1 Rahman Sir
DEFINITE INTEGRAL
Properties of definite integral
𝒃
𝒃
1.
Property I. ∫𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒛)𝒅𝒛
2
Property II. ∫𝒂 𝒇(𝒙)𝒅𝒙 = − ∫𝒃 𝒇(𝒙)𝒅𝒙
𝒃
𝒂
𝒃
𝒄
𝒃
3. Property III. ∫𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒙)𝒅𝒙 + ∫𝒄 𝒇(𝒙)𝒅𝒙. 𝒘𝒉𝒆𝒓𝒆 𝒂 < 𝒄 < 𝒃
𝒃
4.
𝒃
Property IV. ∫𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒂 + 𝒃 − 𝒙)𝒅x
𝒂
𝒂
Special case ∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒂 − 𝒙)𝒅𝒙
5.
𝒂
Property V. ∫−𝒂 𝒇(𝒙)𝒅𝒙 = 𝟎. 𝒊𝒇 𝒇(𝒙)𝒊𝒔 𝒐𝒅𝒅 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏
𝒂
= 2∫𝟎 𝒇(𝒙)𝒅𝒙 if f(x) is even function
𝒂
𝟎
𝒂
Proof : ∫−𝒂 𝒇(𝒙)𝒅𝒙 = ∫−𝒂 𝒇(𝒙)𝒅𝒙 + ∫𝟎 𝒇(𝒙)𝒅𝒙
Let 𝒛 = −𝒙, 𝒕𝒉𝒆𝒏 𝒅𝒛 = −𝒅𝒙 𝒘𝒉𝒆𝒓𝒆, 𝒙 = −𝒂, 𝒛 = 𝒂
𝟎
𝟎
Now ∫–𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(−𝒛)(−𝒅 𝒛)
𝟎
= − ∫ 𝒇(−𝒛)𝒅𝒛
𝒂
𝒂
∫𝟎 𝒇(−𝒛)𝒅𝒛 by property II
𝒂
= ∫𝟎 𝒇(−𝒙)𝒅𝒙 by property I
𝒂
𝒂
𝒂
… (1)
𝑻𝒉𝒖𝒔 , ∫−𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒙)𝒅𝒙 + ∫𝟎 𝒇(−𝒙)𝒅𝒙
This concept is used when f(x) is neither odd nor even.
Case I. when f(x) is odd function, then f(-x) = -f(x)
𝒂
𝒂
𝒂
From (1) ∫−𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒙)𝒅𝒙 + ∫𝟎 𝒇(−𝒙)𝒅𝒙
𝒂
𝒂
𝒂
𝒂
= ∫𝟎 𝒇(𝒙)𝒅𝒙 − ∫𝟎 𝒇(𝒙)𝒅𝒙 as 𝒇(−𝒙) = −𝒇(𝒙)
=0
Case II. when f(x) is even function, then f(-x) = f(x)
𝒂
𝒂
𝒂
From (1) ∫−𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒙)𝒅𝒙 + ∫𝟎 𝒇(−𝒙)𝒅𝒙
= ∫𝟎 𝒇(𝒙)𝒅𝒙 + ∫𝟎 𝒇(𝒙)𝒅𝒙 as 𝒇(−𝒙) = 𝒇(𝒙)
𝒂
𝟐𝒂
= 2 ∫𝟎 𝒇(𝒙)𝒅𝒙
𝒂
6. Property VI: ∫𝟎 𝒇(𝒙)𝒅𝒙 = 𝟐 ∫𝟎 𝒇(𝒙) 𝒅𝒙 𝒊𝒇 𝒇(𝟐𝒂 − 𝒙) = 𝒇(𝒙)
= 0 if 𝒇(𝟐𝒂 − 𝒙) = −𝒇(𝒙)
𝑏
𝑐
𝑏
Type I. Problems based on the property : ∫𝑎 𝑓(𝑥) 𝑑𝑥 = ∫𝑎 𝑓(𝑥) 𝑑𝑥 + ∫𝑐 𝑓(𝑥)
This property is used when
(i)
Integrand f(x) is different in different part of ሾ𝑎, 𝑏ሿ in which the function is to be integrated.
2 Rahman Sir
DEFINITE INTEGRAL
(ii)
(iii)
Integrand f(x) is a modulus function for this put modulus part equal to zero and find the
values of x which is between lower and upper limits.
Integrand f(x) is Greatest Integer function.
1 − 2𝑥, 𝑥 ≤ 0
1
Evaluate ∫−1 𝑓(𝑥) 𝑑𝑥 𝑤ℎ𝑒𝑟𝑒 𝑓(𝑥) = ൜
1 + 2𝑥, 𝑥 ≥ 0
1
0
1
∫−1 𝑓(𝑥) 𝑑𝑥 = ∫−1(1 − 2𝑥) 𝑑𝑥 + ∫0 (1 + 2𝑥) 𝑑𝑥
= ሾ𝑥 − 𝑥 2 ሿ0−1 + ሾ𝑥 + 𝑥 2 ሿ10
= ሾ(0) − (−1 − 1)ሿ + ሾ1 + 1ሿ= 2 + 2 = 4
3
2
Find ∫0 ȁ𝑥𝑐𝑜𝑠𝜋𝑥ȁ 𝑑𝑥
Solution: Put 𝑥𝑐𝑜𝑠𝜋𝑥 = 0
⇒ 𝑥 = 0 𝑜𝑟 𝑐𝑜𝑠𝜋𝑥 = 0
𝑥 = 0 𝑜𝑟 𝜋𝑥 =
𝑜𝑟 𝑥 =
(2𝑛+1)
2
(2𝑛+1)
𝑥=
𝜋
where n = 0,1, 2 ,3..
2
1
3
; 𝑤ℎ𝑒𝑛 𝑛 = 0, 𝑎𝑛𝑑 𝑥 = 𝑤ℎ𝑒𝑛 𝑛 = 1
2
2
1
2
3
2
1
2
I= ∫0 𝑥𝑐𝑜𝑠𝜋𝑥 𝑑𝑥 - ∫ 𝑥𝑐𝑜𝑠𝜋𝑥 𝑑𝑥
1
Now, ∫ 𝑥𝑐𝑜𝑠𝜋𝑥 𝑑𝑥 = 𝑥 ∫ 𝑐𝑜𝑠𝜋𝑥 𝑑𝑥 − 𝜋 ∫ 𝑠𝑖𝑛𝜋𝑥 𝑑𝑥
𝑥
1
= 𝜋 𝑠𝑖𝑛𝜋𝑥 + 𝜋2 𝑐𝑜𝑠𝜋𝑥
1
2
3
2
1
2
Hence, I= ∫0 𝑥𝑐𝑜𝑠𝜋𝑥 𝑑𝑥 - ∫ 𝑥𝑐𝑜𝑠𝜋𝑥 𝑑𝑥
𝑥
1
= ቂ𝜋 𝑠𝑖𝑛𝜋𝑥 + 𝜋2 𝑐𝑜𝑠𝜋𝑥ቃ
1
1
1
2
𝑥
− ቂ𝜋 𝑠𝑖𝑛𝜋𝑥 + 𝜋2 𝑐𝑜𝑠𝜋𝑥ቃ
0
3
1
ቂ2𝜋 − 𝜋2 ቃ − ቂ2𝜋 (−1) − 2𝜋ቃ =
5
1
1+1+3
2𝜋
1
= 2𝜋 − 𝜋2
3
2
∫0 ሾ𝑥ሿ 𝑑𝑥 where ሾ𝑥ሿ is greatest integer function.
