# 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:
β«(π‘ 2 + 3) cos(5π‘)ππ‘ππ
ID. No. 18F17988
Let a=1, and b=8
As b>a so 8>1
π€π π βπππ π πππ£π π‘βππ  πππ’ππ‘πππ ππ¦ πππ‘ππππππ‘πππ ππ¦ ππππ‘π  π. π β« ππβ² = ππ β π β² π β¦ β¦ . πππ (1)
π ππππ π = π‘ 2 + 3, and g β² = absorb 5 t,

to meet g we allure intergrate g β² to finf g to be

by replacing the rates of f g fβ and gβ on eqn (1) we allure allure eqn (2) below:

(π‘ 2 + 3)π ππ5(π‘)
2π‘π ππ(5π‘)
β« ππβ² =
ββ«
ππ‘
5
5
=β«

2π‘π ππ(5π‘)
2π‘π ππ(5π‘)
. ππ‘ β β«
ππ‘
5
5

By integrating partially:
π = π‘; πππ π β² = 1
β« ππβ² = ππ β β« π β² π π€βπππ { β²
cos(5π‘)
π = sin(5π‘) ; π = β
5
Which allure then be
β

π‘πππ (5π‘)
cos (5π‘)
β β«β
ππ‘
5
5

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

sin (5π‘) π‘πππ (5π‘) 2 sin(5π‘) 2π‘πππ (5π‘25)
β
=
β
25
5
125
25
(π‘ 2 + 3)sin (5π‘) 2 sin(5π‘) 2π‘πππ (5π‘)
=
β
+
5
125
25

Also, on integrating β«(π‘ 2 + 3). cos(5π‘) ππππ‘πππππ¦; we allure allure the equations below

sin (5π‘)
5

3

ENGINEERING MATHEMATICS
(π‘ 2 + 3)sin (5π‘) 2 sin(5π‘) 2π‘πππ (5π‘)
(25π‘ 2 + 73 sin(5π‘) + 10π‘πππ (5π‘)
β
+
+π =
+π
5
125
25
125
=

98 sin(5) + 10cos (5)
= β0.7
124
π»ππππ π‘βπ πππ π€ππ ππ  0.7
Question Two

Evaluate the subjoined Integral by using accordant method:
β« (πΈ (βππ¦) βπ + ππ¦ βππ¦ β

2π¦ 3 + π¦
) ππ¦
π¦4 + π¦2 + 1

(Note: Grasp π and b any two bulk of your MEC ID enumerate and π, π>π)
ID. No. 18F17988
Given the equation
π = β«(π

βππ¦

(2π¦ 3 + π¦
ππ¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ . (π)
βπ + π^ β ππ¦ππ¦ β β« 4
π¦ + π¦2 + 1
π΄π π π’ππππ π‘βππ‘
π€ = π + π βππ¦
π§ = π¦ 4 + π¦12

Then on differentiating w and z rates, we allure
ππ€
ππ¦
ππ§
ππ¦

= 0 + (βπ)π βππ¦ β¦ β¦ β¦ β¦ . (π)
= 4π¦ 3 + 2π¦ + 0 β¦ β¦ β¦ β¦ β¦ (π)

On simplifying eqn b and c
ππ€
βπ
ππ§
2

= π βππ¦ ππ¦
= (2π¦ 3 + π¦)ππ¦

Therefore, we allure foreclosure eqn (a), hence we have
3
β1
1
1
2
1
π = ( ) β« βπ§ ππ§ β ( ) β« ππ€ = β (π + π βππ¦ )2 β ππ|π¦ 4 + π¦ 2 + 1| + πΆ
π
2
π€
3π
2

4

ENGINEERING MATHEMATICS
Therefore, if our a and b rates are as follows;
πΌπ π = 1 πππ π = 8
The consummate reresolution allure be as follows
3
2
1
π = β (8 + π βπ¦ )2 β ππ|π¦ 4 + π¦ 2 + 1| + π
3
2

Question Three
The operation produced π, to reach a originate from its cosmical protraction to an production of π meters
Is consecrated by
π
π₯
6
π = β« (3 β
β sec 2 (3π₯) β β8)ππ₯
6π₯
+
9
0 βπ₯

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

π₯

6

π = β«0 ( 3 β 6π₯+9 β π ππ 2 (3π₯) β β8) ππ₯ ππ  πππ (π)π‘βππ;
βπ₯

π

2

π ππ₯

= β«0 π₯ 3 ππ₯ β β«0

3
π₯+
2

π

π

3

5

3

3

β β«0 π ππ 2 (3π₯)ππ₯ β β«0 β8ππ₯ = 5 π 3 β ππ (π + 2) + ππ (2) β

ta n(3π)
3

β

β8 π
Since b=7 the rate of W allure be consecrated as
5
3
3
3
π‘ππ21
π = (7)3 β ππ (7 + ) + ππ ( ) β
β β8(7)
5
2
2
3

π = β21.64 π½
Question Four
Write the subjoined intricate enumerate in Polar fashion:
π§=

5 + β4π
5 β 6π

By rationalizing the eqn consecrated over, by multiplying twain the denominator and the numerator
with (5+6i), we get
5+β4π(5+6π)

25β30π+5β4πβ6β4

β (5β6π)(5+6π) =

25+36

5

ENGINEERING MATHEMATICS

β

(25 β 6β4) + π(30 + 5β4
25 β 6β4
(β30 + 5(β4
=(
)+(
) π β¦ β¦ β¦ β¦ πππ(π)
61
61
61

On collecting affect stipulations in eqn (a)
π=

25 β 12 β30 + 10
13 20
+
π=
β
π
61
61
61 61

Hence to employ into polar fashion, we allure earliest meet the modulus i.e. r of 13/61 β 20/61π
13 2
20 2
169 + 400
β( ) + (β ) = β
= 0.39; π‘βπ’π  ππ’π π π£πππ’π = 0.39
61
61
3721
To estimate the angle:
π¦
20
π = tanβ1 ( ) π€βπππ π¦ = 20 πππ π = 13 = β tanβ1 ( )
π
13
Hence to value the eqn in intricate fashion we allure the fashion: π§ = ππ ππ = 0.39π βππ‘ππ

Question Five
Simplify the subjoined expression:
ββππ + πππ + π(ππ )π + |π + ππ| β π(π + ππ)ββππ
ID. No. 18F17988
Let a=988, b=8, and c=1
= ββ25 + 2π 988 + 7(π 988 )1 + |9 + 8π| β 4(5 + 6π)ββ36
5ββ1 + 2(π 2 )494 + 9(π 2 )494 + |9 + 8π| β 6π(4(5 + 6π))
= 5π + 2(β1)494 + 9(β1)494 + |9 + 8π| β 24π(5 + 6π)
= 5π + 2 + 9 + |9 + 8π| β 120π + 144
= 5π + 11 + |9 + 8π| β 120π + 144
π΄ππ  = β115π + 155 + |9 + 8π|
b. Express (πππ (ππ) + ππ ππ(ππ))π in exponential fashion (4 marks)