College of Engineering, Architecture and Technology
Name: ______________________________________ Course & Year: ______________
Module 1
Course Title
Course Number
Course Description
:
Integration Concepts
:
Integral Calculus
:
Math 122
:
The course introduces the concepts of integration and its
application to some physical problems such as evaluation of areas,
volumes of revolution, force and work. The fundamental formulas
and various techniques of integration are taken up and applied to
both single variable and multi-variable functions. The course also
includes tracing of functions of two variables for a better
appreciation of the interpretation of the double and triple integral as
volume of a three-dimensional region bounded by two or more
surfaces
Total Learning Time :
4 units (4 hours lecture per week)
Pre-requisites
:
Differential Calculus
(if there’s any)
Overview:
Integration is the process of finding for the antiderivative or the inverse of the
derivative of a function. This chapter will discuss indefinite integrals, its definition and
properties. This chapter will also discuss the different integral formulas for algebraic and
transcendental functions.
Learning Outcomes:
At the end of this module, the student should be able to:
1. Demonstrate integration process of algebraic and transcendental functions.
2. Develop mastery on the use of formulas in integrating algebraic and
transcendental functions
3. Evaluate Integrals.
Indicative Content:
This module discusses at least the following topics: The Indefinite Integral, Properties of
Indefinite Integral, The Power Formula, Logarithmic Formula, Exponential Formula,
Trigonometric Functions, Method by Trigonometric Transformations, Inverse
Trigonometric Functions and Hyperbolic Functions.
ENGR. SHAINA D. SUCGANG
1
College of Engineering, Architecture and Technology
Name: ____________________________________
Course & Year: _____________________________
Date Submitted: __________
Rating:
__________
Pre-Assessment:
Evaluate the following integrals.
1. ∫(𝑥 3 − 6)𝑑𝑥
2. ∫
3. ∫
𝑑𝑦
√𝑦
5𝑑𝑧
𝑧5
4. ∫ (𝑎 − 𝑥)3 𝑑𝑥
2
5. ∫ (𝑡 3 + 𝑡) 𝑑𝑡
√
ENGR. SHAINA D. SUCGANG
2
Discussion:
Anti-differentiation
Anti-differentiation or integration is the reverse process to differentiation. For
example, if 𝑓(𝑥) = 2𝑥, we know that this is the derivative of 𝑓(𝑥) = 𝑥 2 .
If we shift the parabola 𝑓(𝑥) = 𝑥 2 by sliding it up or down vertically, all the points
on the curve will still have the same tangent slopes, i.e. derivatives. For example:
All have the same derivative function, 𝑦 ′ = 2𝑥, so a general expression for this
family of curves would be;
𝑦 = 𝑥2 + 𝑐
where c is an arbitrary constant (called the integration constant).
Note: Where possible, check your answer by differentiating, remembering that the
derivative of a constant c, is zero.
In mathematical notation, this anti-derivative is written as
∫ 2𝑥 𝑑𝑥 = 𝑥 2 + 𝑐
The integration symbol “∫ ” is an extended S for “summation”. (You will see why
in Mathematics IM.)
The “𝑑𝑥” part indicates that the integration is with respect to 𝑥. For instance, the
integral ∫ 2𝑥 𝑑𝑡 cannot be found, unless 𝑥 can be rewritten as some function of 𝑡.
Indefinite Integrals
We have been concerned principally with finding the derivative and differential of
a given function. Suppose the process is reversed, i.e., the derivative or differential of a
function is given, and we are asked to find the function. Such a function is called an integral
ENGR. SHAINA D. SUCGANG
3
of the given derivative or differential. The process of finding it is called integration. The
given function is called the integrand.
Let f x  be a given function which is the derivative of a function F x  . We can
express their relation by the equation
F x    f x dx ,
Which is read “ F x  is equal to the integral of f x dx ”. The symbol

