DUM20132 ENGINEERING MATHEMATICS 2

DUM20132 ENGINEERING MATHEMATICS 2 49

f '(x) = vu'− uv'
v2

( ) ( )x2 + 4 (3) − (3x − 1)(2x) 3x2 + 12 − 6x2 − 2x
==
( )x2 + 4 2
( )x2 + 4 2

3=x2 + 1x22 −+ 64x22 + 2x −3x2 + 2x + 12
x2 + 4 2
( ) ( )=

Task E : By using quotient rule, find the derivative of the following functions.

1. y = x3 + 3 2. y = 3x2
x2 + 1 1+ 3x3

x4 + 3x2 − 6x 6x − 9x4

( )x2 + 1 2 ( )1+ 3x3 2

3. y = sin x 4. y = ex
x +1 cos x

3cos3x (x + 1) − sin3 x ex cos x + ex sin x
(x + 1)2 cos2 x

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3.3.3 Chain Rule

Composite of functions is when one function is inside of another function.
For example:

=y (2x − 1)2

We can say that this function, y, was formed by the composition of two other functions, the
inside function, and the outside function.

For=y (2x − 1)2 , the inside function is 2x − 1 and the outside function is z2 . The letter z was

just used to represent a different variable.

To differentiate a function of a function, y = f g(x) , that is to find dy , we need to do two
dx

things:

1. Substitute u = g(x) . This gives us:

y = f(u)

2. Next, we need to use Chain Rule is a rule for differentiating compositions of functions.
The Chain Rule states formally that:

{ }D f (g(x)) ( )= f ' g(x) g'(x) ~ Same as derivative of composite function

or
d=y dy × du
dx du dx

Example 3.5 :

Differentiate=y (3x + 1)2 .

Solution :

y '= 2(3x + 1)1 (3)= 6(3x + 1)= 18x + 6

Example 3.6 :
Use Chain rule to differentiate y= (1− 3x)3
Solution:

u =1− 3x y =u3
du =−3 dy =3u2
dx du
Chain Rule :

d=y dy × du
dx du dx

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d=y 3u2 × (−3)
dx
dy = −9u2
dx
dy =−9(1− 3x)2
dx

Example 3.7 :
Differentiate y = cos x3

Solution:

=u x=3 y cosu

du = 3x2 dy = − sinu
dx du
Chain Rule :

d=y dy × du
dx du dx

dy =− sinu × 3x2
dx
dy = −3x2 sinu
dx
dy = −3x2 sin x3
dx

Task F : By using chain rule, find the derivative of the following functions.

( )=1. y sin 3x2 + x ( )2=. y 7 3x2 + x 3

( )(6x + 1)cos 3x2 + x ( )21(6x + 1) 3x2 + x 2
( )4.=y 5x3 − 3 7
( )=3. y cos 4x2 + 5

( )−8x sin 4x2 + 5 ( )105x2 5x3 − 3 6

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3.3.4 Parametric Differentiation

In the geometry of straight lines, circles etc., we encounter parametric equation in which
variables x and y related to each other by a formula, may be expressed individually in terms
of a third variable, usually t or θ , called a parameter.

For example:
Parametric form: x = h(t) and y = g(t) .

It is often necessary to find the rate of change of a function defined parametrically. That is,
we want to calculate dy .

dx
For that reason, we had to use parametric differentiation applying Chain Rule:
d=y dy × du
dx du dx

The following example will show how this is achieved.

Example 3.8

Find dy when x= t3 − t and y= 4 − t2 .
dx

Solution: y= 4 − t2
dy = −2t
x= t3 − t dt
d=x 3t2 − 1
dt

From the Chain Rule, Note : 
 
d=y dy ÷ dx  d=y dy × dt 
dx dt dt
dx dt dx 
dy  
dy = dt  d=y dy ÷ 1 
dx dx  dx dt dt 

dt  dx 
dy = −2t  
dx 3t2 − 1  d=y dy ÷ dx 
 dx dt dt 

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Example 3.9

Find dy when x = t3 and y= t2 − t .
dx

Solution: y= t2 − t
dy= 2t − 1
x = t3 dt
dx = 3t2
dt

From the Chain Rule,

dy dy
dx
= dt
dx
dt

dy = 2t − 1
dx 3t2

dy

Task G : Find in term of t of the following parametric equations.

