Integration Unit-1 Comprehensive Study Guide – Solved Problems, Methods & Applications

UNIT-1
INTEGRATION

Define an integral
A function 𝐹(𝑥) is called an anti derivative or integral of a function 𝑓(𝑥)
on an interval 𝐼 if
𝐹′(𝑥) = 𝑓(𝑥), for every value of 𝑥 in 𝐼
(i.e) If the derivative of a function 𝐹(𝑥) w.r.to 𝑓(𝑥),then we say
that the integral of 𝑓(𝑥) w.r.to 𝑥 is 𝐹(𝑥).
(i.e) ∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥)
𝑥2−5𝑥+1
• Evaluate ∫ 𝑥
𝑑𝑥.
Solution:

𝑥2 − 5𝑥 + 1 𝑥2 5𝑥 1
∫ 𝑑𝑥 = ∫ ( − + ) 𝑑𝑥
𝑥 𝑥 𝑥 𝑥
1
= ∫ (𝑥 − 5 + ) 𝑑𝑥
𝑥
𝑥2
= − 5𝑥 + 𝑙𝑜𝑔𝑥 + C
2
𝑥2+2𝑥−1
• Evaluate: ∫ 𝑑𝑥.
√𝑥
Solution:
𝑥2 + 2𝑥 − 1 1
−
∫ 𝑑𝑥 = ∫(𝑥2 + 2𝑥 − 1)𝑥 2 𝑑𝑥
√𝑥
1 1 1
= ∫(𝑥2−2 + 2𝑥 1−2 − 𝑥 −2) 𝑑𝑥
3 1 1

= ∫(𝑥2 + 2𝑥2 − 𝑥−2) 𝑑𝑥
3 1 −1
𝑥2+1 𝑥2+1 𝑥 2 +1
=3 + 21 − −1 +C
2 5+ 1 32 + 11 2 + 1
𝑥2 𝑥2 𝑥2
= +2 3 − 1 +C
5
2 2 2
5 3 1
𝑥2 𝑥2 𝑥2
= +2 3 − 1 +C
5
2 2 2

1

, 2 5 4 3 1
= 𝑥2 + 𝑥2 − 2𝑥2 + C
5 3

• Evaluate: ∫ 𝑐𝑜𝑠5𝑥 cos 3𝑥 𝑑𝑥
Solution:
We know that,
1
𝑐𝑜𝑠𝐶 cos 𝐷 = [cos(𝐶 − 𝐷) + cos(𝐶 + 𝐷)]
2
1
∴ ∫ 𝑐𝑜𝑠5𝑥 cos 3𝑥 𝑑𝑥 = ∫ [cos(5𝑥 − 3𝑥) + cos(5𝑥 + 3𝑥)]𝑑𝑥
2
1
= ∫[cos 2𝑥 + cos 8𝑥]𝑑𝑥
2
1
= [∫ 𝑐𝑜𝑠2𝑥 𝑑𝑥 + ∫ 𝑐𝑜𝑠8𝑥 𝑑𝑥]
2
1 𝑠𝑖𝑛2𝑥 𝑠𝑖𝑛8𝑥
= [ + ]+C
2 2 8

• Evaluate: ∫ √1 − 𝑐𝑜𝑠2𝑥 𝑑𝑥 .
Solution:
∫ √1 − 𝑐𝑜𝑠2𝑥 𝑑𝑥 = ∫ √2𝑠𝑖𝑛2𝑥 [since 2𝑠𝑖𝑛2𝜃 = 1 − 𝑐𝑜𝑠2𝜃 ]

= √2 ∫ 𝑠𝑖𝑛𝑥 𝑑𝑥
= √2(−𝑐𝑜𝑠𝑥) + C = −√2𝑐𝑜𝑠𝑥 + C

• Evaluate: ∫ √1 + 𝑠𝑖𝑛2𝑥 𝑑𝑥 .
Solution:
We know that,
𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑥 = 1 𝑎𝑛𝑑 sin 2𝑥 = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
∴ ∫ √1 + 𝑠𝑖𝑛2𝑥 𝑑𝑥 = ∫ √(𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑥) + 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 𝑑𝑥

= ∫ √(𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)2) 𝑑𝑥

= ∫(𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥) 𝑑𝑥
= (−𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥) + 𝐶 = 𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥 + C
𝑠𝑖𝑛𝑥
• Integrate:∫ 𝑑𝑥.
𝑐𝑜𝑠2𝑥
Solution:

