥
8
1
= න 1 + cos 2𝑥 − 2 cos 2𝑥 − 2𝑐𝑜𝑠 2 2𝑥 + 𝑐𝑜𝑠 2 2𝑥 + 𝑐𝑜𝑠 3 2𝑥 𝑑𝑥
8
1
= න 1 − cos 2𝑥 − 𝑐𝑜𝑠 2 2𝑥 + 𝑐𝑜𝑠 3 2𝑥 𝑑𝑥
8
1 1 1
= න − cos 4𝑥 − 𝑠𝑖𝑛2 2𝑥 cos 2𝑥 𝑑𝑥
8 2 2
 Example
1 1 + cos 4𝑥 Let 𝑢 = sin 2𝑥 , then
= න 1 − cos 2𝑥 − + (1 − 𝑠𝑖𝑛2 2𝑥)cos2𝑥 𝑑𝑥 1
8 2 𝑑𝑢 = 2 cos 2𝑥 𝑑𝑥; 𝑛𝑓 =
2
1 1 1 Integrate
= න − cos 2𝑥 − cos 4𝑥 + cos 2𝑥 − 𝑠𝑖𝑛2 2𝑥 cos 2𝑥 𝑑𝑥
8 2 2

1 𝑥 1 1 𝑠𝑖𝑛3 2𝑥 simplify
= − sin 4𝑥 − +𝑐
8 2 8 2 3
𝒙 𝟏 𝒔𝒊𝒏𝟑 𝟐𝒙 Final answer
= − 𝐬𝐢𝐧 𝟒𝒙 − +𝒄
𝟏𝟔 𝟔𝟒 𝟒𝟖
 Example 1

𝟏. න 𝒔𝒊𝒏𝟑 𝒙 𝒄𝒐𝒔𝟑 𝒙 𝒅𝒙

3 3 2 3
𝑠𝑖𝑛2 𝑥 = 1 − 𝑐𝑜𝑠 2 𝑥
න 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥 𝑑𝑥 = න 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥 sin 𝑥 𝑑𝑥

= න(1 − 𝑐𝑜𝑠 2 𝑥) 𝑐𝑜𝑠 3 𝑥 sin 𝑥 𝑑𝑥

Let 𝑢 = cos 𝑥 ; 𝑑𝑢 = − sin 𝑥 𝑑𝑥

= න(𝑐𝑜𝑠 3 𝑥 − 𝑐𝑜𝑠 5 𝑥) sin 𝑥 𝑑𝑥 integrate

= − න( 𝑢3 − 𝑢5 ) 𝑑𝑢

𝑢4 𝑢6
=− + +𝑐
4 6
𝒄𝒐𝒔𝟒 𝒙 𝒄𝒐𝒔𝟔 𝒙 Final answer
=− + +𝒄
𝟒 𝟔
 Example 2
𝟐. න 𝒔𝒊𝒏𝟑 𝒙 𝒄𝒐𝒔𝟑 𝒙 𝒅𝒙 (𝐚𝐥𝐭𝐞𝐫𝐧𝐚𝐭𝐞 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧)

𝑐𝑜𝑠 2 𝑥 = 1 − 𝑠𝑖𝑛2 𝑥
න 𝑠𝑖𝑛3 𝑥 𝑐𝑜𝑠 3 𝑥 𝑑𝑥 = න 𝑠𝑖𝑛3 𝑥 𝑐𝑜𝑠 2 𝑥 cos 𝑥 𝑑𝑥

= න 𝑠𝑖𝑛3 𝑥 (1 − 𝑠𝑖𝑛2 𝑥) cos 𝑥 𝑑𝑥

Let 𝑢 = sin 𝑥 ; 𝑑𝑢 = cos 𝑥 𝑑𝑥

= න(𝑠𝑖𝑛3 𝑥 − 𝑠𝑖𝑛5 𝑥) cos 𝑥 𝑑𝑥 integrate

= න( 𝑢3 − 𝑢5 ) 𝑑𝑢

𝑢4 𝑢6
= − +𝑐
4 6
𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏𝟔 𝒙
𝟒 Final answer
= − +𝒄
𝟒 𝟔
 Example 3
𝒅𝒙
𝟑. න
𝒄𝒔𝒄𝟑 𝟑𝒙
𝑑𝑥 3
=න = න 𝑠𝑖𝑛 3𝑥 𝑑𝑥
𝑐𝑠𝑐 3 3𝑥

= න 𝑠𝑖𝑛2 3𝑥 sin 3𝑥 𝑑𝑥

= න( 1 − 𝑐𝑜𝑠 2 3𝑥) sin 3𝑥 𝑑𝑥 Let 𝑢 = cos 3𝑥

𝑑𝑢 = −3 sin 3𝑥 𝑑𝑥
Integrate
= න sin 3𝑥 𝑑𝑥 − න 𝑐𝑜𝑠 2 3𝑥 sin 3𝑥 𝑑𝑥

1
= න sin 3𝑥 𝑑𝑥 + න 𝑢2 𝑑𝑢
3
𝟏 𝟏 Final answer
= − 𝐜𝐨𝐬 𝟑𝒙 + 𝒄𝒐𝒔𝟑 𝟑𝒙 + 𝒄
𝟑 𝟗
 Example 4

𝟒. න 𝒔𝒊𝒏𝟐 𝟓∅ 𝒅∅ 1 − cos 2𝑥
2
𝑠𝑖𝑛 𝑥 =
2
1 − cos 10∅ Integrate
න 𝑠𝑖𝑛2 5∅ 𝑑∅ = න 𝑑∅
2
1 1 simplify
= න 𝑑∅ − න cos 10∅ 𝑑∅
2 2
1 1 1
= ∅− sin 10∅ + 𝑐
2 2 10
𝟏 𝟏 Final Answer
= ∅− 𝐬𝐢𝐧 𝟏𝟎∅ + 𝒄
𝟐 𝟐𝟎
 Example 5
𝟓. න 𝒔𝒊𝒏𝟓 𝒛 𝒅𝒛
𝑠𝑖𝑛2 𝑧 = 1 − 𝑐𝑜𝑠 2 𝑧
න 𝑠𝑖𝑛5 𝑥 𝑑𝑥 = න 𝑠𝑖𝑛4 𝑧. sin 𝑧 𝑑𝑧
Expand
= ‫𝑛𝑖𝑠 ׬‬2 𝑧 2 . sin 𝑧 𝑑𝑧 Distribute sin z

= න 1 − 𝑐𝑜𝑠 2 𝑧 2
. sin 𝑧 𝑑𝑧 Let 𝑢 = cos 𝑧 ; 𝑑𝑢 = − sin 𝑧 𝑑𝑧

Integrate
= න(1 − 2 𝑐𝑜𝑠 2 𝑧 + 𝑐𝑜𝑠 4 𝑧) sin 𝑧 𝑑𝑧
Simplify
= න sin 𝑧 𝑑𝑧 − 2 න 𝑐𝑜𝑠 2 𝑧 sin 𝑧 𝑑𝑧 + න 𝑐𝑜𝑠 4 𝑧 sin 𝑧 𝑑𝑧

= න sin 𝑧 𝑑𝑧 + 2 න 𝑢2 𝑑𝑢 − න 𝑢4 𝑑𝑢

𝑢3 𝑢5
= − cos 𝑧 + 2 − +𝑐
3 5
𝟐 𝟏 Final answer
= − 𝐜𝐨𝐬 𝒛 + 𝒄𝒐𝒔 𝒛 − 𝒄𝒐𝒔𝟓 𝒛 + 𝒄
𝟑
𝟑 𝟓
 References
Printed References
1. Stewart, J. (2019). Calculus concepts and contexts 4th edition enhanced edition. USA :
Cengage Learning, Inc.
2. Canlapan, R.B.(2017). Diwa senior high school series: basic calculus. DIWA Learning Systems
Inc.
3. Hass, J., Weir, M. & Thomas, G. (2016). University calculus early transcendentals,global
edition. Pearson Education, Inc.Larson,
4. Bacani, J.B.,et.al. (2016). Basic calculus for senior high school. Book AtBP. Publishing Corp.
5. Balmaceda, J.M. (2016). Teaching guide in basic calculus. Commission on Higher Education.
6. Pelias, J.G. (2016). Basic calculus. Rex Bookstore.
7. Molina, M.F. (2016). Integral calculus. Unlimited Books Library Services & Publishing Inc.
8. Larson, R & Edwards, B. (2012). Calculus. Pasig City, Philippines. Cengage Learning Asia Pte
Ltd
 References
Electronic Resources
1. Pulham , John. Differential Calculus. Retrieved August 7, 2020 from
http:// www.maths.abdn.ac.uk/igc/tch/math1002/dif
2. Weistein, Eric. Differential Calculus. Retrieved August 7, 2020 from
http://mathworld.wolfram.com/differential calculus
3. Thomas , Christopher . Introduction to Differential Calculus.
Retrieved August 7, 2020 from http://
www.usyd.edu.aav/stuserv/document/maths_learning-center
4. Purplemath. Retrieved August 7, 2020 from
http://www.purplemath.com/modules/
Powers of Tangents
and Secants
 Introduction

Consider the trigonometric integral of the form,

න 𝑡𝑎𝑛𝑚 𝑣𝑠𝑒𝑐 𝑛 𝑣𝑑𝑥

If 𝑚 = 𝑛 = 1, we evaluate the integral directly. If 𝑚 = 𝑎𝑛𝑦 𝑛𝑢𝑚𝑏𝑒𝑟 𝑎𝑛𝑑 𝑛 = 2,

we evaluate the integral by the ordinary method by substitution.

In this section, we shall consider the following cases:

 Case I

When “m” is any number and “n” is a positive even integer greater than 2, then we may write,

𝑡𝑎𝑛𝑚 𝑣𝑠𝑒𝑐 𝑛 𝑣 = (𝑡𝑎𝑛𝑚 𝑣𝑠𝑒𝑐 𝑛−2 𝑣)𝑠𝑒𝑐 2 𝑣

And then use the identity,

𝑠𝑒𝑐 2 𝑣 = 1 + 𝑡𝑎𝑛2 𝑣

To reduce the given integral to the form,

න(𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠)𝑠𝑒𝑐 2 𝑣𝑑𝑥

Which is now integrable by letting 𝑢 = 𝑡𝑎𝑛𝑣 then 𝑑𝑢 = 𝑠𝑒𝑐 2 𝑣𝑑𝑣

 Case I

𝟏. න 𝒕𝒂𝒏𝟑 𝒙𝒔𝒆𝒄𝟒 𝒙𝒅𝒙

= න 𝑡𝑎𝑛3 𝑥. 𝑠𝑒𝑐 2 𝑥. 𝑠𝑒𝑐 2 𝑥𝑑𝑥

= න 𝑡𝑎𝑛3 𝑥(1 + 𝑡𝑎𝑛2 𝑥)𝑠𝑒𝑐 2 𝑥𝑑𝑥

Let 𝑢 = 𝑡𝑎𝑛𝑥, 𝑡ℎ𝑒𝑛 𝑑𝑢 = 𝑠𝑒𝑐 2 𝑥𝑑𝑥
= න(𝑡𝑎𝑛3 𝑥 +𝑡𝑎𝑛5 𝑥)𝑠𝑒𝑐 2 𝑥𝑑𝑥
Integrate
= න(𝑢3 +𝑢5 )𝑑𝑢

𝑢4 𝑢6
= + +𝑐 𝑡ℎ𝑒𝑛 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑢 = 𝑡𝑎𝑛𝑥
4 6
𝒕𝒂𝒏𝟒 𝒙 𝒕𝒂𝒏𝟔 𝒙
= + +𝒄 Final answer
𝟒 𝟔
 Case II

When “m” is a positive odd integer and “n” is any number.

𝑡𝑎𝑛𝑚 𝑣𝑠𝑒𝑐 𝑛 𝑣 = (𝑡𝑎𝑛𝑚−1 𝑣𝑠𝑒𝑐 𝑛−1 𝑣)𝑠𝑒𝑐𝑣𝑡𝑎𝑛𝑣

Then use the identity,

𝑡𝑎𝑛2 𝑣 = 𝑠𝑒𝑐 2 𝑣 − 1

To reduce the integral to the form,

න 𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 𝑠𝑒𝑐𝑎𝑛𝑡𝑠 𝑠𝑒𝑐𝑣𝑡𝑎𝑛𝑣𝑑𝑥

Which is integrable by letting 𝑢 = 𝑠𝑒𝑐𝑣 then 𝑑𝑢 = sec𝑣𝑡𝑎𝑛𝑣𝑑𝑣

 Case II

𝟐. න 𝒕𝒂𝒏𝟑 𝒙𝒔𝒆𝒄𝟓 𝒙𝒅𝒙

= න 𝑡𝑎𝑛2 𝑥. 𝑠𝑒𝑐 4 𝑥. 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥𝑑𝑥

= න(𝑠𝑒𝑐 2 𝑥 − 1)𝑠𝑒𝑐 4 𝑥. 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥𝑑𝑥

Let 𝑢 = 𝑠𝑒𝑐𝑥, 𝑡ℎ𝑒𝑛
= න(𝑠𝑒𝑐 6 𝑥 − 𝑠𝑒𝑐 4 𝑥)𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥𝑑𝑥 𝑑𝑢 = sec𝑥𝑡𝑎𝑛𝑥𝑑𝑥
Integrate
= න(𝑢6 −𝑢4 )𝑑𝑢

𝑢7 𝑢5
= − +𝑐 𝑡ℎ𝑒𝑛 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑢 = 𝑠𝑒𝑐𝑥
7 5
𝒔𝒆𝒄 𝒙 𝒔𝒆𝒄𝟓 𝒙
𝟕
= − +𝒄 Final answer
𝟕 𝟓
 Case III

When “m” is a positive odd (or even) integer and “n” is zero, we may write

𝑡𝑎𝑛𝑚 𝑣 = 𝑡𝑎𝑛𝑚−2 𝑣𝑡𝑎𝑛2 𝑣

Then use the identity,

𝑡𝑎𝑛2 𝑣 = 𝑠𝑒𝑐 2 𝑣 − 1

to reduce the integral to an integrable form

 Case III

𝟑. න 𝒕𝒂𝒏𝟓 𝒙𝒅𝒙

න 𝑡𝑎𝑛5 𝑥𝑑𝑥 = න 𝑡𝑎𝑛3 𝑥𝑡𝑎𝑛2 𝑥𝑑𝑥

= න 𝑡𝑎𝑛3 𝑥(𝑠𝑒𝑐 2 𝑥 − 1 )𝑑𝑥

= න(𝑡𝑎𝑛3 𝑥𝑠𝑒𝑐 2 𝑥 − 𝑡𝑎𝑛3 𝑥)𝑑𝑥

= න(𝑡𝑎𝑛3 𝑥𝑠𝑒𝑐 2 𝑥 − 𝑡𝑎𝑛𝑥. 𝑡𝑎𝑛2 𝑥)𝑑𝑥

= න[𝑡𝑎𝑛3 𝑥𝑠𝑒𝑐 2 𝑥 − 𝑡𝑎𝑛𝑥(𝑠𝑒𝑐 2 𝑥 − 1 )]𝑑𝑥 For 1st and 2nd term, let u= tan x, then du=
𝑠𝑒𝑐 2 𝑥𝑑𝑥
Integrate
= න[𝑡𝑎𝑛3 𝑥𝑠𝑒𝑐 2 𝑥 − 𝑡𝑎𝑛𝑥𝑠𝑒𝑐 2 𝑥 + 𝑡𝑎𝑛𝑥 ]𝑑𝑥

= න 𝑡𝑎𝑛3 𝑥𝑠𝑒𝑐 2 𝑥𝑑𝑥 − න 𝑡𝑎𝑛𝑥𝑠𝑒𝑐 2 𝑥𝑑𝑥 + න 𝑡𝑎𝑛𝑥𝑑𝑥

 Case III

= න 𝑢3 𝑑𝑢 − න 𝑢𝑑𝑢 + න 𝑡𝑎𝑛𝑥𝑑𝑥
𝑡ℎ𝑒𝑛 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑢 = 𝑡𝑎𝑛𝑥
𝑢4 𝑢2
= − − ln 𝑐𝑜𝑠𝑥 + 𝑐
4 2
𝒕𝒂𝒏𝟒 𝒙 𝒕𝒂𝒏𝟐 𝒙
= − − 𝒍𝒏 𝒄𝒐𝒔𝒙 + 𝒄 Final answer
𝟒 𝟐
 Example 4

𝟒. න(𝐬𝐞𝐜 𝒙 + 𝐭𝐚𝐧 𝒙)𝟐 𝒅𝒙 Expand

= න 𝑠𝑒𝑐 2 𝑥 + 2 sec 𝑥 tan 𝑥 + 𝑡𝑎𝑛2 𝑥 𝑑𝑥

𝑡𝑎𝑛2 𝑥 = 𝑠𝑒𝑐 2 𝑥 − 1

= න 𝑠𝑒𝑐 2 𝑥 + 2 sec 𝑥 tan 𝑥 + 𝑠𝑒𝑐 2 𝑥 − 1 𝑑𝑥

Integrate
= න 2𝑠𝑒𝑐 2 𝑥 + 2 sec 𝑥 tan 𝑥 − 1 𝑑𝑥

= 𝟐 𝐭𝐚𝐧 𝒙 + 𝟐 𝐬𝐞𝐜 𝒙 − 𝒙 + 𝒄 Final answer

 Example 5

𝟓. න 𝐭𝐚𝐧 𝒙 𝒔𝒆𝒄𝟔 𝒙 𝒅𝒙

= න 𝑡𝑎𝑛1/2 𝑥 𝑠𝑒𝑐 4 𝑥 𝑠𝑒𝑐 2 𝑥 𝑑𝑥

𝑠𝑒𝑐 2 𝑥 = 1 + 𝑡𝑎𝑛2 𝑥

= න 𝑡𝑎𝑛1/2 𝑥 (𝑠𝑒𝑐 2 𝑥)2 𝑠𝑒𝑐 2 𝑥 𝑑𝑥

Distribute 𝑡𝑎𝑛1/2 𝑥
= න 𝑡𝑎𝑛1/2 𝑥 (1 + 𝑡𝑎𝑛2 𝑥)2 𝑠𝑒𝑐 2 𝑥 𝑑𝑥
Let 𝑢 = tan 𝑥 ; 𝑑𝑢 = 𝑠𝑒𝑐 2 𝑥 𝑑𝑥
= න 𝑡𝑎𝑛1/2 𝑥 (1 + 2𝑡𝑎𝑛2 𝑥 + 𝑡𝑎𝑛4 𝑥) 𝑠𝑒𝑐 2 𝑥 𝑑𝑥
Integrate
 Example 5

= න(𝑡𝑎𝑛1/2 𝑥 + 2𝑡𝑎𝑛5/2 𝑥 + 𝑡𝑎𝑛9/2 𝑥)𝑠𝑒𝑐 2 𝑥 𝑑𝑥

= න( 𝑢1/2 + 2𝑢5/2 + 𝑢9/2 ) 𝑑𝑢

𝑢3/2 2𝑢7/2 𝑢11/2

= + + +𝑐
3/2 7/2 11/2

𝟐 𝟑/𝟐
𝟒 𝟕/𝟐
𝟐
= 𝒕𝒂𝒏 𝒙 + 𝒕𝒂𝒏 𝒙 + 𝒕𝒂𝒏𝟏𝟏/𝟐 𝒙 + 𝒄 Final answer
𝟑 𝟕 𝟏𝟏
 Example 6

𝟔. න 𝒕𝒂𝒏𝟑 𝒙 𝒔𝒆𝒄𝟑/𝟐 𝒙 𝒅𝒙

𝑡𝑎𝑛2 𝑥 = 𝑠𝑒𝑐 2 𝑥 − 1
= න 𝑡𝑎𝑛2 𝑥 𝑠𝑒𝑐 1/2 𝑥. tan 𝑥 sec 𝑥 𝑑𝑥
Distribute 𝑠𝑒𝑐 1/2 𝑥
= න( 𝑠𝑒𝑐 2 𝑥 − 1)𝑠𝑒𝑐 1/2 𝑥. tan 𝑥 sec 𝑥 𝑑𝑥
Let 𝑢 = sec 𝑥 ; 𝑑𝑢 = sec 𝑥 tan 𝑥 𝑑𝑥

= න( 𝑠𝑒𝑐 5/2 𝑥 − 𝑠𝑒𝑐 1/2 𝑥) tan 𝑥 sec 𝑥 𝑑𝑥 Integrate

1
5/2
= න( 𝑢 − 𝑢2 ) 𝑑𝑢

𝑢7/2 𝑢3/2
= − +𝑐
7/2 3/2
𝟐 𝟕/𝟐
𝟐 𝟑 Final answer
= 𝒔𝒆𝒄 𝒙 − 𝒔𝒆𝒄𝟐 𝒙 + 𝒄
𝟕 𝟑
 Example 7
𝟕. න 𝒕𝒂𝒏𝟒 𝒙 𝒅𝒙 𝑡𝑎𝑛2 𝑥 = 𝑠𝑒𝑐 2 𝑥 − 1

= න 𝑡𝑎𝑛2 𝑥 𝑡𝑎𝑛2 𝑥 𝑑𝑥

= න( 𝑠𝑒𝑐 2 𝑥 − 1)𝑡𝑎𝑛2 𝑥 𝑑𝑥

= න( 𝑠𝑒𝑐 2 𝑥𝑡𝑎𝑛2 𝑥 − 𝑡𝑎𝑛2 𝑥) 𝑑𝑥

Let 𝑢 = tan 𝑥 ; 𝑑𝑢 = 𝑠𝑒𝑐 2 𝑥 𝑑𝑥

= න( 𝑠𝑒𝑐 2 𝑥𝑡𝑎𝑛2 𝑥 − (𝑠𝑒𝑐 2 𝑥 − 1)) 𝑑𝑥 Integrate

But 𝑢 = tan 𝑥 , 𝑠𝑜
2 2 2
= න 𝑠𝑒𝑐 𝑥𝑡𝑎𝑛 𝑥 𝑑𝑥 − න 𝑠𝑒𝑐 𝑥 𝑑𝑥 + න 𝑑𝑥

= න 𝑢2 𝑑𝑢 − න 𝑠𝑒𝑐 2 𝑥 𝑑𝑥 + න 𝑑𝑥

𝑢3
= − tan 𝑥 + 𝑥 + 𝑐
3
𝒕𝒂𝒏𝟑 𝒙 Final answer
= − 𝐭𝐚𝐧 𝒙 + 𝒙 + 𝒄
𝟑
 Example 8
1 + 𝑡𝑎𝑛2 𝑥 = 𝑠𝑒𝑐 2 𝑥
𝟖. න 𝒔𝒆𝒄𝟒 𝒙 𝒅𝒙

න 𝑠𝑒𝑐 4 𝑥 𝑑𝑥 = න 𝑠𝑒𝑐 2 𝑥 𝑠𝑒𝑐 2 𝑥 𝑑𝑥

Let 𝑢 = tan 𝑥 ; 𝑑𝑢 = 𝑠𝑒𝑐 2 𝑥 𝑑𝑥

= න( 1 + 𝑡𝑎𝑛2 𝑥)𝑠𝑒𝑐 2 𝑥 𝑑𝑥 Integrate

2 2 2 But 𝑢 = tan 𝑥 , 𝑠𝑜
= න( 𝑠𝑒𝑐 𝑥 + 𝑡𝑎𝑛 𝑥𝑠𝑒𝑐 𝑥) 𝑑𝑥

= න 𝑠𝑒𝑐 2 𝑥 𝑑𝑥 + න 𝑢2 𝑑𝑢

𝑢3
= tan 𝑥 + +𝑐
3
𝒕𝒂𝒏𝟑 𝒙 Final answer
= 𝐭𝐚𝐧 𝒙 + +𝒄
𝟑
 Example 9

𝟗. න 𝒕𝒂𝒏𝟓 𝒙 𝒅𝒙

−1
sec 𝑥
𝑠𝑒𝑐 𝑥. sec 𝑥 = =1
= න 𝑡𝑎𝑛4 𝑥 . tan 𝑥 . 𝑠𝑒𝑐 −1 𝑥. sec 𝑥 𝑑𝑥 sec 𝑥

𝑡𝑎𝑛2 𝑥 = 𝑠𝑒𝑐 2 𝑥 − 1
= න(𝑡𝑎𝑛2 𝑥)2 . 𝑠𝑒𝑐 −1 𝑥. tan 𝑥 . sec 𝑥 𝑑𝑥
Expand
= න(𝑠𝑒𝑐 2 𝑥 − 1)2 . 𝑠𝑒𝑐 −1 𝑥. tan 𝑥 . sec 𝑥 𝑑𝑥
Distribute 𝑠𝑒𝑐 −1 𝑥

= න( 𝑠𝑒𝑐 4 𝑥 − 2𝑠𝑒𝑐 2 𝑥 + 1) . 𝑠𝑒𝑐 −1 𝑥. tan 𝑥 . sec 𝑥 𝑑𝑥 Let 𝑢 = sec 𝑥 ; 𝑑𝑢 = sec 𝑥 tan 𝑥 𝑑𝑥

Integrate
3 −1
= න( 𝑠𝑒𝑐 𝑥 − 2 sec 𝑥 + 𝑠𝑒𝑐 𝑥) . tan 𝑥 . sec 𝑥 𝑑𝑥
But 𝑢 = sec 𝑥, so
 Example 9

= න( 𝑢3 − 2𝑢 + 𝑢−1 )𝑑𝑢

𝑢4 2𝑢 2
= − + ln 𝑢 + 𝑐
4 2

𝒔𝒆𝒄𝟒 𝒙 Final answer

= − 𝒔𝒆𝒄𝟐 𝒙 + 𝐥𝐧( 𝐬𝐞𝐜 𝒙) + 𝒄
𝟒
 Example 10

𝟏𝟎. න 𝒕𝒂𝒏𝟑 𝒙 𝒔𝒆𝒄𝟑 𝒙 𝒅𝒙

𝑡𝑎𝑛2 𝑥 = 𝑠𝑒𝑐 2 𝑥 − 1
න 𝑡𝑎𝑛3 𝑥 𝑠𝑒𝑐 3 𝑥 𝑑𝑥 = න 𝑡𝑎𝑛2 𝑥 𝑠𝑒𝑐 2 𝑥. tan 𝑥 sec 𝑥 𝑑𝑥
Distribute 𝑠𝑒𝑐 2 𝑥
= න( 𝑠𝑒𝑐 2 𝑥 − 1)𝑠𝑒𝑐 2 𝑥. tan 𝑥 sec 𝑥 𝑑𝑥
Let 𝑢 = sec 𝑥 ; 𝑑𝑢 =
sec 𝑥 tan 𝑥 𝑑𝑥
= න( 𝑠𝑒𝑐 4 𝑥 − 𝑠𝑒𝑐 2 𝑥) tan 𝑥 sec 𝑥 𝑑𝑥
Integrate
= න( 𝑢4 − 𝑢2 ) 𝑑𝑢

𝑢5 𝑢3
= − +𝑐
5 3
𝟏 𝟏 Final answer
= 𝒔𝒆𝒄 𝒙 − 𝒔𝒆𝒄𝟑 𝒙 + 𝒄
𝟓
𝟓 𝟑
Powers of Cotangents
and Cosecants
 Introduction

The technique involved in evaluating the integral

න 𝑐𝑜𝑡 𝑚 𝑣𝑐𝑠𝑐 𝑛 𝑣𝑑𝑥

Where v is a differentiable function of x, is similar to that for evaluating the integral

න 𝑡𝑎𝑛𝑚 𝑣𝑠𝑒𝑐 𝑛 𝑣𝑑𝑥

The identity
𝑐𝑠𝑐 2 𝑣 = 1 + 𝑐𝑜𝑡 2 𝑣 or 𝑐𝑜𝑡 2 𝑣 = 𝑐𝑠𝑐 2 𝑣 − 1
is used to reduce the original expression into an integrable form. It also consists of three
possible cases.
 Case I

When “m” is any number and “n” is a positive even integer greater than 2, then
we may write,
𝑐𝑜𝑡 𝑚 𝑣𝑐𝑠𝑐 𝑛 𝑣 = (𝑐𝑜𝑡 𝑚 𝑣𝑐𝑠𝑐 𝑛−2 𝑣)𝑐𝑠𝑐 2 𝑣
And then use the identity,
𝑐𝑠𝑐 2 𝑣 = 1 + 𝑐𝑜𝑡 2 𝑣
To reduce the given integral to the form,

න(𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 𝑐𝑜𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠)𝑐𝑠𝑐 2 𝑣𝑑𝑥

Which is now integrable by letting 𝑢 = 𝑐𝑜𝑡𝑣 then 𝑑𝑢 = −𝑐𝑠𝑐 2 𝑣𝑑𝑣

 Case I

𝟏. න 𝒄𝒐𝒕𝟑 𝒙𝒄𝒔𝒄𝟒 𝒙𝒅𝒙

= න 𝑐𝑜𝑡 3 𝑥. 𝑐𝑠𝑐 2 𝑥. 𝑐𝑠𝑐 2 𝑥𝑑𝑥

= න 𝑐𝑜𝑡 3 𝑥(1 + 𝑐𝑜𝑡 2 𝑥)𝑐𝑠𝑐 2 𝑥𝑑𝑥

Let 𝑢 = 𝑐𝑜𝑡𝑥, 𝑡ℎ𝑒𝑛 𝑑𝑢 = −𝑐𝑠𝑐 2 𝑥𝑑𝑥
= න(𝑐𝑜𝑡 3 𝑥 +𝑐𝑜𝑡 5 𝑥)𝑐𝑠𝑐 2 𝑥𝑑𝑥
Integrate
= − න(𝑢3 +𝑢5 )𝑑𝑢

𝑢4 𝑢6
=− − +𝑐 𝑡ℎ𝑒𝑛 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑢 = 𝑐𝑜𝑡𝑥
4 6
𝒄𝒐𝒕 𝒙 𝒄𝒐𝒕𝟔 𝒙
𝟒
=− − +𝒄 Final answer
𝟒 𝟔
 Case II

When “m” is a positive odd integer and “n” is any number.

𝑐𝑜𝑡 𝑚 𝑣 𝑐𝑠𝑐 𝑛 𝑣 = (𝑐𝑜𝑡 𝑚−1 𝑣 𝑐𝑠𝑐 𝑛−1 𝑣)𝑐𝑠𝑐𝑣 𝑐𝑜𝑡𝑣
Then use the identity,
𝑐𝑜𝑡 2 𝑣 = 𝑐𝑠𝑐 2 𝑣 − 1
To reduce the integral to the form,

න 𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 𝑐𝑜𝑠𝑒𝑐𝑎𝑛𝑡𝑠 𝑐𝑠𝑐𝑣 𝑐𝑜𝑡𝑣 𝑑𝑥

Which is integrable by letting 𝑢 = 𝑐𝑠𝑐𝑣 then 𝑑𝑢 = −csc𝑣 𝑐𝑜𝑡𝑣 𝑑𝑣

 Case II

𝟐. න 𝒄𝒐𝒕𝟑 𝒙𝒄𝒔𝒄𝟓 𝒙𝒅𝒙

= න 𝑐𝑜𝑡 2 𝑥. 𝑐𝑠𝑐 4 𝑥. 𝑐𝑠𝑐𝑥. 𝑐𝑜𝑡𝑥 𝑑𝑥

= න(𝑐𝑠𝑐 2 𝑥 − 1)𝑐𝑠𝑐 4 𝑥. 𝑐𝑠𝑐𝑥. 𝑐𝑜𝑡𝑥 𝑑𝑥

Let 𝑢 = 𝑐𝑠𝑐𝑥, 𝑡ℎ𝑒𝑛
𝑑𝑢 = −𝑐𝑠𝑐𝑥 𝑐𝑜𝑡𝑥 𝑑𝑥
= න(𝑐𝑠𝑐 6 𝑥 − 𝑐𝑠𝑐 4 𝑥) 𝑐𝑠𝑐𝑥 𝑐𝑜𝑡𝑥 𝑑𝑥
Integrate
= − න(𝑢6 −𝑢4 )𝑑𝑢

𝑢7 𝑢5
=− + +𝑐 𝑡ℎ𝑒𝑛 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑢 = 𝑐𝑠𝑐𝑥
7 5
𝒄𝒔𝒄 𝒙 𝒄𝒔𝒄𝟓 𝒙
𝟕
=− + +𝒄 Final answer
𝟕 𝟓
 Case III

When “m” is a positive odd (or even) integer and “n” is zero, we may write

𝑐𝑜𝑡 𝑚 𝑣 = 𝑐𝑜𝑡 𝑚−2 𝑣 𝑐𝑜𝑡 2 𝑣

Then use the identity,

𝑐𝑜𝑡 2 𝑣 = 𝑐𝑠𝑐 2 𝑣 − 1

to reduce the integral to an integrable form

 Case III

𝟑. න 𝒄𝒐𝒕𝟑 𝒙𝒅𝒙

න 𝑐𝑜𝑡 3 𝑥𝑑𝑥 = න cot 𝑥 𝑐𝑜𝑡 2 𝑥𝑑𝑥

= න cot 𝑥 (𝑐𝑠𝑐 2 𝑥 − 1 )𝑑𝑥

For 1st term, let u= cot x, then

= න(cot 𝑥 𝑐𝑠𝑐 2 𝑥 − cot 𝑥)𝑑𝑥
du= −𝑐𝑠𝑐 2 𝑥 𝑑𝑥. Then Integrate

= න cot 𝑥 𝑠𝑒𝑐 2 𝑥𝑑𝑥 − න cot 𝑥 𝑑𝑥

= − න 𝑢𝑑𝑢 − න 𝑐𝑜𝑡𝑥 𝑑𝑥
𝑡ℎ𝑒𝑛 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑢 = 𝑐𝑜𝑡𝑥
2
𝑢
= − − ln sin 𝑥 + 𝑐
2
𝒄𝒐𝒕𝟐 𝒙
=− − 𝐥𝐧 𝐬𝐢𝐧 𝒙 + 𝒄 Final answer
𝟐
 Example 5
𝟏
𝟓. න 𝟔
𝒅𝒙
𝒔𝒊𝒏 𝟒𝒙
1 6 4𝑥 𝑑𝑥
න 𝑑𝑥 = න 𝑐𝑠𝑐
𝑠𝑖𝑛6 4𝑥

= න 𝑐𝑠𝑐 4 4𝑥. 𝑐𝑠𝑐 2 4𝑥 𝑑𝑥

= න(𝑐𝑠𝑐 2 4𝑥)2 . 𝑐𝑠𝑐 2 4𝑥 𝑑𝑥

Distribute 𝑐𝑠𝑐 2 4𝑥 𝑑𝑥
=න 1 + 𝑐𝑜𝑡 2 4𝑥 2 . 𝑐𝑠𝑐 2 4𝑥 𝑑𝑥

= න(1 + 2𝑐𝑜𝑡 2 4𝑥 + 𝑐𝑜𝑡 4 4𝑥). 𝑐𝑠𝑐 2 4𝑥 𝑑𝑥

 Example 5

Let 𝑢 = cot 4𝑥 ;
= න 𝑐𝑠𝑐 2 4𝑥 𝑑𝑥 + 2 න 𝑐𝑜𝑡 2 4𝑥 𝑐𝑠𝑐 2 4𝑥 𝑑𝑥 −1
2
𝑑𝑢 = −4𝑐𝑠𝑐 4𝑥 𝑑𝑥; 𝑛𝑓 =
4
Integrate
+ න 𝑐𝑜𝑡 4 4𝑥 𝑐𝑠𝑐 2 4𝑥 𝑑𝑥
simplify
2 1
= න 𝑐𝑠𝑐 4𝑥 𝑑𝑥 − න 𝑢 𝑑𝑢 − න 𝑢4 𝑑𝑢
2 2
4 4
1 2 𝑢3 1 𝑢5
= − cot 4𝑥 − . − . + 𝑐
4 4 3 4 5

𝟏 𝟏 𝟏 Final answer
= − 𝐜𝐨𝐭 𝟒𝒙 − 𝒄𝒐𝒕𝟑 𝟒𝒙 − 𝒄𝒐𝒕𝟓 𝟒𝒙 + 𝒄
𝟒 𝟔 𝟐𝟎
 Example 6
𝟔. න( 𝒄𝒔𝒄𝟐𝒙 − 𝟏)𝟐 𝒅𝒙

Expand
= න( 𝑐𝑠𝑐 2 𝑥 − 1)2 𝑑𝑥

= න( 𝑐𝑠𝑐 4 𝑥 − 2𝑐𝑠𝑐 2 𝑥 + 1) 𝑑𝑥
𝑐𝑠𝑐 2 𝑥 = 1 + 𝑐𝑜𝑡 2 𝑥

= න( 𝑐𝑠𝑐 2 𝑥. 𝑐𝑠𝑐 2 𝑥 − 2𝑐𝑠𝑐 2 𝑥 + 1) 𝑑𝑥

= න( (1 + 𝑐𝑜𝑡 2 𝑥)𝑐𝑠𝑐 2 𝑥 − 2𝑐𝑠𝑐 2 𝑥 + 1) 𝑑𝑥

Let 𝑢 = cot 𝑥 ; 𝑑𝑢 = −𝑐𝑠𝑐 2 𝑥 𝑑𝑥
= න(𝑐𝑠𝑐 2 𝑥 + 𝑐𝑜𝑡 2 𝑥𝑐𝑠𝑐 2 𝑥 − 2𝑐𝑠𝑐 2 𝑥 + 1) 𝑑𝑥 Substitute then Integrate

= න(𝑐𝑜𝑡 2 𝑥𝑐𝑠𝑐 2 𝑥 − 𝑐𝑠𝑐 2 𝑥 + 1) 𝑑𝑥

= − න 𝑢2 𝑑𝑢 − න 𝑐𝑠𝑐 2 𝑥 𝑑𝑥 + න 𝑑𝑥

𝑢3
= − + cot 𝑥 + 𝑥 + 𝑐
3
𝟏
= − 𝒄𝒐𝒕𝟑 𝒙 + 𝐜𝐨𝐭 𝒙 + 𝒙 + 𝒄 Final answer
𝟑
 Example 7
𝟕. න 𝒄𝒐𝒕𝟒 𝒙 𝒅𝒙 𝑐𝑜𝑡 2 𝑥 = 𝑐𝑠𝑐 2 𝑥 − 1

= න 𝑐𝑜𝑡 2 𝑥 𝑐𝑜𝑡 2 𝑥 𝑑𝑥

= න( 𝑐𝑠𝑐 2 𝑥 − 1)𝑐𝑜𝑡 2 𝑥 𝑑𝑥

= න( 𝑐𝑠𝑐 2 𝑥𝑐𝑜𝑡 2 𝑥 − 𝑐𝑜𝑡 2 𝑥) 𝑑𝑥

Let 𝑢 = cot 𝑥 ; 𝑑𝑢 = −𝑐𝑠𝑐 2 𝑥 𝑑𝑥

= න( 𝑐𝑠𝑐 2 𝑥𝑐𝑜𝑡 2 𝑥 − (𝑐𝑠𝑐 2 𝑥 − 1)) 𝑑𝑥

Integrate

= න 𝑐𝑠𝑐 2 𝑥𝑐𝑜𝑡 2 𝑥 𝑑𝑥 − න 𝑐𝑠𝑐 2 𝑥 𝑑𝑥 + න 𝑑𝑥 But 𝑢 = cot 𝑥 , 𝑠𝑜

= − න 𝑢2 𝑑𝑢 − න 𝑐𝑠𝑐 2 𝑥 𝑑𝑥 + න 𝑑𝑥

𝑢3
= − + cot 𝑥 + 𝑥 + 𝑐
3
𝒄𝒐𝒕𝟑 𝒙 Final answer
=− + 𝐜𝐨𝐭 𝒙 + 𝒙 + 𝒄
𝟑
 Example 8

𝟖. න 𝒄𝒔𝒄𝟒 𝒙 𝒅𝒙 1 + 𝑐𝑜𝑡 2 𝑥 = 𝑐𝑠𝑐 2 𝑥

න 𝑐𝑠𝑐 4 𝑥 𝑑𝑥 = න 𝑐𝑠𝑐 2 𝑥 𝑐𝑠𝑐 2 𝑥 𝑑𝑥

Let 𝑢 = cot 𝑥 ; 𝑑𝑢 =
= න( 1 + 𝑐𝑜𝑡 2 𝑥)𝑐𝑠𝑐 2 𝑥 𝑑𝑥 −𝑐𝑠𝑐 2 𝑥 𝑑𝑥

Integrate
= න( 𝑐𝑠𝑐 2 𝑥 + 𝑐𝑜𝑡 2 𝑥𝑐𝑠𝑐 2 𝑥) 𝑑𝑥
But 𝑢 = cot 𝑥 , 𝑠𝑜
= න 𝑐𝑠𝑐 2 𝑥 𝑑𝑥 − න 𝑢2 𝑑𝑢

𝑢3
= −cot 𝑥 − +𝑐
3
𝒄𝒐𝒕𝟑 𝒙 Final answer
= −𝐜𝐨𝐭 𝒙 − +𝒄
𝟑
 Example 9
𝟗. න 𝒄𝒐𝒕𝟓 𝒙 𝒅𝒙

csc 𝑥
= න 𝑐𝑜𝑡 4 𝑥 . cot 𝑥 . 𝑐𝑠𝑐 −1 𝑥. csc 𝑥 𝑑𝑥 𝑐𝑠𝑐 −1 𝑥. csc 𝑥 = =1
csc 𝑥

𝑐𝑜𝑡 2 𝑥 = 𝑐𝑠𝑐 2 𝑥 − 1
= න(𝑐𝑜𝑡 2 𝑥)2 . 𝑐𝑠𝑐 −1 𝑥. cot 𝑥 . csc 𝑥 𝑑𝑥
Expand
= න(𝑐𝑠𝑐 2 𝑥 − 1)2 . 𝑐𝑠𝑐 −1 𝑥. cot 𝑥 . csc 𝑥 𝑑𝑥
Distribute 𝑐𝑠𝑐 −1 𝑥

= න( 𝑐𝑠𝑐 4 𝑥 − 2𝑐𝑠𝑐 2 𝑥 + 1) . 𝑐𝑠𝑐 −1 𝑥. cot 𝑥 . csc 𝑥 𝑑𝑥 Let 𝑢 = csc 𝑥 ;

𝑑𝑢 = −csc 𝑥 cot 𝑥 𝑑𝑥
= න( 𝑐𝑠𝑐 3 𝑥 − 2 csc 𝑥 + 𝑐𝑠𝑐 −1 𝑥) cot 𝑥 . csc 𝑥 𝑑𝑥
Integrate

= − න( 𝑢3 − 2𝑢 + 𝑢−1 )𝑑𝑢 But 𝑢 = csc 𝑥, so

𝑢4 2𝑢2
= −( − + ln 𝑢) + 𝑐
4 2
𝒄𝒔𝒄𝟒 𝒙 Final answer
=− + 𝒄𝒔𝒄𝟐 𝒙 − 𝐥𝐧( 𝐜𝐬𝐜 𝒙) + 𝒄
𝟒
 Example 10

𝟏𝟎. න 𝒄𝒐𝒕𝟑 𝒙 𝒄𝒔𝒄𝟑 𝒙 𝒅𝒙

𝑐𝑜𝑡 2 𝑥 = 𝑐𝑠𝑐 2 𝑥 − 1
න 𝑐𝑜𝑡 3 𝑥 𝑐𝑠𝑐 3 𝑥 𝑑𝑥 = න 𝑐𝑜𝑡 2 𝑥 𝑐𝑠𝑐 2 𝑥. cot 𝑥 csc 𝑥 𝑑𝑥
Distribute 𝑐𝑠𝑐 2 𝑥

= න( 𝑐𝑠𝑐 2 𝑥 − 1)𝑐𝑠𝑐 2 𝑥. cot 𝑥 csc 𝑥 𝑑𝑥 Let 𝑢 = csc 𝑥 ;

𝑑𝑢 = − csc 𝑥 cot 𝑥 𝑑𝑥
Integrate
= න( 𝑐𝑠𝑐 4 𝑥 − 𝑐𝑠𝑐 2 𝑥) cot 𝑥 csc 𝑥 𝑑𝑥
simplify
= − න( 𝑢4 − 𝑢2 ) 𝑑𝑢

𝑢5 𝑢3
= −( − ) + 𝑐
5 3
𝟏 𝟏 Final answer
= − 𝒄𝒔𝒄𝟓 𝒙 + 𝒄𝒔𝒄𝟑 𝒙 + 𝒄
𝟓 𝟑
 References

Printed References

1. Stewart, J. (2019). Calculus concepts and contexts 4th edition enhanced edition. USA :
Cengage Learning, Inc.
2. Canlapan, R.B.(2017). Diwa senior high school series: basic calculus. DIWA Learning
Systems Inc.
3. Hass, J., Weir, M. & Thomas, G. (2016). University calculus early transcendentals,global
edition. Pearson Education, Inc.Larson,
4. Bacani, J.B.,et.al. (2016). Basic calculus for senior high school. Book AtBP. Publishing
Corp.
5. Balmaceda, J.M. (2016). Teaching guide in basic calculus. Commission on Higher
Education.
6. Pelias, J.G. (2016). Basic calculus. Rex Bookstore.
7. Molina, M.F. (2016). Integral calculus. Unlimited Books Library Services & Publishing Inc.
8. Larson, R & Edwards, B. (2012). Calculus. Pasig City, Philippines. Cengage Learning Asia
Pte Ltd
 References

Electronic Resources

1. Pulham , John. Differential Calculus. Retrieved August 7, 2020 from http://

www.maths.abdn.ac.uk/igc/tch/math1002/dif
2. Weistein, Eric. Differential Calculus. Retrieved August 7, 2020 from
http://mathworld.wolfram.com/differential calculus
3. Thomas , Christopher . Introduction to Differential Calculus. Retrieved
August 7, 2020 from http://
www.usyd.edu.aav/stuserv/document/maths_learning-center
4. Purplemath. Retrieved August 7, 2020 from
http://www.purplemath.com/modules/
 UNIVERSITY OF SAINT LOUIS
Tuguegarao City

SCHOOL OF ENGINEERING, ARCHITECTURE & IT EDUCATION

Engineering Sciences Department
S.Y. 2020 – 2021

CORRESPONDENCE LEARNING MODALITY

CALC 1024 : Calculus 2 (Integral Calculus)

Prepared by:

EROVITA TERESITA BACUD AGUSTIN, DME

ENGR. NORMAN BOB GOMEZ
ENGR. JENIROSE B. MATIAS
ENGR. PEEJAY PAGUIRIGAN
Instructors

Reviewed by:

ENGR. VICTOR VILLALUZ, MEE

EE Program Chair / ES Department Head

Recommended by:

ENGR. RONALD M. VILLAVERDE, ME – ECE, PECE

Academic Dean

Approved by:

EMMANUEL JAMES P. PATTAGUAN, Ph.D.

