# 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:
∫(𝑡 2 + 3) cos(5𝑡)𝑑𝑡𝑏𝑎
(Note: Take 𝒂 and b any two mass of your MEC ID enumerate, after a while a situation that 𝒂 𝑎 𝑠𝑜 4 > 1
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑏𝑦 𝑢𝑠𝑖𝑛𝑔 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑝𝑎𝑟𝑡𝑠
∫ 𝑓𝑔′ = 𝑓𝑔 − 𝑓 ′ 𝑔
{

𝑓 = 𝑡2 + 3
𝑔′ = cos (5𝑡)
𝑔=
=

sin (5𝑡)
5

(𝑡 2 + 3)𝑠𝑖𝑛5(𝑡)
2𝑡𝑠𝑖𝑛(5𝑡)
−∫
𝑑𝑡
5
5

Now,

2𝑡𝑠𝑖𝑛(5𝑡)
2𝑡𝑠𝑖𝑛(5𝑡)
. 𝑑𝑡 − ∫
𝑑𝑡
5
5

Using integration size intermittently we use
∫ 𝑓𝑔′ = 𝑓𝑔 − ∫ 𝑓 ′ 𝑔 𝑤ℎ𝑒𝑟𝑒 𝑏𝑦 {
𝑓 ′ = 1, 𝑔 = −

𝑓=𝑡
𝑔 = sin(5𝑡)

cos (5𝑡)
𝑑𝑡
5

=

𝑡𝑐𝑜𝑠(5𝑡)
cos (5𝑡)
− ∫−
𝑑𝑡
5
5

After solving the adherence by size, we obtain

3

ENGINEERING MATHEMATICS

=

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

=

25
5
125
25

Therefore,
(𝑡 2 + 3)sin (5𝑡)
2𝑡𝑠𝑖𝑛(5𝑡)
−∫
𝑑𝑡
5
5
=

(𝑡 2 + 3)sin (5𝑡) 2 sin(5𝑡) 2𝑡𝑐𝑜𝑠(5𝑡)

+
5
125
25

Hence
∫(𝑡 2 + 3). cos (5𝑡)
(𝑡 2 + 3)sin (5𝑡) 2 sin(5𝑡) 2𝑡𝑐𝑜𝑠(5𝑡)

+
+𝑐
5
125
25
=

(25𝑡 2 + 73 sin(5𝑡) + 10𝑡𝑐𝑜𝑠(5𝑡)
+𝑐
125
=

98 sin(5) + 10cos (5)
124
= −0.7291 ≈ −0.73
Question Two

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

2𝑦 3 + 𝑦
) 𝑑𝑦
𝑦4 + 𝑦2 + 1

(Note: Take 𝒂 and b any two mass of your MEC ID enumerate and 𝒂, 𝒃>𝟎)
Solution
ID. No. 18f17934
𝐼 = ∫(𝑒

−𝑎𝑦

(2𝑦 3 + 𝑦
𝑑𝑦 … … … … … … … … … … … . 𝑒𝑞𝑛(1)
√𝑏 + 𝑒^ − 𝑎𝑦𝑑𝑦 − ∫ 4
𝑦 + 𝑦2 + 1
𝑎𝑛𝑑 𝑦 4 + 𝑦12 = 𝑈

𝐿𝑒𝑡 𝑏 + 𝑒 −𝑎𝑦 = 𝑉,
Therefore
𝑑𝑣
= 0 + (−𝑎)𝑒 −𝑎𝑦
𝑑𝑦

𝑎𝑛𝑑

𝑑𝑢
= 4𝑦 3 + 2𝑦 + 0
𝑑𝑦

4

ENGINEERING MATHEMATICS
𝑑𝑣
−𝑎

= 𝑒 −𝑎𝑦 𝑑𝑦 =

𝑑𝑢
2

𝑎𝑛𝑑 (2𝑦 3 + 𝑦)𝑑𝑦 =

From eqn (1) we have
−1
1
1
𝐼 = ( ) ∫ √𝑣 𝑑𝑣 − ( ) ∫ 𝑑𝑢
𝑎
2
𝑢
𝐼=−

3
2
1
(𝑏 + 𝑒 −𝑎𝑦 )2 − 𝑙𝑛|𝑦 4 + 𝑦 2 + 1| + 𝐶
3𝑎
2

If a=1 and b=3
Then
3
2
1
𝐼 = − (3 + 𝑒 −𝑎𝑦 )2 − 𝑙𝑛|𝑦 4 + 𝑦 2 + 1| + 𝑐
3
2

Question Three
The performance manufactured 𝑊, to spread a start from its consistent protraction to an extension of 𝑏 meters
Is fond by
𝑏

𝑥
6
𝑊 = ∫ (3 −
− sec 2 (3𝑥) − √8)𝑑𝑥
6𝑥
+
9
𝑥

0
Evaluate W.
(Note: Take 𝒃 as any enumerate from your MEC ID enumerate, 𝒃>𝟎)
Solution
𝑏
𝑥
6
𝑊 = ∫ (3 −
− 𝑠𝑒𝑐 2 (3𝑥) − √8) 𝑑𝑥
6𝑥
+
9
√𝑥
0
𝑏

=∫
0

2
𝑥 3 𝑑𝑥

𝑏

−∫
0

𝑑𝑥
3
𝑥+2

𝑏

− ∫ 𝑠𝑒𝑐

𝑏
2 (3𝑥)𝑑𝑥

0

− ∫ √8𝑑𝑥
0

3 5
3
3
ta n(3𝑏)
= 𝑏 3 − 𝑙𝑛 (𝑏 + ) + 𝑙𝑛 ( ) −
− √8 𝑏
5
2
2
3
Taking b=1 i.e. b>0
5
3
5
3
𝑡𝑎𝑛3
𝑊 = (1)3 − 𝑙𝑛 ( ) + 𝑙𝑛 ( ...