Welcome to

Definite Integration
 Table of contents

Session 01 03 Session 07 135

Definite Integration 04 Triangle Inequality 158
Geometrical Meaning of 05 Domain and Range of a Function 161
Definite Integration

Session 08 166
Session 02 28
Numericals based on substitution 29 Problems based on odd and 170
even functions

Session 03 47
Session 09 194
Walli’s formula 59
Gamma function 66 Queen’s property 204

Session 04 69 Session 10 215

Properties of definite integration 70

Some important results 216

Session 05 89
Session 11 242
Some important results 98
Leibnitz Theorem 251
DI of fractional part function 99
Definite Integral as a Limit of 263
sum
Session 06 108
King’s property 119
 Session 1
Introduction and
Geometrical meaning

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Key Takeaways

Definite Integration

If ! 𝑓 𝑥 𝑑𝑥 = 𝑔 𝑥 + 𝑐 , then

, 𝑏
! 𝑓 𝑥 𝑑𝑥 = 𝑔 𝑥 + 𝑐 +
+

= 𝑔 𝑏 +𝑐 − 𝑔 𝑎 +𝑐

=𝑔 𝑏 −𝑔 𝑎

No ‘c’ is available hence integration is definite.

 Geometrical Meaning of Definite Integration

𝑌 𝑥=𝑎 𝑥=𝑏
,
Above
𝑥 − 𝑎𝑥𝑖𝑠 + 𝑣𝑒 ! 𝑓 𝑥 𝑑𝑥
+

= −2 + 5 − 3 + 4

5 =4

𝑋
𝑎 2 𝑏
Below 3
𝑥 − 𝑎𝑥𝑖𝑠 − 𝑣𝑒

Definite Integration represents algebraic sum of area between 𝑥 = 𝑎 & 𝑥 = 𝑏

 Solve the following:
- 0- 0- 0-
. . . .
𝑖) ! sin 𝑥 𝑑𝑥 𝑖𝑖) ! sin 𝑥 𝑑𝑥 𝑖𝑖𝑖) ! sin 𝑥 𝑑𝑥 𝑖𝑣) ! sin 𝑥 𝑑𝑥
/ / / /
-
. -
.
𝑖) ! sin 𝑥 𝑑𝑥 = − cos 𝑥
0
/
𝜋
1 = − cos − (− cos 0)
2
- = 0 − −1 =1
.

0-
.
𝑖𝑖𝑖) ! sin 𝑥 𝑑𝑥
/ 1 1 1
𝜋 2𝜋 5𝜋
=1+1−1−1+1 1 1 2
=1
 Solve the following:
- - 0- 0- 0-
. . . . .
𝑖) ! sin 𝑥 𝑑𝑥 𝑖𝑖) ! cos 𝑥 𝑑𝑥 𝑖𝑖𝑖) ! sin 𝑥 𝑑𝑥 𝑖𝑣) ! sin 𝑥 𝑑𝑥 𝑣) ! sin 𝑥 𝑑𝑥
/ / / / /

0-
. 0-
.
𝑖𝑣) ! sin 𝑥 𝑑𝑥 =1+1+1+1+1 =5 = 1 =1
𝑣) ! sin 𝑥 𝑑𝑥
/
/

1 1 1 1 1
𝜋 5𝜋
2𝜋
2
 , , ,
! 𝑓 𝑥 𝑑𝑥 ≤ ! 𝑓 𝑥 𝑑𝑥 ≤ ! 𝑓 𝑥 𝑑𝑥
+ + +

,
We can integrate ∫+ 𝑓 𝑥 𝑑𝑥 only when 𝑓 𝑥 is defined in 𝑥 ∈ 𝑎, 𝑏

Example:
2-
1
-
𝑖) ! tan 𝑥 𝑑𝑥 = Not possible as at 𝑥 = , tan 𝑥 is not defined
.
-
1

So we integrate such function as

2- - 2-
3
1 . 1
! tan 𝑥 𝑑𝑥 = ! tan 𝑥 𝑑𝑥 + ! tan 𝑥 𝑑𝑥
- - -
4
1 1 .
 ,
! 𝑓 𝑥 𝑑𝑥 = 𝑔 𝑏 − 𝑔 𝑎 𝑓 𝑥 is continuous in 𝑎, 𝑏
+

If 𝑓 𝑥 is discontinuous at 𝑥 = 𝑐 in 𝑎, 𝑏 , then

, 5! ,
! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑥 𝑑𝑥 + ! 𝑓 𝑥 𝑑𝑥
+ + 5"
 How to Solve Questions having discontinuity in Limits

Solution:
6
𝑑 1
𝐼=! cot 36 𝜋
36 𝑑𝑥 𝑥

𝑖 . 𝑥 = 0 discontinuous -
.
,
𝑑 ,
𝑖𝑖 . ! 𝑓 𝑥 𝑑𝑥 = 𝑓(𝑥) G
+ 𝑑𝑥 +
0
1 /! 1 6 -
= cot 36
G + cot 36 G " −.
𝑥 36 𝑥 /
1 −𝜋
36
= lim (cot − cot 36 −1 ) +
7→/ −ℎ
6
lim (cot 36 1 − cot 36 ( ))
7→/ 7

2- - 2- - -
= (cot 36 −∞ − 1
) + ( 1 − cot 36 (∞)) = 𝜋 − 1
+1−0= .
Key Takeaways

If 𝑔(𝑥) is inverse of 𝑓(𝑥) i.e. 𝑔(𝑓 𝑥 ) = 𝑥 and domain of 𝑓 𝑥 is 𝑥 ∈ 𝑎, 𝑏

& 𝑓 𝑎 = 𝑐 and 𝑓 𝑏 = 𝑑

, 9
then
! 𝑓 𝑥 𝑑𝑥 + ! 𝑔 𝑦 𝑑𝑦 = 𝑏𝑑 − 𝑎𝑐
+ 5
Key Takeaways

𝑓 𝑎 = 𝑐 and 𝑓 𝑏 = 𝑑 𝑔 𝑦 must be inverse function of 𝑓 𝑥 , 𝑔 𝑦 = 𝑓 36 𝑥

, 9
! 𝑓 𝑥 𝑑𝑥 + ! 𝑔 𝑦 𝑑𝑦 = 𝑏𝑑 − 𝑎𝑐
+ 5

, 9
Proof: ∫+ 𝑓 𝑥 𝑑𝑥 + ∫5 𝑔 𝑦 𝑑𝑦 = 𝐼 (say)

𝑓 𝑥 = 𝑔36 𝑥 ⇒ 𝑔 𝑓 𝑥 =𝑥 ⇒𝑔 𝑦 =𝑥

, 9
∴ 𝐼 = ! 𝑓 𝑥 𝑑𝑥 + ! 𝑔 𝑦 𝑑𝑦
+ 5

, 9
= ! 𝑓 𝑥 𝑑𝑥 + ! 𝑔 𝑓 𝑥 𝑑 𝑓 𝑥
+ 5

, , ∵𝑓 𝑥 =𝑑 ⇒𝑥=𝑏
= ! 𝑓 𝑥 𝑑𝑥 + ! 𝑥𝑓 : 𝑥 𝑑𝑥
+ +
∵𝑓 𝑥 =𝑐 ⇒𝑥=𝑎
Key Takeaways

𝑓 𝑎 = 𝑐 and 𝑓 𝑏 = 𝑑 𝑔 𝑦 must be inverse function of 𝑓 𝑥 , 𝑔 𝑦 = 𝑓 36 𝑥

, 9
! 𝑓 𝑥 𝑑𝑥 + ! 𝑔 𝑦 𝑑𝑦 = 𝑏𝑑 − 𝑎𝑐
+ 5

, 9
Proof: ∫+ 𝑓 𝑥 𝑑𝑥 + ∫5 𝑔 𝑦 𝑑𝑦 = 𝐼 (say)

, , ∵𝑓 𝑥 =𝑑 ⇒𝑥=𝑏
∴ 𝐼 = ! 𝑓 𝑥 𝑑𝑥 + ! 𝑥𝑓 : 𝑥 𝑑𝑥
+ +
∵𝑓 𝑥 =𝑐 ⇒𝑥=𝑎
,
𝐼 = ! {𝑓 𝑥 + 𝑥𝑓′ 𝑥 }𝑑𝑥
+
,
= 𝑥𝑓 𝑥 +

= 𝑏. 𝑓 𝑏 − 𝑎. 𝑓 𝑎

= 𝑏𝑑 − 𝑎𝑐
 . <#
Find the value of ! 𝑒 ; 𝑑𝑥 + ! ln 𝑥 𝑑𝑥
6 <

A 𝑒 . − 2𝑒 B 2 − 𝑒2

C 2𝑒 . − 𝑒 D 𝑒 − 2𝑒 .
 . <#
Find the value of ! 𝑒 ; 𝑑𝑥 + ! ln 𝑥 𝑑𝑥
6 <

Solution:

. <#
;
! 𝑒 𝑑𝑥 + ! ln 𝑥 𝑑𝑥
6 <

. <#
𝐼 = ! 𝑒 ; 𝑑𝑥 + ! ln 𝑥 𝑑𝑥
6 <

𝑓 𝑥 𝑔 𝑦

(ln 𝑥 is inverse of 𝑒 ; )

𝐼 = 2. 𝑒 . − 1×𝑒
 . <#
Find the value of ! 𝑒 ; 𝑑𝑥 + ! ln 𝑥 𝑑𝑥
6 <

A 𝑒 . − 2𝑒 B 2 − 𝑒2

C 2𝑒 . − 𝑒 D 𝑒 − 2𝑒 .
Key Takeaways

,
If ! 𝑓 𝑥 𝑑𝑥 = 0 then 𝑎 = 𝑏 .
+

,
If ∫+ 𝑓 𝑥 𝑑𝑥 = 0 and a ≠ 𝑏, then 𝐴6 −𝐴. = 0

root

𝐴6

𝑎 𝐴. 𝑏

,
If ∫+ 𝑓 𝑥 𝑑𝑥 = 0 then for a continuous function 𝑓 𝑥 between
𝑥 = 𝑎 & 𝑥 = 𝑏 there exist atleast one root in (𝑎, 𝑏)
Key Takeaways

NOTE: Root theory is possible only when function is continuous in 𝑎, 𝑏

Root 3 No Root
+𝑣𝑒 Area
−𝑣𝑒 Area
−3
If 2𝑎 + 3𝑏 + 6𝑐 = 0 then prove that 𝑎𝑥 ! + 𝑏𝑥 + 𝑐 = 0 has root in (0,1).

Solution: Let 𝑓(𝑥)= 𝑎𝑥 . + 𝑏𝑥 + 𝑐

6
𝐼 = ! 𝑓 𝑥 𝑑𝑥
/
6
Let 𝐼 = ! 𝑎𝑥 . + 𝑏𝑥 + 𝑐 𝑑𝑥 (I = 0 ⇒ Continuous in (0,1))
/
Polynomial

+; $ ,; # 6 + ,
= 2
+ .
+ 𝑐𝑥 /
= 2
+.+𝑐 − 0

.+42,4=5 / 6
= =
= = = 0 ⇒ ∫/ 𝑓 𝑥 𝑑𝑥 = 0

So, this function has atleast one root in (0,1)

 Key Takeaways

If 𝑓 𝑥 > 0 in 𝑥 ∈ (𝑎, 𝑏) Then, If 𝑓 𝑥 < 0, 𝑥 ∈ (𝑎, 𝑏) Then,

, ,
𝑦 = 𝑓 𝑥 > 0 ⇒ ! 𝑓 𝑥 𝑑𝑥 > 0 𝑦 = 𝑓 𝑥 < 0 ⇒ ! 𝑓 𝑥 𝑑𝑥 < 0
+ +
𝑌
𝑌

𝑎 𝑏
𝑋
−𝑣𝑒

𝑎 𝑋
𝑏

If 𝑓 𝑥 > 0 If 𝑓 𝑥 < 0

Then function is above 𝑥 − axis then Then function is below 𝑥 − axis

area will be positive. then area will be negative.
Key Takeaways

But the converse is not true

2-
.
Example: ! sin 𝑥 𝑑𝑥 = 1 + 1 − 1 = 1
/
2- 1 1
.
! 𝑓(𝑥) 𝑑𝑥 = + 𝑣𝑒 𝜋 𝜋 3𝜋
/ 2 −1 2
2-
But 𝑓 𝑥 > 0 in 0, .
is not true

Function is not positive overall

 Key Takeaways

,
𝑑 𝑓 𝑥
! = 𝑓 𝑏 − 𝑓(𝑎) Provided 𝑓(𝑥) is continuous in (𝑎, 𝑏)
+ 𝑑𝑥

, > !% (,)
! 𝑓 𝑥 ⋅𝑑 𝑔 𝑥 =! 𝑓 𝑥 ⋅ 𝑔: 𝑥 𝑑𝑥
+ > !% (+)

9 > ;
𝑔 𝑥 =𝑎 ⇒ 𝑥 = 𝑔36 (𝑎) = 𝑔: 𝑥
9;

𝑔 𝑥 =𝑏 ⇒ 𝑥 = 𝑔36 (𝑏) ⇒𝑑 𝑔 𝑥 = 𝑔: 𝑥 𝑑𝑥
 6
Evaluate ! 𝑥 . 𝑑 ln 𝑥
36

Solution: 9(AB ;) 6 6
= ⇒ 𝑑(ln 𝑥) = . 𝑑𝑥 ln 𝑥 = 𝑡 −1 < ln 𝑥 < 1
9; ; ;

1
6
1 < ⇒ 𝑑𝑥 = 𝑑𝑡 ⇒ 𝑒 36 < 𝑥 < 𝑒 6
! 𝑥 . 𝑑 ln 𝑥 = ! 𝑥 . 𝑑𝑥 𝑥
36
6 𝑥
<

<
𝑥. 𝑒
= ! 𝑥 𝑑𝑥 =
6 2 1]𝑒
<

<# < !#
= .
− .

.
1.
𝑒 − 𝑒
=
2
 6
𝑑𝑥
Evaluate !
𝑥 . + 2𝑥 + 2
36

A tan36 1 B tan36 2

- -
C tan 1 D tan 2
 6
𝑑𝑥
Evaluate !
𝑥 . + 2𝑥 + 2
36

Solution:
6 6
𝑑𝑥 𝑑𝑥
! 𝑑𝑥 → ! 𝑑𝑥
(𝑥 + 1). +1. 𝑥 . + 𝑎.
36 36

6 ;46 6
⇒ 6 . tan36 ( 6
)| 36

= tan36 (2) − tan36 0

= tan36 (2) − 0

= tan36 (2)
 6
𝑑𝑥
Evaluate !
𝑥 . + 2𝑥 + 2
36

A tan36 1 B tan36 2

- -
C tan 1 D tan 2
 6
.
Evaluate 𝑒 ; 2 − 𝑥.
! 𝑑𝑥
1−𝑥 1 − 𝑥.
36

Solution:
6
.
;
1 − 𝑥. + 1
= !𝑒 𝑑𝑥
1−𝑥 1 − 𝑥.
36
6
.
;
1 − 𝑥. 1
= !𝑒 + 𝑑𝑥
1−𝑥 1 − 𝑥 1 − 𝑥.
36

6
.
1+𝑥 1
= ! 𝑒; + 𝑑𝑥
1−𝑥 1 − 𝑥 1 − 𝑥.
36

𝑓 𝑥 𝑓: 𝑥

64; ⁄#% 6
= 𝑒;. 63;
| 36
= 3𝑒 − . 0 = 3𝑒
<
 Session 2
Definite integration using
substitution

Return To Top
 .
1 ;46
Evaluate: ! 1+𝑥− 𝑒 ; 𝑑𝑥
𝑥
6
.

. .
1 ;46 ;
6 1
Solution: ! 1+𝑥− 𝑒 ; !
𝑑𝑥 = 𝑒 ⋅ 𝑒 1 + 𝑥 −
; 𝑑𝑥
𝑥 𝑥
6 6
. .
.
;
6 6 1 6
= !𝑒 𝑒 ; + 𝑥. 𝑒 ; − . 𝑒 ; 𝑑𝑥
𝑥
6
.
.
6 6 1 6
= ! 𝑒 ; 𝑥. 𝑒 ; + 𝑒 ; − . 𝑒 ; 𝑑𝑥
𝑥
6
.
𝑓 𝑓:
% . % %
6
= 𝑒 ⋅ 𝑥𝑒 ; &
% = 2𝑒 .4# − . 𝑒 .4#
#
'
2
= . 𝑒#
 .
Evaluate: ! 𝑥 ; 1 + 𝑥 + 𝑥 ln 𝑥 𝑑𝑥
6

. .
Solution: ! 𝑥 1 + 𝑥 + 𝑥 ln 𝑥 𝑑𝑥 = ! 𝑥 ; + 𝑥. 𝑥 ; 1 + ln 𝑥 𝑑𝑥
;

6 6

𝑓 𝑥 𝑥. 𝑓′(𝑥)

.
= 𝑥 ⋅ 𝑥; 6

= 2 ⋅ 2. − 1 = 7
 - -
. cos 𝑥 + 2 . sin 𝑥 + 1 '
If 𝐼=! 𝑑𝑥 & 𝐽=! 𝑑𝑥 , then =?
/ sin 𝑥 + 2 cos 𝑥 + 5 / sin 𝑥 + 2 cos 𝑥 + 5 (

𝑓 𝑥 𝑑𝑥
Solution: Note : ∫ ≠!
𝑓 𝑥
𝑑𝑥
∫ 𝑔 𝑥 𝑑𝑥 𝑔 𝑥
- -
. 2 cos 𝑥 + 4 . 2sin 𝑥 + 2
2𝐼 = ! 𝑑𝑥 2𝐽 = ! 𝑑𝑥
/ sin 𝑥 + 2 cos 𝑥 + 5 / sin 𝑥 + 2 cos 𝑥 + 5
- -
. sin 𝑥
+ 2 cos 𝑥 + 5 . 𝜋
2𝐼 + 𝐽 = ! 𝑑𝑥 = ! 1𝑑𝑥 =
/ sin 𝑥 + 2 cos 𝑥 + 5 / 2
-
cos 𝑥 − 2 sin 𝑥
. -
𝐼 − 2𝐽 = ! 𝑑𝑥 = ln sin 𝑥 + 2 cos 𝑥 + 5 .
/
= ln 6 − ln 7
/ sin 𝑥 + 2 cos 𝑥 + 5

Two equations, two unknowns:

-
2𝐼 + 𝐽 = ⋯ (𝑖)
.

𝐼 − 2𝐽 = ln 6 − ln 7 ⋯ (𝑖𝑖)
 - -
cos 𝑥 + 2
. . sin 𝑥 + 1 '
If 𝐼=! 𝑑𝑥 & 𝐽=! 𝑑𝑥 , then =?
/ sin 𝑥 + 2 cos 𝑥 + 5 / sin 𝑥 + 2 cos 𝑥 + 5 (

Solution:
Two equations, two unknowns. From 𝑖 − 𝑖𝑖 ×2 we get:

-
2𝐼 + 𝐽 = .
⋯ (𝑖)
- =
2𝐼 + 𝐽 − 2𝐼 − 4𝐽 = .
− 2 ln D
𝐼 − 2𝐽 = ln 6 − ln 7 ⋯ (𝑖𝑖)
- =
From 𝑖 ×2 + (𝑖𝑖) we get: ⇒ 2𝐼 + 𝐽 − 2𝐼 + 4𝐽 = .
− 2 ln D

4𝐼 + 2𝐽 + 𝐼 − 2𝐽 = 𝜋 + ln 6 − ln 7 *
3. AB
(

⇒𝐽= #
0
)

⇒ 5𝐼 = 𝜋 + ln 6 − ln 7
(
( *"+, (
-4AB )
E -4AB
⇒𝐼= 0
)
⇒F= *
'
( =* )
(
#
!# +,
) 3. AB
# )
'
 0

Evaluate: ! 𝑥 + 2 2𝑥 − 4 + 𝑥 − 2 2𝑥 − 4 𝑑𝑥
1

0
Solution: 𝐼 = ! 𝑥 + 2 2𝑥 − 4 + 𝑥 − 2 2𝑥 − 4 𝑑𝑥
1

𝑓 𝑥 = 𝑥 + 2 2𝑥 − 4 + 𝑥 − 2 2𝑥 − 4

.
𝑓 𝑥 = 2𝑥 + 2 𝑥 . − 8𝑥 + 16

= 2𝑥 + 2 𝑥−4 .

= 2𝑥 + 2 𝑥 − 4 = 4𝑥 − 8
+𝑣𝑒 as 𝑥 ∈ 4,5
⇒𝑓 𝑥 =2 𝑥−2
0
2 2 0 4
𝐼 = ! 2 𝑥 − 2 𝑑𝑥 = 2× 𝑥−2 .
1
= (3 3 − 2 2 )
3 3
1
 .+
𝑑𝑥
Evaluate: !
2𝑎𝑥 − 𝑥 .
/

Solution:

.+ .+
𝑑𝑥 𝑑𝑥
! =!
2𝑎𝑥 − 𝑥 . 𝑎. − 𝑥 − 𝑎 .
/ /

𝑥−𝑎 .+
= sin36
𝑎 /

+ +
= sin36 + − sin36 − +

- -
= .
− −. = 𝜋
 -
.
𝑥 𝑥
Evaluate : ! 𝑒 ; cos sin 𝑥 cos. + sin sin 𝑥 sin. 𝑑𝑥
2 2
/

Solution:
-
.
𝑥 𝑥 cos sin 𝑥 :
= − sin sin 𝑥 ⋅ cos 𝑥
! 𝑒 ; cos sin 𝑥 cos . + sin sin 𝑥 sin. 𝑑𝑥
2 2
/
- :
.
sin sin 𝑥 = cos sin 𝑥 ⋅ cos 𝑥
1 + cos 𝑥 1 − cos 𝑥
= ! 𝑒 ; cos sin 𝑥 + sin sin 𝑥 𝑑𝑥
2 2
/
-
.
1
= ! 𝑒 ; cos sin 𝑥 − sin sin 𝑥 ⋅ cos 𝑥 + cos sin 𝑥 ⋅ cos 𝑥 + sin sin 𝑥 𝑑𝑥
2
/
𝑓 𝑓′ 𝑓′ 𝑓
*
6 ;
= 𝑒 cos sin 𝑥 + sin sin 𝑥 #
/
.
*
6
= 𝑒 # cos 1 + sin 1 − 1 ⋅ 1 + 0
.
 < $%
𝜋 sin(𝜋 ln 𝑥)
Evaluate: !
𝑥
𝑑𝑥
6

Solution:
Let 𝜋 ln 𝑥 = 𝑡 𝜋 ln 1 = 0 & 𝜋 ln 𝑒 26 = 31𝜋
-
⇒ 𝑑𝑥 = 𝑑𝑡 𝒙 𝒕
;

1 0

𝑒 26 31𝜋

< $% 26-
𝜋 sin(𝜋 ln 𝑥)
! 𝑑𝑥 = ! sin 𝑡 𝑑𝑡
6 𝑥 /

26-
= − cos 𝑡 /

= − cos 31𝜋 − −cos 0

= − −1 − (−1) = 2
 AB 0
𝑒 ; 𝑒; − 1
Evaluate : ! 𝑑𝑥
/ 𝑒; + 3

Solution:

Let 𝑒 ; − 1 = 𝑡 .
𝒙 𝒕𝟐 𝒕

⇒ 𝑒 ; ⋅ 𝑑𝑥 = 2𝑡 𝑑𝑡 0 0 0

ln 5 4 2

AB 0 < & < & 36 . .G G #

∫/ < & 42
𝑑𝑥 = ∫/ G # 4642
𝑑𝑡

. . 1
= 2 ∫/ 𝑑𝑡 − 2 ∫/ G # 41
𝑑𝑡

. .×1 G .
=2𝑡 / − .
⋅ tan36 .
/

= 4 − 4(tan36 1 − tan36 0) = 4 − 𝜋
 J !%
𝑒 KLB ;
Evaluate: ! .
𝑑𝑥
/ 1+𝑥

Solution:
-
Let tan36 𝑥 = 𝑡 ⇒ tan36 0 = 0 & tan36 ∞ =
.

𝒙 𝒕

0 0
-
∞ .

1
⇒ 𝑑𝑥 = 𝑑𝑡
1 + 𝑥.

J !% -
𝑒 KLB ; . -
! .
𝑑𝑥 = ! 𝑒 G 𝑑𝑡 = 𝑒. − 1
/ 1+𝑥 /
 -
1 tan. 𝑥 sec . 𝑥 𝑑𝑥
Evaluate : !
/ 1 + tan2 𝑥 .