VP for Academics
For this week, March 15-18, 2021 of this semester, the following shall be your guide for the
different lessons and tasks that you need to accomplish. Be patient read it carefully before
proceeding to the tasks expected of you.

Date Topic Activities/Tasks

Integration by Trigonometric
Substitutions Read Lessons
March 15-18, 2021
Additional Standard Formula
Accomplish the worksheet in the
March 19, 2021 Submission of learning tasks Assessment Portion of this
module

GOD BLESS!
TRIGONOMETRIC SUBSTITUTION
ADDITIONAL STANDARD FORMULAS

Learning Objectives

▪ Learn substitutions involving trigonometric functions that lead to

trigonometric integrals

▪ Use additional standard formulas to solve for the integral of a

given function
WEEK 7: Trigonometric Substitutions

Now that you can evaluate integrals involving powers of trigonometric functions, you can use
trigonometric substitution to evaluate integrals when the integrand contains
𝑎2 − 𝑢 2
𝑎2 + 𝑢 2
𝑢 2 − 𝑎2

Trigonometric Substitution (𝒂 > 𝟎)

1. For integrals involving 𝒂𝟐 − 𝒖𝟐, let
𝒖 = 𝒂𝒔𝒊𝒏𝜽

2. For integrals involving 𝒂𝟐 + 𝒖𝟐, let

𝒖 = 𝒂𝒕𝒂𝒏𝜽

3. For integrals involving 𝒖𝟐 − 𝒂𝟐 , let

𝒖 = 𝒂𝒔𝒆𝒄𝜽

Recall:
𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒔𝒊𝒏𝜽 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆
𝒔𝒊𝒏𝜽 = 𝒄𝒐𝒔𝜽 = 𝒕𝒂𝒏𝜽 = =
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 𝒄𝒐𝒔𝜽 𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆

Steps on How to Integrate Applying Trigonometric Substitution

1. Determine what type of trigonometric substitution are you going to use
2. Identify 𝑎 𝑎𝑛𝑑 𝑢, then solve for 𝑑𝑢. 𝑢 𝑎𝑛𝑑 𝑑𝑢 must be in terms of 𝜃
3. You can draw a right triangle as your basis in obtaining certain trigonometric identity/ies
4. Substitute 𝑎, 𝑢 𝑎𝑛𝑑 𝑑𝑢 to the given problem
5. Simple or rearrange and solve for the integral of the problem
6. Convert the 𝜃′𝑠 back to 𝑥′𝑠.
Examples

𝒅𝒙 Since the integrand contains 𝑎2 − 𝑢2

𝟏. ∫
𝒙𝟐 √𝟗 − 𝒙𝟐 Use 𝒖 = 𝒂𝒔𝒊𝒏𝜽
2
𝑎 =9 𝑢2 = 𝑥 2 (TS 1)
𝑎=3 𝑢=𝑥
𝑢 = 𝑎𝑠𝑖𝑛𝜃 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑥 Differentiate 𝑥 with respect to 𝜃
𝑥 = 3𝑠𝑖𝑛𝜃 → 𝑠𝑖𝑛𝜃 =
3
𝑑𝑥 = 3𝑐𝑜𝑠𝜃𝑑𝜃
√9 − 𝑥 2
𝑐𝑜𝑠𝜃 = → 3𝑐𝑜𝑠𝜃 = √9 − 𝑥 2
3

𝑑𝑥 Substitute 𝑎, 𝑢, 𝑥, 𝑎𝑛𝑑 𝑑𝑥 to the given

∫ problem
𝑥 2 √9 − 𝑥 2
3𝑐𝑜𝑠𝜃𝑑𝜃 Simplify and cancel out common terms
=∫ Cancel 𝑐𝑜𝑠𝜃
(3𝑠𝑖𝑛𝜃 )2 (3𝑐𝑜𝑠𝜃)
3𝑐𝑜𝑠𝜃𝑑𝜃 3 1
=∫ =
27 𝑠𝑖𝑛2 𝜃𝑐𝑜𝑠𝜃 27 9
1 𝑑𝜃 Bring out the constant
= ∫ 1
9 𝑠𝑖𝑛2 𝜃 = − 𝑐𝑠𝑐 2 𝜃
1 𝑠𝑖𝑛2 𝜃
= ∫ 𝑐𝑠𝑐 2 𝜃 𝑑𝜃
9 Integrate
1
= (−𝑐𝑜𝑡𝜃) + 𝐶
9 Convert the 𝜃′𝑠 back to 𝑥′𝑠
1
= − 𝑐𝑜𝑡𝜃 + 𝐶 √9 − 𝑥 2
9 𝑐𝑜𝑠𝜃 3 √9 − 𝑥 2
1 √9 − 𝑥 2 𝑐𝑜𝑡𝜃 = = 𝑥 =
=− ∙ +𝐶 𝑠𝑖𝑛𝜃 𝑥
9 𝑥 3
√𝟗 − 𝒙𝟐 Final answer
=− +𝑪
𝟗𝒙

𝒅𝒙 Since the integrand contains 𝑎2 − 𝑢2

𝟐. ∫ 𝟑 Use 𝒖 = 𝒂𝒔𝒊𝒏𝜽
(𝟏𝟔 − 𝒙𝟐 )𝟐
2 (TS 1)
𝑎 = 16 𝑢2 = 𝑥 2
𝑎=4 𝑢=𝑥 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 = 𝑎𝑠𝑖𝑛𝜃 Differentiate 𝑥 with respect to 𝜃
𝑥
𝑥 = 4𝑠𝑖𝑛𝜃 → 𝑠𝑖𝑛𝜃 =
4
𝑑𝑥 = 4𝑐𝑜𝑠𝜃𝑑𝜃
 √16 − 𝑥 2
𝑐𝑜𝑠𝜃 = → 4𝑐𝑜𝑠𝜃 = √16 − 𝑥 2
4
1 3
(4𝑐𝑜𝑠𝜃 )3 = [(16 − 𝑥 2 )2
]
3
64 𝑐𝑜𝑠 3 𝜃 = (16 − 𝑥 2 )2

𝑑𝑥 Substitute 𝑎, 𝑢, 𝑥, 𝑎𝑛𝑑 𝑑𝑥 to the given

∫ 3 problem
(16 − 𝑥 2 )2
4𝑐𝑜𝑠𝜃𝑑𝜃
=∫ Simplify and cancel out common terms
64 𝑐𝑜𝑠 3 𝜃 Cancel 𝑐𝑜𝑠𝜃
1𝑐𝑜𝑠𝜃𝑑𝜃 4 1
=∫ =
16 𝑐𝑜𝑠 3 𝜃 64 16
1 𝑑𝜃 Bring out the constant
= ∫
16 𝑐𝑜𝑠 2 𝜃 1
1 = 𝑠𝑒𝑐 2 𝜃
= ∫ 𝑠𝑒𝑐 2 𝜃𝑑𝜃 𝑐𝑜𝑠 2 𝜃
16 Integrate
1 Convert the 𝜃 ′𝑠 back to 𝑥 ′ 𝑠
= (𝑡𝑎𝑛𝜃 ) + 𝐶 𝑥
16 𝑠𝑖𝑛𝜃 𝑥
1 𝑥 𝑡𝑎𝑛𝜃 = = 4 =
= ∙ +𝐶 𝑐𝑜𝑠𝜃 √16 − 𝑥 2 √16 − 𝑥 2
16 √16 − 𝑥 2
4
𝒙 Final answer
= +𝑪
𝟏𝟔√𝟏𝟔 − 𝒙𝟐

Since the integrand contains 𝑎2 − 𝑢2

𝟑. ∫(𝟒 − 𝒙𝟐 )−𝟐 𝒅𝒙
Use 𝒖 = 𝒂𝒔𝒊𝒏𝜽
𝑎2 = 4 𝑢2 = 𝑥 2 (TS 1)
𝑎=2 𝑢=𝑥
𝑢 = 𝑎𝑠𝑖𝑛𝜃 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑥 Differentiate 𝑥 with respect to 𝜃
𝑥 = 2𝑠𝑖𝑛𝜃 → 𝑠𝑖𝑛𝜃 =
2
𝑑𝑥 = 2𝑐𝑜𝑠𝜃𝑑𝜃
√4 − 𝑥 2
𝑐𝑜𝑠𝜃 = → 2𝑐𝑜𝑠𝜃 = √4 − 𝑥 2
2
1 −4
(2𝑐𝑜𝑠𝜃 )−2 = [(4 − 𝑥 2 )2 ]
−4 𝑐𝑜𝑠 −2 𝜃 = (4 − 𝑥 2 )−2
−4
= (4 − 𝑥 2 )−2
𝑐𝑜𝑠 2 𝜃

Substitute 𝑎, 𝑢, 𝑥, 𝑎𝑛𝑑 𝑑𝑥 to the given

∫(4 − 𝑥 2 )−2 𝑑𝑥
problem
−4
=∫ ∙ 2𝑐𝑜𝑠𝜃𝑑𝜃
𝑐𝑜𝑠 2 𝜃 Simplify and cancel common terms
 −8𝑐𝑜𝑠𝜃𝑑𝜃 Cancel out 𝑐𝑜𝑠𝜃
=∫
𝑐𝑜𝑠 2 𝜃 Bring out the constant
𝑑𝜃 1
= −8 ∫ = 𝑠𝑒𝑐𝜃
𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃
= −8 ∫ 𝑠𝑒𝑐𝜃𝑑𝜃 Integrate
= −8𝑙𝑛|𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃 | + 𝐶 Convert the 𝜃 ′𝑠 back to 𝑥 ′ 𝑠
2 𝑥 1 2
= −8𝑙𝑛 | + |+𝐶 𝑠𝑒𝑐𝜃 = =
√4 − 𝑥 2 √4 − 𝑥 2 𝑐𝑜𝑠𝜃 √4 − 𝑥 2
2+𝑥 𝑥
= −8𝑙𝑛 | |+𝐶 𝑠𝑖𝑛𝜃 2 𝑥
√4 − 𝑥 2 𝑡𝑎𝑛𝜃 = = =
𝑐𝑜𝑠𝜃 √4 − 𝑥 2 √4 − 𝑥 2
2
𝟐+𝒙 Final answer
= −𝟖𝒍𝒏 | |+𝑪
√𝟒 − 𝒙𝟐

Since the integrand contains 𝑎2 − 𝑢2

𝟒. ∫ √𝟐𝟓 − 𝟒𝒙𝟐 𝒅𝒙
Use 𝒖 = 𝒂𝒔𝒊𝒏𝜽
𝑎2 = 25 𝑢2 = 4𝑥 2 (TS 1)
𝑎=5 𝑢 = 2𝑥
𝑢 = 𝑎𝑠𝑖𝑛𝜃 Substitute 𝑢 𝑎𝑛𝑑 𝑎
2𝑥 Differentiate 𝑥 with respect to 𝜃
2𝑥 = 5𝑠𝑖𝑛𝜃 → 𝑠𝑖𝑛𝜃 =
5
5
𝑥= 𝑠𝑖𝑛𝜃
2
5
𝑑𝑥 = 𝑐𝑜𝑠𝜃𝑑𝜃
2
√25 − 4𝑥 2
𝑐𝑜𝑠𝜃 =
5
5𝑐𝑜𝑠𝜃 = √25 − 4𝑥 2

Substitute 𝑎, 𝑢, 𝑥, 𝑎𝑛𝑑 𝑑𝑥 to the given

∫ √25 − 4𝑥 2 𝑑𝑥
problem
5 Bring out the constant
= ∫(5𝑐𝑜𝑠𝜃) 𝑐𝑜𝑠𝜃𝑑𝜃
2 Integrate applying the rules on Powers of
25 Sines and Cosines (Case III)
= ∫ 𝑐𝑜𝑠 2 𝜃𝑑𝜃 1 + 𝑐𝑜𝑠2𝜃
2 𝑐𝑜𝑠 2 𝜃 =
25 1 + 𝑐𝑜𝑠2𝜃 2
= ∫( ) 𝑑𝜃
2 2
25
= ∫(1 + 𝑐𝑜𝑠2𝜃 )𝑑𝜃
4
25 25 Integrate
= ∫ 𝑑𝜃 + ∫ 𝑐𝑜𝑠2𝜃𝑑𝜃
4 4
25 25 1 Use the identity 𝑠𝑖𝑛2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
= 𝜃+ ∙ 𝑠𝑖𝑛2𝜃 + 𝐶 Convert the 𝜃 ′𝑠 back to 𝑥 ′ 𝑠
4 4 2
 25 25 2𝑥 2𝑥
= 𝜃+ ∙ 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 + 𝐶 𝑠𝑖𝑛𝜃 = → 𝜃 = 𝐴𝑟𝑐𝑠𝑖𝑛 ( )
4 8 5 5
25 25 √25 − 4𝑥 2
= 𝜃 + 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 + 𝐶 𝑐𝑜𝑠𝜃 =
4 4 5
25 2𝑥 25 2𝑥 √25 − 4𝑥 2
= 𝐴𝑟𝑐𝑠𝑖𝑛 ( ) + ∙ ∙ +𝐶
4 5 4 5 5
25 2𝑥 50𝑥√25 − 4𝑥 2
= 𝐴𝑟𝑐𝑠𝑖𝑛 ( ) + +𝐶
4 5 100
𝟐𝟓 𝟐𝒙 𝒙√𝟐𝟓 − 𝟒𝒙𝟐 Final answer
= 𝑨𝒓𝒄𝒔𝒊𝒏 ( ) + +𝑪
𝟒 𝟓 𝟐

𝒅𝒙 Since the integrand contains 𝑎2 + 𝑢2

𝟓. ∫
𝟒𝒙𝟐 + 𝟗 Use 𝒖 = 𝒂𝒕𝒂𝒏𝜽
2
𝑎 =9 𝑢2 = 4𝑥 2 (TS 2)
𝑎=3 𝑢 = 2𝑥
𝑢 = 𝑎𝑡𝑎𝑛𝜃 Substitute 𝑢 𝑎𝑛𝑑 𝑎
2𝑥 Differentiate 𝑥 with respect to 𝜃
2𝑥 = 3𝑡𝑎𝑛𝜃 → 𝑡𝑎𝑛𝜃 =
3
3
𝑥 = 𝑡𝑎𝑛𝜃
2
3
𝑑𝑥 = sec2 𝜃𝑑𝜃
2
√4𝑥 2 + 9
𝑠𝑒𝑐𝜃 =
3
2
4𝑥 +9
𝑠𝑒𝑐 2 𝜃 = → 9 𝑠𝑒𝑐 2 𝜃 = 4𝑥 2 + 9
9

𝑑𝑥 Substitute 𝑎, 𝑢, 𝑥, 𝑎𝑛𝑑 𝑑𝑥 to the given

∫
4𝑥 2 + 9 problem
3
2 sec2 𝜃𝑑𝜃
=∫ Cancel 𝑠𝑒𝑐 2 𝜃
9 𝑠𝑒𝑐 2 𝜃
3 Integrate
= ∫ 𝑑𝜃
18
1 Convert the 𝜃 ′𝑠 back to 𝑥 ′ 𝑠
= 𝜃+𝐶
6 2𝑥 2𝑥
1 2𝑥 𝑡𝑎𝑛𝜃 = → 𝜃 = 𝐴𝑟𝑐𝑡𝑎𝑛 ( )
= 𝐴𝑟𝑐𝑡𝑎𝑛 ( ) + 𝐶 3 3
6 3
𝟏 𝟐𝒙 Final answer
= 𝑨𝒓𝒄𝒕𝒂𝒏 ( ) + 𝑪
𝟔 𝟑

𝒅𝒙 Since the integrand contains 𝑎2 + 𝑢2

𝟔. ∫
𝒙𝟐 √𝒙𝟐 + 𝟏 Use 𝒖 = 𝒂𝒕𝒂𝒏𝜽
2
𝑎 =1 𝑢2 = 𝑥 2 (TS 2)
𝑎=1 𝑢=𝑥
𝑢 = 𝑎𝑡𝑎𝑛𝜃 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑥 Differentiate 𝑥 with respect to 𝜃
𝑥 = 𝑡𝑎𝑛𝜃 → 𝑡𝑎𝑛𝜃 =
1
𝑥 = 𝑡𝑎𝑛𝜃
𝑑𝑥 = sec2 𝜃𝑑𝜃
𝑠𝑒𝑐𝜃 = √𝑥 2 + 1
𝑠𝑒𝑐 2 𝜃 = 𝑥 2 + 1
𝑠𝑒𝑐 2 𝜃 − 1 = 𝑥 2 ∗ 𝑠𝑒𝑐 2 𝜃 − 1 = 𝑡𝑎𝑛2 𝜃
𝑡𝑎𝑛2 𝜃 = 𝑥 2

𝑑𝑥 Substitute 𝑎, 𝑢, 𝑥, 𝑎𝑛𝑑 𝑑𝑥 to the given

∫
𝑥 2 √𝑥 2 + 1 problem
𝑠𝑒𝑐 2 𝜃𝑑𝜃
=∫ Simplify
𝑡𝑎𝑛2 𝜃 ∙ 𝑠𝑒𝑐𝜃
𝑠𝑒𝑐𝜃𝑑𝜃
=∫
𝑡𝑎𝑛2 𝜃
1
𝑑𝜃
=∫ 𝑐𝑜𝑠𝜃
2
𝑠𝑖𝑛 𝜃
𝑐𝑜𝑠 2 𝜃
𝑑𝜃
=∫
𝑠𝑖𝑛2 𝜃
𝑐𝑜𝑠𝜃
𝑐𝑜𝑠𝜃𝑑𝜃
=∫
𝑠𝑖𝑛2 𝜃 Integrate using substitution method
𝑢 = 𝑠𝑖𝑛𝜃
= ∫ 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛−2 𝜃𝑑𝜃
𝑑𝑢 = 𝑐𝑜𝑠𝜃𝑑𝜃
= ∫ 𝑢−2 𝑑𝑢
𝑢−1
= +𝐶
−1
1 1
=− +𝐶 = 𝑐𝑠𝑐𝜃
𝑢 𝑠𝑖𝑛𝜃
1 Convert the 𝜃 ′𝑠 back to 𝑥 ′ 𝑠
=− +𝐶 𝑥
𝑠𝑖𝑛𝜃 𝑡𝑎𝑛𝜃 = 𝑠𝑒𝑐𝜃 = √𝑥 2 + 1
= −𝑐𝑠𝑐𝜃 + 𝐶 1
𝑥 𝑜𝑝𝑝 𝑥
=− +𝐶 𝑠𝑖𝑛𝜃 = =
√𝑥 2 + 1 ℎ𝑦𝑝 √𝑥 2 + 1
𝒙 Final answer
=− +𝑪
√𝒙𝟐 + 𝟏

𝒅𝒙 Since the integrand contains 𝑎2 + 𝑢2

𝟕. ∫
(𝒙 𝟐 + 𝟒)𝟐 Use 𝒖 = 𝒂𝒕𝒂𝒏𝜽
2
𝑎 =4 𝑢2 = 𝑥 2 (TS 2)
𝑎=2 𝑢=𝑥
𝑢 = 𝑎𝑡𝑎𝑛𝜃 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑥 Differentiate 𝑥 with respect to 𝜃
𝑥 = 2𝑡𝑎𝑛𝜃 → 𝑡𝑎𝑛𝜃 =
2
𝑥 = 2𝑡𝑎𝑛𝜃
𝑑𝑥 = 2𝑠𝑒𝑐 2 𝜃𝑑𝜃
𝑠𝑒𝑐𝜃 = √𝑥 2 + 4
𝑠𝑒𝑐 2 𝜃 = 𝑥 2 + 4

𝑑𝑥 Substitute 𝑎, 𝑢, 𝑥, 𝑎𝑛𝑑 𝑑𝑥 to the given

∫
(𝑥 2 + 4)2 problem
2𝑠𝑒𝑐 2 𝜃𝑑𝜃
=∫ Simplify
(𝑠𝑒𝑐 2 𝜃)2
𝑑𝜃 1
= 2∫ = 𝑐𝑜𝑠 2 𝜃
𝑠𝑒𝑐 2 𝜃 𝑠𝑒𝑐 2 𝜃
Integrate applying the rules on Powers of
= 2 ∫ 𝑐𝑜𝑠 2 𝜃 𝑑𝜃
Sines and Cosines (Case III)
1 + 𝑐𝑜𝑠2𝜃 1 + 𝑐𝑜𝑠2𝜃
= 2∫ 𝑑𝜃 𝑐𝑜𝑠 2 𝜃 =
2 2
2
= ∫(1 + 𝑐𝑜𝑠2𝜃)𝑑𝜃
2
= ∫(1 + 𝑐𝑜𝑠2𝜃 )𝑑𝜃

= ∫ 𝑑𝜃 + ∫ 𝑐𝑜𝑠2𝜃𝑑𝜃 Integrate
1
= 𝜃 + 𝑠𝑖𝑛2𝜃 + 𝐶 Use the identity 𝑠𝑖𝑛2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
2
1 Convert the 𝜃 ′𝑠 back to 𝑥 ′ 𝑠
= 𝜃 + ∙ 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 + 𝐶 𝑥 𝑥
2 𝑡𝑎𝑛𝜃 = → 𝜃 = 𝐴𝑟𝑐𝑡𝑎𝑛 ( )
= 𝜃 + 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 + 𝐶 2 2
𝑥 𝑥 1 1
= 𝐴𝑟𝑐𝑡𝑎𝑛 ( ) + ∙ +𝐶 𝑠𝑒𝑐𝜃 = √𝑥 2 + 4 → 𝑐𝑜𝑠𝜃 =
2
√𝑥 + 4
2 2 2
√𝑥 + 4 √𝑥 + 4
𝑥 𝑥 𝑜𝑝𝑝 𝑥
= 𝐴𝑟𝑐𝑡𝑎𝑛 ( ) + 2 +𝐶 𝑠𝑖𝑛𝜃 = =
2 𝑥 +4 ℎ𝑦𝑝 √𝑥 2 + 4
𝒙 𝒙 Final answer
= 𝑨𝒓𝒄𝒕𝒂𝒏 ( ) + 𝟐 +𝑪
𝟐 𝒙 +𝟒

𝒅𝒙 Since the integrand contains 𝑢2 − 𝑎2

𝟖. ∫
𝒙√𝒙𝟐 − 𝟗 Use 𝒖 = 𝒂𝒔𝒆𝒄𝜽
2
𝑎 =9 𝑢2 = 𝑥 2 (TS 3)
𝑎=3 𝑢=𝑥
𝑢 = 𝑎𝑠𝑒𝑐𝜃 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑥 Differentiate 𝑥 with respect to 𝜃
𝑥 = 3𝑠𝑒𝑐𝜃 → 𝑠𝑒𝑐𝜃 =
3
𝑑𝑥 = 3𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃𝑑𝜃
 √𝑥 2 − 9
𝑡𝑎𝑛𝜃 =
3
3𝑡𝑎𝑛𝜃 = √𝑥 2 − 9

𝑑𝑥 Substitute 𝑎, 𝑢, 𝑥, 𝑎𝑛𝑑 𝑑𝑥 to the given

∫
𝑥√𝑥 2 − 9 problem
3𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃𝑑𝜃 Simplify
=∫
3𝑠𝑒𝑐𝜃 ∙ 3𝑡𝑎𝑛𝜃
𝑑𝜃 1
=∫ = ∫ 𝑑𝜃 Integrate
3 3
1 Convert the 𝜃 ′𝑠 back to 𝑥 ′ 𝑠
= 𝜃+𝐶 𝑥 𝑥
3 𝑠𝑒𝑐𝜃 = → 𝜃 = 𝐴𝑟𝑐𝑠𝑒𝑐 ( )
1 𝑥 3 3
= 𝐴𝑟𝑐𝑠𝑒𝑐 ( ) + 𝐶
3 3
𝟏 𝒙 Final answer
= 𝑨𝒓𝒄𝒔𝒆𝒄 ( ) + 𝑪
𝟑 𝟑

𝒙𝟐 𝒅𝒙 Since the integrand contains 𝑢2 − 𝑎2

𝟗. ∫ 𝟑 Use 𝒖 = 𝒂𝒔𝒆𝒄𝜽
(𝒙𝟐 − 𝟐𝟓)𝟐 (TS 3)
2
𝑎 = 25 𝑢2 = 𝑥 2
𝑎=5 𝑢=𝑥 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 = 𝑎𝑠𝑒𝑐𝜃 Differentiate 𝑥 with respect to 𝜃
𝑥
𝑥 = 5𝑠𝑒𝑐𝜃 → 𝑠𝑒𝑐𝜃 =
5
𝑑𝑥 = 5𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃𝑑𝜃
1
√𝑥 2 − 25 √𝑥 2 − 25 = (𝑥 2 − 25)2
𝑡𝑎𝑛𝜃 =
5
5𝑡𝑎𝑛𝜃 = √𝑥 2 − 25
1 3
(5𝑡𝑎𝑛𝜃 )3 = [(𝑥 2 − 25)2 ]
3
125 𝑡𝑎𝑛3 𝜃 = (𝑥 2 − 25)2

𝑥 2 𝑑𝑥 Substitute 𝑎, 𝑢, 𝑥, 𝑎𝑛𝑑 𝑑𝑥 to the given

∫ 3 problem
(𝑥 2 − 25)2 Simplify
(5𝑠𝑒𝑐𝜃 )25𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃𝑑𝜃
=∫
125 𝑡𝑎𝑛3 𝜃
3
𝑠𝑒𝑐 𝜃𝑑𝜃
=∫
𝑡𝑎𝑛2 𝜃 𝑠𝑒𝑐 2 𝜃 − 1 = 𝑡𝑎𝑛2 𝜃
𝑠𝑒𝑐 3 𝜃𝑑𝜃 By long division of polynomials
=∫
𝑠𝑒𝑐 2 𝜃 − 1
𝑠𝑒𝑐 2 𝜃 − 1 = 𝑡𝑎𝑛2 𝜃
 𝑠𝑒𝑐𝜃
= ∫ (𝑠𝑒𝑐𝜃 − ) 𝑑𝜃
𝑠𝑒𝑐 2 𝜃 − 1
𝑠𝑒𝑐𝜃
= ∫ 𝑠𝑒𝑐𝜃𝑑𝜃 − ∫ 𝑑𝜃
𝑡𝑎𝑛2 𝜃
1
= ∫ 𝑠𝑒𝑐𝜃𝑑𝜃 − ∫ 𝑐𝑜𝑠𝜃 𝑑𝜃
𝑠𝑖𝑛2 𝜃
𝑐𝑜𝑠 2 𝜃 For the second term,
𝑐𝑜𝑠𝜃𝑑𝜃 Integrate using substitution method
= ∫ 𝑠𝑒𝑐𝜃𝑑𝜃 − ∫
𝑠𝑖𝑛2 𝜃 𝑢 = 𝑠𝑖𝑛𝜃
= ∫ 𝑠𝑒𝑐𝜃𝑑𝜃 − ∫ 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛−2 𝜃𝑑𝜃 𝑑𝑢 = 𝑐𝑜𝑠𝜃𝑑𝜃

= 𝑙𝑛|𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃 | − ∫ 𝑢−2 𝑑𝑢

𝑢−1
= 𝑙𝑛|𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃 | − +𝐶
−1
1
= 𝑙𝑛|𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃 | + + 𝐶 Convert the 𝜃 ′𝑠 back to 𝑥 ′ 𝑠
𝑢
1 𝑥 √𝑥 2 − 25
= 𝑙𝑛|𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃 | + +𝐶 𝑠𝑒𝑐𝜃 = 𝑡𝑎𝑛𝜃 =
𝑠𝑖𝑛𝜃 5 5
1 1 5
= 𝑙𝑛|𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃 | + +𝐶 =
𝑠𝑖𝑛𝜃 𝑠𝑖𝑛𝜃 √𝑥 2 − 25
𝑥 √𝑥 2 − 25 5
= 𝑙𝑛 | + |+ +𝐶
5 5 √𝑥 2 − 25
𝑥 + √𝑥 2 − 25 5
= 𝑙𝑛 | |+ +𝐶
5 √𝑥 2 − 25
𝒙 + √𝒙𝟐 − 𝟐𝟓 𝟓 Final answer
= 𝒍𝒏 | |+ +𝑪
𝟓 √𝒙𝟐 − 𝟐𝟓

𝒅𝒙 Since the integrand contains 𝑢2 − 𝑎2

𝟏𝟎. ∫
𝒙𝟒 √𝟗𝒙𝟐 − 𝟏 Use 𝒖 = 𝒂𝒔𝒆𝒄𝜽
2
𝑎 =1 𝑢2 = 9𝑥 2 (TS 3)
𝑎=1 𝑢 = 3𝑥
𝑢 = 𝑎𝑠𝑒𝑐𝜃 Substitute 𝑢 𝑎𝑛𝑑 𝑎
3𝑥 Differentiate 𝑥 with respect to 𝜃
3𝑥 = 𝑠𝑒𝑐𝜃 → 𝑠𝑒𝑐𝜃 =
1
1
𝑥 = 𝑠𝑒𝑐𝜃
3
1
𝑑𝑥 = 𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃𝑑𝜃
3
√9𝑥 2 − 1
𝑡𝑎𝑛𝜃 =
1
𝑡𝑎𝑛𝜃 = √9𝑥 2 − 1
 𝑑𝑥 Substitute 𝑎, 𝑢, 𝑥, 𝑎𝑛𝑑 𝑑𝑥 to the given
∫
𝑥 4 √9𝑥 2
−1 problem
1 Simplify
𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃𝑑𝜃
=∫ 3 4
1
(3 𝑠𝑒𝑐𝜃) 𝑡𝑎𝑛𝜃
𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃𝑑𝜃
= 27 ∫
𝑠𝑒𝑐 4 𝜃𝑡𝑎𝑛𝜃
𝑑𝜃 1
= 27 ∫ = 𝑐𝑜𝑠 3 𝜃
𝑠𝑒𝑐 3 𝜃 𝑠𝑒𝑐 3 𝜃
Integrate applying the rules on Powers of
= 27 ∫ 𝑐𝑜𝑠 3 𝜃 𝑑𝜃 Sines and Cosines (Case II)
= 27 ∫ 𝑐𝑜𝑠 2 𝜃𝑐𝑜𝑠𝜃𝑑𝜃
𝑢 = 𝑠𝑖𝑛𝜃
= 27 ∫(1 − 𝑠𝑖𝑛2 𝜃)𝑐𝑜𝑠𝜃𝑑𝜃
𝑑𝑢 = 𝑐𝑜𝑠𝜃𝑑𝜃
= 27 ∫(1 − 𝑢2 )𝑑𝑢
𝑢3 Convert the 𝜃 ′𝑠 back to 𝑥 ′ 𝑠
= 27𝑢 − 27 +𝐶
3 3𝑥 √9𝑥 2 − 1
= 27𝑠𝑖𝑛𝜃 − 9 𝑠𝑖𝑛3 𝜃 + 𝐶 𝑠𝑒𝑐𝜃 = 𝑡𝑎𝑛𝜃 =
3
1 1
2
√9𝑥 2 − 1 √9𝑥 2 − 1 𝑜𝑝𝑝 √9𝑥 − 1
= 27 ∙ − 9( ) +𝐶 𝑠𝑖𝑛𝜃 = =
3𝑥 3𝑥 ℎ𝑦𝑝 3𝑥
9√9𝑥 2 − 1 (9𝑥 2 − 1)3
= − +𝐶
𝑥 3𝑥 3
𝟗√𝟗𝒙𝟐 − 𝟏 (𝟗𝒙𝟐 − 𝟏)𝟑 Final answer
= − +𝑪
𝒙 𝟑𝒙𝟑
WEEK 7: Additional Standard Formulas

𝑑𝑢 1 𝑢
1. ∫ = 𝐴𝑟𝑐𝑡𝑎𝑛 +𝐶
𝑢 2 + 𝑎2 𝑎 𝑎
𝑑𝑢 1 𝑢−𝑎
2. ∫ 2 2
= 𝑙𝑛 | | + 𝐶 ∗∗
𝑢 −𝑎 2𝑎 𝑢+𝑎
𝑑𝑢
3. ∫ = 𝑙𝑛 |𝑢 + √𝑢2 + 𝑎2 | + 𝐶
√𝑢2 + 𝑎2
𝑑𝑢
4. ∫ = 𝑙𝑛 |𝑢 + √𝑢2 − 𝑎2 | + 𝐶
√𝑢2 − 𝑎2
𝑑𝑢 𝑢
5. ∫ = 𝐴𝑟𝑐𝑠𝑖𝑛 + 𝐶
√𝑎2 − 𝑢2 𝑎
𝑢 𝑎2 𝑢
6. ∫ √𝑎 − 𝑢 𝑑𝑢 = √𝑎 − 𝑢 + 𝐴𝑟𝑐𝑠𝑖𝑛 + 𝐶
2 2 2 2
2 2 𝑎
2
𝑢 𝑎
7. ∫ √𝑢2 + 𝑎2 𝑑𝑢 = √𝑢2 + 𝑎2 + 𝑙𝑛 |𝑢 + √𝑢2 + 𝑎2 | + 𝐶
2 2
𝑢 𝑎2
8. ∫ √𝑢 − 𝑎 𝑑𝑢 = √𝑢 − 𝑎 − 𝑙𝑛 |𝑢 + √𝑢2 − 𝑎2 | + 𝐶
2 2 2 2
2 2

𝑑𝑢 1 𝑎+𝑢
∗∗ ∫ = 𝑙𝑛 | |+𝐶
𝑎2 − 𝑢2 2𝑎 𝑎−𝑢

These formulas may be obtained by the method of trigonometric substitutions

Examples

𝒅𝒙 ASF 1
𝟏. ∫ 𝑑𝑢 1 𝑢
𝒙𝟐 + 𝟐𝟓 ∫ 2 = 𝐴𝑟𝑐𝑡𝑎𝑛 +𝐶
𝑢 + 𝑎2 𝑎 𝑎
𝑢2 = 𝑥 2 𝑎2 = 25 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢=𝑥 𝑎=5 Differentiate 𝑢 with respect to 𝑥
𝑑𝑢 = 𝑑𝑥
𝑑𝑥 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫ 2
𝑥 + 25
𝑑𝑢
=∫ 2 Integrate applying ASF 1
𝑢 + 𝑎2
1 𝑢 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
= 𝐴𝑟𝑐𝑡𝑎𝑛 + 𝐶
𝑎 𝑎
𝟏 𝒙 Final answer
= 𝑨𝒓𝒄𝒕𝒂𝒏 + 𝑪
𝟓 𝟓

𝒅𝒙 ASF 2
𝟐. ∫ 𝑑𝑢 1 𝑢−𝑎
(𝒙 + 𝟑)𝟐 − 𝟐𝟓
∫ 2 2
= 𝑙𝑛 | |+𝐶
𝑢 −𝑎 2𝑎 𝑢+𝑎
𝑢2 = (𝑥 + 3)2 𝑎2 = 25 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 =𝑥+3 𝑎=5 Differentiate 𝑢 with respect to 𝑥
𝑑𝑢 = 𝑑𝑥
 𝑑𝑥 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫
(𝑥 + 3)2 − 25
𝑑𝑢
=∫ 2 Integrate applying ASF 2
𝑢 − 𝑎2
1 𝑢−𝑎 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
= 𝑙𝑛 | |+𝐶
2𝑎 𝑢+𝑎
1 (𝑥 + 3) − 5
= 𝑙𝑛 | |+𝐶
2(5) (𝑥 + 3) + 5
𝟏 𝒙−𝟐 Final answer
= 𝒍𝒏 | |+𝑪
𝟏𝟎 𝒙+𝟖

𝑑𝑥 ASF 3
3. ∫ 𝑑𝑢
√4𝑥 2 +9 ∫ = 𝑙𝑛 |𝑢 + √𝑢2 + 𝑎2 | + 𝐶
2
√𝑢 + 𝑎 2
𝑢2 = 4𝑥 2 𝑎2 = 9 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 = 2𝑥 𝑎=3 Differentiate 𝑢 with respect to 𝑥
𝑑𝑢 = 2𝑑𝑥
𝑑𝑢
= 𝑑𝑥
2
𝑑𝑥 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫
√4𝑥 2 + 9
𝑑𝑢
=∫ 2 Integrate applying ASF 3
√𝑢 + 𝑎2
2
1 𝑑𝑢
= ∫
2 √𝑢 + 𝑎2
2
1
= 𝑙𝑛 |𝑢 + √𝑢2 + 𝑎2 | + 𝐶 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
2
𝟏 Final answer
= 𝒍𝒏 |𝟐𝒙 + √𝟒𝒙𝟐 + 𝟗| + 𝑪
𝟐

𝒆𝒙 𝒅𝒙 ASF 4
𝟒. ∫ 𝑑𝑢
√𝒆𝟐𝒙 − 𝟏 ∫ = 𝑙𝑛 |𝑢 + √𝑢2 − 𝑎2 | + 𝐶
√𝑢2 − 𝑎2
𝑢2 = 𝑒 2𝑥 𝑎2 = 1 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 = 𝑒𝑥 𝑎=1 Differentiate 𝑢 with respect to 𝑥
𝑑𝑢 = 𝑒 𝑥 𝑑𝑥
𝑒 𝑥 𝑑𝑥 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫
√𝑒 2𝑥 − 1
𝑑𝑢
=∫ Integrate applying ASF 4
√𝑢2 − 𝑎2
= 𝑙𝑛 |𝑢 + √𝑢2 − 𝑎2 | + 𝐶 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
= 𝑙𝑛 |𝑒 𝑥 + √𝑒 2𝑥 − 1| + 𝐶
= 𝑙𝑛𝑒 𝑥 + 𝑙𝑛√𝑒 2𝑥 − 1 + 𝐶
= 𝑥𝑙𝑛𝑒 + 𝑙𝑛√𝑒 2𝑥 − 1 + 𝐶 𝑙𝑛𝑒 = 1
= 𝒙 + 𝒍𝒏√𝒆𝟐𝒙 − 𝟏 + 𝑪 Final answer
 𝒙𝒅𝒙 ASF 5
𝟓. ∫ 𝑑𝑢 𝑢
√𝟏 − 𝒙𝟒 ∫ = 𝐴𝑟𝑐𝑠𝑖𝑛 + 𝐶
√𝑎2 − 𝑢2 𝑎
𝑢2 = 𝑥 4 𝑎2 = 1 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 = 𝑥2 𝑎=1 Differentiate 𝑢 with respect to 𝑥
𝑑𝑢 = 2𝑥𝑑𝑥
𝑑𝑢
= 𝑥𝑑𝑥
2
𝑥𝑑𝑥 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫
√1 − 𝑥 4
𝑑𝑢
=∫ 2
√𝑎 − 𝑢2
2
1 𝑑𝑢
= ∫ Integrate applying ASF 5
2 √𝑎2 − 𝑢2
1 𝑢
= 𝐴𝑟𝑐𝑠𝑖𝑛 + 𝐶 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
2 𝑎
1 𝑥2
= 𝐴𝑟𝑐𝑠𝑖𝑛 + 𝐶
2 1
𝟏 Final answer
= 𝑨𝒓𝒄𝒔𝒊𝒏𝒙𝟐 + 𝑪
𝟐

ASF 6
𝟔. ∫ √𝟑𝟔 − 𝟗𝒙𝟐 𝒅𝒙
𝑢 𝑎2 𝑢
∫ √𝑎2 − 𝑢2 𝑑𝑢 = √𝑎2 − 𝑢2 + 𝐴𝑟𝑐𝑠𝑖𝑛 + 𝐶
2 2 𝑎
𝑢2 = 9𝑥 2 𝑎2 = 36 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 = 3𝑥 𝑎=6 Differentiate 𝑢 with respect to 𝑥
𝑑𝑢 = 3𝑑𝑥
𝑑𝑢
= 𝑑𝑥
3
Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫ √36 − 9𝑥 2 𝑑𝑥
𝑑𝑢
= ∫ √𝑎 2 − 𝑢 2
3
1
= ∫ √𝑎2 − 𝑢2 𝑑𝑢 Integrate applying ASF 6
3
1 𝑢 𝑎2 𝑢
= [ √𝑎 − 𝑢 + 𝐴𝑟𝑐𝑠𝑖𝑛 ] + 𝐶
2 2
Substitute back 𝑢 𝑎𝑛𝑑 𝑎
3 2 2 𝑎
1 3𝑥 36 3𝑥
= [ √36 − 9𝑥 2 + 𝐴𝑟𝑐𝑠𝑖𝑛 ] + 𝐶 Simplify
3 2 2 6
3𝑥√36 − 9𝑥 2 36 3𝑥
= + 𝐴𝑟𝑐𝑠𝑖𝑛 +𝐶
6 6 6
𝒙√𝟑𝟔 − 𝟗𝒙𝟐 𝟑𝒙 Final answer
= + 𝟔𝑨𝒓𝒄𝒔𝒊𝒏 +𝑪
𝟐 𝟔
 ASF 7
𝟕. ∫ √𝟏𝟔𝒙𝟐 + 𝟐𝟓𝒅𝒙
𝑢 𝑎2
∫ √𝑢2 + 𝑎2 𝑑𝑢 = √𝑢2 + 𝑎2 + 𝑙𝑛 |𝑢 + √𝑢2 + 𝑎2 | + 𝐶
2 2
𝑢2 = 16𝑥 2 𝑎2 = 25 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 = 4𝑥 𝑎=5 Differentiate 𝑢 with respect to 𝑥
𝑑𝑢 = 4𝑑𝑥
𝑑𝑢
= 𝑑𝑥
4
Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given
∫ √16𝑥 2 + 25𝑑𝑥
problem
𝑑𝑢
= ∫ √𝑢 2 + 𝑎 2
4
1
= ∫ √𝑢2 + 𝑎2 𝑑𝑢
4
1 𝑢 𝑎2 Integrate applying ASF 7
= [ √𝑢2 + 𝑎2 + 𝑙𝑛 |𝑢 + √𝑢2 + 𝑎2 |] + 𝐶 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
4 2 2
1 4𝑥 25
= [ √16𝑥 2 + 25 + 𝑙𝑛 |4𝑥 + √16𝑥 2 + 25|] + 𝐶
4 2 2
4𝑥√16𝑥 2 + 25 25
= + 𝑙𝑛 |4𝑥 + √16𝑥 2 + 25| + 𝐶
8 8
𝟐
𝒙√𝟏𝟔𝒙 + 𝟐𝟓 𝟐𝟓 Final answer
= + 𝒍𝒏 |𝟒𝒙 + √𝟏𝟔𝒙𝟐 + 𝟐𝟓| + 𝑪
𝟐 𝟖

ASF 8
𝟖. ∫ √𝟒𝒙𝟐 − 𝟔𝟒𝒅𝒙
∫ √𝑢2 − 𝑎2 𝑑𝑢
𝑢 𝑎2
= √𝑢2 − 𝑎2 − 𝑙𝑛 |𝑢 + √𝑢2 − 𝑎2 | + 𝐶
2 2
𝑢2 = 4𝑥 2 𝑎2 = 64 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 = 2𝑥 𝑎=8 Differentiate 𝑢 with respect to 𝑥
𝑑𝑢 = 2𝑑𝑥
𝑑𝑢
= 𝑑𝑥
2
Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫ √4𝑥 2 − 64𝑑𝑥
𝑑𝑢
= ∫ √𝑢 2 − 𝑎 2
2
1 Integrate applying ASF 8
= ∫ √𝑢2 − 𝑎2 𝑑𝑢
2
1 𝑢 𝑎2 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
= [ √𝑢2 − 𝑎2 − 𝑙𝑛 |𝑢 + √𝑢2 − 𝑎2 |] + 𝐶
2 2 2
1 2𝑥 64
= [ √4𝑥 2 − 64 − 𝑙𝑛 |2𝑥 + √4𝑥 2 − 64|] + 𝐶
2 2 2
2𝑥√4𝑥 2 − 64 64
= − 𝑙𝑛 |2𝑥 + √4𝑥 2 − 64| + 𝐶
4 4
𝒙√𝟒𝒙𝟐 − 𝟔𝟒 Final answer
= − 𝟏𝟔𝒍𝒏 |𝟐𝒙 + √𝟒𝒙𝟐 − 𝟔𝟒| + 𝑪
𝟐
LEARNING ASSESSMENT

ASSESSMENT 7
First semester SY 2020 – 2021

NAME
PROGRAM

General Instructions: Solve the following problem neatly and completely. Show your complete
solution and simplify whenever possible. Box your final answers. Answers/Solutions must be
handwritten in a bond paper.