dx

1. y= t2 + 1 and x = 6t

2t
6

2. y = 3 and x = 5t
t2

3. y= t − 3t2 and x= t − 1 −6
t 5t3

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( )t2 + 1

DUM20132 ENGINEERING MATHEMATICS 2 54

4. =y 3t2 − 1 and =x 7t − 1

6t
7

5. Consider a parametric curve, x = 2 cos 3t and y = 2 sin 2t . Find dy for this curve.
dx

− 2cos 2t
3 sin3t

3.3.5 Implicit Differentiation

Some relationships between two variables x and y do not give y explicitly in terms of x ( or x
explicitly in terms of y) ; but, nevertheless, it is implied that one of the two variables is a function of
the other.
We shall normally assume that y is a function of x.

For example:

x2 + y2 − 4x + 5y − 8 =0

It would be quite difficult to rearrange this so that y explicitly as a function of x. A function given in
this way is said to be defined implicitly.

To differentiate a function in this form, it will be necessary to use Chain Rule discussed before:
d=y dy × du
dx du dx

Example 3.10: dy

Suppose z = y2 . Find .

It follows that: dx

dz = 2y
dy

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From Chain Rule:

d=z dz × dy
dx dy dx
dz =2y × dy =2y dy
dx dx dx
z = y2
d(y2 ) = 2y dy

dx dx

Note that, in order to differentiate y2 with respect to x we have differentiated y2 with respect
to y , and the multiplied by dy , i.e.

dx

d=(y2 ) d (y2 ) × dy
dx dy dx

In general, the rule for differentiating a function of y with respect to x is:

d=[f(y)] d [f(y)] × dy

dx dy dx

Example 3.11

Differentiate 3y2 + 2x =4

Solution:
Differentiate each term with respect to x:

( )d 3y2 + d (2x) =d (4)

dx dx dx

Differentiating function of x to x is straightforward. Differentiating a function of y with respect
to x, should follow the given rule,

( ) d dy 
 dy 3y2 × dx  + 2 =0
 

that is

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6y dy + 2 =0
dx
dy

Rearrange to collect all terms involving together.

dx
6y dy = −2

dx

then

dy = −2
dx 6y
dy = −1
dx 3y

Example 3.12

Differentiate xy2 − 3y =x

Solution :
Differentiate each term with respect to x:

( )d xy2 − d (3y) =d (x)

dx dx dx

Differentiating function of x with respect to x is straightforward. Differentiate a function of y
with respect to x as before. The first term is a product, need to use product rule,

( )y2d (x) + x d y2  − d (3y) × d =1
dx dx  dy dx


So that

y2( )(1)+ x d y2 dy  − d (3y) dy =1
dy dx  dy dx
 


Tidying this up gives

y2 + x2y dy − 3 dy =1
dx dx

dy

Rearrange to collect all terms involving together.

dx

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y2 − 1 =−2xy dy + 3 dy
dx dx

y2 − 1 =(−2xy + 3) dy

dx

finally

dy = y2 − 1
dx −2xy + 3

Example 3.13

Differentiate y2 + x3 − y3 + 6 =3y

Solution:

Differentiate each term with respect to x:

d (y2 ) + d (x3 ) − d (y3 ) + d (6) =d (3y)

dx dx dx dx dx

Differentiating function of x to x is straightforward. Differentiating a function of y with respect
to x, as before,

( ) ( )d y2 × dy + 3x2 − d y3 × dy +=0 d (3y) × dy
dy dx dy dx dy dx

that is

2y dy + 3x2 − 3y2 dy =3 dy
dx dx dx

Rearrange to collect all terms involving dy together.
dx

3x2 =3 dy − 2y dy + 3y2 dy
dx dx dx

then,

( )3x2 = 3 − 2y + 3y2 dy
dx

dy = 3x2

dx 3 − 2y + 3y2

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Example 3.14

Differentiate sin y + x2y3 − cos x =2y

Solution:
Differentiate each term with respect to x:

( )d (sin y) + d x2y3 − d (cos x) =d (2y)
dx dx dx dx

Differentiate a function of y with respect to x, as before. The second term is a product, need to
use product rule,

d + x2 d d (2y) × dy
dx dx
( ) ( )d(siny)× dy +  y3 x2 y3  + sin=x dy dx
dx  
dy

So that

( )cos dy  d dy  =2 dy
y dx +  y3 (2x) + x2 dy y3 dx  + sin x dx



Tidying this up gives

cos y dy + 2y3x + x2 3y2 dy + sin x =2 dy
dx dx dx

dy

Rearrange to collect all terms involving together.

dx

2y3x + sin x = 2 dy − cos y dy − 3x2y2 dy
dx dx dx

( )2y3x + sin x =2 − cos y − 3x2y2 dy
dx

Finally,

dy = 2 2y3x + sin x
dx − cos y − 3x2y2

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Task H: Find the derivative, with respect to x, of each of the following functions.