is called
the integral sign. The equation indicates that we must find a function F x  whose
differential is f x dx or whose derivative is f x  .
As an example, if
dy  5 x 4 dx,
The following are possible solutions:
y  x 5 , y  x 5  1, y  x 5  6, y  x 5  C,
where C is any constant. The last solution is the general function whose differential is
5 x 4 dx . It embraces all the other solutions. This fact is written
 5x dx  x
4
5
 C.
Theorem: If F x  is a function whose differential is f x dx , or whose derivative is f x 
then F x   C is a whole class of functions having the same differential (or derivative).
General Properties of Indefinite Integrals
The following are properties of indefinite integrals.
I.
 dv  v  C
II.
III.
 dv  dw  ...  dz    dv   dw  ...   dz  v  w  ...  z  C
 cdv  c  dv  C  cv  C
The Power Formula
Let us derive and illustrate the first four formulas of integration.
d v  C   dv , hence, we have the formula
I.
 dv  v  C
d v  w  ...  z  C   dv  dw  ...  dz, Thus, we have the formula
ENGR. SHAINA D. SUCGANG
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II.
 dv  dw  ...  dz    dv   dw  ...   dz  v  w  ...  z  C .
d cv  C   cdv, Where c and C are constants and v is a function of a variable, hence,
we have the formula
III.
 cdv  c  dv  cv  C
 v n1

d 
 C   v n dv, where n and C are constants, v is a function of a variable, and
 n 1