Trigonometric Substitution

√9 − 𝑥 2 𝑑𝑥
1. ∫ 𝑑𝑥 6. ∫ 3
𝑥2 (2 + 𝑥 2 )2
𝑥2 𝑑𝑥
2. ∫ 3 7. ∫
(𝑥 2 + 16)2 𝑥 2 √𝑥 − 7
3. ∫ 𝑥 2 √9 − 4𝑥 2 𝑑𝑥 8. ∫ 𝑥 3 √9 + 𝑥 2 𝑑𝑥

𝑥2 𝑥3
4. ∫ 9. ∫
5 √4𝑥 2 + 9
(1 − 𝑥 2 )2
√16 − 𝑒 2𝑥
𝑑𝑥 10. ∫ 𝑑𝑥
5. ∫ 𝑒𝑥
𝑥 3 √𝑥 2 − 4
For this week, March 22-25, 2021 of this semester, the following shall be your guide for the
different lessons and tasks that you need to accomplish. Be patient read it carefully before
proceeding to the tasks expected of you.

Date Topic Activities/Tasks

Integrands Involving Quadratic
Expressions
Read Lessons
March 22-25, 2021
Integration of Rational Functions
of Sinx and Cosx
Accomplish the worksheet in the
March 26, 2021 Submission of learning tasks Assessment Portion of this
module

GOD BLESS!
Integrands Involving Quadratic
Expressions
Integration of Rational
Functions of sinx and cosx

Learning Objectives

▪ Perform “completing the square” on the given quadratic

expressions to evaluate the integral of a given function

▪ Use substitution on the following rational functions of sinx and

cosx to solve for the integral of a given function
WEEK 8: Integrands Involving Quadratic Expressions

To this point we’ve seen quite a few integrals that involve quadratics. A couple of examples are,

𝑥 1
∫ 𝑑𝑥 = 𝑙𝑛|𝑥 2 ± 𝑎| + 𝐶
𝑥2 ±𝑎 2

𝑑𝑥 1 𝑥
∫ = 𝐴𝑟𝑐𝑡𝑎𝑛 ( )+𝐶
𝑥 2 ± 𝑎2 𝑎 𝑎

We also encountered integrals such as

√𝑏2 𝑥 2 − 𝑎2 , √𝑎2 − 𝑏2 𝑥 2 , 𝑎𝑛𝑑 √𝑎2 + 𝑏2 𝑥 2

Which could be solved applying trigonometric substitution. However, all those integrands were
missing an 𝑥 term. They only consist of a quadratic term and a constant only. Some integrals
involving general quadratics are easy to solve using substitution method. Some can also be
evaluated by partial fractions. Unfortunately, these methods won’t work on a lot of integrals. A
simple substitution will only work if the numerator is a constant multiple of the derivative of the
denominator and partial fractions will only work if the denominator can be factored.

The week’s topic focuses on how to deal with integrals involving quadratics when the techniques
that we’ve looked at to this point simply won’t work.
Back in the Trigonometric Substitution section we learned how to deal with square roots that had
a general quadratic in them. That kind of integral is exactly what we are going to need for the other
integrals in this section.

By completing the square, we can evaluate an integral that had a general quadratic in it and convert
it into a form that will allow us to use a known integration technique.Let’s recall ‘completing the
square’ before proceeding. Here is the general completing the square formula that we’ll use.

𝑏 2 𝑏 2 𝑏 2 𝑏2
𝑥 2 + 𝑏𝑥 + 𝑐 = 𝑥 2 + 𝑏𝑥 + ( ) − ( ) + 𝑐 = (𝑥 + ) + 𝑐 −
2 2 2 4

This will always take a general quadratic and written in terms of a squared term and a constant
term. Recall as well that in order to do this, we must have a coefficient of one in front of the 𝑥 2 . If
not, we’ll need to factor out the coefficient before completing the square. In other words,

𝑏 𝑐
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 𝑎 (𝑥 2 + 𝑥 + )
𝑎 𝑎

Examples

𝒅𝒙 By completing the square

𝟏. ∫ 𝑏 = −3 𝑐 = 2
𝒙𝟐
− 𝟑𝒙 + 𝟐
𝑏 2 𝑏2
(𝑥 + ) + 𝑐 −
2 4
 3 2 (−3)2
(𝑥 − ) + 2 −
2 4
2
3 9
(𝑥 − ) + 2 −
2 4

3 2 1
(𝑥 − ) −
2 4
𝑑𝑥 ASF 2
=∫ 𝑑𝑢 1 𝑢−𝑎
3 2 1 ∫ 2 = 𝑙𝑛 | |+𝐶
(𝑥 − 2) − 4 𝑢 −𝑎 2 2𝑎 𝑢+𝑎
3 2 1 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢2 = (𝑥 − ) 𝑎2 = Differentiate 𝑢 with respect to 𝑥
2 4
3 1
𝑢=𝑥− 𝑎=
2 2
𝑑𝑢 = 𝑑𝑥
𝑑𝑥 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
=∫
3 2 1
(𝑥 − 2) − 4
𝑑𝑢
=∫ 2
𝑢 − 𝑎2
1 𝑢−𝑎 Integrate applying ASF 2
= 𝑙𝑛 | |+𝐶 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
2𝑎 𝑢+𝑎
3 1
1 𝑥−2−2
= 𝑙𝑛 | |+𝐶
1 3 1
2 (2) 𝑥−2+2
𝑥−2
= 𝑙𝑛 | |+𝐶
𝑥+1
𝒙−𝟐 Final answer
𝒍𝒏 | |+𝑪
𝒙+𝟏

𝒅𝒙 By completing the square

𝟐. ∫
𝟐𝒙𝟐− 𝟐𝒙 + 𝟏 𝑎 = 2 𝑏 = −2 𝑐 = 1
𝑏 𝑐 Factor out 𝑎
𝑎 (𝑥 2 + 𝑥 + )
𝑎 𝑎
−2 1
2 (𝑥 2 + 𝑥+ ) 1
2 2
1 𝑏 = −1 𝑐 =
2 (𝑥 2 − 𝑥 + ) 2
2

𝑏 2 𝑏2
(𝑥 + ) + 𝑐 −
2 4
−1 2 1 (−1)2
(𝑥 + ) +
2 2 4
1 2 1 1
(𝑥 − ) + −
2 2 4
1 2 1
(𝑥 − ) +
2 4
 𝑑𝑥 ASF 1
=∫ 2 𝑑𝑢 1 𝑢
1 1 ∫ 2 = 𝐴𝑟𝑐𝑡𝑎𝑛 + 𝐶
(𝑥 − 2) + 4 𝑢 +𝑎 2 𝑎 𝑎
1 1 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢2 = (𝑥 − )2 𝑎2 =
2 4 Differentiate 𝑢 with respect to 𝑥
1 1
𝑢=𝑥− 𝑎=
2 2
𝑑𝑢 = 𝑑𝑥
𝑑𝑥 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫
1 2 1
(𝑥 − 2) + 4
𝑑𝑢
∫ 2
𝑢 + 𝑎2
1 𝑢 Integrate applying ASF 1
= 𝐴𝑟𝑐𝑡𝑎𝑛 + 𝐶 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
𝑎 𝑎
1
1 𝑥−
= 𝐴𝑟𝑐𝑡𝑎𝑛 2 +𝐶
1 1
2 2
1
= 2𝐴𝑟𝑐𝑡𝑎𝑛 [2 (𝑥 − )] + 𝐶
2
𝟐𝑨𝒓𝒄𝒕𝒂𝒏(𝟐𝒙 − 𝟏) + 𝑪 Final answer

By completing the square

𝟑. ∫ √𝟑 − 𝟐𝒙 − 𝒙𝟐 𝒅𝒙
ASF 6
= ∫ √4 − (𝑥 + 1)2 𝑑𝑥
∫ √𝑎2 − 𝑢2 𝑑𝑢
𝑢 𝑎2 𝑢
= √ 2 2
𝑎 − 𝑢 + 𝐴𝑟𝑐𝑠𝑖𝑛 + 𝐶
2 2 𝑎
𝑢2 = (𝑥 + 1)2 𝑎2 = 4
𝑢 =𝑥+1 𝑎=2
𝑑𝑢 = 𝑑𝑥
Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫ √4 − (𝑥 + 1)2 𝑑𝑥

= ∫ √𝑎2 − 𝑢2 𝑑𝑢
Integrate applying ASF 6
2
𝑢 𝑎 𝑢 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
= √𝑎2 − 𝑢2 + 𝐴𝑟𝑐𝑠𝑖𝑛 + 𝐶
2 2 𝑎
𝑥+1 (2)2 𝑥+1
= ( ) 2 (
√ 2 − 𝑥+1 + ) 2 𝐴𝑟𝑐𝑠𝑖𝑛 ( )+𝐶
2 2 2
𝑥+1 𝑥+1
= √4 − (𝑥 2 + 2𝑥 + 1) + 2𝐴𝑟𝑐𝑠𝑖𝑛 ( )+𝐶
2 2
𝒙+𝟏 𝒙+𝟏 Final answer
√𝟑 − 𝟐𝒙 − 𝒙𝟐 + 𝟐𝑨𝒓𝒄𝒔𝒊𝒏 ( )+𝑪
𝟐 𝟐
 (𝟐𝒙 + 𝟕)𝒅𝒙 Rewrite the given
𝟒. ∫
𝒙𝟐 + 𝟐𝒙 + 𝟓
[(2𝑥 + 2) + 5]
∫ 2 𝑑𝑥
𝑥 + 2𝑥 + 5
(2𝑥 + 2)𝑑𝑥 𝑑𝑥 Integrate the first term using substitution
∫ 2 +5∫ 2 method
𝑥 + 2𝑥 + 5 𝑥 + 2𝑥 + 5
For the second term, apply completing the
square on its denominator
(2𝑥 + 2)𝑑𝑥 𝑑𝑥 First term
=∫ 2
+ 5∫ 𝑢 = 𝑥 2 + 2𝑥 + 5
𝑥 + 2𝑥 + 5 (𝑥 + 1)2 + 4
𝑑𝑢 𝑑𝑥 𝑑𝑢 = (2𝑥 + 2)𝑑𝑥
=∫ + 5∫ Second term (ASF 1)
𝑢 (𝑥 + 1)2 + 4
𝑑𝑢 𝑑𝑢 𝑑𝑢 1 𝑢
=∫ + 5∫ 2 ∫ 2 2
= 𝐴𝑟𝑐𝑡𝑎𝑛 + 𝐶
𝑢 𝑢 + 𝑎2 𝑢 +𝑎 𝑎 𝑎
𝑢2 = (𝑥 + 1)2 𝑎2 = 4
𝑢 =𝑥+1 𝑎=2
𝑑𝑢 = 𝑑𝑥
1 𝑢 Integrate
= 𝑙𝑛|𝑢| + 5 𝐴𝑟𝑐𝑡𝑎𝑛 + 𝐶
𝑎 𝑎 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
2
5 𝑥+1
= 𝑙𝑛|𝑥 + 2𝑥 + 5| + 𝐴𝑟𝑐𝑡𝑎𝑛 +𝐶
2 2
𝟓 𝒙+𝟏 Final answer
𝒍𝒏|𝒙𝟐 + 𝟐𝒙 + 𝟓| + 𝑨𝒓𝒄𝒕𝒂𝒏 +𝑪
𝟐 𝟐

𝒅𝒙 Rewrite the given by completing the square

𝟓. ∫
√𝟗𝒙𝟐 + 𝟏𝟐𝒙 + 𝟖
𝒅𝒙
=∫
√(𝟗𝒙𝟐 + 𝟏𝟐𝒙 + 𝟒) + 𝟒
𝒅𝒙 ASF 3
=∫ 𝑑𝑢
√(𝟑𝒙 + 𝟐)𝟐 + 4 ∫ = 𝑙𝑛 |𝑢 + √𝑢2 + 𝑎2 | + 𝐶
√𝑢2 + 𝑎2
𝑢2 = (3𝑥 + 2)2 𝑎2 = 4 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 = 3𝑥 + 2 𝑎=2 Differentiate 𝑢 with respect to 𝑥
𝑑𝑢 = 3𝑑𝑥
𝑑𝑢
= 𝑑𝑥
3
𝒅𝒙 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫
√(𝟑𝒙 + 𝟐)𝟐 + 4
𝑑𝑢
=∫ 3
√𝑢 + 𝑎2
2
1 𝑑𝑢 Integrate applying ASF 3
= ∫
3 √𝑢 + 𝑎2
2
1
= 𝑙𝑛 |𝑢 + √𝑢2 + 𝑎2 | + 𝐶 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
2
1
= 𝑙𝑛 |(3𝑥 + 2) + √(𝟑𝒙 + 𝟐)𝟐 + 4| + 𝐶
3
𝟏 Final answer
𝒍𝒏 |(𝟑𝒙 + 𝟐) + √𝟗𝒙𝟐 + 𝟏𝟐𝒙 + 𝟖| + 𝑪
𝟑

(𝟐𝒙 − 𝟏)𝒅𝒙 Rewrite the given by completing the square

𝟔. ∫
√𝟒𝒙𝟐 − 𝟖𝒙 − 𝟕
(2𝑥 − 1)𝑑𝑥
=∫
√(4𝑥 2 − 8𝑥 + 4) − 11
(2𝑥 − 1)𝑑𝑥
=∫
√(2𝑥 − 2)2 − 11
(2𝑥 − 2 + 1)𝑑𝑥 Separate the numerator, group (2𝑥 − 2)
=∫
√(2𝑥 − 2)2− 11
(2𝑥 − 2)𝑑𝑥 𝑑𝑥
=∫ +∫ Transform the first term’s denominator into its
√(2𝑥 − 2)2 − 11 √(2𝑥 − 2)2 − 11 original form
(2𝑥 − 2)𝑑𝑥 𝑑𝑥
=∫ +∫
√4𝑥 2 − 8𝑥 − 7 √(2𝑥 − 2)2 − 11
First term: integrate applying substitution Second term: use ASF 4
method 𝑑𝑢
∫ = 𝑙𝑛 |𝑢 + √𝑢2 − 𝑎2 | + 𝐶
𝑢 = 4𝑥 2 − 8𝑥 − 7 √𝑢2 − 𝑎2
𝑑𝑢 = (8𝑥 − 8)𝑑𝑥 → 𝑑𝑢 = 4(2𝑥 − 2)𝑑𝑥 𝑢2 = (2𝑥 − 2)2 𝑎2 = 11
𝑑𝑢 𝑢 = 2𝑥 − 2 𝑎 = √11
= (2𝑥 − 2)𝑑𝑥
4 𝑑𝑢 = 2𝑑𝑥
𝑑𝑢
1 𝑑𝑢 2
=∫ ∙ +∫
√𝑢 4 √𝑢 − 𝑎2
2
1 1 1 𝑑𝑢
= ∫ 𝑢−2 𝑑𝑢 + ∫
4 2 √𝑢2 − 𝑎2
1 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
1 𝑢2 1
√ 2 2
= [ ] + 𝑙𝑛 |𝑢 + 𝑢 − 𝑎 | + 𝐶
4 1 2
2
1 1 1
= (4𝑥 2 − 8𝑥 − 7)2 + 𝑙𝑛 |(2𝑥 − 2) + √(2𝑥 − 2)2 − 11| + 𝐶
2 2
1 1
= √4𝑥 − 8𝑥 − 7 + 𝑙𝑛 |2𝑥 − 2 + √4𝑥 2 − 8𝑥 − 7| + 𝐶
2
2 2
𝟏 𝟏 Final answer
√𝟒𝒙𝟐 − 𝟖𝒙 − 𝟕 + 𝒍𝒏 |𝟐𝒙 − 𝟐 + √𝟒𝒙𝟐 − 𝟖𝒙 − 𝟕| + 𝑪
𝟐 𝟐

Rewrite the given by completing the square

𝟕. ∫ √𝟏𝟔𝒙𝟐 + 𝟐𝟒𝒙 𝒅𝒙
ASF 8
= ∫ √(16𝑥 2 + 24𝑥 + 9) − 9 𝑑𝑥
∫ √𝑢2 − 𝑎2 𝑑𝑢
= ∫ √(4𝑥 + 3)2 − 9 𝑑𝑥
𝑢 𝑎2
= √ 𝑢 − 𝑎 − 𝑙𝑛 |𝑢 + √𝑢2 − 𝑎2 | + 𝐶
2 2
2 2
𝑢2 = (4𝑥 + 3)2 𝑎2 = 9 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 = 4𝑥 + 3 𝑎=3 Differentiate 𝑢 with respect to 𝑥
𝑑𝑢 = 4𝑑𝑥
𝑑𝑢
= 𝑑𝑥
4
Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫ √(4𝑥 + 3)2 − 9 𝑑𝑥
𝑑𝑢
= ∫ √𝑢 2 − 𝑎 2
4
1
= ∫ √𝑢2 − 𝑎2 𝑑𝑢 Integrate applying ASF 8
4
2
1 𝑢 𝑎 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
= [ √𝑢2 − 𝑎2 − 𝑙𝑛 |𝑢 + √𝑢2 − 𝑎2 |] + 𝐶 Simplify
4 2 2
1 4𝑥 + 3 9
= [ √(4𝑥 + 3)2 − 9 − 𝑙𝑛 |4𝑥 + 3 + √(4𝑥 + 3)2 − 9|] + 𝐶
4 2 2
4𝑥 + 3 9
= √16𝑥 2 + 24𝑥 − 𝑙𝑛 |4𝑥 + 3 + √16𝑥 2 + 24𝑥| + 𝐶
4 8
3 9
= (𝑥 + ) √16𝑥 + 24𝑥 − 𝑙𝑛 |4𝑥 + 3 + √16𝑥 2 + 24𝑥| + 𝐶
2
4 8
𝟑 𝟗 Final answer
= (𝒙 + ) √𝟏𝟔𝒙𝟐 + 𝟐𝟒𝒙 − 𝒍𝒏 |𝟒𝒙 + 𝟑 + √𝟏𝟔𝒙𝟐 + 𝟐𝟒𝒙| + 𝑪
𝟒 𝟖
WEEK 8: Integration of Rational Functions of sinx and cosx
To integrate a rational function involving 𝑠𝑖𝑛𝑥, 𝑐𝑜𝑠𝑥 or both 𝑠𝑖𝑛𝑥 and 𝑐𝑜𝑠𝑥, we shall use the
substitution

𝑥
𝑧 = 𝑡𝑎𝑛 1
2

It can be shown that the substitution given 1 implies that

1 − 𝑧2
𝑐𝑜𝑠𝑥 = 2
1 + 𝑧2
2𝑧
𝑠𝑖𝑛𝑥 = 3
1 + 𝑧2
2𝑑𝑧
𝑑𝑥 = 4
1 + 𝑧2

The substitution of 2 , 3 , or both and of 4 will transform the original integral involving
𝑓 (𝑐𝑜𝑠𝑥 ), 𝑓 (𝑠𝑖𝑛𝑥 ), 𝑜𝑟 𝑓(𝑠𝑖𝑛𝑥, 𝑐𝑜𝑠𝑥) into an integral that is rational in 𝑧.

Examples

𝒅𝒙 Substitute
𝟏. ∫
𝟏 + 𝒄𝒐𝒔𝒙 1 − 𝑧2 2𝑑𝑧
𝑐𝑜𝑠𝑥 = 2
𝑑𝑥 =
1+𝑧 1 + 𝑧2
2𝑑𝑧 Simplify and rearrange the integrand
=∫ 1 + 𝑧2
1 − 𝑧2
1+
1 + 𝑧2
2𝑑𝑧 Cancel out(1 + 𝑧 2 )
=∫ 1 + 𝑧2
(1 + 𝑧 2 ) + 1 − 𝑧 2
1 + 𝑧2
2𝑑𝑧
=∫
2
= ∫ 𝑑𝑧 Integrate
Substitute
=𝑧+𝐶 𝑥
𝑧 = 𝑡𝑎𝑛
2
𝒙 Final answer
= 𝒕𝒂𝒏 +𝑪
𝟐

𝒅𝒙 Substitute
𝟐. ∫ 2𝑧 2𝑑𝑧
𝟐 − 𝒔𝒊𝒏𝒙 𝑠𝑖𝑛𝑥 = 𝑑𝑥 =
1 + 𝑧2 1 + 𝑧2
2𝑑𝑧 Simplify and rearrange the integrand
2
=∫ 1+𝑧
2𝑧
2−
1 + 𝑧2
 2𝑑𝑧
=∫ 1 + 𝑧2 Cancel out(1 + 𝑧 2 )
2(1 + 𝑧 2 ) − 2𝑧
1 + 𝑧2
2𝑑𝑧
=∫
2 + 2𝑧 2 − 2𝑧 Factor out 2
2𝑑𝑧
=∫ 2
2(𝑧 − 𝑧 + 1) Perform ‘completing the square’
𝑑𝑧
=∫
1 3
(𝑧 2 − 𝑧 + 4) + 4
𝑑𝑧 ASF 1
=∫ 𝑑𝑢 1 𝑢
1 2 3 ∫ 2 = 𝐴𝑟𝑐𝑡𝑎𝑛 +𝐶
(𝑧 − 2) + 4 𝑢 + 𝑎2 𝑎 𝑎
1 3 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢2 = (𝑧 − )2 𝑎2 =
2 4 Differentiate 𝑢 with respect to 𝑥
1 √3
𝑢 =𝑧− 𝑎=
2 2
𝑑𝑢 = 𝑑𝑧
𝑑𝑧 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫
1 2 3
(𝑧 − 2) + 4
𝑑𝑢
∫ 2
𝑢 + 𝑎2
1 𝑢 Integrate applying ASF 1
= 𝐴𝑟𝑐𝑡𝑎𝑛 + 𝐶 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
𝑎 𝑎
1 Simplify
1 𝑧−2
= 𝐴𝑟𝑐𝑡𝑎𝑛 +𝐶
√3 √3
2 2
2𝑧 − 1
2
= 𝐴𝑟𝑐𝑡𝑎𝑛 2 + 𝐶
√3 √3
2 Substitute
2 2𝑧 − 1 𝑥
= 𝐴𝑟𝑐𝑡𝑎𝑛 +𝐶 𝑧 = 𝑡𝑎𝑛
√3 √3 2
𝒙 Final answer
𝟐 𝟐 𝒕𝒂𝒏 𝟐 − 𝟏
= 𝑨𝒓𝒄𝒕𝒂𝒏 +𝑪
√𝟑 √𝟑

Substitute
𝟑. ∫ 𝒔𝒆𝒄𝒙𝒅𝒙
1 − 𝑧2 2𝑑𝑧
𝑐𝑜𝑠𝑥 = 𝑑𝑥 =
1 + 𝑧2 1 + 𝑧2
1
𝑠𝑒𝑐𝑥 =
𝑐𝑜𝑠𝑥
1 + 𝑧 2 2𝑑𝑧 Simplify
=∫ ∙ Cancel out (1 + 𝑧 2 )
1 − 𝑧2 1 + 𝑧2
2𝑑𝑧 ASF 2**
=∫
1 − 𝑧2
 𝑑𝑢 1 𝑎+𝑢
∫ = 𝑙𝑛 | |+𝐶
𝑎2 −𝑢 2 2𝑎 𝑎−𝑢
𝑢2 = 𝑧 2 𝑎2 = 1
𝑢=𝑧 𝑎=1
𝑑𝑢 = 𝑑𝑧
2𝑑𝑧 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
=∫
1 − 𝑧2
𝑑𝑢
= 2∫ 2 Integrate applying ASF 2**
𝑎 − 𝑢2
1 𝑎+𝑢 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
= 2 [ 𝑙𝑛 | |] + 𝐶
2𝑎 𝑎−𝑢
1 1+𝑧 Simplify
= 2[ 𝑙𝑛 | |] + 𝐶
2(1) 1−𝑧
1+𝑧 Substitute
= 𝑙𝑛 | |+𝐶 𝑥
1−𝑧 𝑧 = 𝑡𝑎𝑛
2
𝑥
1 + 𝑡𝑎𝑛 2
= 𝑙𝑛 | 𝑥| + 𝐶
1 − 𝑡𝑎𝑛 2
𝒙 Final answer
𝟏 + 𝒕𝒂𝒏 𝟐
𝒍𝒏 | 𝒙| + 𝑪
𝟏 − 𝒕𝒂𝒏
𝟐

𝒅𝒙 Substitute
𝟒. ∫
𝒔𝒊𝒏𝒙 + 𝒄𝒐𝒔𝒙 + 𝟑 1 − 𝑧2 2𝑧 2𝑑𝑧
𝑐𝑜𝑠𝑥 = 2
, 𝑠𝑖𝑛𝑥 = 2
, 𝑑𝑥 =
1+𝑧 1+𝑧 1 + 𝑧2
2𝑑𝑧 Simplify and rearrange the integrand
=∫ 1 + 𝑧2
2𝑧 1 − 𝑧2
2 + +3
1+𝑧 1 + 𝑧2
2𝑑𝑧
=∫ 1 + 𝑧2 Cancel out (1 + 𝑧 2 )
2𝑧 + 1 − 𝑧 2 + 3(1 + 𝑧 2 )
1 + 𝑧2
2𝑑𝑧
=∫
2𝑧 + 1 − 𝑧 2 + 3 + 3𝑧 2
2𝑑𝑧
=∫ 2
2𝑧 + 2𝑧 + 4
2𝑑𝑧 Factor out 2
=∫ 2
2(𝑧 + 𝑧 + 2)
𝑑𝑧 Perform ‘completing the square’
=∫
1 7
(𝑧 2 + 𝑧 + 4) + 4
𝑑𝑧 ASF 1
=∫ 𝑑𝑢 1 𝑢
1 2 7 ∫ 2 = 𝐴𝑟𝑐𝑡𝑎𝑛 + 𝐶
(𝑧 + 2) + 4 𝑢 +𝑎 2 𝑎 𝑎
1 7 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢2 = (𝑧 + )2 𝑎2 =
2 4 Differentiate 𝑢 with respect to 𝑥
 1 √7
𝑢 =𝑧+ 𝑎=
2 2
𝑑𝑢 = 𝑑𝑧
𝑑𝑧 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫
1 2 7
(𝑧 + 2) + 4
𝑑𝑢
∫ 2
𝑢 + 𝑎2
1 𝑢 Integrate applying ASF 1
= 𝐴𝑟𝑐𝑡𝑎𝑛 + 𝐶 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
𝑎 𝑎
1 Simplify
1 𝑧+2
= 𝐴𝑟𝑐𝑡𝑎𝑛 +𝐶
√7 √7
2 2
2𝑧 + 1
2
= 𝐴𝑟𝑐𝑡𝑎𝑛 2 + 𝐶
√7 √7
2
2 2𝑧 + 1 Substitute
= 𝐴𝑟𝑐𝑡𝑎𝑛 +𝐶 𝑥
√7 √7 𝑧 = 𝑡𝑎𝑛
2
𝒙 Final answer
𝟐 𝟐 𝒕𝒂𝒏 𝟐 + 𝟏
= 𝑨𝒓𝒄𝒕𝒂𝒏 +𝑪
√𝟕 √𝟕

𝒅𝒙 Substitute
𝟓. ∫ 2𝑧 2𝑑𝑧
𝟒 + 𝟐𝒔𝒊𝒏𝒙 𝑠𝑖𝑛𝑥 = 𝑑𝑥 =
1+𝑧 2 1 + 𝑧2
2𝑑𝑧 Simplify and rearrange the integrand
=∫ 1 + 𝑧2
2𝑧
4 + 2( )
1 + 𝑧2
2𝑑𝑧
=∫ 1 + 𝑧2 Cancel out (1 + 𝑧 2 )
4(1 + 𝑧 2 ) + 4𝑧
1 + 𝑧2
2𝑑𝑧
=∫
4 + 4𝑧 2 + 4𝑧
2𝑑𝑧 Factor out 4
=∫ 2
4(𝑧 + 𝑧 + 1)
1 𝑑𝑧 Perform ‘completing the square’
= ∫ 2
2 𝑧 +𝑧+1
1 𝑑𝑧
= ∫
2 (𝑧 2 + 𝑧 + 1) + 3
4 4
1 𝑑𝑧 ASF 1
= ∫ 𝑑𝑢 1 𝑢
2 1 2 3 ∫ 2 = 𝐴𝑟𝑐𝑡𝑎𝑛 + 𝐶
(𝑧 + 2) + 4 𝑢 +𝑎 2 𝑎 𝑎
2
1 2 3 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 = (𝑧 + ) 𝑎2 = Differentiate 𝑢 with respect to 𝑥
2 4
 1 √3
𝑢 =𝑧+ 𝑎=
2 2
𝑑𝑢 = 𝑑𝑧
1 𝑑𝑧 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
∫
2 1 2 7
(𝑧 + 2) + 4 Integrate applying ASF 1
1 𝑑𝑢
∫ 2
2 𝑢 + 𝑎2
1 1 𝑢 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
= [ 𝐴𝑟𝑐𝑡𝑎𝑛 ] + 𝐶
2 𝑎 𝑎
1 Simplify
1 1 𝑧+2
= [ 𝐴𝑟𝑐𝑡𝑎𝑛 ]+𝐶
2 √3 √3
2 2
2𝑧 + 1
1 2
= [ 𝐴𝑟𝑐𝑡𝑎𝑛 2 ] + 𝐶
2 √3 √3
2
1 2𝑧 + 1 Substitute
= 𝐴𝑟𝑐𝑡𝑎𝑛 +𝐶 𝑥
√3 √3 𝑧 = 𝑡𝑎𝑛
2
𝒙 Final answer
𝟏 𝟐 𝒕𝒂𝒏 𝟐 + 𝟏
= 𝑨𝒓𝒄𝒕𝒂𝒏 +𝑪
√𝟑 √𝟑

𝒅𝒙 Substitute
𝟔. ∫
𝟏 − 𝟐𝒄𝒐𝒔𝒙 1 − 𝑧2 2𝑑𝑧
𝑐𝑜𝑠𝑥 = 𝑑𝑥 =
1 + 𝑧2 1 + 𝑧2
2𝑑𝑧 Simplify and rearrange the integrand
=∫ 1 + 𝑧2
1 − 𝑧2
1 − 2( )
1 + 𝑧2
2𝑑𝑧
=∫ 1 + 𝑧2 Cancel out (1 + 𝑧 2 )
1 + 𝑧 − 2(1 − 𝑧 2 )
2

1 + 𝑧2
2𝑑𝑧
=∫
1 + 𝑧 − 2 + 2𝑧 2
2
2𝑑𝑧 Factor out 2
=∫ 2
3𝑧 − 1
𝑑𝑧 ASF 2
= 2∫ 2 𝑑𝑢 1 𝑢−𝑎
3𝑧 − 1 ∫ 2 = 𝑙𝑛 | |+𝐶
𝑢 −𝑎 2 2𝑎 𝑢+𝑎
𝑢2 = 3𝑧 2 𝑎2 = 1 Substitute 𝑢 𝑎𝑛𝑑 𝑎
𝑢 = √3𝑧 𝑎=1 Differentiate 𝑢 with respect to 𝑥
𝑑𝑢 = √3𝑑𝑧
𝑑𝑧 Substitute 𝑎, 𝑢, 𝑎𝑛𝑑 𝑑𝑢 to the given problem
2∫ 2
3𝑧 − 1
 𝑑𝑢 Integrate applying ASF 2
= 2∫
𝑢2− 𝑎2
1 𝑢−𝑎 Substitute back 𝑢 𝑎𝑛𝑑 𝑎
= 2 [ 𝑙𝑛 | |] + 𝐶
2𝑎 𝑢+𝑎
1 √3𝑧 − 1
= 2[ 𝑙𝑛 | |] + 𝐶
2(1) √3𝑧 + 1
√3𝑧 − 1 Substitute
= 𝑙𝑛 | |+𝐶 𝑥
√3𝑧 + 1 𝑧 = 𝑡𝑎𝑛
2
𝒙 Final answer
√𝟑 𝒕𝒂𝒏 𝟐 − 𝟏
= 𝒍𝒏 | 𝒙 |+𝑪
√𝟑 𝒕𝒂𝒏 𝟐 + 𝟏

𝒅𝒙 Substitute
𝟕. ∫
𝟏 + 𝒔𝒊𝒏𝒙 + 𝒄𝒐𝒔𝒙 1 − 𝑧2 2𝑧 2𝑑𝑧
𝑐𝑜𝑠𝑥 = , 𝑠𝑖𝑛𝑥 = , 𝑑𝑥 =
1 + 𝑧2 1 + 𝑧2 1 + 𝑧2
2𝑑𝑧 Simplify and rearrange the integrand
=∫ 1 + 𝑧2
2𝑧 1 − 𝑧2
1+ +
1 + 𝑧2 1 + 𝑧2
2𝑑𝑧 Simplify and rearrange the integrand
=∫ 1 + 𝑧2 Cancel out (1 + 𝑧 2 )
1 + 𝑧 + 2𝑧 + 1 − 𝑧 2
2
1 + 𝑧2
2𝑑𝑧 Factor out 2
=∫
2𝑧 + 2
2𝑑𝑧
=∫
2(𝑧 + 1)
𝑑𝑧 Integrate by substitution method
=∫
𝑧+1 𝑢 =𝑧+1
𝑑𝑢 = 𝑑𝑧
𝑑𝑢
=∫
𝑢
= 𝑙𝑛|𝑢| + 𝐶 Substitute back 𝑢
= 𝑙𝑛|𝑧 + 1| + 𝐶 Substitute
𝑥
𝑧 = 𝑡𝑎𝑛
2
𝒙 Final answer
= 𝒍𝒏 |𝒕𝒂𝒏 + 𝟏| + 𝑪
𝟐
References
Printed References
1. Stewart, J. (2019). Calculus concepts and contexts 4th edition enhanced edition. USA :
Cengage Learning, Inc.
2. Canlapan, R.B.(2017). Diwa senior high school series: basic calculus. DIWA Learning
Systems Inc.
3. Hass, J., Weir, M. & Thomas, G. (2016). University calculus early transcendentals,global
edition. Pearson Education, Inc.Larson,
4. Bacani, J.B.,et.al. (2016). Basic calculus for senior high school. Book AtBP. Publishing
Corp.
5. Balmaceda, J.M. (2016). Teaching guide in basic calculus. Commission on Higher
Education.
6. Pelias, J.G. (2016). Basic calculus. Rex Bookstore.
7. Molina, M.F. (2016). Integral calculus. Unlimited Books Library Services & Publishing
Inc.
8. Larson, R & Edwards, B. (2012). Calculus. Pasig City, Philippines. Cengage Learning
Asia Pte Ltd

Electronic Resources
1. Pulham , John. Differential Calculus. Retrieved August 7, 2020 from http://
www.maths.abdn.ac.uk/igc/tch/math1002/dif
2. Weistein, Eric. Differential Calculus. Retrieved August 7, 2020 from
http://mathworld.wolfram.com/differential calculus
3. Thomas , Christopher . Introduction to Differential Calculus. Retrieved August 7, 2020
from http:// www.usyd.edu.aav/stuserv/document/maths_learning-center
4. Purplemath. Retrieved August 7, 2020 from http://www.purplemath.com/modules/
5. https://tutorial.math.lamar.edu/classes/calcii/integralswithquadratics.aspx

Learning Materials
1. Worksheet
2. Module
LEARNING ASSESSMENT

ASSESSMENT 8
First semester SY 2020 – 2021

NAME
PROGRAM

General Instructions: Solve the following problem neatly and completely. Show your complete
solution and simplify whenever possible. Box your final answers. Answers/Solutions must be
handwritten in a bond paper.

Part I: Integrands Involving Quadratic Expressions

1. ∫ √𝑥 2 − 8𝑥 + 25𝑑𝑥
𝑑𝑥
2. ∫
√12𝑥 − 4𝑥 2 − 5
3. ∫ √4𝑥 2 − 12𝑥 + 5𝑑𝑥
(4𝑥 − 1)𝑑𝑥
4. ∫
√4𝑥 2 − 9
(𝑥 + 3)
5. ∫ 𝑑𝑥
6𝑥 − 𝑥 2
(𝑥 + 1)𝑑𝑥
6. ∫ 2
2𝑥 + 6𝑥 + 9
(5𝑥 + 2)𝑑𝑥
7. ∫
√𝑥 2 + 2𝑥 + 10
Part II: Rational Functions of sinx and cosx
8. ∫ 𝑐𝑠𝑐𝑑𝑥
𝑑𝑥
9. ∫
4 + 5𝑐𝑜𝑠𝑥
𝑑𝑥
10. ∫
4𝑠𝑖𝑛𝑥 − 3𝑐𝑜𝑠𝑥
 UNIVERSITY OF SAINT LOUIS
Tuguegarao City

SCHOOL OF ENGINEERING, ARCHITECTURE & IT EDUCATION

Engineering Sciences Department
S.Y. 2020 – 2021

CORRESPONDENCE LEARNING MODALITY

CALC 1024: Calculus 2 (Integral Calculus)

Prepared by:

EROVITA TERESITA BACUD AGUSTIN, DME

ENGR. NORMAN BOB GOMEZ
ENGR. JENIROSE B. MATIAS
ENGR. PEEJAY PAGUIRIGAN
Instructors

Reviewed by:

ENGR. VICTOR VILLALUZ, MEE

EE Program Chair / ES Department Head

Recommended by:

ENGR. RONALD M. VILLAVERDE, ME – ECE, PECE

Academic Dean

Approved by:

EMMANUEL JAMES P. PATTAGUAN, Ph.D.