2

, 𝑠𝑖𝑛𝑥 𝑠𝑖𝑛𝑥
∫ 𝑑𝑥 = ∫ 𝑑𝑥
𝑐𝑜𝑠2𝑥 (𝑐𝑜𝑠𝑥)(𝑐𝑜𝑠𝑥)
𝑠𝑖𝑛𝑥 1
= ∫( )( ) 𝑑𝑥
𝑐𝑜𝑠𝑥 𝑐𝑜𝑠𝑥
= ∫ 𝑡𝑎𝑛𝑥 𝑠𝑒𝑐𝑥 𝑑𝑥 = 𝑠𝑒𝑐𝑥 + 𝐶
• Integrate: ∫ 𝑐𝑜𝑠3𝑥 𝑑𝑥.
Solution:
We know that,
𝑐𝑜𝑠 3𝑥 = 4𝑐𝑜𝑠3𝑥 − 3 𝑐𝑜𝑠 𝑥
⟹ 4𝑐𝑜𝑠3𝑥 = 𝑐𝑜𝑠 3𝑥 + 3 𝑐𝑜𝑠 𝑥
1
⟹ 𝑐𝑜𝑠3𝑥 = (𝑐𝑜𝑠 3𝑥 + 3 𝑐𝑜𝑠 𝑥)
4
1
∴ ∫ 𝑐𝑜𝑠3 𝑥 𝑑𝑥 = ∫(𝑐𝑜𝑠 3𝑥 + 3 𝑐𝑜𝑠 𝑥) 𝑑𝑥
4
1 𝑠𝑖𝑛 3𝑥
= [ + 3 𝑠𝑖𝑛𝑥] + C
4 3
• Integrate: ∫ 𝑠𝑖𝑛3𝑥 cos 2𝑥 𝑑𝑥.
Solution:
We know that,
1
𝑠𝑖𝑛𝐶 sin 𝐷 = [sin(𝐶 + 𝐷) + sin(𝐶 − 𝐷)]
2
1
∴ ∫ 𝑠𝑖𝑛3𝑥 cos 2𝑥 𝑑𝑥 = ∫ [sin(3𝑥 + 2𝑥) + sin(3𝑥 − 2𝑥)]𝑑𝑥
2
1
= ∫ [sin(3𝑥 + 2𝑥) + sin(3𝑥 − 2𝑥)]𝑑𝑥
2
1
= ∫[sin 5𝑥 + sin 𝑥]𝑑𝑥
2
1
= [∫ sin 5𝑥 𝑑𝑥 + ∫ sin 𝑥 𝑑𝑥]
2
1 −𝑐𝑜𝑠5𝑥
= [ − 𝑐𝑜𝑠𝑥] + C
2 5
−1 𝑐𝑜𝑠5𝑥
= [ + 𝑐𝑜𝑠𝑥] + C
𝑑𝑥
2 5
• Integrate:∫ .
𝑥2+2𝑥+5
Solution:
𝑥2 + 2𝑥 + 5 = 𝑥2 + 2𝑥 + 1 + 4 = (𝑥 + 1)2 + 22



3

, 𝑑𝑥 𝑑𝑥
∴ ∫ =∫
𝑥2 + 2𝑥 + 5 22 + (𝑥 + 1)2
We know that,
∫ 2 1 2 1 𝑥
𝑎 + 𝑥 𝑑𝑥 = 𝑎 tan−1 ( ) + 𝐶
𝑎
𝑑𝑥 𝑑𝑥 1 𝑥+1
∴ ∫ =∫ = tan −1 ( )+𝐶
𝑥 + 2𝑥 + 5
2 2 + (𝑥 + 1)
2 2 2 2
𝑑𝑥
• Integrate:∫ .
𝑥2+2𝑥+10
Solution:
𝑥2 + 2𝑥 + 10 = 𝑥2 + 2𝑥 + 1 + 9 = (𝑥 + 1)2 + 32
𝑑𝑥 𝑑𝑥
∴ ∫ 2 =∫ 2
𝑥 + 2𝑥 + 10 3 + (𝑥 + 1)2
We know that,
∫ 2 1 2 1 𝑥
𝑎 + 𝑥 𝑑𝑥 = 𝑎 tan−1 ( ) + 𝐶
𝑎
𝑑𝑥 𝑑𝑥 1 𝑥+1
∴ ∫ =∫ = tan −1 ( )+C
𝑥 + 2𝑥 + 10
2 3 + (𝑥 + 1)
2 2 3 3
𝑑𝑥
• Evaluate ∫
sin2 𝑥 cos2 𝑥
Solution:
𝑑𝑥 (sin2 𝑥 + cos2 𝑥)𝑑𝑥
∫ 2 = ∫
sin 𝑥 cos2 𝑥 sin2 𝑥 cos2 𝑥
sin2 𝑥 cos2 𝑥 𝑑𝑥
= (∫ 2 + ) 𝑑𝑥
sin 𝑥 cos2 𝑥 sin2 𝑥 cos2 𝑥
= ∫ sec2 𝑥𝑑𝑥 + ∫ cosec2 𝑥 𝑑𝑥
= 𝑡𝑎𝑛𝑥 − 𝑐𝑜𝑡𝑥 + C
• Evaluate ∫ 𝑠𝑖𝑛7𝑥 . 𝑐𝑜𝑠5𝑥𝑑𝑥
Solution:
We know that,
1
𝑠𝑖𝑛𝐶𝑠𝑖𝑛𝐷 = [sin(𝐶 + 𝐷) + sin(𝐶 − 𝐷)]
2
1
∴ 𝑠𝑖𝑛7𝑥 . 𝑐𝑜𝑠5𝑥𝑑𝑥 = [sin(7𝑥 + 5𝑥) + sin(7𝑥 − 5𝑥)]
2
1
= [sin(12𝑥) + sin(2𝑥)]
2
1
∴ ∫ 𝑠𝑖𝑛7𝑥 . 𝑐𝑜𝑠5𝑥𝑑𝑥 = ∫[sin(12𝑥) + sin(2𝑥)]𝑑𝑥
2

4