is not equal to -1, hence, we obtain the formula
v n1
n
v
dv

C
IV.
n  1

n 1
Formula IV is called the power formula for integration.
Simple Trigonometric Functions
Consider the differentials
d  cos v  C   sin v dv
d sin v  C   cos v dv
d tan v  C   sec 2 v dv
d  cot v  C   csc 2 v dv
d sec v  C   sec v tan v dv
d  csc v  C   csc v cot v dv
From the above differentials, the following integration formulas are verified.
V.
VI.
VII.
VIII.
IX.
X.
 sin v dv   cos v  C
 cos v dv  sin v  C
 sec v dv  tan v  C
 csc v dv   cot v  C
 sec v tan v dv  sec v  C
 csc v cot v dv   csc v  C
2
2
ENGR. SHAINA D. SUCGANG
5
Let us consider the following cases:
I.
Odd Powers of Sine and Cosine
Let ∫ 𝑠𝑖𝑛𝑚 𝑢 𝑐𝑜𝑠 𝑛 𝑢 𝑑𝑢, where either of the two powers 𝑚 or 𝑛 is a
positive odd integer or both of the two powers 𝑚 and 𝑛 are positive odd integers.
Let 𝑛 be a positive odd integer and in the form ∫ 𝑠𝑖𝑛𝑚 𝑢 𝑐𝑜𝑠 𝑛−1 𝑢 cos 𝑢 𝑑𝑢
and putting,
𝑐𝑜𝑠 2 𝑢 = 1 − 𝑠𝑖𝑛2 𝑢
The process continues and a basic power formula is used to evaluate the
integral. A similar process works for an odd power of sine using,
𝑠𝑖𝑛2 𝑢 = 1 − 𝑐𝑜𝑠 2 𝑢
II.
Even Powers of Sine and Cosine
Let ∫ 𝑠𝑖𝑛𝑚 𝑢 𝑐𝑜𝑠 𝑛 𝑢 𝑑𝑢, where both of the two powers 𝑚 and 𝑛 are
positive even integers.
When both of the two powers are positive even integers, use the following
identities.
1
𝑠𝑖𝑛2 𝑢 = 2 (1 − cos 2𝑢)
1
𝑐𝑜𝑠 2 𝑢 = 2 (1 + cos 2𝑢)
1
sin 𝑢 cos 𝑢 = 2 sin 2𝑢
III.
Powers of Tangent and Cotangent
Let ∫ 𝑡𝑎𝑛𝑚 𝑢 𝑑𝑢 𝑜𝑟 𝑐𝑜𝑡 𝑚 𝑢 𝑑𝑢, where, 𝑚 is either an odd or an even
integer.
Put,
𝑡𝑎𝑛2 𝑢 = 𝑠𝑒𝑐 2 𝑢 − 1
𝑐𝑜𝑡 2 𝑢 = 𝑐𝑠𝑐 2 𝑢 − 1
IV.
Powers of Secant and Cosecant
Let ∫ 𝑠𝑒𝑐 𝑚 𝑢 𝑑𝑢 𝑜𝑟 𝑐𝑠𝑐 𝑚 𝑢 𝑑𝑢, where, 𝑚 is an even integer.
Put,
𝑠𝑒𝑐 2 𝑢 = 1 + 𝑡𝑎𝑛2 𝑢
𝑐𝑠𝑐 2 𝑢 = 1 + 𝑐𝑜𝑡 2 𝑢
ENGR. SHAINA D. SUCGANG
6
V.
Products of Tangent and Secant and Cotangent and Cosecant
Let ∫ 𝑡𝑎𝑛𝑚 𝑢 𝑠𝑒𝑐 𝑛 𝑢 𝑑𝑢 and ∫ 𝑐𝑜𝑡 𝑚 𝑢 𝑐𝑠𝑐 𝑛 𝑢 𝑑𝑢, where, 𝑚 is either an
odd or an even integer and 𝑛 is an even integer.
Put,
𝑠𝑒𝑐 2 𝑢 = 1 + 𝑡𝑎𝑛2 𝑢
𝑐𝑠𝑐 2 𝑢 = 1 + 𝑐𝑜𝑡 2 𝑢
VI.
Products of Sine and Cosine
For the product of sine and cosine of different angles, use the following
∫ sin(𝑚𝑢) cos(𝑛𝑢) 𝑑𝑢 =
∫ sin(𝑚𝑢) sin(𝑛𝑢) 𝑑𝑢 =
∫ cos(𝑚𝑢) cos(𝑛𝑢) 𝑑𝑢 =
cos(𝑚−𝑛)𝑢
2(𝑚−𝑛)
sin(𝑚−𝑛)𝑢
2(𝑚−𝑛)
−
cos(𝑚−𝑛)𝑢
2(𝑚−𝑛)
−
cos(𝑚+𝑛)𝑢
2(𝑚+𝑛)
sin(𝑚+𝑛)𝑢
+
2(𝑚+𝑛)
+𝐶
cos(𝑚+𝑛)𝑢
2(𝑚+𝑛)
+𝐶
+𝐶
Logarithmic Function
Consider the differential
dv
d ln v  C  
v
Therefore, we have
dv
XI.
 v  ln v  C
𝑑𝑢
The logarithmic formula is, ∫ 𝑢 = ln u + c, u > 0
The integral of any quotient when the numerator is the derivative of the denominator is
equal to the logarithm of the numerator.
Exponential Function
Consider the differential
d e v  C  e v dv , and


 av

d 
 C   a v dv
 ln a

Thus,
v
v
XII.
 e dv  e  C , and
ENGR. SHAINA D. SUCGANG
7
VI’
av
 a dv  ln a  C
v
Inverse Trigonometric Functions
Let us consider the differential,
v
dv


d  arcsin  C  
, and
2
a


a  v2
v
dv
1

d  arctan  C   2
2
a
a
 a v
Thus,
dv
v
XIII. 
 arcsin  C
a
a2  v2
dv
1
v
XIV.  2
 arctan  C
2
a
a
a v
Hyperbolic Functions
I.
II.
III.
IV.
V.
VI.
VII.
VIII.
 sec h v dv  tanh v  C
 cosh v dv  sinh v  C
 sinh v dv  cosh v  C
 csc h v dv   coth v  C
 tanh v dv  ln cosh v  C
 sec h v tanh v dv   sec h v  C
 coth v dv  ln sinh v  C
 csc h v cot h v dv   csc h v  C
2
2
Constant of Integration
All of integrations include the constant of integration, C. Any pure constant always
disappears during the differentiation because the differential of a constant is zero. The
constant of integration may assume any constant value, and in order to fix its value, more
conditions must be given.
Suppose we wish to find a function whose differential is 3x 2  4 x  6 dx , and which shall
have a value 10 when x=1.

