MAT1613 ASSIGNMENT 2 2022

MAT1613
ASSIGNMENT 2
2022

,Solution:


3
−
(1 + √𝑦) 2
𝐼=∫ 𝑑𝑦
√𝑦



𝐿𝑒𝑡: 𝑢 = 1 + √𝑦

𝑑𝑢 1
=
𝑑𝑦 2√𝑦

𝑑𝑦 = 2√𝑦 ∙ 𝑑𝑢




3
−
(1 + √𝑦) 2
𝐼=∫ 𝑑𝑦
√𝑦
3
(𝑢)−2
=∫ ∙ 2√𝑦 ∙ 𝑑𝑢
√𝑦
3
= 2 ∫ 𝑢−2 𝑑𝑢

3
𝑢−2+1
=2∙ +𝐶
3
−2 + 1
1
𝑢 −2
=2∙ +𝐶
1
(− 2)
1
= −4𝑢−2 + 𝐶
1
−
= −4(1 + √𝑦 ) 2 +𝐶

, Solution:


1
𝐼=∫ 𝑑𝑥 , 𝑥 > 1
3𝑥 ∙ sec(𝜋 ln(𝑥))
𝐿𝑒𝑡: 𝑢 = 𝜋 ln(𝑥)
𝑑𝑢 𝜋
=
𝑑𝑥 𝑥
𝑥
𝑑𝑥 = 𝑑𝑢
𝜋


1
𝐼=∫ 𝑑𝑥
3𝑥 ∙ sec(𝜋 ln(𝑥))
1 𝑥
𝐼=∫ ∙ 𝑑𝑢
3𝑥 ∙ sec(𝑢) 𝜋
1 1
𝐼= ∫ ∙ 𝑑𝑢
3𝜋 sec(𝑢)
1 1
𝐼= ∫ ∙ 𝑑𝑢
3𝜋 ( 1
)
cos(𝑢)
1
𝐼= ∫ cos(𝑢) 𝑑𝑢
3𝜋
1
𝐼= sin(𝑢) + 𝐶
3𝜋




Solution:


4
𝑑𝑥
𝐼=∫
√𝑥 − 2𝑥
3

𝑑𝑥
𝐸𝑣𝑎𝑙𝑢𝑎𝑡𝑒 ∫
√𝑥 − 2𝑥