Now, f(x) = 0. When 𝑥 ∈ (0,1)
1
− 𝜋2
3
2
1
2
3 Rahman Sir
DEFINITE INTEGRAL
3
f(x) = 1. When 𝑥 ∈ (1, 2 )
3
2
3
1
3
3
1
∫0 ሾ𝑥ሿ 𝑑𝑥 = ∫0 0 𝑑𝑥 + ∫12 1 𝑑𝑥 = 0 + ሾ𝑥ሿ12 = 2 − 1 = 2
𝜋
𝑥
𝑥
Example: The integral of ∫0 ට1 + 4 sin2 2 − 4 sin 2 is
2𝜋
(𝐴) 𝜋 − 4 (B)
3
− 4 − 4ඥ3
𝜋
(C) 4ඥ3 − 4
𝜋
𝑥
(D) 4ඥ3 − 4 − 3
𝑥
Solution: Let I= ∫0 ට1 + 4 sin2 2 − 4 sin 2 dx
𝜋
2
𝑥
= ∫0 ටቀ2 sin 2 − 1ቁ dx
𝜋
𝑥
∵ ඥ(𝑥)2 = ȁ𝑥ȁ
= ∫0 ቚ2 sin 2 − 1ቚ dx
𝑥
𝑥
1
𝑥
𝜋
𝜋
Put 2 sin 2 − 1 = 0 𝑜𝑟 sin 2 = 2 or 2 = 6 𝑜𝑟 𝑥 = 3
𝜋
𝑥
𝜋
𝑥
When 𝑥 ∈ ቀ0, 3 ቁ 𝑡ℎ𝑒𝑛 2 sin 2 − 1 ≤ 0 and When 𝑥 ∈ ቀ 3 , 𝜋ቁ 𝑡ℎ𝑒𝑛 2 sin 2 − 1 ≥ 0
𝜋
𝜋
𝑥
𝑥
𝐼 = − ∫03 (2 sin 2 − 1) dx + ∫𝜋 (2 sin 2 − 1) dx
3
𝜋
3
𝑥
𝑥
𝜋
= − ቂ −4 cos 2 − 𝑥ቃ + ቂ −4 cos 2 − 𝑥ቃ𝜋
0
ξ3
3
𝜋
ξ3
𝜋
= ቀ4 × 2 + 3 − 4ቁ + ቀ 0 − 𝜋 + 4 × 2 + 3 ቁ
2𝜋
𝜋
= 4ඥ3 − 4 + 3 − 𝜋 = 4ඥ3 − 4 − 3
𝜋
𝐻𝑒𝑛𝑐𝑒, the correct option is (D) 4ඥ3 − 4 − 3
Type II. Problems based on the property:
𝑏
𝑏
∫𝑎 𝑓(𝑥) 𝑑𝑥 = ∫𝑎 𝑓(𝑎 + 𝑏 − 𝑥) 𝑑𝑥
𝑎
𝑎
or ∫0 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥) 𝑑𝑥 special case
𝑇ℎ𝑖𝑠 property is used when 𝑓(𝑥) + 𝑓(𝑎 + 𝑏 − 𝑥) becomes 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑏𝑙𝑒 function of x i.e.,
(𝑖) f(x) is trigonometric function of complementary angles.
(ii) Integrand is of the form x f(x) so that x is eliminated and the function becomes 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑏𝑙𝑒
function of x.
(iii) Integrand is a logarithmic function.
4 Rahman Sir
DEFINITE INTEGRAL
Working rule:
First we assume the Integration as
𝑏
I = ∫𝑎 𝑓(𝑥) 𝑑𝑥 … (1)
𝑏
I = ∫𝑎 𝑓(𝑎 + 𝑏 − 𝑥) 𝑑𝑥…. (2)
Adding (1) and (2) we get
𝑏
∫𝑎 ሾ𝑓(𝑥) + 𝑓(𝑎 + 𝑏 − 𝑥)ሿ 𝑑𝑥 which becomes 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑏𝑙𝑒 function of x.
𝜋 𝑥𝑠𝑖𝑛𝑥
Example: Evaluate: ∫0 1+cos2 𝑥 𝑑𝑥
𝜋
Let 𝐼 = ∫0 1+cos2 𝑥 𝑑𝑥
𝑥𝑠𝑖𝑛𝑥
… (1)
𝜋 (𝜋−𝑥)sin(𝜋−𝑥)
𝜋 𝜋𝑠𝑖𝑛𝑥
𝜋 −𝑥𝑠𝑖𝑛𝑥
Or 𝐼 = ∫0 1+cos2 (𝜋− 𝑥) dx = ∫0 1+cos2 𝑥 dx+ ∫0 1+cos2 𝑥 dx … (2)
Adding (1) and (2) we get
𝜋 𝜋𝑠𝑖𝑛𝑥
2I= ∫0 1+cos2 𝑥dx
𝑙𝑒𝑡 𝑐𝑜𝑠𝑥 = 𝑡 , 𝑡ℎ𝑒𝑛 − 𝑠𝑖𝑛𝑥𝑑𝑥 = 𝑑𝑡 , 𝑤ℎ𝑒𝑛 𝑥 = 0 𝑡 = 𝑐𝑜𝑠0 = 1 𝑎𝑛𝑑 𝑤ℎ𝑒𝑛 𝑥 = 𝜋 𝑡 = −1
−1 −𝑑𝑡
1+𝑡 2
1
2
1
1
𝐼 = 2 𝜋 ∫1
𝑑𝑡
= 2 𝜋 ∫−1 1+𝑡 2
1 𝑑𝑡
= 2 𝜋 ∫0 1+𝑡 2
𝒃
𝒂
∵ ∫𝒂 𝒇(𝒙)𝒅𝒙 = − ∫𝒃 𝒇(𝒙)𝒅𝒙
∵ 𝑓(𝑡)𝑖𝑠 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝜋
𝜋2
𝜋ሾtan−1 𝑥ሿ10 = 𝜋ൣtan−1(1 ) − tan−1 0൧ = 𝜋 ቀ 4 ቁ = 4
𝜋
𝑑𝑥
Evaluate: ∫𝜋3 1+ 𝑡𝑎𝑛𝑥
ξ
6
𝜋
𝑑𝑥
3
𝜋
1+ ξ𝑡𝑎𝑛𝑥
6
Let 𝐼 = ∫
𝜋
= ∫
ξ𝑐𝑜𝑠𝑥𝑑𝑥
ξ𝑠𝑖𝑛𝑥
Or 𝐼 = ∫𝜋3 𝑐𝑜𝑠𝑥+
ξ
6
𝜋
ξ𝑐𝑜𝑠𝑥𝑑𝑥
3
𝜋
𝑐𝑜𝑠𝑥+ ξ𝑠𝑖𝑛𝑥
ξ
6
… (1)
𝜋 𝜋
𝜋
ටcos( + −𝑥)𝑑𝑥
3 6
3
𝜋
𝜋 𝜋
𝜋 𝜋
6 ට𝑐𝑜𝑠( + −𝑥)+ ට𝑠𝑖𝑛( + −𝑥)
3 6
3 6
𝑜𝑟 𝐼 = ∫
𝜋
ξ𝑠𝑖𝑛𝑥𝑑𝑥
𝑜𝑟 𝐼 = ∫𝜋3 𝑐𝑜𝑠𝑥+
ξ𝑠𝑖𝑛𝑥
ξ
… (2)
6
Adding (1) and (2) we get
5 Rahman Sir
DEFINITE INTEGRAL