VP for Academics
For this week, March 29-31, 2021 of this semester, the following shall be your guide for the
different lessons and tasks that you need to accomplish. Be patient read it carefully before
proceeding to the tasks expected of you.

Date Topic Activities/Tasks

Read Lessons
March 29-31, 2021 Integration by Parts

GOD BLESS!
Integration by Parts

Learning Objectives

Use the formula for integration by parts to solve the integral of a

given problem
WEEK 9: Integration by Parts

One of the most widely used techniques of integration is integration by parts, obtained from the
formula for the derivative of the product of two functions. If 𝑓 and 𝑔 are differentiable functions,
then

𝐷𝑥 [𝑓 (𝑥 )𝑔(𝑥 )] = 𝑓 (𝑥 )𝑔′(𝑥 ) + 𝑔(𝑥 )𝑓′(𝑥)

𝑓 (𝑥 )𝑔′(𝑥 ) = 𝐷𝑥 [𝑓 (𝑥 )𝑔(𝑥 ) − 𝑔(𝑥 )𝑓′(𝑥)

Integrating on each side of this equation, we obtain

∫ 𝑓(𝑥 )𝑔′(𝑥 )𝑑𝑥 = ∫ 𝐷𝑥 [𝑓(𝑥 )𝑔(𝑥 )]𝑑𝑥 − ∫ 𝑔(𝑥 )𝑓 ′(𝑥 )𝑑𝑥

∫ 𝑓 (𝑥 )𝑔′(𝑥 )𝑑𝑥 = 𝑓 (𝑥 )𝑔(𝑥 ) − ∫ 𝑔(𝑥 )𝑓 ′(𝑥 )𝑑𝑥 (1)

We call (1) the formula for integration by parts. For computational purposes, a more convenient
way of writing this formula is obtained by letting

𝑢 = 𝑓 (𝑥 ) 𝑎𝑛𝑑 𝑣 = 𝑔(𝑥)
Then
𝑑𝑢 = 𝑓 ′(𝑥 )𝑑𝑥 𝑎𝑛𝑑 𝑑𝑣 = 𝑔′(𝑥 )𝑑𝑥
so that (1) becomes
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢

This formula expresses the integral ∫ 𝑢 𝑑𝑣 in terms of another integral ∫ 𝑣 𝑑𝑢. By a suitable choice
of 𝑢 and 𝑑𝑣, it may be easier to evaluate the second integral than the first. When choosing the
substitutions for 𝑢 and 𝑑𝑣, we usually want 𝑑𝑣 to be the most complicated factor of the integrand
that can be integrated directly and 𝑢 to be a function whose derivative is a simpler function.
Examples

Identify 𝑢 and 𝑑𝑣
𝟏. ∫ 𝒙𝒄𝒐𝒔𝒙𝒅𝒙
𝑢=𝑥 𝑣 = 𝑠𝑖𝑛𝑥 Use the formula for integration by parts
𝑑𝑢 = 𝑑𝑥 𝑑𝑣 = 𝑐𝑜𝑠𝑥𝑑𝑥
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢
Substitute 𝑢, 𝑣, 𝑑𝑢, 𝑎𝑛𝑑 𝑑𝑣
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢
Integrate
∫ 𝑥𝑐𝑜𝑠𝑥𝑑𝑥 = 𝑥𝑠𝑖𝑛𝑥 − ∫ 𝑠𝑖𝑛𝑥𝑑𝑥
= 𝑥𝑠𝑖𝑛𝑥 − (−𝑐𝑜𝑠𝑥 ) + 𝐶 Simplify
= 𝒙𝒔𝒊𝒏𝒙 + 𝒄𝒐𝒔𝒙 + 𝑪 Final answer

Identify 𝑢 and 𝑑𝑣
𝟐. ∫ 𝒍𝒏𝒙𝒅𝒙
𝑢 = 𝑙𝑛𝑥 𝑣=𝑥 Use the formula for integration by parts
1
𝑑𝑢 = 𝑑𝑥 𝑑𝑣 = 𝑑𝑥 ∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢
𝑥
Substitute 𝑢, 𝑣, 𝑑𝑢, 𝑎𝑛𝑑 𝑑𝑣
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢
1 Simplify and cancel out common terms before
∫ 𝑙𝑛𝑥𝑑𝑥 = (𝑙𝑛𝑥 ) ∙ (𝑥 ) − ∫(𝑥) ∙ ( ) 𝑑𝑥
𝑥 integrating
Cancel out 𝑥 on the second term
Integrate
= 𝑥𝑙𝑛𝑥 − ∫ 𝑑𝑥
= 𝒙𝒍𝒏𝒙 − 𝒙 + 𝑪 Final answer

𝒙𝑨𝒓𝒄𝒔𝒊𝒏𝒙 Identify 𝑢 and 𝑑𝑣

𝟑. ∫ 𝒅𝒙
√𝟏 − 𝒙𝟐
𝑢 = 𝐴𝑟𝑐𝑠𝑖𝑛𝑥 𝑣 = −√ 1 − 𝑥 2 𝑥 𝑑𝑢
∫ 𝑑𝑥 − 2 1 1
1 𝑥 √1 − 𝑥 2 = ∫ 1 = − ∫ 𝑢−2 𝑑𝑢
𝑑𝑢 = 𝑑𝑥 𝑑𝑣 = 𝑑𝑥 2
√1 − 𝑥 2 √1 − 𝑥 2 𝑢2
𝑢 = 1 − 𝑥2 1
1 𝑢2
𝑑𝑢 = −2𝑥𝑑𝑥 =− ∙ = −√ 1 − 𝑥 2
𝑑𝑢 2 1
− = 𝑥𝑑𝑥 2
2
Substitute 𝑢, 𝑣, 𝑑𝑢, 𝑎𝑛𝑑 𝑑𝑣
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢
𝑥𝐴𝑟𝑐𝑠𝑖𝑛𝑥 1
∫ 𝑑𝑥 = (𝐴𝑟𝑐𝑠𝑖𝑛𝑥 ) ∙ (−√1 − 𝑥 2 ) − ∫ (−√1 − 𝑥 2 ) ∙ 𝑑𝑥
√1 − 𝑥 2 √1 − 𝑥 2
√1 − 𝑥 2 Cancel out common terms
= −√1 − 𝑥 2 𝐴𝑟𝑐𝑠𝑖𝑛𝑥 + ∫ 𝑑𝑥
√1 − 𝑥 2
 Integrate
= −√1 − 𝑥 2 𝐴𝑟𝑐𝑠𝑖𝑛𝑥 + ∫ 𝑑𝑥

= −√𝟏 − 𝒙𝟐 𝑨𝒓𝒄𝒔𝒊𝒏𝒙 + 𝒙 + 𝑪 Final answer

Rewrite the integrand

𝟒. ∫ 𝒔𝒆𝒄𝟑 𝒙𝒅𝒙
Identify 𝑢 and 𝑑𝑣
= ∫ 𝑠𝑒𝑐𝑥𝑠𝑒𝑐 2 𝑥𝑑𝑥
𝑢 = 𝑠𝑒𝑐𝑥 𝑣 = 𝑡𝑎𝑛𝑥
𝑑𝑢 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥𝑑𝑥 𝑑𝑣 = 𝑠𝑒𝑐 2 𝑥 𝑑𝑥
Substitute 𝑢, 𝑣, 𝑑𝑢, 𝑎𝑛𝑑 𝑑𝑣
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢

∫ 𝑠𝑒𝑐 3 𝑥𝑑𝑥 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 − ∫(𝑡𝑎𝑛𝑥)(𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥)𝑑𝑥

Use the identity
= 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 − ∫ 𝑡𝑎𝑛2 𝑥𝑠𝑒𝑐𝑥𝑑𝑥
𝑡𝑎𝑛2 𝑥 = 𝑠𝑒𝑐 2𝑥 −1
= 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 − ∫(𝑠𝑒𝑐 2𝑥 −1)𝑠𝑒𝑐𝑥𝑑𝑥
Integrate 𝑠𝑒𝑐𝑥 (𝑡ℎ𝑖𝑟𝑑 𝑡𝑒𝑟𝑚) with respect to 𝑥
= 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 − ∫ 𝑠𝑒𝑐 3 𝑥𝑑𝑥 + ∫ 𝑠𝑒𝑐𝑥𝑑𝑥
Transpose − ∫ 𝑠𝑒𝑐 3 𝑥𝑑𝑥
∫ 𝑠𝑒𝑐 3 𝑥𝑑𝑥 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 + 𝑙𝑛|𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 | − ∫ 𝑠𝑒𝑐 3 𝑥𝑑𝑥

∫ 𝑠𝑒𝑐 3 𝑥𝑑𝑥 + ∫ 𝑠𝑒𝑐 3 𝑥𝑑𝑥 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 + 𝑙𝑛|𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 | + 𝐶

Divide both sides by 2
2 ∫ 𝑠𝑒𝑐 3 𝑥𝑑𝑥 = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 + 𝑙𝑛|𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 | + 𝐶
1
∫ 𝑠𝑒𝑐 3 𝑥𝑑𝑥 = (𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 + 𝑙𝑛|𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 |) + 𝐶
2
𝟏 Final answer
= (𝒔𝒆𝒄𝒙𝒕𝒂𝒏𝒙 + 𝒍𝒏|𝒔𝒆𝒄𝒙 + 𝒕𝒂𝒏𝒙|) + 𝑪
𝟐

Identify 𝑢 and 𝑑𝑣
𝟓. ∫ 𝒆−𝒙 𝒄𝒐𝒔𝟐𝒙𝒅𝒙
𝑢 = 𝑐𝑜𝑠2𝑥 𝑣 = −𝑒 −𝑥
𝑑𝑢 = −2𝑠𝑖𝑛2𝑥𝑑𝑥 𝑑𝑣 = 𝑒 −𝑥 𝑑𝑥
Substitute 𝑢, 𝑣, 𝑑𝑢, 𝑎𝑛𝑑 𝑑𝑣
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢
Simplify
∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥 = (𝑐𝑜𝑠2𝑥 ) ∙ (−𝑒 −𝑥 ) − ∫(−𝑒 −𝑥 ) ∙ (−2𝑠𝑖𝑛2𝑥)𝑑𝑥
For the term 2 ∫ 𝑒 −𝑥 𝑠𝑖𝑛2𝑥𝑑𝑥,
= −𝑒 −𝑥 𝑐𝑜𝑠2𝑥 − 2 ∫ 𝑒 −𝑥 𝑠𝑖𝑛2𝑥𝑑𝑥
apply IBP (Integration by Parts)
𝑢 = 𝑠𝑖𝑛2𝑥 𝑣 = −𝑒 −𝑥
∗∗ ∫ 𝑒 −𝑥 𝑠𝑖𝑛2𝑥𝑑𝑥
𝑑𝑢 = 2𝑐𝑜𝑠2𝑥𝑑𝑥 𝑑𝑣 = 𝑒 −𝑥 𝑑𝑥
∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥 = −𝑒 −𝑥 𝑐𝑜𝑠2𝑥 − 2 ∫ 𝑒 −𝑥 𝑠𝑖𝑛2𝑥𝑑𝑥
Substitute 𝑢, 𝑣, 𝑎𝑛𝑑 𝑑𝑢
∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥 = −𝑒 −𝑥 𝑐𝑜𝑠2𝑥 − 2 [𝑢𝑣 − ∫ 𝑣𝑑𝑢]

∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥 = −𝑒 −𝑥 𝑐𝑜𝑠2𝑥 − 2 [(𝑠𝑖𝑛2𝑥 ) ∙ (−𝑒 −𝑥 ) − ∫(−𝑒 −𝑥 )2𝑐𝑜𝑠2𝑥𝑑𝑥]

Transpose
∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥 = −𝑒 −𝑥 𝑐𝑜𝑠2𝑥 + 2𝑒 −𝑥 𝑠𝑖𝑛2𝑥 − 4 ∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥
−4 ∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥
4 ∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥 + ∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥 = −𝑒 −𝑥 𝑐𝑜𝑠2𝑥 + 2𝑒 −𝑥 𝑠𝑖𝑛2𝑥 + 𝐶
Divide both sides by 5
5 ∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥 = −𝑒 −𝑥 𝑐𝑜𝑠2𝑥 + 2𝑒 −𝑥 𝑠𝑖𝑛2𝑥 + 𝐶
1
∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥𝑑𝑥 = (−𝑒 −𝑥 𝑐𝑜𝑠2𝑥 + 2𝑒 −𝑥 𝑠𝑖𝑛2𝑥 ) + 𝐶
5
𝟏 Final answer
= (−𝒆−𝒙 𝒄𝒐𝒔𝟐𝒙 + 𝟐𝒆−𝒙 𝒔𝒊𝒏𝟐𝒙) + 𝑪
𝟓

Expand the integrand before integrating

𝟔. ∫(𝒆𝒙 + 𝟐𝒙)𝟐 𝒅𝒙
Integrate separately
∫(𝑒 2𝑥 + 4𝑥𝑒 𝑥 + 4𝑥 2 )𝑑𝑥
Apply IBP on the third term
= ∫ 𝑒 2𝑥 𝑑𝑥 + ∫ 4𝑥 2 𝑑𝑥 + ∫ 4𝑥𝑒 𝑥 𝑑𝑥
𝑢=𝑥 𝑣 = 𝑒𝑥
∗∗ ∫ 4𝑥𝑒 𝑥 𝑑𝑥 = 4 ∫ 𝑥𝑒 𝑥 𝑑𝑥
𝑑𝑢 = 𝑑𝑥 𝑑𝑣 = 𝑒 𝑥 𝑑𝑥
= ∫ 𝑒 2𝑥 𝑑𝑥 + ∫ 4𝑥 2 𝑑𝑥 + 4 ∫ 𝑥𝑒 𝑥 𝑑𝑥
Substitute 𝑢, 𝑣, 𝑎𝑛𝑑 𝑑𝑢
= ∫ 𝑒 2𝑥 𝑑𝑥 + ∫ 4𝑥 2 𝑑𝑥 + 4 [𝑢𝑣 − ∫ 𝑣𝑑𝑢]
Simplify and integrate
= ∫ 𝑒 2𝑥 𝑑𝑥 + ∫ 4𝑥 2 𝑑𝑥 + 4 [𝑥𝑒 𝑥 − ∫ 𝑒 𝑥 𝑑𝑥]

= ∫ 𝑒 2𝑥 𝑑𝑥 + ∫ 4𝑥 2 𝑑𝑥 + 4 𝑥𝑒 𝑥 − 4 ∫ 𝑒 𝑥 𝑑𝑥
1 2𝑥 𝑥3
= 𝑒 + 4 + 4𝑥𝑒 𝑥 − 4𝑒 𝑥 + 𝐶
2 3
𝟏 𝟐𝒙 𝟒 𝟑 Final answer
= 𝒆 + 𝒙 + 𝟒𝒙𝒆𝒙 − 𝟒𝒆𝒙 + 𝑪
𝟐 𝟑

Identify 𝑢 and 𝑑𝑣
𝟕. ∫ 𝒔𝒊𝒏𝒙 𝒍𝒏(𝟏 + 𝒔𝒊𝒏𝒙)𝒅𝒙
𝑢 = 𝑙𝑛|1 + 𝑠𝑖𝑛𝑥 | 𝑣 = −𝑐𝑜𝑠𝑥
1
𝑑𝑢 = ∙ 𝑐𝑜𝑠𝑥𝑑𝑥 𝑑𝑣 = 𝑠𝑖𝑛𝑥𝑑𝑥
1 + 𝑠𝑖𝑛𝑥
 Substitute 𝑢, 𝑣, 𝑑𝑢, 𝑎𝑛𝑑 𝑑𝑣
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢
1
∫ 𝑠𝑖𝑛𝑥 𝑙𝑛(1 + 𝑠𝑖𝑛𝑥 )𝑑𝑥 = (𝑙𝑛|1 + 𝑠𝑖𝑛𝑥 |) ∙ (−𝑐𝑜𝑠𝑥 ) − ∫(−𝑐𝑜𝑠𝑥) ∙ ( ∙ 𝑐𝑜𝑠𝑥) 𝑑𝑥
1 + 𝑠𝑖𝑛𝑥
𝑐𝑜𝑠 2 𝑥 Use the identity
= −𝑐𝑜𝑠𝑥 𝑙𝑛|1 + 𝑠𝑖𝑛𝑥 | + ∫ 𝑑𝑥 𝑐𝑜𝑠 2 𝑥 = 1 − 𝑠𝑖𝑛2 𝑥
1 + 𝑠𝑖𝑛𝑥
1 − 𝑠𝑖𝑛2 𝑥 1 − 𝑠𝑖𝑛2 𝑥 = (1 + 𝑠𝑖𝑛𝑥)(1 − 𝑠𝑖𝑛𝑥)
= −𝑐𝑜𝑠𝑥 𝑙𝑛|1 + 𝑠𝑖𝑛𝑥 | + ∫ 𝑑𝑥
1 + 𝑠𝑖𝑛𝑥
(1 + 𝑠𝑖𝑛𝑥)(1 − 𝑠𝑖𝑛𝑥) Cancel out (1 + 𝑠𝑖𝑛𝑥)
= −𝑐𝑜𝑠𝑥 𝑙𝑛|1 + 𝑠𝑖𝑛𝑥 | + ∫ 𝑑𝑥
1 + 𝑠𝑖𝑛𝑥
Integrate
= −𝑐𝑜𝑠𝑥 𝑙𝑛|1 + 𝑠𝑖𝑛𝑥 | + ∫(1 − 𝑠𝑖𝑛𝑥 )𝑑𝑥
= −𝒄𝒐𝒔𝒙 𝒍𝒏|𝟏 + 𝒔𝒊𝒏𝒙| + 𝒙 + 𝒄𝒐𝒔𝒙 + 𝑪 Final answer

Identify 𝑢 and 𝑑𝑣
𝟖. ∫ 𝒙𝟑 𝒄𝒐𝒔𝒙𝒅𝒙
𝑢 = 𝑥3 𝑣 = 𝑠𝑖𝑛𝑥
𝑑𝑢 = 3𝑥 2 𝑑𝑥 𝑑𝑣 = 𝑐𝑜𝑠𝑥𝑑𝑥
Substitute 𝑢, 𝑣, 𝑑𝑢, 𝑎𝑛𝑑 𝑑𝑣
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢

∫ 𝑥 3 𝑐𝑜𝑠𝑥𝑑𝑥 = 𝑥 3 𝑠𝑖𝑛𝑥 − ∫(𝑠𝑖𝑛𝑥) ∙ (3𝑥 2 )𝑑𝑥

Apply IBP on the second term
= 𝑥 3 𝑠𝑖𝑛𝑥 − 3 ∫ 𝑥 2 𝑠𝑖𝑛𝑥𝑑𝑥
𝑢 = 𝑥2 𝑣 = −𝑐𝑜𝑠𝑥
∗∗ ∫ 𝑥 2 𝑠𝑖𝑛𝑥𝑑𝑥
𝑑𝑢 = 2𝑥𝑑𝑥 𝑑𝑣 = 𝑠𝑖𝑛𝑥𝑑𝑥
= 𝑥 3 𝑠𝑖𝑛𝑥 − 3 ∫ 𝑥 2 𝑠𝑖𝑛𝑥𝑑𝑥
Substitute 𝑢, 𝑣, 𝑎𝑛𝑑 𝑑𝑢
= 𝑥 3 𝑠𝑖𝑛𝑥 − 3 [𝑢𝑣 − ∫ 𝑣𝑑𝑢]
Simplify
= 𝑥 3 𝑠𝑖𝑛𝑥 − 3 [(𝑥 2 ) ∙ (−𝑐𝑜𝑠𝑥) − ∫(−𝑐𝑜𝑠𝑥) ∙ (2𝑥 )𝑑𝑥]

= 𝑥 3 𝑠𝑖𝑛𝑥 + 3𝑥 2 𝑐𝑜𝑠𝑥 − 3 ∫ 2𝑥𝑐𝑜𝑠𝑥𝑑𝑥

Apply IBP on the third term
= 𝑥 3 𝑠𝑖𝑛𝑥 + 3𝑥 2 𝑐𝑜𝑠𝑥 − 6 ∫ 𝑥𝑐𝑜𝑠𝑥𝑑𝑥
𝑢=𝑥 𝑣 = 𝑠𝑖𝑛𝑥
∗∗ ∫ 𝑥𝑐𝑜𝑠𝑥𝑑𝑥
𝑑𝑢 = 𝑑𝑥 𝑑𝑣 = 𝑐𝑜𝑠𝑥𝑑𝑥
= 𝑥 3 𝑠𝑖𝑛𝑥 + 3𝑥 2 𝑐𝑜𝑠𝑥 − 6 ∫ 𝑥𝑐𝑜𝑠𝑥𝑑𝑥

= 𝑥 3 𝑠𝑖𝑛𝑥 + 3𝑥 2 𝑐𝑜𝑠𝑥 − 6 [𝑢𝑣 − ∫ 𝑣𝑑𝑢]

= 𝑥 3 𝑠𝑖𝑛𝑥 + 3𝑥 2 𝑐𝑜𝑠𝑥 − 6 [𝑥𝑠𝑖𝑛𝑥 − ∫ 𝑠𝑖𝑛𝑥𝑑𝑥]

 Integrate the last term
= 𝑥 3 𝑠𝑖𝑛𝑥 + 3𝑥 2 𝑐𝑜𝑠𝑥 − 6𝑥𝑠𝑖𝑛𝑥 + 6 ∫ 𝑠𝑖𝑛𝑥𝑑𝑥
= 𝒙𝟑 𝒔𝒊𝒏𝒙 + 𝟑𝒙𝟐 𝒄𝒐𝒔𝒙 − 𝟔𝒙𝒔𝒊𝒏𝒙 − 𝟔𝒄𝒐𝒔𝒙 + 𝑪 Final answer

Identify 𝑢 and 𝑑𝑣
𝟗. ∫ 𝒙(𝟐𝒙 + 𝟏)𝟏𝟎 𝒅𝒙
1 𝑑𝑢 1
𝑢=𝑥 𝑣= (2𝑥 + 1)11 ∫(2𝑥 + 1)10 𝑑𝑥 = ∫ 𝑢10 = ∫ 𝑢10 𝑑𝑢
22 2 2
𝑑𝑢 = 𝑑𝑥 𝑑𝑣 = (2𝑥 + 1)10 𝑑𝑥 𝑢 = 2𝑥 + 1 1 𝑢11
𝑑𝑢 = 2𝑑𝑥 = ∙
2 11
𝑑𝑢 1
= 𝑑𝑥 = (2𝑥 + 1)11
2 2
Substitute 𝑢, 𝑣, 𝑑𝑢, 𝑎𝑛𝑑 𝑑𝑣
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢
1 1
∫ 𝑥 (2𝑥 + 1)10 𝑑𝑥 = (𝑥 ) ∙ (2𝑥 + 1)11 − ∫ [ (2𝑥 + 1)11 𝑑𝑥]
22 22
11
𝑥 (2𝑥 + 1) 1 Integrate the second term by using substitution
= − ∫(2𝑥 + 1)11 𝑑𝑥 method
22 22
𝑢 = 2𝑥 + 1
∗∗ ∫(2𝑥 + 1)11 𝑑𝑥
𝑑𝑢 = 2𝑑𝑥
𝑑𝑢
= 𝑑𝑥
2
𝑥 (2𝑥 + 1)11 1
= − ∫(2𝑥 + 1)11 𝑑𝑥
22 22
𝑥 (2𝑥 + 1)11 1 𝑑𝑢 Bring out the constant
= − ∫ 𝑢11
22 22 2
( )11
𝑥 2𝑥 + 1 1 Integrate
= − ∫ 𝑢11 𝑑𝑢
22 44
𝑥 (2𝑥 + 1)11 1 𝑢12 Simplify and substitute back 𝑢 = 2𝑥 + 1
= − ∙ +𝐶
22 44 12
𝒙(𝟐𝒙 + 𝟏)𝟏𝟏 (𝟐𝒙 + 𝟏)𝟏𝟐 Final answer
= − +𝑪
𝟐𝟐 𝟓𝟐𝟖

𝒆𝒙 𝒙𝒅𝒙 Identify 𝑢 and 𝑑𝑣

𝟏𝟎. ∫
(𝟏 + 𝒙 )𝟐
1 1 𝑑𝑢
𝑢 = 𝑒𝑥𝑥 𝑣=− 2
𝑑𝑥 = ∫ 2 = ∫ 𝑢−2 𝑑𝑢
1+𝑥 (1 + 𝑥 ) 𝑢
1 𝑢 =1+𝑥 𝑢 −1
1
𝑑𝑢 = (𝑒 𝑥 + 𝑥𝑒 𝑥 )𝑑𝑥 𝑑𝑣 = 𝑑𝑥 = =−
(1 + 𝑥 ) 2 𝑑𝑢 = 𝑑𝑥 −1 1+𝑥
Substitute 𝑢, 𝑣, 𝑑𝑢, 𝑎𝑛𝑑 𝑑𝑣
∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢
 𝑒 𝑥 𝑥𝑑𝑥 𝑥 ) (−
1 1
∫ = ( 𝑒 𝑥 ∙ ) − ∫ (− ) ∙ (𝑒 𝑥 + 𝑥𝑒 𝑥 )𝑑𝑥
(1 + 𝑥 )2 1+𝑥 1+𝑥
1 1 Simplify and cancel out common terms
= (𝑒 𝑥 𝑥 ) ∙ (− ) + ∫( ) ∙ (𝑒 𝑥 )(1 + 𝑥)𝑑𝑥
1+𝑥 1+𝑥
𝑥𝑒 𝑥 Integrate the second term
=− + ∫ 𝑒 𝑥 𝑑𝑥
1+𝑥
𝑥𝑒 𝑥 Factor out 𝑒 𝑥
=− + 𝑒𝑥 + 𝐶
1+𝑥
𝑥
−𝑥
=𝑒 ( + 1) + 𝐶
1+𝑥
−𝑥 + (1 + 𝑥) Simplify further
= 𝑒𝑥 ( )+𝐶
1+𝑥
1
= 𝑒𝑥 ( )+𝐶
1+𝑥
𝒆𝒙 Final answer
= +𝑪
𝟏+𝒙
For this week, April 5-9, 2021 of this semester, the following shall be your guide for the different
lessons and tasks that you need to accomplish. Be patient read it carefully before proceeding to the
tasks expected of you.

Date Topic Activities/Tasks

Read Lessons
April 5-8, 2021 Integration of Rational Fractions

Accomplish the worksheet in the

April 9, 2021 Submission of learning tasks Assessment Portion of this
module

GOD BLESS!
Integration of
Rational Fractions

Learning Objectives

▪ Recognize simple and repeated linear factors in a rational function

▪ Integrate a rational function using the method of partial fractions

Integration of Rational Fractions

By definition, a rational fraction is a quotient of two polynomials. Let N(x) and D(x) be two
polynomials having real coefficient and no common factor other than one. Then a rational fraction
may be generally represented by

𝑁(𝑥)
𝐷(𝑥)

the degree of the numerator N(x) is less than the degree of the denominator D(x), then N(x)/D(x)
is called a proper rational fraction. Otherwise, it is an improper rational fraction.

• Proper rational fraction Examples

𝑥+2
1. 𝑥 2 −𝑥+1
𝑥 2 −3𝑥+1
2. 𝑥 5 −3𝑥
𝑥 7 +𝑥 3 −3𝑥+5𝑥+1
3. 𝑥 9 +𝑥 8 +𝑥 7 −4𝑥 2 +3𝑥+1
• Improper rational fraction Examples
𝑥 3 +5𝑥
1.
𝑥+1
𝑥 8 +8
2. 𝑥 7 +3𝑥
3𝑥 8 +1
3. 2𝑥 3 +𝑥 2 −3𝑥+1

In evaluating the integral of a proper rational fraction, it is important that we know how to represent
the given fraction as the sum of simpler fractions. These simpler fractions are called partial
fractions. For example,

𝑥 2 + 9𝑥 + 2 2 3 4
= + −
(𝑥 − 1)(𝑥 + 1)(𝑥 + 2) 𝑥 − 1 𝑥 + 1 𝑥 + 2

The three simple fractions at the right side of the equation are the partial fractions. Thus, to
integrate a proper rational fraction, we simply integrate its equivalent partial fractions.

If N(x)/D(x) is an improper rational fraction, it should first reduce by division into a polynomial
and a proper rational fraction, that is,

𝑁(𝑥) 𝑅(𝑥)
= 𝑄 (𝑥 ) +
𝐷(𝑥) 𝐷(𝑥)

where:

Q(x) is a polynomial
 𝑅(𝑥)
is a proper rational fraction, and it should be resolved or broken up into partial fraction.
𝐷(𝑥)
For instance,
𝑥3 4𝑥
=𝑥+
𝑥 2 −4 𝑥 2 −4

4𝑥
=𝑥+ (𝑥+2)(𝑥−2)

2 2
=𝑥+ +
𝑥+2 𝑥−2

A factor of the type 𝑎𝑥 + 𝑏 is called a linear factor and if (ax + b) n, where n ≥ 2, occurs in D(x),
we say that ax + b is a repeated linear factor of the denominator. A factor of the type ax 2 + bx + c
is called a quadratic factor and it is said to be irreducible if it can not be factored further into real
linear factors. If 𝑎𝑥2 + 𝑏𝑥 + 𝑐 is irreducible and if (ax2 + bx + c) n occurs in D(x), then we say
that 𝑎𝑥2 + 𝑏𝑥 + 𝑐 is a repeated and irreducible quadratic factor.

We shall now study how to break up (or resolve) a proper rational fraction into partial
fractions. A proper rational fraction N(x)/D(x) can be broken up (or resolved) into a sum of partial
fractions subject to the following cases:

Case 1:

When the factors of D(x) are all linear and none is repeated.

To each non repeated linear factor ax + b, there corresponds a partial fraction of the form

𝐴
𝑎𝑥 + 𝑏
4𝑥 𝐴 𝐵
Example: (𝑥+2)(𝑥−2) = + 𝑥−2
𝑥+2

where A and B are the constant to be determined

Case 2:

When the factors of D(x) are all linear and some are repeated.

To each repeated linear factor (ax + b) n there corresponds the sum of n partial fractions
of the form

𝐴 𝐵 𝐿
+ 2
+⋯+
𝑎𝑥 + 𝑏 (𝑎𝑥 + 𝑏) (𝑎𝑥 + 𝑏)𝑛

where A, B, ..., L are the constants to be determined

 2 𝐴 𝐵 𝐶
Example: (𝑥+1)(𝑥+2)2
= (𝑥+1) + (𝑥+2)2 + (𝑥+2)
where A, B, and C are the constant to be determined

Case 3:

When D(x) contains irreducible quadratic factors (or simply imaginary roots)

To each non repeated irreducible quadratic factor ax2+ bx + c, there corresponds the partial
fraction of the form

𝐴(2𝑎𝑥 + 𝑏) + 𝐵
𝑎𝑥 2 + 𝑏𝑥 + 𝑐
where A and B are constants to be determined and 2ax + b is the derivative of ax2+bx+c
2𝑥+1 𝐴 𝐵(2𝑥+2)+𝐶
Example: (3𝑥−1)(𝑥2 +2𝑥+2) = 3𝑥−1 + 𝑥 2 +2𝑥+2

where A, B, and C are the constant to be determined

Case 4:

When D(x) contains irreducible quadratic factors and some are repeated.

To each repeated quadratic factor (𝑎𝑥 2 + 𝑏𝑥 + 𝑐 )𝑛 there corresponds the sum of n partial
fractions of the form

𝐴(2𝑎𝑥 + 𝑏) + 𝐵 𝐶 (2𝑎𝑥 + 𝑏) + 𝐷 𝐾 (2𝑎𝑥 + 𝑏) + 𝐿

2
+ 2 2
+ ⋯+
𝑎𝑥 + 𝑏𝑥 + 𝑐 (𝑎𝑥 + 𝑏𝑥 + 𝑐 ) (𝑎𝑥 2 + 𝑏𝑥 + 𝑐 )𝑛

Where A, B, C, D, . . . , K, L are constants to be determined.

Our next problem is how to determine the constants. There are two general methods for finding
such constants. These are

Method 1:

By assigning particular values to x in order to obtain equations involving at least one of the
constants

Method 2:

By equating the coefficients of equal powers of x to obtain simultaneous equations involving

the constants
Examples

12𝑥 + 18 Case 1
1. ∫ 𝑑𝑥
(𝑥 + 2)(𝑥 + 4)(𝑥 − 1) roots of the denominator are all
linear and none is repeated
12𝑥 + 18 𝐴 𝐵 𝐶
= + +
(𝑥 + 2)(𝑥 + 4)(𝑥 − 1) 𝑥+2 𝑥+4 𝑥−1
12𝑥 + 18 = 𝐴(𝑥 + 4)(𝑥 − 1) + 𝐵(𝑥 + 2)(𝑥 − 1) + multiply both sides of the
𝐶 (𝑥 + 2)(𝑥 + 4) equation by the LCD which is
(𝑥 + 2)(𝑥 + 4)(𝑥 − 1)
Using Method 1
12𝑥 + 18 = 𝐴(𝑥 + 4)(𝑥 − 1) to determine the value of A we set
12(−2) + 18 = 𝐴(−2 + 4)(−2 − 1) 𝑥 = −2, terms with (𝑥 + 2)
𝐴 = 1 automatically becomes zero
12𝑥 + 18 = 𝐵(𝑥 + 2)(𝑥 − 1)
12(−4) + 18 = 𝐵(−4 + 2)(−4 − 1) to determine the value of B we set
𝐵 = −3 𝑥 = −4, terms with (𝑥 + 4)
12𝑥 + 18 = 𝐶 (𝑥 + 2)(𝑥 + 4) automatically becomes zero
12(1) + 18 = 𝐶 (1 + 2)(1 + 4)
𝐶 = 2 to determine the value of C we set
𝑥 = 1, terms with (𝑥 − 1)
automatically becomes zero
Using Method 2 we start with expanding the
12𝑥 + 18 = 𝐴(𝑥 + 4)(𝑥 − 1) + 𝐵(𝑥 + 2)(𝑥 − 1) + equation
𝐶 (𝑥 + 2)(𝑥 + 4)
12𝑥 + 18 = 𝐴(𝑥 2 + 3𝑥 − 4) + 𝐵(𝑥 2 + 𝑥 − 2) +
𝐶(𝑥 2 + 6𝑥 + 8)
12𝑥 + 18 = 𝑥 2 (𝐴 + 𝐵 + 𝐶) + 𝑥(3𝐴 + 𝐵 + 6𝐶) +
𝑥 0 (−4𝐴 − 2𝐵 + 8𝐶)
𝑥 2: 𝐴 + 𝐵 + 𝐶 = 0 we equate coefficients of equal
𝑥 1 : 3𝐴 + 𝐵 + 6𝐶 = 12 powers of x
𝑥 0 : − 4𝐴 − 2𝐵 + 8𝐶 = 18 Solve for the value of A, B, and C
𝐴= 1 either by elimination or using
𝐵 = −3 your calculator
𝐶= 2
12𝑥 + 18 𝐴 𝐵 𝐶
= + +
(𝑥 + 2)(𝑥 + 4)(𝑥 − 1) 𝑥+2 𝑥+4 𝑥−1
12𝑥 + 18 1 3 2 substitute the respective values of
= − +
(𝑥 + 2)(𝑥 + 4)(𝑥 − 1) 𝑥+2 𝑥+4 𝑥−1 A B and C
1 3 2 Integrate
∫ 𝑑𝑥 − ∫ 𝑑𝑥 + ∫ 𝑑𝑥
𝑥+2 𝑥+4 𝑥−1
𝒍𝒏|𝒙 + 𝟐| − 𝟑𝒍𝒏|𝒙 + 𝟒| + 𝟐𝒍𝒏|𝒙 − 𝟏| + 𝑪 Final answer
 𝒙−𝟓 Case 1
𝟐. ∫ 𝒅𝒙 roots of the denominator are all linear and
𝒙𝟐 − 𝒙 − 𝟐
none is repeated
𝑥−5 denominator can be written as shown w/c falls
∫ 𝑑𝑥 under case 1
(𝑥 − 2)(𝑥 + 1)
or we can simply identify the nature of roots
using table shown below to know which case
the quadratic equation falls

𝑥−5 𝐴 𝐵 multiply both sides of the equation by the

= + LCD which is (𝑥 − 2)(𝑥 + 1)
(𝑥 − 2)(𝑥 + 1) 𝑥 − 2 𝑥 + 1
𝑥 − 5 = 𝐴(𝑥 + 1) + 𝐵(𝑥 − 2)
Using Method 1
𝑥 − 5 = 𝐴(𝑥 + 1) to determine the value of A we set 𝑥 = 2,
2 − 5 = 𝐴(2 + 1) terms with (𝑥 − 2) automatically becomes
𝐴 = −1 zero

𝑥 − 5 = 𝐵(𝑥 − 2) to determine the value of B we set 𝑥 = −1,

−1 − 5 = 𝐵(−1 − 2) terms with (𝑥 + 1) automatically becomes
𝐵 = 2 zero
Using Method 2
𝑥 − 5 = 𝐴(𝑥 + 1) + 𝐵(𝑥 − 2)
𝑥 − 5 = 𝑥 1 (𝐴 + 𝐵) + 𝑥 0 (𝐴 − 2𝐵)

𝑥1: 𝐴 + 𝐵 = 1 we equate coefficients of equal powers of x

𝑥 0 : 𝐴 − 2𝐵 = −5 Solve for the value of A, and B either by
𝐴 = −1 elimination or using your calculator
𝐵= 2
𝑥−5 𝐴 𝐵
= +
(𝑥 − 2)(𝑥 + 1) 𝑥 − 2 𝑥 + 1
𝑥−5 −1 2 substitute the respective values of A B and C
= +
(𝑥 − 2)(𝑥 + 1) 𝑥 − 2 𝑥 + 1
−1 2 Integrate
∫ 𝑑𝑥 + ∫ 𝑑𝑥
𝑥−2 𝑥+1
−𝒍𝒏|𝒙 − 𝟐| + 𝟐𝒍𝒏|𝒙 + 𝟏| + 𝑪 Final Answer
 𝟐𝒅𝒙 Case 2
𝟑. ∫ When the factors of D(x) are all linear
(𝒙 + 𝟏)(𝒙 + 𝟐)𝟐
and some are repeated.
2 𝐴 𝐵 𝐶
2
= + +
(𝑥 + 1)(𝑥 + 2) 𝑥 + 1 𝑥 + 2 (𝑥 + 2)2
2
2 = 𝐴(𝑥 + 2) + 𝐵(𝑥 + 1)(𝑥 + 2) + 𝐶(𝑥 + 1) multiply both sides of the equation by
the LCD which is (𝑥 + 1)(𝑥 + 2)2
Using Method 1 to determine the value of A we set 𝑥 =
2 = 𝐴(𝑥 + 2)2 −1, terms with (𝑥 + 1) automatically
2 = 𝐴(−1 + 2)2 becomes zero
𝐴= 2
2 = 𝐶 (𝑥 + 1) to determine the value of C we set 𝑥 =
2 = 𝐶 (−2 + 1) −2, terms with (𝑥 + 2) automatically
𝐶 = −2 becomes zero
Important note for case 2:
since in the equation, Constant 𝐵 also contains 𝑥 + 2, therefore if we set 𝑥 = −2, 𝐵 will also
cancel or becomes zero.
𝑑 𝑑 In order to find B, we first differentiate
(2) = [𝐴(𝑥 + 2)2 + 𝐵(𝑥 + 1)(𝑥 + 2) both sides of the equation then proceed
𝑑𝑥 𝑑𝑥
(
+𝐶 𝑥+1 ] ) with setting the value of 𝑥 = −2
0 = 2𝐴(𝑥 + 2) + 𝐵[(𝑥 + 1) + (𝑥 + 2)] + 𝐶 𝑑 𝑑𝑣 𝑑𝑢
recall:𝑑𝑥 (𝑢𝑣 ) = 𝑢 𝑑𝑥 + 𝑣 𝑑𝑥
0 = 2𝐴(𝑥 + 2) + 𝐵(𝑥 + 1) + 𝐵(𝑥 + 2) + 𝐶
we set 𝑥 = −2, terms with (𝑥 + 2)
0 = 𝐵(𝑥 + 1) + 𝐶 automatically becomes zero and
0 = 𝐵(−2 + 1) − 2 substitute the value of B which is −2
𝐵 = −2
Using Method 2
2 = 𝐴(𝑥 + 2)2 + B(𝑥 + 1)(𝑥 + 2) + 𝐵(𝑥 + 1) we start with expanding the equation
( 2 ) ( 2 ) ( )
2 = 𝐴 𝑥 + 4𝑥 + 4 + 𝐵 𝑥 + 3𝑥 + 2 + 𝐶 𝑥 + 1
2 = 𝑥 2 (𝐴 + B) + 𝑥 1 (4𝐴 + 3𝐵 + 𝐶 ) + 𝑥 0 (4𝐴 +
2𝐵 + 𝐶 )