Solution:

T get the function, integrate the differential,
2
3
2
 3x  4 x  6dx  x  2 x  6 x  C
ENGR. SHAINA D. SUCGANG
8
From the given conditions, this result is 10 when x=1. Thus,
10  1  2  6  C :
C 5
Therefore the required function is x 3  2 x 2  6 x  5.
The geometric meaning of the constant of integration may be understood from the
following problem:
Find the equation of the curve whose slope at any point is twice its abscissa at said point.
dy
Solution:
The slope at any curve at any point is given by . Thus,
dx
dy
 2 x or dy  2 xdx . By means of integration,
dx
2
 dy  2 xdx or y  x  C
Since C may assume any value, this equation only represents the family
of parabolas opening upwards, with vertices on the y-axis. If for example, the parabola
passes thru the point 2,1 , then y=1 when x=2. Thus, 1  4  C : C  3
Therefore, the required equation is y  x 2  3 .
The physical meaning of the constant of integration may be interpreted from the following
problem in motion.
Derive the equations of motion of a particle moving in a straight vertical line under the
influence of the acceleration of gravity “g”, which is constant.
Solution:
We have learned that
dv
dv
a  , or g  . Thus,
dt
dt
dv  g dt or v  gt  C
To get the value of C, suppose when t=0 sec, v=2 ft per sec. Then,
2  0C :
C2
Therefore, v  gt  2
ds
ds
We also know that v  , or gt  2  . Thus,
dt
dt
1 2
ds  gt  2dt or s  gt  2t  C
2
To get the value of C, suppose s=3 when t=0. Then,
C  3 . The distance equation becomes
1
s  gt 2  2t  3
2
ENGR. SHAINA D. SUCGANG
9
Definite Integral
After studying the fundamental formulas and other methods of integration, let us
focus our attention to the physical applications of integration in solving various problems
such as areas, volumes of solids, lengths of curves, centroids, moments of inertia, fluid
pressure and work done by a variable force. For this purpose, we will employ definite
integration using the definite integral.
Assume a given function f x  to be continuous between x  a and x  b and let
F x  be an anti-derivative or integral of f x  . When x  a, F x  F a and when
x  b, F x  F b. The difference F b  F a  or change in the value of F x  in the
interval x  a, x  b is called the definite integral of f x  between the limits x  a and
x  b , or simply the integral of f x  from a to b.
b
Definite integral will be denoted by the symbol
 f x dx where the numbers a and
a
b are called the lower and upper limits of integration, respectively. The value of the definite
b
integral is given by the equation  f x dx  F x a  F b  F a 
b
a
The above equation, can be stated as “the definite integral is the value of the indefinite
integral at the upper limit minus its values at the lower limit.”
Note: The definite integral is independent of C.
The value of definite integral is not a function of the variable of integration but a
Constant number.
1.
0
2
 x4

x  2 x dx    x 2   4  4  0  8
4
0

2

3



t
1
1
1 
 1
2.  cos dt   2 1  cos t  dt  t  sin t 02    1  0    2
0
2
2
2
22
 4
0
2
2
2
3. Evaluate  x 3 ln x dx .
1
Solution: By means of integrating by parts. Let
u  ln x,
dv  x 3 dx
dx
1
du  ,
v  x4
x
4
Therefore,

