# SOLUTION: MASC 0009 2 Middle East College Engineering Mathematics Questions

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Engineering Science1

Engineering Mathematics
[By Name and I.D]

Module Name
Name of Faculty
Session
Date

2

ENGINEERING MATHEMATICS
Question One

Evaluate the subjoined integral:
∫(𝑡 2 + 3) cos(5𝑡)𝑑𝑡𝑏𝑎
ID. No. 18F17988
Let a=1, and b=8
As b>a so 8>1
𝑤𝑒 𝑠ℎ𝑎𝑙𝑙 𝑠𝑜𝑙𝑣𝑒 𝑡ℎ𝑖𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑖𝑛𝑡𝑒𝑟𝑔𝑟𝑎𝑡𝑖𝑛𝑔 𝑏𝑦 𝑝𝑎𝑟𝑡𝑠 𝑖. 𝑒 ∫ 𝑓𝑔′ = 𝑓𝑔 − 𝑓 ′ 𝑔 … … . 𝑒𝑞𝑛 (1)
𝑠𝑖𝑛𝑐𝑒 𝑓 = 𝑡 2 + 3, and g ′ = absorb 5 t,

to meet g we allure intergrate g ′ to finf g to be

by replacing the rates of f g f’ and g’ on eqn (1) we allure allure eqn (2) below:

(𝑡 2 + 3)𝑠𝑖𝑛5(𝑡)
2𝑡𝑠𝑖𝑛(5𝑡)
∫ 𝑓𝑔′ =
−∫
𝑑𝑡
5
5
=∫

2𝑡𝑠𝑖𝑛(5𝑡)
2𝑡𝑠𝑖𝑛(5𝑡)
. 𝑑𝑡 − ∫
𝑑𝑡
5
5

By integrating partially:
𝑓 = 𝑡; 𝑎𝑛𝑑 𝑓 ′ = 1
∫ 𝑓𝑔′ = 𝑓𝑔 − ∫ 𝑓 ′ 𝑔 𝑤ℎ𝑒𝑟𝑒 { ′
cos(5𝑡)
𝑔 = sin(5𝑡) ; 𝑔 = −
5
Which allure then be

𝑡𝑐𝑜𝑠(5𝑡)
cos (5𝑡)
− ∫−
𝑑𝑡
5
5

After substituting the equation over by magnitude, we allure get
=

sin (5𝑡) 𝑡𝑐𝑜𝑠(5𝑡) 2 sin(5𝑡) 2𝑡𝑐𝑜𝑠(5𝑡25)

=

25
5
125
25
(𝑡 2 + 3)sin (5𝑡) 2 sin(5𝑡) 2𝑡𝑐𝑜𝑠(5𝑡)
=

+
5
125
25

Also, on integrating ∫(𝑡 2 + 3). cos(5𝑡) 𝑝𝑎𝑟𝑡𝑖𝑎𝑙𝑙𝑦; we allure allure the equations below

sin (5𝑡)
5

3

ENGINEERING MATHEMATICS
(𝑡 2 + 3)sin (5𝑡) 2 sin(5𝑡) 2𝑡𝑐𝑜𝑠(5𝑡)
(25𝑡 2 + 73 sin(5𝑡) + 10𝑡𝑐𝑜𝑠(5𝑡)

+
+𝑐 =
+𝑐
5
125
25
125
=

98 sin(5) + 10cos (5)
= −0.7
124
𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑠 0.7
Question Two

Evaluate the subjoined Integral by using accordant method:
∫ (𝐸 (−𝑎𝑦) √𝑏 + 𝑒𝑦 −𝑎𝑦 −

2𝑦 3 + 𝑦
) 𝑑𝑦
𝑦4 + 𝑦2 + 1

(Note: Grasp 𝒂 and b any two bulk of your MEC ID enumerate and 𝒂, 𝒃>𝟎)
ID. No. 18F17988
Given the equation
𝑆 = ∫(𝑒

−𝑎𝑦

(2𝑦 3 + 𝑦
𝑑𝑦 … … … … … … … … … … … . (𝑎)
√𝑏 + 𝑒^ − 𝑎𝑦𝑑𝑦 − ∫ 4
𝑦 + 𝑦2 + 1
𝐴𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑡ℎ𝑎𝑡
𝑤 = 𝑏 + 𝑒 −𝑎𝑦
𝑧 = 𝑦 4 + 𝑦12

Then on differentiating w and z rates, we allure
𝑑𝑤
𝑑𝑦
𝑑𝑧
𝑑𝑦

= 0 + (−𝑎)𝑒 −𝑎𝑦 … … … … . (𝑏)
= 4𝑦 3 + 2𝑦 + 0 … … … … … (𝑐)

On simplifying eqn b and c
𝑑𝑤
−𝑎
𝑑𝑧
2

= 𝑒 −𝑎𝑦 𝑑𝑦
= (2𝑦 3 + 𝑦)𝑑𝑦

Therefore, we allure foreclosure eqn (a), hence we have
3
−1
1
1
2
1
𝑆 = ( ) ∫ √𝑧 𝑑𝑧 − ( ) ∫ 𝑑𝑤 = − (𝑏 + 𝑒 −𝑎𝑦 )2 − 𝑙𝑛|𝑦 4 + 𝑦 2 + 1| + 𝐶
𝑎
2
𝑤
3𝑎
2