𝑥 2: 𝐴 + 𝐵 = 0 we equate coefficients of equal powers

𝑥 1 : 4𝐴 + 3𝐵 + 𝐶 = 0 of x
𝑥 0 : 4𝐴 + 2𝐵 + 𝐶 = 2
𝐴= 2
𝐵 = −2
𝐶 = −2
2 2 2 2 substitute the respective values of A B
2
= − −
(𝑥 + 1)(𝑥 + 2) 𝑥 + 1 𝑥 + 2 (𝑥 + 2)2 and C
2 2 2 Integrate
∫ 𝑑𝑥 − ∫ 𝑑𝑥 − ∫ 𝑑𝑥
𝑥+1 (𝑥 + 2)2 𝑥+2
𝟐 Final Answer
𝟐𝒍𝒏|𝒙 + 𝟏| − 𝟐𝒍𝒏|𝒙 + 𝟐| + +𝑪
(𝒙 + 𝟐)
 −𝟐𝒙𝟐 + 𝟔𝒙 + 𝟖 Case 2
𝟒. ∫ 𝒅𝒙 When the factors of D(x) are all linear and
𝒙𝟐 (𝒙 + 𝟐)
some are repeated.
−2𝑥 2 + 6𝑥 + 8 𝐴 𝐵 𝐶
= + +
𝑥 2 (𝑥 + 2) 𝑥 + 2 𝑥 𝑥2
−2𝑥 2 + 6𝑥 + 8 = 𝐴(𝑥 2 ) + 𝐵(𝑥)(𝑥 + 2) + multiply both sides of the equation by the
𝐶(𝑥 + 2) LCD which is 𝑥 2 (𝑥 + 2)
Using Method 1
−2𝑥 2 + 6𝑥 + 8 = 𝐴(𝑥 2 ) to determine the value of A we set 𝑥 = −2,
−2(−2)2 + 6(−2) + 8 = 𝐴((−2)2 ) terms with (𝑥 + 2) automatically becomes
𝐴 = −3 zero
−2𝑥 2 + 6𝑥 + 8 = 𝐶 (𝑥 + 2)
−2(0)2 + 6(0) + 8 = 𝐶 (0 + 2) to determine the value of C we set 𝑥 = 0,
𝐶 = 4 terms multiplied by 𝑥 or 𝑥 2 automatically
becomes zero
𝑑 differentiate both sides of the equation then
(−2𝑥 2 + 6𝑥 + 8)
𝑑𝑥 proceed with setting the value of x= 0
𝑑 𝑑 𝑑𝑣 𝑑𝑢
recall:𝑑𝑥 (𝑢𝑣 ) = 𝑢 𝑑𝑥 + 𝑣 𝑑𝑥
= [𝐴(𝑥 2 ) + 𝐵(𝑥)(𝑥 + 2) + 𝐶 (𝑥 + 2)]
𝑑𝑥
−4𝑥 + 6 = 2𝐴(𝑥 ) + 𝐵[(𝑥 ) + (𝑥 + 2)] + 𝐶
we set x = 0, terms multiplied by (x or x2)
−4𝑥 + 6 = 2𝐴(𝑥 ) + 𝐵(𝑥 ) + 𝐵(𝑥 + 2) + 𝐶
automatically becomes zero and substitute the
value of B which is 4
6 = 𝐵(𝑥 + 2) + 𝐶
6 = 𝐵(0 + 2) + 4
𝐵 = 1
Using Method 2 we start with expanding the
equation
−2𝑥 2 + 6𝑥 + 8 = 𝐴(𝑥 2 ) + 𝐵(𝑥)(𝑥 + 2) + 𝐶 (𝑥 + 2)
−2𝑥 2 + 6𝑥 + 8 = 𝐴(𝑥 2 ) + 𝐵(𝑥 2 + 2𝑥) + 𝐶 (𝑥 + 2) we equate coefficients of equal
−2𝑥 2 + 6𝑥 + 8 = 𝑥 2 (𝐴 + 𝐵) + 𝑥 1 (2𝐵 + 𝐶 ) + 𝑥 0 (2𝐶) powers of x
2
𝑥 : 𝐴 + 𝐵 = −2 Solve for the value of A, B, and C using
1
𝑥 : 2𝐵 + 𝐶 = 6 either elimination and substitution method or
0 using your calculator
𝑥 : 2𝐶 = 8
𝐴 = −3
𝐵= 1
𝐶= 4
−2𝑥 2 + 6𝑥 + 8 −3 1 4 Substitute the respective values of A B and C
= + +
𝑥 2 (𝑥 + 2) 𝑥 + 2 𝑥 𝑥2
−3 1 4 Integrate
∫ 𝑑𝑥 + ∫ 𝑑𝑥 + ∫ 2 𝑑𝑥
𝑥+2 𝑥 𝑥
𝟒 Final answer
−𝟑𝒍𝒏|𝒙 + 𝟐| + 𝒍𝒏|𝒙| − + 𝑪
𝒙
 𝟐𝒙 + 𝟏
𝟓. ∫ 𝒅𝒙
(𝟑𝒙 − 𝟏)(𝒙𝟐 + 𝟐𝒙 + 𝟐)
2𝑥 + 1 𝐴 𝐵(2𝑥 + 2) + 𝐶
2
= + 2
(3𝑥 − 1)(𝑥 + 2𝑥 + 2) 3𝑥 − 1 𝑥 + 2𝑥 + 2
2
𝑥 + 2𝑥 + 2 is a non-repeated irreducible quadratic
factor or simply the nature of its roots are imaginary or
complex numbers
this problem is under case 3

When D(x) contains irreducible quadratic factors (or

simply imaginary roots)
you can check the nature of the roots using
2𝑥 + 1 = 𝐴 (𝑥 2 + 2𝑥 + 2) + 𝐵 (2𝑥 + 2 )(3𝑥 − 1) + 𝐶 (3𝑥 − 1)
multiply both sides of the equation by the LCD which is Note: For case 3, we can’t use
2
(3𝑥 − 1)(𝑥 + 2𝑥 + 2) method 1 to find the value of the
constants, because of the presence
of imaginary or complex roots
Using Method 2
2𝑥 + 1 = 𝐴 (𝑥 2 + 2𝑥 + 2) + 𝐵 (2𝑥 + 2 )(3𝑥 − 1) + 𝐶 (3𝑥 − 1)
2𝑥 + 1 = 𝐴 (𝑥 2 + 2𝑥 + 2) + 𝐵 (6𝑥 2 + 4𝑥 − 2 ) + 𝐶 (3𝑥 − 1)
2𝑥 + 1 = 𝑥 2 (𝐴 + 6𝐵) + 𝑥 1 (2𝐴 + 4𝐵 + 3𝐶 ) + 𝑥 0 (2𝐴 − 2𝐵 − 𝐶)
3
2 𝐴=
𝑥 : 𝐴 + 6𝐵 = 0 5
𝑥 1 : 2𝐴 + 4𝐵 + 3𝐶 = 2 −1
0 𝐵=
𝑥 : 2𝐴 − 2𝐵 − 𝐶 = 1 10
2
𝐶=
5
2𝑥 + 1 𝐴 𝐵(2𝑥 + 2) + 𝐶 Substitute A, B, and C
= + 2
(3𝑥 − 1)(𝑥 2 + 2𝑥 + 2) 3𝑥 − 1 𝑥 + 2𝑥 + 2
3 −1 2
2𝑥 + 1 (2𝑥 + 2) +
= 5 + 10 5
2
(3𝑥 − 1)(𝑥 + 2𝑥 + 2) 3𝑥 − 1 2
𝑥 + 2𝑥 + 2
3 −1 2 Integrate
(2𝑥 + 2)
∫ 5 𝑑𝑥 + ∫ 210 𝑑𝑥 + ∫ 2 5 𝑑𝑥 Note: third integral which is
3𝑥 − 1 𝑥 + 2𝑥 + 2 𝑥 + 2𝑥 + 2 2 1
∫ 𝑑𝑥
5 (𝑥 + 1)2 + 1
1
which is in the form of ∫ 𝑢2 +1 𝑑𝑢 ,
use ASF 1
𝟏 𝟏 𝟐 Final answer
𝒍𝒏|𝟑𝒙 − 𝟏| − 𝒍𝒏|𝒙𝟐 + 𝟐𝒙 + 𝟐| + 𝐚𝐫𝐜𝐭𝐚𝐧(𝒙 + 𝟏) + 𝑪
𝟓 𝟏𝟎 𝟓
 𝟓𝒙𝟐 − 𝒙 + 𝟏𝟕 Case 3
𝟔. ∫ 𝒅𝒙 When D(x) contains irreducible quadratic
(𝒙 + 𝟐)(𝒙𝟐 + 𝟗)
factors (or simply imaginary roots)
2
5𝑥 − 𝑥 + 17 𝐴 𝐵(2𝑥 ) + 𝐶 multiply both sides of the equation by the
2
= + 2 LCD which is (𝑥 + 2)(𝑥 2 + 9)
(𝑥 + 2)(𝑥 + 9) 𝑥+2 𝑥 +9
5𝑥 2 − 𝑥 + 17 = 𝐴(𝑥 2 + 9) + 𝐵(2𝑥 )(𝑥 + 2) + 𝐶 (𝑥 + 2)
Using Method 2
5𝑥 2 − 𝑥 + 17 = 𝐴(𝑥 2 + 9) + 𝐵(2𝑥 )(𝑥 + 2) + 𝐶 (𝑥 + 2)
5𝑥 2 − 𝑥 + 17 = 𝐴(𝑥 2 + 9) + 𝐵(2𝑥 2 + 4𝑥 ) + 𝐶 (𝑥 + 2)
5𝑥 2 − 𝑥 + 17 = 𝑥 2 (𝐴 + 2𝐵) + 𝑥 1 (4𝐵 + 𝐶 ) + 𝑥 0 (9𝐴 + 2𝐶 )
𝑥 2 : 𝐴 + 2𝐵 = 5 Solve for the value of A, B, and C using
𝑥 1 : 4𝐵 + 𝐶 = −1 either elimination and substitution method
0 or using your calculator
𝑥 : 9𝐴 + 2𝐶 = 17
𝐴= 3
𝐵= 1
𝐶 = −5
5𝑥 2 − 𝑥 + 17 𝐴 𝐵(2𝑥 ) + 𝐶 Substitute the respective values of A, B,
= + and C
(𝑥 + 2)(𝑥 2 + 9) 𝑥+2 𝑥2 + 9
2
5𝑥 − 𝑥 + 17 3 1(2𝑥 ) − 5
2
= +
(𝑥 + 2)(𝑥 + 9) 𝑥+2 𝑥2 + 9
3 1(2𝑥) 5 Integrate
∫ 𝑑𝑥 + ∫ 2 𝑑𝑥 − ∫ 2 𝑑𝑥 note: the integral of the third term which is
𝑥+2 𝑥 +9 𝑥 +9
1
5∫ 2 𝑑𝑥
𝑥 +9
1
is in the form of ∫ 𝑢2 +1 𝑑𝑢 ,use ASF 1
𝟓 𝒙 Final answer
𝟑𝒍𝒏|𝒙 + 𝟐| + 𝒍𝒏|𝒙𝟐 + 𝟗| − 𝒂𝒓𝒄𝒕𝒂𝒏 ( ) + 𝑪
𝟑 𝟑

𝒙𝟓 + 𝟐𝒙𝟑 − 𝟑𝒙 Case 4
𝟕. ∫ 𝒅𝒙 When 𝐷(𝑥) contains irreducible quadratic factors
( 𝒙 𝟐 + 𝟏)𝟑
and some are repeated
5 3
𝑥 + 2𝑥 − 3𝑥 𝐴(2𝑥) + 𝐵 𝐶 (2𝑥 ) + 𝐷 𝐸(2𝑥 ) + 𝐹
∫ 2 3
𝑑𝑥 = + +
(𝑥 + 1) 𝑥2 + 1 (𝑥 + 1)2 (𝑥 + 1)3
𝑥 5 + 2𝑥 3 − 3𝑥 = [𝐴(2𝑥 ) + 𝐵] (𝑥 2 + 1)2 + [𝐶 (2𝑥 ) + 𝐷 ](𝑥 2 + 1) + [𝐸 (2𝑥 ) + 𝐹 ]
𝑥 5 + 2𝑥 3 − 3 = 𝐴(2𝑥)(𝑥 4 + 2𝑥 2 + 1) + 𝐵(𝑥 4 + 2𝑥 2 + 1) + 𝐶(2𝑥)(𝑥 2 + 1) + 𝐷(𝑥 2 + 1) + 𝐸(2𝑥) + 𝐹
𝑥 5 + 2𝑥 3 − 3𝑥 = 𝐴(2𝑥 5 + 4𝑥 3 + 2𝑥) + 𝐵(𝑥 4 + 2𝑥 2 + 1) + 𝐶(2𝑥 3 + 2𝑥) + 𝐷(𝑥 2 + 1) + 𝐸(2𝑥) + 𝐹
𝑥 5 + 2𝑥 3 − 3𝑥 = 𝑥 5 (2𝐴) + 𝑥 4 (𝐵) + 𝑥 3 (4𝐴 + 2𝐶) + 𝑥 2 (2𝐵 + 𝐷) + 𝑥 1 (2𝐴 + 2𝐶 + 2𝐸) + 𝑥 0 (𝐵 + 𝐷 + 𝐹)
1
𝑥 5 : 1 = 2𝐴 𝐴=
2
𝑥 4: 0 = 𝐵 𝐵=0
𝑥 3 : 2 = 4𝐴 + 2𝐶 𝐶=0 Substitute 𝐴 = 1/2
𝑥 2 : 0 = 2𝐵 + 𝐷 𝐷=0 Substitute 𝐵 = 0
𝑥 1 : −3 = 2𝐴 + 2𝐶 + 2𝐸 𝐸 = 0 Substitute 𝐴 = 1/2 and 𝐶 = 0
𝑥 0: 0 = 𝐵 + 𝐷 + 𝐹 𝐹=0 Substitute 𝐵 = 0 and 𝐷 = 0
5 3
𝑥 + 2𝑥 − 3𝑥 𝐴(2𝑥) + 𝐵 𝐶 (2𝑥 ) + 𝐷 𝐸(2𝑥 ) + 𝐹
∫ 2 3
𝑑𝑥 = + +
(𝑥 + 1) 𝑥2 + 1 (𝑥 + 1)2 (𝑥 + 1)3
1
𝑥 5 + 2𝑥 3 − 3𝑥 ( ) (2𝑥) + 0 (0)(2𝑥 ) + 0 (0)(2𝑥 ) + (0)
∫ 2 3
𝑑𝑥 = 2 2 + +
(𝑥 + 1) 𝑥 +1 (𝑥 + 1)2 (𝑥 + 1)3
5 3
𝑥 + 2𝑥 − 3𝑥 𝑥
∫ 𝑑𝑥 =
(𝑥 2 + 1)3 𝑥2 + 1
5 3
𝑥 + 2𝑥 − 3𝑥 𝑥𝑑𝑥 Integrate using substitution method
∫ 𝑑𝑥 = ∫ 𝑢 = 𝑥2 + 1
(𝑥 2 + 1)3 𝑥2 + 1
𝑑𝑢 = 2𝑥𝑑𝑥
𝑑𝑢
= 𝑥𝑑𝑥
2
1 𝑑𝑢
=∫ ∙
𝑢 2
1 𝑑𝑢
= ∫
2 𝑢
1
= 𝑙𝑛𝑢 + 𝐶
2
𝟏 Final answer
𝒍𝒏|𝒙𝟐 + 𝟏| + 𝑪
𝟐

𝒙𝟒 + 𝟐𝒙𝟑 + 𝟏𝟏𝒙𝟐 + 𝟖𝒙 + 𝟏𝟔 Case 4

𝟖. ∫ 𝟐 𝟐
𝒅𝒙
𝒙 (𝒙 + 𝟒)
𝑥 + 2𝑥 + 11𝑥 2 + 8𝑥 + 16
4 3
𝐴 𝐵(2𝑥 ) + 𝐶 𝐷 (2𝑥 ) + 𝐸 Multiply both sides by
∫ 2 2
𝑑𝑥 = + + 2 (𝑥 )(𝑥 2 + 4)2
𝑥 (𝑥 + 4) 𝑥 (𝑥 2 + 4) (𝑥 + 4)2

𝑥 4 + 2𝑥 3 + 11𝑥 2 + 8𝑥 + 16 = 𝐴(𝑥 2 + 4)2 + 𝐵(2𝑥)(𝑥)(𝑥 2 + 4) + 𝐶(𝑥)(𝑥 2 + 4) + 𝐷(2𝑥)(𝑥) + 𝐸(𝑥)

𝑥 4 + 2𝑥 3 + 11𝑥 2 + 8𝑥 + 16 = 𝐴(𝑥 4 + 8𝑥 2 + 16) + 𝐵(2𝑥 4 + 8𝑥 2 ) + 𝐶(𝑥 3 + 4𝑥) + 𝐷(2𝑥 2 ) + 𝐸(𝑥)
𝑥 4 + 2𝑥 3 + 11𝑥 2 + 8𝑥 + 16 = 𝑥 4 (𝐴 + 2𝐵) + 𝑥 3 (𝐶) + 𝑥 2 (8𝐴 + 8𝐵 + 2𝐷) + 𝑥 1 (4𝐶 + 𝐸) + 𝑥 0 (16𝐴)
𝑥 4 : 1 = 𝐴 + 2𝐵 𝐴=1
𝑥 3: 2 = 𝐶 𝐵=0
2 𝐶=2
𝑥 : 11 = 8𝐴 + 8𝐵 + 2𝐷
𝑥 1 : 8 = 4𝐶 + 𝐸 𝐷 = 3/2
0 𝐸=0
𝑥 : 16 = 16𝐴
𝑥 4 + 2𝑥 3 + 11𝑥 2 + 8𝑥 + 16 𝐴 𝐵(2𝑥 ) + 𝐶 𝐷 (2𝑥 ) + 𝐸 Substitute A, B, C, D, E,
∫ 𝑑𝑥 = + + 2
𝑥 (𝑥 2 + 4)2 𝑥 (𝑥 2 + 4) (𝑥 + 4)2 and F
3 Simplify then integrate
1 (0)(2𝑥 ) + 2 (2) (2𝑥 ) + 0
= + +
𝑥 (𝑥 2 + 4) (𝑥 2 + 4)2
𝑑𝑥 2𝑑𝑥 3𝑥𝑑𝑥 For the second term, use ASF 1
=∫ +∫ 2 +∫ 2 2 For the third term, apply substitution method
𝑥 (𝑥 + 4) (𝑥 + 4)
 1 𝑥 3
= 𝑙𝑛|𝑥 | + 2 ( 𝐴𝑟𝑐𝑡𝑎𝑛 ) − 2
+𝐶
2 2 2(𝑥 + 4)
𝒙 𝟑 Final answer
𝒍𝒏|𝒙| + 𝑨𝒓𝒄𝒕𝒂𝒏 − +𝑪
𝟐 𝟐(𝒙 𝟐 + 𝟒)
References
Printed References
1. Stewart, J. (2019). Calculus concepts and contexts 4th edition enhanced edition. USA :
Cengage Learning, Inc.
2. Canlapan, R.B.(2017). Diwa senior high school series: basic calculus. DIWA Learning
Systems Inc.
3. Hass, J., Weir, M. & Thomas, G. (2016). University calculus early transcendentals,global
edition. Pearson Education, Inc.Larson,
4. Bacani, J.B.,et.al. (2016). Basic calculus for senior high school. Book AtBP. Publishing
Corp.
5. Balmaceda, J.M. (2016). Teaching guide in basic calculus. Commission on Higher
Education.
6. Pelias, J.G. (2016). Basic calculus. Rex Bookstore.
7. Molina, M.F. (2016). Integral calculus. Unlimited Books Library Services & Publishing
Inc.
8. Larson, R & Edwards, B. (2012). Calculus. Pasig City, Philippines. Cengage Learning
Asia Pte Ltd

Electronic Resources
1. Pulham , John. Differential Calculus. Retrieved August 7, 2020 from http://
www.maths.abdn.ac.uk/igc/tch/math1002/dif
2. Weistein, Eric. Differential Calculus. Retrieved August 7, 2020 from
http://mathworld.wolfram.com/differential calculus
3. Thomas , Christopher . Introduction to Differential Calculus. Retrieved August 7, 2020
from http:// www.usyd.edu.aav/stuserv/document/maths_learning-center
4. Purplemath. Retrieved August 7, 2020 from http://www.purplemath.com/modules/

Learning Materials
1. Worksheet
2. Module
LEARNING ASSESSMENT

ASSESSMENT 10
First semester SY 2020 – 2021

NAME
PROGRAM

General Instructions: Solve the following problem neatly and completely. Show your complete
solution and simplify whenever possible. Box your final answers. Answers/Solutions must be
handwritten in a bond paper.

(3𝑥 2 + 8𝑥 − 12)𝑑𝑥
1. ∫
𝑥 3 + 7𝑥 2 + 12𝑥
(𝑥 3 − 12𝑥 + 4)𝑑𝑥
2. ∫
(𝑥 2 − 3𝑥 + 2)2
3𝑡 2 𝑑𝑡
3. ∫ 4
𝑡 + 5𝑡 2 + 4
5𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃𝑑𝜃
4. ∫
𝑠𝑖𝑛2 𝜃 + 3𝑠𝑖𝑛𝜃 − 4
𝑑𝑥
5. ∫
(9 − 𝑥)√𝑥
(2𝑥 2 + 1)𝑑𝑥
6. ∫ 3
𝑥 − 3𝑥 + 2
(𝑥 2 − 7)𝑑𝑥
7. ∫ 3
𝑥 − 12𝑥 + 16
(𝑥 4 + 1)𝑑𝑥
8. ∫ 2
𝑥 (𝑥 + 1)2
10𝑑𝑥
9. ∫ 3
4𝑥 − 4𝑥 2 + 5𝑥
(𝑥 − 1)(2𝑥 3 + 2𝑥 2 + 3𝑥 + 2)𝑑𝑥
10. ∫
𝑥 3 (𝑥 + 1)
(4𝛽2 − 3𝛽 + 6)𝑑𝛽
11. ∫
𝛽4 (𝛽 − 2)
(4𝑥 + 5)𝑑𝑥
12. ∫ 2
𝑥 + 4𝑥 + 20
 UNIVERSITY OF SAINT LOUIS
Tuguegarao City

SCHOOL OF ENGINEERING, ARCHITECTURE & IT EDUCATION

Engineering Sciences Department
S.Y. 2020 – 2021

CORRESPONDENCE LEARNING MODALITY

CALC 1024: Calculus 2 (Integral Calculus)

Prepared by:

EROVITA TERESITA BACUD AGUSTIN, DME

ENGR. NORMAN BOB GOMEZ
ENGR. JENIROSE B. MATIAS
ENGR. PEEJAY PAGUIRIGAN
Instructors

Reviewed by:

ENGR. VICTOR VILLALUZ, MEE

EE Program Chair / ES Department Head

Recommended by:

ENGR. RONALD M. VILLAVERDE, ME – ECE, PECE

Academic Dean

Approved by:

EMMANUEL JAMES P. PATTAGUAN, Ph.D.

VP for Academics

This document is a property of University of Saint Louis – Tuguegarao. It must not be reproduced or transmitted in any form, in whole or in part, without expressed written permission.
For this week, April 12-16, 2021 of this semester, the following shall be your guide for the different
lessons and tasks that you need to accomplish. Be patient read it carefully before proceeding to the
tasks expected of you.

Date Topic Activities/Tasks

The Differential
Derivatives from Parametric Read Lessons
April 12-15, 2021
Equations
Radius and Center of Curvature
Accomplish the worksheet in the
Submission of learning tasks
April 16, 2021 Assessment Portion of this
module

GOD BLESS!

This document is a property of University of Saint Louis – Tuguegarao. It must not be reproduced or transmitted in any form, in whole or in part, without expressed written permission.
The Differential
Derivatives from Parametric Equations

Learning Objectives

▪ Find the differential of a function using differentiation formulas

▪ Estimate a propagated error using a differential
▪ Use differential in approximating function values
▪ Solve the derivative of parametric equations
▪ Compute the radius and center of curvature

This document is a property of University of Saint Louis – Tuguegarao. It must not be reproduced or transmitted in any form, in whole or in part, without expressed written permission.
WEEK 11: The Differential

When the tangent line to the graph of 𝑓 at the point (𝑐, 𝑓 (𝑐 ))

𝒚 = 𝒇(𝒄) + 𝒇′(𝒄)(𝒙 − 𝒄) *Tangent line at (𝑐, 𝑓(𝑐 ))
is used as an approximation of the graph of 𝑓, the quantity 𝑥 − 𝑐 is called the change in 𝑥, and is
denoted as ∆𝑥. When ∆𝑥 is small, the change in 𝑦 (denoted as ∆𝑦) can be approximated as
∆𝒚 = 𝒇(𝒄 + ∆𝒙) − 𝒇(𝒄) *Actual change in 𝑦
∆𝒚 ≈ 𝒇′(𝒄)∆𝒙 *Approximate change in 𝑦

For such an approximation, the quantity ∆𝑥 is traditionally

denoted by 𝑑𝑥 and is called the differential of 𝑥. The
expression 𝑓 ′(𝑥 )𝑑𝑥 is denoted by 𝑑𝑦 and is called the
differential of 𝑦.

Definition of Differentials
Let 𝑦 = 𝑓(𝑥) represent a function that is differentiable on an open interval containing 𝑥. The
differential of 𝑥 (denoted as 𝑑𝑥) is any nonzero real number. The differential of 𝑦 (denoted as 𝑑𝑦)
is 𝑑𝑦 = 𝑓 ′(𝑥 )𝑑𝑥

In many types of applications, the differential of 𝑦 can be used as an approximation of the change
in 𝑦. That is,
∆𝒚 ≈ 𝒅𝒚 or ∆𝒚 ≈ 𝒇′(𝒙)𝒅𝒙

Error Propagation
Physicists and engineers tend to make liberal use of the approximation of ∆𝑦 by 𝑑𝑦. One way this
occurs in practice is in the estimation of errors propagated by physical measuring devices. For
example, if you let 𝑥 represent the measured value of a variable and let 𝑥 + ∆𝑥 represent the exact
value, then ∆𝑥 is the error in measurement. Finally, if the measured value of 𝑥 is used to compute
another value 𝑓(𝑥), then the difference between 𝑓(𝑥 + ∆𝑥) and 𝑓(𝑥) is the propagated error.
𝒇(𝒙 + ∆𝒙) − 𝒇(𝒙) = ∆𝒚

𝑓 (𝑥 + ∆𝑥 ) Exact value
𝑓(∆𝑥) Measurement error
𝑓 (𝑥 ) Measured value
∆𝑦 Propagated error

Calculating Differentials

Each of the differentiation rules can be written in differential form. For example, let 𝑢 and 𝑣 be
differentiable functions of 𝑥. By the definition of differentials, you have
𝑑𝑢 = 𝑢′𝑑𝑥 and 𝑑𝑣 = 𝑣′𝑑𝑥

This document is a property of University of Saint Louis – Tuguegarao. It must not be reproduced or transmitted in any form, in whole or in part, without expressed written permission.
So, you can write the differential form of the Product Rule as
𝑑 Differential of 𝑢𝑣
𝑑 [𝑢𝑣 ] = [𝑢𝑣 ]𝑑𝑥
𝑑𝑥
= [𝑢𝑣 ′ + 𝑣𝑢′ ]𝑑𝑥 Product Rule
′ ′
= 𝑢𝑣 𝑑𝑥 + 𝑣𝑢 𝑑𝑥
= 𝑢𝑑𝑣 + 𝑣𝑑𝑢

Differential Formulas
Let 𝑢 and 𝑣 be differentiable functions of 𝑥
Constant multiple 𝑑 [𝑐𝑢] = 𝑐𝑑𝑢
Sum or difference 𝑑 [𝑢 ± 𝑣 ] = 𝑑𝑢 ± 𝑑𝑣
Product 𝑑 [𝑢𝑣 ] = 𝑢𝑑𝑣 + 𝑣𝑑𝑢
Quotient 𝑢 𝑣𝑑𝑢 − 𝑢𝑑𝑣
𝑑[ ] =
𝑣 𝑣2

Finding Differentials

Function Derivative Differential

1. 𝑦 = 𝑥 2 𝑑𝑦 𝑑𝑦 = 2𝑥𝑑𝑥
= 2𝑥
𝑑𝑥
2. 𝑦 = √𝑥 𝑑𝑦 1 𝑑𝑥
= 𝑑𝑦 =
𝑑𝑥 2√𝑥 2√𝑥
3. 𝑦 = 2𝑠𝑖𝑛𝑥 𝑑𝑦 𝑑𝑦 = 2𝑐𝑜𝑠𝑥𝑑𝑥
= 2𝑐𝑜𝑠𝑥
𝑑𝑥
4. 𝑦 = 𝑥𝑐𝑜𝑠𝑥 𝑑𝑦 𝑑𝑦 = (−𝑥𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 )𝑑𝑥
= −𝑥𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥
𝑑𝑥
1 𝑑𝑦 1 𝑑𝑥
5. 𝑦 = =− 2 𝑑𝑦 = −
𝑥 𝑑𝑥 𝑥 𝑥2

This notation is called the Leibniz notation for derivative and differentials, named after the German
mathematician Gottfried Wilhelm Leibniz. The beauty of this notation is that it provides an easy
way to remember several important calculus formulas by making it seem as though the formulas
were derived from algebraic manipulations of differentials. For instance, in Leibniz notation, the
Chain Rule
𝑑𝑦 𝑑𝑦 𝑑𝑢
=
𝑑𝑥 𝑑𝑢 𝑑𝑥
would appear to be true because the 𝑑𝑢′𝑠 divide out. Even though this reasoning is incorrect, the
notation does help on remember the chain rule.

Differentials can be used to approximate function values. To do this for the function given by 𝑦 =
𝑓(𝑥), use the formula
𝒇(𝒙 + ∆𝒙) ≈ 𝒇(𝒙) + 𝒅𝒚 = 𝒇(𝒙) + 𝒇′ (𝒙)𝒅𝒙
Note: This formula is equivalent to the tangent line approximation

Reference: Larson, R & Edwards, B. (2018). Calculus 11th Ed. Boston, MA USA. Cengage Learning

This document is a property of University of Saint Louis – Tuguegarao. It must not be reproduced or transmitted in any form, in whole or in part, without expressed written permission.
Examples

Find the differential 𝑑𝑦 of the given function

𝟏. 𝒚 = 𝒙𝟑 − 𝟒𝒙𝟐 + 𝟓𝒙 Original function

𝑑𝑦 = 𝑑(𝑥 3 − 4𝑥 2 + 5𝑥 ) Differentiate the function using the rules for
differentiation
𝒅𝒚 = (𝟑𝒙𝟐 − 𝟖𝒙 + 𝟓)𝒅𝒙 Final answer (differential form)

𝟐𝒙 Original function
𝟐. 𝒚 =
𝟑𝒙 − 𝟏
2𝑥 Differentiate applying quotient rule
𝑑𝑦 = 𝑑 ( ) 𝑢 𝑣𝑑𝑢 − 𝑢𝑑𝑣
3𝑥 − 1 𝑑( ) =
𝑣 𝑣2
(3𝑥 − 1)(2𝑑𝑥 ) − (2𝑥)(3𝑑𝑥) Simplify and factor out 𝑑𝑥
𝑑𝑦 =
(3𝑥 − 1)2
(𝟔𝒙 − 𝟐 − 𝟔𝒙)𝒅𝒙 Final answer (differential form)
𝒅𝒚 =
(𝟑𝒙 − 𝟏)𝟐

𝟑. 𝒚 = 𝒙𝒍𝒏𝒙 Original function

𝑑𝑦 = 𝑑(𝑥𝑙𝑛𝑥 ) Differentiate applying product rule
𝑑 (𝑢𝑣 ) = 𝑢𝑑𝑣 + 𝑣𝑑𝑢
1 Simplify and factor out 𝑑𝑥
𝑑𝑦 = (𝑥 ) ( ) 𝑑𝑥 + (𝑙𝑛𝑥 )(1)𝑑𝑥
𝑥
𝑑𝑦 = 𝑑𝑥 + 𝑙𝑛𝑥𝑑𝑥
𝒅𝒚 = (𝟏 + 𝒍𝒏𝒙)𝒅𝒙 Final answer (differential form)

𝟒. 𝒚 = √𝟗 − 𝒙𝟐 Original function
1 Differentiate applying power rule
𝑑𝑦 = 𝑑 (√9 − 𝑥 2 ) = 𝑑 [(9 − 𝑥 2 )2 ]
1 1 Simplify
𝑑𝑦 = ( ) (9 − 𝑥 2 )−2 (−2𝑥 )𝑑𝑥
2
1 1 Cancel out 2
𝑑𝑦 = ( ) (−2𝑥)(9 − 𝑥 2 )−2 𝑑𝑥
2
1 1
𝑑𝑦 = −𝑥(9 − 𝑥 2 )−2 𝑑𝑥 Bring down (9 − 𝑥 2 )−2
𝑥𝑑𝑥
𝑑𝑦 = − 1
(9 − 𝑥 2 )2
𝒙𝒅𝒙 Final answer (differential form)
𝒅𝒚 = −
√𝟗 − 𝒙𝟐

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 𝑑𝑦
Find 𝑑𝑥 by use of differentials

𝟓. 𝒙𝒚 + 𝒔𝒊𝒏𝒙 = 𝒍𝒏𝒚 Original function

1 Group terms containing 𝑑𝑦 and 𝑑𝑥
𝑥𝑑𝑦 + 𝑦𝑑𝑥 + 𝑐𝑜𝑠𝑥𝑑𝑥 = 𝑑𝑦
𝑦
1 Factor out 𝑑𝑦 and 𝑑𝑥
𝑥𝑑𝑦 − 𝑑𝑦 = −𝑦𝑑𝑥 − 𝑐𝑜𝑠𝑥𝑑𝑥
𝑦
1
𝑑𝑦 (𝑥 − ) = −𝑑𝑥(𝑦 + 𝑐𝑜𝑠𝑥 )
𝑦
𝑑𝑦 −(𝑦 + 𝑐𝑜𝑠𝑥)
=
𝑑𝑥 1
(𝑥 − 𝑦)
𝑑𝑦 −(𝑦 + 𝑐𝑜𝑠𝑥) Simplify
=
𝑑𝑥 𝑥𝑦 − 1
( 𝑦 )
𝒅𝒚 −𝒚(𝒚 + 𝒄𝒐𝒔𝒙) Final answer
=
𝒅𝒙 𝒙𝒚 − 𝟏

𝟔. 𝒙𝒚 + 𝑨𝒓𝒄𝒕𝒂𝒏(𝒙𝒚) = 𝟎 Original function

1 Recall
𝑥𝑑𝑦 + 𝑦𝑑𝑥 + (𝑥𝑑𝑦 + 𝑦𝑑𝑥 ) = 0 𝑑 1 𝑑𝑢
1 + (𝑥𝑦)2 (𝐴𝑟𝑐𝑡𝑎𝑛𝑢) =
𝑑𝑥 1 + 𝑢2 𝑑𝑥
𝑥𝑑𝑦 + 𝑦𝑑𝑥 Group terms containing 𝑑𝑦 and 𝑑𝑥
𝑥𝑑𝑦 + 𝑦𝑑𝑥 + =0
1 + 𝑥2𝑦2
𝑥𝑑𝑦 𝑦𝑑𝑥
𝑥𝑑𝑦 + 2 2
= −𝑦𝑑𝑥 −
1+𝑥 𝑦 1 + 𝑥 2𝑦 2
𝑥𝑑𝑦(1 + 𝑥 2 𝑦 2 ) + 𝑥𝑑𝑦 −𝑦𝑑𝑥(1 + 𝑥 2 𝑦 2 ) − 𝑦𝑑𝑥 Cancel the denominator
2 2
= 2 2
1+𝑥 𝑦 1+𝑥 𝑦
𝑥𝑑𝑦 1 + 𝑥 𝑦 + 𝑥𝑑𝑦 = −𝑦𝑑𝑥 1 + 𝑥 2 𝑦 2 ) − 𝑦𝑑𝑥
( 2 2) ( Factor out 𝑥𝑑𝑦 and −𝑦𝑑𝑥
𝑥𝑑𝑦(1 + 𝑥 2 𝑦 2 + 1) = −𝑦𝑑𝑥(1 + 𝑥 2 𝑦 2 + 1)
𝑑𝑦 −𝑦(1 + 𝑥 2 𝑦 2 + 1) Cancel (1 + 𝑥 2 𝑦 2 + 1)
=
𝑑𝑥 𝑥(1 + 𝑥 2 𝑦 2 + 1)
𝒅𝒚 𝒚 Final answer
=−
𝒅𝒙 𝒙

𝟕. 𝟒𝒙 + 𝟒𝒚 = 𝟑𝟐 Original function
Recall:
𝑑 𝑢 𝑑𝑢
(𝑎 ) = 𝑎𝑢 (𝑙𝑛𝑎)
𝑑𝑥 𝑑𝑥
4𝑥 (𝑙𝑛4)(1)𝑑𝑥 + 4𝑦 (𝑙𝑛4)(1)𝑑𝑦 = 0
4𝑥 (𝑙𝑛4)𝑑𝑥 + 4𝑦 (𝑙𝑛4)𝑑𝑦 = 0 Transpose the term containing 𝑑𝑥
4𝑦 (𝑙𝑛4)𝑑𝑦 = −4𝑥 (𝑙𝑛4)𝑑𝑥

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 𝑑𝑦 −4𝑥 (𝑙𝑛4) Cancel (𝑙𝑛4)
= 𝑦
𝑑𝑥 4 (𝑙𝑛4)
𝑑𝑦 −4𝑥 Simplify applying law of exponent
= 𝑦
𝑑𝑥 4
𝒅𝒚 Final answer
= −𝟒𝒙−𝒚
𝒅𝒙

Application of Differentials

8. Compute √37 approximately by use of differentials.

Let
√37 = 𝑦 + 𝑑𝑦

37 = 𝑥 + 𝑑𝑥 𝑥 is a perfect square nearest to 37

37 = 36 + 1 𝑥 = 36 𝑑𝑥 = 1
Let
𝑦 = √𝑥 Differentiate 𝑦
1 1
𝑑𝑦 = (𝑥 )−2 𝑑𝑥
2
𝑑𝑥 Substitute 𝑥 = 36 𝑎𝑛𝑑 𝑑𝑥 = 1
𝑑𝑦 =
2√𝑥
1 1 1
𝑑𝑦 = = =
2√36 2(6) 12
1
√37 = 𝑦 + 𝑑𝑦 Substitute 𝑦 = √𝑥 𝑎𝑛𝑑 𝑑𝑦 = 12
1
√37 = √36 +
12
1
√37 = 6 +
12
𝟕𝟑 Final answer
√𝟑𝟕 = = 𝟔. 𝟖𝟎𝟑
𝟏𝟐
5
9. Compute√31.6 approximately by use of differentials.
Let
5
√31.6 = 𝑦 + 𝑑𝑦
31.6 = 𝑥 + 𝑑𝑥 𝑥 is a perfect fifth root nearest to 31.6
31.6 = 32 − 0.4 𝑥 = 32 𝑑𝑥 = −0.4
Let
𝑦 = 5√𝑥 Differentiate 𝑦
1 −4
𝑑𝑦 = 𝑥 5 𝑑𝑥
5
𝑑𝑥 Substitute 𝑥 = 32 𝑎𝑛𝑑 𝑑𝑥 = −0.4
𝑑𝑦 = 5
5√𝑥 4
−0.4
𝑑𝑦 = 5 = −0.0459
5√24

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 5
√31.6 = 𝑦 + 𝑑𝑦 Substitute 𝑦 = 5√𝑥 𝑎𝑛𝑑 𝑑𝑦 = −0.0459
5 5
√31.6 = √32 + (−0.0459)
5
√31.6 = 2 − 0.0459
𝟓
√𝟑𝟏. 𝟔 = 𝟏. 𝟗𝟓𝟒𝟏 Final answer

10. If 𝑦 = 𝑥 3 + 2𝑥 2 − 3, find the approximate value of y when 𝑥 = 2.01

𝑦 = 𝑥 3 + 2𝑥 2 − 3 Differentiate 𝑦
2
𝑑𝑦 = (3𝑥 + 4𝑥 )𝑑𝑥
When 𝑥 = 2 Substitute 𝑥 = 2
𝑦 = 𝑥 3 + 2𝑥 2 − 3
𝑦 = (2)3 + 2(2)2 − 3
𝑦 = 13
When 𝑥 = 2 𝑎𝑛𝑑 𝑑𝑥 = 0.01 Substitute 𝑥 = 2 𝑎𝑛𝑑 𝑑𝑥 = 0.01
( 2 )
𝑑𝑦 = 3𝑥 + 4𝑥 𝑑𝑥
𝑑𝑦 = [3(2)2 + 4(2)](0.01) To solve for the approximate value
𝑑𝑦 = 0.20 𝑦 + 𝑑𝑦
𝑦 + 𝑑𝑦 = 13 + 0.20
𝒚 + 𝒅𝒚 = 𝟏𝟑. 𝟐𝟎 Final answer (Required Approximation)

11. If 𝑦 = (2𝑥 − 1)4 , find the approximate value of y when 𝑥 = 0.98

𝑦 = (2𝑥 − 1)4 Differentiate 𝑦
𝑑𝑦 = 4(2𝑥 − 1)3 (2)𝑑𝑥
𝑑𝑦 = 8(2𝑥 − 1)3 𝑑𝑥
When 𝑥 = 1 Substitute 𝑥 = 1
𝑦 = (2𝑥 − 1)4
𝑦 = [(2)(1) − 1]4
𝑦=1
When 𝑥 = 1 𝑎𝑛𝑑 𝑑𝑥 = −0.02 Substitute 𝑥 = 1 𝑎𝑛𝑑 𝑑𝑥 = −0.02
( ) 3
𝑑𝑦 = 8 2𝑥 − 1 𝑑𝑥
𝑑𝑦 = 8[(2)(1) − 1]3 (−0.02) To solve for the approximate value
𝑑𝑦 = −0.16 𝑦 + 𝑑𝑦
𝑦 + 𝑑𝑦 = 1 + (−0.16)
𝒚 + 𝒅𝒚 = 𝟎. 𝟖𝟒 Final answer (Required Approximation)

12. Find the approximate percentage error in the computed volume V of a cube of edge x cm if
an error of 2% is made in measuring an edge.
𝑑𝑉 𝑅𝑒𝑐𝑎𝑙𝑙: 𝑉𝑐𝑢𝑏𝑒 = 𝑥 3
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝐸𝑟𝑟𝑜𝑟 (𝑅𝐸) = 𝑑𝑥
𝑉 (100) = 2%
𝑑𝑉 𝑥
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 (𝑃𝐸) = (100)
𝑉
𝑉 = 𝑥3 Differentiate 𝑉
2
𝑑𝑉 = 3𝑥 𝑑𝑥
𝑑𝑉 Substitute 𝑉 𝑎𝑛𝑑 𝑑𝑉
𝑃𝐸 = (100)
𝑉

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 3𝑥 2 𝑑𝑥 Simplify
𝑃𝐸 = (100)
𝑥3
3𝑑𝑥 𝑑𝑥
𝑃𝐸 = (100) (100) = 2%
𝑥 𝑥
𝑃𝐸 = 3(2%)
𝑷𝑬 = 𝟔% Final answer

13. The circumference of a circle is 100 cm. if the radius is increased by 0.1 cm, find the
approximate increase in the area.
𝐶 = 100 𝑐𝑚
𝑑𝑟 = 0.1 𝑐𝑚
𝑑𝐴 =?
𝐶 = 2𝜋𝑟 Substitute 𝐶 = 100 𝑐𝑚 to solve for 𝑟
100 = 2𝜋𝑟
50
𝑟= 𝑐𝑚
𝜋
𝐴 = 𝜋𝑟 2 Differentiate 𝐴
𝑑𝐴 = 2𝜋𝑟𝑑𝑟 50
Substitute 𝑟 = 𝜋 𝑐𝑚 𝑎𝑛𝑑 𝑑𝑟 = 0.1 𝑐𝑚
50
𝑑𝐴 = 2𝜋 ( 𝑐𝑚) (0.1 𝑐𝑚)
𝜋
𝒅𝑨 = 𝟏 𝒔𝒒. 𝒄𝒎. Final answer

14. The radius of a sphere is measured to be 4 cm with an error of 0.002 cm. Find the relative
and percentage error in the computed volume
𝑑𝑉 𝑅𝑒𝑐𝑎𝑙𝑙:
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝐸𝑟𝑟𝑜𝑟 (𝑅𝐸) = 4
𝑉 𝑉𝑠𝑝ℎ𝑒𝑟𝑒 = 𝜋𝑟 3
𝑑𝑉 3
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 (𝑃𝐸) = (100)
𝑉 𝑟 = 4 𝑐𝑚 𝑑𝑟 = 0.002 𝑐𝑚
4 3
𝑉 = 𝜋𝑟
3
4
𝑉 = 𝜋(4)3
3
256𝜋
𝑉= 𝑐𝑢. 𝑐𝑚
3
4 Cancel 3
𝑑𝑉 = (3)𝜋𝑟 2 𝑑𝑟
3
𝑑𝑉 = 4𝜋𝑟 2 𝑑𝑟
𝑑𝑉 = 4𝜋(4)2 (0.002)
𝑑𝑉 = 0.4021 𝑐𝑢. 𝑐𝑚
𝑑𝑉
𝑅𝐸 =
𝑉
0.4021
𝑅𝐸 =
256𝜋
3

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 𝑹𝑬 = 𝟎. 𝟎𝟎𝟏𝟓 To solve for Percentage Error (PE), multiply
RE by 100
𝑃𝐸 = (𝑅𝐸)(100)
𝑃𝐸 = (0.0015)(100)
𝑷𝑬 = 𝟎. 𝟏𝟓%

15. For a right circular cylinder of height 25 cm, the radius was measured as 20 cm with an
error of 0.05 cm. Find the approximate percentage error in the computed volume.
𝑑𝑉 𝑅𝑒𝑐𝑎𝑙𝑙:
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 (𝑃𝐸) = (100)
𝑉 𝑉∟𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 𝜋𝑟 2 ℎ
ℎ = 25 𝑐𝑚 𝑟 = 20 𝑐𝑚 𝑑𝑟 = 0.5 𝑐𝑚 𝑃𝐸 =?
𝑉 = 𝜋𝑟 2 ℎ Substitute ℎ = 25 𝑐𝑚 𝑎𝑛𝑑 𝑟 = 20 𝑐𝑚
2
𝑉 = 𝜋(20) (25)
𝑉 = 10000𝜋 𝑐𝑢. 𝑐𝑚
𝑉 = 𝜋𝑟 2 ℎ Differentiate 𝑉 with respect to 𝑟
𝑑𝑉 = 2𝜋𝑟ℎ𝑑𝑟 Substitute
𝑟 = 20 𝑐𝑚, ℎ = 25 𝑐𝑚, 𝑎𝑛𝑑 𝑑𝑟 = 0.05 𝑐𝑚
𝑑𝑉 = 2𝜋(20)(25)(0.05)
𝑑𝑉 = 50𝜋 𝑐𝑢. 𝑐𝑚
𝑑𝑉 Substitute 𝑑𝑉 = 50𝜋 𝑎𝑛𝑑 𝑉 = 10000𝜋
𝑃𝐸 = (100)
𝑉
10000𝜋
𝑃𝐸 = (100)
50𝜋
𝑷𝑬 = 𝟎. 𝟓 % Final answer

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WEEK 11: Derivatives from Parametric Equation; Radius and Center of Curvature

In analytic geometry, we have learned that a curve may also be described analytically by a pair of
equations of the form
𝑥 = 𝑔(𝑡), 𝑦 = ℎ(𝑡)
These equations are called parametric equations of the curve and the third variable 𝑡 is called a
parameter. For example, the circle 𝑥 2 + 𝑦 2 = 𝑎2 may be represented by the parametric equations
𝑥 = 𝑎𝑐𝑜𝑠𝑡, 𝑦 = 𝑎𝑠𝑖𝑛𝑡
where the parameter 𝑡 is the angle between the 𝑥 − 𝑎𝑥𝑖𝑠 and the radius to the point (𝑥, 𝑦).

