# SOLUTION: MASC 0009 American Academy of English Engineering Mathematics Problems

Attached.

Engineering Mathematics
Name
Institutional Affiliation
Course
Date

1

2

ENGINEERING MATHEMATICS
Question One

Evaluate the subjoined integral:
β«(π‘ 2 + 3) cos(5π‘)ππ‘ππ
(Note: Take π and b any two mass of your MEC ID enumerate, after a while a situation that π π π π 4 > 1
πβπππππππ ππ¦ π’π πππ πΌπ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
{

π = π‘2 + 3
πβ² = cos (5π‘)
π=
=

sin (5π‘)
5

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

Now,
β«

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

Using integration size intermittently we use
β« ππβ² = ππ β β« π β² π π€βπππ ππ¦ {
π β² = 1, π = β

π=π‘
π = sin(5π‘)
β²

cos (5π‘)
ππ‘
5

=
β

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

After solving the adherence by size, we obtain

3

ENGINEERING MATHEMATICS

=

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

Therefore,
(π‘ 2 + 3)sin (5π‘)
2π‘π ππ(5π‘)
ββ«
ππ‘
5
5
=

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

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

(25π‘ 2 + 73 sin(5π‘) + 10π‘πππ (5π‘)
+π
125
=

98 sin(5) + 10cos (5)
124
= β0.7291 β β0.73
Question Two

Evaluate the subjoined Integral by using befitting 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: Take π and b any two mass of your MEC ID enumerate and π, π>π)
Solution
ID. No. 18f17934
πΌ = β«(π

βππ¦

(2π¦ 3 + π¦
ππ¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ . πππ(1)
βπ + π^ β ππ¦ππ¦ β β« 4
π¦ + π¦2 + 1
πππ π¦ 4 + π¦12 = π

πΏππ‘ π + π βππ¦ = π,
Therefore
ππ£
= 0 + (βπ)π βππ¦
ππ¦

πππ

ππ’
= 4π¦ 3 + 2π¦ + 0
ππ¦

4

ENGINEERING MATHEMATICS
ππ£
βπ

= π βππ¦ ππ¦ =

ππ’
2

πππ (2π¦ 3 + π¦)ππ¦ =

From eqn (1) we have
β1
1
1
πΌ = ( ) β« βπ£ ππ£ β ( ) β« ππ’
π
2
π’
πΌ=β

3
2
1
(π + π βππ¦ )2 β ππ|π¦ 4 + π¦ 2 + 1| + πΆ
3π
2

If a=1 and b=3
Then
3
2
1
πΌ = β (3 + π βππ¦ )2 β ππ|π¦ 4 + π¦ 2 + 1| + π
3
2

Question Three
The performance manufactured π, to spread a start from its consistent protraction to an extension of π meters
Is fond by
π

π₯
6
π = β« (3 β
β sec 2 (3π₯) β β8)ππ₯
6π₯
+
9
π₯
β
0
Evaluate W.
(Note: Take π as any enumerate from your MEC ID enumerate, π>π)
Solution
π
π₯
6
π = β« (3 β
β π ππ 2 (3π₯) β β8) ππ₯
6π₯
+
9
βπ₯
0
π

=β«
0

2
π₯ 3 ππ₯

π

ββ«
0

ππ₯
3
π₯+2

π

β β« π ππ

π
2 (3π₯)ππ₯

0

β β« β8ππ₯
0

3 5
3
3
ta n(3π)
= π 3 β ππ (π + ) + ππ ( ) β
β β8 π
5
2
2
3
Taking b=1 i.e. b>0
5
3
5
3
π‘ππ3
π = (1)3 β ππ ( ) + ππ ( ...