4

ENGINEERING MATHEMATICS
Therefore, if our a and b rates are as follows;
𝐼𝑓 𝑎 = 1 𝑎𝑛𝑑 𝑏 = 8
The consummate reresolution allure be as follows
3
2
1
𝑆 = − (8 + 𝑒 −𝑦 )2 − 𝑙𝑛|𝑦 4 + 𝑦 2 + 1| + 𝑐
3
2

Question Three
The operation produced 𝑊, to reach a originate from its cosmical protraction to an production of 𝑏 meters
Is consecrated by
𝑏
𝑥
6
𝑊 = ∫ (3 −
− sec 2 (3𝑥) − √8)𝑑𝑥
6𝑥
+
9
0 √𝑥

Evaluate W.
Since our I.D enumerate is ID. No. 18F17988, we allure grasp b=7
Therefore, consecrated the equation:
𝑏

𝑥

6

𝑊 = ∫0 ( 3 − 6𝑥+9 − 𝑠𝑒𝑐 2 (3𝑥) − √8) 𝑑𝑥 𝑎𝑠 𝑒𝑞𝑛 (𝑎)𝑡ℎ𝑒𝑛;
√𝑥

𝑏

2

𝑏 𝑑𝑥

= ∫0 𝑥 3 𝑑𝑥 − ∫0

3
𝑥+
2

𝑏

𝑏

3

5

3

3

− ∫0 𝑠𝑒𝑐 2 (3𝑥)𝑑𝑥 − ∫0 √8𝑑𝑥 = 5 𝑏 3 − 𝑙𝑛 (𝑏 + 2) + 𝑙𝑛 (2) −

ta n(3𝑏)
3

√8 𝑏
Since b=7 the rate of W allure be consecrated as
5
3
3
3
𝑡𝑎𝑛21
𝑊 = (7)3 − 𝑙𝑛 (7 + ) + 𝑙𝑛 ( ) −
− √8(7)
5
2
2
3

𝑊 = −21.64 𝐽
Question Four
Write the subjoined intricate enumerate in Polar fashion:
𝑧=

5 + √4𝑖
5 − 6𝑖

By rationalizing the eqn consecrated over, by multiplying twain the denominator and the numerator
with (5+6i), we get
5+√4𝑖(5+6𝑖)

25−30𝑖+5√4𝑖−6√4

⇒ (5−6𝑖)(5+6𝑖) =

25+36

5

ENGINEERING MATHEMATICS

(25 − 6√4) + 𝑖(30 + 5√4
25 − 6√4
(−30 + 5(√4
=(
)+(
) 𝑖 … … … … 𝑒𝑞𝑛(𝑎)
61
61
61

On collecting affect stipulations in eqn (a)
𝑍=

25 − 12 −30 + 10
13 20
+
𝑖=

𝑖
61
61
61 61

Hence to employ into polar fashion, we allure earliest meet the modulus i.e. r of 13/61 − 20/61𝑖
13 2
20 2
169 + 400
√( ) + (− ) = √
= 0.39; 𝑡ℎ𝑢𝑠 𝑜𝑢𝑟 𝑟 𝑣𝑎𝑙𝑢𝑒 = 0.39
61
61
3721
To estimate the angle:
𝑦
20
𝜃 = tan−1 ( ) 𝑤ℎ𝑒𝑟𝑒 𝑦 = 20 𝑎𝑛𝑑 𝑛 = 13 = − tan−1 ( )
𝑛
13
Hence to value the eqn in intricate fashion we allure the fashion: 𝑧 = 𝑟𝑒 𝑖𝜃 = 0.39𝑒 −𝑖𝑡𝑎𝑛

Question Five
Simplify the subjoined expression:
√−𝟐𝟓 + 𝟐𝒊𝒂 + 𝒃(𝒊𝒂 )𝒄 + |𝟗 + 𝟖𝒊| − 𝟒(𝟓 + 𝟔𝒊)√−𝟑𝟔
ID. No. 18F17988
Let a=988, b=8, and c=1
= √−25 + 2𝑖 988 + 7(𝑖 988 )1 + |9 + 8𝑖| − 4(5 + 6𝑖)√−36
5√−1 + 2(𝑖 2 )494 + 9(𝑖 2 )494 + |9 + 8𝑖| − 6𝑖(4(5 + 6𝑖))
= 5𝑖 + 2(−1)494 + 9(−1)494 + |9 + 8𝑖| − 24𝑖(5 + 6𝑖)
= 5𝑖 + 2 + 9 + |9 + 8𝑖| − 120𝑖 + 144
= 5𝑖 + 11 + |9 + 8𝑖| − 120𝑖 + 144
𝐴𝑛𝑠 = −115𝑖 + 155 + |9 + 8𝑖|
b. Express (𝑐𝑜𝑠(𝑎𝜃) + 𝑖𝑠𝑖𝑛(𝑎𝜃))𝑏 in exponential fashion (4 marks)