2
1
2
1
 1 2
x ln x dx   x 4 ln x   x 3 dx
4
1 4 1
3
ENGR. SHAINA D. SUCGANG
10
2
1 
1
  x 4 ln x  x 4 
16 1
4
1
1
 16 ln 2  0   16  1
4
16
15
 4 ln 2 
16
Properties of the Definite Integral
Certain properties of definite integrals are as follows:
1
 f xdx   f xdx
2  f xdx   f xdx   f xdx
3  f xdx   f t dt   f z dz
b
a
a
b
c
b
b
a
b
a
d
c
f
a
c
e
Integrals of Even and Odd Function
A function is called an even function if it remains unchanged when x is replaced by  x .
f  x   f x 
An odd function is one whose sign changes when x is replaced by  x.
f  x   f x
In integrating an even or odd function, the following properties will be followed:
Theorem I: If f x  is an even function, then  f x  dx  2 f x  dx
a
a
a
a
0
Theorem II: If f x  is an odd function, then  f x dx  0
a
Properties of the Definite Integral
Certain properties of definite integrals are as follows:
1
 f xdx   f xdx
2  f xdx   f xdx   f xdx
3  f xdx   f t dt   f z dz
b
a
a
b
c
b
b
a
b
a
d
c
f
a
c
e
ENGR. SHAINA D. SUCGANG
11
Integrals of Even and Odd Functions
A function is called an even function if it remains unchanged when x is replaced by  x .
f  x   f x 
An odd function is one whose sign changes when x is replaced by  x. f  x   f x
In integrating an even or odd function, the following properties will be followed:
Theorem I: If f x  is an even function, then  f x  dx  2 f x  dx
Theorem II: If f x  is an odd function, the
a
a
a
a
0
 f xdx  0
a
Wallis’ Formula:
Consider the integral


2
0
sin m x cos n x dx
2 
2







m

1
m

3
.....
or
n

1
n

3
......
or




1 
1

m
n
2
This will yield into  sin x cos x dx 
 
0
2

m  n m  n  2......or 1


Where m and n are integers  0.


, If m and n are both even,
2
  1, If either one or both are odd,
And that the lower and upper limits are 0 and

respectively.
2
The above expression is called Wallis’ Formula.
ENGR. SHAINA D. SUCGANG
12
Exercises/Drills:
1.
1
If 𝑦 ′ = 𝑥, then 𝑦 = 2 𝑥 2 + 𝑐
Solution:
1
(Check: 𝑦 ′ = 2 ∙ 2𝑥 + 0 = 𝑥 )

1
If 𝑦 ′ = 𝑥 2 , then 𝑦 = 3 𝑥 3 + 𝑐
2.
Solution:
1
 (Check: 𝑦 ′ = 3 ∙ 3𝑥 2 + 0 = 𝑥 2 )
3.∫ 𝑥 4 𝑑𝑥 =
𝑥5
5
+𝐶
Solution:
=
𝑥5
+𝐶
5
4.  4 x 4  3 x  5dx = 4 x 4 dx  3 xdx  5 dx
Solution:
4 x 5 3x 2
=

 5x  C
5
2
5.
 x
2

2
 2 dx
Solution:


  x 4  4 x 2  4 dx
=  x 4 dx  4 x 2 dx  4 dx
x5 4x3

 4x  C
=
5
3
ENGR. SHAINA D. SUCGANG
13
 x
6.

3
 5 xdx
2
Solution:
Let v  x 2  5
dv  d x 2  5
 2xdx


We introduce 2 after the integral sign, and neutralize it by putting its reciprocal
before the integral sign.
Hence,
x
=
7.
2
 x
2



3
3
1
 5 xdx     x 2  5 2 xdx
2
 5
C
8
4
.  sin  cos d
Solution:
Let v  sin 
dv  d sin    cos d
Then,  sin  cos  d    cos  sin  d
=
8.
 cos 2 
C
2
Evaluate  sin 4 x dx
Solution:
Let v  4x
dv  4dx
Insert 4 after the integral sign and
1
before the integral sign. Thus
4
1
 sin 4 x dx   4  sin 4 x 4dx
1
=  cos 4 x  C
4
ENGR. SHAINA D. SUCGANG
14
1
2
9.
Evaluate
x
 cos 3dx
Solution:
x
1
, dv  dx
3
3
x
x1
 cos 3 dx  3 cos 3  3  dx
x
= 3 sin  C
3
Let v 
10.
Evaluate
 x sec
2
3x 2 dx
Solution:
Let v  3x 2 , dv  6 x
1
2
2
2
2
 x sec 3x dx  6  6 x sec 3x dx
1
= tan 3 x 2  C
6
11.
1
Evaluate  csc 2 ln x dx
x
Solution:
Let v  ln x, dv 
1
 x csc
12.
2
dx
x
ln x dx   cot ln x  C
Evaluate  e 2 x sec e 2 x tan e 2 x 2 e 2 x dx
Solution:
Let v  e 2 x , dv  2e 2 x dx
1
2x
2x
2x
2x
2x
 e sec e dx  2  sec tan e 2e dx
1
= sec e 2 x  C
2
ENGR. SHAINA D. SUCGANG
15
13.
Evaluate  sin 2 x cos 3 x dx .
Solution:
cos 3 x  cos 2 x cos x
= 1  sin 2 x cos x