Derivatives in Parametric Form

Let 𝑦 = 𝑓(𝑥) be a function whose parametric representation is given in the form
𝑥 = 𝑔(𝑡), 𝑦 = ℎ(𝑡)

𝑑𝑥
= 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑥 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑡
𝑑𝑡
𝑑𝑦
= 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑦 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑡
𝑑𝑡
Then evidently, the rate of change of y with respect to x of a function defined by
𝑥 = 𝑔(𝑡), 𝑦 = ℎ(𝑡) will be given by
𝑑𝑦
𝑑𝑦 𝑑𝑡
=
𝑑𝑥 𝑑𝑥
𝑑𝑡
Next, we consider the problem of finding the second derivative of a function defined by the
parametric equations above. Second derivative is defined as
𝑑2𝑦 𝑑 𝑑𝑦
2
= ( )
𝑑𝑥 𝑑𝑥 𝑑𝑥
and by Chain Rule, we may write the equation above in the form
𝑑 2 𝑦 𝑑 𝑑𝑦 𝑑𝑡
= ( )∙
𝑑𝑥 2 𝑑𝑡 𝑑𝑥 𝑑𝑥

Radius and Center of Curvature

The concept of derivative is related to the tangent of a curve. Another concept of geometric interest
is that of curvature. Consider the curve shown below. Let s be the length of the arc of the curve
between a fixed-point A and a variable point P. denote the slope-angle of the tangent T to the curve
P by ∅. The rate of change of ∅ with respect to s is called the curvature of the curve at P and is
denoted as K. that is
𝑑∅
𝐾=
𝑑𝑠
The reciprocal of the curvature is called the radius of
curvature and is denoted by R. that is

1
𝑅=
𝐾

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 3
2 2
𝑑𝑦
[1 + (𝑑𝑥 ) ]
𝑅=
𝑑2𝑦
| 2|
𝑑𝑥

𝑑𝑦 2
The absolute value symbol is not needed in the numerator since (𝑑𝑥) > 0. To simplify notation,
we may write it in the form
3
[1 + (𝑦 ′)2 ]2
𝑅=
|𝑦′′|
𝑑𝑦
𝑦′ =
𝑑𝑥
𝑑2𝑦
𝑦 ′′ = 2
𝑑𝑥

If the equation of the curve is given by 𝑥 = 𝑔(𝑦), then the defining equation for R takes the form
3
[1 + (𝑥 ′ )2 ]2
𝑅=
|𝑥 ′′|
𝑑𝑥
𝑥′ =
𝑑𝑦
′′
𝑑2𝑥
𝑥 =
𝑑𝑦 2

When the equation of the curve is given parametrically in the form

𝑥 = 𝑔(𝑡), 𝑦 = ℎ(𝑡)
3
[(𝑔′)2 + (ℎ′ )2 ]2
𝑅=
|𝑔′ℎ′′ − 𝑔′′ ℎ′ |
𝑑𝑥 𝑑2𝑥
𝑔′ = 𝑔′′ =
𝑑𝑡 𝑑𝑡 2
𝑑𝑦 𝑑2𝑦
ℎ′ = ℎ′′ = 2
𝑑𝑡 𝑑𝑡

Center of Curvature

Through any point 𝑃(𝑥, 𝑦) on the curve 𝑦 = 𝑓(𝑥), we can construct a tangent circle whose radius
r is equal to the radius of curvature R of the curve at P as shown in the figure below. This unique
circle is called the circle of curvature and its center is called the center of curvature. This center of
curvature which has coordinates (ℎ, 𝑘) lies on the normal to the curve at P. We shall show how
the coordinates (ℎ, 𝑘) can be expressed in terms of the coordinates (𝑥, 𝑦) of P.

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The equation of the circle of curvature is
(𝑥 − ℎ ) 2 + (𝑦 − 𝑘 )2 = 𝑅 2
[ 1 + ( 𝑦 ′ ) 2 ]3
(𝑥 − ℎ ) 2 + (𝑦 − 𝑘 )2 =
(𝑦 ′′)2
Since the slope of the tangent at P is 𝑦′, then the slope of the
normal is −1/𝑦′. The equation of the normal is
1
𝑦 − 𝑘 = − ′ (𝑥 − ℎ )
𝑦
Solving simultaneously for ℎ and 𝑘, we obtain
𝑦 ′[1 + (𝑦 ′)2 ]
ℎ=𝑥−
𝑦 ′′
1 + (𝑦 ′ )2
𝑘 =𝑦+
𝑦′′

Examples

𝑑𝑦 𝑑2𝑦
𝐹𝑖𝑛𝑑 𝑎𝑛𝑑 2
𝑑𝑥 𝑑𝑥

𝟏. 𝒙 = 𝒕𝟑 + 𝟏, 𝒚 = 𝒕𝟐 + 𝟏
𝑑𝑥 𝑑𝑦
Solve for and
𝑑𝑡 𝑑𝑡
𝑑𝑥 𝑑𝑦
= 3𝑡 2 Solve for 𝑑𝑥
𝑑𝑡
𝑑𝑦
= 2𝑡
𝑑𝑡
𝑑𝑦
𝑑𝑦 𝑑𝑡
=
𝑑𝑥 𝑑𝑥
𝑑𝑡
𝑑𝑦 2𝑡 Simplify
= 2
𝑑𝑥 3𝑡
𝒅𝒚 𝟐
=
𝒅𝒙 𝟑𝒕
𝑑2𝑦 𝑑 𝑑𝑦 𝑑𝑡 𝑑𝑡 1
= ( )∙ =
𝑑𝑥 2 𝑑𝑡 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑𝑡
𝑑2𝑦 𝑑 2 1 Differentiate applying quotient rule
= ( ) ∙ ( )
𝑑𝑥 2 𝑑𝑡 3𝑡 3𝑡 2
𝑑2𝑦 2 1
2
= (− 2 ) ∙ ( 2 )
𝑑𝑥 3𝑡 3𝑡
𝒅𝟐 𝒚 𝟐
=− 𝟒
𝒅𝒙𝟐 𝟗𝒕

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 𝟐. 𝒙 = 𝒖𝟑 + 𝟏, 𝒚 = 𝟒𝒖𝟐 − 𝟒𝒖
𝑑𝑥 𝑑𝑦
Solve for 𝑑𝑢 and 𝑑𝑢
𝑑𝑥 𝑑𝑦
= 3𝑢2 Solve for 𝑑𝑥
𝑑𝑢
𝑑𝑦
= 8𝑢 − 4
𝑑𝑢
𝑑𝑦
𝑑𝑦 𝑑𝑢
=
𝑑𝑥 𝑑𝑥
𝑑𝑢
𝑑𝑦 8𝑢 − 4
=
𝑑𝑥 3𝑢2
𝒅𝒚 𝟒(𝟐𝒖 − 𝟏)
=
𝒅𝒙 𝟑𝒖𝟐
2 𝑑𝑢 1
𝑑 𝑦 𝑑 𝑑𝑦 𝑑𝑢
= ( )∙ =
𝑑𝑥 2 𝑑𝑢 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑𝑢
𝑑2𝑦 𝑑 4(2𝑢 − 1) 1 Differentiate applying quotient rule
2
= [ 2
] ∙ ( 2)
𝑑𝑥 𝑑𝑢 3𝑢 3𝑢
2 2
𝑑 𝑦 4 (𝑢 )(2) − (2𝑢 − 1)(2𝑢) 1 Simplify
2
= [ 2 2
] ∙ ( 2)
𝑑𝑥 3 (𝑢 ) 3𝑢
2 2 2
𝑑 𝑦 4 2𝑢 − 4𝑢 + 2𝑢 1
2
= [ 4
] ∙ ( 2)
𝑑𝑥 3 𝑢 3𝑢
2 2 2
𝑑 𝑦 4 2𝑢 − 4𝑢 + 2𝑢 Combine similar terms
= ( )
𝑑𝑥 2 9 𝑢6
𝑑 2 𝑦 4 −2𝑢2 + 2𝑢 Factor out 2𝑢
= ( )
𝑑𝑥 2 9 𝑢6
𝑑 2 𝑦 4 (2𝑢)(−𝑢 + 1)
= [ ]
𝑑𝑥 2 9 𝑢 ∙ 𝑢5
𝒅𝟐 𝒚 𝟖(𝟏 − 𝒖)
=
𝒅𝒙𝟐 𝟗𝒖𝟓

3. 𝑥 = 1 + 𝑐𝑜𝑠𝑡, 𝑦 = sin2 𝑡
𝑑𝑥 𝑑𝑦
Solve for and
𝑑𝑡 𝑑𝑡
𝑑𝑥 Solve for 𝑑𝑥
𝑑𝑦
= −𝑠𝑖𝑛𝑡
𝑑𝑡
𝑑𝑦
= 2𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡
𝑑𝑡
𝑑𝑦
𝑑𝑦 𝑑𝑡
=
𝑑𝑥 𝑑𝑥
𝑑𝑡
𝑑𝑦 2𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡 Simplify
=
𝑑𝑥 −𝑠𝑖𝑛𝑡

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 𝒅𝒚
= −𝟐𝒄𝒐𝒔𝒕
𝒅𝒙
𝑑2𝑦 𝑑 𝑑𝑦 𝑑𝑡 𝑑𝑡 1
= ( )∙ =
𝑑𝑥 2 𝑑𝑡 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑𝑡
𝑑2𝑦 𝑑 1 Differentiate first before simplifying the
= ( −2𝑐𝑜𝑠𝑡 ) ∙ ( ) problem
𝑑𝑥 2 𝑑𝑡 −𝑠𝑖𝑛𝑡
𝑑2𝑦 1
2
= (2𝑠𝑖𝑛𝑡) ∙ ( )
𝑑𝑥 −𝑠𝑖𝑛𝑡
𝑑2𝑦 2𝑠𝑖𝑛𝑡 Cancel out similar terms
= ( )
𝑑𝑥 2 −𝑠𝑖𝑛𝑡
𝒅𝟐 𝒚
= −𝟐
𝒅𝒙𝟐

Radius of Curvature

𝝅
𝟒. 𝒚 = 𝒔𝒊𝒏𝒙 𝒂𝒕 ( , 𝟏)
𝟐
3
′ 2
[ 1 + ( 𝑦 ) ]2 𝑑𝑦 𝑑2𝑦
𝑦′ = 𝑦 ′′ =
𝑅= 𝑑𝑥 𝑑𝑥 2
|𝑦′′|
′
𝑦 = 𝑐𝑜𝑠𝑥 First derivative of y
𝜋 Substitute 𝑥 = 2
𝜋
𝑦 ′ = cos
2
𝑦′ = 0
𝑦 ′′ = −𝑠𝑖𝑛𝑥 Second derivative of y
𝜋 𝜋
Substitute 𝑥 = 2
𝑦 ′′ = − 𝑠𝑖𝑛
2
𝑦 ′′ = −1
Substitute 𝑦 ′𝑎𝑛𝑑 𝑦′′
3 3
[1 + (0)2 ]2 (1)2
𝑅= =
|−1| 1
𝑹 = 𝟏 𝒖𝒏𝒊𝒕 Final answer

𝟓. 𝒙 = 𝒆𝒚 − 𝟐𝒚 𝒂𝒕 (𝟏, 𝟎)
3
[1 + (𝑥 ′ )2 ]2 𝑑𝑥 𝑑2𝑥
𝑥′ = 𝑥 ′′ =
𝑅= 𝑑𝑦 𝑑𝑦 2
|𝑥 ′′ |
′ 𝑦
𝑥 =𝑒 −2 First derivative of x
𝑥′ = 𝑒0 − 2 Substitute 𝑦 = 0
𝑥 ′ = −1
𝑥 ′′ = 𝑒 𝑦 Second derivative of x
𝑥 ′′ = 𝑒 0 Substitute 𝑦 = 0
𝑥 ′′ = 1
Substitute 𝑥 ′ 𝑎𝑛𝑑 𝑥′′

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 3
[1 + (−1)2 ]2
𝑅=
|1|
3
(2)2
𝑅=
1
𝑅 = √23
𝑹 = 𝟐√𝟐 units Final answer

𝟔. 𝒙 = 𝒕𝟐 − 𝟐𝒕, 𝒚 = 𝟏 − 𝟒𝒕 𝒂𝒕 𝒕 = 𝟏
[(𝑔′)2 + (ℎ′ )2 ]2
3 𝑑𝑥 𝑑 2 𝑥 ′ 𝑑𝑦 𝑑2𝑦
𝑔′ = 𝑔′′ = ℎ = ℎ′′ =
𝑅= 𝑑𝑡 𝑑𝑡 2 𝑑𝑡 𝑑𝑡 2
|𝑔′ℎ′′ − 𝑔′′ℎ′ |
′
𝑔 = 2𝑡 − 2 𝑔′′ = 2 Substitute 𝑡 = 1
𝑔′ = 2(1) − 2
𝑔′ = 0
ℎ′ = −4 ℎ′′ = 0 Substitute 𝑔′, 𝑔′′, ℎ′, 𝑎𝑛𝑑 ℎ′′
3
[(0)2 + (−4)2 ]2
𝑅=
|(0)(0) − (2)(−4)|
3
(16)2
𝑅=
8
𝑹 = 𝟖 𝒖𝒏𝒊𝒕𝒔 Final answer

Center of Curvature

𝟏 𝒙 Solve for (ℎ, 𝑘)

𝟕. 𝒚 = (𝒆 + 𝒆−𝒙 ) 𝒂𝒕 (𝟎, 𝟏)
𝟐
𝑦 ′[1 + (𝑦 ′)2 ] 1 + (𝑦 ′ )2
ℎ=𝑥− 𝑘 =𝑦+
𝑦 ′′ 𝑦′′
1 Substitute 𝑥 = 0
𝑦 ′ = (𝑒 𝑥 − 𝑒 −𝑥 )
2
1
𝑦 = (𝑒 0 − 𝑒 0 )
′
2
𝑦′ = 0
1 Substitute 𝑥 = 0
𝑦 ′′ = (𝑒 𝑥 + 𝑒 −𝑥 )
2
1 1
𝑦 ′′ = (𝑒 0 + 𝑒 0 ) = (1 + 1)
2 2
𝑦 ′′ = 1
𝑦 ′[1 + (𝑦 ′)2 ] Substitute 𝑥, 𝑦 ′, 𝑎𝑛𝑑 𝑦 ′′
ℎ=𝑥−
𝑦 ′′
0[1 + (0)2
ℎ=0−
1
ℎ=0

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 1 + (𝑦 ′ )2 Substitute 𝑦, 𝑦 ′, 𝑎𝑛𝑑 𝑦 ′′
𝑘 =𝑦+
𝑦′′
1 + (0)2
𝑘 =1+
1
𝑘=2
(𝒉, 𝒌) = (𝟎, 𝟐) Center of curvature

8. 𝑦 = 𝑙𝑛𝑥 𝑎𝑡 (1, 0) Solve for (ℎ, 𝑘)

𝑦 ′[1 + (𝑦 ′)2 ] 1 + (𝑦 ′ )2
ℎ=𝑥− 𝑘 =𝑦+
𝑦 ′′ 𝑦′′
1 Substitute 𝑥 = 1
𝑦′ =
𝑥
1
𝑦′ = = 1
1
1 Substitute 𝑥 = 1
𝑦 ′′ = − 2
𝑥
′′
1
𝑦 =− = −1
(1)2
𝑦 ′[1 + (𝑦 ′)2 ] Substitute 𝑥, 𝑦 ′, 𝑎𝑛𝑑 𝑦 ′′
ℎ=𝑥−
𝑦 ′′
1[1 + (1)2 ]
ℎ=1−
−1
2
ℎ=1+
1
ℎ=3
1 + (𝑦 ′ )2 Substitute 𝑦, 𝑦 ′, 𝑎𝑛𝑑 𝑦 ′′
𝑘 =𝑦+
𝑦′′
1 + (1)2
𝑘 =0+
−1
𝑘 = −2
(𝒉, 𝒌) = (𝟑, −𝟐) Center of curvature

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LEARNING ASSESSMENT

ASSESSMENT 10
First semester SY 2020 – 2021

NAME
PROGRAM

General Instructions: Solve the following problem neatly and completely. Show your complete
solution and simplify whenever possible. Box your final answers. Answers/Solutions must be
handwritten in a bond paper.

𝑑𝑦
Find 𝑑𝑥 by use of differentials
𝑦
1. 2 ln(𝑥 2 + 𝑦 2 ) 𝐴𝑟𝑐𝑡𝑎𝑛
𝑥
2. 𝑒 𝑦 = sin (𝑥 − 𝑦)
3
3. Find the approximate value of √215 by use of differentials
2
4. Find the approximate value of (63.4)3 by use of differentials
5. If an error of 1.5% is made in measuring the side of an equilateral triangle, find the percentage
error made in the computed area.
6. In a right circular cone, the radius of the base is half as long as the altitude. If an error of 2%
is made in measuring the radius, find the percentage error made in the computed volume.
7. The diameter of a circle is to be measured and its area computed. If the diameter can be
measured with a maximum error of 0.002 cm and the allowable error in the area is 0.01 sq cm,
find the diameter of the largest circle for which the specifications are met.
8. A circular hole 4 inches in diameter and 1 foot deep in a block of iron is drilled out to increase
its diameter to 4.1 in. Find the approximate volume of the metal removed.
𝑑𝑦 𝑑2 𝑦
Find 𝑑𝑥 and 𝑑𝑥2 and simplify whenever possible
9. 𝑥 = 𝑐𝑜𝑠∅ + ∅𝑠𝑖𝑛∅, 𝑦 = 𝑠𝑖𝑛∅ − ∅𝑐𝑜𝑠∅
10. 𝑥 = 1 − 𝑙𝑛𝑡, 𝑦 = 𝑡 − 𝑙𝑛𝑡
Find the radius of curvature at the given point
𝜋
11. 𝑥 = 2𝑠𝑖𝑛𝑡, 𝑦 = 𝑐𝑜𝑠2𝑡 𝑎𝑡 𝑡 =
6
12. 𝑦 2 = 4𝑥 𝑎𝑡 (1, 2)
8
13. 𝑥 = 2 𝑎𝑡 (2, 0)
𝑦 +4
Find the center of curvature at the point indicated
𝜋
14. 𝑦 = 𝑠𝑖𝑛𝑥 𝑎𝑡 ( , 1)
2
1 𝜋 1
15. 𝑦 = 𝑡𝑎𝑛2𝑥 𝑎𝑡 ( , )
2 8 2

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Partial Differentiation

Learning Objectives

▪ Find and use partial derivatives of a function of two variables

▪ Find and use partial derivatives of a function of three or more variables
▪ Find higher-order partial derivatives of a function of two or three
variables
▪ Find partial derivatives implicitly

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For this week, April 19-23, 2021 of this semester, the following shall be your guide for the different
lessons and tasks that you need to accomplish. Be patient read it carefully before proceeding to the
tasks expected of you.

Date Topic Activities/Tasks

Read Lessons
April 19-22, 2021 Partial Differentiation

Accomplish the worksheet in the

Submission of learning tasks
April 23, 2021 Assessment Portion of this
module

GOD BLESS!

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WEEK 12: Partial Differentiation

Partial Derivative

Let 𝑧 = 𝑓(𝑥, 𝑦) be a function of two independent variables 𝑥 and 𝑦. If 𝑦 is held constant, then 𝑧
becomes temporarily a function of the single variable 𝑥. From this point of view, we can compute
the derivative of 𝑧 with respect to 𝑥 by employing the rules for ordinary differentiation of functions
with single variable. The derivative found in this manner is called the partial derivative of 𝑧 with
respect to 𝑥 and the process involved is called partial differentiation. The derivative of 𝑧 with
respect to 𝑥 is denoted by any of the following symbols:
𝜕𝑧 𝜕𝑓
, , 𝑧 , 𝑓 (𝑥, 𝑦) , 𝑓𝑥
𝜕𝑥 𝜕𝑥 𝑥 𝑥
Similarly, if 𝑥 is held constant, then 𝑧 becomes temporarily a function of 𝑦. As a result, we can
compute the partial derivative of 𝑧 with respect to 𝑦 and this derivative may be denoted by any of
the following symbols:
𝜕𝑧 𝜕𝑓
, , 𝑧 , 𝑓 (𝑥, 𝑦) , 𝑓𝑦
𝜕𝑦 𝜕𝑦 𝑦 𝑦
𝜕𝑧 𝜕𝑧
It should be noted that the symbol 𝜕𝑥 (𝑜𝑟 ) cannot be thought of as a fraction since neither of
𝜕𝑦
𝜕
the symbols 𝜕𝑧 and 𝜕𝑥 (𝑜𝑟 𝜕𝑧 𝑎𝑛𝑑 𝜕𝑦) has a separate meaning. The symbol alone means to
𝜕𝑥
𝜕
differentiate partially with respect to 𝑥 whatever follows it. The symbol is interpreted in like
𝜕𝑦
manner.

It should be noted that before performing any partial differentiation of functions of several
variable, it is important to know first which of the variables are considered or held constants.

Partial Derivatives of Higher Order

Continued differentiation of two (or more) variables gives rise to partial derivatives of higher
𝜕𝑧
order. If 𝑧 = 𝑓(𝑥, 𝑦), the 𝜕𝑥 is the first partial derivative of 𝑧 with respect to 𝑥. Then the partial
𝜕𝑧
derivative of 𝜕𝑥 with respect to 𝑥 (𝑜𝑟 𝑦) is called the second partial derivative of 𝑧. The partial
𝜕𝑧
derivative of 𝜕𝑦 with respect to 𝑦 (𝑜𝑟 𝑥) is also called the second partial derivative of 𝑧. Hence,
for a function defined by 𝑧 = 𝑓 (𝑥, 𝑦), the following notations for the second derivatives are used:
𝜕 𝜕𝑧 𝜕2𝑧 𝜕 2𝑓
( ) = 2 = 2 = 𝑧𝑥𝑥 = 𝑓𝑥𝑥
𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥
𝜕 𝜕𝑧 𝜕2𝑥 𝜕 2𝑓
( )= = = 𝑧𝑥𝑦 = 𝑓𝑥𝑦
𝜕𝑦 𝜕𝑥 𝜕𝑦𝜕𝑥 𝜕𝑦𝜕𝑥
𝜕 𝜕𝑧 𝜕2𝑧 𝜕 2𝑓
( )= = = 𝑧𝑦𝑥 = 𝑓𝑦𝑥
𝜕𝑥 𝜕𝑦 𝜕𝑥𝜕𝑦 𝜕𝑥𝜕𝑦
𝜕 𝜕𝑧 𝜕2𝑧 𝜕2𝑓
( ) = 2 = 2 = 𝑧𝑦𝑦 = 𝑓𝑦𝑦
𝜕𝑦 𝜕𝑦 𝜕𝑦 𝜕𝑦

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 𝜕2 𝑧
Notice that the symbol means that 𝑧 is differentiated first with respect to 𝑥 and then with
𝜕𝑦𝜕𝑥
respect to 𝑦. The symbol 𝑧𝑥𝑦 means the same thing. However, observe that the subscripts
(𝑥 𝑎𝑛𝑑 𝑦) are written in the reverse order. That is, in the subscript notation, the partial
differentiations are to be performed in the order in which the subscripts are written, starting from
left to right. For example, 𝑧𝑥𝑦𝑦 means to differentiate 𝑧 with respect to 𝑥, next with respect to 𝑦
and then with respect to 𝑦 again. Note that
𝜕3𝑧
𝑧𝑥𝑦𝑦 = 2
𝜕𝑦 𝜕𝑥
If the partial derivatives of 𝑧 = 𝑓(𝑥, 𝑦) are continuous functions, then it can be shown that
𝑧𝑥𝑦 = 𝑧𝑦𝑥
The relation above says that it is immaterial whether we differentiate first with respect to 𝑥 and
then with respect to 𝑦 or the other way around. Hence, for continuous partial derivatives, 𝑧 has
only three distinct second partial derivatives, namely
𝑧𝑥𝑥 , 𝑧𝑥𝑦 = 𝑧𝑦𝑥 , 𝑎𝑛𝑑 𝑧𝑦𝑦

Total Differentiation

We define the total differential 𝑑𝑧 of a function 𝑧 = 𝑓(𝑥, 𝑦) by

𝜕𝑧 𝜕𝑧
𝑑𝑧 = 𝑑𝑥 + 𝑑𝑦
𝜕𝑥 𝜕𝑦
A function of more than two independent variables is defined in the same manner. Thus if 𝑢 =
𝑓 (𝑥, 𝑦, 𝑧), then its total differential 𝑑𝑢 is
𝜕𝑢 𝜕𝑢 𝜕𝑢
𝑑𝑢 = 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧
𝜕𝑥 𝜕𝑦 𝜕𝑧

Total Derivative

Let a function be defined by the equation

𝑧 = 𝑓 (𝑥, 𝑦) 1
Where 𝑥 and 𝑦 are both functions of another variable 𝑡, that is
𝑥 = 𝑔 (𝑡 ) , 𝑦 = ℎ (𝑡 ) 2
When the values of (2) are substituted in (1), 𝑧 becomes a function of one variable (i.e., t) and its
𝑑𝑧
derivative can be found in the usual manner. However, in some instances, that method appears
𝑑𝑡
more difficult to perform. An alternative solution will be given here.

We know from the preceding section that the total differential of 𝑧 = 𝑓(𝑥, 𝑦) is
𝜕𝑧 𝜕𝑧
𝑑𝑧 = 𝑑𝑥 + 𝑑𝑦
𝜕𝑥 𝜕𝑦
Now dividing both sides or members of the equation above by 𝑑𝑡, we get
𝑑𝑧 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦
= ∙ + ∙
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡

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 𝑑𝑧 𝑑𝑧
The symbol 𝑑𝑡 is called the total derivative of 𝑧 with respect to 𝑡. If t represents the time, then ,
𝑑𝑡
𝑑𝑥 𝑑𝑦
, and are the time rates of change of 𝑧, 𝑥, 𝑎𝑛𝑑 𝑦 respectively.
𝑑𝑡 𝑑𝑡

If 𝑢 = 𝑓(𝑥, 𝑦, 𝑧), the total derivative of 𝑢 with respect to 𝑡 is

𝑑𝑢 𝜕𝑢 𝑑𝑥 𝜕𝑢 𝑑𝑦 𝜕𝑢 𝑑𝑧
= ∙ + ∙ + ∙
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡 𝜕𝑧 𝑑𝑡
If we let 𝑡 = 𝑥, then we have the relation
𝑑𝑧 𝜕𝑧 𝜕𝑧 𝑑𝑦
= + ∙
𝑑𝑥 𝜕𝑥 𝜕𝑦 𝑑𝑥
𝑑𝑧
The symbol 𝑑𝑥 is called the total derivative of 𝑧 with respect to 𝑥. If 𝑢 = 𝑓(𝑥, 𝑦, 𝑧), the total
derivative of 𝑢 with respect to 𝑥 is given by
𝑑𝑢 𝜕𝑢 𝜕𝑢 𝑑𝑦 𝜕𝑢 𝑑𝑧
= + ∙ + ∙
𝑑𝑥 𝜕𝑥 𝜕𝑦 𝑑𝑥 𝜕𝑧 𝑑𝑥

Partial Differentiation of Implicit Functions

𝑑𝑦
Last semester, we learned how to find 𝑑𝑥 of an implicit function defined by the equation
𝑓 (𝑥, 𝑦) = 0 (1)
𝑑𝑦
Now, we shall employ partial differentiation to find 𝑑𝑥 of the equation above. For this purpose, we
𝑑𝑦
shall develop a formula for 𝑑𝑥 in terms of partial derivatives.

Suppose we have an equation of the form

𝑧 = 𝑓 (𝑥, 𝑦) (2)
Then
𝑑𝑧 𝜕𝑓 𝜕𝑓 𝑑𝑦
= + ∙ (3)
𝑑𝑥 𝜕𝑥 𝜕𝑦 𝑑𝑥
𝑑𝑧
If the values of 𝑥 and y in (2) will satisfy (1), then 𝑧 = 0 and consequently, = 0. Hence (3)
𝑑𝑥
becomes
𝜕𝑓 𝜕𝑓 𝑑𝑦
+ =0 (4)
𝜕𝑥 𝜕𝑦 𝑑𝑥
𝑑𝑦
Solving for 𝑑𝑥 from (4), we get
𝜕𝑓
𝑑𝑦 − 𝜕𝑥 𝜕𝑓
= , ≠0
𝑑𝑥 𝜕𝑓 𝜕𝑦
𝜕𝑦

This document is a property of University of Saint Louis – Tuguegarao. It must not be reproduced or transmitted in any form, in whole or in part, without expressed written permission.
Examples

𝜕𝑢 𝜕𝑢 𝜕𝑢 To solve the problem, treat other variables as

Find 𝜕𝑥 , 𝜕𝑦 , 𝑎𝑛𝑑 𝜕𝑧 constant
1. 𝑢 = 𝑥 2 + 𝑦 2 + 𝑧 2
Note: Derivative of a constant is zero
𝜕𝑢 𝜕 2 Treat 𝑦 and 𝑧 as constants
= (𝑥 + 𝑦 2 + 𝑧 2 )
𝜕𝑥 𝜕𝑥
𝝏𝒖
= 𝟐𝒙
𝝏𝒙
𝜕𝑢 𝜕 2 Treat 𝑥 and 𝑧 as constants
= (𝑥 + 𝑦 2 + 𝑧 2 )
𝜕𝑦 𝜕𝑦
𝝏𝒖
= 𝟐𝒚
𝝏𝒚
𝜕𝑢 𝜕 2 Treat 𝑥 and 𝑦 as constants
= (𝑥 + 𝑦 2 + 𝑧 2 )
𝜕𝑧 𝜕𝑧
𝝏𝒖
= 𝟐𝒛
𝝏𝒛
𝜕𝑢 𝜕𝑢 𝜕𝑢 To solve the problem, treat other variables as
Find 𝜕𝑥 , 𝜕𝑦 , 𝑎𝑛𝑑 𝜕𝑧 constant
2. 𝑢 = 𝑥𝑙𝑛𝑦 + 𝑦𝑙𝑛𝑧 + 𝑧𝑙𝑛𝑥 Note: Derivative of a constant is zero
𝜕𝑢 𝜕 Treat 𝑦 and 𝑧 as constants
= (𝑥𝑙𝑛𝑦 + 𝑦𝑙𝑛𝑧 + 𝑧𝑙𝑛𝑥 ) 𝑑 1 𝑑𝑢
𝜕𝑥 𝜕𝑥 (ln 𝑢) =
𝑑𝑥 𝑢 𝑑𝑥
𝝏𝒖 𝒛
= 𝒍𝒏𝒚 +
𝝏𝒙 𝒙
𝜕𝑢 𝜕 Treat 𝑥 and 𝑧 as constants
= (𝑥𝑙𝑛𝑦 + 𝑦𝑙𝑛𝑧 + 𝑧𝑙𝑛𝑥 )
𝜕𝑦 𝜕𝑦
𝝏𝒖 𝒙
= + 𝒍𝒏𝒛
𝝏𝒚 𝒚
𝜕𝑢 𝜕 Treat 𝑥 and 𝑦 as constants
= (𝑥𝑙𝑛𝑦 + 𝑦𝑙𝑛𝑧 + 𝑧𝑙𝑛𝑥 )
𝜕𝑧 𝜕𝑧
𝝏𝒖 𝒚
= + 𝒍𝒏𝒙
𝝏𝒛 𝒛

𝜕2 𝑧 𝜕2𝑧 𝜕2 𝑧 To solve the problem, treat other variables as

Find 𝜕𝑥2 , 𝜕𝑦𝜕𝑥 , 𝑎𝑛𝑑 𝜕𝑦 2 constant
3. 𝑧 = 𝑥 2 𝑠𝑖𝑛𝑦 + 𝑦 2 𝑐𝑜𝑠𝑥 Note: Derivative of a constant is zero
𝜕𝑧 𝜕 2 Treat 𝑦 as constant
= (𝑥 𝑠𝑖𝑛𝑦 + 𝑦 2 𝑐𝑜𝑠𝑥)
𝜕𝑥 𝜕𝑥
𝜕𝑧
= 2𝑥𝑠𝑖𝑛𝑦 − 𝑦 2 𝑠𝑖𝑛𝑥
𝜕𝑥
𝜕2𝑧 𝜕 Treat 𝑦 as constant
2
= (2𝑥𝑠𝑖𝑛𝑦 − 𝑦 2 𝑠𝑖𝑛𝑥)
𝜕𝑥 𝜕𝑥

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 𝝏𝟐 𝒛
= 𝟐𝒔𝒊𝒏𝒚 − 𝒚𝟐 𝒄𝒐𝒔𝒙
𝝏𝒙𝟐
𝜕2𝑧 𝜕 Treat 𝑥 as constant
= (2𝑥𝑠𝑖𝑛𝑦 − 𝑦 2 𝑠𝑖𝑛𝑥)
𝜕𝑦𝜕𝑥 𝜕𝑦
𝝏𝟐 𝒛
= 𝟐𝒙𝒄𝒐𝒔𝒚 − 𝟐𝒚𝒔𝒊𝒏𝒙
𝝏𝒚𝝏𝒙
𝜕𝑧 𝜕 2 Treat 𝑥 as constant
= (𝑥 𝑠𝑖𝑛𝑦 + 𝑦 2 𝑐𝑜𝑠𝑥)
𝜕𝑦 𝜕𝑦
𝜕𝑧
= 𝑥 2 𝑐𝑜𝑠𝑦 + 2𝑦𝑐𝑜𝑠𝑥
𝜕𝑦
𝜕2𝑧 𝜕 2 Treat 𝑥 as constant
2
= (𝑥 𝑐𝑜𝑠𝑦 + 2𝑦𝑐𝑜𝑠𝑥)
𝜕𝑦 𝜕𝑦
𝝏𝟐 𝒛
= −𝒙𝟐 𝒔𝒊𝒏𝒚 + 𝟐𝒄𝒐𝒔𝒙
𝝏𝒚𝟐

Find 𝑧𝑥𝑦𝑦 Differentiate z with respect to x, next with

3. 𝑧 = 𝑥𝑒 2𝑦 + 𝑦𝑒 2𝑥 respect to y and then with respect to y again
𝑧𝑥 = 𝑒 2𝑦 + 2𝑦𝑒 2𝑥 In z, treat y as constant
𝑧𝑥𝑦 = 2𝑒 2𝑦 + 2𝑒 2𝑥 In 𝑧𝑥 , treat x as constant
𝒛𝒙𝒚𝒚 = 𝟒𝒆𝟐𝒚 In 𝑧𝑥𝑦 , treat x as constant

Find 𝑧𝑥𝑦𝑥 Differentiate z with respect to x, next with

4. 𝑧 = (𝑥 + 𝑦)3 respect to y and then with respect to x again
𝑧𝑥 = 3(𝑥 + 𝑦)2 In z, treat y as constant
𝑧𝑥𝑦 = 6(𝑥 + 𝑦) = 6𝑥 + 6𝑦 In 𝑧𝑥 , treat x as constant
𝒛𝒙𝒚𝒙 = 𝟔 In 𝑧𝑥𝑦 , treat y as constant

5. Find the total differential of 𝜕𝑧 𝜕𝑧

𝑑𝑧 = 𝑑𝑥 + 𝑑𝑦
𝑧 = √𝑥 2 + 𝑦 2 𝜕𝑥 𝜕𝑦
1
𝑧 = (𝑥 2 + 𝑦 2 )2
𝜕𝑧 𝜕𝑧
𝑑𝑧 = 𝑑𝑥 + 𝑑𝑦
𝜕𝑥 𝜕𝑦
1 1 1 1 Simplify
𝑑𝑧 = (𝑥 2 + 𝑦 2 )−2 (2𝑥 )𝑑𝑥 + (𝑥 2 + 𝑦 2 )−2 (2𝑦)𝑑𝑦
2 2 Cancel out common terms
𝑥 𝑦
𝑑𝑧 = 1+ 1
(𝑥 + 𝑦 )2 (𝑥 2 + 𝑦 2 )2
2 2
𝒙+𝒚
𝒅𝒛 =
√𝒙𝟐 + 𝒚𝟐

6. Each edge of a rectangular tank is increased by 0.5 ft. if originally, the height is 4 ft and the
base is a square of side 2 ft, find the approximate increase in the volume.

