LEC 20 [Applied Calculus] B.Sc.

Engineering 1st Semester

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Engr. Riaz Ahmad Rana Assistant Prof. FOE, UCP, Lahore Page 1 of 14
LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

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Engr. Riaz Ahmad Rana Assistant Prof. FOE, UCP, Lahore Page 2 of 14
LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

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Engr. Riaz Ahmad Rana Assistant Prof. FOE, UCP, Lahore Page 3 of 14
LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

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Engr. Riaz Ahmad Rana Assistant Prof. FOE, UCP, Lahore Page 4 of 14
LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

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Engr. Riaz Ahmad Rana Assistant Prof. FOE, UCP, Lahore Page 5 of 14
LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

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Engr. Riaz Ahmad Rana Assistant Prof. FOE, UCP, Lahore Page 6 of 14
LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

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LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

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LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

Some Standard Integrals.

Derivatives Antiderivative
𝑑(𝑥𝑛 ) 𝑛−1 𝑛 𝑥𝑛+1
1. = 𝑛𝑥 1. ∫ 𝑥 𝑑𝑥 = + 𝑐, 𝑛 ≠ 1
𝑑𝑥 𝑛+1

𝑑(𝑎𝑥+𝑏) 𝑛 (𝑎𝑥+𝑏)𝑛+1
2. = 𝑛(𝑎𝑥 + 𝑏)𝑛−1 . (𝑎) 2.∫(𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 = +𝑐
𝑑𝑥 𝑎.(𝑛+1)

𝑑(𝑥)
3. =1 3. ∫ 1𝑑𝑥 = 𝑥 + 𝑐
𝑑𝑥

𝑑(𝑆𝑖𝑛𝑥 )
4. = 𝐶𝑜𝑠𝑥 4. ∫ 𝐶𝑜𝑠𝑥𝑑𝑥 = 𝑆𝑖𝑛𝑥 + 𝑐
𝑑𝑥

𝑑(𝐶𝑜𝑠𝑥 )
5. = −𝑆𝑖𝑛𝑥 5. ∫ 𝑆𝑖𝑛𝑥𝑑𝑥 = −𝐶𝑜𝑠𝑥 + 𝑐
𝑑𝑥

𝑑(𝑇𝑎𝑛𝑥 )
6. = 𝑆𝑒𝑐 2 𝑥 6. ∫ 𝑆𝑒𝑐 2 𝑥𝑑𝑥 = 𝑇𝑎𝑛𝑥 + 𝑐
𝑑𝑥

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LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

𝑑(𝐶𝑜𝑡𝑥)
7. = −𝐶𝑜𝑠𝑒𝑐 2 𝑥 7. ∫ 𝐶𝑜𝑠𝑒𝑐 2 𝑥𝑑𝑥 = −𝐶𝑜𝑡𝑥 + 𝑐
𝑑𝑥

𝑑(𝑆𝑒𝑐𝑥)
8. = 𝑆𝑒𝑐𝑥𝑇𝑎𝑛𝑥 8. ∫ 𝑆𝑒𝑐𝑥𝑇𝑎𝑛𝑥𝑑𝑥 = 𝑆𝑒𝑐𝑥 + 𝑐
𝑑𝑥

𝑑(𝐶𝑜𝑠𝑒𝑐𝑥)
9. = −𝐶𝑜𝑠𝑒𝑐𝑥𝐶𝑜𝑡𝑥 9.∫ 𝐶𝑜𝑠𝑒𝑐𝑥𝐶𝑜𝑡𝑥𝑑𝑥 = −𝐶𝑜𝑠𝑒𝑐𝑥 + 𝑐
𝑑𝑥

𝑑(𝑆𝑖𝑛 −1 𝑥) 1 1
10. = 10. ∫ 𝑑𝑥 = sin−1 𝑥 + 𝑐
𝑑𝑥 √1−𝑥2 √1−𝑥2

𝑑 (tan −1 𝑥) 1 1
11. = 11. ∫ 𝑑𝑥 = tan −1 𝑥 + 𝑐
𝑑𝑥 1+𝑥2 1+𝑥2

𝑑(sec −1 𝑥) 1 1
12. = 12. ∫ 𝑑𝑥 = sec −1 𝑥 + 𝑐
𝑑𝑥 𝑥√𝑥2 −1 𝑥√𝑥2 −1

𝑑(𝑒 𝑥 )
13. = 𝑒𝑥 13. ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝑐
𝑑𝑥

𝑑(𝑎𝑥 ) 𝑥 𝑥 𝑎𝑥
14. = 𝑎 𝑙𝑛𝑎 14. ∫ 𝑎 𝑑𝑥 = +𝑐
𝑑𝑥 𝑙𝑛𝑎

𝑑(𝑙𝑛𝑥) 1 1
15. = 15. ∫ 𝑑𝑥 = 𝑙𝑛𝑥 + 𝑐
𝑑𝑥 𝑥 𝑥

Note: If in any of the above formula, argument is other than ‘x’ then
first, treat it as ‘x’, use the standard result and then divide derivative of
the argument.
𝑒. 𝑔 ∫ 𝐶𝑜𝑠𝑥𝑑𝑥 = 𝑆𝑖𝑛𝑥 + 𝑐
𝑆𝑖𝑛5𝑥
∫ 𝐶𝑜𝑠5𝑥𝑑𝑥 = +𝑐
5