𝜋
3
𝜋
6
2I = ∫ 𝑑𝑥
𝜋
3
𝜋
6
𝜋
𝜋
𝜋
= ሾ𝑥ሿ = 3 − 6 = 6
Hence, I =
𝜋
12
𝜋
4
Evaluate ∫0 log(1 + 𝑡𝑎𝑛𝑥 ) 𝑑𝑥
𝜋
4
Solution: let I = ∫0 log(1 + 𝑡𝑎𝑛𝑥 ) 𝑑𝑥 … (1)
𝜋
4
𝜋
Or I = ∫0 log 1 + tan( 4 − 𝑥) 𝑑𝑥
𝜋
4
𝜋
4
= I = ∫0 log(1 +
𝜋
4
= ∫ log(
0
tan −𝑡𝑎𝑛𝑥
𝜋
4
1+tan .𝑡𝑎𝑛𝑥
) 𝑑𝑥
1 + 𝑡𝑎𝑛𝑥 + 1 −𝑡𝑎𝑛𝑥
) 𝑑𝑥
1 + 𝑡𝑎𝑛𝑥
𝜋
4
2
= ∫0 log(1+𝑡𝑎𝑛𝑥) 𝑑𝑥
𝜋
4
𝜋
4
Or I = ∫0 𝑙𝑜𝑔2𝑑𝑥 − ∫0 log(1 + 𝑡𝑎𝑛𝑥 ) 𝑑𝑥 …. (2)
Adding (1) and (2) we get
𝜋
4
𝜋
4
𝜋
2 I = ∫0 𝑙𝑜𝑔2𝑑𝑥 = 𝑙𝑜𝑔2ሾ𝑥ሿ0 = 4 log 2
1
𝜋
Or I = 8 log 2
2−𝑥
𝑬𝒗𝒂𝒍𝒖𝒂𝒕𝒆: ∫−1 𝑙𝑜𝑔 ቚ2+𝑥ቚ 𝑑𝑥
1
2−𝑥
Solution: Let I = ∫−1 𝑙𝑜𝑔 ቚ2+𝑥ቚ 𝑑𝑥 … (1)
1
2−(1−1−𝑥)
𝑏
Or I= ∫−1 𝑙𝑜𝑔 ฬ2+(1−1−𝑥)ฬ 𝑑𝑥
1
2−(0−𝑥)
= ∫−1 𝑙𝑜𝑔 ฬ2+(0−𝑥)ฬ 𝑑𝑥
1
2+𝑥
…
Or I = ∫−1 𝑙𝑜𝑔 ቚ2−𝑥ቚ 𝑑𝑥
(2)
Adding (1) and (2) we get
1
2−𝑥
2+𝑥
𝑏
∵ ∫𝑎 𝑓(𝑥) 𝑑𝑥 = ∫𝑎 𝑓(𝑎 + 𝑏 − 𝑥) 𝑑𝑥
2I = ∫−1 𝑙𝑜𝑔 ቚ2+𝑥ቚ + 𝑙 𝑜𝑔 ቚ2−𝑥ቚ 𝑑𝑥
6 Rahman Sir
DEFINITE INTEGRAL
1
1
2I = ∫−1 𝑙𝑜𝑔1dx =∫−1 0 𝑑𝑥 = 0
Or I = 0
1+cos2 𝑥
𝐼1
𝐹𝑖𝑛𝑑
. 𝐼𝑓 𝐼1 = ∫sin2 𝑥
𝐼
2
1+cos2 𝑥
Solution: 𝐼1 = ∫sin2 𝑥
𝑥𝑓ሼ𝑥(2 − 𝑥)ሽ𝑑𝑥 and
1+cos2 𝑥
𝐼2 = ∫sin2 𝑥
𝑓ሼ𝑥(2 − 𝑥)ሽ𝑑𝑥
𝑥𝑓(2 − 𝑥)𝑑𝑥
Let F(x) = 𝑥𝑓ሼ𝑥(2 − 𝑥)ሽ
Now, 𝑎 + 𝑏 − 𝑥 = (1 + cos2 𝑥 + sin2 𝑥 − 𝑥) = (2 − 𝑥)
∴ 𝐹(𝑎 + 𝑏 − 𝑥) = (2 − 𝑥)𝑓ሼ(2 − 𝑥)𝑥ሽ
1+cos2 𝑥
Hence, ∫sin2 𝑥
1+cos2 𝑥
𝐼1 = ∫sin2 𝑥
1+cos2 𝑥
𝑥𝑓ሼ𝑥(2 − 𝑥)ሽ𝑑𝑥 = ∫sin2 𝑥
1+cos2 𝑥
2𝑓ሼ𝑥(2 − 𝑥)ሽ𝑑𝑥 − ∫sin2 𝑥
(𝑥 − 2)𝑓ሼ𝑥(2 − 𝑥)ሽ𝑑𝑥
𝑥𝑓ሼ𝑥(2 − 𝑥)ሽ𝑑𝑥
1+cos2 𝑥
𝑜𝑟 𝐼1 =
∫
2𝑓ሼ𝑥(2 − 𝑥)ሽ𝑑𝑥 − 𝐼1
sin2 𝑥
1+cos2 𝑥
Or 2𝐼1 = 2∫sin2 𝑥
𝑓ሼ𝑥(2 − 𝑥)ሽ𝑑𝑥
Or 2 𝐼1 = 2 𝐼2 ⇒ 𝐼1 = 𝐼2
𝐼
or 𝐼1 = 1
2
ൣ𝑥 2 ൧
10
Example: The Value of integral ∫4
(A) 6
ሾ𝑥 2 −28𝑥+196ሿ+ ሾ𝑥 2 ሿ
(B) 3
ൣ𝑥 2 ൧
10
Solution: Let I = ∫4
(C ) 7
ሾ𝑥 2 −28𝑥+196ሿ+ ሾ𝑥 2 ሿ
ൣ𝑥 2 ൧
10
Or I = ∫4
ሾ(𝑥−14)2 ሿ+ ሾ𝑥 2 ሿ
𝑑𝑥 is
(D)
1
3
𝑑𝑥
𝑑𝑥 … (1)
ൣ𝑥 2 ൧
Let 𝑓(𝑥) = ሾ(𝑥−14)2 ሿ+ ሾ𝑥 2 ሿ
Now, 𝑓(𝑎 + 𝑏 − 𝑥) = 𝑓(10 + 4 − 𝑥) = 𝑓(14 − 𝑥)
ൣ(14−𝑥)2 ൧
Therefore, 𝑓(14 − 𝑥) = ሾ(𝑥)2 ሿ+ ൣ(14−𝑥)2 ൧
10
Or I = ∫4
ൣ(14−𝑥)2 ൧
ሾ(𝑥)2 ሿ+ ൣ(14−𝑥)2 ൧
Adding (1) and (2) we get,
𝑏
𝑏
𝑑𝑥 … (2) using the property ∫𝑎 𝑓(𝑥) 𝑑𝑥 = ∫𝑎 𝑓(𝑎 + 𝑏 − 𝑥) 𝑑𝑥
7 Rahman Sir
DEFINITE INTEGRAL
10
2I = ∫4 1 𝑑𝑥 = ሾ𝑥ሿ10
4 = 10 − 4 = 6
6
Or I = 2 = 3
Hence, the correct option is (B)
𝜋
2 𝑐𝑜𝑠𝑥−𝑠𝑖𝑛𝑥𝑥
0 10−𝑥 2 +𝜋𝑥
2
Example: The Value of integral ∫
𝜋
(B) 𝜋
(A) 2
𝜋
2 𝑐𝑜𝑠𝑥−𝑠𝑖𝑛𝑥
0 10−𝑥 2 +𝜋𝑥
2
Solution: Let I = ∫
(C ) 0
=∫
= ∫0
𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥
𝜋2
𝜋𝑥 𝜋2
10−൬ −𝜋𝑥+𝑥 2 ൰− +
4
2
4
𝜋
2 𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥
0 10−𝑥 2 +𝜋𝑥
2
𝑑𝑥
𝑑𝑥
… (2)
𝑑𝑥
=∫
Or I
(D) 4𝜋
…(1)
𝑑𝑥
𝜋
𝜋
𝜋
cos( −𝑥)−sin( −𝑥)
2
2
2
𝜋
0
𝜋ቀ −𝑥ቁ
𝜋
10−( −𝑥)2 + 2
2
2
𝜋
2
𝑑𝑥 is
Adding (1) and (2) we get,
𝜋
2I = ∫02
0
10−𝑥 2 +
𝜋𝑥
2
𝑑𝑥 = 0
Or I = 0
Type III. Problems based on the property.