Example 3.15 1. 2y2= 7 − x

x2 + y2 =25

d (x2 ) + d (y2 ) =d (25) −1
dx dx dx 4y
2 x+ d (y2 ) dy =0
3. 5 cos=y 3x + 2y
dy dx
2 x+ 2y dy =0

dx
dy = −2x = − x
dx 2y y

2. 4xy + 2y =x2

x − 2y −3
2x + 1 2 + 5 sin y

4. 2x2 + 2xy − y3 =−70 5. Given a function of curve as

2x3 − 5y2 =5 . Find the gradient
dy dy

function, . Hence, evaluate at

dx dx

point (2,-1).

4x + 2y −4
3y2 − 2x 5

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3.3.6 Higher Order Derivatives

Let y = f ( x) be a continuous function defined at some interval. The first derivative f is f ' . If

f ' is differentiable, then its derivative is denoted as f '' and it is called the second derivative of
f . As long as the function and its derivative is differentiable, we can continue the process of
differentiating the function to obtain the third, fourth, fifth and even higher derivative of f .

Theorem f '(x) Symbol
First derivative f ''(x)
Second derivative f '''(x) dy
Third derivative fiv ( x) dx
Fourth derivative d2 y
fn (x) dx2
nth derivative d3 y
dx3
d4 y
dx4

dn y
dxn

Example 3.16 :

Find f '''(x) , given that f ( x) = 6x−9 .

Solution:

f ' ( x) = −54x−10
f '' ( x) = 540x−11
f ''' ( x) =−5940x−12 =− 5x91420

Example 3.17 :

Given=h((t) 3 sin 2t − 4t5 , find h '' (t) .

Solution:

=h((t) 3 sin 2t − 4t5
h ' (t=) 3(cos 2t)(2) − 20t=4 6 cos 2t − 20t4
h ' (t) =6(− sin 2t)(2) − 80t3 =−12 sin 2t − 80t3

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Revision Unit 3

3.1 INTRODUCTION OF DIFFERENTIATION 16.If A = (3 − r )(2 − r ) , find dA .

1. Using the first principle, find the derivative r dr
of =y 3x + 2 .
( )( )17.Differentiate 3x + x2 x3 −1 with the
2. Find the derivative of =y 6x2 − 5 x by
using the first principle. respect to x.

3.2 DERIVATIVE OF STANDARD FUNCTIONS 18.Find d  x x 3  .
dx  − 
3. Differentiate y = sin 3x with the respect of
x. 19.If y = 2x +1 , find dy .
3x − 2 dx
4. Differentiate A = cos 8x with the respect
of x. ( )20.If f(=x ) x3 − 2 6 , find f ' (x) .

5. Differentiate M = e9x with the respect of x. 21.If=f( x ) 3 (2x + 5 )4 , find f ' (1) .

( )6. Differentiate y= tan 5u − 8u3 + ln 3u2 22.Given that f (x) = 18 − 4x , find f' (2) .
3x3
with the respect of u.
23.The trajectory of a particle in a plane as a
( )7. Differentiate=y ln 5x2 − 3 + e4x with the function of the time in second is given by
the parametric equations x = 3t2 + 2t − 3
respect of x.
and=y 2t3 + 2 . Find dy as a function of t.
8. Differentiate y = 56 − 2x +13x2 . dx

9. Differentiate y =x − x2 +100x−4 . 24.The parametric equations for a curve are
43 given by x = θ − sin θ and y =1 − cos θ .

10.Differentiate=y 16 − 7 . Find dy as a function of θ .
x4 x dx

11.Differentiate y = 100x5 −15 x3 . 25. Find dy =if y sin x + cos y .
5 x2 dx

12.Differentiat=e y 14 + x3 . 26. Use implicit differentiation to find dy to
2x4 5 dx

the curve x2 + y2 = 3x3 − 4x2 y .

