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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
′
(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) ∫
𝑥
2
− 5 𝑥+
𝑥
Solution:
2
2
𝑥
2
+2𝑥− 1
√
𝑥
= − 5 𝑥 + 𝑙𝑜𝑔𝑥 + C
Solution:
2
2
1
2
𝑑𝑥
3
1
2
1
1
2 − 𝑥
1
1
2 ) 𝑑𝑥
2
2
− 𝑥
−
2
) 𝑑𝑥
3
2
1
2
− 1
2
+ C
5
2
5
2
3
2
3
2
1
2
+ C
1
2
+ C
−
2− 1− −
+1 +
5 4
3 1
2
2
− 2 𝑥
2
𝑐𝑜𝑠 5 𝑥 cos 3 𝑥 𝑑𝑥
Solution:
We know that,
𝑐𝑜𝑠𝐶 cos 𝐷 =
[cos(𝐶 − 𝐷) + cos(𝐶 + 𝐷)]
∴ ∫ 𝑐𝑜𝑠 5 𝑥 cos 3 𝑥 𝑑𝑥 = ∫
[cos( 5 𝑥 − 3 𝑥) + cos(5𝑥 + 3 𝑥)]𝑑𝑥
[
cos 2 𝑥 + cos 8 𝑥
]
[∫ 𝑐𝑜𝑠 2 𝑥 𝑑𝑥 + ∫ 𝑐𝑜𝑠 8 𝑥 𝑑𝑥]
[
Solution:
] + C
2
[
since 2 𝑠𝑖𝑛
2
]
+ C = −√ 2 𝑐𝑜𝑠𝑥 + C
Solution:
We know that,
2
2
𝑥 = 1 𝑎𝑛𝑑 sin 2 𝑥 = 2 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
2
2
2
𝑠𝑖𝑛𝑥
𝑐𝑜𝑠
2
𝑥
= (−𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥) + 𝐶 = 𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥 + C
Solution:
− 1
− 1
2
2
2
We know that,
2
2
tan
− 1
= tan ( ) + 𝐶
2
2
2
𝑑𝑥
𝑥
2
+2𝑥+
Solution:
2
2
2
2
2
2
2
We know that,
2
2
𝑑𝑥 =
tan
− 1
= tan ( ) + C
2
𝑑𝑥
sin
2
𝑥 cos
2
𝑥
2
2
Solution:
sin
2
𝑥 cos
2
(sin
2
𝑥 + cos
2
sin
2
𝑥 cos
2
sin
2
𝑥 cos
2
sin
2
𝑥 cos
2
sin
2
𝑥 cos
2
= ∫ sec
2
𝑥𝑑𝑥 + ∫ cosec
2
= 𝑡𝑎𝑛𝑥 − 𝑐𝑜𝑡𝑥 + C
- Evaluate ∫ 𝑠𝑖𝑛 7 𝑥. 𝑐𝑜𝑠 5 𝑥𝑑𝑥
Solution:
We know that,
[
sin
]
[sin( 7 𝑥 + 5 𝑥) + sin(7𝑥 − 5 𝑥)]
[
sin
]
∫[sin( 12 𝑥) + sin(2𝑥)]𝑑𝑥
[
] + C
[𝑐𝑜𝑠 2 𝑥 +
Integration by Method of substitution
] + C
2 𝑥
1+𝑥
2
Solution:
Let 𝑢 = 1 + 𝑥
