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MAT2691 EXAM PACK PAST QUESTIONS AND SOLUTIONS
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MAT2691 EXAM PACK PAST QUESTIONS AND SOLUTIONS QUESTION 1.1 Determine the following integrals ∫(
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MAT2691 EXAM PACKPAST QUESTIONS
AND SOLUTIONS
,QUESTION 1.1
Determine the following integrals
∫(𝑎𝑥2 + 𝑏𝑥 + 𝑐)√2𝑎3 + 3𝑏𝑥2 + 6𝑐𝑥 + 𝑑 𝑑𝑥
SOLUTION
Let 𝑢 = 2𝑎𝑥3 + 3𝑏𝑥2 + 6𝑐𝑥 + 𝑑
𝑑𝑢
Hint:
= 6𝑎𝑥2 + 6𝑏𝑥 + 6𝑐
𝑑𝑥 The main concept to master
in Que 1.1 is the basic
integration. Since 𝑎𝑥2 +
𝑑𝑢 = 6(𝑎𝑥2 + 6𝑏𝑥 + 𝑐) 𝑏𝑥 + 𝑐 is a derivative of
𝑑𝑥
2𝑎3 + 3𝑏𝑥2 + 6𝑐𝑥 + 𝑑 so it
𝑑𝑢
Therefore 𝑑𝑥 = makes it easy to integrate
6(𝑎𝑥2+𝑏𝑥+𝑐
using substitution.
𝑑𝑢
∫(𝑎𝑥2 + 𝑏𝑥 + 𝑐) √𝑢
6(𝑎𝑥2 + 𝑏𝑥 + 𝑐)
1
∫ √𝑢 𝑑𝑢
6
3
1 2
=6 × 3 𝑢2 + 𝑐
2 3
=18 𝑢 2 + 𝑐
1
=9 √(2𝑎𝑥3 + 3𝑏𝑥2 + 6𝑐𝑥 + 𝑑)3 + 𝑐 ANS.
1.2
√𝑡𝑎𝑛22𝑥 − 9
∫ 𝑑𝑥
𝑐𝑜𝑠22𝑥
1
=∫ √𝑡𝑎𝑛2 2𝑥 − 9 𝑑𝑥
𝑐𝑜𝑠2 2
𝑥
=∫ 𝑠𝑒𝑐2√𝑡𝑎𝑛2 2𝑥 − 9 𝑑𝑥
=∫ 𝑠𝑒𝑐2𝑥 √(tan 2𝑥)2 − 32 𝑑𝑥
From the standard
2 (𝑥 ) (𝑥)
=∫ ′(𝑥) √[𝑓(𝑥)]2 − 𝑎2 𝑑𝑥 = − 𝑎 𝑎𝑟𝑐 cos ℎ (𝑓 +𝑓 √[𝑓(𝑥)]2 − 𝑎2 +C
2 𝑎 2
Since the derivative of 𝑡𝑎𝑛2𝑥 = 𝑠𝑒𝑐2 2𝑥
, 2|Page MAT2691 OCT/NOV2010
𝑡𝑎𝑛2 2𝑥−9 𝑑𝑥
∫√ 𝑐𝑜𝑠2 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
∫ 𝑠𝑖𝑛3 𝑥. 𝑐𝑜𝑠 3𝑥 𝑑𝑥 question easily the function has to
0
𝜋
be simplified first. Usually with this
type of questions we have to
∫04 𝑠𝑖𝑛3 𝑥((𝑐𝑜𝑠2𝑥)(𝑐𝑜𝑠𝑥))𝑑𝑥 From {𝑐𝑜𝑠2𝑥 = 1 − 𝑠𝑖𝑛2𝑥} simplify the function and leave it in
𝜋
the form
=∫04 𝑠𝑖𝑛3 𝑥((1 − 𝑠𝑖𝑛2x)(cosx))dx
𝜋
=∫04 𝑠𝑖𝑛3 𝑥𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛5𝑥 𝑐𝑜𝑠 𝑥 𝑑𝑥 In this form it can easily be solved
𝜋 𝜋 using the standard equations.
𝑠𝑖𝑛4 𝑥 𝑠𝑖𝑛6 𝑥
4 4
=[ ]0 − [ ]0
4 6
𝜋
4 𝜋 𝑠𝑖𝑛6 ( )
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
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