This document is a property of University of Saint Louis – Tuguegarao. It must not be reproduced or transmitted in any form, in whole or in part, without expressed written permission.
 𝑠 = 2 𝑓𝑡 ℎ = 4 𝑓𝑡 Apply the concepts of total differentiation to
𝑑𝑠 = 𝑑ℎ = 0.5 𝑓𝑡 𝑑𝑉 =? solve the problem
𝑉 = 𝑠 2ℎ
𝜕𝑉 𝜕𝑉
𝑑𝑉 = 𝑑𝑠 + 𝑑ℎ
𝜕𝑠 𝜕ℎ
𝑑𝑉 = 2𝑠ℎ𝑑𝑠 + 𝑠 2 𝑑ℎ Substitute the given values
𝑑𝑉 = 2(2)(4)(0.5) + (2)2 (0.5)
𝒅𝑽 = 𝟏𝟎 𝒇𝒕𝟑

7. Find the relative error in the computed volume of a right circular cone if there is an error of
one percent in measuring the base radius and altitude of the cone.
𝑑𝑟 𝑑ℎ Apply the concepts of total differentiation to
(100) = 1% (100) = 1%
𝑟 ℎ solve the problem
𝑑𝑉 𝑑𝑉 1
𝑅𝐸 = = 𝑉𝑐𝑜𝑛𝑒 = 𝜋𝑟 2 ℎ
𝑉 1 2 3
𝜋𝑟 ℎ
3
𝜕𝑉 𝜕𝑉
𝑑𝑟 + 𝑑ℎ
𝑅𝐸 = 𝜕𝑟 𝜕ℎ
1 2
3 𝜋𝑟 ℎ
2 1 Factor out
𝜋𝑟ℎ𝑑𝑟 + 3 𝜋𝑟 2 𝑑ℎ
𝑅𝐸 = 3 1
1 2 𝜋𝑟
3
3 𝜋𝑟 ℎ
1 Simplify
𝜋𝑟[2ℎ𝑑𝑟 + 𝑟𝑑ℎ]
𝑅𝐸 = 3
1
3 𝜋𝑟(𝑟ℎ)
2ℎ𝑑𝑟 + 𝑟𝑑ℎ Separate the numerator
𝑅𝐸 =
𝑟ℎ
2ℎ𝑑𝑟 𝑟𝑑ℎ Cancel out common terms
𝑅𝐸 = +
𝑟ℎ 𝑟ℎ
2𝑑𝑟 𝑑ℎ 𝑑𝑟 𝑑ℎ
𝑅𝐸 = + = 0.01 = 0.01
𝑟 ℎ 𝑟 ℎ
𝑅𝐸 = 2(0.01) + 0.01
𝑹𝑬 = 𝟎. 𝟎𝟑

8. Find the total derivative of 𝑧 = 𝑥𝑦 with respect to t and in terms of t when 𝑥 = 𝑠𝑖𝑛𝑡, 𝑦 = 𝑐𝑜𝑠𝑡
𝑧 = 𝑥𝑦 𝑥 = 𝑠𝑖𝑛𝑡, 𝑦 = 𝑐𝑜𝑠𝑡 Apply the concepts of total derivative to solve
the problem
𝜕𝑧 𝑑𝑥
= 𝑦 = 𝑐𝑜𝑠𝑡 = 𝑐𝑜𝑠𝑡
𝜕𝑥 𝑑𝑡
𝜕𝑧 𝑑𝑦
= 𝑥 = 𝑠𝑖𝑛𝑡 = −𝑠𝑖𝑛𝑡
𝜕𝑦 𝑑𝑡
𝑑𝑧 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦 Substitute
= ∙ + ∙
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡

This document is a property of University of Saint Louis – Tuguegarao. It must not be reproduced or transmitted in any form, in whole or in part, without expressed written permission.
 𝑑𝑧
= 𝑐𝑜𝑠𝑡 ∙ 𝑐𝑜𝑠𝑡 + 𝑠𝑖𝑛𝑡 ∙ −𝑠𝑖𝑛𝑡
𝑑𝑡
𝑑𝑧 𝑐𝑜𝑠 2 𝑡 − 𝑠𝑖𝑛2 𝑡 = 𝑐𝑜𝑠2𝑡
= 𝑐𝑜𝑠 2 𝑡 − 𝑠𝑖𝑛2 𝑡
𝑑𝑡
𝒅𝒛
= 𝒄𝒐𝒔𝟐𝒕
𝒅𝒕

9. Find the total derivative of 𝑢 = 𝑥𝑙𝑛𝑦 + 𝑧𝑙𝑛𝑥 + 𝑦𝑙𝑛𝑧 with respect to t and in terms of t when
𝑥 = 𝑡, 𝑦 = 𝑡 2 , 𝑧 = 4𝑡
𝑢 = 𝑥𝑙𝑛𝑦 + 𝑧𝑙𝑛𝑥 + 𝑦𝑙𝑛𝑧 Apply the concepts of total derivative to solve
2
𝑥 = 𝑡, 𝑦 = 𝑡 , 𝑧 = 4𝑡 the problem
𝜕𝑢 𝑧 4𝑡 𝑑𝑥
= 𝑙𝑛𝑦 + = 𝑙𝑛𝑡 2 + = 𝑙𝑛𝑡 2 + 4 =1
𝜕𝑥 𝑥 𝑡 𝑑𝑡
𝜕𝑢 𝑥 𝑡 1 𝑑𝑦
= + 𝑙𝑛𝑧 = 2 + 𝑙𝑛4𝑡 = + 𝑙𝑛4𝑡 = 2𝑡
𝜕𝑦 𝑦 𝑡 𝑡 𝑑𝑡
𝜕𝑢 𝑦 𝑡2 𝑡 𝑑𝑧
= 𝑙𝑛𝑥 + = 𝑙𝑛𝑡 + = 𝑙𝑛𝑡 + =4
𝜕𝑧 𝑧 4𝑡 4 𝑑𝑡
𝑑𝑢 𝜕𝑢 𝑑𝑥 𝜕𝑢 𝑑𝑦 𝜕𝑢 𝑑𝑧 Substitute
= ∙ + ∙ + ∙
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡 𝜕𝑧 𝑑𝑡
𝑑𝑢 1 𝑡
= (𝑙𝑛𝑡 2 + 4)(1) + ( + 𝑙𝑛4𝑡) (2𝑡) + (𝑙𝑛𝑡 + ) (4)
𝑑𝑡 𝑡 4
𝑑𝑢 Simplify
= 𝑙𝑛𝑡 2 + 4 + 2 + 2𝑡𝑙𝑛4𝑡 + 4𝑙𝑛𝑡 + 𝑡
𝑑𝑡
𝑑𝑢
= 2𝑙𝑛𝑡 + 6 + 2𝑡𝑙𝑛4𝑡 + 4𝑙𝑛𝑡 + 𝑡
𝑑𝑡
𝒅𝒖
= 𝟔𝒍𝒏𝒕 + 𝟔 + 𝟐𝒕𝒍𝒏𝟒𝒕 + 𝒕
𝒅𝒕
𝑑𝑦 Apply concepts of partial differentiation of
10. Find 𝑑𝑥
implicit functions
𝑥 3 + 𝑦 3 − 3𝑥𝑦 = 0
𝜕𝑓
𝑑𝑦 − 𝜕𝑥
=
𝑑𝑥 𝜕𝑓
𝜕𝑦
𝜕𝑓
= 3𝑥 2 − 3𝑦
𝜕𝑥
𝜕𝑓
= 3𝑦 2 − 3𝑥
𝜕𝑦
𝑑𝑦 3𝑥 2 − 3𝑦 Factor out 3
=− 2
𝑑𝑥 3𝑦 − 3𝑥
𝑑𝑦 3(𝑥 2 − 𝑦) Simplify
=−
𝑑𝑥 3(𝑦 2 − 𝑥)
𝒅𝒚 𝒙𝟐 − 𝒚
=− 𝟐
𝒅𝒙 𝒚 −𝒙

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 𝑑𝑦 Apply concepts of partial differentiation of
11. Find 𝑑𝑥
implicit functions
𝑥𝑙𝑛𝑦 + 𝑦𝑙𝑛𝑥 = 0
𝜕𝑓
𝑑𝑦 − 𝜕𝑥
=
𝑑𝑥 𝜕𝑓
𝜕𝑦
𝜕𝑓 𝑦
= 𝑙𝑛𝑦 +
𝜕𝑥 𝑥
𝜕𝑓 𝑥
= + 𝑙𝑛𝑥
𝜕𝑦 𝑦
𝑦 Simplify
𝑑𝑦 𝑙𝑛𝑦 + 𝑥
= −𝑥
𝑑𝑥
𝑦 + 𝑙𝑛𝑥
𝑥𝑙𝑛𝑦 + 𝑦
𝑑𝑦 𝑥
=−
𝑑𝑥 𝑥 + 𝑦𝑙𝑛𝑥
𝑦
𝒅𝒚 𝒚(𝒙𝒍𝒏𝒚 + 𝒚)
=−
𝒅𝒙 𝒙(𝒙 + 𝒚𝒍𝒏𝒙)

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References
Printed References
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Asia Pte Ltd

Electronic Resources
1. Pulham , John. Differential Calculus. Retrieved August 7, 2020 from http://
www.maths.abdn.ac.uk/igc/tch/math1002/dif
2. Weistein, Eric. Differential Calculus. Retrieved August 7, 2020 from
http://mathworld.wolfram.com/differential calculus
3. Thomas , Christopher . Introduction to Differential Calculus. Retrieved August 7, 2020
from http:// www.usyd.edu.aav/stuserv/document/maths_learning-center
4. Purplemath. Retrieved August 7, 2020 from http://www.purplemath.com/modules/
5. Larson, R & Edwards, B. (2018). Calculus 11th Ed. Boston, MA USA. Cengage Learning

Learning Materials
1. Worksheet
2. Module

This document is a property of University of Saint Louis – Tuguegarao. It must not be reproduced or transmitted in any form, in whole or in part, without expressed written permission.
LEARNING ASSESSMENT

ASSESSMENT 12
First semester SY 2020 – 2021

NAME
PROGRAM

General Instructions: Solve the following problem neatly and completely. Show your complete
solution and simplify whenever possible. Box your final answers. Answers/Solutions must be
handwritten in a bond paper.

𝜕𝑢 𝜕𝑢 𝜕𝑢
Find 𝜕𝑥 , 𝜕𝑦 , 𝑎𝑛𝑑 𝜕𝑧
𝑦 𝑧 𝑥
1. 𝑢 = + +
𝑧 𝑥 𝑦
2. 𝑢 = 𝑦𝑧 + 𝑥𝑧 + 𝑥𝑦
𝜕2 𝑧 𝜕2𝑧 𝜕2 𝑧
Find 𝜕𝑥2 , 𝜕𝑦𝜕𝑥 , 𝑎𝑛𝑑 𝜕𝑦 2
3. 𝑧 = 𝑙𝑛√𝑥 2 + 𝑦 2
4. 𝑧 = 𝑥𝑒 𝑥𝑦
Show that 𝑧𝑥𝑥 + 𝑧𝑦𝑦 = 0
5. 𝑧 = ln (𝑥 2 + 𝑦 2 )
6. 𝑧 = 𝑒 −𝑥 sin (𝑦 + 𝑎)
Verify that 𝑧𝑥𝑦 = 𝑧𝑦𝑥
𝑥+𝑦
7. 𝑧 =
𝑥−𝑦
8. 𝑧 = ln(𝑥𝑦)3
𝑥
9. Find the total differential of 𝑧 = 𝐴𝑟𝑐𝑡𝑎𝑛 𝑦
10. If the hypotenuse of one side of a right triangle are 13 meters and 5 meters, respectively,
find the approximate increase in the other side if each of the given sides is increased by 0.02
meter, the triangle being kept a right triangle.
11. Find the total derivative of 𝑧 = sin (𝑥𝑦) with respect to t and in terms of t when
𝑥 = 𝑡, 𝑦 = 𝑒 𝑡
12. Find the total derivative of 𝑢 = 𝑥𝑦 + 𝑥𝑧 + 𝑦𝑧 with respect to t and in terms of t when
𝑥 = 𝑡 2, 𝑦 = 𝑡 3, 𝑧 = 𝑡 4
13. The radius of a right circular cylinder decreases at the rate of 2 in/sec and the height increases
at the rate of 4 in/sec. Find the rate and change of the volume when the radius is 8 in and the
altitude is 12 in.
𝑑𝑦 𝑦
14. Find when 2𝐴𝑟𝑐𝑡𝑎𝑛 = ln(𝑥 2 + 𝑦 2 )
𝑑𝑥 𝑥
𝑑𝑦 𝑥𝑦
15. Find 𝑑𝑥 when 𝑒 + 𝑠𝑖𝑛𝑥𝑦 = 0

This document is a property of University of Saint Louis – Tuguegarao. It must not be reproduced or transmitted in any form, in whole or in part, without expressed written permission.
 UNIVERSITY OF SAINT LOUIS
Tuguegarao City

SCHOOL OF ENGINEERING, ARCHITECTURE & IT EDUCATION

Engineering Sciences Department
S.Y. 2020 – 2021

CORRESPONDENCE LEARNING MODALITY

CALC 1024: Calculus 2 (Integral Calculus)

Prepared by:

EROVITA TERESITA BACUD AGUSTIN, DME

ENGR. NORMAN BOB GOMEZ
ENGR. JENIROSE B. MATIAS
ENGR. PEEJAY PAGUIRIGAN
Instructors

Reviewed by:

ENGR. VICTOR VILLALUZ, MEE

EE Program Chair / ES Department Head

Recommended by:

ENGR. RONALD M. VILLAVERDE, ME – ECE, PECE

Academic Dean

Approved by:

EMMANUEL JAMES P. PATTAGUAN, Ph.D.

VP for Academics

WEEK 13 CALC 1024: CALCULUS 2

 Weekly Study and Assessment Guide
For this week, May 5 – 9, 2021 of this semester, the following shall be your guide for the different lessons and
tasks that you need to accomplish. Be patient read it carefully before proceeding to the tasks expected of you.

Date Topic Activities/Tasks

Riemann Sum
Read Lessons
May 5 – 9, 2021 Area Under the Curve
Area Between Two Curves

GOD BLESS!

WEEK 13: Riemann Sum, Area Under the Curve &Area Between Two Curves

LEARNING OUTCOMES
At the end of the lesson, the students should be able to:
1. Find the area of the region using Riemann Sum approximation;
2. Find the area of the region under the curve using integration;
3. Find the area of the region between two curves using integration; and
4. Find the area of the region between intersecting curves using integration.

Approximating the Area of a Region Using Riemann Sums

To calculate the area of geometric figures such as rectangles, squares, triangles, and circles, you are guided by
specific formulas.

In this particular lesson, you will be introduced to several problems that involve that finding the area of an
irregular figure or a region that is bounded by curves, such as the shaded region within the closed interval [2, 5]
shown in the figure.

The area of this enclosed region bounded by definite lower and upper intervals along the curve and the x – axis
can easily be obtained.

Reference: https://www.desmos.com/calculator/tgyr42ezjq
WEEK 13 CALC 1024: CALCULUS 2
Let us approximate areas such as this using the Riemann sums. As illustrated, the interval [2, 5] is divided into
six (6) rectangles of width 0.5. This width is denoted as Δx.

b−a
∆x =
n

where b is the upper limit

a is the lower limit
n is the number of rectangles or subintervals

b−a 5−2
∆x = = = 0.50
n 6

Reference: https://www.desmos.com/calculator/tgyr42ezjq

Labeling the endpoints of each rectangle as x1, x2, x3, …and so on, the area of the shaded region can be expressed
using the summation
6

∑ f(xi )∆x = f(x1 )∆x + f(x2 )∆x + f(x3 )∆x + f(x4 )∆x + f(x5 )∆x + f(x6 )∆x.
i=1

Generally, to approximate the area under a curve above the x – axis and bounded by the interval [a, b], divide
the interval into n rectangles and then find the summation of the areas of the rectangles. In symbol,
n

∑ f(xi )∆x.
i=1
This is called the Riemann Sum. The Riemann sum is an approximation of the actual area under the curve. To
evaluate the Riemann Sum, you can either use the left endpoints or the right endpoints to be the values of xi.

WEEK 13 CALC 1024: CALCULUS 2

Left – Hand and Right – Hand Riemann Sums
If a region under the curve is defined by a continuous function f(x) and bounded by the interval [a, b] such that
𝑎 ≤ 𝑏, then the Riemann sum that approximates the area of this region may be determined by dividing the
interval [a, b] into subintervals and using the summation
n

∑ f(xi )∆x.
i=1
b−a
where ∆x = and xi are the endpoints of the subinterval.
n

The left – hand Riemann sum L approximates the area of the said region using the left endpoints of the subintervals
while the right – hand Riemann sum R approximates the area using the right endpoints.

Right – Hand Approximation Left – Hand Approximation

EXAMPLE 1
Compute the Left – hand and Right – hand Riemann sum approximations of the area under the curve f(x) = x 2
bounded by the x – axis and the lines x = 1 and x = 5 using four rectangles.

Because the area is defined over the interval [1, 5], then a = 1 and b = 5. So,
b−a 5−1 4
∆x = = = =1
n 4 4

The width of each rectangle is 1 unit. Thus, the interval [1, 5] is divided into the following subintervals:
[1, 2], [2, 3], [3, 4], [4, 5].

To get the left – hand approximation of the area, take the left endpoints, i.e. x1 = 1, x2 = 2, x3 = 3 and x4 = 4, and
evaluate the function at each. Then
f(x1 ) = f(1) = 12 = 1
f(x2 ) = f(2) = 22 = 4
f(x3 ) = f(3) = 32 = 9
f(x4 ) = f(4) = 42 = 16
WEEK 13 CALC 1024: CALCULUS 2
 4

L = ∑ f(xi )∆x = f(x1 )∆x + f(x2 )∆x + f(x3 )∆x + f(x4 )∆x.
i=1
4

L = ∑ f(xi )∆x = (1 ∗ 1) + (4 ∗ 1) + (9 ∗ 1) + (16 ∗ 1)

i=1
4

L = ∑ f(xi )∆x = 1 + 4 + 9 + 16
i=1
4

L = ∑ f(xi )∆x = 30 square units

i=1

To get the right – hand approximation of the area, take the right endpoints, i.e. x1 = 2, x2 = 3, x3 = 4 and x4 = 5,
and evaluate the function at each. Then
f(x1 ) = f(2) = 22 = 4
f(x2 ) = f(3) = 32 = 9
f(x3 ) = f(4) = 42 = 16
f(x4 ) = f(5) = 52 = 25
4

R = ∑ f(xi )∆x = f(x1 )∆x + f(x2 )∆x + f(x3 )∆x + f(x4 )∆x.
i=1
4

R = ∑ f(xi )∆x = (4 ∗ 1) + (9 ∗ 1) + (16 ∗ 1) + (25 ∗ 1)

i=1
4

R = ∑ f(xi )∆x = 4 + 9 + 16 + 25
i=1
4

R = ∑ f(xi )∆x = 54 square units

i=1

EXAMPLE 2
1
Compute the Left – hand Riemann sum approximations of the area under the curve f(x) = 4 x 3 + 1 bounded by
the x – axis and the lines x = -2 and x = 2 using five (5) rectangles.

Given that a = – 2 and b = 2,

b − a 2 − (−2) 4
∆x = = = = 0.80
n 5 5

WEEK 13 CALC 1024: CALCULUS 2

The width of each rectangle is 0.80 unit. Thus, the interval [–2, 2] is divided into the following subintervals:
[–2, –1.2], [–1.2, –0.4], [–0.4, 0.4], [0.4, 1.2] and [1.2, 2].

To compute the left – hand approximation of the area, use the left endpoints, i.e. x 1 = –2, x2 = –1.2, x3 = –0.4,
x4 = 0.4 and x5 = 1.2, and evaluate the function at those points.
1
f(𝑥1 ) = f(−2) = (−2)3 + 1 = −1
4
1
f(𝑥2 ) = f(−1.2) = (−1.2)3 + 1 = 0.568
4
1
f(𝑥3 ) = f(−0.4) = (−0.4)3 + 1 = 0.984
4
1
f(𝑥4 ) = f(0.4) = (0.4)3 + 1 = 1.016
4
1
f(𝑥5 ) = f(1.2) = (1.2)3 + 1 = 1.432
4
5

L = ∑ f(xi )∆x.
i=1

L = f(x1 )∆x + f(x2 )∆x + f(x3 )∆x + f(x4 )∆x + f(x5 )∆x
L = f(−2) ∗ 0.8 + f(−1.2) ∗ 0.8 + f(−0.4) ∗ 0.8 + f(0.4) ∗ 0.8 + f(1.2) ∗ 0.8
L = −1 ∗ 0.8 + 0.568 ∗ 0.8 + 0.984 ∗ 0.8 + 1.016 ∗ 0.8 + 1.432 ∗ 0.8
L = 0.8(−1 + 0.568 + 0.984 + 1.016 + 1.432)
L = 0.8(3)
L = 2.4 square units

Left – hand approximation

WEEK 13 CALC 1024: CALCULUS 2

EXAMPLE 3
Compute the Right – hand Riemann sum approximation of the area under the curve f(x) = √𝑥 − 2 bounded by
the x – axis and the lines x = 4 and x = 16 using six (6) rectangles.

Given that a = 4 and b = 16,

b − a 16 − 4 12
∆x = = = =2
n 6 6

The width of each rectangle is 2 units. So, the six subintervals of the interval [4, 16] are: [4, 6], [6, 8], [8, 10], [10,
12], [12, 14] and [14, 16].

To compute the right – hand approximation of the area, use the right endpoints, i.e., x1 = 6, x2 = 8, x3 = 10,
x4 = 12, x5 = 14 and x6 = 16 and evaluate the function at those points.
f(𝑥1 ) = f(6) = √6 − 2 = 2
f(𝑥2 ) = f(8) = √8 − 2 = √6
f(𝑥3 ) = f(10) = √10 − 2 = √8 = 2√2
f(𝑥4 ) = f(12) = √12 − 2 = √10
f(𝑥5 ) = f(14) = √14 − 2 = √12 = 2√3
f(𝑥6 ) = f(16) = √16 − 2 = √14
6

L = ∑ f(xi )∆x.
i=1

L = f(x1 )∆x + f(x2 )∆x + f(x3 )∆x + f(x4 )∆x + f(x5 )∆x + f(x6 )∆x
L = f(6) ∗ 2 + f(8) ∗ 2 + f(10) ∗ 2 + f(12) ∗ 2 + f(14) ∗ 2 + f(16) ∗ 2
L = 2 ∗ 2 + √6 ∗ 2 + 2√2 ∗ 2 + √10 ∗ 2 + 2√3 ∗ 2 + √14 ∗ 2
L = 2(2 + √6 + 2√2 + √10 + 2√3 + √14)
L = 2(17.646)
L = 35.292 square units

WEEK 13 CALC 1024: CALCULUS 2

Definition of the Definite Integral
Suppose that the function f is continuous and nonnegative on the closed interval [a, b]. The exact area of the
region bounded by the graph of f, the x – axis, and the lines x = a and x = b may be determined by taking the limit
of the Riemann Sum as the number of n subintervals approaches infinity. This exact area is denoted using the
definite integral
b
∫ f(x)dx
a
and thus,

b n

∫ f(x)dx = lim ∑ f(xi )∆x

a n→∞
i=1

EXAMPLE 1
Evaluate the definite integral of
6
∫ 2xdx
0

6
By definition, the value of ∫0 2xdx is the area under the graph of f(x) = 2x bounded by the x – axis and the
lines x = 0 and x = 4. By graphing, you can get the value of this definite integral.

From the graph shown below, the region representing the integral is a triangle. The base of the triangle is 4 units
and the height is 8 units. Thus,
6
1 1
∫ 2xdx = bh = (4)(8) = 16 square units
0 2 2

WEEK 13 CALC 1024: CALCULUS 2

EXAMPLE 2
Evaluate the definite integral of
3
∫ √9 − 𝑥 2 dx
0

The function 𝑓(𝑥) = √9 − 𝑥 2 is half of a circle with center at the point of origin and a radius of 3 units as
shown in the figure below. The shaded region represents the given definite integral and is half of a semicircle.
Thus,
3
1 9
∫ √9 − x 2 dx = π (32 ) = π square units
0 4 4

The Fundamental Theorem of Calculus

Let f be a nonnegative and continuous function on the interval [a, b]. Then you have,
b
∫ f(x)dx = 𝐹(𝑏) − 𝐹(𝑎)
a
where F is an antiderivative of f for all x in [a, b].

The following are some notes that will further enhance your understanding of the fundamental theorem of
calculus and definite integrals:
1. When using the fundamental theorem of calculus, use the following notation:
b
∫ f(x)dx = 𝐹(𝑏) − 𝐹(𝑎)
a

2. The constant of integration C can be disregarded in evaluating definite integrals because

|F(x) + C|ba = (F(b) + C) − (F(a) + C) = F(b) + C − F(a) − C = F(b) − F(a)

3. The definite integral does not necessarily denote area under a curve. It only does when the curve is
above the x – axis on the closed interval [a, b].

4. The properties of the indefinite integrals can be used for definite integrals with the following additions:
𝑎
∫ 𝑓(𝑥)𝑑𝑥 = 0
𝑎
𝑏 𝑎
∫ 𝑓(𝑥)𝑑𝑥 = − ∫ 𝑓(𝑥)𝑑𝑥 = 0
𝑎 𝑏
WEEK 13 CALC 1024: CALCULUS 2
AREA OF THE REGION UNDER F

CASE 1
When f(x) ≥ 0, the area, A, enclosed by the curve y = f(x) and the x – axis in the interval [a, b] is
b
A = ∫ f(x)dx
a

CASE 2
When f(x) ≤ 0, the area, A, enclosed by the curve y = f(x) and the x – axis in the interval [a, b] is
b
A = − ∫ f(x)dx
a

WEEK 13 CALC 1024: CALCULUS 2

EXAMPLE 1
Find the area enclosed by the given curve and the x – axis as:
𝑦 = 𝑥 2 − 4x.

STEP 1 Determine the x – intercepts (set y = 0)

0 = 𝑥 2 − 4x
0 = x(x − 4)
x = 0 → y = 02 − 4(0) = 0, (0,0)
2
x = 4 → y = 4 − 4(4) = 0, (4,0)

STEP 2 Sketch the graph

STEP 3 Solve for the area

4
A = − ∫ (x 2 − 4x)dx
0
𝑥 3 4𝑥 2 4
A = −| − |
3 2 0
(4)3 4(4)2 (0)3 4(0)2
A = − [( − )−( − )]
3 2 3 2
32
A= square units
3

WEEK 13 CALC 1024: CALCULUS 2

EXAMPLE 2
Find the area enclosed by the given curve and the x – axis as:
𝑦 = 𝑥 2 − 4𝑥 + 4.

STEP 1 Determine the x – intercepts (set y = 0)

0 = x 2 − 4x + 4
0 = (x − 2)
x = 2 → y = 22 − 4(2) + 4 = 0, P(2,0)

STEP 2 Sketch the graph

STEP 3 Solve for the area

4
A = ∫ (x 2 − 4x + 4)dx
0
𝑥 3 4𝑥 2 4
A=| − + 4𝑥|
3 2 0
3 2
(4) 4(4) (0)3 4(0)2
A=( − + 4(4)) − ( − + 4(0))
3 2 3 2
16
A= square units
3

WEEK 13 CALC 1024: CALCULUS 2

EXAMPLE 3
Find the area enclosed by the given curve and the x – axis as:
𝑦 = 𝑥 2 − 4𝑥 − 5.

STEP 1 Determine the x – intercepts (set y = 0)

0 = 𝑥 2 − 4x − 5
0 = (x − 5)(x + 1)
x = 5 → y = 52 − 4(5) − 5 = 0, (5, 0)
x = −1 → y = (−1)2 − 4(−1) − 5 = 0, (−1,0)

STEP 2 Sketch the graph

STEP 3 Solve for the area

4
A = − ∫ (x 2 − 4x − 5)dx
0
𝑥 3 4𝑥 2 4
A = −| − − 5𝑥|
3 2 0
3 2
(4) 4(4) (0)3 4(0)2
A = − [( − − 5(4)) − ( − − 5(0))]
3 2 3 2
92
A= square units
3

WEEK 13 CALC 1024: CALCULUS 2

EXAMPLE 4
Find the area enclosed by the given curve and the x – axis as:
𝑦 = 4 − 𝑥2.

STEP 1 Determine the x – intercepts (set y = 0)

0 = 4 − x2
0 = (2 − x)(2 + x)
x = 2 and x = −2
(−2, 0) and (2, 0)

STEP 2 Sketch the graph

STEP 3 Solve for the area

2
A = ∫ (4 − x 2 )dx
−2
𝑥3 2
A = |4x − |
3 −2
(2)3 (−2)3
A = (4(2) − ) − (4(−2) − )
3 3
32
A= square units
3

WEEK 13 CALC 1024: CALCULUS 2

EXAMPLE 5
Find the area enclosed by the given curves.
y = 4 − 2x, y=x and y = 0

STEP 1 Determine the points of intersection by comparison

y=y
4 − 2x = x
−3x = −4
4 4 4 4
x = 3 → y = 3 (3, 3)

STEP 2 Sketch the graph

STEP 3 Solve for the area

4/3 2
A1 = ∫ 𝑥dx A2 = ∫ (4 − 2𝑥)dx
0 4/3
𝑥 2 4/3
A1 = | | 2𝑥 2 2
2 0 A2 = |4𝑥 − |
4 2 4/3
(3)2 (0)2
A1 = | − | 4
2 2 A2 = square units
9
8
A1 = square units A = A1 + A2
9
8 4 12 4
A= + = or square units
9 9 9 3

WEEK 13 CALC 1024: CALCULUS 2

EXAMPLE 6
Find the area enclosed by the given curves.
y = x3 x = −2 x=2

STEP 1 Determine the x – intercept/s (set y = 0)

y = x3
0 = x3
𝑥 = 0 (0, 0)

STEP 2 Sketch the graph

WEEK 13 CALC 1024: CALCULUS 2

STEP 3 Solve for the area
0 2
A1 = − ∫ x 3 dx A2 = ∫ x 3 dx
−2 0

𝑥4 0 𝑥4 2
A1 = − | | A2 = | |
4 −2 4 0

(0)4 (−2)4 0 (2)4 (0)4 2

A1 = − | − | A2 = | − |
4 4 −2 4 4 0

A1 = 4 square units A2 = 4 square units

A = A1 + A2
A= 4+4

A = 8 square units

EXAMPLE 7
Find the area enclosed by the curve 2x 2 + 4x + y = 0 and the x – axis.

STEP 1 Determine the vertex of the parabola:

2x 2 + 4x + y = 0
y = −2x 2 − 4x
y′ = −4x − 4 (Evaluate the derivative)
0 = −4x − 4
4x = −4 → x = −1

If x = −1, y = −2(−1)2 − 4(−1) = 2; V(−1, 2)

STEP 2 Determine the x – intercept/s (set y = 0)

2x 2 + 4x + y = 0
2x 2 + 4x + 0 = 0
2x(x + 2) = 0
x = 0 and x = −2
Hence, the x – intercepts are (0, 0) and (−2, 0).

STEP 3 Sketch the graph

WEEK 13 CALC 1024: CALCULUS 2

STEP 4 Solve for the area
0
A = ∫ f(x)) dx
−2
0
A = ∫ (−2x 2 − 4x) dx
−2

8
A= square units
3

EXAMPLE 8
Find the area enclosed by the curve y 2 + 2x − 2y − 3 = 0 and the y – axis.

STEP 1 Determine the vertex of the parabola by completing the square:

y 2 + 2x − 2y − 3 = 0
y 2 − 2y + 1 = −2x + 3 + 1
(y − 1)2 = −2x + 4
(y − 1)2 = −2(x − 2)
Hence the vertex is located at V(2, 1) .

STEP 2 Determine the y - intercepts of the parabola by setting x = 0:

y 2 + 2x − 2y − 3 = 0
y 2 + 2(0) − 2y − 3 = 0
y 2 − 2y − 3 = 0
(y − 3)(y + 1) = 0
y = 3 and y = −1
Hence, the y – intercepts are (0, − 1) and (0, 3).

STEP 3 Sketch the graph

WEEK 13 CALC 1024: CALCULUS 2
STEP 4 Solve x in terms of y
y 2 + 2x − 2y − 3 = 0
2x = −y 2 + 2y + 3
1
x = 2 (−y 2 + 2y + 3)
1
Hence, f(y) = 2 (−y 2 + 2y + 3).

STEP 5 Solve for the area

3
A = ∫ f(y)) dy
−1
3
1
A = ∫ ( (−y 2 + 2y + 3) ) dy
−1 2

16
A= square units
3

EXAMPLE 9
Find the area enclosed by the curve y = x 3 −6x 2 + 8x and the x – axis.

STEP 1 Determine the maximum and minimum points:

y = x 3 −6x 2 + 8x
𝑦 ′ = 3x 2 − 12x + 8
0 = 3x 2 − 12x + 8
x = 0.85 and x = 3.15
If x = 0.85 → y = (0.85)3 − 6(0.85)2 + 8(0.85) = 3.08 (0.85, 3.08)
If x = 3.15 → y = (3.15)3 − 6(3.15)2 + 8(3.15) = −3.08 (3.15, -3.08)

WEEK 13 CALC 1024: CALCULUS 2

STEP 2 Determine the x - intercepts of the parabola by setting x = 0:
y = x 3 −6x 2 + 8x
0 = x 3 −6x 2 + 8x
0 = 𝑥(x 2 − 6𝑥 + 8)
0 = x(x − 2)(x − 4)
x = 0, x = 2 and x = 4
Hence, the x – intercepts are (0, 0), (2, 0) and (4, 0).

STEP 3 Sketch the graph

STEP 5 Solve for the area

2 4
A1 = ∫ f(x) dx A2 = ∫ f(x) dx
0 2

2 4
A1 = ∫ (x 3 2
−6x + 8x) dx A2 = ∫ (x 3 −6x 2 + 8x) dx
0 2

A1 = 4 square units A2 = 4 square units

Hence,
A = A1 + A2

A= 4+4

A= 8 square units

WEEK 13 CALC 1024: CALCULUS 2

AREA BETWEEN TWO CURVES

Area of the Region Between Two Curves

If f(x) and g(x) are continuous on [a, b], then the area of the region bounded by the graphs of f and g and the
vertical lines x = a and x = b is
b
A = ∫ [f(x) − g(x)]dx
a

WEEK 13 CALC 1024: CALCULUS 2

Area of the Region Between f and g
CASE I
Let the upper function be f(x) and the lower function as g(x) as shown in the figure. The function g(x) is at the
lower end of the vertical strip while f(x) is on the upper end. The area of the region between the curves at interval
[a, b] using vertical differential strips can be expressed as:
b
A = ∫ [f(x) − g(x)]dx 𝐍𝐨𝐭𝐞: yAbove : f(x) and yBelow : g(x)
a

CASE II
Let the right function be f(y) and the left function as g(y) as shown in the figure.
The function g(y) is at the left side of the horizontal strip while f(y) is at the right side.
The area of the region between the curves at interval [a,b] using horizontal differential strips can be expressed
as:
b
A = ∫ [f(y) − g(y)]dy 𝐍𝐨𝐭𝐞: xRight : f(y) and xLeft : g(y)
a

WEEK 13 CALC 1024: CALCULUS 2

EXAMPLE 1
Find the area enclosed by the given curve and the given line.
y = x 2 − 1 and y = x + 1

STEP 1 Determine the points of intersection by comparison

𝑦=𝑦
x2 − 1 = x + 1
x2 − x − 2 = 0
(x − 2)(x + 1) = 0
x = 2 → y = (2)2 − 1 = 3, (2, 3)
2
x = −1 → y = (−1) − 1 = 0, (−1, 0)
Note:
The x – coordinates of the points of intersection of y = x 2 − 1 and 𝑦 = x + 1 are x = −1 and x = 2, the
interval over which the integral should be calculated is [−1, 2] .
2
A = ∫ (yAbove − yBelow )dx
−1
STEP 2 Sketch the graph

WEEK 13 CALC 1024: CALCULUS 2

STEP 3 Solve for the area
2
A = ∫ (yAbove − yBelow )dx
−1
2
A = ∫ [(x + 1) − (x 2 − 1)]dx
−1
2
A = ∫ (−x 2 + 𝑥 + 2)dx
−1
2
x3 x2
A = |− + + 2x|
3 2 −1
23 22 (−1)3 (−1)2
A = (− + + 2(2)) − (− + + 2(−1))
3 2 3 2
9
A= square units
2

EXAMPLE 2
Find the area enclosed by the given curve and the given line.
y = x 2 and y = 2x + 3

STEP 1 Determine the points of intersection by comparison

𝑦=𝑦
x 2 = 2x + 3
x 2 − 2x − 3 = 0
(x − 3)(x + 1) = 0
x = 3 → y = (3)2 = 9, (3, 9)
x = −1 → y = (−1)2 = 1, (−1, 1)
Note:
The x – coordinates of the points of intersection of y = x 2 and 𝑦 = 2x + 3 are x = −1 and x = 3, the interval
over which the integral should be calculated is [−1, 3] .
3
A = ∫ (yAbove − yBelow )dx
−1
STEP 2 Sketch the graph
WEEK 13 CALC 1024: CALCULUS 2
STEP 3 Solve for the area
3
A = ∫ (yAbove − yBelow )dx
−1
3
A = ∫ [(2x + 3) − (x 2 )]dx
−1
3
2
x3
A = |x + 3x − |
3 −1
33 (−1)3
A = (32 + 3(3) − ) − ((−1)2 + 3(−1) − )
3 3
32
A= square units
3

EXAMPLE 3
Find the area enclosed by the given curves:
x = 𝑦 2 and x = 𝑦 3

STEP 1 Determine the points of intersection by comparison

𝑥=𝑥
y2 = y3
y3 − y2 = 0
y(y 2 − 1) = 0
y(y − 1)(y + 1) = 0
y = 0 → x = (0)2 = 0, (0, 0)
2
y = 1 → x = (1) = 1, (1, 1)
Note:
The x – coordinates of the points of intersection of x = 𝑦 2 and x = 𝑦 3 are x = 0 and x = 1, the interval over
which the integral should be calculated is [0, 1] .

WEEK 13 CALC 1024: CALCULUS 2

STEP 2 Sketch the graph

STEP 3 Solve for the area

Note:
yAbove : x = y 2 → y = √x
3
yBelow : x = y 3 → y = √x
1
A = ∫ (yAbove − yBelow ) dx
0
1
3
A = ∫ (√x − √x)d𝑥
0
1
A= square unit
12

EXAMPLE 4
Find the area enclosed by the given curves:
x = 12y 2 − 12y 3 and x = 2y 2 − 2y

STEP 1 Determine the point/s of intersection by comparison

x=x
12y 2 − 12y 3 = 2y 2 − 2y
12y 3 − 10y 2 − 2y = 0
2y(6y 2 − 5y − 1) = 0
1
y(y − 1)(6y + 1) = 0 → y = 0, y = 1, y = − 6
y = 1 → x = 12(1)2 − 12(1)3 = 0, (0, 1)

STEP 2 Sketch the graph

WEEK 13 CALC 1024: CALCULUS 2

STEP 3 Solve for the area
1
A = ∫ (xRight − xLeft ) dy
0
1
A = ∫ [(12y 2 − 12y 3 ) − (2y 2 − 2y)]dy
0
4
A = 3 square units

EXAMPLE 5
Find the area enclosed by the given curves:
y = x 3 , y = 8 and x = 0

STEP 1 Determine the point/s of intersection by comparison

y=y
x3 = 8
x=2
If x = 2 → y = (2)3 = 8, (2, 8)

STEP 2 Sketch the graph

WEEK 13 CALC 1024: CALCULUS 2

STEP 3 Solve for the area
2
A = ∫ (yAbove − yBelow ) dy
0
2
A = ∫ (8 − x 3 )dx
0
A = 12 square units

EXAMPLE 6
Find the area enclosed by the given curves and the given lines:
y = x 2 , y = −2x 4 , x = −1 and x = 1

STEP 1 Determine the points of intersection by comparison

𝑦=𝑦
x 2 = −2x 4
2x 4 − x 2 = 0
√2 √2
x 3 − 𝑥 = 0 → x = 0, x = ,x = −
2 2
WEEK 13 CALC 1024: CALCULUS 2
 x = 0 → y = (0)2 = 0, (0, 0)

STEP 2 Sketch the graph

STEP 3 Solve for the area

1
A = ∫ (yAbove − yBelow ) dx
−1
1
A = ∫ (x 2 − (−2x 4 ))d𝑥
−1
22
A = 15 square units

EXAMPLE 7
Find the area enclosed by the given curves and the given lines:
y = Ln x 3 , y = Ln x and x = e

STEP 1 Determine the points of intersection by comparison

𝑦=𝑦
Ln x 3 = Lnx
x3 = x
x3 − 𝑥 = 0
WEEK 13 CALC 1024: CALCULUS 2
 x(x + 1)(x − 1) = 0
x = 0 → y = Ln(0)3 = undefined
x = −1 → y = Ln(−1)3 = undefined
x = 1 → y = Ln(1)3 = 0 (1,0)

STEP 2 Sketch the graph

STEP 3 Solve for the area

e
A = ∫ (yAbove − yBelow ) dx
1
e
A = ∫ (Lnx 3 − Lnx)dx
1
A = 2 square units

References
Printed References
1. Stewart, J. (2019). Calculus concepts and contexts 4th edition enhanced edition. USA : Cengage
Learning, Inc.
WEEK 13 CALC 1024: CALCULUS 2
 2. Canlapan, R.B.(2017). Diwa senior high school series: basic calculus. DIWA Learning Systems Inc.
3. Hass, J., Weir, M. & Thomas, G. (2016). University calculus early transcendentals,global edition.
Pearson Education, Inc.Larson,
4. Bacani, J.B.,et.al. (2016). Basic calculus for senior high school. Book AtBP. Publishing Corp.
5. Balmaceda, J.M. (2016). Teaching guide in basic calculus. Commission on Higher Education.
6. Pelias, J.G. (2016). Basic calculus. Rex Bookstore.
7. Molina, M.F. (2016). Integral calculus. Unlimited Books Library Services & Publishing Inc.
8. Larson, R & Edwards, B. (2012). Calculus. Pasig City, Philippines. Cengage Learning Asia Pte Ltd

Electronic Resources
1. Pulham , John. Differential Calculus. Retrieved August 7, 2020 from http://
www.maths.abdn.ac.uk/igc/tch/math1002/dif
2. Weistein, Eric. Differential Calculus. Retrieved August 7, 2020 from
http://mathworld.wolfram.com/differential calculus
3. Thomas , Christopher . Introduction to Differential Calculus. Retrieved August 7, 2020 from http://
www.usyd.edu.aav/stuserv/document/maths_learning-center
4. Purplemath. Retrieved August 7, 2020 from http://www.purplemath.com/modules/

WEEK 13 CALC 1024: CALCULUS 2

 UNIVERSITY OF SAINT LOUIS
Tuguegarao City

SCHOOL OF ENGINEERING, ARCHITECTURE & IT EDUCATION

Engineering Sciences Department
S.Y. 2020 – 2021

CORRESPONDENCE LEARNING MODALITY

CALC 1024: Calculus 2 (Integral Calculus)

Prepared by:

EROVITA TERESITA BACUD AGUSTIN, DME

ENGR. NORMAN BOB GOMEZ
ENGR. JENIROSE B. MATIAS
ENGR. PEEJAY PAGUIRIGAN
Instructors

Reviewed by:

ENGR. VICTOR VILLALUZ, MEE

EE Program Chair / ES Department Head

Recommended by:

ENGR. RONALD M. VILLAVERDE, ME – ECE, PECE

Academic Dean

Approved by:

EMMANUEL JAMES P. PATTAGUAN, Ph.D.

VP for Academics

WEEK 14 CALC 1024: CALCULUS 2

 Weekly Study and Assessment Guide
For this week, May 12 – 16, 2021 of this semester, the following shall be your guide for the different lessons and
tasks that you need to accomplish. Be patient read it carefully before proceeding to the tasks expected of you.

Date Topic Activities/Tasks

Area of the Region in Polar Read Lessons
May 12 – 15, 2021
Coordinates

May 16, 2021 Submission of Learning Tasks

GOD BLESS!

WEEK 14: Area of the Region in Polar Coordinates

LEARNING OUTCOME
At the end of the lesson, the students should be able to solve for area of the region in polar coordinates.