2
2
= log 𝑢 + 𝐶 = log( 1 + 𝑥
2
𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥
Solution:
Let 𝑢 = 𝑠𝑖𝑛𝑥
Solution:
𝑠𝑒𝑐
2
𝑥
𝑡𝑎𝑛𝑥
= log 𝑢 + 𝐶 = log(sin 𝑥) + 𝐶
Let 𝑢 = 𝑡𝑎𝑛𝑥
2
2
= log 𝑢 + 𝐶 = log(tan 𝑥) + 𝐶
𝑑𝑥
(1+𝑥
2
)tan
− 1
𝑥
Solution:
Let 𝑢 = tan
− 1
2
2
)tan
− 1
𝑐𝑜𝑠𝑥−𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥+𝑠𝑖𝑛𝑥
= log 𝑢 + C = log(tan
− 1
𝑥) + C
Solution:
Let 𝑢 = 𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥
2 𝑥
2
2
2
𝑢 𝑑𝑢 = 𝑡𝑎𝑛 𝑢 + C = 𝑡𝑎𝑛(𝑙𝑜𝑔 𝑥) + C
2 𝑙𝑜𝑔𝑥
𝑥
3
Solution:
2 𝑙𝑜g𝑥 𝑥
3
𝑙𝑜g𝑥
2
𝑥
3
3
Put 𝑡 = 𝑥
3
2
2
2
𝑥
3
𝑡
𝑡
+ C =
𝑥
3
+ C
𝑒
𝑡𝑎𝑛𝑥
𝑐𝑜𝑠 𝑥
Solution:
Let 𝑡 = 𝑡𝑎𝑛𝑥 ⟹ 𝑑𝑡 = 𝑠𝑒𝑐
2
𝑡𝑎𝑛𝑥
2
𝑡𝑎𝑛𝑥
2
𝑡
𝑡
+ C = 𝑒
𝑡𝑎𝑛𝑥
+ C
𝑙𝑜𝑔𝑥
d𝑥
𝑥
Solution:
Let 𝑡 = 𝑙𝑜𝑔𝑥 ⟹ 𝑑𝑡 =
2
+ C =
2
+ C
𝑥
2
Solution:
Let 𝑢 = 𝑥
2
𝑥
2
𝑢
𝑢
𝑢
+ C =
𝑥
2
+ C
𝑠𝑖𝑛(𝑙𝑜𝑔𝑥)
𝑥
Solution:
Let 𝑡 = 𝑙𝑜𝑔𝑥 ⟹ 𝑑𝑡 =
3
𝑥
𝑑𝑥 = ∫ 𝑠𝑖𝑛𝑡 𝑑𝑡 = −𝑐𝑜𝑠𝑡 + C = − cos(𝑙𝑜𝑔𝑥) + C
𝑑𝑥
𝑥.𝑙𝑜𝑔𝑥
Solution:
Let 𝑡 = 𝑙𝑜𝑔𝑥 ⟹ 𝑑𝑡 =
1
𝑥
= ∫ 𝑑𝑡 = 𝑙𝑜𝑔𝑡 + C
𝑐𝑜𝑡(𝑥)
𝑙𝑜𝑔(𝑠𝑖𝑛𝑥)
Solution:
Let 𝑡 = 𝑙𝑜𝑔(𝑠𝑖𝑛𝑥) and 𝑢 = 𝑠𝑖𝑛𝑥
= 𝑙𝑜𝑔𝑡 + C = 𝑙𝑜𝑔(𝑙𝑜𝑔(𝑠𝑖𝑛𝑥)) + C
2
Solution:
Here, ∫
2
2
2
Let 𝑢 = 2 𝑥 ⟹ 𝑑𝑢 = 2 𝑑𝑥 ⟹ 𝑑𝑥 =
𝑑𝑢
2
2
2
2
= [ {
2
2
2
− 1
)}]
[2𝑥
2
− 1
)]
Integration by parts
Let u & v be two functions of x then
2
Evaluate ∫
Solution:
Put 𝑡 = 𝑥
2
Now 𝑢 = 𝑥, 𝑑𝑣 = 𝑒
−𝑥
−𝑥
2
−𝑥
2
−𝑥
[
−𝑥
−𝑥
]
2
−𝑥
+ 2 [−𝑥𝑒
−𝑥
−𝑥
)] + 𝐶 = −𝑒
−𝑥
[𝑥
2
+ 2 𝑥 + 2 ] + 𝐶
2
3 𝑥
Solution:
Applying integration by parts,
Let 𝑢 = 𝑥
2
3 𝑥
3 𝑥
2
3 𝑥
2
3 𝑥
2
3 𝑥
3 𝑥
3 𝑥
Again applying integration by parts