Solution:
- 𝜋
1 + tan2 𝑥 = 𝑡 At 𝑥 = 1 ; 𝑡 = 1 + tan2 = 2 & At 𝑥 = 0; 𝑡 = 1
4
3tan. 𝑥 ⋅ sec . 𝑥 𝑑𝑥 = 𝑑𝑡

9G
𝒙 𝒕
⇒ tan. 𝑥 ⋅ sec . 𝑥 𝑑𝑥 =
2
- 0 1
1 tan. 𝑥 sec . 𝑥 𝑑𝑥 𝑑𝑡 .
! =! -
/ 1 + tan2 𝑥 .
6 3𝑡
. 2
1

6 6 .
=−
2 G 6

6 6
= −2 .
−1

6
==
 -
1 sin. 𝑥 ⋅ cos . 𝑥
Evaluate : ! 𝑑𝑥
/ sin2 𝑥 + cos 2 𝑥 .

Solution:
Divide both 𝑁 M & 𝐷 M by cos = 𝑥
.
- sin 𝑥 cos . 𝑥 1
1 cos . 𝑥 ⋅ ⋅
! cos 𝑥 cos . 𝑥 𝑑𝑥
.
.
/ sin2 𝑥 cos 2 𝑥
+
cos 2 𝑥 cos 2 𝑥
-
.
1 tan 𝑥⋅ sec . 𝑥
=! 𝑑𝑥
/ tan2 𝑥 + 1 .

-
.
1 tan 𝑥 ⋅ sec . 𝑥
=! 𝑑𝑥 Let tan 𝑥 = 𝑡 ⇒ sec . 𝑥 = 𝑑𝑡
/ tan2 𝑥 + 1 .

1 6 3𝑡 . ⋅ 𝑑𝑡 6 6 6 6
= ! 2 .
𝑑𝑥 = − $ ==
3 / 𝑡 +1 2 G 46 /
 2
𝑑𝑥
Evaluate: !
.
𝑥−1 5−𝑥

Solution:

Let 𝑥 = 1 ⋅ cos . 𝜃 + 5 ⋅ sin. 𝜃 Upper limit

⇒ 𝑑𝑥 = 2 cos 𝜃 ⋅ − sin 𝜃 𝑑𝜃 + 10 sin 𝜃 ⋅ cos 𝜃 𝑑𝜃 3 = 1 ⋅ cos . 𝜃 + 5 ⋅ sin. 𝜃

⇒ 𝑑𝑥 = 8 sin 𝜃 ⋅ cos 𝜃 𝑑𝜃
⇒ 2 = 1 1 − sin. 𝜃 + 5 ⋅ sin. 𝜃
⇒ 𝑑𝑥 = 4 sin 2𝜃 𝑑𝜃
⇒ 4 sin. 𝜃 = 2
Lower limit
6 -
. . ⇒ sin. 𝜃 = ⇒ 𝜃 =
2 = 1 ⋅ cos 𝜃 + 5 ⋅ sin 𝜃 . 1

⇒ 2 = 1 1 − sin. 𝜃 + 5 ⋅ sin. 𝜃

⇒ 4 sin. 𝜃 = 1
6 -
⇒ sin. 𝜃 = 1 ⇒ 𝜃 = =
 2
𝑑𝑥
Evaluate: !
.
𝑥−1 5−𝑥

Solution:

Let 𝑥 = 1 ⋅ cos . 𝜃 + 5 ⋅ sin. 𝜃

⇒ 𝑑𝑥 = 4 sin 2𝜃 𝑑𝜃
6 - 6 -
Lower limit ⇒ sin. 𝜃 = 1 ⇒ 𝜃 = =
Upper limit ⇒ sin. 𝜃 = . ⇒ 𝜃 = 1
-
2 1
𝑑𝑥 4 sin 2𝜃 𝑑𝜃
! = !
𝑥−1 5−𝑥 -
4 sin. 𝜃 ⋅ 4 cos . 𝜃
.
=
- -
1 1
4 sin 2𝜃 𝑑𝜃
=! = ! 2𝑑𝜃
2 sin 2𝜃
- -
= =
- - -
=2 − =
1 = =
 ! 𝑥−1
Evaluate 3 𝑑𝑥
) 3−𝑥
!

Solution:
.
𝑥−1
𝐼= ! 𝑑𝑥
2 3−𝑥
.

𝑥−𝛼
Format: Substitute 𝑥 = 𝛼 cos . 𝜃 + 𝛽 sin. 𝜃
𝛽−𝑥

So, 𝑥 = 1 ⋅ cos . 𝜃 + 3 ⋅ sin. 𝜃

⇒ 𝑑𝑥 = − sin 2𝜃 + 3 sin 2𝜃 𝑑𝜃

⇒ 𝑑𝑥 = 2 sin 2𝜃 𝑑𝜃

𝑥 − 1 = cos . 𝜃 + 3 sin. 𝜃 − 1

= 3 sin. 𝜃 − 1 − cos . 𝜃 = 2 sin. 𝜃

3 − 𝑥 = 3 − cos . 𝜃 + 3 sin. 𝜃 = 3 cos . 𝜃 − cos . 𝜃 = 2 cos . 𝜃

 ! 𝑥−1
Evaluate 3 𝑑𝑥
) 3−𝑥
!

Solution:

2
Lower limit : = cos . 𝜃 + 3 sin. 𝜃
.

⇒ 3 = 2cos . 𝜃 + 6 1 − cos . 𝜃
.
3
⇒ cos 𝜃 =
⇒ 4 cos . 𝜃 = 3 4
𝜋
⇒𝜃=
6
Upper limit : 2 = cos . 𝜃 + 3 sin. 𝜃

⇒ 2 = cos . 𝜃 + 3(1 − cos . 𝜃)

1
⇒ 2 cos . 𝜃 = 1 ⇒ cos . 𝜃 =
2
𝜋
⇒𝜃=
4
 ! 𝑥−1
Evaluate 3 𝑑𝑥
) 3−𝑥
!

Solution:
-
1 2 sin. 𝜃
∴ 𝐼 = 2! × sin 2𝜃 𝑑𝜃
- 2 cos . 𝜃
=
-
1
= 4 ! sin. 𝜃 𝑑𝜃
-
=
-
1 1 cos 2𝜃
= 4! − 𝑑𝜃
- 2 2
=
-
𝜃 sin 2𝜃 1
⇒𝐼=4 −
2 4 -
=

𝜋 𝜋 4
=2 − − (sin 90° − sin 60°)
4 6 4
- 2
⇒𝐼 = =
− 1− .
Key Takeaways

Invalid
substitution
.
𝑑𝑥 6
𝐼=! Put 𝑥 = G →This substitution is wrong
.
3. 4 + 𝑥
6
𝑑𝑥 = − G # As in −2 to 2 → it is discontinuous
Method-I
. %
−𝑑𝑡 6 -
𝐼=! .
= − . tan36 2𝑡 #
% = −1
3. 4𝑡 + 1
3
#

Method-II
.
𝑑𝑥 6 ; . -
𝐼=! . = − . tan36 . =
3. 4 + 𝑥 3. 1
 Session 3
Walli's formula and Gamma
function

Return To Top
 6

Evaluate ! 𝑥𝑒 3; 𝑑𝑥
/

Solution:
6

𝐼 = ! 𝑥𝑒 3; 𝑑𝑥
/
𝑢 𝑣
6 6
= −𝑥 𝑒 3; / − ∫/ (1 ⋅ ∫ 𝑒 3; ⋅ 𝑑𝑥 ) 𝑑𝑥
,
36
= −1 ⋅ 𝑒 − 0 + ! 𝑒 -. ⋅ 𝑑𝑥
+
6 6 6 6
= − < − 𝑒 3; / = −< − <
−1

.
⇒𝐼 =1−<
 -
1
If 𝑎N =! tanN 𝑥 𝑑𝑥 then 𝑎. + 𝑎1 , 𝑎2 + 𝑎0 , 𝑎1 + 𝑎= are in which progression ?
/
-
1
Solution: 𝑎N = ! tanN 𝑥 𝑑𝑥
/
-
1
𝑎N4. = ! tanN4. 𝑥 𝑑𝑥
/
- -
1 1
∴ 𝑎N + 𝑎N4. = ! tanN 𝑥 𝑑𝑥 + ! tanN4. 𝑥 𝑑𝑥
/ /
- -
1 1
⇒ 𝑎N + 𝑎N4. = ! tanN 𝑥 1 + tan. 𝑥 𝑑𝑥 = ! tanN 𝑥 sec . 𝑥 𝑑𝑥
/ /
𝑥 𝑡
Let 𝑡 = tan 𝑥 ⇒ 𝑑𝑡 = sec . 𝑥 𝑑𝑥
0 0
-
1
1
 -
1
If 𝑎N = ! tanN 𝑥 then 𝑎. + 𝑎1 , 𝑎2 + 𝑎0 , 𝑎1 + 𝑎= are in which progression ?
/

Solution:
6 1
N
𝑡 N46 1
∴ 𝑎N + 𝑎N4. = ! 𝑡 𝑑𝑡 = =
𝑛+1 /
𝑛+1
/

1 1 1
⇒ 𝑎. + 𝑎1 = , 𝑎2 + 𝑎0 = , 𝑎1 + 𝑎= =
3 4 5
6 6 6
𝑖. 𝑒. , 2 , 1 , 0 are in H.P.

∴ 𝑎. + 𝑎1 , 𝑎2 + 𝑎0 , 𝑎1 + 𝑎= are in H.P.
 *
If 𝑎N = ∫/- tanN 𝑥 𝑑 𝑥 − 𝑥 , 𝑤ℎ𝑒𝑟𝑒 . denotes G.I.F then 𝑎N + 𝑎N4. = ?

Solution: *
/ ⇒ 𝑎N + 𝑎N4. = ∫ - tan1 𝑥 1 + tan3 𝑥 𝑑𝑥
0 +
*
𝑎N = ! tan1 𝑥 𝑑 𝑥 − 0 = ∫+ tan1 𝑥 sec 3 𝑥 𝑑𝑥
-

+
Let 𝑡 = tan 𝑥 ⇒ 𝑑𝑡 = sec . 𝑥 𝑑𝑥
- 2.61
𝑥 ∈ 0, 1 ⇒ 𝑥 ∈ 0,
1 𝑥 𝑡

0 0
⇒ 𝑥 ∈ 0,0.8
-
1
1
⇒ 𝑥 =0
6 1
* G ."% 6
⇒ 𝑎N = ∫+ tan 𝑥 𝑑𝑥 & 𝑎N4.
- 1 ∴ 𝑎N + 𝑎N4. = ! 𝑡 N 𝑑𝑡 = = N46
N46 /
/
*
6
= ∫+ tan123 𝑥 𝑑𝑥
-
∴ 𝑎N + 𝑎N4. = N46
 /
4 1 1
sin4 𝜃 − cos4 𝜃 − cos3 𝜃 sin 𝜃 + cos 𝜃 + cos3 𝜃 3++5 𝑎+ 𝑏 − 1+ 𝑐
If * 𝑑𝜃 = ; 𝑎, 𝑏, 𝑐, 𝑑 ∈ 𝐼4
sin 𝜃 3++6 × cos 𝜃 3++6 𝑑
/
0

then 𝑎 + 𝑏 + 𝑐 + 𝑑 = ?
Solution:
/
4 4 4 3 3 3++5
Let 𝐼 = * sin 𝜃 − cos 𝜃 − cos 𝜃 sin 𝜃 + cos 𝜃 + cos 𝜃 𝑑𝜃
3++6
sin 𝜃 3++6
× cos 𝜃
/
0

/
44 4 3 3 3++5
= * sin 𝜃 − cos 𝜃 − cos 𝜃 sin 𝜃 + cos 𝜃 + cos 𝜃 𝑑𝜃
3
sin 𝜃 3++5
sin 𝜃 3
cos 𝜃 3++5
cos 𝜃
/
0
/
4 4 4 3 3 3++5
= * sin 𝜃 − cos 𝜃 − cos 𝜃 × sin 𝜃 + cos 𝜃 + cos 𝜃 𝑑𝜃
3 3
sin 𝜃 cos 𝜃 sin 𝜃 cos 𝜃
/
0
/
4
= * tan 𝜃 sec 𝜃 − cot 𝜃 cosec 𝜃 − cosec 3𝜃 × sec 𝜃 + cosec 𝜃 + cot 𝜃 3++5 𝑑𝜃

/
0
 /
4 1 1
sin4 𝜃 − cos4 𝜃 − cos3 𝜃 sin 𝜃 + cos 𝜃 + cos3 𝜃 3++5 𝑎+ 𝑏 − 1+ 𝑐
If * 𝑑𝜃 = ; 𝑎, 𝑏, 𝑐, 𝑑 ∈ 𝐼4
sin 𝜃 3++6 × cos 𝜃 3++6 𝑑
/
0

then 𝑎 + 𝑏 + 𝑐 + 𝑑 = ?
Solution:
Let 𝑡 = sec 𝜃 + cosec 𝜃 + cot 𝜃 𝜃 𝑡
-
⇒ 𝑑𝑡 = tan 𝜃 sec 𝜃 − cot 𝜃 cosec 𝜃 − cosec . 𝜃 2
2+ 3
.4 2 -
G #//0
1+ 8
1
∴𝐼 = ! 𝑡 .//D
𝑑𝑡 = .//P
64 P

#//0 #//0
.4 2 3 64 P
⇒𝐼= … (𝑖)
.//P
. .
+4 , 3 64 5
Also, 𝐼 = … (𝑖𝑖)
9

From (𝑖) and (𝑖𝑖),

𝑎 = 2, 𝑏 = 3, 𝑐 = 8, 𝑑 = 2008 ⇒ 𝑎 + 𝑏 + 𝑐 + 𝑑 = 2021
 J
𝑑𝑥 𝜋
If ! = , then 𝑎 = ?
. 5050
1
/ 𝑎. + 𝑥 − 𝑥

6 36
Solution: Let 𝑥 = G
⇒ 𝑑𝑥 = G#
𝑑𝑡
7
𝑑𝑥 𝜋
Let 𝐼 = * 3 = ⋯ (𝑖) 𝑥 𝑡
1 5050
+ 𝑎3 + 𝑥−
𝑥
0 ∞
1+ 7 1
− 3 𝑑𝑡 𝑑𝑡
𝑡 𝑡3 ∞ 0
𝐼 =* 1 3 =*
1 3
7 𝑎3 + 𝑎3 + 𝑡 − 𝑡
𝑡 −𝑡 +

7 1
𝑑𝑥
𝑥3
⇒𝐼 =* 3 ⋯ (𝑖𝑖) Substitute 𝑡 = 𝑥 , 𝑑𝑡 = 𝑑𝑥
1
+ 𝑎3 + 𝑥−𝑥

1 7
3 𝑑𝑥 1+
= * 𝑥
𝑖 + 𝑖𝑖 → 2𝐼 1 3
+ 𝑎 3 + 𝑥 −
𝑥
 J
𝑑𝑥 𝜋
If ! = , then 𝑎 = ?
. 5050
1
/ 𝑎. + 𝑥 − 𝑥

Solution:
𝑥 𝑡
6 6
Let 𝑥 − = 𝑡 ⇒ 1 + # 𝑑𝑥 = 𝑑𝑡 0 −∞
; ;

7 ∞ ∞
𝑑𝑡
∴ 2𝐼 = *
𝑎3 + 𝑡3
-7

6 G J
⇒ 2𝐼 = + tan36 +
3J

6
= (tan36 ∞ − tan36 (−∞))
+

6 - - -
⇒ 2𝐼 = + =
+ . . +

𝜋 𝜋
⇒𝐼 = = From question ⇒ 𝑎 = 2525
2𝑎 5050
 6
1−𝑥
Evaluate ! 𝑑𝑥
1+𝑥 𝑥 + 𝑥. + 𝑥2
/

Solution: ,
1−𝑥
𝐼= * 𝑑𝑥
1+𝑥 𝑥 + 𝑥3 + 𝑥4
+
,
1 − 𝑥3
= * 𝑑𝑥 Multiplying numerator and denominator by (1 + 𝑥)
𝑥 3 + 2𝑥 + 1 𝑥 + 𝑥3 + 𝑥4
+

, 1
1− 𝑥 𝑡
∴ 𝐼 = −* 𝑥3 𝑑𝑥
1 1
𝑥+𝑥+2 𝑥+𝑥+1
+
0 ∞
6 6
Let 𝑡 . = ; + 𝑥 + 1 ⇒ 2𝑡 𝑑𝑡 = 1 − # 𝑑𝑥 1 3
;
4 4
2𝑡 𝑑𝑡 𝑡 𝑑𝑡 3
∴𝐼= −* 3 = −2 * = −2 tan36 𝑡 J
𝑡 +1 𝑡 𝑡3+1 𝑡
7 7
𝜋
⇒ 𝐼 = −2 tan36 3 − tan36 ∞ =
3
 ,

5050 * 1 − 𝑥 8+ ,++
𝑑𝑥
+
Find the value of ,

* 1 − 𝑥 8+ ,+, 𝑑𝑥

+
,

Solution: Let 𝐼 = * 1 − 𝑥 8+ ,+, 𝑑𝑥 Cancellation of terms under Integration is not possible

+

1 ,
0/ 6/6
= 1−𝑥 𝑥 8+ ,++ −50𝑥 06 ⋅ 𝑥 𝑑𝑥
/ − * 101 1 − 𝑥
+
,

= 0 − 5050 * 1 − 𝑥 8+ ,++
−𝑥 8+ 𝑑𝑥
+
,

𝐼 = − 5050 * 1 − 𝑥 8+ ,++ 1 − 𝑥 8+ − 1 𝑑𝑥
+
, ,

= − 5050 * 1 − 𝑥 8+ ,+, 𝑑𝑥
+ 5050 * 1 − 𝑥 8+ ,++ 𝑑𝑥

+ +
,

= − 5050𝐼 + 5050 * 1 − 𝑥 8+ ,++

𝑑𝑥
+
 ,

5050 * 1 − 𝑥 8+ ,++
𝑑𝑥
+
Find the value of ,

* 1 − 𝑥 8+ ,+, 𝑑𝑥

Solution:
,

𝐼=
=−− 5050𝐼 + 5050 * 1 − 𝑥 8+ ,++ 𝑑𝑥

+
,

⇒ 5051𝐼 = 5050 * 1 − 𝑥 8+ ,++ 𝑑𝑥

+
,

5050 * 1 − 𝑥 8+ ,++ 𝑑𝑥

⇒ +
= 5051
𝐼
,
5050 * 1 − 𝑥 8+ ,++ 𝑑𝑥
+
⇒ = 5051
,
* 1 − 𝑥 8+ ,+, 𝑑𝑥
+
 Key Takeaways

Walli's Formula
- -
. .
Ø If 𝐼N = ! sinN 𝑥 𝑑𝑥 ; ! cos N 𝑥 𝑑𝑥 , 𝑛 ∈ 𝕀 and 𝑛 ≥ 2
/ /

N36 N32 N30 ⋯×6 , 𝑛 is odd

N N3. N31 ⋯×6
𝐼N =
*
N36 N32 N30 ⋯× , 𝑛 is even
#
N N3. N31 ⋯×.

Example:
-
.
10 ⋅ 8 ⋅ 6 ⋅ 4 ⋅ 2 256
! cos66 𝑥 𝑑𝑥 = ×1 =
11 ⋅ 9 ⋅ 7 ⋅ 5 ⋅ 3 ⋅ 1 11×63
/
 - -
. .

Evaluate 𝑖 ! cos D 𝑥 𝑑𝑥 𝑖𝑖 ! sinP 𝑥 𝑑𝑥

/ /

Solution:
- -
. . 𝜋
7 − 1 7 − 3 7 − 5 ×1 8−1 8−3 8−5 × 8−7 ×2
𝑖 ! cos D 𝑥 𝑑𝑥 = 𝑖𝑖 ! sinP 𝑥 𝑑𝑥 =
7 7−2 7−4 7−6 8 8−2 8−4 8−6
/ /
6×4×2 𝜋
= 7×5×3×1× 2
7×5×3×1
=
8×6×4×2
16
= 35𝜋
35 =
256
 2
𝑥2
Evaluate ! 𝑑𝑥
3−𝑥
/

A
.DR
B −
.DR
P P

C 2=R D SR
P P
 2
𝑥2
Evaluate ! 𝑑𝑥
3−𝑥
/

Solution:
2
𝑥2
𝐼=! 𝑑𝑥
3−𝑥
/ 𝑥 𝜃
Let 𝑥 = 3 sin. 𝜃 ⇒ 𝑑𝑥 = 6 sin 𝜃 cos 𝜃 𝑑𝜃 0 0
-
3 .
-
.
3 sin. 𝜃 2
∴ 𝐼= ! ⋅ 6 sin 𝜃 cos 𝜃 𝑑𝜃
3 − 3 sin. 𝜃
/

- -
. .
27 sin= 𝜃
⇒𝐼= ! .
⋅ 6 sin 𝜃 cos 𝜃 𝑑𝜃 = 3 ! sin2 𝜃 ⋅ 6 sin 𝜃 𝑑𝜃
3 cos 𝜃
/ /
 2
𝑥2
Evaluate ! 𝑑𝑥
3−𝑥
/

Solution:
- -
. .
27 sin= 𝜃
⇒𝐼= ! .
⋅ 6 sin 𝜃 cos 𝜃 𝑑𝜃 = 3 ! sin2 𝜃 ⋅ 6 sin 𝜃 𝑑𝜃
3 cos 𝜃
/ /
-
.
= 18 ! sin1 𝜃 𝑑𝜃
/

2 6 -
= 18 ⋅ ⋅
1 . .

.D-
⇒𝐼 =
P
 6
𝑥2
Evaluate ! 𝑑𝑥
1 − 𝑥.
/

A
.
B −
.
P P

C . D −
.
2 2
 6
𝑥2
Evaluate ! 𝑑𝑥
1 − 𝑥.
/

Solution:
6
𝑥2
𝐼= ! 𝑑𝑥
1 − 𝑥.
/
𝑥 𝜃

Let 𝑥 = sin 𝜃 ⇒ 𝑑𝑥 = cos 𝜃 𝑑𝜃 0 0

-
1 .

- -
. .
sin2 𝜃 cos 𝜃
∴ 𝐼 =! 𝑑𝜃 = ! sin2 𝜃 𝑑𝜃
1 − sin. 𝜃 /
/
.
= 2⋅6 ⋅ 1 Using Walli’s formula
.
=2
Key Takeaways

Gamma Function
R
.

Ø If IW,B = ! sinW x cos B x dx , 𝑚, 𝑛 ∈ ℕ and 𝑚 + 𝑛 ≥ 2

/

W36 W32 W30 ⋯ B36 B32 B30 ⋯ R

W4B W4B3. W4B31 ⋯
⋅ .
, when both 𝑚, 𝑛 are even
𝐼U,N =
W36 W32 W30 ⋯ B36 B32 B30 ⋯
W4B W4B3. W4B31 ⋯
⋅ 1 , otherwise
 -
.
Evaluate ! sinD 𝑥 ⋅ cos P 𝑥 𝑑𝑥
/

Solution:
-
.
𝐼 = ! sinD 𝑥 𝑑𝑥 ⋅ cos P 𝑥 𝑑𝑥
/

Using Gamma function

7 −71− 1 7 −75−×5 8×−81− 1

7 −73− 3 8 −83− 3
8 −85− 5
8 −87− 7
= ⋅1
15 15 − 2 15 − 4 15 − 6 15 − 8 15 − 10 15 − 12
15 15 − 2 15 − 4 15 − 6 15 − 8 15 − 10 15 − 12 15 − 14 15 − 14

@×A×!×B×C×)×D
= ⋅1
DC×D)×DD×E×B×C×)×D

D@
=
DEC×))
 +
0
Evaluate ! 𝑥 . 𝑎 − 𝑥 𝑑𝑥
/

Solution: + -
0
.
Let 𝐼 = ! 𝑥 . 𝑎 − 𝑥 𝑑𝑥
= 2𝑎1 ! sin= 𝜃 ⋅ cos . 𝜃 𝑑𝜃
/
Let 𝑥 = 𝑎 sin. 𝜃 ⇒ 𝑑𝑥 = 𝑎 2sin 𝜃𝑐𝑜𝑠𝜃 𝑑𝜃 /

𝑥 𝜃 =36 =32 =30 × .36 -

= 2𝑎1 P×=×1×.
⋅ .
0 0
Using Gamma function
-
𝑎 . 0⋅2⋅6×6 - 0-+ -
/
⇒ 𝐼 = 2𝑎1 P×=×1×. ⋅ .
= 6.P
3
8
∴ 𝐼 = ! 𝑎3 sin8 𝜃 𝑎 − 𝑎 sin3 𝜃 𝑎 2sin 𝜃 𝑐𝑜𝑠𝜃 𝑑𝜃
+
-
.
⇒ 𝐼 = ! 𝑎2 sin0 𝜃 ⋅ cos 𝜃 2 𝑎 sin 𝜃 cos 𝜃 𝑑𝜃
/
 Session 4
Properties of Definite
Integration

Return To Top
Key Takeaways

Definite Integration

Property 1
: : :
Ø *𝑓 𝑥 𝑑𝑥 = * 𝑓 𝑡 𝑑𝑡 = * 𝑓 𝜙 𝑑𝜙
9 9 9

Change of variable makes no difference in the value of definite integral.

