MAT1613 ASSIGNMENT 2 2021

MAT1613

ASSIGNMENT 2 2021

QUESTION 1
𝑒 √𝑥
∫ 𝑑𝑥
√𝑥

𝐿𝑒𝑡: 𝑢 = √𝑥

𝑑𝑢 1
=
𝑑𝑥 2√𝑥

𝑑𝑥 = 2√𝑥 𝑑𝑢

𝑒 √𝑥 𝑒𝑢
∫ 𝑑𝑥 = ∫ ∙ 2√𝑥 𝑑𝑢
√𝑥 √𝑥

= 2 ∫ 𝑒 𝑢 𝑑𝑢

= 2𝑒 𝑢 + 𝐶 ∴ 𝑢 = √𝑥

= 2𝑒 √𝑥 + 𝐶

ANSWER:[2]



QUESTION 2
8
∫ 3 𝑑𝑥
𝑥[ln(𝑥) + 1]2

𝐿𝑒𝑡: 𝑢 = ln(𝑥) + 1

𝑑𝑢 1
= , 𝑑𝑥 = 𝑥 𝑑𝑢
𝑑𝑥 𝑥
8 8
∫ 3 𝑑𝑥 = ∫ 3⁄ 𝑥 𝑑𝑢
𝑥[ln(𝑥) + 1]2 𝑥∙𝑢 2


8
=∫ 3⁄ 𝑑𝑢
𝑢 2

, 3⁄
= 8 ∫ 𝑢− 2 𝑑𝑢

1
𝑢− ⁄2
=8 +𝐶
(− 1⁄2)

16
=− 1⁄ +𝐶
𝑢 2

16
=− +𝐶 ∴ 𝑢 = ln(𝑥) + 1
√𝑢
16
=− +𝐶
√ln(𝑥) + 1

ANSWER:[4]



QUESTION 3
3𝑡 + 2 3𝑡 + 3 − 3 + 2
∫ 𝑑𝑡 = ∫ 𝑑𝑡
𝑡+1 𝑡+1
3𝑡 + 3 −3 + 2
= ∫( + ) 𝑑𝑡
𝑡+1 𝑡+1

3(𝑡 + 1) 1
= ∫( − ) 𝑑𝑡
𝑡+1 𝑡+1

1
= ∫ (3 − ) 𝑑𝑡
𝑡+1
1
= ∫ 3 𝑑𝑡 − ∫ 𝑑𝑡
𝑡+1

= 3𝑡 − ln(𝑡 + 1) + 𝐶

ANSWER:[1]



QUESTION 4
5𝑡√𝑡
∫ 𝑑𝑡
1 + 𝑡5

𝐿𝑒𝑡: 𝑢 = √𝑡

𝑑𝑢 1
=
𝑑𝑡 2√𝑡

𝑑𝑡 = 2√𝑡 𝑑𝑢 = 2𝑢 𝑑𝑢

, 2
5𝑡√𝑡 5(√𝑡) √𝑡
∫ 5
𝑑𝑡 = ∫ 10 𝑑𝑡
1+𝑡 1 + (√𝑡)

5𝑡√𝑡 5(𝑢)2 𝑢
∫ 𝑑𝑡 = ∫ 2𝑢 𝑑𝑢
1 + 𝑡5 1 + (𝑢)10

𝑢4
= 10 ∫ 𝑑𝑢
1 + 𝑢10

𝑑𝑣 1
𝐿𝑒𝑡: 𝑣 = 𝑢5 = 5𝑢4 , 𝑑𝑢 = 𝑑𝑣
𝑑𝑢 5𝑢4

5𝑡√𝑡 𝑢4
∫ 𝑑𝑡 = 10 ∫ 𝑑𝑢
1 + 𝑡5 1 + 𝑢10

5𝑡√𝑡 𝑢4 1
∫ 5
𝑑𝑡 = 10 ∫ 2
∙ 4 𝑑𝑣
1+𝑡 1 + 𝑣 5𝑢

1
= 2∫ 𝑑𝑣
1 + 𝑣2

= 2 tan−1 (𝑣) + 𝐶 ∴ 𝑣 = 𝑢5

= 2 tan−1 (𝑢5 ) + 𝐶 ∴ 𝑢 = √𝑡
5
= 2 tan−1 ((√𝑡) ) + 𝐶

5⁄
= 2 tan−1 (𝑡 2) + 𝐶

ANSWER:[4]



QUESTION 5
𝑥+1
∫ 𝑑𝑥
√3 − 2𝑥 − 𝑥 2
𝑑𝑢 𝑑𝑢 𝑑𝑢
𝐿𝑒𝑡: 𝑢 = 3 − 2𝑥 − 𝑥 2 , = −2 − 2𝑥, 𝑑𝑥 = =−
𝑑𝑥 −2 − 2𝑥 2(𝑥 + 1)

𝑥+1 𝑥+1 𝑑𝑢
∫ 𝑑𝑥 = ∫ ∙−
√3 − 2𝑥 − 𝑥2 √𝑢 2(𝑥 + 1)

1 1
=− ∫ 𝑑𝑢
2 √𝑢
1
1 𝑢 ⁄2
=− ( )+𝐶
2 1⁄
2
1⁄
= −𝑢 2 +𝐶 ∴ 𝑢 = 3 − 2𝑥 − 𝑥 2

, 1⁄
= −(3 − 2𝑥 − 𝑥 2 ) 2 +𝐶

ANSWER:[2]



QUESTION 6
2
∫ 𝑑𝑥
10 + 6𝑥 + 𝑥 2

𝐶𝑜𝑚𝑝𝑙𝑒𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒

6 2 6 2
𝑥 2 + 6𝑥 + 10 = 𝑥 2 + 6𝑥 + ( ) + 10 − ( )
2 2

6 2
= (𝑥 + ) + 10 − 32
2

= (𝑥 + 3)2 + 10 − 9

= (𝑥 + 3)2 + 1

2 2
∫ 2
𝑑𝑥 = ∫ 𝑑𝑥
10 + 6𝑥 + 𝑥 (𝑥 + 3)2 + 1

𝑑𝑢
𝐿𝑒𝑡: 𝑢 = 𝑥 + 3 , = 1 , 𝑑𝑢 = 𝑑𝑥
𝑑𝑥
2 2
∫ 2
𝑑𝑥 = ∫ 𝑑𝑥
10 + 6𝑥 + 𝑥 (𝑥 + 3)2 + 1

2
=∫ 𝑑𝑥
𝑢2 + 1

= 2 tan−1 (𝑢) + 𝐶

= 2 tan−1 (𝑥 + 3) + 𝐶

ANSWER:[4]



QUESTION 7
𝜋
√
4
∫ [sin(2𝑥 2 )]2 ∙ 𝑥 ∙ cos(2𝑥 2 ) 𝑑𝑥
0


𝐸𝑣𝑎𝑙𝑢𝑎𝑡𝑒 ∫[sin(2𝑥 2 )]2 ∙ 𝑥 ∙ cos(2𝑥 2 ) 𝑑𝑥




𝑑𝑢 𝑑𝑢
𝐿𝑒𝑡: 𝑢 = sin(2𝑥 2 ) , = 4𝑥 cos(2𝑥 2 ) , 𝑑𝑥 =
𝑑𝑥 4𝑥 cos(2𝑥 2 )