𝑎
(i)
∫−𝑎 𝑓(𝑥) 𝑑𝑥 = 0, if f(x) is odd function. i.e., 𝑓(−𝑥) = −𝑓(𝑥)
(ii)
∫−𝑎 𝑓(𝑥) 𝑑𝑥 = 2 ∫0 𝑓(𝑥) 𝑑𝑥 if f(x) is even function. i.e., 𝑓(−𝑥) = 𝑓(𝑥)
(iii)
∫−𝑎 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑥) 𝑑𝑥 + ∫0 𝑓(−𝑥) 𝑑𝑥 when f(x) is neither odd nor even function i.e.,
𝑎
𝑎
𝑎
𝑎
𝑎
some part is odd and some part is even
1
𝑬𝒗𝒂𝒍𝒖𝒂𝒕𝒆 ∫−1 sin5 𝑥 cos4 𝑥 𝑑𝑥
1
Solution: Let I = ∫−1 sin5 𝑥 cos 4 𝑥 𝑑𝑥. Let 𝑓(𝑥) = sin5 𝑥 𝑐𝑜𝑠 4 𝑥
𝑓(−𝑥) = sin5(−𝑥) cos 4 (−𝑥) = − sin5 𝑥 cos4 𝑥 = −𝑓(𝑥)
𝑎
𝐼 = 0 , ∵ ∫−𝑎 𝑓(𝑥) 𝑑𝑥 = 0, if f(x) is odd function. i.e., 𝑓(−𝑥) = −𝑓(𝑥)
𝑎
𝑎−𝑥
𝑎
𝑎−𝑥
𝑬𝒗𝒂𝒍𝒖𝒂𝒕𝒆 ∫−𝑎 ට𝑎+𝑥 dx
Let I = ∫−𝑎 ට𝑎+𝑥 dx
DEFINITE INTEGRAL
𝑎
𝑎−𝑥
𝑑𝑥
= ∫−𝑎 2
ඥ
𝑎 −𝑥2
𝑎
𝑎
𝑎
= ∫−𝑎 2
ඥ
𝑎 −𝑥 2
8 Rahman Sir
𝑑𝑥 − ∫−𝑎
−𝑥
ඥ𝑎2 −𝑥 2
𝑑𝑥
= 𝐼1 − 𝐼2
𝐼2 = 0 𝑎𝑠
𝑎
I= ∫−𝑎
−𝑥
𝑎
𝑑𝑥
ඥ𝑎2 −𝑥 2
𝑎
= 2 ∫0
is odd function
ඥ𝑎2 −𝑥 2
𝑎
ඥ𝑎2 −𝑥 2
𝑑𝑥 as
𝑎
ඥ𝑎2 −𝑥 2
is even function
𝑥 𝑎
𝜋
= 2a ቂsin−1 𝑎ቃ = 2𝑎 sin−1 1 = 2𝑎. 2 = 𝜋𝑎
0
𝜋
𝑑𝑥
Evaluate ∫ 2𝜋 1+𝑒 𝑠𝑖𝑛𝑥
−
2
This function is neither odd nor even
𝜋
𝑑𝑥
2
𝜋
𝑠𝑖𝑛𝑥
1+𝑒
−
2
Solution: Let I = ∫
𝜋
1
1
= ∫02 ቂ1+𝑒 𝑠𝑖𝑛𝑥 +
ቃ 𝑑𝑥
1+𝑒 sin(−𝑥)
𝑎
𝑎
𝑎
∫−𝑎 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑥) 𝑑𝑥 + ∫0 𝑓(−𝑥) 𝑑𝑥 when f(x) is neither odd nor even function i.e.,
some part is odd and some part is even
𝜋
2
1
= ∫0 ቂ1+𝑒 𝑠𝑖𝑛𝑥 +
𝜋
2
1
= ∫0 ൤1+𝑒 𝑠𝑖𝑛𝑥 +
𝜋
2
1
1+𝑒 −sin(𝑥)
𝑒 𝑠𝑖𝑛𝑥
1+𝑒 sin(𝑥)
𝜋
ቃ 𝑑𝑥
൨ 𝑑𝑥
𝜋
= ∫0 𝑑𝑥 = 2 − 0 = 2
𝜋
2 𝑐𝑜𝑠𝑥
𝜋
− 1+𝑒 𝑥
2
Evaluate ∫
𝑑𝑥
[CBSE(F)2015]
This function is neither odd nor even
𝜋
2 𝑐𝑜𝑠𝑥
𝜋
− 1+𝑒 𝑥
2
∴ ∫
𝑎
𝜋
2
𝑐𝑜𝑠𝑥
cos(−𝑥)
𝑑𝑥 = ∫0 ቂ1+𝑒 𝑥 + 1+𝑒 −𝑥 ቃ 𝑑𝑥
𝑎
𝑎
∫−𝑎 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑥) 𝑑𝑥 + ∫0 𝑓(−𝑥) 𝑑𝑥 when f(x) is neither odd nor even function i.e.,
some part is odd and some part is even
9 Rahman Sir
DEFINITE INTEGRAL
𝜋
2
𝑐𝑜𝑠𝑥
= ∫0 ቈ1+𝑒 𝑥 +
𝜋
𝑐𝑜𝑠𝑥
= ∫02 ቂ1+𝑒 𝑥 +
𝜋
= ∫02
cos(𝑥)
1
𝑒
1+ 𝑥
ex cos(𝑥)
1+𝑒 𝑥
(1+𝑒 𝑥 )𝑐𝑜𝑠𝑥
1+𝑒 𝑥
቉ 𝑑𝑥
ቃ 𝑑𝑥
𝑑𝑥
𝜋
𝜋
𝜋
= ∫02 𝑐𝑜𝑠𝑥 𝑑𝑥 = ሾ𝑠𝑖𝑛𝑥ሿ02 = sin ቀ2 ቁ − 𝑠𝑖𝑛0 = 1
𝜋
N.C.E.R.T. Example 36: Evaluate ∫02 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 𝑑𝑥
𝜋
Solution: Let I= ∫02 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 𝑑𝑥 … (1)
𝜋
𝜋
Or I = ∫02 logsin ቀ 2 − 𝑥ቁ 𝑑𝑥
𝜋
Or I = ∫02 𝑙𝑜𝑔𝑐𝑜𝑠𝑥 𝑑𝑥 … (2)
Adding (1) and (2)
𝜋
𝜋
2I = ∫02 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 𝑑𝑥 + ∫02 𝑙𝑜𝑔𝑐𝑜𝑠𝑥 𝑑𝑥