3.3 TECHNIQUES OF DIFFERENTIATION

13.If y = 2x3 − 5 x2 − 9x + 8 , find the value of 3.4 HIGHER ORDER DIFFERENTIATION
dy when x = −1 .
dx 27.If=y (2x + 3 )4 , find d2 y .
dx2

14.If y =(4x − 3 )(3x + 5 ) , find dy . 28.If f ( x) = (2 3 4 ) , find f '' ( x ) .
x−
dx

15.If v = 2t4 − 3t2 + 7 , find dv . 29. Find f ''' (2 ) if f=(x) 4 x4 − 12 x3
t2 dt
53

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ANSWER: 17. 5 x4 +12x3 − 2x − 3

1. 3 18. −3
−3
2. 12x − 5 ( x )2

3. 3 cos 3x 19. −7

4. −8 sin 8x (3x − 2 )2

5. 9e9x ( )20.18x2 x3 − 2 5

6. 5 sec2 5u − 24u2 + 2 21. 8232
u 22. − 19

7. 10x + 4e4x 24
5x2 − 3 23. 3t2

8. −2 + 36x 2t +1
24. 1 + sin θ
9. 1 − 2 x − 400x−5
43 1 − cos θ
25. cos x
10. − 64 + 7
x5 x2 1 + sin y

11. 60x2 − 3 9x2 − 8xy − 2x
2y + 4x2
28 3 26.
x5 5
12. − + x2

13. 6x2 −10x − 9,value =7 27. 48 (2x + 3 )2

14. 24x +11 28. 24 )3

(2x − 4

15. 4 t − 14 29. f ''' =96 x − 24,value =14.4
t3 5

16. −6 +1
r2

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Why do we need to study integration?
Often, we know the relationship involving the rate of change of two
variables, but we may need to know the direct relationship between the two
variables.
For example, we may know the velocity of an object at a particular time, but
we may want to know the position of the object at that time. To find this
direct relationship, we need to use the process which is opposite to
differentiation. This is called integration (or antidifferentiation).
The processes of integration are used in many applications.
The Petronas Towers in Kuala Lumpur experience high forces due to winds. Integration was used
to design the building for strength.

The Sydney Opera House is a very unusual design based on slices
out of a ball. Many differential equations (one type of integration)
were solved in the design of this building.
Historically, one of the first uses of integration was in finding
the volumes of wine-casks (which have a curved surface). We learn
to find the volume of these objects later (in Unit 5: Application of
Integration).
After completing the unit, students should be able to:
1. Find the integrals of functions by considering integration as a reverse of differentiation.
2. Integrate indefinite integrals by rule of integration.
3. Integrate composite functions by using substitution method.
4. Integrate product of functions using integration by part.
5. Integrate quotient of functions using partial fraction method.
6. Evaluate the definite integrals.

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4.1 Integration as a Reverse Function of Differentiation

We wish to perform the opposite process to differentiation. This is called “antidifferentiation” or
“antiderivative” and later, we will call it "integration".

Differentiate

Function Derivative

Anti - differentiate

ANTIDERIVATIVE

If F’(x) = f(x), then F(x) is an antiderivative of f(x).

Example :
a) If F(x) = 10x, then F’(x) = 10
 F(x) is the antiderivative of f(x) = 10

b) If F(x) = x2, then F’(x) = 2x
 F(x) is the antiderivative of f(x) = 2x

In the above example, F(x) = x2 is not the only function whose derivative is f(x) = 2x.
 G(x) = x2 + 2 has 2x as the derivative
 H(x) = x2 – 7 also has 2x as the derivative

 For any real number, C , the function F(x) = x2 + C has f(x) as an antiderivative.

The family of all antiderivatives of f is indicated by

This is called the
indefinite integral
and is the most
general
antiderivative of f

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4.2 Indefinite Integral Of Function DUM20132 ENGINEERING MATHEMATICS 2 65
INDEFINITE INTEGRAL
f(x) = integral function
For any continuous function, f(x), F(x) = integrated function
C = constant of integration
for any real number C.

