MAT2691 ASSIGNMENT 2 2025

MAT2691
ASSIGNMENT 2
2025

, QUESTION 1


1.1).


2
(√x − 4) (√x − 4)(√x − 4)
∫ 3 dx = ∫ 1 dx
√x x3
1
x − 8x 2 + 16
=∫ 1 dx
x3
1
x 8x 2 16
= ∫( 1 − 1 + 1 ) dx
x3 x3 x3
2 1 1
= ∫ (x 3 − 8x 6 + 16x −3 ) dx

2 1 1
x (3+1) 8x (6+1) 16x (−3+1)
= − + +C
2 1 1
(3 + 1) (6 + 1) (− 3 + 1)
5 7 2
x (3) 8x (6) 16x (3)
= − + +C
5 7 2
(3) (6) (3)

3 5 48 7 2
= x 3 − x 6 + 24x 3 + C
5 7




1.2).


x3
∫ dx
1 + x4


du du
Let ∶ u = 1 + x 4 ⇒ = 4x 3 ⇒ dx =
dx 4x 3


x3 x 3 du
∫ dx = ∫ ∙
1 + x4 u 4x 3

, x3 1 1
∫ dx = ∫ du
1 + x4 4 u
x3 1
∫ dx = ln|u| + C ∴ u = 1 + x4
1 + x4 4
x3 1
∫ 4
dx = ln|1 + x 4 | + C
1+x 4




1.3).


π π
2 2

∫ sin5 (x) cos 3 (x) dx = ∫ sin5 (x) cos2 (x) cos(x) dx
0 0
π
2

= ∫ sin5(x) [1 − sin2 (x)] cos(x) dx
0
π
2

= ∫[sin5 (x) − sin7 (x)] cos(x) dx
0
π
2

= ∫[sin5 (x) cos(x) − sin7 (x) cos(x)] dx
0
π
sin6(x) sin8 (x) 2
=[ − ]
6 8 0

π π
sin6 (2) sin8 (2 ) sin6 (0) sin8 (0)
=[ − ]−[ − ]
6 8 6 8

(1)6 (1)8 (0)6 (0)8
=[ − ]−[ − ]
6 8 6 8
1 1
= [ − ] − [0 − 0]
6 8
1
=
24