𝜋
Or 2I = ∫02 𝑙𝑜𝑔𝑠𝑖𝑛𝑥. 𝑐𝑜𝑠𝑥 𝑑𝑥 ∵ 𝑙𝑜𝑔𝐴 + 𝑙𝑜𝑔𝐵 = 𝑙𝑜𝑔𝐴𝐵
𝜋
2
Or 2I = ∫0 log
(2𝑠𝑖𝑛𝑥.𝑐𝑜𝑠𝑥)
2
𝑑𝑥
𝜋
𝜋
Or 2I = ∫02 𝑙𝑜𝑔𝑠𝑖𝑛2𝑥 𝑑𝑥 − 𝑙𝑜𝑔2 ∫02 𝑑𝑥
𝜋
2I = 𝐼1 − 2 𝑙𝑜𝑔2
𝜋
Now 𝐼1 = ∫02 𝑙𝑜𝑔𝑠𝑖𝑛2𝑥 𝑑𝑥
𝜋
Put 2x = t , then 2 dx = 𝑑𝑡 , 𝑤ℎ𝑒𝑛 𝑥 = 0 𝑡 = 0 𝑎𝑛𝑑 𝑤ℎ𝑒𝑛 𝑥 = 2 , 𝑡 = 𝜋
1
𝜋
𝐼1 = 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛𝑡 𝑑𝑡
2
𝜋
= 2 ∫02 𝑙𝑜𝑔𝑠𝑖𝑛𝑡 𝑑𝑡
𝜋
𝟐𝒂
𝜋
= ∫02 𝑙𝑜𝑔𝑠𝑖𝑛𝑡 𝑑𝑡 = ∫02 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 𝑑𝑥
𝜋
Hence, 2I = I− 2 𝑙𝑜𝑔2
𝜋
Or I = − 2 𝑙𝑜𝑔2
𝒂
since, ∫𝟎 𝒇(𝒙)𝒅𝒙 = 𝟐 ∫𝟎 𝒇(𝒙) 𝒅𝒙 𝒊𝒇 𝒇(𝟐𝒂 − 𝒙) = 𝒇(𝒙)
𝒃
𝒃
[ ∵ ∫𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒛)𝒅𝒛]
10 Rahman Sir
DEFINITE INTEGRAL
EXERCISE 7.11
By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.
1.
𝜋
2
∫0 cos2 𝑥 dx
𝜋
Let I = ∫02 cos2 𝑥 dx … (1)
Solution:
𝜋
𝜋
𝜋
Or I = ∫02 cos 2 ( 2 − 𝑥) dx = ∫02 sin2 𝑥 dx .. (2)
Adding (1) and (2) we get
𝜋
𝜋
2 I = ∫02 (cos2 𝑥 + sin2 𝑥) dx = ∫02 𝑑𝑥
𝜋
𝜋
𝜋
Or I = ሾ𝑥ሿ02 = ቀ2 − 0 ቁ = 2
2.
𝜋
2
ඥ𝑠𝑖𝑛𝑥
∫0 ඥ𝑠𝑖𝑛𝑥+
dx
ξ𝑐𝑜𝑠𝑥
𝜋
ඥ𝑠𝑖𝑛𝑥
Solution: Let I = ∫02
ඥ𝑠𝑖𝑛𝑥+
𝒂
dx … (1)
𝜋
𝜋
2
Or I = ∫0
ξ𝑐𝑜𝑠𝑥
ටsin( 2 −𝑥)
𝜋
dx
𝜋
ටsin( 2 −𝑥)+ ටcosቀ 2 − 𝑥ቁ
𝒄
𝝅
𝝅
∵ ∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒂 − 𝒙)𝒅𝒙 𝒂𝒏𝒅 𝒔𝒊𝒏 ቀ 𝟐 − 𝒙ቁ = 𝒄𝒐𝒔𝒙 and 𝒄𝒐𝒔 ቀ 𝟐 − 𝒙ቁ = 𝒔𝒊𝒏𝒙
𝜋
2
Or I = ∫0
ξ𝑐𝑜𝑠𝑥
dx
ඥ𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥+
ξ
…
(2)
Adding (1) and (2) we get
𝜋
𝜋
2I = ∫02 𝑑𝑥 = ቀ2 − 0ቁ
𝜋
Or I = 4
3.
𝜋
2
∫0
3
sin2 𝑥
3
dx
3
sin2 𝑥+cos2 𝑥
3
𝜋
2
Solution: Let I = ∫0
sin2 𝑥
3
3
𝜋
2
Or I = ∫0
𝒂
𝒄
3
… (1)
dx
sin2 𝑥+cos2 𝑥
𝜋
2
sin2 ( − 𝑥)
3
𝜋
2
3
𝜋
2
)
sin2 ቀ −𝑥ቁ+cos2 ( − 𝑥
𝝅
𝝅
∵ ∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒂 − 𝒙)𝒅𝒙 𝒂𝒏𝒅 𝒔𝒊𝒏 ቀ 𝟐 − 𝒙ቁ = 𝒄𝒐𝒔𝒙 and 𝒄𝒐𝒔 ቀ 𝟐 − 𝒙ቁ = 𝒔𝒊𝒏𝒙
11 Rahman Sir
DEFINITE INTEGRAL
3
𝜋
2
= ∫0
Or I
cos2 𝑥
3
…. (2)
3
cos2 𝑥+sin2 𝑥
Adding (1) and (2) we get
𝜋
2
𝜋
2I = ∫0 𝑑𝑥 = ቀ2 − 0ቁ
𝜋
Or I = 4
𝜋
2
4.
cos5 𝑥
∫0 sin5 𝑥 +cos5 𝑥 dx
𝜋
2
cos5 𝑥
… (1)
Let I = ∫0 sin5 𝑥 +cos5 𝑥 dx
Solution:
𝜋
Or I = ∫02
𝒂
𝜋
2
cos5 ( − 𝑥)
𝜋
2
𝜋
2
sin5 ( − 𝑥 )+cos5 ( − 𝑥)
𝑑𝑥
𝒄
𝝅
𝝅
∵ ∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒂 − 𝒙)𝒅𝒙 𝒂𝒏𝒅 𝒔𝒊𝒏 ቀ 𝟐 − 𝒙ቁ = 𝒄𝒐𝒔𝒙 and 𝒄𝒐𝒔 ቀ 𝟐 − 𝒙ቁ = 𝒔𝒊𝒏𝒙
𝜋
sin5 𝑥
… (2)
Or I = ∫02 cos5 𝑥 +sin5 𝑥 dx
Adding (1) and (2) we get
𝜋
2
𝜋
2I = ∫0 𝑑𝑥 = ቀ − 0ቁ
2
𝜋
Or I = 4
5.