Let 𝑢 = 𝑥 , 𝑑𝑣 = 𝑒
3 𝑥
3 𝑥
∴ Equation (1) ⇒
2
3 𝑥
2
3 𝑥
3 𝑥
3 𝑥
3 𝑥
2
3 𝑥
3 𝑥
3 𝑥
2
3 𝑥 2 𝑥𝑒
3 𝑥
3 𝑥
2
3 𝑥
3 𝑥
3 𝑥
2
sin
− 1
Solution:
Use Integration by parts method
Put 𝑢 = sin
− 1
2
1
√1−𝑥
2
𝑥
3
3
3
2
sin
− 1
3
sin
− 1
2
3
3
= sin
− 1
2
To find ∫ 𝑑𝑥
2
Put 𝑡 = 1 − 𝑥
2
2
3
2
2
2
∫ [
− √𝑡] 𝑑𝑡
3
2 1
3
[
] = − [
2 ]
2
∴ Equation
gives
3
2
2
2
sin
− 1
3
3
sin
− 1
[−
2
3
2
2
]
3
= sin
− 1
2
2
2
0
0
0
1
2
𝜋
2
2
0
𝜋
2
0
𝜋
0
𝜋
2
= [𝑥 − ( )]
0
[(
) − 𝑠𝑖𝑛0)]
[(
(𝑠𝑖𝑛𝜋 − 0 )]
1 1
1+𝑥
2
[𝑠𝑖𝑛𝑐𝑒 sin 𝜋 = 0]
Solution:
We know that,
2
𝑑𝑥 = tan
− 1
𝑥 + C
1
2
𝑑𝑥 = [tan
− 1
𝑥]
1
= tan
− 1
− tan
− 1
𝜋
𝑐𝑜𝑠𝑥
Evaluate:
2
0 1+𝑠𝑖𝑛𝑥
Solution:
When 𝑥 = 0, 𝑢 = 1
When 𝑥 =
𝜋
2
0
[
]
2
1
1
1
0
−𝑎
= 𝑙𝑜𝑔 2 − 𝑙𝑜𝑔 1 = 𝑙𝑜𝑔 2 [since 𝑙𝑜𝑔 1 = 0]
2
lo g x d x.
1
Solution:
Using Integration by parts, ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢,
Let 𝑢 = log 𝑥, 𝑑𝑣 = 𝑑𝑥.
2 2
∴ ∫ 𝑙𝑜𝑔𝑥 𝑑𝑥 = [𝑥𝑙𝑜𝑔𝑥]
2
1 1
= [ 2 𝑙𝑜𝑔 2 − 1 𝑙𝑜𝑔 1 ] − [𝑥]
2
[since log1 = 0]
4
ta n x d x.
0
2
Solution:
We know that,
+ C
𝜋
4
0
[
𝜋
)]
4
= − [𝑙𝑜𝑔 (𝑐𝑜𝑠
) − 𝑙𝑜𝑔(cos 0)]
= − [𝑙𝑜𝑔 (
) − 𝑙𝑜𝑔1]
a
e
x
d x.
− a
= −[𝑙𝑜𝑔 1 − 𝑙𝑜𝑔√2 − 𝑙𝑜𝑔1]
Solution:
We know that,
𝑥
𝑥
+ C
𝑎
𝑥
𝑑𝑥 = [𝑒
𝑥
]
𝑎
−𝑎
1
0
0
[
]
2
𝜋
3
sin
3
Solution:
We know that, 𝑠𝑖𝑛 3 𝑥 = 3 𝑠𝑖𝑛𝑥 − 4 𝑠𝑖𝑛
3
3
𝜋
3 𝜋
3
∫ sin
0
0
𝜋
[−3𝑐𝑜𝑠𝑥 +
3
]
= [−3𝑐𝑜𝑠 +
0
− (−3𝑐𝑜𝑠 0 + 𝑐𝑜𝑠0)]
[−3 (
] =
[
+ 3]
[
𝜋
] =
2
Solution:
𝜋
2
We know that,
𝜋
2
[𝑠𝑖𝑛(𝐴 + 𝐵) − sin(𝐴 − 𝐵)]
[
− sin(3𝑥 − 2 𝑥)
]
0 0
𝜋
2
[
− sin(𝑥)
]
0
𝜋
2
[
+ 𝑐𝑜𝑠𝑥]
0
0
[
cos (5.