FUNCTIONS FORMULA EXAMPLES

a) ∫ a d=x ax + C a) ∫ 3 d=x 3x + C

* a and C are constant

∫b) xn=dx xn+1 + C , n ≠ −1 ∫b) x=4 dx x4+1 + C

n+1 4 +1

∫c) axn=dx axn+1 + C , n ≠ −1 = x5 + C
5
n+1
BASIC ∫c) 3x=4 dx 3x4+1 + C
FUNCTIONS
4 +1
d) ∫ (ax + b)n dx
= 3x5 + C
(ax + b)n+1 5
= (n + 1)(a) + C ; n ≠ −1
d) ∫ (3x + 2)4 =dx (3x + )2 4+1
(4 + 1)(3) + C

= (3x + 2)5 +C

15

∫ ∫e) 1 dx = x−4 dx
x4

= x−4+1 + C
= −4 + 1
x−3 + C
−3

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Task A : Integrate the following functions. DUM20132 ENGINEERING MATHEMATICS 2 66

1. ∫ 3 dx 2. ∫ −2 dx

3. ∫ x4 dx 4. ∫ −x6 dx

5. ∫ x dx 6. ∫ 1 dx
x2

7. ∫ 1 dx 8. ∫ (2x5 + 3x − 7) dx
4x5

9. ∫ (2x − 2)(x2 + 1) dx ∫10. x4 − x2 dx

x4

11. ∫ (x + 1)3 dx 12. ∫ (3 − 5x)9 dx

13. ∫ −4(3x + 2)3 dx 14. ∫ 1 dx
3−x

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FUNCTIONS FORMULA EXAMPLES

a) ∫ 1 dx = ln x + C a) ∫ 2 dx = 2 ln x + C
x x

b) ∫ 3 dx = 3 ln x + C
2x 2

LOGARITHMIC 1 1 ln 1 1 ln
FUNCTIONS ax + b a 3x + 3

b) ∫ dx = ax + b +C c) ∫ 2 dx = 3x +2 +C

d)

∫ 4 2 dx = (4) 1  ln 3x + 2 + C
3x +
3

= 4 ln 3x + 2 + C
3

∫a) e x dx = e x + C ∫a) 2e x dx = 2e x + C
∫b) e3x+2 dx = 1 e3x+2 + C
EXPONENTIAL
FUNCTIONS 3

∫b) eax+b dx = 1 eax+b + C

a

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Task B : Integrate the following functions.

1. ∫ 3 dx 2. ∫  3 + 6x  dx
5x  x 

3. ∫ x 1 1 dx 4. ∫ 1 3 dx
+ 2x +

5. ∫ 4 6 dx 6. ∫ −10 dx
− 3x 6 − 5x

∫7. e5x dx ∫8.  e2 x + 1  dx
 ex 
( )∫9. e5x + 12 dx
∫10. 2 + x10  dx
 3ex 

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FUNCTIONS FORMULA EXAMPLES

a) ∫ cos x dx = sin x + C a)

∫ cos (2x + 3) dx = 1 sin (2x + 3) + C
2

∫TRIGONOMETRIC b) sin x dx = − cos x + C b) ∫ sin 4θ dθ = − 1 cos 4θ +C
FUNCTIONS 4

∫c) sec2 x dx = tan x + C ∫c) 3sec2 x dx = 3 tan x + C

Task C : Integrate the following functions. 2. ∫ 5cos x dx
4. ∫ sin(4x) dx
1. ∫ cos(4x) dx 6. ∫ −10 sin(2x) dx
8. ∫ 2sec2 x dx
3. ∫ 1 cos ( πx ) dx
2

5. ∫ −4 sin x dx

7. ∫ sin  π x  dx
 2 

( )9. ∫ sec2 3x + 6 sin x dx ( )∫10. 5 sec2 x − cos 4x dx

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4.3 Techniques of Integration
4.3.1 Integration by Substitution

FORMULA

STEP: EXAMPLE: EXAMPLE:

1) Make a choice of ∫ ( )(a) x2 + 1 50 2x dx ∫(b) sin2 x cos x dx
‘u’
u = x2 + 1 u = sin x
2) Differentiate ‘u’ with
respect to x du = 2x du = cos x
dx dx
3) Compute for ‘dx’
du = 2x(dx) ⇒ dx = du =du cos x (dx) ⇒=dx du
4) Make the 2x
substitution cos x
for ‘u’ and ‘dx’
( )∫ x2 + 1 50 2x dx ∫ sin2 x cos x dx
5) Evaluate the
resulting integral ∫= u50 (2x )  du  = ∫ u2 (cos x )  du 
 2x   cos x 
6) Replace ‘u’ so that  
the final answer is
in terms of ‘x’ ∫= u50du = ∫ u2du

= u51 + C = u3 + C
51 3

= ( )x2 + 1 51 +C = sin3 x + C
3
51

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Task D : Integrate the followings by using substitution method.