5
∫−5ȁ𝑥 + 2ȁ dx
Solution:
put 𝑥 + 2 = 0 , ⇒ 𝑥 = −2
−2
5
I= ∫−5 −(𝑥 + 2) 𝑑𝑥 + ∫−2(𝑥 + 2)𝑑𝑥
−2
𝑥2
= − ቂ 2 + 2𝑥ቃ
−5
5
𝑥2
+ ቂ 2 + 2𝑥ቃ
−2
25
25
= ቂ−(2 − 4) + ቀ 2 − 10ቁቃ + ቂቀ 2 + 10ቁ − (2 − 4)ቃ
25
25
= ቀ 2 − 8ቁ + ቀ12 + 2 ቁ = 29
8
∫2 ȁ𝑥 − 5ȁ dx
Solution: let 𝑥 − 5 = 0 , 𝑜𝑟 𝑥 = 5
It can be seen that (x − 5) ≤ 0 on [2, 5] and (x − 5) ≥ 0 on [5, 8].
6.
5
8
I= ∫2 −(𝑥 − 5) 𝑑𝑥 + ∫5 (𝑥 −)𝑑𝑥
5
𝑥2
𝑥2
= − ቂ 2 − 5𝑥ቃ + ቂ 2 − 5𝑥ቃ
2
25
8
5
25
= − ቂቀ 2 − 25ቁ − (2 − 10)ቃ + ቂ(32 − 40) − ቀ 2 − 25ቁቃ
12 Rahman Sir
DEFINITE INTEGRAL
25
25
= − 2 + 25 − 8 −8 − 2 + 25 = −25 + 25 + 25 − 16 = 9
7.
1
∫0 𝑥(1 − 𝑥)𝑛 dx
Solution:
1
∫0 𝑥(1 − 𝑥)𝑛 dx
Let I =
1
𝒂
I= ∫0 (1 − 𝑥)(𝑥)𝑛 dx
𝒂
∵ ∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒂 − 𝒙)𝒅𝒙
1
= ∫0 (𝑥 𝑛 − 𝑥 𝑛+1 )𝑑𝑥
𝑥 𝑛+1
1
𝑥 𝑛+2
= ቂ 𝑛+1 − 𝑛+2 ቃ
1
0
(𝑛+2)−(𝑛+1)
1
= ቂ𝑛+1 − 𝑛+2ቃ = (𝑛+1)(𝑛+2)
1
= (𝑛+1)(𝑛+2)
8.
𝜋
4
∫0 log(1 + 𝑡𝑎𝑛𝑥) dx
𝜋
4
Solution: Let I = ∫0 log(1 + 𝑡𝑎𝑛𝑥) dx … (1)
𝜋
4
𝜋
= ∫0 log ቀ1 + tan ‫ ۃ‬− 𝑥‫ۄ‬ቁ dx
4
𝜋
4
𝜋
4
tan −𝑡𝑎𝑛𝑥
=∫0 log ቈ1 + ቆ
ቇ቉ dx
𝜋
1+tan 𝑡𝑎𝑛𝑥
4
𝜋
4
1−𝑡𝑎𝑛𝑥
= ∫0 log ቂ1 + 1+𝑡𝑎𝑛𝑥ቃ 𝑑𝑥
𝜋
= ∫04 log ቂ
1+𝑡𝑎𝑛𝑥+1−𝑡𝑎𝑛𝑥
𝜋
1+𝑡𝑎𝑛𝑥
ቃ 𝑑𝑥
2
= ∫04 log ቂ1+𝑡𝑎𝑛𝑥ቃ 𝑑𝑥
𝜋
𝜋
Or I = log2 ∫04 𝑑𝑥 − ∫04 log(1 + 𝑡𝑎𝑛𝑥) dx … (2)
Adding (1) and (2) we get,
𝜋
𝜋
2I = log2 ∫04 𝑑𝑥 = 4 𝑙𝑜𝑔2
𝜋
Or I = 8 𝑙𝑜𝑔2
9.
2
∫0 𝑥ඥ2 − 𝑥 dx
Solution:
2
Let I = ∫0 𝑥ඥ2 − 𝑥 dx
DEFINITE INTEGRAL
2
= ∫0 (2 − 𝑥)ඥ2 − (2 − 𝑥) dx
2
= ∫0 (2 − 𝑥)ξ𝑥
2
3
2
= 2∫0 ξ𝑥 𝑑𝑥 − ∫0 𝑥 2 𝑑𝑥
3
4
5
2
= ൤3 𝑥 2 − 5 𝑥 2 ൨
4
2
0
2
= ቂ3 × 2ඥ2 − 5 × 4 × ඥ2ቃ
8ξ 2
8ξ2
= 3 − 5 =
10.
40ξ2−24ξ2
16ξ2
=
15
15
𝜋
2
∫0 (2𝑙𝑜𝑔𝑠𝑖𝑛𝑥 − 𝑙𝑜𝑔𝑠𝑖𝑛2𝑥) dx
𝜋
2
Let I = ∫0 (2𝑙𝑜𝑔𝑠𝑖𝑛𝑥 − 𝑙𝑜𝑔𝑠𝑖𝑛2𝑥) dx
Solution:
𝜋
2
𝜋
𝜋
I = ∫0 (2𝑙𝑜𝑔𝑠𝑖𝑛( 2 − 𝑥) − 𝑙𝑜𝑔𝑠𝑖𝑛2 ቄ2 − 𝑥ቅ) dx
Or
𝒂
𝒂
∵ ∫ 𝒇(𝒙)𝒅𝒙 = ∫ 𝒇(𝒂 − 𝒙)𝒅𝒙
𝟎
𝟎
𝜋
2
… (2)
𝐼 = ∫0 (2𝑙𝑜𝑔𝑐𝑜𝑠𝑥 − 𝑙𝑜𝑔𝑠𝑖𝑛2𝑥)
𝑎𝑑𝑑𝑖𝑛𝑔 (1)𝑎𝑛𝑑 (2) we get
𝜋
2
2𝐼 = ∫0 (2𝑙𝑜𝑔𝑠𝑖𝑛𝑥 + 2𝑙𝑜𝑔𝑐𝑜𝑠𝑥 − 2𝑙𝑜𝑔𝑠𝑖𝑛2𝑥)
𝜋
2
𝜋
2
2𝐼 = 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥𝑑𝑥 − 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛2𝑥 𝑑𝑥
𝜋
2
= 2 ∫0 log
(2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥𝑑𝑥)
2
𝜋
2
− 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛2𝑥 𝑑𝑥
𝜋
2
𝜋
2
𝜋
2
= 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛2𝑥 𝑑𝑥 − 2 𝑙𝑜𝑔2 ∫0 𝑑𝑥 − 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛2𝑥 𝑑𝑥
𝜋
2
= −2 𝑙𝑜𝑔2 ∫0 𝑑𝑥
𝜋
1
= −2 ቀ2 ቁ 𝑙𝑜𝑔2 = −𝜋𝑙𝑜𝑔2 = 𝜋 log 2
𝜋
1
𝑜𝑟 𝐼 = 2 log 2
𝜋
2
𝜋
−
2
11. ∫ sin2 𝑥 dx
𝜋
Solution: Let
As sin
2 (−𝑥)
I= ∫ 2𝜋 sin2 𝑥 dx
−
2
= sin2 𝑥 , 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, sin2 𝑥 is an even function
13 Rahman Sir
14 Rahman Sir
DEFINITE INTEGRAL
𝒂
𝒂
It is known that if f(x) is an even function, then ∫−𝒂 𝒇(𝒙)𝒅𝒙 = 2∫𝟎 𝒇(𝒙)𝒅𝒙
𝜋