Example:
/
3 /
𝜋
• * sin 𝑥 𝑑𝑥 = − cos 𝑥 3
+ = − cos − cos 0 = − 0 − 1 = 1
2
+
/ /
3 / 3
𝜋
• * sin 𝑧 𝑑𝑧 = − cos 𝑧 3
+ = − cos − cos 0 = − 0 − 1 = 1 = * sin 𝑥 𝑑𝑥
2
+ +
Key Takeaways

Definite Integration

Property 2

: 9
Ø * 𝑓 𝑥 𝑑𝑥 = − * 𝑓 𝑥 𝑑𝑥
9 :

Inter-Change in limit gives −ve value of definite integral.

 7
K
If 𝐼I = * 𝑒 3; 𝑥 N36 𝑑𝑥 , then ∫J 𝑒 LMN 𝑥 ILD𝑑𝑥 = ?
+

Solution: 7

Let 𝐼 = * 𝑒 3X; 𝑥 N36 𝑑𝑥 Let 𝜆𝑥 = 𝑡 ⇒ 𝑑𝑥 =

9G
+
X

7
G N36 9G
= * 𝑒 3G 𝑥 𝑡
X X
+
0 0

∞ ∞

7
1
= 1 * 𝑒 3G 𝑡 N36 𝑑𝑡
𝜆
Using Property 1
+

E
𝐼 = X..
 0
P Q YZ[ \ 3
If 𝐹 𝑥 = 𝑥 > 0 and $
! 𝑒 ;<= . 𝑑𝑥 = 𝐹 𝑘 − 𝐹 1 , then 𝑘 = ?
PN N 𝑥
,

Solution: 0
3 !
* 𝑒 ;<= . 𝑑𝑥 = 𝐹 𝑘 − 𝐹 1 ⋯ 𝑖 9
𝐹 𝑥 =
< 12, &
𝑥>0
𝑥 9; ;
,

0 0
3 ;<= .! 3𝑥 3 ;<= .!
* 𝑒 𝑑𝑥 = * 4 𝑒 𝑑𝑥 Let 𝑡 = 𝑥 2 ⇒ 𝑑𝑡 = 3𝑥 . 𝑑𝑥
𝑥 𝑥 < 12, &
, , ⇒𝐹 𝑥 =∫ ;
𝑑𝑥
>0
𝑒 ;<= ? 𝑥 𝑡
=* 𝑑𝑡
𝑡
,
>0 1 1
𝑒 ;<= .
=* 𝑑𝑥
𝑥 64
, 4
=1
= 𝐹 𝑥 1

= 𝐹 64 − 𝐹 1

∴ From 𝑖 , 𝑘 = 64
 J
log 𝑥
Evaluate ! 𝐹 𝑥 N + 𝑥 3N 𝑑𝑥
1 + 𝑥.
/
Solution: 7
log 𝑥
Let 𝐼 = * 𝐹 𝑥 1 + 𝑥 -1 𝑑𝑥
1 + 𝑥3
+

+
− log 𝑡 1 6 6
∴ 𝐼 = * 𝐹 𝑡 -1 + 𝑡 1 1
− 3 𝑑𝑡
𝑡
Let 𝑥 = ⇒ 𝑑𝑥 = − 𝑑𝑡
1+ 3 G G#
7 𝑡
At 𝑥 → 0, 𝑡 → ∞ and At 𝑥 → ∞, 𝑡 → 0
+
log 𝑡
= * 𝐹 𝑡 -1 + 𝑡 1 𝑑𝑡
1 + 𝑡3
7

7 : :
log 𝑥
= −* 𝐹 𝑥 -1 + 𝑥1 𝑑𝑥 * 𝑓 𝑡 𝑑𝑡 = * 𝑓 𝑥 𝑑𝑥
1 + 𝑥3
+ 9 9

⇒ 𝐼 = −𝐼 ⇒ 2𝐼 = 0 ⇒ 𝐼 = 0
 @
1 1
Evaluate * sin 𝑥 − 𝑑𝑥
𝑥 𝑥
,
@

A 𝑒

6
B <

C 0

D 1
 @
1 1
Evaluate * sin 𝑥 − 𝑑𝑥
𝑥 𝑥
,
@

Solution: @
1 1
Let 𝐼 = * 𝑥 sin 𝑥 − 𝑥 𝑑𝑥
,
@
,
@
1 1 6 6
⇒ 𝐼 = * 𝑡 sin 𝑡 − 𝑡 −
𝑡3
𝑑𝑡 Let 𝑥 = G
⇒𝑑𝑥 = − G # 𝑑𝑡
@
,
@
1 1 𝑥 𝑡
⇒ 𝐼 = * sin − 𝑡 − 𝑑𝑡
𝑡 𝑡
@
𝑒 1/𝑒
@
1 1
⇒ 𝐼 = − * sin 𝑡 − 𝑑𝑡 6 𝑒
𝑡 𝑡
, <
@
@
1 1
∴ 𝐼 = − * sin 𝑥 − 𝑑𝑥
𝑥 𝑥
,
@
⇒ 2𝐼 = 0 ⇒ 𝐼 = 0
 @
1 1
Evaluate * sin 𝑥 − 𝑑𝑥
𝑥 𝑥
,
@

A 𝑒

6
B <

C 0

D 1
 .
D log 𝑡
If 𝐹 𝑥 = 𝑓 𝑥 + 𝑓 and 𝑓 𝑥 = ! 𝑑𝑡 , then find 𝐹 𝑒
N 1+𝑡
,
Solution:
1
⇒𝐹 𝑥 =𝑓 𝑥 +𝑓
𝑥
. ,/.
log 𝑡 log 𝑡
⇒𝐹 𝑥 = * 𝑑𝑡 + * 𝑑𝑡
1+𝑡 1+𝑡
, ,
. . 1 6 6
log 𝑡 log 𝑧 1 Let 𝑡 = ⇒ 𝑑𝑡 = − 𝑑𝑧
⇒𝐹 𝑥 =* 𝑑𝑡 + *
1
− 3 𝑑𝑧 ] ]# 𝑡 𝑧
1+𝑡 𝑧
, , 1+𝑧
. .
log 𝑧
1 . . 1 1
log 𝑡 log 𝑡 log 𝑡
⇒𝐹 𝑥 = * 𝑑𝑡 − * 𝑑𝑧 = * 𝑑𝑡 + * 𝑑𝑡
1+𝑡 𝑧 1+𝑧 1+𝑡 𝑡 1+𝑡 6
,
.
, , ,
;
𝑥
𝑡 log 𝑡 + log 𝑡
⇒𝐹 𝑥 = * 𝑑𝑡
𝑡 1+𝑡
,
.
log 𝑡
⇒𝐹 𝑥 = * 𝑑𝑡
𝑡
,

# A^_ < # 6
A^_ ;
⇒𝐹 𝑥 = ∴𝐹 𝑒 = .
=.
.
 Key Takeaways
𝑌

Property 3 (Splitting Limit Based)

Ø If 𝑓 𝑥 is Continuous/Discontinuous at a point or points in
𝑎, 𝑏 say 𝑐6 , 𝑐. , 𝑐2 then,
𝑋
: B" B# B! : 𝑂 𝑎 𝑐$ 𝑐% 𝑐& 𝑏
* 𝑓 𝑥 𝑑𝑥 = * 𝑓 𝑥 𝑑𝑥 + * 𝑓 𝑥 𝑑𝑥 + * 𝑓 𝑥 𝑑𝑥 + * 𝑓 𝑥 𝑑𝑥
9 9 B" B# B!

Ø If 𝑐 is outside of 𝑎, 𝑏 , then also property will work.

𝑌
:
B :
* 𝑓 𝑥 𝑑𝑥 = * 𝑓 𝑥 𝑑𝑥 + * 𝑓 𝑥 𝑑𝑥
9 B
9

𝑋
𝑂 𝑎 𝑏 𝑐
Key Takeaways
This represents the area
under curve between
Proof: 𝑎&𝑏
𝑌 𝑦 = 𝑓(𝑥)
From figure

: B B

* 𝑓 𝑥 𝑑𝑥 = * 𝑓 𝑥 𝑑𝑥 − * 𝑓 𝑥 𝑑𝑥
( )
9 9 :
j 𝑓 𝑥 𝑑𝑥 j 𝑓 𝑥 𝑑𝑥
' (
: B :
𝑋
𝑂 𝑎 𝑏 𝑐
⇒ * 𝑓 𝑥 𝑑𝑥 = * 𝑓 𝑥 𝑑𝑥 + * 𝑓 𝑥 𝑑𝑥
9 9 B This represents the
area under curve
Remark : between 𝑏 & 𝑐

We use this property in discontinuous functions also , like

;
𝑥, 𝑥 , , sgn 𝑥 , where represents Greatest Integer Function and
;

represents Fractional Part Function.

 0 0 -,

If ! 𝑓 𝑥 𝑑𝑥 = 4 and ! 3 − 𝑓 𝑥 𝑑𝑥 = 7, then ! 𝑓 𝑥 𝑑𝑥 = ?
-, 3 3

Solution: 0

* 3 − 𝑓 𝑥 𝑑𝑥 = 7
3
0

⇒ (3𝑥)03 − * 𝑓 𝑥 𝑑𝑥 = 7
3
0

⇒ * 𝑓 𝑥 𝑑𝑥 = −1 … (𝑖)
3
−1 2 4
-, 0 -,
: B :
* 𝑓 𝑥 𝑑𝑥 = * 𝑓 𝑥 𝑑𝑥 + * 𝑓 𝑥 𝑑𝑥 ?
3 3 0 * 𝑓 𝑥 𝑑𝑥 = * 𝑓 𝑥 𝑑𝑥 + * 𝑓 𝑥 𝑑𝑥
0 0
9 9 B
= * 𝑓 𝑥 𝑑𝑥 − * 𝑓 𝑥 𝑑𝑥 ?
: 9
3 -,
* 𝑓 𝑥 𝑑𝑥 = − * 𝑓 𝑥 𝑑𝑥
= −1 − 4 From (𝑖) 9 :

= −5
 3
𝑥 ; 0≤𝑥≤1
If 𝑓 𝑥 = , then ! 𝑓 𝑥 𝑑𝑥 = ?
𝑥! ; 1 < 𝑥 ≤ 2 +

Solution:
3 , 3

* 𝑓 𝑥 𝑑𝑥 = * 𝑓 𝑥 𝑑𝑥 + * 𝑓 𝑥 𝑑𝑥
+ + ,

, 3

= * 𝑥𝑑𝑥 + * 𝑥 3 𝑑𝑥
+ ,

$ 6 .
. ;$
= 2
𝑥 # + 2 6
/

. P 6
=2 1−0 + 2
−2

S
=2=3
 ,++ ,++ ,

If * 𝑓 𝑥 𝑑𝑥 = 𝑎, then c * 𝑓 𝑟 − 1 + 𝑥 𝑑𝑥 = ?
+ CD, +

Solution: ,++ , , , , ,

Let 𝐼 = c * 𝑓 𝑟 − 1 + 𝑥 𝑑𝑥 = * 𝑓 𝑥 𝑑𝑥 + * 𝑓 𝑥 + 1 𝑑𝑥 + * 𝑓 𝑥 + 2 𝑑𝑥 + ⋯ + * 𝑓 𝑥 + 99 𝑑𝑥
CD, + + + + +

𝑥+1=𝑡 𝑥 + 2 = 𝑢 …. 𝑥 + 99 = 𝑧
⇒ 𝑑𝑥 = 𝑑𝑡 ⇒ 𝑑𝑥 = 𝑑𝑢 ⇒ 𝑑𝑥 = 𝑑𝑧
𝑥 𝑡 𝑥 𝑢 𝑥 𝑧
0 1 0 2 0 99
1 2 1 3 1 100
, 3 4 ,++

∴ 𝐼 = * 𝑓 𝑥 𝑑𝑥 + * 𝑓 𝑡 𝑑𝑡 + * 𝑓 𝑢 𝑑𝑢 + ⋯ + * 𝑓 𝑧 𝑑𝑧
+ , 3 66
, 3 4 ,++
: :
⇒ 𝐼 = * 𝑓 𝑥 𝑑𝑥 + * 𝑓 𝑥 𝑑𝑥 + * 𝑓 𝑥 𝑑𝑥 + ⋯ + * 𝑓 𝑥 𝑑𝑥
* 𝑓 𝑥 𝑑𝑥 = * 𝑓 𝑡 𝑑𝑡
+ , 3 66
9 9
,++

= * 𝑓 𝑥 𝑑𝑥 =𝑎
+
 6

Evaluate ! 𝑒 ; 𝑑𝑥
36
Solution:
𝑥 ,𝑥 ≥ 0
𝑥 =ƒ
−𝑥 , 𝑥 < 0

𝑒; , 𝑥 ≥ 0
Turning point of 𝑒 ; is 𝑥 = 0, 𝑒 ; = ƒ
𝑒 3; , 𝑥 < 0
, + ,

∴ * 𝑒 . 𝑑𝑥 = * 𝑒 -. 𝑑𝑥 + * 𝑒 . 𝑑𝑥
-, -, +

/ 6
= − 𝑒 3; 36 + 𝑒; /

= − 𝑒 / − 𝑒 6 + (𝑒 6 − 𝑒 / )
,

⇒ * 𝑒 . 𝑑𝑥 = 2(𝑒 − 1)
-,
 0

Evaluate ! 𝑥 − 3 𝑑𝑥
/
Solution:
Method 1 Method 2
8

* 𝑥 − 3 𝑑𝑥 𝑥 − 3 ,𝑥 ≥ 3
𝑥−3 =ƒ
+ − 𝑥 − 3 ,𝑥 < 3
Turning Point is at 𝑥 = 3
8 4 8 𝑦=− 𝑥−3
3 𝑦=𝑥−3
* 𝑥 − 3 𝑑𝑥 = * − 𝑥 − 3 𝑑𝑥 + * 𝑥 − 3 𝑑𝑥 Area = ×3×3
4
+ + 4 3

2 0 3
;# ;# 2 Area = ×2×2
= − + 3𝑥 + − 3𝑥 4
. / . 2

S .0 S
= − +9 − 0 + − 15 − −9 0 3 2
. . . 3 5

8
S 0 S S 62 6 6 62
= + − + = +2= * 𝑥 − 3 𝑑𝑥 = ×3×3 + ×2×2 =
. . . . . . . .
+
 6

Evaluate ! 3𝑥 − 1 𝑑𝑥
/
Solution:
6
3𝑥 − 1 , 𝑥 ≥ 2
3𝑥 − 1 = „ 6
− 3𝑥 − 1 , 𝑥 <
2

6
Plot the graph of 𝑦 = 𝑥 −
2
𝑌

3 4 4
3 Area = × ×
4 5 5
𝑦=− 𝑥− 𝑦= 𝑥−
3
5
3 3 3 5
Area = × ×
4 5 5
3
5
4
3 5
5
𝑋
0 3 3 4 1
5 5 5

, ,
% % # #
1 × × 0
* 3𝑥 − 1 𝑑𝑥 = 3 * 𝑥 − 𝑑𝑥 = 3 $ $
+ $ $
=
3 . . =
+ +
 6

Evaluate ! 𝑥 . + 𝑥 + 1 𝑑𝑥
/
Solution:

𝑥 . + 𝑥 + 1 is a quadratic expression.

𝑎 = 1 ,𝑏 = 1 ,𝑐 = 1

𝐷 = 1. − 4 = −3 < 0, 𝑎 > 0

⇒ 𝑥. + 𝑥 + 1 > 0

, ,
3
𝑥4 𝑥3 1 1 11
∴ 𝐼 = * 𝑥 + 𝑥 + 1 𝑑𝑥 = + +𝑥 = + +1 −0=
3 2 +
3 2 6
+
 .

Evaluate ! 𝑥 . − 3𝑥 + 2 𝑑𝑥
6
Solution: 3 3

𝐼=* 𝑥3 − 3𝑥 + 2 𝑑𝑥 = * 𝑥 − 1 𝑥 − 2 𝑑𝑥
, ,

+ − +
𝑥 − 3𝑥 + 2 is negative in the interval (1, 2)
.

1 2
3

∴ 𝐼 = − * 𝑥 3 − 3𝑥 + 2 𝑑𝑥
,

3
𝑥 4 3𝑥 3
=− − + 2𝑥
3 2 ,

P 6 2 6
𝐼=− −6+4− + −2 =
2 2 . =
 Session 5
Definite Integration of G.I.F &
fractional part function

Return To Top
 @
ln 𝑥
Evaluate ! 𝑥 𝑑𝑥
,
@

@ 6
ln 𝑥 𝑥 is +ve from < to 𝑒
Given, 𝐼 = * 𝑑𝑥
𝑥 𝑌
, 6
@ ln 𝑥 ∈ ln < , ln 𝑒 ≡ −1,1 ln 𝑥
@
ln 𝑥 𝑦=−
=* 𝑑𝑥 For turning point,ln 𝑥 = 0 𝑥
𝑥
, ln 𝑥
@ ⇒𝑥=1 𝑦=
, @ 𝑥
− ln 𝑥 ln 𝑥
=* 𝑑𝑥 + * 𝑑𝑥
𝑥 𝑥
6 𝑒
,
@
, 0 1 𝑋
<
6 1 𝑒
=− ln 𝑥 .
%
6
+ . ln 𝑥 .
6
For turning point, ln 𝑥 = 0
. 6

6 . 6 . 6 . . ∴ Turning Point is at 𝑥 = 1
= − . ln 1 − ln < + . ln 𝑒 − ln 1

6 6
= −. 0 − 1 + . 1 − 0

=1
 Q

Evaluate 3 ln 𝑥 𝑑𝑥
D
Q

𝑌
< 6 < 𝑦 = − ln 𝑥
𝐼 = ! ln 𝑥 𝑑𝑥 = ! − ln 𝑥 𝑑𝑥 + ! ln 𝑥 𝑑𝑥
6 6
𝑦 = ln 𝑥
6
< <

1 6 𝑒
= − 𝑥 ln 𝑥 − 𝑥 6 + 𝑥 ln 𝑥 − 𝑥 < 0 1 𝑋
1 <
<

1 1 For turning point, ln 𝑥 = 0

∴𝐼 =− 0−1 − −1 − + 𝑒−𝑒 − 0−1
𝑒 𝑒
∴ Turning Point is at 𝑥 = 1
2
=2−
𝑒
Integration of important function and area related to it.

! sin 𝑥 𝑑𝑥 Area = 2
/ 0 𝜋
-
.
-
! sin 2𝑥 𝑑𝑥 . Area = 1
/ 0
-
2
.
! sin 3𝑥 𝑑𝑥 Area = 2
-
/ 0
2
6

! sin 𝜋𝑥 𝑑𝑥 .
- Area =
-
/ 0 -

! sin 2𝜋𝑥 𝑑𝑥 . 6
- Area = =
.- -
/ 0 .-
 Integration of important function and area related to it.

-
Example: y = cos 𝑥
.

D E F G
∫CD cos F 𝑥 𝑑𝑥 = - =
E
.

−1 0 1

* *

− , = −1,1
#
*
#
*
# #

-
So, one loop of 𝑦 = cos . 𝑥 lies in between −1,1
 /

Evaluate ! 𝑥 sin 𝜋𝑥 𝑑𝑥
36

Solution: Given, As we know that, 𝑦 = sin 𝜋𝑥

/
−1 < 𝑥 < 0
𝐼 = ! 𝑥 sin 𝜋𝑥 𝑑𝑥
36
/
−𝜋 < 𝜋𝑥 < 0 −1 1
0
= ! 𝑥 sin 𝜋𝑥 𝑑𝑥
sin 𝜋𝑥 is −ve
36
+
3 l^m -; 0 cos 𝜋𝑥 𝑥 sin 𝜋𝑥 is +ve
= 𝑥 + !1 ⋅ 𝑑𝑥
- 36 𝜋
-,

/
− cos 𝜋𝑥 / sin 𝜋𝑥
= 𝑥 +
𝜋 36 𝜋. 36

cos 𝜋 sin 𝜋 1
= 0 − 0 − −1. − − . = 0− −
𝜋 𝜋 𝜋

1
=
𝜋
 N
In general ![𝑥]𝑑𝑥 = 0 + 1 + 2 + ⋯ (𝑛 − 1) N N36 , where [.] denotes G.I.F
=
.
/

𝑛 must be integer

N
In general ! 𝑥 𝑑𝑥 = N ; 𝑛 must be integer , where {.} denotes fractional part function
.
/
 8 0

Evaluate (𝑎) ![𝑥] 𝑑𝑥 (𝑏) ! 𝑥 𝑑𝑥

+ /

(where [.] denotes G.I.F and {.} denotes fractional part function)

Given, 8
0 0

𝐼 = ![𝑥] 𝑑𝑥 ! 𝑥 𝑑𝑥 = !(𝑥 − 𝑥 )𝑑𝑥

8 + / /
, 3 4 0 8
0 0
![𝑥] 𝑑𝑥 = ! 0 𝑑𝑥 + ! 1 𝑑𝑥 + ! 2 𝑑𝑥 + ! 3 𝑑𝑥 + ! 4 𝑑𝑥
+
= ! 𝑥 𝑑𝑥 − ! 𝑥 𝑑𝑥
+ , 3 4 0
/ /
= 0 +(𝑥)6. +2 𝑥 2
. +3 𝑥 1
2 +4 𝑥 0
1 ;#
0
0×(036)
= . /
− .
= 0 + 2 − 1 + 2 3 − 2 + 3 4 − 3 + 4(5 − 4)
.0 0×1
= −
= 0 + 1 + 2 + 3 + 4 = 10 . .

0
=.
 0

Evaluate ! 𝑥 𝑑𝑥 , where {.} denotes fractional part function

/
0
Solution:
Given, 𝐼 = ! 𝑥 𝑑𝑥
/
Method - 2 : By Area
𝑌

𝑋
0 1 2 3 4 5

6 6
Area bounded by one triangle is = 1×1 =
. .

