MAT2691 EXAM PACK 2023

MAT2691
EXAM
PACK 2023

,QUESTION 1.1

Determine the following integrals

∫ ( 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 ) √2𝑎 3 + 3𝑏𝑥 2 + 6𝑐𝑥 + 𝑑 𝑑𝑥
SOLUTION

Let 𝑢 = 2𝑎𝑥 + 3𝑏𝑥 + 6𝑐𝑥 + 𝑑
Hint:
3 2


𝑑𝑢
= 6𝑎𝑥 2 + 6𝑏𝑥 + 6𝑐
𝑑𝑥 The main concept to maste
in Que 1.1 is the basic
integration. Since 𝑎𝑥 2 +
𝑑𝑢 = (𝑎𝑥
6 2 + 6𝑏𝑥 + 𝑐) 𝑏𝑥 + 𝑐 is a derivative of
𝑑𝑥
2𝑎 3 + 3𝑏𝑥 2 + 6𝑐𝑥 + 𝑑
𝑑𝑢
Therefore 𝑑𝑥 = makes it easy to integrate
6(𝑎𝑥 2 +𝑏𝑥+𝑐
using substitution.
𝑑𝑢
∫ ( 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 ) √𝑢
6(𝑎𝑥 2 + 𝑏𝑥 + 𝑐)
1
∫ √ 𝑢 𝑑𝑢
6
3
1 2
= × 𝑢2 + 𝑐
6 3
2 3
= 𝑢 +𝑐
18 2

1 3 2 3
= √(2𝑎𝑥 + 3𝑏𝑥 + 6𝑐𝑥 + 𝑑) +𝑐 ANS.
9

1.2

√𝑡𝑎𝑛 2 2𝑥 − 9
∫ 𝑑𝑥
𝑐𝑜𝑠 2 2𝑥
1
=∫ √𝑡𝑎𝑛 2 2𝑥 − 9 𝑑𝑥
𝑐𝑜𝑠 2 2𝑥



= ∫ 𝑠𝑒𝑐 2 √𝑡𝑎𝑛 2 2𝑥 − 9 𝑑𝑥

= ∫ 𝑠𝑒𝑐 2 𝑥 √(tan 2𝑥) 2 −3 2 𝑑𝑥

From the standard
2 (𝑥) (𝑥 )
= ∫ ′(𝑥) √ [𝑓(𝑥)] 2 − 𝑎 2 𝑑𝑥 = − 𝑎 𝑎𝑟𝑐 cos ℎ ( 𝑓 + 𝑓 √[𝑓(𝑥)] 2 − 𝑎 2 + C
2 𝑎 2

Since the derivative of 𝑡𝑎𝑛2𝑥 = 𝑠𝑒𝑐 2 2𝑥

, 2|Page MAT2691 OCT/NOV2010
𝑡𝑎𝑛 2 2𝑥−9 𝑑𝑥
∫√ 𝑐𝑜𝑠 2 2𝑥



=− 9 𝑎𝑟𝑐 cos ℎ 𝑡𝑎𝑛2𝑥 + 𝑡𝑎𝑛2𝑥 √ (𝑡𝑎𝑛2𝑥 ) − 9) + 𝐶
4 3 4

1.3

∫
𝑠𝑖𝑛ℎ𝑥
𝑑𝑥
Hint:
1 + 𝑐𝑜𝑠ℎ𝑥
Que 1.3. The derivative of 1+
From
𝑐𝑜𝑠ℎ𝑥 is 𝑠𝑖𝑛ℎ𝑥

𝑓′(𝑥) Since now we know that the
∫ = 𝑙𝑛 | 𝑓(𝑥) | + 𝑐 question consist of the function and
𝑓(𝑥)
its derivative, therefore one of the
Since the derivative of 1 + cosh 𝑥 = sinh 𝑥 standard formulae has to be used
i.e.

=∫ 𝑠𝑖𝑛ℎ𝑥 𝑑𝑥 𝑙𝑛= |1 + cos ℎ 𝑥 | + 𝑐 ANS.
1+𝑐𝑜𝑠ℎ𝑥

1.4
𝜋
4 Que 1.4 To be able to integrate this
∫ 𝑠𝑖𝑛 3 𝑥. 𝑐𝑜𝑠 3 𝑥 𝑑𝑥 question easily the function has to
0
𝜋
be simplified first. Usually with this
type of questions we have to
∫04 𝑠𝑖𝑛 3 𝑥( ( 𝑐𝑜𝑠 2 𝑥 )( 𝑐𝑜𝑠𝑥 ) )𝑑𝑥 From { 𝑐𝑜𝑠 2 𝑥 = 1 − 𝑠𝑖𝑛 2 𝑥} simplify the function and leave it in
𝜋
the form
= ∫04 𝑠𝑖𝑛 3 𝑥((1 − 𝑠𝑖𝑛 2 x)(cosx))dx
𝜋
= ∫04 𝑠𝑖𝑛 3 𝑥𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛 5 𝑥 𝑐𝑜𝑠 𝑥 𝑑𝑥 In this form it can easily be solved
𝜋 𝜋 using the standard equations.
𝑠𝑖𝑛 4 𝑥 𝑠𝑖𝑛 6 𝑥
4 4


=[ ]0 −[ ]
4 6 0

𝜋
4 𝜋 𝑠𝑖𝑛 6 ( )
4

= [𝑠𝑖𝑛 ( )− ]
4 6
1
=
24

=0.042 ANS.

1.5

𝑑𝑥
∫
𝑥2 + 4𝑥 + 20

, By completing the square on the denominator.
3|Page MAT2691 OCT/NOV2010