∴ 𝐼 = 2 ∫02 sin2 𝑥 𝑑𝑥
𝜋
1
= ∫02 (1 − 𝑐𝑜𝑠2𝑥) 𝑑𝑥 = ቂ𝑥 − 2 𝑠𝑖𝑛2𝑥ቃ
𝜋
𝜋
2
0
𝜋
= ቂ 2 − 0 − (0 − 0)ቃ = 2
𝜋
𝑥
12. ∫0 1+𝑠𝑖𝑛𝑥 dx
𝜋
𝑥
… (1)
I = ∫0 1+𝑠𝑖𝑛𝑥 dx
Solution: Let
𝜋
(𝜋−𝑥)
I = ∫0 1+sin(𝜋−𝑥) dx
𝜋 (𝜋−𝑥
=∫0 1+𝑠𝑖𝑛𝑥) dx
Or I
… (2)
Adding (1) and (2) we get,
𝜋
1
2I = 𝜋 ∫0 1+𝑠𝑖𝑛𝑥 dx
𝜋
1−𝑠𝑖𝑛𝑥
2I = 𝜋 ∫0 (1+𝑠𝑖𝑛𝑥)(1−𝑠𝑖𝑛𝑥) dx
𝜋
1−𝑠𝑖𝑛𝑥
= 𝜋 ∫0 (1−sin2 𝑥 ) dx
𝜋 1−𝑠𝑖𝑛𝑥
= 𝜋 ∫0 cos2 𝑥 dx
𝜋
= 𝜋 ∫0 (sec 2 𝑥 − 𝑡𝑎𝑛𝑥𝑠𝑒𝑐𝑥) 𝑑𝑥
= 𝜋 ሾ𝑡𝑎𝑛𝑥 − 𝑠𝑒𝑐𝑥ሿ𝜋0
= 𝜋ሾ𝑡𝑎𝑛𝜋 − 𝑠𝑒𝑐𝜋ሿ − 𝜋ሾ𝑡𝑎𝑛0 − 𝑠𝑒𝑐0ሿ
= 𝜋(0 + 1) − 𝜋(0 − 1) = 2𝜋
𝑜𝑟 𝐼 = 𝜋
𝜋
13. ∫−2𝜋 sin7 𝑥 dx
2
Solution:
Let
𝜋
2
𝜋
−
2
I = ∫ sin7 𝑥 dx
As sin7 (−𝑥) = (− sin 𝑥 )7 = − sin7 𝑥 , therefore, sin7 𝑥 is an odd function.
𝒂
It is known that if f(x) is an odd function, then ∫−𝒂 𝒇(𝒙)𝒅𝒙 =0
15 Rahman Sir
DEFINITE INTEGRAL
𝜋
2
𝜋
−
2
∴ I = ∫ sin7 𝑥 dx = 0
2𝜋
14. ∫0 cos5 𝑥 dx
Solution:
Let
2𝜋
I = ∫0 cos 5 𝑥 dx
… (1)
𝑐𝑜𝑠 5 (2𝜋 − 𝑥) = cos5 𝑥
𝟐𝒂
𝒂
It is known that ∫𝟎 𝒇(𝒙)𝒅𝒙 = 𝟐 ∫𝟎 𝒇(𝒙) 𝒅𝒙 𝒊𝒇 𝒇(𝟐𝒂 − 𝒙) = 𝒇(𝒙)
𝜋
Or I = 2 ∫0 cos5 𝑥 dx
𝒂
⇒ 𝐼 = 2(0) ∵ cos(𝜋 − 𝑥) = −𝑐𝑜𝑠𝑥 and ∫𝟎 𝒇(𝒙) 𝒅𝒙 = 𝟎, 𝒊𝒇 𝒇(𝒂 − 𝒙) = −𝒇(𝒙
𝜋
𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥
15. ∫02 1+𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 dx
𝜋
Solution: :
Let
𝒂
𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥
I = ∫02 1+𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
… (1)
𝒂
∵ ∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒂 − 𝒙)𝒅𝒙
𝜋
𝐼 = ∫02
𝜋
2
𝜋
2
𝜋
𝜋
1+𝑠𝑖𝑛( −𝑥)𝑐𝑜𝑠( −𝑥)
2
2
sin( −𝑥)−𝑐𝑜𝑠( −𝑥)
𝜋
2 𝑐𝑜𝑠𝑥−𝑠𝑖𝑛𝑥
… (2)
𝐼 = ∫0 1+𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
Adding (1) and (2) we get,
2I = 0
Or I = 0
𝜋
16. ∫0 log(1 + 𝑐𝑜𝑠𝑥) dx
𝜋
let I =∫0 log(1 + 𝑐𝑜𝑠𝑥)
Solution:
Or I =
𝜋
∫0 log(1 + cos(𝜋 − 𝑥))
… (1)
𝒂
𝜋
= ∫0 log(1 − 𝑐𝑜𝑠𝑥) dx … (2)
∵ cos(𝜋 − 𝑥) = −𝑐𝑜𝑠𝑥
Adding (1) and (2) we get
𝜋
2𝐼 = ∫0 log(1 − cos2 𝑥) dx
𝜋
𝜋
Or 2I = ∫0 𝑙𝑜𝑔𝑠𝑖𝑛2 𝑥𝑑𝑥 = 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 𝑑𝑥
𝜋
Or I = ∫0 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 𝑑𝑥
… (3)
𝜋
2
Or I = 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 𝑑𝑥 as ∵ sin(𝜋 − 𝑥) = 𝑠𝑖𝑛𝑥
𝟐𝒂
𝒂
∫ 𝒇(𝒙)𝒅𝒙 = 𝟐 ∫ 𝒇(𝒙) 𝒅𝒙 𝒊𝒇 𝒇(𝟐𝒂 − 𝒙) = 𝒇(𝒙)
𝟎
𝟎
𝜋
2
Or I = 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 𝑑𝑥
𝒂
∵ ∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒂 − 𝒙)𝒅𝒙
… (4)
16 Rahman Sir
DEFINITE INTEGRAL
𝜋
2
𝒂
𝜋
Or I = 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛( 2 − 𝑥) 𝑑𝑥
𝜋
2
Or I = 2 ∫0 𝑙𝑜𝑔𝑐𝑜𝑠𝑥 𝑑𝑥
𝒂
∵ ∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒂 − 𝒙)𝒅𝒙
… (5)
Adding (4) and (5) we get,
𝜋
2
2I = 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥
𝜋
2
1
Or 2I =2 ∫0 log ( 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 ) 2 𝑑𝑥