Now, area bounded by all triangle is

0
6
! 𝑥 𝑑𝑥 = 5× = 0
. .
/
Key Takeaways

Some important results:

N

! 𝑥 𝑑𝑥 𝑛−1
𝑛 2
/ = =𝑛−1
N
𝑛
2
! 𝑥 𝑑𝑥
/

;
𝑥
! 𝑥 𝑑𝑥 =
2
/
66.D 66
11 66.D
! 𝑥 𝑑𝑥 = ! 𝑥 𝑑𝑥 = =
2 .
/ /

G
#
In general ! 𝑥 𝑑𝑥 = G
+
G
. .
/
 66.D

Evaluate ! 𝑥 𝑑𝑥
/
66.D
Given, 𝐼 = ! 𝑥 𝑑𝑥
/
𝑌

1
𝑦 = 0.7

1 2 3 4 5 6 7 8 9 10 11 11.7 𝑋
0
66.D

∴ ! 𝑥 𝑑𝑥 = 66 + /.D×/.D = 66.1S O𝑅 [66.D]

+
66.D × 66.D
. . . . .
/
G
#
In general ! 𝑥 𝑑𝑥 = G
+
G
. .
/
 6

Evaluate ! 3𝑥 𝑑𝑥
/

Solution: Given,
6 𝑥 →0−1
𝐼 = ! 3𝑥 𝑑𝑥
/ 3𝑥 → 0 − 3
6 .
2 2 6

= ! 3𝑥 𝑑𝑥 + ! 3𝑥 𝑑𝑥 + ! 3𝑥 𝑑𝑥 3𝑥 → 0 − 1 − 2 − 3
/ 6 .
2 2
1 2
. 6 . 𝑥 →0− − −1
=0+ 2
−2 +2 1−2 3 3

=1
 2
.
Evaluate ! 𝑥 . 𝑑𝑥 , where [.] denotes G.I.F
/

Solution: Given,
2
. 3
𝑥 →0−
𝐼 = ! 𝑥 . 𝑑𝑥 2
/
2 9
6 . . 𝑥. → 0 −
4
= ! 0 𝑑𝑥 + ! 1 𝑑𝑥 + ! 2 𝑑𝑥
/ 6 9
. 𝑥. → 0 − 1 − 2 −
2
4
=0+ 2−1 +2 − 2
.
3
=2− 2 𝑥 →0−1− 2−
2
 1

Evaluate ! 𝑥 𝑑𝑥 , where [.] denotes G.I.F

/

Solution: Given, 1

𝐼 = ! 𝑥 𝑑𝑥 𝑥 →0−4
/
𝑥 →0−2
6 1

= ! 0 𝑑𝑥 + ! 1 𝑑𝑥
𝑥 →0−1−2
/ 6

=0+ 4−1 𝑥 →0−1−4

=3
 6

Evaluate ! sin 𝑥 + 2𝑥 𝑑𝑥
/

Solution: Given,
6
𝐼 = ! sin 𝑥 + 2𝑥 𝑑𝑥 𝑥 →0−1
/
6 6
When 𝑥 → . − 1 When 𝑥 → 0 − . 2𝑥 → 0 − 2
𝑥 =0 𝑥 =0
Whereas 2𝑥 → 1 − 2 Whereas 2𝑥 → 0 − 1 2𝑥 → 0 − 1 − 2

2𝑥 = 1 2𝑥 = 0
1
𝑥 →0− −1
6
2
. 6

= ! sin 0 + 0 𝑑𝑥 + ! sin 1 𝑑𝑥
/ 6
.
6
= 0 + sin 1 × 𝑥 6
.
6
= sin 1
.
 6

Evaluate ! 3𝑥 − 7 𝑑𝑥 , where {.} denotes fractional part function

/

Solution: Given,
6 𝑥 →0−1
𝐼 = ! 3𝑥 − 7 𝑑𝑥
/ 3𝑥 → 0 − 3
6

= ! 3𝑥 − 3𝑥 − 7 𝑑𝑥 3𝑥 → 0 − 1 − 2 − 3
/
6
6 1 2
2; #
= .
− 7𝑥 − ! 3𝑥 𝑑𝑥 𝑥 →0− − −1
/ 3 3
/
6 .
2 6
2
2
= . − 7 − ! 0𝑑𝑥 − ! 1 𝑑𝑥 − ! 2 𝑑𝑥
/ 6 .
2 2
66 6 .
=− −0− −
. 2 2

62
=− .
 4
𝑑𝑥
Evaluate ! 𝑥 3 + 1 + 𝑥 3 − 2𝑥 𝑥 , where [.] denotes G.I.F
,

Solution: Given,
4
𝑑𝑥
𝐼= * 𝑥 →1−3
𝑥3 + 1 + 𝑥 3 − 2𝑥 𝑥
,
4
𝑑𝑥 𝑥 →1−2−3
=*
1 + 𝑥 3 + 𝑥 3 − 2𝑥 𝑥
,
4
𝑑𝑥
=* 3
1+ 𝑥− 𝑥
,
3 4
𝑑𝑥 𝑑𝑥
=* 3 +* 3
1+ 𝑥−1 1+ 𝑥−2
, 3

. 2
= tan36 𝑥 − 1 6 + tan36 𝑥 − 2 .

-
= 2 tan36 1 − 2 tan36 0 = .
 8

Evaluate ! sgn 𝑒 ; + 𝑒 3; + 2 + cos 𝑥 𝑑𝑥

+

Solution: Given,
8

𝐼 = ! sgn 𝑒 ; + 𝑒 3; + 2 + cos 𝑥 𝑑𝑥
+

+ + + +
0

= ! 1 𝑑𝑥
/

=5
 7
2
Evaluate ! 𝑑𝑥
𝑒.
+

Solution: Given,
7
2 𝑦 = 2𝑒 3; 𝑥 = ln 2
𝐼 = ! . 𝑑𝑥
𝑒
+
7
E= 3
2𝑒 3; = 1
-.
= ! 2×𝑒 -. 𝑑𝑥 + ! 2×𝑒 𝑑𝑥
⇒ 𝑒; = 2
+ E= 3

⇒ 𝑥 = ln 2
= ln 2 ×1 (0,2)
1
= ln 2
0 𝑦 = [2𝑒 3; ]
 Session 6
King’s Property

Return To Top
 <
𝑥 .; 𝑒 ;
Evaluate ! − ln 𝑥 𝑑𝑥 .
6 𝑒 𝑥

3 1 1
A − − 3
2 𝑒 2e

3 1
B 2
−𝑒− 3
2e

1 1
C 2
−𝑒− 3
e

1 1 1
D − + − 3
2 𝑒 2e
 <
𝑥 .; 𝑒 ;
Evaluate ! − ln 𝑥 𝑑𝑥
6 𝑒 𝑥

Solution : <
𝑥 .; 𝑒 ; ; ; 𝑥 𝑡
𝐼=! − ln 𝑥 𝑑𝑥 Let =𝑡
<
6 𝑒 𝑥
1 1/𝑒
;
6 ⇒ 𝑥. ln = ln𝑡
.
1 1 <
=! 𝑡 − ⋅ 𝑑𝑡 𝑒 1
6/< 𝑡 𝑡 6 6 ; 6
⇒ 𝑥. ;/q × q + ln <
= G 𝑑𝑡
6
1 ;
=! 𝑡− 𝑑𝑡 ⇒ 𝑡(1 + ln ) = 𝑑𝑡
6/< 𝑡. <

6
𝑡.
1
6 ⇒ ln 𝑥 𝑑𝑥 = G 𝑑𝑡
= +
2 𝑡 6/<

1 1 3 1
= − .+1−𝑒 = −𝑒− .
2 2e 2 2e
 -
Evaluate ! [2 sin 𝑥]𝑑𝑥 (where [⋅] is G.I.F function)
/

𝑌
Solution : From graph of [2 sin 𝑥], 2 sin 𝑥 = 1
D
- ⇒ sin 𝑥 =
For 0 < 𝑥 < , 2 sin 𝑥 = 0 !
=
g Cg
⇒ 𝑥 = @, @
- 0-
For ≤𝑥< , 2 sin 𝑥 = 1 2
= =
𝑦 = [2 sin 𝑥]
0- 1 𝑦 = 2 sin 𝑥
For =
< 𝑥 < 𝜋, 2 sin 𝑥 = 0

𝑋
0 𝜋 4𝜋 5𝜋
4𝜋 4𝜋 2𝜋 𝑥= 𝑥=
∴𝐼= ×1 = = 6 6 6
6 6 3
 60/
Evaluate ! tan36 𝑥 𝑑𝑥 where [.] is G.I.F function.
/

𝑌
Solution : From graph of [tan36 𝑥],

g 𝑦 = tanLD 𝑥
For 0 < 𝑥 < tan 1 , tan36 𝑥 = 0 !

1
For tan1 ≤ 𝑥 < 150, [tan36 𝑥] = 1
𝑋
0 tan 1 150
60/
∴𝐼=! tan36 𝑥 𝑑𝑥
/ g
−
!
KLB 6 60/
=! tan 36
𝑥 𝑑𝑥 + ! tan36 𝑥 𝑑𝑥 tanLD 𝑥 = 1
/ KLB 6
𝑥 = tan 1
= 0 + 150 − tan 1 ×1

= 150 − tan 1
 6/-
Evaluate ! cot 36 𝑥 𝑑𝑥 where [.] is G.I.F function.
/

From graph of [cot 36 𝑥], 𝑌

𝑦 = cot LD 𝑥
𝜋
For 0 < 𝑥 < co𝑡 1 , cot 36 𝑥 = 1

g
For cot1 ≤ 𝑥 < 10𝜋, [cot 36 𝑥] = 0
!
1
6/-
∴𝐼=! cot 36 𝑥 𝑑𝑥 𝑋
/
0 cot 1 10𝜋
l^K 6 6/-
36
=! cot 𝑥 𝑑𝑥 + ! cot 36 𝑥 𝑑𝑥 cot LD 𝑥 = 1
/ l^K 6

= cot 1 ×1 + 0 𝑥 = cot 1

= cot 1
 6/-
Evaluate ! cot 36 𝑥 𝑑𝑥 where [.] is G.I.F function.
/

A cot1 − tan1

B tan1

C cot1

D 1
 6
𝜋𝑥
Evaluate ! 𝑥 1 + cos + 1 𝑑𝑥 , where [.] is G.I.F function.
3. 2

𝑌
Solution :
6 -;
𝐼 = ∫3. 𝑥 1 + cos + 1 𝑑𝑥 𝜋𝑥
. 1 𝑦 = cos
2
- -
cos 𝑥 → 1 loop → − ,
. .
−2
𝑋
* *
0 1
-
cos . 𝑥 → 1 loop ≡ − ,
#
*
#
*
≡ −1, 1 −1
. .
2 units
- −1
For 𝑥 → −2 to −1, cos . 𝑥 → −1 to 0

-
⇒ 1 + cos . 𝑥 → 0 to 1 ⇒ 1 + cos - 𝑥 = 0
.
 6
𝜋𝑥
Evaluate ! 𝑥 1 + cos + 1 𝑑𝑥 , where [.] is G.I.F function.
3. 2

Solution : 𝑌
-
For 𝑥 → −1 to 0, cos . 𝑥 → 0 to 1 𝜋𝑥
1 𝑦 = cos
2
-
⇒ 1 + cos . 𝑥 → 1 to 2

-
−2
⇒ 1 + cos 𝑥 = 1 𝑋
. 0 1
−1
For 𝑥 → 0 to 1,
-
cos 𝑥 → 1 to 0 2 units
.
−1
-
⇒ 1 + cos 𝑥 → 2 to 1
.

-
⇒ 1 + cos 𝑥 = 1
.
 6
𝜋𝑥
Evaluate ! 𝑥 1 + cos + 1 𝑑𝑥 , where [.] is G.I.F function.
3. 2

Solution : 36 6 𝑌
𝜋𝑥 𝜋𝑥
𝐼=! 𝑥 1 + cos + 1 𝑑𝑥 + ! 𝑥 1 + cos + 1 𝑑𝑥
2 2 𝜋𝑥
3. 36
1 𝑦 = cos
2
36 6
⇒𝐼=! 𝑥×0 + 1 𝑑𝑥 + ! 𝑥×1 + 1 𝑑𝑥
3. 36
−2
36 6 𝑋
0 1
⇒ 𝐼 = ! 1 ⋅ 𝑑𝑥 + ! 𝑥 + 1 𝑑𝑥 −1
3. 36
2 units
/ 6
⇒𝐼= 𝑥 36
+ ! −1 + 1 𝑑𝑥 + ! 0 + 1 𝑑𝑥 ⋯ −1
3.
36 /
(G.I.F is discontinuous at integer)
⇒ 𝐼 = −1 − −2 + 0 + 1 − 0
 D
1
Evaluate !6 𝑥 + 𝑑𝑥 where [.] is G.I.F function.
2
.

Solution : D
1
𝐼 = !6 𝑥 + 𝑑𝑥
2
.

6
D
. 6
∴𝐼=! 𝑡 𝑑𝑡 Let 𝑥 + . = 𝑡 ⇒ 𝑑𝑥 = 𝑑𝑡 𝑥 𝑡
6
6
6
1
. 2 D .
.
⇒ 𝐼 = ! 1𝑑𝑥 + ! 2𝑑𝑥 ⋯ + ! 7𝑑𝑥 6
6 . D 7 7
.

6
D
. 2 1 .
= 𝑥 6 +2 𝑥 . +3 𝑥 2 + ⋯7 𝑥 D

D 1S
= 1 + 2 + 3 + ⋯+ 6 + . = .
 Key Takeaways

Property 4
, ,
Ø ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑎 + 𝑏 − 𝑥 𝑑𝑥 → King Property
+ +

Proof:

, +
! 𝑓 𝑥 𝑑𝑥 = − ! 𝑓 𝑎 + 𝑏 − 𝑡 𝑑𝑡 Let 𝑥 = 𝑎 + 𝑏 − 𝑡 ⇒ 𝑑𝑥 = −𝑑𝑡
+ ,

𝑥 𝑡
,
= ! 𝑓 𝑎 + 𝑏 − 𝑡 𝑑𝑡 ⋯ Property 2 𝑎 𝑏
+

𝑏 𝑎
,
= ! 𝑓 𝑎 + 𝑏 − 𝑥 𝑑𝑥 ⋯ Property 1
+
 King Property (Auxiliary)

+ +
Ø ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑎 − 𝑥 𝑑𝑥
/ /

Except in . , . , sgn , defined function, see if king property is helping or not.

Example: Example:

. -/.
𝐼 = ! 𝑓 𝑥 𝑑𝑥 𝐼=! 𝑓 𝑥 𝑑𝑥
/ /

. -/.
𝜋
⇒ 𝐼 = ! 𝑓 2 − 𝑥 𝑑𝑥 ⇒𝐼=! 𝑓 − 𝑥 𝑑𝑥
/ / 2
 /
3
Evaluate * log 4 + 3 sin 𝑥
𝑑𝑥
4 + 3 cos 𝑥
+

Solution : -
.
Let 𝐼 = 4 + 3 sin 𝑥
! log 𝑑𝑥 ⋯ 𝑖 = Using Property 4,
4 + 3 cos 𝑥
/

-
. 𝜋
4 + 3 sin 2 − 𝑥 𝜋
⇒ 𝐼 = ! log 𝜋 𝑑𝑥 𝑥→ −𝑥
4 + 3 cos − 𝑥 2
/ 2

-
.
4 + 3 cos 𝑥
⇒ 𝐼 = ! log 𝑑𝑥 ⋯ 𝑖𝑖 =
4 + 3 sin 𝑥
/

-
.
4 + 3 sin 𝑥 4 + 3 cos 𝑥
𝑖 + 𝑖𝑖 ⇒ 2𝐼 = ! log + log 𝑑𝑥
4 + 3 cos 𝑥 4 + 3 sin 𝑥
/
 /
3
Evaluate * log 4 + 3 sin 𝑥
𝑑𝑥
4 + 3 cos 𝑥
+

Solution :
-
.
4 + 3 sin 𝑥 4 + 3 sin 𝑥
⇒ 2𝐼 = ! log − log 𝑑𝑥
4 + 3 cos 𝑥 4 + 3 cos 𝑥
/

⇒ 2𝐼 = 0

⇒𝐼=0
 -
1
Evaluate ! 𝑑𝑥
1 + 3l^m ;
/

Solution :
-
1
Let 𝐼 = ! 𝑑𝑥
1 + 3l^m ;
/

-
1
=! 𝑑𝑥 Using Property 4, 𝑥 → 𝜋 − 𝑥
1 + 3l^m(-3;)
/

-
1
=! 𝑑𝑥
1 + 33 l^m ;
/

-
1
=! 𝑑𝑥
1
/ 1 + 3l^m ;
 -
1
Evaluate ! 𝑑𝑥
1 + 3l^m ;
/

Solution :
-
3l^m ;
⇒𝐼=! 𝑑𝑥 ⋯ 𝑖𝑖
1 + 3l^m ;
/

Adding 𝑖 and 𝑖𝑖

-
1 + 3l^m ; -
2𝐼 = ! 𝑑𝑥 = 𝑥 /
1 + 3l^m ;
/

⇒ 2𝐼 = 𝜋

-
⇒𝐼=
.
 /
3
Evaluate * sin 8𝑥 ⋅ log cot 𝑥 𝑑𝑥 ?
cot 2𝑥
+

Solution : -
.
Let 𝐼 = ! sin 8𝑥 ⋅ log cot 𝑥 𝑑𝑥 ? ⋯ 𝑖 =Using Property 4, 𝑥 → - − 𝑥
cot 2𝑥 .
/

-
. sin 𝜋 𝜋
8 −𝑥 ⋅ log cot −𝑥
⇒𝐼= ! 2 2
𝜋 𝑑𝑥 ?
/
cot 2 2 − 𝑥

-
.
sin 4𝜋 − 8𝑥 ⋅ log tan 𝑥
=! 𝑑𝑥 ?
cot 𝜋 − 2𝑥
/
 /
3
Evaluate * sin 8𝑥 ⋅ log cot 𝑥 𝑑𝑥 ?
cot 2𝑥
+

Solution :
-
.
sin 8𝑥 ⋅ log tan 𝑥 4𝜋 − 8𝑥 → 𝐼𝑉 G7 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡 → 𝑠𝑖𝑛𝑒 − 𝑣𝑒
⇒𝐼=! 𝑑𝑥 ? ⋯ 𝑖𝑖 =
cot 2𝑥 𝜋 − 2𝑥 → 𝐼𝐼N9 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡 → 𝑐𝑜𝑡 − 𝑣𝑒
/

- -
. .
sin 8𝑥 ⋅ log cot 𝑥 sin 8𝑥 ⋅ log tan 𝑥
𝑖 + 𝑖𝑖 ⇒ 2𝐼 = ! 𝑑𝑥 + ! 𝑑𝑥 ?
cot 2𝑥 cot 2𝑥
/ /

-
.
sin 8𝑥
⇒ 2𝐼 = ! log cot 𝑥 + log tan 𝑥 𝑑𝑥
cot 2𝑥
/

-
.
sin 8𝑥
⇒ 2𝐼 = ! log 1 𝑑𝑥 ⇒ 2𝐼 = 0 ⇒ 𝐼 = 0
cot 2𝑥
/
 -

Evaluate ! 𝑒 mrB
# ;
⋅ tan. 𝑥 ⋅ cos 2 2𝑛 + 1 𝑥 𝑑𝑥 , 𝑛 ∈ 𝐼?
/

Solution : -
Let 𝐼 = ! 𝑒 mrB
# ;
⋅ tan. 𝑥 ⋅ cos 2 2𝑛 + 1 𝑥 𝑑𝑥 ? ⋯ 𝑖 Using Property 4, 𝑥 → 𝜋 − 𝑥
/
-
#
⇒ 𝐼 = ! 𝑒 mrB -3;
⋅ tan. 𝜋 − 𝑥 ⋅ (cos 2 2𝑛 + 1 𝜋 − 𝑥 ) 𝑑𝑥 ?
/

-
#
⇒ 𝐼 = ! 𝑒 mrB ;
⋅ tan. 𝑥 ⋅ cos( 2𝑛 + 1 𝜋 − 2𝑛 + 1 𝑥) 2
𝑑𝑥 ?
/
-
#
⇒ 𝐼 = − ! 𝑒 mrB ;
⋅ tan. 𝑥 ⋅ cos 2𝑛 + 1 𝑥 2
𝑑𝑥 ?
/
⋯ cos 2𝑛 + 1 𝜋 − 𝑥 = − cos 𝑥 , 𝑛 ∈ 𝐼

⇒ 𝐼 = −𝐼? ⇒ 2𝐼 = 0 ⇒ 𝐼 = 0
 /
3 /
ijk N 𝜋 𝜋
If 𝑓 𝑥 = N
∀ 𝑥 ∈ 0, 𝜋 . Prove that 2
*𝑓 𝑥 ⋅𝑓
2
− 𝑥 𝑑𝑥 = * 𝑓 𝑥 𝑑𝑥 ?
+ +

Solution :
mrB ; - l^m ;
Given, 𝑓 𝑥 = ∀ 𝑥 ∈ 0, 𝜋 Also, 𝑓 .
−𝑥 = *
; 3;
#

/ /
3 3
𝜋 sin 𝑥 cos 𝑥 𝜋 sin 2𝑥
L.H.S = * ⋅ 𝜋 𝑑𝑥 = * 𝑑𝑥 ?
2 𝑥 − 𝑥 2𝑥 𝜋 − 2𝑥
+ 2 +

/
3
𝜋 − 2𝑥 + 2𝑥 sin 2𝑥
=* 𝑑𝑥 ? (adding and subtracting 2𝑥)
2𝑥 𝜋 − 2𝑥
+

/ /
3 3
𝜋 − 2𝑥 sin 2𝑥 2𝑥 sin 2𝑥
=* 𝑑𝑥 + * 𝑑𝑥 ?
2𝑥 𝜋 − 2𝑥 2𝑥 𝜋 − 2𝑥
+ +
 /
3 /
ijk N 𝜋 𝜋
If 𝑓 𝑥 = N
∀ 𝑥 ∈ 0, 𝜋 . Prove that 2
*𝑓 𝑥 ⋅𝑓
2
− 𝑥 𝑑𝑥 = * 𝑓 𝑥 𝑑𝑥 ?
+ +

Solution : / /
3 3
sin 2𝑥 sin 2𝑥 𝜋
=* 𝑑𝑥 + * 𝑑𝑥 ? 𝑥→ −𝑥
2𝑥 𝜋 − 2𝑥 2
+ +

/ /
3 𝜋3
sin 2𝑥 sin 2 2 − 𝑥
=* 𝑑𝑥 + * 𝑑𝑥 ?
2𝑥 𝜋
+ + 𝜋−2 2−𝑥

/ / /
3 3 3
sin 2𝑥 sin 2𝑥 sin 2𝑥
=* 𝑑𝑥 + * 𝑑𝑥 ?= 2 * 𝑑𝑥 ? Let 2𝑥 = 𝑡, 2𝑑𝑥 = 𝑑𝑡
2𝑥 2𝑥 2𝑥
+ + +
𝑥 𝑡
/ /
sin 𝑡 sin 𝑥 0 0
== * 𝑑𝑡==
? * 𝑑𝑥 ? = R.H.S
𝑡 𝑥
+ + /
𝜋
3
 -/.
cos 𝑥 − sin 𝑥
Evaluate ! 𝑑𝑥 ?
1 + sin 𝑥 cos 𝑥
/

-
A .

-
B 1

C 0

D 1
 -/.
cos 𝑥 − sin 𝑥
Evaluate ! 𝑑𝑥 ?
1 + sin 𝑥 cos 𝑥
/

-
A .

-
B 1

C 0

D 1
 -/.
cos 𝑥 − sin 𝑥
Evaluate ! 𝑑𝑥 ?
1 + sin 𝑥 cos 𝑥
/

Solution : -/.
cos 𝑥 − sin 𝑥
Let 𝐼 = ! 𝑑𝑥 ? ⋯ 𝑖 -
Using Property 4, 𝑥 → − 𝑥
1 + sin 𝑥 cos 𝑥 .
/

-/.
sin 𝑥 − cos 𝑥 ⋯ 𝑖𝑖
⇒𝐼=! 𝑑𝑥 ?
1 + cos 𝑥 sin 𝑥
/

-/. -/.
cos 𝑥 − sin 𝑥 sin 𝑥 − cos 𝑥
𝑖 + 𝑖𝑖 ⇒ 2𝐼 = ! 𝑑𝑥 + ! 𝑑𝑥 ?
1 + sin 𝑥 cos 𝑥 1 + cos 𝑥 sin 𝑥
/ /

-/.

⇒ 2𝐼 = ! 0 𝑑𝑥 ?
/

⇒ 2𝐼 = 0 ⇒ 𝐼 = 0
 -/.
𝑎 sin 𝑥 + 𝑏 cos 𝑥
Evaluate ! 𝑑𝑥 ?
sin 𝑥 + cos 𝑥
/

Solution : -/.
𝑎 sin 𝑥 + 𝑏 cos 𝑥
Let 𝐼 = ! 𝑑𝑥 ?⋯ 𝑖
sin 𝑥 + cos 𝑥
/
-/.
𝑎 cos 𝑥 + 𝑏 sin 𝑥
⇒𝐼=! 𝑑𝑥 ? ⋯ 𝑖𝑖 Using Property 4, 𝑥 → - − 𝑥
cos 𝑥 + sin 𝑥 .
/

-/.
𝑎 sin 𝑥 + cos 𝑥 + 𝑏 sin 𝑥 + cos 𝑥
𝑖 + 𝑖𝑖 ⇒ 2𝐼 = ! 𝑑𝑥 ?
sin 𝑥 + cos 𝑥
/

-/.
sin 𝑥 + cos 𝑥 (𝑎 + 𝑏)
2𝐼 = ! 𝑑𝑥 ?
(sin 𝑥 + cos 𝑥)
/

𝜋 𝜋
⇒ 2𝐼 = 𝑎 + 𝑏 ⇒𝐼 = 𝑎+𝑏
2 4
 -/.
sinN 𝑥
Evaluate ! 𝑑𝑥 ?
sinN 𝑥 + cos N 𝑥
/

Solution : -/.
Let 𝐼 = sinN 𝑥 -
! N N
𝑑𝑥 ? ⋯ 𝑖 Using Property 4, 𝑥 → . − 𝑥
sin 𝑥 + cos 𝑥
/

-/.
cos N 𝑥
⇒𝐼=! 𝑑𝑥 ? ⋯ 𝑖𝑖
cos N 𝑥 + sinN 𝑥
/

-/.
sinN 𝑥 + cos N 𝑥
𝑖 + 𝑖𝑖 ⇒ 2𝐼 = ! 𝑑𝑥 ?
sinN 𝑥 + cos N 𝑥
/

-/.
𝜋/2
⇒ 2𝐼 = ! 1 𝑑𝑥 = 𝑥
/ 0

- E
⇒ 2𝐼 = .
⇒𝐼=
G
 Session 7
Application of King’s
Property

Return To Top
 -
.
sin2 𝑥
The value of ! 𝑑𝑥 is :
sin 𝑥 + cos 𝑥
/
JEE Main April 2019
𝜋−2 𝜋−1
A B
8 2

𝜋−1 𝜋−2
C D
4 4

Solution : -
.
sin2 𝑥
Let 𝐼 = ! 𝑑𝑥 ⋯ (𝑖)
sin 𝑥 + cos 𝑥
/
-
. , ,
cos 2 𝑥 ⋯ ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑎 + 𝑏 − 𝑥 𝑑𝑥
⇒𝐼=! 𝑑𝑥 ⋯ (𝑖𝑖)
sin 𝑥 + cos 𝑥 + +
/
-
.
sin2 𝑥 + cos 2 𝑥
𝑖 + 𝑖𝑖 : ⇒ 2𝐼 = ! 𝑑𝑥
sin 𝑥 + cos 𝑥
/
 -
.
sin2 𝑥
The value of ! 𝑑𝑥 is :
sin 𝑥 + cos 𝑥
/
JEE Main April 2019
-
Solution : .
⇒ 2𝐼 = ! ( sin. 𝑥 + cos . 𝑥 − sin 𝑥 ⋅ cos 𝑥) 𝑑𝑥
/
⋯ 𝑎2 + 𝑏2 = 𝑎 + 𝑏 𝑎. + 𝑏. − 𝑎𝑏
-
.
sin 2𝑥
⇒ 2𝐼 = ! 1 − 𝑑𝑥
2
/
-s
1 cos 2𝑥 .
I= 𝑥+
2 4 /

𝜋−1
⇒𝐼=
4

∴ Option 𝐶 is correct option.