EXAMPLE-1
I =∫ 𝑠𝑖𝑛𝑥 4 . 𝑐𝑜𝑠𝑥 𝑑𝑥

SOLUTION:
I =∫ 𝑠𝑖𝑛4 𝑥 . 𝑐𝑜𝑠𝑥 𝑑𝑥

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Engr. Riaz Ahmad Rana Assistant Prof. FOE, UCP, Lahore Page 10 of 14
LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

Put u= sinx du =cox dx

I =∫ 𝑢 4 𝑑𝑥
𝑢5
I= +𝑐
5

𝑠𝑖𝑛5 𝑥
= +𝑐
5

EXAMPLE-2
I = ∫ 𝑐𝑜𝑠 (7𝜃 + 5) 𝑑𝜃

SOLUTION:
I = ∫ 𝑐𝑜𝑠 (7𝜃 + 5) 𝑑𝜃
𝑑𝑢
Put u = 7𝜃 + 5 du = 7 d 𝜃 d𝜃 =
7
𝑑𝑢
I = ∫ 𝑐𝑜𝑠𝑢 .
7
1
= ∫ 𝑐𝑜𝑠𝑢 𝑑𝑢
7
1
= 𝑠𝑖𝑛𝑢 + 𝑐
7
1
= sin(7𝜃 + 5) + 𝑐
7

EXAMPLE-3
𝑙𝑛𝑥
I=∫ 𝑑𝑥
𝑥

SOLUTION:
𝑙𝑛𝑥
I=∫ 𝑑𝑥
𝑥
1
= ∫ 𝑙𝑛𝑥 . dx
𝑥

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Engr. Riaz Ahmad Rana Assistant Prof. FOE, UCP, Lahore Page 11 of 14
LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

1
Put u = lnx du = dx
𝑥

I = ∫ 𝑢 . 𝑑𝑢
𝑢2 (𝑙𝑛𝑥) 2
= +𝑐 = +𝑐
2 2

EXAMPLE-4
I = ∫ 𝑥 2 𝑠𝑖𝑛𝑥 3dx

SOLUTION:
I = ∫ 𝑠𝑖𝑛𝑥 3 . 𝑥 2 𝑑𝑥
𝑑𝑢
Put u = 𝑥3 du = 3𝑥 2 𝑑𝑥 𝑥 2 𝑑𝑥 =
3
𝑑𝑢
I = ∫ 𝑠𝑖𝑛𝑢.
3
1
I = ∫ 𝑠𝑖𝑛𝑢. 𝑑𝑢
3
1
= (−𝑐𝑜𝑢 ) + 𝑐
3

1
= - cosu+𝑐
3

1
=- 𝑐𝑜𝑠𝑥 3 + 𝑐
3

EXAMPLE-5
I = ∫(𝑥 2 + 2𝑥 − 3)2 . (𝑥 + 1)𝑑𝑥

SOLUTION:
I= ∫(𝑥 2 + 2𝑥 − 3)2 . (𝑥 + 1)𝑑𝑥

Put 𝑥 2 + 2𝑥 − 3 = u du = 2(𝑥 + 1)𝑑𝑥

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LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

(𝑥 + 1)𝑑𝑥 = 𝑑𝑢/2
𝑑𝑢
I = ∫ 𝑢2 .
2
1
= ∫ 𝑢 2 𝑑𝑢
2

1 𝑢3
= . +𝑐
2 3
1
= (𝑥 2 + 2𝑥 − 3 )3 + 𝑐
6

EXAMPLE-6
2𝑧
I=∫ 1 𝑑𝑧
( 𝑧 2 +1) ⁄3

SOLUTION:
2𝑧
I=∫ 1 𝑑𝑧
( 𝑧 2 +1) ⁄3

Put u = 𝑧2 + 1 du = 2z dz
𝑑𝑢
I=∫3
√𝑢

−1
I = ∫𝑢 3 𝑑𝑢
2
𝑢 ⁄3
I= 2⁄ +𝑐
3

⁄ 2
(𝑧 2 +1) 3 3 2⁄
= 2⁄ + 𝑐= (𝑧 2 + 1) 3 +𝑐
3 2

EXAMPLE-7
I = ∫ 28(7𝑥 − 2)−5 𝑑𝑥

SOLUTION:

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Engr. Riaz Ahmad Rana Assistant Prof. FOE, UCP, Lahore Page 13 of 14
LEC 20 [Applied Calculus] B.Sc. Engineering 1st Semester

I = ∫ 28(7𝑥 − 2)−5 𝑑𝑥

= 28∫ (7𝑥 − 2)−5 𝑑𝑥

𝑑𝑢
Put u = (7𝑥 − 2) du = 7dx dx =
7
1
So, I = 28∫ 𝑢 −5 𝑑𝑢
7
1
= 28. ∫ 𝑢 −5 𝑑𝑢
7

𝑢−4
= 4. + 𝑐 = -(7𝑥 − 2)−4 + 𝑐
−4

EXAMPLE-8
I = ∫ 𝑡 3 (1 + 𝑡 4 )3 𝑑𝑡

SOLUTION

𝐼 = ∫(1 + 𝑡 4 )3 𝑡 3 𝑑𝑡

𝑑𝑢
Put u =1 + 𝑡 4 du = 4𝑡 3dt 𝑡 3dt =
4
1
So, I = ∫ 𝑢 3 𝑑𝑢
4
1
= ∫ 𝑢 3 𝑑𝑢
4

1 𝑢4 1
= . + 𝑐= (1 + 𝑡 4 )4 + 𝑐
4 4 16

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Engr. Riaz Ahmad Rana Assistant Prof. FOE, UCP, Lahore Page 14 of 14