𝜋
2
𝜋
2
2I = 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛2𝑥 𝑑𝑥 − 2 𝑙𝑜𝑔2 ∫0 𝑑𝑥
2I = 𝐼1 − 𝜋𝑙𝑜𝑔2
𝜋
2
𝐼1 =2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛2𝑥 𝑑𝑥
let 2𝑥 = 𝑡 𝑡ℎ𝑒𝑛 2𝑑𝑥 = 𝑑𝑡 ,
𝑤ℎ𝑒𝑛 𝑥 = 0 𝑡 = 0, 𝑤ℎ𝑒𝑛 𝑥 =
𝜋
2
𝜋
𝑡ℎ𝑒𝑛 𝑡 = 𝜋
2
𝜋
𝐼1 = 2 ∫0 𝑙𝑜𝑔𝑠𝑖𝑛𝑡 𝑑𝑡 = ∫0 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 𝑑𝑥
𝑏
= 𝐼 𝑓𝑟𝑜𝑚 (3)
or
2I =𝐼 − 𝜋 𝑙𝑜𝑔2
Or I = −𝜋 log 2
𝑎
𝑥
17. ∫0 𝑥+ξ 𝑎−𝑥 dx
ξ
ξ
𝑎
let I = ∫0
Solution:
𝑎
I = ∫0
ξ𝑥
dx
ξ𝑥+ ξ𝑎−𝑥
ξ𝑎−𝑥
ξ𝑎−𝑥+ ඥ𝑎−(𝑎−𝑥)
𝑎
Or I = ∫0
Adding (1) and (2) we get,
𝑎
2I = ∫0 𝑑𝑥 = ሾ𝑥ሿ𝑎0 = 𝑎
𝑎
Or I = 2
4
18. ∫0 ȁ𝑥 − 1ȁ dx
…. (1)
dx
ξ𝑎−𝑥
dx
ξ𝑎−𝑥+ ξ𝑥
𝑏
as ∫𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓൫𝑦൯𝑑𝑦
… (2)
17 Rahman Sir
DEFINITE INTEGRAL
4
Let I = ∫0 ȁ𝑥 − 1ȁ dx
Solution:
Put 𝑥 − 1 = 0 𝑡ℎ𝑒𝑛 𝑥 = 1
It can be seen that, (x − 1) ≤ 0 when 0 ≤ x ≤ 1 and (x − 1) ≥ 0 when 1 ≤ x ≤ 4
1
4
∴ 𝐼 = ∫0 (1 − 𝑥) 𝑑𝑥 + ∫1 (𝑥 − 1) 𝑑𝑥
𝑥2
1
𝑥2
4
= ቂ𝑥 − 2 ቃ + ቂ 2 − 𝑥ቃ
0
1
1
1
= ቂቀ1 − 2ቁ − (0)ቃ + ቂ(8 − 4) − ቀ2 − 1ቁቃ
1
1
=2 +4+2=5
𝑎
𝑎
19. ∫0 𝑓(𝑥) 𝑔(𝑥) dx = 2 ∫0 𝑓(𝑥) 𝑑𝑥 , 𝑖𝑓 𝑓 𝑎𝑛𝑑 𝑔 𝑎𝑟𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑎𝑠 𝑓(𝑥) = 𝑓(𝑎 − 𝑥) and
𝑔(𝑥) + 𝑔(𝑎 − 𝑥) = 4
𝑎
… (1)
Let I= ∫0 𝑓(𝑥) 𝑔(𝑥) dx
Solution:
𝒂
𝒂
∵ ∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒂 − 𝒙)𝒅𝒙
𝑎
𝑎
𝑜𝑟 𝐼 = ∫0 𝑓(𝑎 − 𝑥) 𝑔(𝑎 − 𝑥)dx = ∫0 𝑓(𝑥) 𝑔(𝑎 − 𝑥) dx
𝑎
𝑜𝑟 𝐼 = ∫0 𝑓(𝑥) 𝑔(𝑎 − 𝑥) dx
𝐴𝑑𝑑𝑖𝑛𝑔 (1)𝑎𝑛𝑑 (2) we get,
∵ 𝑓(𝑥) = 𝑓(𝑎 − 𝑥)
… (2)
𝑎
𝑎
2𝐼 = ∫0 ൣ𝑓(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔(𝑎 − 𝑥)൧ dx = ∫0 ൣ𝑔(𝑥) + 𝑔(𝑎 − 𝑥)൧ 𝑓(𝑥)𝑑𝑥
𝑎
2I = ∫0 4 𝑓(𝑥)𝑑𝑥
∵ 𝑔(𝑥) + 𝑔(𝑎 − 𝑥) = 4
𝑎
Or I = 2 ∫0 𝑓(𝑥) 𝑑𝑥
Choose the correct answer in Exercises 20 and 21.
𝜋
2
𝜋
−
2
20. ∫ (𝑥 3 + 𝑥𝑐𝑜𝑠𝑥 + tan5 𝑥 + 1) dx
(A) 0
(C ) 𝜋
(B) 2
𝜋
2
𝜋
−
2
Solution: ∫ (𝑥 3 + 𝑥𝑐𝑜𝑠𝑥 + tan5 𝑥 + 1) dx
𝜋
2
𝜋
−
2
3
5
𝜋
2
𝜋
−
2
= ∫ (𝑥 + 𝑥𝑐𝑜𝑠𝑥 + tan 𝑥) 𝑑𝑥 + ∫ (1)dx
Let f(x) = 𝑥 3 + 𝑥𝑐𝑜𝑠𝑥 + tan5 𝑥.
F(-x)= (−𝑥)3 + (−𝑥) cos(−𝑥) + tan5(−𝑥)
= −𝑥 3 − 𝑥𝑐𝑜𝑠𝑥 − tan5 𝑥 = −(𝑥 3 + 𝑥𝑐𝑜𝑠𝑥 + tan5 𝑥) = −𝑓(𝑥)
∴ 𝑓(𝑥)𝑖𝑠 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
(D) 1
18 Rahman Sir
DEFINITE INTEGRAL
𝜋
2
𝜋
−
2
𝑜𝑟 ∫ (𝑥 3 + 𝑥𝑐𝑜𝑠𝑥 + tan5 𝑥) dx = 0
𝜋
2
𝜋
−
2
𝜋
𝜋
𝐻𝑒𝑛𝑐𝑒, 𝐼 = ∫ (1)dx = ቂ 2 ቃ − (− 2 ) = 𝜋
Hence, the correct option is (C)
𝜋
4+3𝑠𝑖𝑛𝑥
21. ∫02 𝑙𝑜𝑔 ቚ4+3𝑐𝑜𝑠𝑥ቚ dx
3
(A) 2
Solution:
(B) 4
𝜋
2
(D) −2
(C ) 0
4+3𝑠𝑖𝑛𝑥
Let I = ∫0 𝑙𝑜𝑔 ቚ4+3𝑐𝑜𝑠𝑥ቚ dx … (1)
𝜋
2
Or I = ∫0 𝑙𝑜𝑔 ቤ
𝜋
2
𝜋
2
𝜋
4+3cos( −𝑥)
2
4+3sin( −𝑥)
4+3𝑐𝑜𝑠𝑥
Or I = ∫0 𝑙𝑜𝑔 ቚ4+3𝑠𝑖𝑛𝑥 ቚ dx
ቤ dx
…. (2)
Adding (1) and (2) we get,
𝜋
2I = ∫02 𝑙𝑜𝑔1 𝑑𝑥 = 0
Or, I = 0
Hence, the correct option is (C)
𝒂
𝒂
∵ ∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒂 − 𝒙)𝒅𝒙
DEFINITE INTEGRAL
19 Rahman Sir