Let 𝑓 and 𝑔 be continuous functions on 0, 𝑎 such that 𝑓 𝑥 = 𝑓(𝑎 − 𝑥)
+

and 𝑔 𝑥 + 𝑔 𝑎 − 𝑥 = 4 , then ! 𝑓 𝑥 𝑔 𝑥 𝑑𝑥 is equal to :

/
JEE Main Jan 2019

+ +
A 4 ! 𝑓 𝑥 𝑑𝑥 B ! 𝑓 𝑥 𝑑𝑥
/ /

+ +
C 2 ! 𝑓 𝑥 𝑑𝑥 D −3 ! 𝑓 𝑥 𝑑𝑥
/ /
 Let 𝑓 and 𝑔 be continuous functions on 0, 𝑎 such that 𝑓 𝑥 = 𝑓(𝑎 − 𝑥)
+

and 𝑔 𝑥 + 𝑔 𝑎 − 𝑥 = 4 , then ! 𝑓 𝑥 𝑔 𝑥 𝑑𝑥 is equal to :

/
JEE Main Jan 2019
+
Solution :
Let 𝐼 = ! 𝑓 𝑥 ⋅ 𝑔 𝑥 𝑑𝑥 ⋯ (𝑖)
/
+ , ,
⋯ (𝑖𝑖) ⋯ ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑎 + 𝑏 − 𝑥 𝑑𝑥
⇒ 𝐼 = ! 𝑓 𝑎 − 𝑥 ⋅ 𝑔 𝑎 − 𝑥 𝑑𝑥
+ +
/
+

𝑖 + 𝑖𝑖 : ⇒ 2𝐼 = ! 𝑓 𝑥 𝑔 𝑥 + 𝑓 𝑎 − 𝑥 𝑔 𝑎 − 𝑥 𝑑𝑥
+ /
⋯ 𝑓 𝑥 = 𝑓(𝑎 − 𝑥)
⇒ 2𝐼 = ! 𝑓(𝑥) 𝑔(𝑥) + 𝑔 𝑎 − 𝑥 𝑑𝑥
/
+ +

= 4 ! 𝑓(𝑥) 𝑑𝑥 ⇒ 𝐼 = 2 ! 𝑓(𝑥) 𝑑𝑥
/ /

∴ Option 𝐶 is correct option.

Let 𝑓 and 𝑔 be continuous functions on 0, 𝑎 such that 𝑓 𝑥 = 𝑓(𝑎 − 𝑥)
+

and 𝑔 𝑥 + 𝑔 𝑎 − 𝑥 = 4 , then ! 𝑓 𝑥 𝑔 𝑥 𝑑𝑥 is equal to :

/
JEE Main Jan 2019

+ +
A 4 ! 𝑓 𝑥 𝑑𝑥 B ! 𝑓 𝑥 𝑑𝑥
/ /

+ +
C
A 2 ! 𝑓 𝑥 𝑑𝑥 D −3 ! 𝑓 𝑥 𝑑𝑥
/ /
 -

! cot 𝑥 𝑑𝑥 , where [.] denotes G.I.F. , is equal to :

/ IIT JEE 2009
Solution :
-

Let 𝐼 = ! cot 𝑥 𝑑𝑥 ⋯ (𝑖)

/
-

⇒ 𝐼 = ! − cot 𝑥 𝑑𝑥 ⋯ (𝑖𝑖) , ,
/ ⋯ ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑎 + 𝑏 − 𝑥 𝑑𝑥
+ +
-

𝑖 + 𝑖𝑖 : 2𝐼 = ! cot 𝑥 + − cot 𝑥 𝑑𝑥
/
- 0, 𝑥 ∈ 𝕀
= ! −1. 𝑑𝑥 ⋯ 𝑥 + −𝑥 =
/ −1, 𝑥 ∉ 𝕀
⇒ 2𝐼 = −𝜋
-
⇒ 𝐼 = −.
-

! cot 𝑥 𝑑𝑥 , where [.] denotes G.I.F. , is equal to :

/ IIT JEE 2009

A 1 B −1

- -
A
CC −. D .
 -
1
Evaluate ! log (1 + tan 𝜃) 𝑑𝜃
/
-
Solution : 1

𝐼 = ! log (1 + tan 𝜃) 𝑑𝜃 ⋯ (𝑖)

/
-
1
𝜋
⇒ 𝐼 = ! log 1 + tan −𝜃 𝑑𝜃
4
/ , ,
⋯ ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑎 + 𝑏 − 𝑥 𝑑𝑥
+ +
-
1
1 − tan 𝜃
⇒ 𝐼 = ! log 1 + 𝑑𝜃
1 + tan 𝜃
/
-
1
1 + tan 𝜃 + 1 − tan 𝜃
= ! log 𝑑𝜃
1 + tan 𝜃
/
 -
1
Evaluate ! log (1 + tan 𝜃) 𝑑𝜃
/
-
Solution : 1
2
= ! log 𝑑𝜃
1 + tan 𝜃
/
- -
1 1

⇒ 𝐼 = ! log 2 𝑑𝜃− ! log 1 + tan 𝜃 𝑑𝜃 ⋯ (𝑖𝑖)

/ /

On adding 𝑖 𝑎𝑛𝑑 (𝑖𝑖),

-
1
2𝐼 = log 2 ! 𝑑𝜃
/
-
⇒ 2𝐼 = log 2 (𝜃)/1

𝜋 𝜋
⇒ 2𝐼 = log 2 ⇒ 𝐼 = log 2
4 8
 g
!
Evaluate 3 log tan 𝜃 𝑑𝜃
J
- -
Solution : . .
𝐼 = ! log tan 𝜃 𝑑𝜃 ⋯ (𝑖) ⇒ 𝐼 = ! log cot 𝜃 𝑑𝜃 ⋯ (𝑖𝑖)
/ / , ,
⋯ ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑎 + 𝑏 − 𝑥 𝑑𝑥
- + +
.

𝑖 + 𝑖𝑖 ⇒ 2𝐼 = ! log (tan 𝜃 cot 𝜃) 𝑑𝜃

Point to remember
/

⇒ 2𝐼 = 0 ⇒ 𝐼 = 0 - -
. .
! log tan 𝜃 𝑑𝜃 = ! log cot 𝜃 𝑑𝜃 = 0
/ /
 64 0
.
𝑥. + 1 1
Evaluate ! ln 1 + 𝑥 − 𝑑𝑥
𝑥1 − 𝑥. + 1 𝑥
6

64 0
Solution : .
𝑥. + 1 1
Let 𝐼 = ! ln 1 + 𝑥 − 𝑑𝑥
𝑥1 − 𝑥. + 1 𝑥
6
64 0
. 1
1+ 1
= ! 𝑥. ln 1 + 𝑥 − 𝑑𝑥
1 𝑥 ⋯ Dividing by 𝑥 .
6 𝑥. −1+ .
𝑥
6 1
Let 𝑥 − ; = tan 𝜃 ⇒ 1 + 𝑑𝑥 = sec . 𝜃 𝑑𝜃
𝑥.

𝑥 𝜃

1 0

1+ 5 𝜋
2 4
 64 0
.
𝑥. + 1 1
Evaluate ! ln 1 + 𝑥 − 𝑑𝑥
𝑥1 − 𝑥. + 1 𝑥
6

-
Solution : 1
sec . 𝜃
⇒𝐼=! ln 1 + tan 𝜃 𝑑𝜃
1 + tan. 𝜃
/
-
1
= ! ln 1 + tan 𝜃 𝑑𝜃
/
𝜋
⇒𝐼= ln 2
8
 J
𝑥
Evaluate ! 𝑑𝑥
1 + 𝑥 1 + 𝑥.
/

Solution :
J
𝑥
Let 𝐼 =! 𝑑𝑥
1 + 𝑥 1 + 𝑥.
/

Let 𝑥 = tan 𝜃 ⇒ 𝑑𝑥 = sec . 𝜃 𝑑𝜃

𝑥 𝜃

0 0

∞ 𝜋
2
- -
. .
tan 𝜃 ⋅ sec . 𝜃 𝑑𝜃 tan 𝜃 𝑑𝜃
𝐼=! = !
1 + tan 𝜃 1 + tan. 𝜃 1 + tan 𝜃
/ /
- -
. .
sin 𝜃 𝑑𝜃 - sinU 𝑥 𝜋
⇒𝐼=! =1 ⋯! U 𝑑𝑥 =
sin 𝜃 + cos 𝜃 U
sin 𝑥 + cos 𝑥 4
/ /
Results :
,
𝐹 𝑥 𝑏−𝑎
Ø ! 𝑑𝑥 =
𝐹 𝑥 +𝐹 𝑎+𝑏−𝑥 2
+

Proof:
,
𝐹 𝑥
𝐼= ! 𝑑𝑥 ⋯ (𝑖) Let 𝑥 → 𝑎 + 𝑏 − 𝑥
𝐹 𝑥 +𝐹 𝑎+𝑏−𝑥
+

,
𝐹 𝑎+𝑏−𝑥
=! 𝑑𝑥 ⋯ (𝑖𝑖)
𝐹 𝑎+𝑏−𝑥 +𝐹 𝑥
+

,
𝐹 𝑥 +𝐹 𝑎+𝑏−𝑥
Adding (𝑖) and (𝑖𝑖): 2𝐼 = ! 𝑑𝑥 ⇒ 2𝐼 = (𝑥),+
𝐹 𝑎+𝑏−𝑥 +𝐹 𝑥
+

,3+
⇒ 2𝐼 = 𝑏 − 𝑎 ⇒ 𝐼 =
.
 1 1 1
log 𝑥 𝑥 𝑒;
Evaluate (𝑖) ! 𝑑𝑥 (𝑖𝑖) ! 𝑑𝑥 (𝑖𝑖𝑖) ! 𝑑𝑥
log 𝑥 + log(6 − 𝑥) 𝑥+ 6−𝑥 𝑒 ; + 𝑒 =3;
. . .

Solution :
1
log 𝑥
Let I=! 𝑑𝑥
log 𝑥 + log(4 + 2 − 𝑥)
.

,
4−2 𝑓 𝑥 𝑏−𝑎
= ⋯! 𝑑𝑥 =
2 𝑓 𝑥 +𝑓 𝑎+𝑏−𝑥 2
+

=1
 1 1 1
log 𝑥 𝑥 𝑒;
Evaluate (𝑖) ! 𝑑𝑥 (𝑖𝑖) ! 𝑑𝑥 (𝑖𝑖𝑖) ! 𝑑𝑥
log 𝑥 + log(6 − 𝑥) 𝑥+ 6−𝑥 𝑒 ; + 𝑒 =3;
. . .

Solution :
1
𝑥
I=! 𝑑𝑥
𝑥+ 4+2−𝑥
.
,
4−2 𝑓 𝑥 𝑏−𝑎
= ⋯! 𝑑𝑥 =
2 𝑓 𝑥 +𝑓 𝑎+𝑏−𝑥 2
+

=1
 1 1 1
log 𝑥 𝑥 𝑒;
Evaluate (𝑖) ! 𝑑𝑥 (𝑖𝑖) ! 𝑑𝑥 (𝑖𝑖𝑖) ! 𝑑𝑥
log 𝑥 + log(6 − 𝑥) 𝑥+ 6−𝑥 𝑒 ; + 𝑒 =3;
. . .

Solution :
1
𝑒;
Let 𝐼 = ! ; 𝑑𝑥
𝑒 + 𝑒 14.3;
.
,
4−2 𝑓 𝑥 𝑏−𝑎
= ⋯! 𝑑𝑥 =
2 𝑓 𝑥 +𝑓 𝑎+𝑏−𝑥 2
+

=1
 1
log 𝑥 .
The integral ! 𝑑𝑥 is equal to :
log 𝑥 . + log 36 − 12𝑥 + 𝑥 .
.
JEE MAIN 2015
Solution :
1
log 𝑥 .
Let 𝐼 = ! 𝑑𝑥
log 𝑥 . + log 36 − 12𝑥 + 𝑥 .
.
1 1
log 𝑥 . 2log |𝑥|
=! 𝑑𝑥 = ! 𝑑𝑥
log 𝑥 . + log 𝑥 − 6 . 2log |𝑥| + 2 log |𝑥 − 6|
. .
1 ,
log 𝑥 𝑓 𝑥 𝑏−𝑎
⇒𝐼=! 𝑑𝑥 ⋯! 𝑑𝑥 =
log 𝑥 + log(6 − 𝑥) 𝑓 𝑥 +𝑓 𝑎+𝑏−𝑥 2
. +

4−2
⇒𝐼=
2
⇒𝐼=1
 1
log 𝑥 .
The integral ! 𝑑𝑥 is equal to :
log 𝑥 . + log 36 − 12𝑥 + 𝑥 .
.
JEE MAIN 2015

A 2 B 4

C 1 D 6
 -
𝑥 sin 𝑥
Evaluate! 𝑑𝑥
1 + cos . 𝑥
/

Solution : -
𝑥 sin 𝑥
Let 𝐼 = ! 𝑑𝑥 ⋯ 𝑖
1 + cos . 𝑥
/
-
𝜋 − 𝑥 sin 𝜋 − 𝑥 , ,
⇒ 𝐼 =! 𝑑𝑥 ⋯ 𝑖𝑖 ⋯ ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑎 + 𝑏 − 𝑥 𝑑𝑥
1 + cos . 𝜋 − 𝑥 + +
/

𝑖 + 𝑖𝑖 ⇒

- -
𝑥 sin 𝑥 𝜋 − 𝑥 sin 𝑥
2𝐼 = ! 𝑑𝑥 + ! 𝑑𝑥
1 + cos . 𝑥 1 + cos . 𝑥
/ /

-
𝜋 sin 𝑥
⇒ 2𝐼 = ! 𝑑𝑥
1 + cos . 𝑥
/
 -
𝑥 sin 𝑥
Evaluate ! 𝑑𝑥
1 + cos . 𝑥
/

Solution : Let cos 𝑥 = 𝑡

𝑥 𝑡

− sin 𝑥 𝑑𝑥 = 𝑑𝑡 0 1
36
𝜋 −1
𝜋 𝑑𝑡
⇒𝐼=− !
2 1 + 𝑡.
6

𝜋 −1
=− tan36 𝑡
2 1

𝜋 𝜋 𝜋
=− − −
2 4 4

𝜋.
=
4
 DIRECT METHOD for removal of 𝑥 in King’s Property

Solution :

When problem contains function such that 𝑓 𝑎 − 𝑥 = 𝑓 𝑥 (Applied King’s

upper limit + lower limit
Property) , then put that 𝑥 which we want to remove =
2
OR

When function remains as it is except 𝑥 that we want to remove on doing king’s approach.
 -
.
𝑥 sin 𝑥 cos 𝑥 𝑑𝑥
Evaluate !
sin1 𝑥 + cos 1 𝑥
/

Solution :
-
.
𝑥 sin 𝑥 cos 𝑥 𝑑𝑥
Let 𝐼 = !
sin1 𝑥 + cos 1 𝑥
/

-
. *
𝜋 sin 𝑥 cos 𝑥 𝑑𝑥 4/ -
⇒𝐼= ! 1 ⋯ ∵𝑓 𝑎−𝑥 =𝑓 𝑥 , 𝑥 = #
=
4 sin 𝑥 + cos 1 𝑥 . 1
/

-
.
𝜋 tan 𝑥 sec . 𝑥 𝑑𝑥
⇒𝐼= ! ⋯ dividing by cos 1 𝑥
4 1 + tan. 𝑥 .
/
 -
.
𝑥 sin 𝑥 cos 𝑥 𝑑𝑥
Evaluate !
sin1 𝑥 + cos 1 𝑥
/

Solution : Let tan. 𝑥 = 𝑡

J
𝜋 𝑑𝑡
∴𝐼= ! .
⇒ 2 tan 𝑥 sec . 𝑥 𝑑𝑥 = 𝑑𝑡
4×2 1 + 𝑡
/
𝑥 𝑡

𝜋 J 0 0
= tan36 𝑡 /
8
𝜋 ∞
2
𝜋 𝜋
= −0
8 2

𝜋.
=
16
 -
1
𝑥 ⋅ 𝑑𝑥 𝜋 ln 2
If 𝐼 = ! = then 𝑏 =?
1 + sin 2𝑥 + cos 2𝑥 𝑏
/
-
Solution : 1
𝑥 ⋅ 𝑑𝑥 𝜋 ln 2
Given 𝐼 = ! =
1 + sin 2𝑥 + cos 2𝑥 𝑏
/
-
1 *
𝜋 𝑑𝑥 4/ -
L.H.S = ! ⋯ ∵𝑓 𝑎−𝑥 =𝑓 𝑥 , 𝑥 = -
=
8 1 + sin 2𝑥 + cos 2𝑥 . P
/
-
1
𝜋 𝑑𝑥
= !
8 2 cos . 𝑥 + 2 sin𝑥 cos 𝑥
/
-
1
𝜋 𝑑𝑥 ⋯ taking cos . 𝑥 common
⇒𝐼= !
8 2 cos . 𝑥 (1 + tan 𝑥)
/
-
1
𝜋 sec . 𝑥 𝑑𝑥
𝐼= !
8 2 (1 + tan 𝑥)
/
 -
1
𝑥 ⋅ 𝑑𝑥 𝜋 ln 2
If 𝐼 = ! = then 𝑏 =?
1 + sin 2𝑥 + cos 2𝑥 𝑏
/

Solution : Let 1 + tan 𝑥 = 𝑡⇒ sec . 𝑥 𝑑𝑥 = 𝑑𝑡

𝑥 𝑡

0 1

𝜋 2
4

.
𝜋 𝑑𝑡 - .
∴𝐼= ! = 6= ln 𝑡 6
16 𝑡
6
-
= 6= ln 2

∴ 𝑏 = 16
 -
1
𝑥
Evaluate ! 𝑑𝑥
cos 𝑥 cos 𝑥 + sin 𝑥
/

Solution :
-
1
𝑥
𝐼 = 2! 𝑑𝑥
2cos . 𝑥 + 2 sin 𝑥 cos 𝑥
/
-
1
𝑥
= 2! 𝑑𝑥
1 + cos 2𝑥 + sin 2𝑥
/

𝜋
= 2× ln 2 ⋯ from previous question
16

𝜋
= ln 2
8
 63t 63t
E
If 𝐼6 = ! 𝑥 ⋅ 𝐹 𝑥 1 − 𝑥 𝑑𝑥 and 𝐼. = ! 𝐹 𝑥 1 − 𝑥 𝑑𝑥 , then % = ?
E
#
t t

63t
Solution :
Given 𝐼6 = ! 𝑥 ⋅ 𝐹 𝑥 1 − 𝑥 𝑑𝑥
t

Let 𝑔 𝑥 = 𝐹 𝑥 1 − 𝑥 … (𝑖)

Now 𝑔 𝐾 + 1 − 𝐾 − 𝑥 = 𝑔(1 − 𝑥)

𝑔(1 − 𝑥) = 𝐹 1−𝑥 1− 1−𝑥 = 𝐹 1 − 𝑥 ⋅ 𝑥 from (𝑖) = 𝑔 𝑥

63t
𝐾+1−𝐾 u463u
⇒ 𝐼6 = ! 𝐹 𝑥 1 − 𝑥 𝑑𝑥 ⋯ ∵𝑓 𝑎−𝑥 =𝑓 𝑥 , 𝑥 =
2 .
t

1
⇒ 𝐼6 = 𝐼
2 .
𝐼6 1
∴ =
𝐼. 2
 64mql# G 64mql# G
E
If 𝐼6 = ! 𝑥 ⋅ 𝑔 𝑥 2 − 𝑥 𝑑𝑥 and 𝐼. = ! 𝑔 𝑥 2 − 𝑥 𝑑𝑥 , then % = ?
E
#
3KLB# G 3KLB# G

64mql# G
Solution :
𝐼6 = ! 𝑥 ⋅ 𝑔 𝑥 2 − 𝑥 𝑑𝑥
3KLB# G

UL + LL = 2
𝑥 →2−𝑥

Let ℎ 𝑥 = 𝑔 𝑥 2 − 𝑥
ℎ 1 + sec . 𝑡 − tan. 𝑡 − 𝑥 = ℎ(2 − 𝑥)

=𝑔 2−𝑥 2− 2−𝑥

=𝑔 𝑥 2−𝑥

=ℎ 𝑥
 64mql# G 64mql# G
E
If 𝐼6 = ! 𝑥 ⋅ 𝑔 𝑥 2 − 𝑥 𝑑𝑥 and 𝐼. = ! 𝑔 𝑥 2 − 𝑥 𝑑𝑥 , then % = ?
E
#
3KLB# G 3KLB# G

Solution :
64mql# G
. .
1 + sec 𝑡 − tan 𝑡 64mql# G3KLB# G
⇒ 𝐼6 = × ! 𝑔 𝑥 2 − 𝑥 𝑑𝑥 ⋯ ∵ 𝑓 𝑎 − 𝑥 = 𝑓 𝑥 , 𝑥 =
2 .
3KLB# G

⇒ 𝐼6 = 𝐼.

𝐼6
∴ =1
𝐼.
 Session 8
Odd and even Function
property

Return To Top
 +

If 𝜙 𝑎 − 𝑥 = 𝜙 𝑥 then ! 𝑥 ⋅ 𝜙 𝑎 − 𝑥 𝑑𝑥 =?
/

+
Solution :
! 𝑥 ⋅ 𝜙 𝑎 − 𝑥 𝑑𝑥
/ 𝜙 𝑎− 𝑎−𝑥 =𝜙 𝑥 =𝜙 𝑎−𝑥

+
𝑎 So apply removal of 𝑥
⇒ ! 𝜙 𝑎 − 𝑥 𝑑𝑥
2
/

+
𝑎
⇒ ! 𝜙 𝑥 𝑑𝑥
2
/
 .-
The value of ! sin 2𝑥 (1 + cos 3𝑥) 𝑑𝑥
/ where 𝑡 denotes greatest integer function is :

A 2𝜋 B 𝜋 C −2𝜋 D −𝜋
Solution : .-
, ,
𝐼 = ! sin 2𝑥 (1 + cos 3𝑥) 𝑑𝑥
! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑎 + 𝑏 − 𝑥 𝑑𝑥
/
+ +
.-
𝑥 → 2𝜋 − 𝑥
𝐼 = ! − sin 2𝑥 (1 + cos 3𝑥) 𝑑𝑥 ⋯ (𝑖𝑖)
/

𝑖 + 𝑖𝑖 ⇒
.-

2𝐼 = ! sin 2𝑥 1 + cos 3𝑥 + − sin 2𝑥 1 + cos 3𝑥 𝑑𝑥 𝑥 + −𝑥 = −1, 𝑥 ∉ 𝐼

/
.-

2𝐼 = ! −1 𝑑𝑥 ⇒ 𝐼 = −𝜋
/
Properties of Definite integration ( Even – Odd function )

0 , if 𝑓 𝑥 is odd

+
+
2 ! 𝑓 𝑥 𝑑𝑥 , if 𝑓 𝑥 is even
Ø ! 𝑓 𝑥 𝑑𝑥 =
/
3+
+

! 𝑓 𝑥 + 𝑓 −𝑥 𝑑𝑥 , if 𝑓 𝑥 is neither even nor odd

/

−𝑎 𝑓 𝑥 = even
𝑎
−𝑎 𝑎
𝑓 𝑥 = odd
 D
! !
Evaluate ∫LD 𝑥 + 1 𝑥 + 1 + 𝑥 + 2 𝑥 + 2 𝑑𝑥 , where
. denotes fractional part function
6

Solution : ! 𝑥 . + 1 𝑥 + 1 + 𝑥 . + 2 𝑥 + 2 𝑑𝑥
36
6

= ! 𝑥 . 𝑥 + 𝑥 . 𝑥 𝑑𝑥
36
6

= ! 2 𝑥 . 𝑥 𝑑𝑥
36
6 + +

= 2 ! 𝑥 . 𝑥 + 𝑥 . −𝑥 𝑑𝑥 ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑥 + 𝑓 −𝑥 𝑑𝑥
/ 3+ /

6 6
2
= 2 ! 𝑥. 𝑥 + −𝑥 𝑑𝑥 = 2 ! 𝑥 . 𝑑𝑥 =
3
/ /
 D

Evaluate 3 sinLD 𝑥 𝑑𝑥
LD

Solution :

D 𝑓 𝑥 = sin36 𝑥
3 sinLD 𝑥 𝑑𝑥 = 0
LD −1 1

𝑓 𝑥 = odd
 !
𝑥
Evaluate 3 𝑑𝑥
𝑥
L!