+ 𝑐𝑜𝑠 0 − 𝑐𝑜𝑠0]
[
cos
+0] [since cos
= cos (360 +
) = cos
]
𝜋
2
Solution:
Let 𝑢 = 𝑠𝑖𝑛𝑥 when 𝑥 = 0; 𝑢 = 𝑠𝑖𝑛 0 = 0
⟹ 𝑑𝑢 = 𝑐𝑜𝑠𝑥𝑑𝑥 when 𝑥 =
𝜋
𝜋
2 2
1 1
1
1
1
2
3
1
2 2 2
2
𝑑𝑢 = [
] = [
]
= [1 − 0] =
0 0
0
0
Evaluate
3
2
− 2
Solution:
3
2
𝑑𝑥 = [4𝑥 −
− 2
3
3
]
− 2
= (4 × 3 −
3
3
1 sin
− 1
𝑥𝑑𝑥
0
Solution:
√ 1−𝑥
2
Let 𝑡 = sin
− 1
1
√(1−𝑥
2
)
When 𝑥 = 0 ⟹ 𝑡 = sin
− 1
When 𝑥 = 1 ⟹ 𝑡 = sin
− 1
1
sin
− 1
𝜋
2
𝜋
2
2
2
0
0 0
1
1
0
1
1
2
√
2 √
2
√𝑥
𝑡
𝑡
1
1 1
√
2
= 2 [
𝑡
√
𝑡
𝑑𝑡]
𝜋
2
𝑠𝑖𝑛𝑥𝑑𝑥
2
1
= 2 [√2𝑒
√
1
[
𝑡
]
√
]
= 2 [√2𝑒
√
− 𝑒 − [𝑒
√
− 𝑒]]
√
0 1+cos 𝑥
Solution:
𝜋 𝑠𝑖𝑛𝑥
Let 𝐼 = ∫
2
1+cos
2
𝑥
Let 𝑡 = 𝑐𝑜𝑠𝑥 ⟹ 𝑑𝑡 = −𝑠𝑖𝑛𝑥 𝑑𝑥
When 𝑥 = 0 ⟹ 𝑡 = 𝑐𝑜𝑠0 = 1
When 𝑥 =
0
∴ I = ∫
2
= −[tan
− 1
𝑡]
0
1
[
tan
− 1
0 − tan
− 1
]
2 𝑑𝑥
𝑥
2
+5𝑥+
= − [0 − ] =
Solution:
Let
Put 𝑥 = −2; 1 = 𝐴
Put 𝑥 = −3; 1 = 0 + 𝐵(− 3 + 2 )
0 2 0 0 0 0
𝜋
2
2
Solution:
Use Integration by parts method
Let 𝑢 = 𝑥
2
𝜋
2
2
[
2
𝜋
𝜋
2
)]
2
0
0
0
𝜋
2
2
0
𝜋
2
0
Again Using Integration by parts method,
Let 𝑢 = 𝑥, 𝑑𝑣 = 𝑐𝑜𝑠𝑥𝑑𝑥
𝜋
𝜋
𝜋
2
0
= 2 [
2
2
𝑠𝑖𝑛𝑥𝑑𝑥]
0
= 2 [(
𝜋
2
]
1
2
𝑥
Solution:
Use Integration by parts method
Let 𝑢 = 𝑥
2
𝑥
𝑥
1
1
2
𝑥
𝑑𝑥 = [𝑥
2
𝑥
]
1
𝑥
0
0
