Summary MAT2691 ASSIGNMENT 2 2025 SOLUTIONS

MAT2691 ASSIGNMENT 2 2025
SOLUTIONS

Question 1

1.1
2
(√𝑥 − 4)
∫ 3 𝑑𝑥
√𝑥
(√𝑥 − 4)(√𝑥 − 4)
=∫ 1 𝑑𝑥
𝑥3


𝑥 − 8√𝑥 + 16
=∫ 1 𝑑𝑥
𝑥3
2 1 1
= ∫ 𝑥 3 𝑑𝑥 − 8 ∫ 𝑥 6 𝑑𝑥 + 16 ∫ 𝑥 −3 𝑑𝑥

5 7 2
𝑥 3 8𝑥 6 16𝑥 3
= − + +𝐶
5 7 2
3 6 3
7
3 5 48𝑥 6 2
= 𝑥3 − + 24𝑥 3 + 𝐶
5 7
1.2

𝑥3
∫ 𝑑𝑥 Let 𝑢 = 1 + 𝑥 4
1 + 𝑥4
1 𝑑𝑢 1 1 𝑑𝑢
∫ . = ∫ 𝑑𝑢 = 4𝑥 3
𝑢 4 4 𝑢 𝑑𝑥
𝑑𝑢
1 ∴ 𝑥 3 𝑑𝑥 =
= ln 𝑢 + 𝐶 4𝑥 3
4
1
= ln(1 + 𝑥 4 ) + 𝐶
4
1.3
𝜋 𝜋
2 2
∫ sin5 𝑥 cos3 𝑥 𝑑𝑥 = ∫ sin 𝑥 sin4 𝑥 cos 𝑥 cos 2 𝑥 𝑑𝑥
0 0

, 𝜋
2
= ∫ sin 𝑥 sin4 𝑥 cos 𝑥 cos2 𝑥 𝑑𝑥
0

𝑑𝑢
Let 𝑢 = sin 𝑥 = cos 𝑥 ∴ 𝑑𝑢 = cos 𝑥 𝑑𝑥
𝑑𝑥

cos2 𝑥 = 1 − sin2 𝑥
𝜋 𝜋
2 2
= ∫ 𝑢. 𝑢2 . 𝑢2 (1 − 𝑢2 )𝑑𝑢 = ∫ 𝑢5 (1 − 𝑢2 )𝑑𝑢
0 0

1
= ∫ 𝑢5 − 𝑢7 𝑑𝑢
0

1
𝑢6 𝑢8
=[ − ]
6 8 0

16 18 06 08
=[ − ]−[ − ]
6 8 6 8
New limits:
1
=
24 𝑢 = sin 0 = 0
𝜋
𝑢 = sin =1
2
1.4

∫ cosh−1 𝑥 𝑑𝑥

Let 𝑢 = cosh−1 𝑥 and 𝑑𝑣 = 1 𝑑𝑥 hence 𝑣 = 𝑥
𝑑𝑢 1
=
𝑑𝑥 √𝑥 − 1
2


∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢

1
∫ cosh−1 𝑥 𝑑𝑥 = cosh−1 𝑥 . 𝑥 − ∫ 𝑥. 𝑑𝑥
√𝑥 2 − 1


1 1
∫ cosh−1 𝑥 𝑑𝑥 = 𝑐 cosh−1 𝑥 − ∫ 2𝑥 (𝑥 2 − 1)−2 𝑑𝑥
2

1 √𝑥 2 − 1
= 𝑥 cosh−1 𝑥 − [ ]+𝐶
2 1
2

𝑥 cosh−1 𝑥 − √𝑥 2 − 1 + 𝐶