JEE Mathematics/ Lakshmikanta Satapathy/ Definite integrals part 8/ JEE question on definite integral involving integration by parts solved with complete explanation

1. Physics Helpline
L K Satapathy
Definite Integrals 8

2. Physics Helpline
L K Satapathy
Definite Integrals - 8
Question : The value of is equal to
22
2
cos
1 x
x x
dx
e




2 2
2 22 2
( ) 2 ( ) 2 ( ) ( )
4 4
a b c e d e
 
 
    
Answer :
22
2
cos
1 x
x x
I dx
e





[Putting x = – t ]
22
2
cos
( )
1 t
t t
dt
e




 

The given integral
22
2
cos
1 t
t t
dt
e






. . . (1)
( ) ( )
b a
a b
f x dx f x dx
 
  
 
 

3. Physics Helpline
L K Satapathy
Definite Integrals - 8
22
2
cos
1 x
x x
I dx
e




 

2
2
2
cosx xdx



 
22
2
cos
1
x
x
x e x
I dx
e



 

[Replacing t by x ]
[ Multiplying numerator
& denominator by ]x
e
. . . (2)
[ Adding equations (1) & (2) ]
22
2
cos (1 )
2
1
x
x
x x e
I dx
e




 


4. Physics Helpline
L K Satapathy
Definite Integrals - 8
2
22
0
0
sin 2 sinI x x x xdx


     
2
2
0
2 2 cosI x xdx

   [ Since is an even function]
2
2
0
cosI x xdx

  
2
cosx x
Integrating by parts :
2
& cosx u xdx dv  2 & sindu xdx v x  
udv uv vdu  
22
0
2 sin
4
I x xdx


    . . . (3)

5. Physics Helpline
L K Satapathy
Definite Integrals - 8
Correct option = (a)
Again integrating by parts :sinx xdx
& sin & cosx u xdx dv du dx v x     
sin ( cos ) cosx xdx x x xdx     
cos cosx x xdx   
cos sinx x x  
   
2
2 2
0 0
0
sin cos sin 0 1 1x xdx x x x

 
      
22 2
0
(3) 2 sin 2
4 4
[ ]I x xd sx An

 
    

6. Physics Helpline
L K Satapathy
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