∫1. (1+ 2x)5 dx (u = 1 + 2x) ∫ ( )2. x2 5x3 − 4 6 dx (u = 5x3 – 4)

Answer: (1+ 2x)6 + C 5x3 − 4 7
+C
12 ( )Answer:

105

∫ ( )3. 7 (u = 1 + x2) 4. e3x dx
2x 1+ x2 (u = 1 + e3x)
dx ∫ ( )1+ e3x 2

( )1+ x2 8 Answer: ( )1+ e3x −1 +C

Answer: 8 +C −3

∫5. sin7 x cos x dx (u = sin x) ∫ ( )6. x x2 − 1 20 dx (=u x2 − 1 )

Answer: sin8 x + C Answer: ( )x2 − 1 21 +C
8
42
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4.3.2 Integration by Partial Fraction ∫Example: Find  x+6 3  dx
a) Distinct Linear Factor:  2x2 − x − 

f(x) = A + B + C x +=6 A + B
x−a x−b x−c 2x2 − x − 3 2x − 3 x + 1
(x − a)(x −b)(x − c)

x + 6= A (x + 1) + B(2x − 3)

When x =−1⇒ B =−1

When x = 3 ⇒ A =3
2

∴ ∫  x+6 3 dx =∫  2x3− 3 − x 1  dx
 2x2 − x − + 1

= 3ln 2x − 3 − ln x + 1 + C

2

b) Repeated Linear Factor: ∫Example: Find x dx
(x + 2)2

f(x) =A1 + A2 + A3 + ... + An x= A + B

(x + 2)2 x + 2 (x + 2)2

(x − a)n x − a (x − a)2 (x − a)3 (x − a)n x= A (x + 2) + B

When x =−2⇒ B =−2
2A + B = 0 ⇒ A = 1

x  1 2 
= +  dx
∫ ∫∴ + 2)2 dx  x 2 − + 2)2 
(x  (x 

= ln x + 2 + 2 + C
x+2

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Task E : Integrate the followings by using partial fraction.

∫1.  (1− ) 1 2)  dx ∫2.  3x +1  dx
   2x2 − x − 1
x (3x −

Answer: −ln 1− x + ln 3x − 2 + C Answer: 1 ln 2x + 1 + 4 ln x − 1 + C

63

∫3.  3x + 2  dx ∫4.  x2 − 6x + 3 
 x2 + 3x +    dx
2  (x − 2)3 
 

Answer: 4ln x + 2 − ln x + 1 + C Answer: ln x − 2 + 2 + 5 + C

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DUM20132 ENGINEERING MATHEMATICS 2 74

4.3.3 Integration by Part

FORMULA

The procedures:
i) Split the function into two simpler functions, one called ‘u’ and the other ‘dv’
ii) ‘u’ should be a function which becomes a simpler function after differentiation.
iii) ‘dv’ must be a function that is possible to integrate to obtain ‘v’.
iv) The priority of equating ‘u’ :

I = Inverse Trigo Function ( Eg: sin−1 x, cos−1 x )

L = Logarithm Function ( Eg: log x, ln 2x )
A = Algebraic Function ( Eg: 3x2,5x + 1)

T = Trigonometric Function ( Eg: sin x, cos x )
E = Exponential Function ( Eg: ex,e2x )

Example : ∫b) x ex dx

a) ∫ x cos x dx u =x ⇒ du =dx

u =x ⇒ du =dx dv = exdx ⇒ ∫ dv = ∫ exdx

d=v cos x dx ⇒ ∫ d=v ∫ cos x dx v = ex

v = sin x ∫ ud=v uv − ∫ v du

∫ ud=v uv − ∫ v du = (x)(ex ) − ∫ ex dx
= (x)(sin x) − ∫ sin x dx
= x ex − ex + C
= x sin x − (− cos x) + C

= x sin x + cos x + C

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Task F : Integrate the followings by using integration by part.