Thus,
 sin
2


x cos 3 x dx   sin 2 x 1  sin 2 x cos x dx
=  sin 2 x cos x dx   sin 4 x cos x dx
sin 3 x sin 5 x

C
3
5
Evaluate  sin 2  cos 2  dx
=
14.
Solution:
=  sin  cos   d
2
1

=   sin 2 d
2

2
1
=    sin 2 2 d
2
1
=  1  cos 4  d
8
1
1
=  d   cos 4 d
8
8
1
1
=   sin 4  C
8
32
15.
Evaluate  sin 4  d
Solution:
=  sin 2   d
2
=
1
1  cos 2 2 d
4


1
1  2 cos 2  cos 2 2 d

4
1 
1 1

=  1  2 cos 2   cos 4  d
4 
2 2

=
ENGR. SHAINA D. SUCGANG
16
3
1
1
=   sin 2  sin 4  C
8
4
32
16.
Evaluate  cot 3 x dx
Solution:
=  cot x cot 2 x dx
=  cot x csc 2 x dx
=  cot x csc 2 x dx   cot x dx
=
17.
cot 2 x
 ln sin x  C
2
Evaluate  sec 4 x tan 4 x dx
Solution:
=  tan 4 x sec 2 x sec 2 x dx
=  tan 4 xtan 2 x  1sec 2 x dx
=  tan 6 x sec 2 x dx   tan 4 x sec 2 x dx
=
18.
tan 7 x tan 5 x

C
7
5
Evaluate  sec 5 x tan 3 x dx
Solution:
=  sec 4 x tan 2 x sec x tan x dx
=  sec 6 x sec x tan x dx   sec 4 x sec x tan x dx
=
sec 7 x sec 5 x

C
7
5
ENGR. SHAINA D. SUCGANG
17
 x
2
19.
Evaluate
3
 2 x dx
0
Solution:
2
 x4

   x2 
4
0
 4 40 8

2
20.
t
Evaluate  cos 2 dt
2
0
Solution:

1 2
1  cos t  dt
2 0

1
 t  sin t 02
2
1 
 1
   1  0    2 
22
 4

21.
2
Evaluate  x 3 ln x dx .
1
Solution:
By means of integrating by parts.
Let:
u  ln x,
dv  x 3 dx
dx
1
du  ,
v  x4
x
4
Therefore,

2
1
2
1
 1 2
x ln x dx   x 4 ln x   x 3 dx
4
1 4 1
3
2
1 
1
  x 4 ln x  x 4 
16 1
4
1
1
 16 ln 2  0   16  1
4
16
15
 4 ln 2 
16
ENGR. SHAINA D. SUCGANG
18
22.
Evaluate
16

1  x dx
0
Solution:
Using Algebraic Substitution,
Let
1  x  z,
x  z 1
x  z  1 ,
dx  2z  1 dz
Changing the limits,
If x  0,
z 1
If x  16, z  5
Therefore,
2
16