A 1 B 2

0 D 4
C
 !
𝑥
Evaluate 3 𝑑𝑥
𝑥
L!

!
Solution : 𝑥
3 𝑑𝑥 = 0
𝑥
L!

−2
0 2
 -
1
𝑥 D − 3𝑥 0 + 3𝑥 2 − 𝑥 + 1
Evaluate ! 𝑑𝑥
cos . 𝑥
-
3
1

-
Solution : 1
𝑥 D − 3𝑥 0 + 3𝑥 2 − 𝑥 + 1
! 𝑑𝑥
cos . 𝑥
-
3
1
- - - - -
1 1 0 1 2 1 1
𝑥D 𝑥 𝑥 𝑥 1
= ! 𝑑𝑥 − 3 ! 𝑑𝑥 + 3 ! 𝑑𝑥 − ! 𝑑𝑥 + ! 𝑑𝑥
cos . 𝑥 cos . 𝑥 cos . 𝑥 cos . 𝑥 cos .𝑥
- - - - -
3 3 3 3 3
1 1 1 1 1

-
1
= ! sec . 𝑥 𝑑𝑥
-
3
1
-
1
= tan 𝑥 - =2
3
1
 D
2𝑥
Evaluate 3 sinLD 𝑑𝑥
1 + 𝑥!
LD

A 2 B tan 1

0 D − tan 1
C
 D
2𝑥
Evaluate 3 sinLD 𝑑𝑥
1 + 𝑥!
LD

A 2 B tan 1

0 D − tan 1
C
 D
2𝑥
Evaluate 3 sinLD 𝑑𝑥
1 + 𝑥!
LD

Solution : 6
2𝑥
! sin36 𝑑𝑥
1 + 𝑥.
36
6

= 2 ! 2 tan36 𝑥 𝑑𝑥
36
−1 0 1
=0
 D
!
1−𝑥
Evaluate 3 sec 𝑥 ln 𝑑𝑥
1+𝑥
D
L!

Solution :
𝑓 𝑥 = sec 𝑥

𝑓 −𝑥 = sec −𝑥 = sec 𝑥 = 𝑓 𝑥

63;
𝑔 𝑥 = ln 64;

64; 63;
𝑔 −𝑥 = ln 63;
= − ln 64;
= −𝑔 𝑥
6
.
1−𝑥
! sec 𝑥 ln 𝑑𝑥 = 0
1+𝑥
6
3
.
Even × Odd=Odd
 -
3
P

Evaluate ! 𝑥 6/ sin 𝑥 2 𝑑𝑥
-
P

- -
3
P P
Solution : ! 𝑥 6/ sin 𝑥 2 𝑑𝑥 = − ! 𝑥 6/ sin 𝑥 2 𝑑𝑥 = 0
- -
P 3
P

even odd

→ even× odd

→ odd
 6

Evaluate ! 1 − 𝑥 + 𝑥 . − 1 + 𝑥 + 𝑥 . 𝑑𝑥
36

Solution : 𝑓 𝑥 = 1 − 𝑥 + 𝑥. − 1 + 𝑥 + 𝑥.

𝑓 −𝑥 = 1 + 𝑥 + 𝑥. − 1 − 𝑥 + 𝑥.

⇒ 𝑓 −𝑥 = − 1 − 𝑥 + 𝑥. − 1 + 𝑥 + 𝑥.

⇒ 𝑓 −𝑥 = −𝑓 𝑥 → odd

! 1 − 𝑥 + 𝑥 . − 1 + 𝑥 + 𝑥 . 𝑑𝑥 = 0
36
 -
.

Evaluate ! sin 𝑥 𝑓 cos 𝑥 𝑑𝑥

-
3
.

-
Solution : .
! sin 𝑥 𝑓 cos 𝑥 𝑑𝑥
-
3
.

odd even

→ odd× even

→ odd

-
.
! sin 𝑥 𝑓 cos 𝑥 𝑑𝑥 = 0
-
3
.
 !

If 𝑓 𝑥 is even, then evaluate 3 𝑥 )𝑓 𝑥 + 𝑥𝑓 nn 𝑥 + 2 𝑑𝑥

L!
.
.
Solution : 2
! 𝑥 𝑓 𝑥 + 𝑥𝑓 ::
𝑥 + 2 𝑑𝑥 ∴ ! 𝑥 2 𝑓 𝑥 + 𝑥𝑓 :: 𝑥 + 2 𝑑𝑥
3.
3.

.
odd even even
= ! 2 𝑑𝑥
even odd 3.

.
[∵ odd× even=odd ]
= 2 ! 2 𝑑𝑥 = 8
And /

⇒ 𝑓 𝑥 = even

⇒ 𝑓′ 𝑥 = odd

⇒ 𝑓′′ 𝑥 = even
 If 𝑓 𝑥 is even and 𝑔 𝑥 & ℎ 𝑥 are odd, then
+ .N
𝑓 𝑥 + 𝑓 −𝑥 .N36
! ℎ 𝑥 − ℎ −𝑥 𝑑𝑥 = ?
𝑔 𝑥 − 𝑔 −𝑥
3+

Solution : 𝑓 𝑥 + 𝑓 −𝑥 → Even

𝑔 𝑥 − 𝑔 −𝑥 → Odd

ℎ 𝑥 − ℎ −𝑥 → Odd

even .N
→ odd .N36
odd

→ even×odd

→ odd
+ .N
𝑓 𝑥 + 𝑓 −𝑥 .N36
∴ ! ℎ 𝑥 − ℎ −𝑥 𝑑𝑥 = 0
𝑔 𝑥 − 𝑔 −𝑥
3+
 D
cos 𝑥 𝑥 sin 𝑥 𝑥 ) cos 𝑥
If 𝐹 𝑥 = 𝑥 ! sec 𝑥 𝑥 + sin 𝑥 then 𝐼 = 3 𝐹 𝑥 + 𝐹 nn 𝑥 ⋅ 𝑥 ! + 1 𝑑𝑥 = ?
1 2 tan 𝑥 LD

6
cos 𝑥 𝑥 sin 𝑥 𝑥 2 cos 𝑥
Solution : 𝐹 𝑥 = 𝑥. sec 𝑥 𝑥 + sin 𝑥 𝐼 = ! 𝐹 𝑥 + 𝐹 :: 𝑥 ⋅ 𝑥 . + 1 𝑑𝑥 =
= ?0
1 2 tan 𝑥 36

cos 𝑥 𝑥 sin 𝑥 −𝑥 2 cos 𝑥

𝐹 −𝑥 = 𝑥 . sec 𝑥 −(𝑥 + sin 𝑥) odd odd even
1 2 − tan 𝑥

cos 𝑥 𝑥 sin 𝑥 𝑥 2 cos 𝑥 → (odd + odd) ×even

= − 𝑥. sec 𝑥 𝑥 + sin 𝑥 = −𝐹 𝑥
1 2 tan 𝑥 → odd×even

⇒ 𝐹 𝑥 = odd
→ odd
⇒ 𝐹′ 𝑥 = even

⇒ 𝐹′′ 𝑥 = odd
 1 2𝑥 3𝑥 ! )

If 𝑓 𝑥 = 3 𝑎 27 and 3 𝑓(𝑥) 𝑑𝑥 = 0 , then 𝑎 =?

1 3 9 J

2
1 2𝑥 3𝑥 .
Solution : 𝑓 𝑥 = 3 𝑎
! 𝑓 𝑥 𝑑𝑥 = 0 27
/ 1 3 9

𝑥 𝑥. 𝑥2 3 3 9 27
⇒ 3 𝑎 27 =0 ⇒ 3 𝑎 27 − 0 = 0
1 3 9 / 1 3 9

1 3 9
⇒3 3 𝑎 27 = 0
1 3 9

∴𝑎∈ℝ
 6
2𝑥 22. + 𝑥 SSP + 4𝑥 6==P ⋅ sin 𝑥 =S6
Evaluate ! 𝑑𝑥
1 + 𝑥 ===
36

6
2𝑥 22. + 𝑥 SSP + 4𝑥 6==P ⋅ sin 𝑥 =S6
Solution : 𝐼 = ! 𝑑𝑥
1 + 𝑥 ===
36

6 6 6
𝑥 22.
𝑥 +𝑥 22. 4𝑥 6==P ⋅ sin 𝑥 =S6
SSP
= ! 𝑑𝑥 + ! 𝑑𝑥 + ! 𝑑𝑥
1 + 𝑥 === 1 + 𝑥 === 1 + 𝑥 ===
36 36 36

6
6 6
4𝑥 6==P ⋅ sin 𝑥 =S6
𝑥 22. ! 𝑑𝑥 → Odd = 0
𝑥 22. (1 + 𝑥 === ) 1 + 𝑥 ===
= ! 𝑑𝑥 + ! 𝑑𝑥 + 0 36
1 + 𝑥 222 .
1 + 𝑥 ===
36 36

6 6
𝑥 22.
= ! 𝑑𝑥 + ! 𝑥 22. 𝑑𝑥
1 + 𝑥 222 .
36 36
 6
2𝑥 22. + 𝑥 SSP + 4𝑥 6==P ⋅ sin 𝑥 =S6
Evaluate ! 𝑑𝑥
1 + 𝑥 ===
36

6 6
𝑥 22. Put 𝑥 222 = 𝑡
Solution : 𝐼 = ! 𝑑𝑥 + ! 𝑥 22. 𝑑𝑥
1 + 𝑥 222 .
36 36 333𝑥 22. 𝑑𝑥 = 𝑑𝑡
6 6
22.
𝐼 = 1 ! 333 𝑥 𝑑𝑥 + ! 𝑥 22. 𝑑𝑥
333 1+ 𝑥 222 .
36 36
6 1
𝐼= 1 ! 1 𝑥 222
𝑑𝑡 +
333 1 + 𝑡. 333 36
36

6 6 .
= 222 tan36 𝑡 36 + 222

6 - .
= 222 .
+ 222
 ,
3
1−𝑥
Evaluate * 𝑥 + ln
1+𝑥
𝑑𝑥 , where [.] denotes G.I.F
,
-
3

A 1 B ln 2

−1 D 1
C −
2
 ,
3
1−𝑥
Evaluate * 𝑥 + ln
1+𝑥
𝑑𝑥 , where [.] denotes G.I.F
,
-
3

1−𝑥
Solution : 6 6 𝑓 𝑥 = ln
. . 1+𝑥
1−𝑥
= ! 𝑥 𝑑𝑥 + ! ln 𝑑𝑥
1+𝑥
6 6 1+𝑥
3
.
3
. 𝑓 −𝑥 = ln
1−𝑥

odd 1−𝑥
6 𝑓 −𝑥 = − ln
/ . 1+𝑥
= ! −1𝑑𝑥 + ! 0𝑑𝑥
6 /
𝑓 −𝑥 = −𝑓 𝑥 → odd
3
.
6
= −.
 6 1
𝑑
Evaluate ! 𝑥 1 + 𝑒 36/; , where [.] denotes G.I.F
𝑑𝑥
36

Solution : 6 1
𝑑
! 𝑥 1 + 𝑒 36/;
𝑑𝑥
36

/ 1 6 1
𝑑 36/; 𝑑
= ! −𝑥 1+𝑒 +! 𝑥 1 + 𝑒 36/;
𝑑𝑥 𝑑𝑥
36 /

/ 1 6 1
𝑑 36/; 𝑑 36/;
= !0 ⋅ 1+𝑒 +!0 ⋅ 1 + 𝑒
𝑑𝑥 𝑑𝑥
36 /

=0
 -
.
.
Evaluate ! 1 − 𝑥 cos 𝑥 sin 𝑥 𝑑𝑥
-
3
.

-
.
! 1 − 𝑥 . cos 𝑥 sin 𝑥 𝑑𝑥 = 𝐸 − 𝐸 ⋅ (𝐸×𝑂)
-
3
.
=0
 -
.
Evaluate ! cos2 𝜃 1 + sin 𝜃 . 𝑑𝜃
-
3
.
-
.
Popular NENO Form
! cos 2 𝜃 1 + sin 𝜃 . 𝑑𝜃 𝐹 𝑥 = 1 + sin 𝑥
-
3
.
- - - 𝐹 𝑥 = 1 + 𝑎;
. . .
= ! cos 2 𝜃 𝑑𝜃 + ! 2 sin 𝜃 cos 2 𝜃 𝑑𝜃 + ! cos 2 𝜃 sin. 𝜃 𝑑𝜃 + +
3
-
3
-
3
- ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑥 + 𝑓 −𝑥 𝑑𝑥
. . . 3+ /
- -
. .
2 .
= 2 ! cos 2 𝜃 𝑑𝜃 + 0 + 2 ! cos 𝜃 sin 𝜃 𝑑𝜃
/ /

. .
= 2× ×1 + 2× ×1
2×6 0×2×6

1 1 .1
= 2 + 60 = 60
 g
𝑥!
Evaluate 3 𝑑𝑥
1 + sin 𝑥 + 1 + sin! 𝑥
Lg

-
Solution : 𝑥.
! 𝑑𝑥 → Neither even nor odd
1 + sin 𝑥 + 1 + sin. 𝑥
3-

-
1 1
= ! 𝑥. + 𝑑𝑥
1 + sin 𝑥 + 1 + sin. 𝑥 1 − sin 𝑥 + 1 + sin. 𝑥
/

-
.
1 − sin 𝑥 + 1 + sin. 𝑥 + 1 + sin 𝑥 + 1 + sin. 𝑥
= !𝑥 . 𝑑𝑥
/ 1 + 1 + sin. 𝑥 − sin. 𝑥

- -
.
2 1 + 1 + sin. 𝑥 .
2 1 + 1 + sin. 𝑥 𝜋2
= !𝑥 𝑑𝑥 = ! 𝑥 𝑑𝑥 =
1 + 2 1 + sin. 𝑥 + 1 + sin. 𝑥 − sin. 𝑥 2 1 + 1 + sin. 𝑥 3
/ /
 Session 9
Queen's property, some
special integrals

Return To Top
 -
2𝑥 1 + sin 𝑥
Evaluate ! 𝑑𝑥 .
1 + cos . 𝑥
3-

Solution: - Let cos 𝑥 = 𝑡 𝒙 𝒕

2𝑥 1 + sin 𝑥
! 𝑑𝑥 → Neither even nor odd ⇒ sin 𝑥 𝑑𝑥 = −𝑑𝑡
1 + cos . 𝑥 0 1
3-

- - 36 𝜋 −1
2𝑥 𝑥 sin 𝑥 𝑑𝑡
= ! 𝑑𝑥 + 2 ! 𝑑𝑥 = −2𝜋 !
1 + cos . 𝑥 1 + cos . 𝑥 1 + 𝑡.
3- 3- 6
odd even
36
= −2𝜋 tan36 𝑡 6
-
𝑥 sin 𝑥
= 0 + 2×2 ! 𝑑𝑥 → Removal of 𝑥 𝜋 𝜋
1 + cos . 𝑥 = − 2𝜋 − − = 𝜋.
/ 4 4
-
𝜋 sin 𝑥 -4/ -
= 4× ! 𝑑𝑥 𝑥= =
2 1 + cos . 𝑥 . .
/
 -
cos . 𝑥
Evaluate ! ;
𝑑𝑥 .
3- 1 + 𝑎

+ +
Solution: -
cos . 𝑥
! ;
𝑑𝑥 Break is not possible. So, trying ! 𝐹 𝑥 𝑑𝑥 = ! 𝐹 𝑥 + 𝐹 −𝑥 𝑑𝑥
3- 1 + 𝑎 3+ /

-
cos . 𝑥 cos . 𝑥
=! ;
+ 𝑑𝑥
/ 1+𝑎 1 + 𝑎3;

-
1 .
𝑎;
= ! cos 𝑥 + 𝑑𝑥
/ 1 + 𝑎; 1 + 𝑎;

-
1 + 𝑎;
.
= ! cos 𝑥 𝑑𝑥
/ 1 + 𝑎;

-
1 + cos 2𝑥 ; 6 mrB .; - - -
=! 𝑑𝑥 = +.⋅ = +0 −0=
/ 2 . . / . .
 -
sin 𝑛𝑥
If 𝐼N = ! 𝑑𝑥 , 𝑛 = 0, 1, 2, ⋯ , then :
3- 1 + 𝜋 ; sin 𝑥
IIT JEE 2009

6/

A 𝐼N = 𝐼N4. B š 𝐼.U46 = 10𝜋

Uv6

6/

C š 𝐼.U = 0 D 𝐼N = 𝐼N46
Uv6
 -
sin 𝑛𝑥
If 𝐼N = ! 𝑑𝑥 , 𝑛 = 0, 1, 2, ⋯ , then :
3- 1 + 𝜋 ; sin 𝑥
IIT JEE 2009
- + +
sin 𝑛𝑥
Solution: 𝐼N = ! 𝑑𝑥 ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑥 + 𝑓 −𝑥 𝑑𝑥
3- 1 + 𝜋 ; sin 𝑥 3+ /

-
sinsin
𝑛𝑥𝑛𝑥
sin 𝑛𝑥 −sin
−sin
sin𝑛𝑥𝑛𝑥
𝑛𝑥
𝐼N = ! + +
+ 𝑑 𝑑𝑥
/ 1+ 𝜋+;𝜋𝜋;sin
11+ ; sin
𝑥 𝑥𝑥 1 +
sin 11+ 𝜋𝜋3;
𝜋+3; (−(−
3; sinsin
(sin 𝑥)𝑥)
𝑥)

-
sin 𝑛𝑥 𝜋 ; ⋅ sin 𝑛𝑥
𝐼N = ! + 𝑑𝑥
/ 1 + 𝜋 ; sin 𝑥 1 + 𝜋 ; sin 𝑥

-
sin 𝑛𝑥
𝐼N = ! 𝑑𝑥
/ sin 𝑥
 -
sin 𝑛𝑥
If 𝐼N = ! 𝑑𝑥 , 𝑛 = 0, 1, 2, ⋯ , then :
3- 1 + 𝜋 ; sin 𝑥
IIT JEE 2009
-
sin 𝑛𝑥
Solution: 𝐼N = ! 𝑑𝑥
/ sin 𝑥

- 𝐶−𝐷 𝐶+𝐷
sin(𝑛 + 2)𝑥 − sin 𝑛𝑥
𝐼N4. − 𝐼N = ! 𝑑𝑥 sin 𝐶 − sin 𝐷 = 2 sin cos
/ sin 𝑥 2 2

-
2 sin 𝑥 . cos 𝑛 + 1 𝑥
𝐼N4. − 𝐼N = ! 𝑑𝑥
/ sin 𝑥

-
sin 𝑛 + 1 𝑥
𝐼N4. − 𝐼N = 2 =0
𝑛+1 /

𝐼N4. = 𝐼N
 -
sin 𝑛𝑥
If 𝐼N = ! 𝑑𝑥 , 𝑛 = 0, 1, 2, ⋯ , then :
3- 1 + 𝜋 ; sin 𝑥
IIT JEE 2009

-
Solution: sin 𝑛𝑥 𝐼N4. = 𝐼N
𝐼N = ! 𝑑𝑥
/ sin 𝑥

-
sin 3𝑥
Put 𝑛 = 3 ⇒ 𝐼2 = ! 𝑑𝑥 =𝜋
/ sin 𝑥
6/
∴ 𝐼2 = 𝐼0 = ⋯ = 𝜋 ⇒ š 𝐼.U46 = 10𝜋
Uv6

-
sin 2𝑥 -
Put 𝑛 = 2 ⇒ 𝐼. = ! 𝑑𝑥 = ! 2 cos 𝑥 𝑑𝑥 = 2 sin 𝑥 /𝜋 = 0
/ sin 𝑥 /

6/
∴ 𝐼. = 𝐼1 = 𝐼= = ⋯ = 0 ⇒ š 𝐼.U = 0
Uv6
 -
sin 𝑛𝑥
If 𝐼N = ! 𝑑𝑥 , 𝑛 = 0, 1, 2, ⋯ , then :
3- 1 + 𝜋 ; sin 𝑥
IIT JEE 2009

6/

A 𝐼N = 𝐼N4. B š 𝐼.U46 = 10𝜋

Uv6

6/

C š 𝐼.U = 0 D 𝐼N = 𝐼N46
Uv6
 𝑒 opi N ⋅ sin 𝑥 , 𝑥 ≤2 )
If 𝐹 𝑥 = then 3 𝐹(𝑥)𝑑𝑥 = _____?
2 , otherwise L!

Solution:
𝑒 l^m ; ⋅ sin 𝑥 , −2 ≤ 𝑥 ≤ 2
𝐹 𝑥 =
2 , otherwise

2 . 2
l^m ;
! 𝐹(𝑥)𝑑𝑥 = ! 𝑒 ⋅ sin 𝑥 𝑑𝑥 + ! 2 𝑑𝑥
3. 3. .

2
= odd + 2 𝑥 .

=2
 gr
!LN opi N A
If 𝑓 𝑥 = and 𝑔 𝑥 = ln 𝑥 , 𝑥 > 0 then the value of 3 𝑔 𝑓 𝑥 𝑑𝑥 is:
!qN opi N
LgrA

JEE Main April 2019

A ln 1 B ln 3 C ln 𝑒 D ln 2

Solution:
.3; l^m ; +
𝑔 𝑓 𝑥 = ln .4; l^m ; 2 ! 𝑓 𝑥 𝑑𝑥 , if 𝑓 𝑥 is even
+ /
! 𝑓 𝑥 𝑑𝑥 =
.4; l^m ; .3; l^m ; 3+
𝑔 𝑓 −𝑥 = ln .3; l^m ;
= −ln .4; l^m ; 0 , if 𝑓 𝑥 is odd

⇒ 𝑔 𝑓 −𝑥 = −𝑔 𝑓 𝑥 → 𝑔(𝑓 𝑥 ) is odd

-s
1
! 𝑔 𝑓 𝑥 𝑑𝑥 = 0 = ln 1
3-s1
Property 6
To reduce the limit by half ( Known as Queen’s Property )

+
2 ! 𝑓 𝑥 𝑑𝑥 , If 𝑓 2𝑎 − 𝑥 = 𝑓 𝑥
.+ /
! 𝑓 𝑥 𝑑𝑥 =
/

0 , If 𝑓 2𝑎 − 𝑥 = −𝑓 𝑥
 !s s
Prove that 3 𝑓 𝑥 𝑑𝑥 = 3 𝑓 𝑥 + 𝑓 2𝑎 − 𝑥 𝑑𝑥.
J J

Solution: L.H.S +

.+
2 ! 𝑓 𝑥 𝑑𝑥 , if 𝑓 2𝑎 − 𝑥 = 𝑓(𝑥)
.+ /
= ! 𝑓 𝑥 𝑑𝑥 ! 𝑓 𝑥 𝑑𝑥 =
/
/
0 , if 𝑓 2𝑎 − 𝑥 = −𝑓(𝑥)
+ .+
= ! 𝑓 𝑥 𝑑𝑥 + ! 𝑓 𝑥 𝑑𝑥
/ +
Put 𝑥 = 2𝑎 − 𝑡
+ / 𝒙 𝒕
= ! 𝑓 𝑥 𝑑𝑥 − ! 𝑓 2𝑎 − 𝑡 𝑑𝑡 ⇒ 𝑑𝑥 = −𝑑𝑡
/ +
𝑎 𝑎
+ +
= ! 𝑓 𝑥 𝑑𝑥 + ! 𝑓 2𝑎 − 𝑥 𝑑𝑥 By property 1 & 2 2𝑎 0
/ /

+
= ! 𝑓 𝑥 + 𝑓 2𝑎 − 𝑥 𝑑𝑥 = R.H.S
/
 !g
Evaluate 3 cos A 𝑥 𝑑𝑥.
J

Solution: .-
! cos 1 𝑥 𝑑𝑥
/

By Queen’s Property,

-
𝐼 = 2 ! cos 1 𝑥 𝑑𝑥 ∵ cos 1 2𝜋 − 𝑥 = cos 1 𝑥
/

Again, applying Queen’s property

-
.
𝐼 = 4 ! cos 1 𝑥 𝑑𝑥 (∵ cos 1 𝜋 − 𝑥 = cos 1 𝑥
/

2×6 - 2-
By Walli’s theorem, 𝐼 = 4× 1 × . × . = 1
 @Ag
Evaluate 3 sin@ 𝑥 𝑑𝑥.
J

Solution: -
sin 𝑥 has even power , so it undergoes reduction until upper limit becomes ,
.