∫1. 2x ex dx ∫2. x cos3x dx

Answer: 2x ex − 2ex + C Answer: x sin3x + cos3x + C

∫3. (2x + 1)cos x dx 39

∫4. (2x + 3)(ln x)dx

.

Answer: (2x + 1) sin x + 2cos x + C ( )Answer: x2 + 3x ln x − x2 − 3x + C
2
∫5. ln3x dx
∫6. x e−2x dx

Answer: x ln3x − x + C Answer: − xe−2x − e−2x + C
24
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4.4 Definite Integral
An integral with limits is called a definite integral

F(x) is the integral of f(x) ;

F(b) is the value of the integral
at the upper limit, x = b ;

F(a) is the value of the integral
at the lower limit, x = a
*Note that it does not involve a constant of integration and
it gives us a definite value (a number) at the end of the calculation.

Example: Evaluate the followings.

2 2 25

1. ∫ 6x2 dx 2. ∫ (1+ 2x + x2 ) dx 3. ∫ x dx
0 −1 1

[ ]= 2 2 25
2x3 0  x3  ∫25 1 3
= x + x2 + 3  −1 dx = 2 x 2
 x2
( )= 2(2)3 − 0 1 3

1

= 2(8) =  2 + 4 + 8  −  − 1+ 1− 1  = 2  [25]32 − 1
3
= 16  3  3

=9 = 82 2 @ 82.67
3

∫4.3  x2 − 1  dx ∫5. −1  x − 1 2 dx
2  x2  −2  x 

∫ ( )3 dx = x3 − x −1  3 x −1  −1
x2 − x−2   ∫ ( )−1
 3 − 1  x2 − 2 + x−2 dx = x3 − 2x + 
2  
2 −2  3 − 1

x3 13 − 2

=  3 + x  x3 1  −1
 
2 =  3 − 2x − x 
 
1 (27) 1 8 1 − 2
=  3 + 3  −  3 + 2 
=  − 1 + 2 + 1 −  − 8 + 4 + 1 
3 3 2
= 6 1 @ 6.17
6 = 2 2 − 15 = 5 @ 0.83
3 66

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Task G : Evaluate the followings. DUM20132 ENGINEERING MATHEMATICS 2 77

9 8

1. ∫ x dx 2. ∫ 3 x dx
4 1

Ans : 12.67 Ans : 11.25

∫3. 9 1 dx 27 1 −1
1x
∫4.12 x3 dx

Ans : 4 Ans : 6

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4 DUM20132 ENGINEERING MATHEMATICS 2 78

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

Ans : 31 Ans : -11.5

∫7.3 x 2 − 1 dx ∫8.2 3x2 + 2 x dx
1 x2 1 x2

Ans : 1.33 Ans : 4.1716
Ans : -0.25
2

9. ∫ x (x − 1)(x − 2) dx
1

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Revision Unit 4

4.1 INTEGRATION AS A ∫12.Find sin5x dx . ∫28.Find x 2 ln x dx .
REVERSE FUNCTION OF
DIFFERENTIATION 13.Find ∫29.Find xe−x dx .

1. Given y = 1 x 2 , find dy ∫ 2 sec 3x tan3x dx . ∫30.Find x sin4x dx .
2 dx
∫14.Find sec 2 5 x dx .
∫. Hence, find x dx .

2. If y = (4x + 3)3 , then ∫15.Find 2csc 2 3x dx . ∫31.Find x +3 dx .
∫16.Find cos 6x dx . x2 −4
find dy .
dx ∫32.Find 2x + 3 dx .
+
x(x 1)

Hence, show that 17.Find dx

∫12(4x + 3)2 dx = 4(4x + 3)3 ∫33.Find (x + 1)(x − 1) .

∫ cos ec4x cot 4x dx .

4.2 INDEFINITE INTEGRAL ∫18.Find x cos x dx 4.4 DEFINITE INTEGRAL
∫19.Find e x dx .
3. Find ∫20.Find e5x dx . 2
∫21.Find 8e−3x dx .
∫ (2x + 1)(5x − 4) dx . ∫22.Find 55x dx . ∫34.Evaluate x(1 − x )2 dx .
0

∫4. Find  2 − 3  dx . ∫35.Evaluate 2 4 + 2x dx .
 x2 x4  1 x3

∫5. Find x 2 6 + 3  dx . 36.Calculate the value
 x6 
∫1 (4 x )(4 )
+ − x dx .
−1 x4
∫6. Find 2x2 +9 dx .
x4