Let it be required to find the area A of the region R bounded by the polar curve 𝑟 = 𝑓(𝜃) and the lines 𝜃 = 𝛼
and 𝜃 = 𝛽. For simplicity, inscribe in R n equal circular sectors, each of central angle ∆𝜃. A typical sector is
1
shown in the figure below. Recalling that the area of a circular sector of radius a and central angle 𝜑 is 2 𝑎2 𝜑,
then the sum of the areas of the sectors in R will approximate the area bounded by 𝑟 = 𝑓(𝜃), 𝜃 = 𝛼 𝑎𝑛𝑑 𝜃 = 𝛽 .

This sum, in sigma notation is

𝑛
1 1 1 1 1
∑ (𝑟𝑖 )2 ∆𝜃 = (𝑟1 )2 ∆𝜃 + (𝑟2 )2 ∆𝜃 + (𝑟3 )2 ∆𝜃 + ⋯ + (𝑟𝑛 )2 ∆𝜃.
2 2 2 2 2
𝑖=1

WEEK 14 CALC 1024: CALCULUS 2

It is reasonable to expect that as n increases, this sum comes closer and closer to the exact area of R. Thus we
defined the area of R, as the limit of the sum of the areas of the sectors as n tends to infinity. That is,
𝑛
1
𝐴 = Lim ∑ (𝑟𝑖 )2 ∆𝜃
𝑛→∞ 2
𝑖=1
and by the Fundamental Theorem, we have
1 𝛽
𝐴 = ∫ 𝑟 2 𝑑𝜃
2 𝛼

If the region R is bounded by 𝑟 = 𝑓1 (𝜃), 𝑟 = 𝑓2 (𝜃), 𝜃 = 𝛼 𝑎𝑛𝑑 𝜃 = 𝛽, then the area of R can be shown to be

1 𝛽
𝐴 = ∫ [(𝑟2 )2 − (𝑟1 )2 ] 𝑑𝜃
2 𝛼
where 𝑟1 and 𝑟2 are the respective radius vectors of the inner and outer curves.

WEEK 14 CALC 1024: CALCULUS 2

EXAMPLE 1
Find the area bounded by the lemniscate 𝑟 2 = 4𝑐𝑜𝑠2𝜃.

STEP 1: Sketch the region.

Note: You can use this site https://www.desmos.com/calculator/xbaqwuxedc to graph the given
function.

STEP 2: Determine the values of 𝜃

𝑟2 = 𝑟2
4𝑐𝑜𝑠2𝜃 = 02
4𝑐𝑜𝑠2𝜃 = 0
𝑐𝑜𝑠2𝜃 = 0
2𝜃 = 𝐴𝑟𝑐𝑐𝑜𝑠(0)
𝜋 3𝜋
2𝜃 = 𝑎𝑛𝑑
2 2
𝜋 3𝜋
𝜃 = 𝑎𝑛𝑑
4 4
𝜋
STEP 3: By symmetry the total area is four times the area between 𝜃 = 0 𝑎𝑛𝑑 𝜃 = 4 .
1 𝛽
𝐴 = 4 ∗ ∫ 𝑟 2 𝑑𝜃
2 𝛼
𝜋
1 4
𝐴 = 4 ∗ ∫ (4𝑐𝑜𝑠2𝜃)2 𝑑𝜃
2 0
𝜋
4
𝐴 = 2 ∫ 16(𝑐𝑜𝑠2𝜃)2 𝑑𝜃
0

𝐴 = 4𝜋 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠

WEEK 14 CALC 1024: CALCULUS 2

EXAMPLE 2
Find the area bounded by the circle 𝑟 = 16𝑠𝑖𝑛𝜃.

STEP 1: Sketch the region

Note: You can use this site https://www.desmos.com/calculator/xbaqwuxedc to graph the given
function.

STEP 2: Determine the values of 𝜃

𝑟=𝑟
16𝑠𝑖𝑛𝜃 = 0
𝑠𝑖𝑛𝜃 = 0
𝜃 = 𝐴𝑟𝑐𝑠𝑖𝑛(0)
𝜃 = 0 𝑎𝑛𝑑 𝜋

STEP 3: The area is

1 𝛽
𝐴 = ∫ 𝑟 2 𝑑𝜃
2 𝛼

1 𝜋
𝐴 = ∫ (16𝑠𝑖𝑛𝜃)2 𝑑𝜃
2 0

𝐴 = 64𝜋 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠

WEEK 14 CALC 1024: CALCULUS 2

EXAMPLE 3
Find the area bounded by the rose 𝑟 = 4𝑠𝑖𝑛3𝜃.

STEP 1: Sketch the region

Note: You can use this site https://www.desmos.com/calculator/xbaqwuxedc to graph the given
function.

STEP 2: Determine the values of 𝜃, 𝑟 = 4𝑠𝑖𝑛3𝜃 𝑎𝑛𝑑 𝑟 = 0

𝑟=𝑟
4𝑠𝑖𝑛3𝜃 = 0
𝑠𝑖𝑛3𝜃 = 0
3𝜃 = 𝐴𝑟𝑐𝑠𝑖𝑛(0)
𝜋
𝜃 = 0 𝑎𝑛𝑑
3
𝜋
STEP 3: By symmetry the total area is three times the area between 𝜃 = 0 𝑎𝑛𝑑 𝜃 = 3 .
1 𝛽
𝐴 = 3 ∗ ∫ 𝑟 2 𝑑𝜃
2 𝛼
𝜋
1 3
𝐴 = 3 ∗ ∫ (4𝑠𝑖𝑛3𝜃)2 𝑑𝜃
2 0

𝐴 = 4𝜋 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠

WEEK 14 CALC 1024: CALCULUS 2

EXAMPLE 4
Find the area bounded by the cardioid 𝑟 = 2(1 − 𝑠𝑖𝑛𝜃).

STEP 1: Sketch the region

Note: You can use this site https://www.desmos.com/calculator/xbaqwuxedc to graph the given
function.

STEP 2: Determine the values of 𝜃, 𝑟 = 2(1 − 𝑠𝑖𝑛𝜃) 𝑎𝑛𝑑 𝑟 = 0

𝑟=𝑟
2(1 − 𝑠𝑖𝑛𝜃) = 0
1 − 𝑠𝑖𝑛𝜃 = 0
𝑠𝑖𝑛𝜃 = 1
𝜃 = 𝐴𝑟𝑐𝑠𝑖𝑛(1)
𝜋
𝜃=
2
𝜋 3𝜋
STEP 3: By symmetry the total area is two times the area between 𝜃 = 𝑎𝑛𝑑 𝜃 = .
2 2
𝛽
1
𝐴 = 2 ∗ ∫ 𝑟 2 𝑑𝜃
2 𝛼

3𝜋
1 2
𝐴 = 2 ∗ ∫ (2(1 − 𝑠𝑖𝑛𝜃) )2 𝑑𝜃
2 𝜋
2

𝐴 = 6𝜋 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠

WEEK 14 CALC 1024: CALCULUS 2

EXAMPLE 5
Find the area bounded by the limacon 𝑟 = 2 + 𝑐𝑜𝑠𝜃

STEP 1: Sketch the region

STEP 2: By symmetry the total area is two times the area between 𝜃 = 0 𝑎𝑛𝑑 𝜃 = 𝜋.
1 𝛽
𝐴 = 2 ∗ ∫ 𝑟 2 𝑑𝜃
2 𝛼

1 𝜋
𝐴 = 2 ∗ ∫ (2 + 𝑐𝑜𝑠𝜃 )2 𝑑𝜃
2 0

9𝜋
𝐴= 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
2

WEEK 14 CALC 1024: CALCULUS 2

EXAMPLE 6
Determine the area of the region inside the circle 𝑟 = 3𝑠𝑖𝑛𝜃 and outside the limacon 𝑟 = 2 − 𝑠𝑖𝑛𝜃.

STEP 1: Sketch the region

Note: You can use this site https://www.desmos.com/calculator/xbaqwuxedc to graph the given
function.

STEP 2: Determine the values of 𝜃. Since 𝑟 = 3𝑠𝑖𝑛𝜃 𝑎𝑛𝑑 𝑟 = 2 − 𝑠𝑖𝑛𝜃, then by comparison
𝑟=𝑟
3𝑠𝑖𝑛𝜃 = 2 − 𝑠𝑖𝑛𝜃
4𝑠𝑖𝑛𝜃 = 2
1
𝑠𝑖𝑛𝜃 =
2
1
𝜃 = 𝐴𝑟𝑐𝑠𝑖𝑛 ( )
2
𝜋 5𝜋
𝜃 = 𝑎𝑛𝑑
6 6
𝜋 𝜋
STEP 3: The area of the region between 𝜃 = 𝑎𝑛𝑑 𝜃 = is:
6 2
1 𝛽
𝐴= ∫ [(𝑟2 )2 − (𝑟1 )2 ] 𝑑𝜃
2 𝛼

5𝜋
1 6
𝐴 = ∫ [(3𝑠𝑖𝑛𝜃 )2 − (2 − 𝑠𝑖𝑛𝜃 )2 ] 𝑑𝜃
2 𝜋
6

𝐴 = 5.196 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠

WEEK 14 CALC 1024: CALCULUS 2

EXAMPLE 7
Determine the area of the region inside the small loop of the limacon 𝑟 = 1 + 2𝑠𝑖𝑛𝜃.

STEP 1: Sketch the region

Note: You can use this site https://www.desmos.com/calculator/xbaqwuxedc to graph the given
function.

STEP 2: Determine the values of 𝜃. Since 𝑟 = 1 + 2𝑠𝑖𝑛𝜃 𝑎𝑛𝑑 𝑟 = 0, then by comparison

𝑟=𝑟
1 + 2𝑠𝑖𝑛𝜃 = 0
2𝑠𝑖𝑛𝜃 = −1
1
𝑠𝑖𝑛𝜃 = −
2
1
𝜃 = 𝐴𝑟𝑐𝑠𝑖𝑛 (− )
2
7𝜋 11𝜋
𝜃= 𝑎𝑛𝑑
6 6
7𝜋 11𝜋
STEP 3: Solve the area of the region between 𝜃 = 𝑎𝑛𝑑 𝜃 = .
6 6
𝛽
1
𝐴= ∫ 𝑟 2 𝑑𝜃
2 𝛼

11𝜋
1 6
𝐴 = ∫ (1 + 2𝑠𝑖𝑛𝜃 )2 𝑑𝜃
2 7𝜋
6

𝐴 = 0.5435 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠

WEEK 14 CALC 1024: CALCULUS 2

EXAMPLE 8
Determine the area that lies inside 𝑟 = 3 + 2𝑠𝑖𝑛𝜃 and outside r = 2.

STEP 1: Sketch the region

STEP 2: Determine the values of 𝜃 for which the two curves intersect. Since 𝑟 = 3 + 2𝑠𝑖𝑛𝜃 𝑎𝑛𝑑 𝑟 = 2,
then by comparison
𝑟=𝑟
3 + 2𝑠𝑖𝑛𝜃 = 2
2𝑠𝑖𝑛𝜃 = 2 − 3
2𝑠𝑖𝑛𝜃 = −1
1
𝑠𝑖𝑛𝜃 = −
2
1
𝜃 = 𝐴𝑟𝑐𝑠𝑖𝑛 (− )
2
7𝜋 11𝜋
𝜃= 𝑎𝑛𝑑
6 6
7𝜋 11𝜋
STEP 3 From the illustration below, if 𝑎𝑛𝑑 are used as limits of integration, the required area
6 6
𝜋 7𝜋
will not be enclosed, hence use − 6 𝑎𝑛𝑑 .
6

1 𝛽
𝐴 = ∫ [(𝑟2 )2 − (𝑟1 )2 ] 𝑑𝜃
2 𝛼
7𝜋
1 6
𝐴= ∫ [(3 + 2𝑠𝑖𝑛𝜃)2 − (2)2 ] 𝑑𝜃
2 −𝜋
6
𝐴 = 24.187 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
WEEK 14 CALC 1024: CALCULUS 2
EXAMPLE 9
Determine the area common to the circle r = 2 and the cardioid 𝑟 = 2(1 − 𝑐𝑜𝑠𝜃).

STEP 1: Sketch the region

Note: You can use this site https://www.desmos.com/calculator/xbaqwuxedc to graph the given
function.

STEP 2: Determine the values of 𝜃 for which the two curves intersect. Since 𝑟 = 2(1 − 𝑐𝑜𝑠𝜃)𝑎𝑛𝑑 𝑟 = 2,
then by comparison
𝑟=𝑟
2(1 − 𝑐𝑜𝑠𝜃) = 2
1 − 𝑐𝑜𝑠𝜃 = 1
−𝑐𝑜𝑠𝜃 = 0
𝑐𝑜𝑠𝜃 = 0
𝜃 = 𝐴𝑟𝑐𝑐𝑜𝑠(0)
𝜋 3𝜋
𝜃 = 𝑎𝑛𝑑
2 2

STEP 3 Compute first for the area of A1 as shown below:

1 𝛽
𝐴1 = ∫ [(𝑟2 )2 − (𝑟1 )2 ] 𝑑𝜃
2 𝛼
𝜋
1 2 2
𝐴1 = ∫ [(2)2 − (2(1 − 𝑐𝑜𝑠𝜃)) ] 𝑑𝜃
2 −𝜋
2
𝐴1 = 4.858 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
WEEK 14 CALC 1024: CALCULUS 2
STEP 4 Compute for the area of the shaded region as shown below:

𝐴 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 − 𝐴1

𝐴 = 𝜋𝑟 2 − 𝐴1
𝐴 = 𝜋(2)2 − 4.858
𝐴1 = 7.708 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠

WEEK 14 CALC 1024: CALCULUS 2

 References
Printed References
1. Stewart, J. (2019). Calculus concepts and contexts 4th edition enhanced edition. USA : Cengage
Learning, Inc.
2. Canlapan, R.B.(2017). Diwa senior high school series: basic calculus. DIWA Learning Systems Inc.
3. Hass, J., Weir, M. & Thomas, G. (2016). University calculus early transcendentals,global edition.
Pearson Education, Inc.Larson,
4. Bacani, J.B.,et.al. (2016). Basic calculus for senior high school. Book AtBP. Publishing Corp.
5. Balmaceda, J.M. (2016). Teaching guide in basic calculus. Commission on Higher Education.
6. Pelias, J.G. (2016). Basic calculus. Rex Bookstore.
7. Molina, M.F. (2016). Integral calculus. Unlimited Books Library Services & Publishing Inc.
8. Larson, R & Edwards, B. (2012). Calculus. Pasig City, Philippines. Cengage Learning Asia Pte Ltd

Electronic Resources
1. Pulham , John. Differential Calculus. Retrieved August 7, 2020 from http://
www.maths.abdn.ac.uk/igc/tch/math1002/dif
2. Weistein, Eric. Differential Calculus. Retrieved August 7, 2020 from
http://mathworld.wolfram.com/differential calculus
3. Thomas , Christopher . Introduction to Differential Calculus. Retrieved August 7, 2020 from http://
www.usyd.edu.aav/stuserv/document/maths_learning-center
4. Purplemath. Retrieved August 7, 2020 from http://www.purplemath.com/modules/

WEEK 14 CALC 1024: CALCULUS 2

 ASSESSMENT 2
Final Period
Second Semester School Year 2020 – 2021

Name
Year/Program

TEST 1
Find the area of the region enclosed by the graph of the given equation.
1. r = 3cosθ
2. r = 2 − sinθ
3. r = 4cos3θ
4. r = 4sin2 θ
5. r 2 = 4sin2θ
6. r = 4sin2 θcosθ

TEST II
Find the area of the region enclosed by one loop of the graph of the given equation.
7. r = 3cos2θ
8. r = 1 + 3sinθ

TEST 1
Find the area of the intersection of the regions enclosed by the graphs of the two given equations.
9. r = 2 and r = 3 − 2cosθ
10. r = 3sin2θ and 3cos2θ

WEEK 14 CALC 1024: CALCULUS 2

Finals Week 3
Volume of Solid of Revolution
If a plane area is rotated about a line in its plane, the solid thus generated is called a solid of
revolution, and the line about which the area is rotated is called an axis of rotation (AOR).
The volume of the solid of revolution can be found by integration and this will be discussed in the
following sections

The Circular Disk Method

Consider a simple plane figure such as the rectangle in figure below. If this rectangle is revolved
about the x-axis, the solid of revolution generated is a circular disk

Consider the area A under the curve C of a certain function f and between the line ℎ = 𝑎 and ℎ =
𝑏. Assume that f is continuous on [a, b]. If the area 𝐴 is revolved about the horizontal line 𝐻, a
solid of revolution is generated. Our problem is to calculate the Volume V of this solid of
revolution

ℎ=𝑎 ℎ=𝑏
INTEGRAL CALCULUS 1
Divide the interval [a, b] into n equal subintervals, each of length ∆ℎ. Then draw n rectangles with
the lower bases resting on the line H and the corresponding upper bases touching the curve C. A
typical rectangle is shown below. If the area A is revolved about H. each rectangular element will
sweep out a cylindrical element or circular disk. Note that the length and width of the rectangular
element become the radius and height of the disk respectively

𝑥=𝑎 𝑥=𝑏
The sum of the volume of the disks will approximate the volume V of the solid of revolution that
is
𝑛

𝑉 = ∑ 𝜋𝑙𝑖2 ∆ℎ = 𝜋𝑙12 ∆ℎ + 𝜋𝑙22 ∆ℎ + 𝜋𝑙32 ∆ℎ + ⋯

𝑖=1

The volume of the solid can be regarded as the limit of the sum as intended to infinity. Hence
𝑛

𝑉 = lim ∑ 𝜋𝑙𝑖2 ∆ℎ
n→∞
𝑖=1

Then by the fundamental theorem, we can write it as

𝒃
𝑽 = 𝝅 ∫ 𝒍𝟐 𝒅𝒉
𝒂

where: 𝑎 = 𝑡ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡

𝑏 = 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 𝑙𝑖𝑚𝑖𝑡

𝑙 = 𝑓(𝑥) − 𝑔(𝑥)

𝑓(𝑥) = 𝑖𝑠 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 𝑐𝑢𝑟𝑣𝑒 𝑜𝑟 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑔𝑖𝑜𝑛

𝑔(𝑥) = 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 𝑐𝑢𝑟𝑣𝑒 𝑜𝑟 𝑡ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑔𝑖𝑜𝑛

INTEGRAL CALCULUS 2
Find the volume of the solid generated by revolving the area bounded by the given curves about
the indicated axis

Example 1: 𝑦 = 𝑥 2 − 2𝑥, 𝑥 − 𝑎𝑥𝑖𝑠, 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠

Step 1: Determine points of intersection
𝑥 − 𝑎𝑥𝑖𝑠 → 𝑦 = 0
𝑦 = 𝑥 2 − 2𝑥
𝑥 2 − 2𝑥 = 0
𝑥 = 2, 0

Step 2: Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

𝑁𝑜𝑡𝑒:
𝑦𝑎𝑏𝑜𝑣𝑒 : 𝑦 = 0
𝑦𝑏𝑒𝑙𝑜𝑤 : 𝑦 = 𝑥 2 − 2𝑥
𝑏
𝑉 = 𝜋 ∫𝑎 (𝑦𝑎𝑏𝑜𝑣𝑒 − 𝑦𝑏𝑒𝑙𝑜𝑤 )2 𝑑𝑥
𝟐 𝟐
𝑽 = 𝝅 ∫ (𝟎 − (𝒙𝟐 − 𝟐𝒙)) 𝒅𝒙
𝟎
𝟐
𝑽 = 𝝅 ∫ (𝒙𝟐 − 𝟐𝒙)𝟐 𝒅𝒙
𝟎
𝟏𝟔
𝑽= 𝝅 𝒐𝒓 𝟑. 𝟑𝟓𝟏 𝒖𝒏𝒊𝒕𝒔𝟑
𝟏𝟓

INTEGRAL CALCULUS 3
Example 2: 𝑥 + 𝑦 = 5, 𝑦 = 0, 𝑥 = 0, 𝑎𝑏𝑜𝑢𝑡 𝑦 = 0
Step 1: Determine points of intersection
𝑥 + 𝑦 = 5, 𝑥 = 0, 𝑦 = 5, 𝑃1 (0, 5)
𝑥 + 𝑦 = 5 , 𝑦 = 0, 𝑥 = 5, 𝑃2 (5, 0)
𝑥 = 0, 𝑦 = 0 𝑃3 (0, 0)

Step 2 Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

𝑁𝑜𝑡𝑒:
𝑦𝑎𝑏𝑜𝑣𝑒 : 𝑦 = 5 − 𝑥
𝑦𝑏𝑒𝑙𝑜𝑤 : 𝑦 = 0
𝑏
𝑉 = 𝜋 ∫𝑎 (𝑦𝑎𝑏𝑜𝑣𝑒 − 𝑦𝑏𝑒𝑙𝑜𝑤 )2 𝑑𝑥
𝟓
𝟐
𝑽 = 𝝅 ∫ ((𝟓 − 𝒙) − 𝟎) 𝒅𝒙
𝟎
𝟓
𝑽 = 𝝅 ∫ (𝟓 − 𝒙)𝟐 𝒅𝒙
𝟎
𝑽 = 𝟒𝟏. 𝟔𝟔𝟕𝝅 𝒐𝒓 𝟏𝟑𝟎. 𝟗 𝒖𝒏𝒊𝒕𝒔𝟑

INTEGRAL CALCULUS 4
Example 3: 𝑦 = 𝑥 3 , 𝑦 = 8, 𝑥 = 0, 𝑎𝑏𝑜𝑢𝑡 𝑦 − 𝑎𝑥𝑖𝑠
Step 1: Determine points of intersection
𝑦 = 𝑥 3 , 𝑦 = 8, 𝑥 3 = 8 = 2 𝑃1 (2,8)
3
𝑦 = 𝑥 , 𝑥 = 0, 𝑦 = 03 = 0 𝑃2 (0,0)
𝑦 = 8, 𝑥 = 0 𝑃3 (0,8)

Step 2 Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

𝑁𝑜𝑡𝑒:
𝑥𝑟𝑖𝑔ℎ𝑡 : 𝑦 = 𝑥 3 → 𝑥 = 3√𝑦
𝑥𝑙𝑒𝑓𝑡 : 𝑥 = 0
𝑏 2
𝑉 = 𝜋 ∫𝑎 (𝑥𝑟𝑖𝑔ℎ𝑡 − 𝑥𝑙𝑒𝑓𝑡 ) 𝑑𝑦
𝟖 𝟐
𝑽 = 𝝅 ∫ (( 3√𝑦) − 𝟎) 𝒅𝒚
𝟎
𝟐 𝟐
𝑽 = 𝝅 ∫ 𝒚𝟑 𝒅𝒚
𝟎
𝑽 = 𝟏𝟗. 𝟐𝝅 𝒐𝒓 𝟔𝟎. 𝟑𝟐 𝒖𝒏𝒊𝒕𝒔𝟑

INTEGRAL CALCULUS 5
Example 4: 𝑦 = ln(𝑥) , 𝑥 = 𝑒, 𝑦 = 0; 𝑎𝑏𝑜𝑢𝑡 𝑥 = 𝑒
Step 1: Determine points of intersection
𝑦 = ln (𝑥), 𝑦 = 0, ln(𝑥) = 0, 𝑥=1 𝑃1 (1,0)
𝑦 = ln (x), 𝑥 = 𝑒, 𝑦 = ln(𝑒) , 𝑦 = 1 𝑃2 (𝑒, 1)
𝑥 = 𝑒, 𝑦 = 0 𝑃3 (𝑒, 0)

Step 2 Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

𝑁𝑜𝑡𝑒:
𝑦𝑎𝑏𝑜𝑣𝑒 : 𝑦 = ln (𝑥)
𝑦𝑏𝑒𝑙𝑜𝑤 : 𝑦 = 0
𝑏
𝑉 = 𝜋 ∫𝑎 (𝑦𝑎𝑏𝑜𝑣𝑒 − 𝑦𝑏𝑒𝑙𝑜𝑤 )2 𝑑𝑥
𝒆
𝑽 = 𝝅 ∫ (𝒍𝒏(𝒙) − 𝟎)𝟐 𝒅𝒙
𝟏
𝒆
𝟐
𝑽 = 𝝅 ∫ (𝒍𝒏(𝒙)) 𝒅𝒙
𝟏
𝑽 = 𝟎. 𝟕𝟏𝟖𝝅 𝒐𝒓 𝟐. 𝟐𝟔 𝒖𝒏𝒊𝒕𝒔𝟑

INTEGRAL CALCULUS 6
Example 5: 𝑥𝑦 = 4, 𝑥 = 2, 𝑦 = 4; 𝑎𝑏𝑜𝑢𝑡 𝑦 = 4
Step 1: Determine points of intersection
4
𝑥𝑦 = 4, 𝑥 = 2, 𝑦 = 2, 𝑦=2 𝑃1 (2,2)
4
𝑥𝑦 = 4, 𝑦 = 4, 𝑥 = 4, 𝑥=1 𝑃2 (1,4)
𝑥 = 2, 𝑦 = 4 𝑃3 (2,4)

Step 2 Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

𝑁𝑜𝑡𝑒:
𝑥𝑟𝑖𝑔ℎ𝑡 : 𝑥 = 2
4
𝑥𝑙𝑒𝑓𝑡 : 𝑥 = 𝑦
𝑏 2
𝑉 = 𝜋 ∫𝑎 (𝑥𝑟𝑖𝑔ℎ𝑡 − 𝑥𝑙𝑒𝑓𝑡 ) 𝑑𝑦
𝟒
4 𝟐
𝑽 = 𝝅 ∫ (𝟐 − ) 𝒅𝒚
𝟐 𝑦
𝑽 = 𝟎. 𝟗𝟎𝟗𝟔𝝅 𝒐𝒓 𝟐. 𝟖𝟓𝟖 𝒖𝒏𝒊𝒕𝒔𝟑

INTEGRAL CALCULUS 7
Example 6: 𝑦 = sin(𝑥) , 𝑥 = 0, 𝑦 = 1; 𝑎𝑏𝑜𝑢𝑡 𝑦 = 1
Step 1: Determine points of intersection
𝜋
𝑦 = sin (x), 𝑦 = 1, sin(𝑥) = 1, 𝑥 = 2 𝑜𝑟 1.571 𝑃1 (1.571,1)
𝑦 = sin (x), 𝑥 = 0, 𝑦 = sin (0), 𝑦=0 𝑃2 (0,0)
𝑦 = 1, 𝑥 = 0 𝑃3 (0,1)

Step 2 Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

𝑁𝑜𝑡𝑒:
𝑦𝑎𝑏𝑜𝑣𝑒 : 𝑦 = 1
𝑦𝑏𝑒𝑙𝑜𝑤 : 𝑦 = sin (𝑥)
𝑏
𝑉 = 𝜋 ∫𝑎 (𝑦𝑎𝑏𝑜𝑣𝑒 − 𝑦𝑏𝑒𝑙𝑜𝑤 )2 𝑑𝑥
𝝅
𝟐 𝟐
𝑽 = 𝝅 ∫ (𝟏 − 𝒔𝒊𝒏(𝒙)) 𝒅𝒙
𝟎
𝑽 = 𝟎. 𝟑𝟓𝟔𝝅 𝒐𝒓 𝟏. 𝟏𝟏𝟗 𝒖𝒏𝒊𝒕𝒔𝟑

INTEGRAL CALCULUS 8
Example 7: 𝑦 = 𝑥 2 , 𝑥 = 1, 𝑥 = 2, 𝑦 = 0 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠
Step 1: Determine points of intersection
𝑦 = 𝑥 2 , 𝑥 = 1, 𝑦 = 12 , 𝑦=1 𝑃1 (1, 1)
2
𝑦 = 𝑥 , , 𝑥 = 2, 𝑦 = 22 , 𝑦=4 𝑃2 (2, 4)
𝑥 = 1, 𝑦 = 0 𝑃3 (1, 0)
𝑥 = 2, 𝑦 = 0 𝑃3 (2, 0)
Step 2 Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

𝑁𝑜𝑡𝑒:
𝑦𝑎𝑏𝑜𝑣𝑒 : 𝑦 = 𝑥 2
𝑦𝑏𝑒𝑙𝑜𝑤 : 𝑦 = 0
𝑏
𝑉 = 𝜋 ∫𝑎 (𝑦𝑎𝑏𝑜𝑣𝑒 − 𝑦𝑏𝑒𝑙𝑜𝑤 )2 𝑑𝑥

𝟐
𝑽 = 𝝅 ∫ (𝒙𝟐 − 𝟎)𝟐 𝒅𝒙
𝟏
𝑽 = 𝟔. 𝟐𝝅 𝒐𝒓 𝟏𝟗. 𝟒𝟖 𝒖𝒏𝒊𝒕𝒔𝟑

INTEGRAL CALCULUS 9
Example 8: 𝑦 = 𝑥 3 − 8, 𝑡ℎ𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑎𝑥𝑒𝑠; 𝑎𝑏𝑜𝑢𝑡 𝑦 = 0
Step 1: Determine points of intersection
𝑦 = 𝑥 3 − 8, 𝑥 = 0, 𝑦 = 03 − 8, 𝑦 = −8 𝑃1 (0, −8)
3
𝑦 = 𝑥 − 8, , 𝑦 = 0, 𝑥 3 = 8, 𝑥=2 𝑃2 (2, 0)
𝑥 = 0, 𝑦 = 0 𝑃3 (0, 0)

Step 2 Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

𝑁𝑜𝑡𝑒:
𝑦𝑎𝑏𝑜𝑣𝑒 : 𝑦 = 0
𝑦𝑏𝑒𝑙𝑜𝑤 : 𝑦 = x 3 − 8
𝑏
𝑉 = 𝜋 ∫𝑎 (𝑦𝑎𝑏𝑜𝑣𝑒 − 𝑦𝑏𝑒𝑙𝑜𝑤 )2 𝑑𝑥
𝟐 𝟐
𝑽 = 𝝅 ∫ (𝟎 − (𝒙𝟑 − 𝟖)) 𝒅𝒙
𝟎
𝑽 = 𝟖𝟐. 𝟐𝟖𝟔𝝅 𝒐𝒓 𝟐𝟓𝟖. 𝟓𝟏 𝒖𝒏𝒊𝒕𝒔𝟑

INTEGRAL CALCULUS 10
 ASSESSMENT 14
First semester SY 2020 – 2021

NAME
PROGRAM

General Instructions: Solve the following problem neatly and completely. Show the graph and
your complete solution. Box your final answers. Answers/Solutions must be handwritten in a bond
paper.

Find the volume of the solid generated by revolving the area bounded by the given curves about
the indicated axis

1. x + y = 6, x = 3, x − axis about y = 0

2. 𝑦 2 = 4𝑎𝑥, 𝑥 = 𝑎, 𝑎𝑏𝑜𝑢𝑡 𝑥 = 𝑎

3. y 2 = 2x + 4, y 2 = 4 − 2x; about the y − axis

4. 𝑥 + 𝑦 = 6, 𝑦 = 3, 𝑥 = 0; 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑦 − 𝑎𝑥𝑖𝑠

5. 𝑦 = 2 arctan(0.2𝑥) , 𝑦 = 0, 𝑥 = 0, 𝑥 = 5 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠

2
6. 𝑦 = 𝑒 −𝑥 , 𝑦 = 0, 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠
𝑥 𝑥
7. 𝑦 = 𝑒 −2 + 𝑒 2 , 𝑦 = 0, 𝑥 = −1, 𝑥 = 2, 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠

INTEGRAL CALCULUS 11
Finals Week 4
The Washer Method
If the lower end of the rectangle does not rest on the x- axis or a line, the solid of revolution
generated by revolving about x-axis is a circular ring or washer

Assuming that the functions 𝑓(𝑥) and 𝑔(𝑥) are continuous and non-negative on the interval 𝑎
and 𝑏. Consider a region that is bounded by two curves 𝑦 = 𝑔(𝑥) and 𝑦 = 𝑓(𝑥) between 𝑥 = 𝑎
and 𝑥 = 𝑏.

INTEGRAL CALCULUS 12
The volume of the solid formed by revolving the region about the 𝑥 − 𝑎𝑥𝑖𝑠 is
𝑏
𝑉 = 𝜋 ∫ (𝑓 (𝑥 )2 − 𝑔(𝑥 )2 )𝑑𝑥
𝑎
At point x on the x-axis, a perpendicular cross section of the solid washer-shape with the inner
radius 𝑟 = 𝑔(𝑥) and the outer radius 𝑅 = 𝑓(𝑥).

The volume of the solid generated by revolving about the y-axis a region between curves x=f(y)
and x=g(y), where 𝑔(𝑦) ≤ 𝑓(𝑦)𝑎𝑛𝑑 𝑐 ≤ 𝑦 ≤ 𝑑 is given by the formula
𝑑
𝑉 = 𝜋 ∫ (𝑓(𝑦)2 − 𝑔(𝑦)2 )𝑑𝑦
𝑐

INTEGRAL CALCULUS 13
Determine the volume of the solid generated by revolving the area bounded by the given curves
about the indicated axis of revolution
Example 1:
𝑦 = 𝑥2, 𝑥 = 3, 𝑦 = 0 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑦 − 𝑎𝑥𝑖𝑠
Step 1: Identify points of intersection
𝑦 = 𝑥2, 𝑥 = 3 𝑦 = 32 = 9 𝑃1 (3,9)
2
𝑦=𝑥 , 𝑦=0 𝑦 = 02 = 0 𝑃2 (0,0)
𝑥 = 3, 𝑦 = 0 𝑃3 (3,0)

Step 2: Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

Note:
𝑥𝑟𝑖𝑔ℎ𝑡 : 𝑥 = 3
𝑥𝑙𝑒𝑓𝑡 : 𝑥 = √𝑦

INTEGRAL CALCULUS 14
 9
2
𝑉 = 𝜋 ∫ ((3)2 − (√𝑦) ) 𝑑𝑦
0
𝑉 = 40.5𝜋 𝑜𝑟 127.23 𝑢𝑛𝑖𝑡𝑠 3

Example 2:
𝑥 + 𝑦 = 14 𝑥 = 2, 𝑦 = 2 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠
Step 1: Identify points of intersection
𝑥 + 𝑦 = 14, 𝑥 = 2 𝑦 = 14 − 2 = 12 𝑃1 (2,12)
𝑥 + 𝑦 = 14, 𝑦 = 2 x= 14 − 2 = 12 𝑃2 (12,2)
𝑥 = 2, 𝑦 = 2 𝑃3 (2,2)

Step 2: Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

INTEGRAL CALCULUS 15
 Note:
𝑦𝑎𝑏𝑜𝑣𝑒 : 𝑦 = 14 − 𝑥
𝑦𝑏𝑒𝑙𝑜𝑤 : 𝑦 = 2
12
𝑉 = 𝜋 ∫ ((14 − 𝑥)2 − (2)2 )𝑑𝑥
2
𝑉 = 533.33𝜋 𝑜𝑟 1675.52 𝑢𝑛𝑖𝑡𝑠 3

Example 3:
𝑦 = 𝑥, 𝑥 = 2, 𝑦 = 0 𝑎𝑏𝑜𝑢𝑡 𝑥 = 0
Step 1: Identify point of intersection
𝑦 = 𝑥, 𝑥 = 2 𝑦=2 𝑃1 (2,2)
𝑦 = 𝑥, 𝑦 = 0 𝑥=0 𝑃2 (0,0)
𝑥 = 2, , 𝑦 = 0 𝑃3 (2,0)

Step 2: Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

INTEGRAL CALCULUS 16
Note:
𝑥𝑟𝑖𝑔ℎ𝑡 : 𝑥 = 2
𝑥𝑙𝑒𝑓𝑡 : 𝑥 = 𝑦

2
𝑉 = 𝜋 ∫ ((2)2 − (𝑦)2 )𝑑𝑦
0
𝑉 = 5.33𝜋 𝑜𝑟 16.76 𝑢𝑛𝑖𝑡𝑠 3

Example 4

𝑦 = 2 − 𝑥2, 𝑦 = 2 − √𝑥 𝑎𝑏𝑜𝑢𝑡 𝑦 = 2
Step 1: Identify point of intersection
𝑦 = 2 − 𝑥2, 𝑦 = 2 2 − 𝑥2 = 2 𝑥=0 𝑃1 (0,2)
𝑦 = 2 − 𝑥 2 , 𝑦 = 2 − √𝑥 2 − 𝑥 2 = 2 − √𝑥 𝑥=1
@𝑥 = 1, 𝑦 = 1 𝑃2 (1,1)
Step 2: Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

Note:
𝑦𝑎𝑏𝑜𝑣𝑒 : 𝑦 = 2 − 𝑥 2
INTEGRAL CALCULUS 17
 𝑦𝑏𝑒𝑙𝑜𝑤 : 𝑦 = 2 − √𝑥
1
2
𝑉 = 𝜋 ∫ ((2 − 𝑥 2 )2 − (2 − √𝑥) ) 𝑑𝑥
0
𝑉 = 1.03𝜋 𝑜𝑟 3.25𝑢𝑛𝑖𝑡𝑠 3

Example 5:
𝑥 2 + 𝑦 2 = 25, 𝑥 + 𝑦 = 5(𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑎𝑟𝑒𝑎) 𝑎𝑏𝑜𝑢𝑡 𝑥 = 0
Step 1: Identify points of intersection
𝑥 2 + 𝑦 2 = 25, 𝑦 = 5 − 𝑥 𝑥 2 + (5 − 𝑥)2 = 25 𝑥 = 0,5
@𝑥 = 0, 𝑦 = 5 𝑃1 (0,5)
@𝑥 = 5, 𝑦 = 0 𝑃2 (5,0)

Step 2: Sketch the graph

Consider the smaller area or the area to the right of the line 𝑥 + 𝑦 = 5
Step 3: Solve the volume generated from rotating the designated axis

INTEGRAL CALCULUS 18
 Note:
𝑥𝑟𝑖𝑔ℎ𝑡 : 𝑥 = √25 − 𝑦 2
𝑥𝑙𝑒𝑓𝑡 : 𝑥 = 5 − 𝑦
5 2
𝑉 = 𝜋 ∫ ((√25 − 𝑦 2 ) − (5 − 𝑦)2 ) 𝑑𝑦
0
𝑉 = 41.67𝜋 𝑜𝑟 130.90 𝑢𝑛𝑖𝑡𝑠 3

Example 6:
𝑦 = 𝑥3, 𝑥 = 2, 𝑥 − 𝑎𝑥𝑖𝑠 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑦 − 𝑎𝑥𝑖𝑠
Step 1: Identify points of intersection
𝑦 = 𝑥3, 𝑥 = 2 𝑦 = 23 = 8 𝑃1 (2,8)
3
𝑦=𝑥 , 𝑥=0 𝑦 = 03 = 0 𝑃2 (0,0)
Step 2: Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

INTEGRAL CALCULUS 19
 Note:
𝑥𝑟𝑖𝑔ℎ𝑡 : 𝑥 = 2
𝑥𝑙𝑒𝑓𝑡 : 𝑥 = 3√𝑦
8
2
𝑉 = 𝜋 ∫ ((2)2 − ( 3√𝑦) ) 𝑑𝑦
0
𝑉 = 12.8𝜋 𝑜𝑟 40.21 𝑢𝑛𝑖𝑡𝑠 3

Example 7:
𝑦 2 = 4𝑥, 𝑥 2 = 4𝑦 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠
Step 1: Identify points of intersection
𝑥2
𝑦 2 = 4𝑥, 𝑥 2 = 4𝑦 ( )2 = 4𝑥 𝑥 = 0,4
4
@𝑥 = 4, 𝑦 = 4 𝑃1 (4,4)
@𝑥 = 0, 𝑦 = 0 𝑃1 (0,0)
Step 2: Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

INTEGRAL CALCULUS 20
 Note:
𝑦𝑎𝑏𝑜𝑣𝑒 : 𝑦 = √4𝑥
𝑥2
𝑦𝑏𝑒𝑙𝑜𝑤 : 𝑦 = 4

4 2
2 𝑥2
𝑉 = 𝜋 ∫ ((√4𝑥) − ( ) ) 𝑑𝑥
0 4
𝑉 = 19.2𝜋 𝑜𝑟 60.32𝑢𝑛𝑖𝑡𝑠 3

Example 8:
𝑦 2 = 8𝑥, 𝑦 = 2𝑥 𝑎𝑏𝑜𝑢𝑡 𝑦 = 4
Step 1: Identify points of intersection
𝑦 2 = 8𝑥, 𝑦 = 2𝑥 (2𝑥)2 = 8𝑥 𝑥 = 0,2
@𝑥 = 2, 𝑦 = 4 𝑃1 (2,4)
@𝑥 = 0, 𝑦 = 0 𝑃1 (0,0)
Step 2: Sketch the graph

Step 3: Solve the volume generated from rotating the designated axis

INTEGRAL CALCULUS 21
Note:
𝑦𝑎𝑏𝑜𝑣𝑒 : 𝑦 = √8𝑥
𝑦𝑏𝑒𝑙𝑜𝑤 : 𝑦 = 2𝑥
2
2
𝑉 = 𝜋 ∫ ((√8𝑥) − (2𝑥)2 ) 𝑑𝑥
0
𝑉 = 5.33𝜋 𝑜𝑟 16.76𝑢𝑛𝑖𝑡𝑠 3

INTEGRAL CALCULUS 22
 ASSESSMENT 15
First semester SY 2020 – 2021

NAME
PROGRAM

General Instructions: Solve the following problem neatly and completely. Show the graph and
your complete solution. Box your final answers. Answers/Solutions must be handwritten in a bond
paper.

Find the volume of the solid generated by revolving the area bounded by the given curves about
the indicated axis of revolution

1. 𝑦 2 = 4𝑎𝑥, 𝑥 = 𝑎, 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑦 − 𝑎𝑥𝑖𝑠

2. 𝑦 2 = 𝑥, 𝑦 = 𝑥 − 2, 𝑎𝑏𝑜𝑢𝑡 𝑥 = 0

3. 𝑥 2 + 𝑦 2 = 4; 𝑎𝑏𝑜𝑢𝑡 𝑥 = 2

4. 𝑦 = 4𝑥 − 𝑥 2 , 𝑦 = 𝑥; 𝑎𝑏𝑜𝑢𝑡 𝑥 = 3

5. 𝑦 = √2𝑥, 𝑥 2 = 𝑦 𝑎𝑏𝑜𝑢𝑡 𝑦 = 0

1
6. 𝑦 = , 𝑥 = 5, 𝑦 = 4, 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠
√𝑥+1

2
7. 𝑦 = 𝑥 3 , 𝑥 = 4, 𝑦 = 0; 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑦 − 𝑎𝑥𝑖𝑠

REFERENCES:
washer method lesson reference: Volume of a Solid of Revolution: Disks and Washers
(math24.net)
reference book used: Differential and Integral Calculus by Feliciano
disk and washer method graphing application used: GeoGebra Classic

INTEGRAL CALCULUS 23