∵ sin= 𝑥 is +𝑣𝑒 in all quadrants.

Shortcut way,

=1- wxqB
=
𝐼=! sin 𝑥 𝑑𝑥 ! sinwxqB 𝑥 𝑑𝑥
/ /

- -
6.P
. .
=! sin= 𝑥 𝑑𝑥 = 128 ! sin= 𝑥 𝑑𝑥
/ /

0×2×6 -
= 128× = × 1 × . × . = 20𝜋
 g
Evaluate 3 𝑥 sinA 𝑥 𝑑𝑥.
J

Solution: -
1 𝑓 𝑥 = sin1 𝑥
! 𝑥 sin 𝑥 𝑑𝑥.
/

𝑓 𝜋 − 𝑥 = sin1 𝜋 − 𝑥 = sin1 𝑥
𝜋 - 1
= ! sin 𝑥 𝑑𝑥 By Removal of 𝑥
2 /

-
𝜋 .
= ×2 ! sin1 𝑥 𝑑𝑥 By Queen’s Property
2 /

2×6 -
= 𝜋× 1×. × . By Walli’s theorem

2- #
= 6=
 !g 𝑥 ⋅ sin!I 𝑥
Evaluate 3 𝑑𝑥.
J sin!I 𝑥 + cos !I 𝑥

Solution: .-
𝑥 ⋅ sin.N 𝑥
𝐼=! 𝑑𝑥 Removal of 𝑥
/ sin.N 𝑥 + cos .N 𝑥

.-
sin.N 𝑥
= 𝜋! 𝑑𝑥
/ sin.N 𝑥 + cos .N 𝑥
-
1×
. sin.N 𝑥
= 𝜋! 𝑑𝑥 Using property 6
/ sin.N 𝑥 + cos .N 𝑥
-
. sin.N 𝑥
= 4𝜋 ! .N 𝑥 + cos .N 𝑥
𝑑𝑥 Using property 4
/ sin

𝜋
= 4𝜋×
4

= 𝜋.
 g
Evaluate 3 𝑥 sin! sin 𝑥 + cos ! cos 𝑥 𝑑𝑥.
J

Solution: -
𝐼 = ! 𝑥 sin. sin 𝑥 + cos . cos 𝑥 𝑑𝑥
/

-4/ -
𝑓 𝜋−𝑥 =𝑓 𝑥 Removal of 𝑥 → 𝑥 = =
. .

𝜋 -
= ! sin. sin 𝑥 + cos . cos 𝑥 𝑑𝑥
2 /

Here, 𝑓 𝜋 − 𝑥 = 𝑓 𝑥

-
𝜋 .
∴ 𝐼 = 2× ! sin. sin 𝑥 + cos . cos 𝑥 𝑑𝑥
2 /
 g
Evaluate 3 𝑥 sin! sin 𝑥 + cos ! cos 𝑥 𝑑𝑥.
J

Solution: -
.
𝐼 = 𝜋 ! sin. sin 𝑥 + cos . cos 𝑥 𝑑𝑥 ⋯ 𝑖
/

By King’s Property

-
.
𝐼 = 𝜋 ! sin. cos 𝑥 + cos . sin 𝑥 𝑑𝑥 ⋯ 𝑖𝑖
/

Adding 𝑖 and 𝑖𝑖
-
. *
-#
2𝐼 = 𝜋 ! 2 𝑑𝑥 ⇒𝐼=𝜋 𝑥 # ⇒𝐼=
/ / .
 g
!
Evaluate 3 ln sin 𝑥 𝑑𝑥 .
J

Solution: g
-
.
-
.
Put 2𝑥 = 𝑡
! ⇒ 2𝐼 = ! ln sin 2𝑥 𝑑𝑥 − ! ln 2 𝑑𝑥 9G
𝐼 = 3 ln sin 𝑥 𝑑𝑥 ⋯ 𝐴 / /
⇒ 𝑑𝑥 =
.
J
1 - 𝜋 𝒙 𝒕
Apply King’s property and add ⇒ 2𝐼 = ! ln sin 𝑡 𝑑𝑡 − ln 2 ×
2 / 2
- 0 0
. 𝜋 𝑓 𝜋−𝑡 =𝑓 𝑡
𝐼 = ! ln sin − 𝑥 𝑑𝑥 -
/ 2 𝜋
Applying 𝑃6 & 𝑃1 .
-
. -
⇒ 𝐼 = ! ln cos 𝑥 𝑑𝑥 ⋯ 𝐵 1 . 𝜋
/
⇒ 2𝐼 = ×2 ! ln sin 𝑡 𝑑𝑡 − ln 2
2 / 2

𝐴 + 𝐵 , -
⇒ 2𝐼 = 𝐼 − . ln 2
-
.
2𝐼 = ! ln sin 𝑥 cos 𝑥 𝑑𝑥 -
/ ⇒ 𝐼 = − . ln 2
Results:

- -
. . 𝜋
v ! ln sin 𝑥 𝑑𝑥 = ! ln cos 𝑥 𝑑𝑥 = − ln 2
/ / 2

- -
. . 𝜋
v ! ln sec 𝑥 𝑑𝑥 = ! ln cosec 𝑥 = ln 2
/ / 2

-
. 𝜋
v ! ln sin 2𝑥 𝑑𝑥 = − ln 2
/ 2
 g 1 1
Evaluate 𝐼 = 3 𝑒 opi N 2 sin cos 𝑥 + 3 cos cos 𝑥 sin 𝑥 𝑑𝑥.
J 2 2

Solution: -
l^m ;
1 -
1
𝐼=! 𝑒 2 sin cos 𝑥 sin 𝑥 𝑑𝑥 + ! 𝑒 l^m ; 3 cos cos 𝑥 sin 𝑥 𝑑𝑥
/ 2 / 2

𝑓 𝜋 − 𝑥 = −𝑓 𝑥 𝑓 𝜋 − 𝑥 = +𝑓 𝑥

∴ This term becomes 0

-
. 1
𝐼 = 2 ! 𝑒 l^m ; 3 cos cos 𝑥 sin 𝑥 𝑑𝑥 / 6
/ 2 G
𝑡 𝑡
= −6 ! 𝑒 cos 𝑑𝑡 = 6 ! 𝑒 G cos 𝑑𝑡
6 2 / 2

𝒙 𝒕
Use IBP here
Put cos 𝑥 = 𝑡
⇒ − sin 𝑥 𝑑𝑥 = 𝑑𝑡 0 1 .1 6 < 6
= 𝑒 cos + sin −1
0 . . .
-
0
.
 Session 10
Periodic Function Property

Return To Top
 Results
- -
. . 𝜋
v ! ln sin 𝑥 𝑑𝑥 = ! ln cos 𝑥 𝑑𝑥 = − ln 2
/ / 2
- -
. . 𝜋
v ! ln sec 𝑥 𝑑𝑥 = ! ln cosec 𝑥 = ln 2
/ / 2

-
. 𝜋
v ! ln sin 2𝑥 𝑑𝑥 = − ln 2
/ 2

- -
. .
v ! ln tan 𝑥 𝑑𝑥 = ! ln cot 𝑥 𝑑𝑥 = 0
/ /
 6
Evaluate ! ln 𝑥 𝑑𝑥
/

𝐼 = 𝑥 ln 𝑥 − 𝑥 6 ln 𝑥 ∞
0 lim 𝑥 ln 𝑥 = lim
;→/ ;→/ 1 ∞
= 1×0 − 1 − → 0×∞ − 0 𝑥
1
indeterminate = lim 𝑥
;→/ 1
form − .
𝑥
= −1 − 0 = lim −𝑥
;→/

= −1
=0

Point to remember

In definite integration, if we get indeterminate form by substituting

any limit , then use concept of limit to get the answer.
 6
sin36 𝑥
Evaluate 𝐼 = ! 𝑑𝑥
/ 𝑥

Let sin36 𝑥 = 𝑡 ⇒ 𝑥 = sin 𝑡 ⇒ 𝑑𝑥 = cos 𝑡 𝑑𝑡

𝑥 𝑡

0 0
-
1
.

-
. 𝑡
𝐼= ! × cos 𝑡 𝑑𝑡
/ sin 𝑡
-
.
= ! 𝑡 cot 𝑡 𝑑𝑡
/
 6
sin36 𝑥
Evaluate 𝐼 = ! 𝑑𝑥
/ 𝑥

-
-
.
𝐼 = 𝑡 ⋅ ln sin 𝑡 .
− ! ln sin 𝑡 𝑑𝑡 ln sin 𝑡 ∞
/
/ lim 𝑡 ln sin 𝑡 = lim 1
G→/ G→/ ∞
- -
𝑡
⇒ 𝐼 = ×0 − 0× ln 0 + ln 2 cot 𝑡
. . - = lim
. 𝜋 G→/ 1
⋯ ! ln sin 𝑡 𝑑𝑡 = − ln 2 − .
2 𝑡
/
G
⇒ 𝐼 = 0 − 0 + ln 2
- = lim cos 𝑡 ⋅ ⋅𝑡
G→/ mrB G
.
-
⇒ 𝐼 = . ln 2 =1⋅1⋅0

=0
 -
1
Evaluate 𝐼 = ! log sin 2𝑥 𝑑𝑥
/

-
1
𝐼 = ! log sin 2𝑥 𝑑𝑥
/

𝑑𝑡
Let 2𝑥 = 𝑡 ⇒ 𝑑𝑥 =
2

𝑥 𝑡

0 0
- -
1 .

-
1 .
= ! log sin 𝑡 𝑑𝑡
2 /
-
𝜋 . 𝜋
= − ln 2 ⋯ ! ln sin 𝑡 𝑑𝑡 = − ln 2
4 / 2
 -
.
Evaluate 𝐼 = ! log tan 𝑥 + cot 𝑥 𝑑𝑥
/

-
. 1
𝐼 = ! log tan 𝑥 + 𝑑𝑥
/ tan 𝑥
-
. 1 + tan 𝑥
= ! log 𝑑𝑥
/ tan 𝑥
-
. sec . 𝑥
= ! log 𝑑𝑥
/ tan 𝑥

- -
. .
= ! log sec . 𝑥 𝑑𝑥 − ! log tan 𝑥 𝑑𝑥
/ /
-
.
= 2 ! log sec 𝑥 𝑑𝑥
/

-
= 2. . log 2 = 𝜋 log 2
 J
1 𝑑𝑥
Evaluate 𝐼 = ! log 𝑥 +
/ 𝑥 1 + 𝑥.

Put 𝑥 = tan 𝜃 ⇒ 𝑑𝑥 = sec . 𝜃 𝑑𝜃 -

. sec . 𝜃
𝐼 = ! log 𝑑𝜃
/ tan 𝜃
𝑥 𝜃 - -
. .
0 0 = ! log sec . 𝜃 𝑑𝜃 − ! log tan 𝜃 𝑑𝜃
/ /
- -
∞ .
. = 2 ! log sec 𝜃 𝑑𝜃
/

𝐼6 -
1 = 2. log 2 = 𝜋 log 2
.
-
. sec . 𝜃
= ! log tan 𝜃 + cot 𝜃 ⋅ 𝑑𝜃
/ 1 + tan. 𝜃

-
. 1
𝐼 = ! log tan 𝜃 + 𝑑𝜃
/ tan 𝜃
 -
1
Evaluate 𝐼 = ! - log sin 𝑥 + cos 𝑥 𝑑𝑥
3
1

-
1
Let 𝐼 = ! - log sin 𝑥 + cos 𝑥 𝑑𝑥
3
1
-
1 𝜋 -
= ! log 2 sin +𝑥 𝑑𝑥 Let 𝑡 = + 𝑥 ⇒ 𝑑𝑡 = 𝑑𝑥
3
- 4 1
1

𝑥 𝑡
-
−1 0
- - - -
. 1 .
= ! log sin 𝑡 𝑑𝑡 + ! log 2 𝑑𝑡 1 .
/ 2 /

𝜋 𝜋 𝜋
= − log 2 + log 2 = − log 2
2 4 4
 -
. 𝑝𝑥 . + 𝑞𝑥 + 𝑟
Evaluate 𝐼 = ! log 𝑎 + 𝑏 sin 𝑥 𝑑𝑥, (𝑎 + 𝑏 > 0)
-
3 𝑝𝑥 . − 𝑞𝑥 + 𝑟
.

𝑝𝑥 . + 𝑞𝑥 + 𝑟
Consider 𝑓 𝑥 = log
𝑝𝑥 . − 𝑞𝑥 + 𝑟

𝑝𝑥 . − 𝑞𝑥 + 𝑟 𝑝𝑥 . + 𝑞𝑥 + 𝑟
𝑓 −𝑥 = log = −log = −𝑓 𝑥 → odd
𝑝𝑥 . + 𝑞𝑥 + 𝑟 𝑝𝑥 . − 𝑞𝑥 + 𝑟

odd even
- - -
. 𝑝𝑥 . + 𝑞𝑥 + 𝑟 . .
∴ 𝐼 = ! - log + ! log 𝑎 + 𝑏 𝑑𝑥 + ! log sin 𝑥 𝑑𝑥
3 𝑝𝑥 . − 𝑞𝑥 + 𝑟 3
-
3
-
. . .

- -
. .
⇒ 𝐼 = 0 +2 ! log 𝑎 + 𝑏 𝑑𝑥 + 2 ! log sin 𝑥 𝑑𝑥

-
-
⇒ 𝐼 = 𝜋 log 𝑎 + 𝑏 + 2× − . ln 2 . 𝜋
⋯ ! ln sin 𝑥 𝑑𝑥 = − ln 2
/ 2
⇒ 𝐼 = 𝜋 log 𝑎 + 𝑏 − 𝜋 ln 2
 -
.
Evaluate 𝐼 = ! 2 cos . 𝑥 ln sin 2𝑥 𝑑𝑥
/

-
A
-
− . ln 2 B .
ln 2

-
C .
ln 4 D -
− . ln 4
 -
.
Evaluate 𝐼 = ! 2 cos . 𝑥 ln sin 2𝑥 𝑑𝑥
/

- -
. .
𝜋
𝐼 = ! 2 cos . 𝑥 ln sin 2𝑥 𝑑𝑥 ⋯ 𝑖 𝐼 = ! ln sin 2𝑥 𝑑𝑥 = − ln 2
2
/ /
- -
. .
𝜋 𝜋
𝐼 = ! 2 sin. 𝑥 ln sin 2𝑥 𝑑𝑥 ⋯ 𝑖𝑖 𝑥 → −𝑥 ! ln sin 2𝑥 𝑑𝑥 = − ln 2
2 2
/ /

Adding 𝑖 and 𝑖𝑖
-
.
2𝐼 = ! 2 cos . 𝑥 + sin. 𝑥 ln sin 2𝑥 𝑑𝑥
/
-
.
2𝐼 = ! 2 ln sin 2𝑥 𝑑𝑥
/
 6/-
Evaluate 𝐼 = ! sin 𝑥 𝑑𝑥
-

From diagram, Period (𝑇) = 𝜋

𝑦 = sin 𝑥
S-
⋯ Shifting graph as
⇒𝐼=! sin 𝑥 𝑑𝑥
/ shown in diagram ⋯
- Ny y
= 9× ! sin 𝑥 𝑑𝑥 ⋯ ! 𝑓 𝑥 𝑑𝑥 = 𝑛 ! 𝑓 𝑥 𝑑𝑥 0 𝜋 2𝜋 3𝜋 9𝜋 10𝜋
/ / /
-
= 9×2
⋯ ! | sin 𝑥 |𝑑𝑥 = 2
/
= 18
⋯
0 𝜋 2𝜋 3𝜋 8𝜋 9𝜋
 Key Takeaways

Property 7
Periodic function-based properties : If 𝑓 𝑥 is a periodic function with period 𝑇
Ny y

Ø ! 𝑓 𝑥 𝑑𝑥 = 𝑛 ! 𝑓 𝑥 𝑑𝑥 , 𝑛 ∈ ℤ
/ /

,4Ny ,

Ø ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓 𝑥 𝑑𝑥
+4Ny +

+4y y
Ø ∫+ 𝑓 𝑥 𝑑𝑥 = ∫/ 𝑓 𝑥 𝑑𝑥 , 𝑎 ∈ ℝ ( It is independent of 𝑎 )

Ny y

Ø ! 𝑓 𝑥 𝑑𝑥 = 𝑛 − 𝑚 ! 𝑓 𝑥 𝑑𝑥
Uy /
Key Takeaways

Property 7
Periodic function-based properties : If 𝑓 𝑥 is a periodic function with period 𝑇
Ny y
v ! 𝑓 𝑥 𝑑𝑥 = 𝑛 ! 𝑓 𝑥 𝑑𝑥 ,𝑛 ∈ 𝕀 𝑌
/ /

⋯
𝑋
0 𝑇 2𝑇 3𝑇 (𝑛 − 1)𝑇 𝑛𝑇
+4y y
v ! 𝑓 𝑥 𝑑𝑥 = ! 𝑓(𝑥) 𝑑𝑥 ( It is independent of 𝑎 )
+ /

𝑋
0 𝑎 𝑇 𝑎+𝑇
 -
Evaluate 𝐼 = ! log(1 + cos 𝑥) 𝑑𝑥
/

-
𝑥
𝐼 = ! log 2 cos . 𝑑𝑥
/ 2
𝑥 𝑡
- -
𝑥 ;
Let . = 𝑡 ⇒ 𝑑𝑥 = 2𝑑𝑡
= ! log 2 𝑑𝑥 + 2 ! log cos 𝑑𝑥 0 0
/ / 2
-
- 𝜋
. .
= 𝜋 log 2 + 4 ! log cos 𝑡 𝑑𝑡
/
-
𝜋 . 𝜋
= 𝜋 log 2 + 4× − log 2 ⋯ ! log cos 𝑥 𝑑𝑥 = − log 2
2 / 2

= −𝜋 log 2
 6=-
2
Evaluate 𝐼 = ! sin 𝑥 𝑑𝑥
/

-
0-4
2
𝐼=! sin 𝑥 𝑑𝑥
/
-
0- 0-4
2
=! sin 𝑥 𝑑𝑥 + ! sin 𝑥 𝑑𝑥
/ 0-
- Ny y
- 2
= 5× ! sin 𝑥 𝑑𝑥 + ! sin 𝑥 𝑑𝑥 ⋯ ! 𝑓 𝑥 𝑑𝑥 = 𝑛 ! 𝑓 𝑥 𝑑𝑥
/ / / /
-
2 𝜋 -
= 5×2 + ! sin 𝑥 𝑑𝑥 𝑥 ∈ 0, ⇒ sin 𝑥 = sin 𝑥 ⋯ ! | sin 𝑥 |𝑑𝑥 = 2
/ 3
/
-
2
⇒ 𝐼 = 10 + (− cos 𝑥 /
)
6
= 10 + − .
−1

6 .6
= 10 + . ∴𝐼=
.
 1//-
1 − cos 2𝑥
Evaluate 𝐼=! 𝑑𝑥
/ 2

1//-
2 sin. 𝑥
𝐼=! 𝑑𝑥
/ 2
1//-
=! sin 𝑥
/
- Ny y
= 400× ! sin 𝑥 ⋯ ! 𝑓 𝑥 𝑑𝑥 = 𝑛 ! 𝑓 𝑥 𝑑𝑥
/ / /

-
= 400×2 = 800 ⋯ ! | sin 𝑥 |𝑑𝑥 = 2
/
Let 𝑥 & 𝑥 denote the fractional part of 𝑥 and the greatest integer ≤ 𝑥 respectively of
N N
a real number 𝑥. If ! 𝑥 𝑑𝑥, ! [𝑥] 𝑑𝑥 and 10 𝑛. − 𝑛 , 𝑛 ∈ ℕ, 𝑛 > 1 are three consecutive
/ /

terms of a G.P., then 𝑛 is equal to:

JEE MAIN 2020

We know that
N
𝑛
(𝑖) ! 𝑥 𝑑𝑥 =
/ 2
N 𝑛 𝑛−1
(𝑖𝑖) ! 𝑥 𝑑𝑥 = 0 + 1 + 2 + 3 ⋯ + 𝑛 − 1 =
/ 2
N N
Given that ! 𝑥 𝑑𝑥, ! [𝑥] 𝑑𝑥 and 10 𝑛. − 𝑛 , 𝑛 ∈ ℕ, 𝑛 > 1 are in 𝐺. 𝑃.
/ /

In G.P: 𝑏. = 𝑎𝑐
Let 𝑥 & 𝑥 denote the fractional part of 𝑥 and the greatest integer ≤ 𝑥 respectively of
N N
a real number 𝑥. If ! 𝑥 𝑑𝑥, ! [𝑥] 𝑑𝑥 and 10 𝑛. − 𝑛 , 𝑛 ∈ ℕ, 𝑛 > 1 are three consecutive
/ /

terms of a G.P., then 𝑛 is equal to:

JEE MAIN 2020

.
𝑛. 𝑛 − 1 .
𝑛
∴ 𝑏 = 𝑎𝑐 ⇒ = ×10 𝑛. − 𝑛 ⋯ 𝑓𝑟𝑜𝑚 𝑖 𝑎𝑛𝑑 (𝑖𝑖)
4 2
⇒ 𝑛. 𝑛 − 1 .
= 20𝑛. 𝑛 − 1
⇒ 𝑛. − 22𝑛 + 21 = 0

⇒ (𝑛 − 21)(𝑛 − 1) = 0

⇒ 𝑛 = 21 ⋯ (∵ 𝑛 > 1)
 6/
N
Evaluate š ! 𝑥 − 𝑥 𝑑𝑥 , where [.] denotes Greatest integer function
Nv6 N36

6 . 2 6/
𝐼 = ! 𝑥 − 𝑥 𝑑𝑥 + ! 𝑥 − 𝑥 𝑑𝑥+ ! 𝑥 − 𝑥 𝑑𝑥 + ⋯ + ! 𝑥 − 𝑥 𝑑𝑥
/ 6 . S

6/ 6/
=! 𝑥 − 𝑥 𝑑𝑥 = ! 𝑥 𝑑𝑥 ⋯ ∵𝑥− 𝑥 = 𝑥
/ /
N
𝑛
=5 ⋯ ! 𝑥 𝑑𝑥 =
/ 2
 6/ 6/.D
2; 2;
Evaluate , 𝑎 𝐼 =! 𝑑𝑥 𝑏 𝐼 =! 𝑑𝑥
/ 2; / 2;

where [.] denotes Greatest integer function

6/ 6/ 6/.D
𝑎 𝐼 = ! 2;3 ; 𝑑𝑥 𝑏 𝐼=! 2 ;3 ;
𝑑𝑥 + ! 2;3 ; 𝑑𝑥
/ / 6/
6/
= ! 2 ; 𝑑𝑥 ⋯ ∵𝑥− 𝑥 = 𝑥 10 /.D
/ 𝐼= + ! 2;3 ; 𝑑𝑥
ln 2 /
6
⇒ 𝐼 = 10 ! 2 ; 𝑑𝑥 ⋯ (∵ 𝑇 = 1) 10 /.D
/ 𝐼= + ! 2;3/ 𝑑𝑥
6
ln 2 /
⇒ 𝐼 = 10 ! 2; 𝑑𝑥 ⋯ For 𝑥 ∈ 0, 1 , {𝑥} = 𝑥
/ 10 (2/.D − 1) 9 + 2/.D
𝐼= + =
2; 6 ln 2 ln 2 ln 2
𝐼 = 10× G
ln 2 /
2 1 10
⇒ 𝐼 = 10 − =
ln 2 ln 2 ln 2
 16
. 𝑒 .;
Evaluate 𝐼 = ! 𝑑𝑥 , where [.] denotes Greatest integer function
/ 𝑒 .;

16
.
𝐼 =! .;3 .;
𝑒 𝑑𝑥 ⋯ ∵ 𝑥 − 𝑥 = 𝑥
/
6
16×
. 6
𝐼= ! 𝑒 {.;} 𝑑𝑥 Take ∶ 𝑇 = .
/
6
. 6
𝐼 = 41 ! 𝑒 .; 𝑑𝑥 ⋯ For 𝑥 ∈ 0, . , {2𝑥} = 2𝑥
/
%
< #& <36
= 41× |/ = 41
#
. .
 N-4| 𝜋
Evaluate 𝐼 = ! cos 𝑥 𝑑𝑥 𝑛∈ℕ, <𝑉<𝜋
/
2

N- N-4|
𝐼 =! cos 𝑥 𝑑𝑥 + ! cos 𝑥 𝑑𝑥 cos 𝑥 ∶ 𝑇 = 𝜋
/ N-

- | -
𝐼 = 𝑛 ! cos 𝑥 𝑑𝑥 + ! cos 𝑥 𝑑𝑥 𝑛 ! cos 𝑥 𝑑𝑥 = 2𝑛
/ / /
-
|
.
= 2𝑛 + ! cos 𝑥 𝑑𝑥 + ! − cos 𝑥 𝑑𝑥 ∵ 𝑉 is in 2N9 quadrant
-
/
.
|
= 2𝑛 + 1 − sin 𝑥 - = 2𝑛 + 1 − sin 𝑉 − 1 = 2𝑛 + 2 − sin 𝑉
.
 .N-
Evaluate 𝐼 =! max sin 𝑥 , sin36 sin 𝑥 𝑑𝑥
/

Graph of max sin 𝑥 , sin36 sin 𝑥

𝑌
sin36 sin 𝑥

sin 𝑥
2𝜋
𝑋
0 𝜋
sin 𝑥

This will repeat after 2𝜋

∴ Period 𝑇 = 2𝜋
 .N-
Evaluate 𝐼 =! max sin 𝑥 , sin36 sin 𝑥 𝑑𝑥
/

𝑌
- sin36 sin 𝑥
.N-
𝐼= ! max sin 𝑥 , sin36 sin 𝑥 𝑑𝑥 .
/
.-
⇒ 𝐼 = 𝑛 ! max sin 𝑥 , sin36 sin 𝑥 𝑑𝑥 2𝜋
/ - 𝑋
0 𝜋
- .-
. sin 𝑥
⇒ 𝐼 = 𝑛 ! sin36 sin 𝑥 𝑑𝑥 + ! sin 𝑥 𝑑𝑥
/ -

.-
1 𝜋
= 𝑛 . 𝜋. + ! sin 𝑥 𝑑𝑥
2 2 -

𝜋.
=𝑛 −2
4
If 𝑓 𝑥 + 1 + 𝑓 𝑥 + 7 = 0 ∀ 𝑥 ∈ ℝ, then minimum possible integer value of 𝑡 for which
+4G
! 𝑓 𝑥 𝑑𝑥 will be independent of 𝑎.
+

It can be independent only when 𝑡 is period 𝑓 𝑥 .