∫7. Find (2x + 5)3 dx . 4.3 TECHNIQUES OF 37.Calculate the value
INTEGRATION
∫1 1 dx .
∫23.Find (5x − 12)9 dx .
∫8. Find 10(1 − 2x )4 dx . −1(3 − x )2

∫9. Find 3(1 4 x )5 dx . ∫24.Find x(2x + 9)4 dx . 1
−
∫38.Find (2 + x)(1 − x) dx .
−2

12 ∫25.Find 1− x dx . ∫39.Find 3  x + 1 2 dx .
1 x 
∫10.Find (2x − 5)3 dx . (1 + x)3

∫11.Find (3 + x )(3 − x ) ∫26.Find xe5x+1 dx . 2

x2 dx ∫40.Find (2x − 3)4 dx .
1
∫27.Find ln4x dx .

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ANSWER: 14. tan5 x +c 28. x3 ln x − x3 + c
5 39
x2
∫1. x dx = 2 +C 2 cot 3 x x 1
3 ex ex
15. − + c 29. − − +c

2. 16. sin6 x +c 30. −x cos4 x + sin4 x + c
6 4 16
∫12 (4 x + 3 )2 dx =4 (4 x + 3 )3 + C

3. 10 x3 − 3 x2 −4 x + C 17. − csc4 x +c 31.
32 4
5 ln(x − 2 )− 1 ln(x + 2 )+ c
−2 1 18. ; x sin x + cos x + c
x x3 44
4. + +C
32. 3 ln x − ln(x + 1 ) + c
19. e x + c

5. 2 x3 − 1 +C 20. e5 x + c 33.
x3 5
− 1 ln(x + 1 ) + 1 ln(x − 1 ) + c

22

6. − 2− 3 +C 21. − 8 e−3 x + c
x x3 3
34. 2
3
7. (2x + 5)4 + C
22. 5 5 x +c 35. 9
8 ln3125 2

8. − (1 − 2 x )5 + C 23. (5 x − 12 )10 + c 36. − 26
3
50

9. 1 +C 24.

3 (1 − x)4 (2 x +9 )6 − 9 (2 x +9 )5 + c 37. 1
2
3 24 20
10. − − +C
(2 x 5 )2
9
11. − 9 − x + C 25. − (1 1 )2 +1 +c
x +x 1+x 38.

12. − cos5 x + c 2
5
26. xe5 x +1 − e5 x +1 + c 39. 13 1
5 25 3

27. x ln4 x − x +c 40. 1
4 5
13. 2 sec3 x + c
3

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ENGINEERING MATHEMATICS 2

Engineering Mathematics 2 is an essential guide for diploma students in MARA Japan
Industrial Institute. This book fulfills the requirements of the latest syllabus, where it focuses on
four main topics: Matrices, Vector, Differentiation and Integration. This book serves as a main
resource for students to gain more understanding about mathematics and to excel in their
studies.

Key Features

 Comprehensive content which is based on
the Mara Education Institution syllabus.

 Descriptive notes with ample worked examples.
 Concise summary of the main topics in each

chapter for quick revision.

Nur Amalina binti Zainal has been teaching mathematics for over 10 years. She has
a great passion in teaching Mathematics and be involved in innovation projects. Her
background is in Bachelor of Science (Hons) Management Mathematics helps to drive her
passion in solving daily problems in the most effective way .
Fathiyah binti Jamaludin has taught mathematics subject for more than 10 years. Her
expertise is in calculus and holds a Bachelor of Science (Hons) in Mathematics. She has
received numbers of awards on her excellent work and expertise.
Nurhafiza binti Mohd Hassim has taught mathematics for more than 10 years of teaching
experience, specialising in engineering mathematics. She holds a Bachelor of Civil
Engineering.

Sajidah binti Abdul Rahman has been teaching mathematics for over 24 years.
Her background is in Master of Applied Statistics and Bachelor of Mathematics.
Teaching mathematics is her passion for all time. She also teach numerous
students from other institutions in bussiness mathematics, additional mathematics for
secondary form and Cambridge Mathematics for secondary level.

MARA JAPAN INDUSTRIAL INSTITUTE
Lot 2333 Jalan Kajang Seremban,
43700 Beranang, Selangor.
Web: mjii.kktm.edu.my