0
1
1  x dx  2 z 2 z  1 dz
5
1
5
2 5 2 3 
2 z 2  z 2   34.33
3 1
5
3
23.
Evaluate

a
0
2 2
 23

 a  x 3  dx




Solution:
Using Trigonometric Substitution,
Let
1
3
1
3
x  a sin 
x  a sin 3 
dx  3a sin 2  cos d
Changing the limits,
If x  0,
sin   0, or   0
If x  a,
sin   1, or  
3
Therefore,

a
0

2
3
2 2
2

 23

 2
2
 a  x 3  dx   2  a 3  a 3 sin 2   3a sin 2  cos d



0 




 3a
2


2
0
cos 4  sin 2  d
ENGR. SHAINA D. SUCGANG
19

3a 2
2

1
 1  cos 2  4 sin
2
0
2
2 d



3 
 a 2  2 sin 2 2 d   2 sin 2 2 cos 2 d 
0
8 0


3 1
 a 2   2 1  cos 4 d 
8 2 0


2
0

sin 2 2 cos 2 d 


3 1
1
1
2
 a 2    sin 4  sin 3 2 
8 2
8
6
0
3 2  1     3a 2
 a     0 
8 2  2  
32
24.
Evaluate


sin 4 x cos 6 x dx
2
0
Solution:
By Wallis’ formula, m and n are both even.

Therefore,
25.
Evaluate

sin 4 x cos 6 x dx 
2
0

31531     3
 
108642  2  512

2
0
cos 3 x sin 6 x dx
Solution:
Since m is odd
Therefore,
26.


2
0
cos 3 x sin 6 x dx 
Evaluate

2531 1  2
97531
63

2
0
sin 6 x cos x dx
Solution:


2
0
sin 6 x cos x dx 
5.3.11 1  1
7.5.3.1
ENGR. SHAINA D. SUCGANG
7
20
27.
Evaluate 

2
0
cos 8 x dx
Solution:


2
0
cos 8 x dx 
7.5.3.11     35
 
8.6.4.2  2 
ENGR. SHAINA D. SUCGANG
256
21
College of Engineering, Architecture and Technology
Name: ____________________________________
Course & Year: _____________________________
Date Submitted: __________
Rating:
__________
Evaluation:
Evaluate the following integrals and check by differentiation. Write your solutions on the
space provided.
1.  5 xdx
2.
 x  3 dx
4
ENGR. SHAINA D. SUCGANG
3.
 5  x dx
4.
 a  bx dx
.
2
22
 8 x
5.
6.
 sin

4
 5 x 2  2 dx
7.
 cos
3
4 x dx
8.
 sin
ENGR. SHAINA D. SUCGANG
3
2
4 x dx
 d
.
23
9.
 cos
2
1
 d
3
10.  sin 5 3 x dx
ENGR. SHAINA D. SUCGANG
4dx
11.
 3x  2
12.
 1  5x
2dx
24
13.
x
xdx
2
4
14.
x
2
15.
2 x  3dx
x
2
 3x  1
x  1dx
 2 x  16
ENGR. SHAINA D. SUCGANG
25
Evaluate Problems 1-10 by algebraic substitution.
1.

a
2.

4
dx
0
2 x
3.
0
2
4
0

x 3 a 2  x 2 dx
cos x dx
ENGR. SHAINA D. SUCGANG
26
4.
a
 y
0
y 3 dy
2
 a2 
x2  a2
dx
x
5.

2a
6.

4
x dx
0
2 x
a
2
ENGR. SHAINA D. SUCGANG
27
7.
8.
9.
4
dx
2
x x2 1

a
x 3 dx
x2  a2
 
0

4
0
ln


a2 
 Put x 

v 



z  2 dz
ENGR. SHAINA D. SUCGANG
28
10.

2a
a
a2  x2
dx
x4
Solve problems 11-18 by trigonometric substitution.
11. Prob. 1
12. Prob. 5
ENGR. SHAINA D. SUCGANG
29
13. Prob. 6
14. Prob. 7
15. Prob. 8
ENGR. SHAINA D. SUCGANG
30
Additional Reading:
Calculus Differential and Integral by Ignacio J. Sevilla, Carlos G. Cruz, Eduardo Calayan,
Demtrio A. Quirino Jr., Jose M. Mijares
References:
H.J. Terano. (2015). A Simplified Text in Differential Calculus. Camarines Sur Plytechnic
Colleges
ENGR. SHAINA D. SUCGANG
31