2023
Assign
ment 2
MAT2691-MATHEMATICS II [ENGINEERING]
SEMESTER 02
ASSIGNMENT 02 SOLUTIONS
Operational Research
P.hD..Msc..Hons..Bsc in
Mathematics & Physics
BlessCademy Tutorials
074 244 4909
, Question 1
1.1
𝜋
∫ 3 𝑑𝑥
𝑥 √ln 𝑥
𝜋 𝑑𝑢 1
=∫ 1 𝑑𝑥 Let 𝑢 = ln 𝑥 𝑑𝑥
=𝑥 ∴ 𝑑𝑥 = 𝑑𝑢 𝑥
𝑥[ln 𝑥]3
𝜋
=∫ 1 𝑑𝑢 𝑥
𝑥[𝑢]3
1
= 𝜋∫ 1 𝑑𝑢
𝑢3
1
= 𝜋 ∫ 𝑢−3 𝑑𝑢
2 2
𝜋𝑢3 3𝜋[ln 𝑥]3
= 2 +𝐶 = 2
+𝐶
3
1.2
𝑒
sin−1 𝑥 𝑑𝑢 1
∫04 √1−𝑥2 𝑑𝑥 Let 𝑢 = sin−1 𝑥 = 𝑑𝑥 = √1 − 𝑥 2 𝑑𝑢
𝑑𝑥 √1−𝑥 2
𝑒
𝑢
= ∫04 . √1 − 𝑥 2 𝑑𝑢
√1−𝑥 2
𝑒
= ∫04 𝑢 𝑑𝑢
𝑥=0 𝑢=0
𝑒 𝑒
𝑥= 𝑢 = sin−1
4 4
𝑒
sin−1
= ∫0 4 𝑢 𝑑𝑢
𝑒
sin−1
𝑢2 4
= [2]
0
𝑒 2 𝑒 2
(sin−1 ) (sin−1 )
4 4
=[ 2
− 2
]
𝑒 2
(sin−1 )
4
= 2
Assign
ment 2
MAT2691-MATHEMATICS II [ENGINEERING]
SEMESTER 02
ASSIGNMENT 02 SOLUTIONS
Operational Research
P.hD..Msc..Hons..Bsc in
Mathematics & Physics
BlessCademy Tutorials
074 244 4909
, Question 1
1.1
𝜋
∫ 3 𝑑𝑥
𝑥 √ln 𝑥
𝜋 𝑑𝑢 1
=∫ 1 𝑑𝑥 Let 𝑢 = ln 𝑥 𝑑𝑥
=𝑥 ∴ 𝑑𝑥 = 𝑑𝑢 𝑥
𝑥[ln 𝑥]3
𝜋
=∫ 1 𝑑𝑢 𝑥
𝑥[𝑢]3
1
= 𝜋∫ 1 𝑑𝑢
𝑢3
1
= 𝜋 ∫ 𝑢−3 𝑑𝑢
2 2
𝜋𝑢3 3𝜋[ln 𝑥]3
= 2 +𝐶 = 2
+𝐶
3
1.2
𝑒
sin−1 𝑥 𝑑𝑢 1
∫04 √1−𝑥2 𝑑𝑥 Let 𝑢 = sin−1 𝑥 = 𝑑𝑥 = √1 − 𝑥 2 𝑑𝑢
𝑑𝑥 √1−𝑥 2
𝑒
𝑢
= ∫04 . √1 − 𝑥 2 𝑑𝑢
√1−𝑥 2
𝑒
= ∫04 𝑢 𝑑𝑢
𝑥=0 𝑢=0
𝑒 𝑒
𝑥= 𝑢 = sin−1
4 4
𝑒
sin−1
= ∫0 4 𝑢 𝑑𝑢
𝑒
sin−1
𝑢2 4
= [2]
0
𝑒 2 𝑒 2
(sin−1 ) (sin−1 )
4 4
=[ 2
− 2
]
𝑒 2
(sin−1 )
4
= 2