𝑓 𝑥+1 +𝑓 𝑥+7 =0

⇒ 𝑓 𝑥 + 1 = −𝑓 𝑥 + 6 + 1 ⇒ 𝑓 𝑥 + 7 = −𝑓 𝑥 + 6 + 7 ⋯ 𝑅𝑒𝑝𝑙𝑎𝑐𝑖𝑛𝑔 𝑥 𝑏𝑦 𝑥 + 6
⇒ −𝑓 𝑥 + 1 = −𝑓 𝑥 + 13 𝑥 + 1 → 𝑋

⇒ 𝑓 𝑋 = 𝑓 𝑋 + 12 ⇒ 𝑡 = 12
∴ For 𝑡 = 12 property will work & we will get terms independent of 𝑎
 Session 11
Definite Integration as limit of
sum, Leibnitz rule

Return To Top
 Key Takeaways

Property 8 : Inequality Based Property

𝑦 = 𝑓(𝑥)
𝑌
𝑦 = 𝑔(𝑥)
Ø If 3 functions 𝑓 𝑥 , 𝑔 𝑥 , ℎ 𝑥 are such that

𝑦 = ℎ(𝑥)
𝑓 𝑥 > 𝑔 𝑥 > ℎ 𝑥 , ∀ 𝑥 ∈ 𝑎, 𝑏 , then

, , ,
𝑂 𝑎 𝑏 𝑋
! 𝑓 𝑥 𝑑𝑥 > ! 𝑔 𝑥 𝑑𝑥 > ! ℎ 𝑥 𝑑𝑥
+ + +
 Key Takeaways

Property 8 : Inequality Based Property

𝑌
Ø If 𝑚 & 𝑀 are the minimum and maximum value of 𝑓(𝑥)
respectively in 𝑎, 𝑏 , then 𝑀
𝑦 = 𝑓(𝑥)
, , ,
! 𝑚 𝑑𝑥 ≤ ! 𝑓 𝑥 𝑑𝑥 ≤ ! 𝑀 𝑑𝑥
+ + + 𝑚

,
𝑂 𝑎 𝑏 𝑋
𝑚(𝑏 − 𝑎) ≤ ! 𝑓 𝑥 𝑑𝑥 ≤ 𝑀(𝑏 − 𝑎)
+
 Note:
𝑌 𝑦 = 𝑓(𝑥)

If 𝑓(𝑥) is strictly increasing in 𝑎, 𝑏 , then 𝑓(𝑏)

,
𝑓 𝑎 𝑏 − 𝑎 ≤ ! 𝑓 𝑥 𝑑𝑥 ≤ 𝑓 𝑏 𝑏 − 𝑎 .
+
, 𝑓(𝑎)
Ø Step 1 ∶ Check the monotonicity of the ! 𝑓 𝑥 𝑑𝑥 is
+ 𝑂 𝑎 𝑏 𝑋
in the interval 𝑎, 𝑏 by finding 𝑓 : 𝑥

Ø Step 2 ∶ Use Inequality Theorem

 6 6
𝑑𝑥 𝑑𝑥
If 𝐼D = ! & 𝐼! = ! , then prove that 𝐼D < 𝐼!.
1 + 𝑥. 𝑥
/ /

𝑥 ∈ 0,1

⇒ 1 + 𝑥. > 𝑥.

⇒ 1 + 𝑥. > 𝑥

1 1
⇒ <
1 + 𝑥. 𝑥

6 6
𝑑𝑥 𝑑𝑥
⇒! <!
1 + 𝑥. 𝑥
/ /

⇒ 𝐼6 < 𝐼.
 6 6
1 + 𝑥P 1 + 𝑥S
If 𝐼D = ! 𝑑𝑥 & 𝐼! = ! 𝑑𝑥 , then check whether 𝐼D > 𝐼! is TRUE .
1 + 𝑥1 1 + 𝑥2
/ /

𝑥 ∈ 0,1

⇒ 𝑥2 > 𝑥1
⇒ 𝑥P > 𝑥S

⇒ 1 + 𝑥P > 1 + 𝑥S ⇒ 1 + 𝑥2 > 1 + 𝑥1

1 + 𝑥P 1 + 𝑥S 1 1
∴ > ⇒ <
1 + 𝑥1 1 + 𝑥2 1 + 𝑥2 1 + 𝑥1

6 6
1 + 𝑥P 1 + 𝑥S
⇒! 𝑑𝑥 > ! 𝑑𝑥
1 + 𝑥1 1 + 𝑥2
/ /

⇒ 𝐼6 > 𝐼. True
 ! CLN @
Prove that 1 ≤ ∫JELN}
𝑑𝑥 ≤C

Solution :
5−𝑥 𝑥−5 Always check monotonic behaviour of 𝑓 𝑥 by 𝑓 : 𝑥 .
𝑓 𝑥 = =
9 − 𝑥. 𝑥. − 9
𝒙 𝑓(𝑥)
𝑥 . − 9 − 𝑥 − 5 ⋅ 2𝑥
:
𝑓 𝑥 = 0
− + −
𝑥. − 9 . 0
S
0 1 2 9
− 𝑥 . − 10𝑥 + 9 2
= 2 0
𝑀
𝑥. − 9 .
6
1 .
𝑚
− 𝑥−9 𝑥−1
=
𝑥. − 9 .
. .
1 5−𝑥 3 5−𝑥 6
2−0 ≤! 𝑑𝑥 ≤ 2−0 ⇒1≤! 𝑑𝑥 ≤
2 9 − 𝑥. 5 9 − 𝑥. 5
/ /
 -
.
𝜋 𝜋
Prove that ≤ ! sin 𝑥 6/
𝑑𝑥 ≤
128 4
-
1

6/
𝜋 𝜋 1 1
𝑚=𝑓 = sin6/ = =
4 4 2 32 𝑀
𝑚

𝜋 𝜋
𝑀=𝑓 = sin6/ = 1 6/
=1
2 2 - -
1 .

- 𝜋 𝜋
1 𝜋 𝜋 P 𝜋 𝜋 , in , sin 𝑥 is increasing
− ≤ ! sin 𝑥 6/
≤1 − 𝑚(𝑏 − 𝑎) ≤ ! 𝑓 𝑥 𝑑𝑥 ≤ 𝑀(𝑏 − 𝑎) 4 2
32 2 4 - 2 4 +
1

-
𝜋 .
6/
𝜋
⇒ ≤ ! sin 𝑥 ≤
128 - 4
1
 <
𝑒−1 𝑑𝑥 𝑒−1
Prove that ≤! ≤
3 2 + ln 𝑥 2
6

1
ln 𝑥 is an increasing funtion ⇒ is a decreasing funtion 𝑀
2 + ln 𝑥

1 1
𝑀=𝑓 1 = =
2 + ln 1 2 𝑚

1 1 1 𝑒
𝑚=𝑓 𝑒 = =
2 + ln 𝑒 3

< ,
1 𝑑𝑥 1
𝑒−1 ≤! ≤ 𝑒−1 𝑚(𝑏 − 𝑎) ≤ ! 𝑓 𝑥 𝑑𝑥 ≤ 𝑀(𝑏 − 𝑎)
3 6 2 + ln 𝑥 2 +

<
𝑒−1 𝑑𝑥 𝑒−1
⇒ ≤! ≤
3 6 2 + ln 𝑥 2
Leibnitz Theorem

7(;)
Ø If 𝐹 𝑥 = ! 𝑓 𝑡 𝑑𝑡 , then
>(;)

𝑑
𝐹 𝑥 = 𝑓 ℎ 𝑥 . ℎ: 𝑥 − 𝑓 𝑔 𝑥 . 𝑔 : 𝑥
𝑑𝑥

Ø If limits of integration are variable, then we use Leibnitz Theorem.

 ;#

If 𝑓 𝑥 = ! cos 𝑡 𝑑𝑡 ,then 𝑓′(𝑥) equals

;

;# 7(;)
𝑓 𝑥 = ! cos 𝑡 𝑑𝑡 , differentiate w.r.t 𝑥 If 𝐹 𝑥 = ! 𝑓 𝑡 𝑑𝑡 , then
; >(;)
𝑑
𝐹 𝑥 = 𝑓 ℎ 𝑥 . ℎ: 𝑥 − 𝑓 𝑔 𝑥 . 𝑔 : 𝑥
𝑑𝑥
𝑓 : 𝑥 = cos 𝑥 . ⋅ 2𝑥 − cos 𝑥 ⋅ 1

𝑓 : 𝑥 = 2𝑥 ⋅ cos 𝑥 . − cos 𝑥
 N D 1
Let 3 𝑓(𝑡) 𝑑𝑡 = 𝑥 ! + 3 𝑡 !𝑓 𝑡 𝑑𝑡 , then , 𝑓 n =?
2
J N
JEE Main Jan 2019

24 18
A B
25 25

4 6
C D
5 25
 N D 1
Let 3 𝑓(𝑡) 𝑑𝑡 = 𝑥 ! + 3 𝑡 !𝑓 𝑡 𝑑𝑡 , then , 𝑓 n =?
2
J N
JEE Main Jan 2019
Solution :
; 6
! 𝑓(𝑡) 𝑑𝑡 = 𝑥 + ! 𝑡 . 𝑓 𝑡 𝑑𝑡
.
/ ;
7(;)
Differentiate w.r.t 𝑥 ⇒ 𝑓 𝑥 = 2𝑥 + 0 − 𝑥 . 𝑓 𝑥 ⋅ 1 If 𝐹 𝑥 = ! 𝑓 𝑡 𝑑𝑡 , then
>(;)
𝑑
⇒ 𝑓 𝑥 = 2𝑥 − 𝑥 . 𝑓 𝑥 𝐹 𝑥 = 𝑓 ℎ 𝑥 . ℎ: 𝑥 − 𝑓 𝑔 𝑥 . 𝑔 : 𝑥
𝑑𝑥
2𝑥
⇒𝑓 𝑥 =
1 + 𝑥.

Differentiate w.r.t 𝑥

2 1 + 𝑥 . − 4𝑥 . 2 − 2𝑥 . 1 24
:
𝑓 𝑥 = = 𝑓: =
1 + 𝑥. . 1 + 𝑥. . 2 25
 N D 1
Let 3 𝑓(𝑡) 𝑑𝑡 = 𝑥 ! + 3 𝑡 !𝑓 𝑡 𝑑𝑡 , then , 𝑓 n =?
2
J N
JEE Main Jan 2019

24 18
A B
25 25

4 6
C D
5 25
 𝜋
Let 𝑓: ℝ → ℝ be a differentiable function such that 𝑓 0 = 0, 𝑓 =3
0 2
1 g
and 𝑓 n 0 = 1. If 𝑔 𝑥 = * 𝑓 2 𝑡 cosec 𝑡 − cot 𝑡 cosec 𝑡 𝑓 𝑡 𝑑𝑡 for 𝑥 ∈ 0, ! ,
/
then lim 𝑔 𝑥 = _______.
N→J
JEE Advanced 2017
Solution :
-
𝜋 .
𝑓 0 = 0, 𝑓 = 3 and 𝑓 : 0 = 1. 𝑔 𝑥 = ! 𝑓 : 𝑡 cosec 𝑡 − cot 𝑡 cosec 𝑡 𝑓 𝑡 𝑑𝑡
2 ;
-
.
𝑔 𝑥 = ! 𝑓 : 𝑡 cosec 𝑡 − cot 𝑡 cosec 𝑡 𝑓 𝑡 𝑑𝑡
;

-
. 𝑑
⇒𝑔 𝑥 =! 𝑓 𝑡 cosec 𝑡 𝑑𝑡
; 𝑑𝑡

𝜋 𝜋 𝑓 𝑥
⇒𝑔 𝑥 =𝑓 cosec − 𝑓 𝑥 cosec 𝑥 ⇒ 𝑔 𝑥 = 3 −
2 2 sin 𝑥
𝑓 𝑥 𝑓: 𝑥 1
lim 𝑔 𝑥 = 3 − lim ⇒ 3 − lim =3− =2
;→/ ;→/ sin 𝑥 ;→/ cos 𝑥 1
Let a function 𝑓 ∶ 0,5 → ℝ be continuous , 𝑓 1 = 3 and 𝐹 be defined as:
N t
𝐹 𝑥 =3 𝑡 !𝑔 𝑡 𝑑𝑡 , where 𝑔 𝑡 = 3 𝑓(𝑢)𝑑𝑢 .Then for the function 𝐹 the point
D D
𝑥 = 1 is :
JEE Main Jan 2020
Solution :
; G
𝐹 𝑥 = ! 𝑡 . 𝑔 𝑡 𝑑𝑡 𝑔 𝑡 = ! 𝑓(𝑢)𝑑𝑢
6 6
Differentiate w.r.t 𝑥
𝐹: 𝑥 = 𝑥 . 𝑔 𝑥 𝑔: 𝑡 = 𝑓 𝑡
Since, 𝑔 1 = 0 ⇒ 𝐹 : 1 = 0
⇒ 𝑔: 1 = 𝑓 1 = 3
Differentiate again w.r.t 𝑥

𝐹 :: 𝑥 = 2𝑥𝑔 𝑥 + 𝑥 . 𝑔′(𝑥)

𝐹 :: 1 = 2𝑔 1 + 𝑔: 1 = 3 > 0
 Let 𝑓: 𝑅 → 𝑅 be a continuously differentiable function such that 𝑓 2 = 6,
v(N)
D
𝑓n 2 = . If 3 4𝑡 )𝑑𝑡 = 𝑥 − 2 𝑔 𝑥 , then lim 𝑔(𝑥) is equal to :
Au @ N→!
JEE Main Apr 2019

A 18 B 36

C 12 D 24
 Let 𝑓: 𝑅 → 𝑅 be a continuously differentiable function such that 𝑓 2 = 6,
v(N)
D
𝑓n 2 = . If 3 4𝑡 )𝑑𝑡 = 𝑥 − 2 𝑔 𝑥 , then lim 𝑔(𝑥) is equal to :
Au @ N→!
JEE Main Apr 2019
Solution :
~(;)
! 4𝑡 2 𝑑𝑡
lim lim
𝑔 𝑥 = ;→. =
0/0 form Applying L.H. rule
;→.
𝑥−2
2
4 𝑓 𝑥 . 𝑓: 𝑥 2
= lim =4 𝑓 2 . 𝑓: 2
;→. 1

1. = $ .6
=
1P

= 18
 Let 𝑓: 𝑅 → 𝑅 be a continuously differentiable function such that 𝑓 2 = 6,
v(N)
D
𝑓n 2 = . If 3 4𝑡 )𝑑𝑡 = 𝑥 − 2 𝑔 𝑥 , then lim 𝑔(𝑥) is equal to :
Au @ N→!
JEE Main Apr 2019

A 18 B 36

C 12 D 24
 1 1 1 1
lim + + +⋯ =?
I→K 𝑛+1 𝑛+2 𝑛+3 6𝑛

Solution :
1 1 1 1
= lim + + +⋯
I→K 𝑛+1 𝑛+2 𝑛+3 𝑛 + 5𝑛

1
𝑇w =
𝑛+𝑟
CI CI/I
1 1
⇒ lim i ⇒ lim i 𝑟
I→K 𝑛+𝑟 I→K
wxD wxD/I 𝑛 1 + 𝑛

C 𝑑𝑥
=3 C = ln 6 − ln 1 = ln 6
= ln 𝑥 + 1 J
J 1+𝑥
 When 𝑥 = 𝑎 + ℎ, 𝑦 = 𝑓 𝑎 + 2ℎ
Key Takeaways
𝑦 = 𝑓(𝑥)
Definite Integral as Limit of Sum
N
,
Ø ! 𝑓 𝑥 𝑑𝑥 = 7→/
lim ℎ š 𝑓(𝑎 + 𝑟ℎ)
+ •v6
𝑏 − 𝑎 = 𝑛ℎ
Area of strips = ℎ 𝑓 𝑎 + ℎ + 𝑓 𝑎 + 2ℎ + ⋯ + 𝑓 𝑎 + 𝑛ℎ
𝑏 = 𝑎 + 𝑛ℎ
N
𝑎 𝑎 + ℎ 𝑎 + 2ℎ 𝑎+ 𝑛−1 ℎ 𝑏
= lim ℎ š 𝑓 𝑎 + 𝑟ℎ
7→/
•v6

Special case : If 𝑎 = 0 , 𝑏 = 1
6
𝑛ℎ = 1 ⇒ ℎ = N

N N
, 6 1 𝑟
lim ℎ š 𝑓(𝑎 + 𝑟ℎ) ⇒ ! 𝑓 𝑥 𝑑𝑥 = N→J
! 𝑓 𝑥 𝑑𝑥 = 7→/ lim š𝑓
/ 𝑛 𝑛
+ •v6 •v6

• 6
Ø Replace N → 𝑥 ; N
→ 𝑑𝑥 ; ∑ → ∫
 ) I I I I
lim 1+ + + + ⋯ =?
I→K I Iq) Iq@ IqE Iq) ILD

Solution :
N36
3 𝑛
= lim š
N→J 𝑛 𝑛 + 3𝑟
•v/

6
63 v6
N
3 1
= lim š
N→J 𝑛 1 + 3𝑟
• /
v v/
N N 𝑛

6 6
𝑑𝑥 2 1 + 3𝑥
= 3! = 3× = 2×2 − 2×1 = 2
/ 1 + 3𝑥 3 /
 N N36
𝑛 𝑛
Let 𝑆I = š . and 𝑇I = š for 𝑛 = 1 , 2 , 3 , ⋯ . Then
𝑛 + 𝑘𝑛 + 𝑘 . 𝑛. + 𝑘𝑛 + 𝑘 .
uv6 uv/

N IIT JEE 2008

𝑛
Solution : Let 𝑃 = lim 𝑆N = N→J
lim š
N→J 𝑛. + 𝑘𝑛 + 𝑘 .
uv6 -
N A 𝑆N <
1 𝑛 2 2
𝑃 = lim š u 6
N→J 𝑛 𝑘 𝑘 . → 𝑥; → 𝑑𝑥
uv6 1 + + N N
𝑛 𝑛 -
C 𝑇N < 2 2
6
1
= ! .
𝑑𝑥
/ 1+𝑥+ 𝑥
B 𝑆N > 2
-
2
6
1
=! . 𝑑𝑥
3 1 -
/
4
+ 𝑥+
2 D 𝑇N > 2 2

. .;46 6 -
= tan36 =
2 2 2 2
/
 N N36
𝑛 𝑛
Let 𝑆I = š . and 𝑇I = š for 𝑛 = 1 , 2 , 3 , ⋯ . Then
𝑛 + 𝑘𝑛 + 𝑘 . 𝑛. + 𝑘𝑛 + 𝑘 .
uv6 uv/
IIT JEE 2008
Solution :
-
𝑃 = lim 𝑆N = 2 2
N→J

-
𝑆N < lim 𝑆N 𝑆N < 2
N→J 2

0 1
N36
Let 𝑄 = lim 𝑇N = lim š 𝑛
N→J N→J
𝑛. + 𝑘𝑛 + 𝑘 .
uv/

N36
1 𝑛
𝑄 = lim š
N→J 𝑛 𝑘 𝑘 .
uv/ 1 + +
𝑛 𝑛
 N N36
𝑛 𝑛
Let 𝑆I = š . and 𝑇I = š for 𝑛 = 1 , 2 , 3 , ⋯ . Then
𝑛 + 𝑘𝑛 + 𝑘 . 𝑛. + 𝑘𝑛 + 𝑘 .
uv6 uv/

Solution :
N36
𝑛
Let 𝑄 = lim 𝑇N = lim š
N→J N→J 𝑛. + 𝑘𝑛 + 𝑘 . 1 3
uv/ → 𝑥; → 𝑑𝑥
2 2
N36
1 𝑛
𝑄 = lim š
N→J 𝑛 𝑘 𝑘 .
uv/ 1 + +
𝑛 𝑛
6
1
=! .
𝑑𝑥
/ 1+𝑥+ 𝑥

-
=2 2
 N N36
𝑛 𝑛
Let 𝑆I = š . and 𝑇I = š for 𝑛 = 1 , 2 , 3 , ⋯ . Then
𝑛 + 𝑘𝑛 + 𝑘 . 𝑛. + 𝑘𝑛 + 𝑘 .
uv6 uv/

Solution : -
A 𝑆N < 2 2
-
𝑄 = lim 𝑇N = 2 2
N→J
-
C 𝑇N < 2 2

𝑇N > lim 𝑇N
N→J
B 𝑆N > 2
-
0 2
1
-
𝑇N > 2 2
-
D 𝑇N > 2 2
 .N
1 𝑟
lim š =?
N→J 𝑛 𝑛 . + 𝑟.
•v6

Solution :
.N
N
1 𝑟
= lim š
N→J 𝑛 . .
6 𝑛 +𝑟
•v
N

.N
N
1 𝑟
= lim š
N→J 𝑛 𝑟 .
6
•v 𝑛 1+ 𝑛
N

.
𝑥 .
=! 𝑑𝑥 = 1 + 𝑥. = 5−1
1+𝑥 . /
/
 >
𝑛! ;
lim =?
;→= 𝑛

Solution :
6 6
𝑛! N 1 2 3 𝑛 N
𝑦 = lim = lim ⋯
N→J 𝑛N N→J 𝑛 𝑛 𝑛 𝑛

1 1 2 3 𝑛
⇒ log 𝑦 = lim log ⋯
N→J 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛N = 𝑛×𝑛×𝑛× ⋯×𝑛

𝑛! = 1×2×3× ⋯×𝑛
1 1 2 3 𝑛
⇒ log 𝑦 = lim log + log + log + ⋯ log
N→J 𝑛 𝑛 𝑛 𝑛 𝑛

N/N 6
𝑟 1
⇒ log 𝑦 = lim š log ⋅ ⇒ log 𝑦 = ! log 𝑥 𝑑𝑥
N→J 𝑛 𝑛 /
•v6/N

6
⇒ log 𝑦 = 𝑥 log 𝑥 − 𝑥 / ⇒ log 𝑦 = −1 ⇒ 𝑦 = 𝑒 36