JEE Mathematics/ Lakshmikanta Satapathy/ Questions on Indefinite Integration Part 13, taken from previous Board Papers solved by the method of substitution using standard integrals

1. Physics Helpline
L K Satapathy
Indefinite Integrals 13

2. Physics Helpline
L K Satapathy
Indefinite Integrals - 13
Answer-1
sec 1.I x dx 
(1 cos )(1 cos )
.
cos (1 cos )
x x
dx
x x
 


1 cos
.
cos
x
dx
x

 
2
sin
.
cos cos
x
dx
x x



1
1 .
cos
dx
x
 
  
 


3. Physics Helpline
L K Satapathy
Indefinite Integrals - 13
cos sin . sin .Put x t x dx dt x dx dt      Substitution :
2
2 1 1
4 4
dt dt
I
t t t t
    
   
 
2 2
1 1 1
log
2 2 2
t t C
     
          
     
21
log
2
t t t C
 
     
 
2 2
1 1
2 2
dt
t
 
   
    
   

2 2
2 2
log
dx
using x x a
x a
  


2
[ ]
1
log cos cos cos
2
x x x C Ans
 
     
 

4. Physics Helpline
L K Satapathy
Indefinite Integrals - 13
Answer-2
Substitution : sin cos .Put x t x dx dt  
2
2sin2 cos
.
6 cos 4sin
x x
I dx
x x


 
2
(4sin 1)cos
.
sin 4sin 5
x x
dx
x x


 
2
(4 1)
.
4 5
t
I dt
t t

 
 
1 22 2
2 4
2 . 7 ( )
4 5 4 5
t dt
dt I I say
t t t t

   
    
2 2
sin2 2sin cos
cos 1 sin
x x x
x x

 
2
4sin cos cos
.
6 (1 sin ) 4sin
x x x
dx
x x


  
2 2
4 8 7 2(2 4) 7
. .
4 5 4 5
t t
dt dt
t t t t
   
 
    

5. Physics Helpline
L K Satapathy
Indefinite Integrals - 13
2 1
1 2 2log 4 5 7tan ( 2)I I I t t t C
        
2 1
2log sin 4sin 5 7tan (sin 2) [ ]Anx x x C s
     
1 2
2 4
2 .
4 5
t
I dt
t t

 
 
2 2 2
7 7
4 5 4 4 1
dt dt
And I
t t t t
 
     
2
4 5 (2 4)Put t t u t dt du     
2
1 2 2log 2log 4 5
du
I u t t
u
     
1
2 2
7 7tan ( 2)
( 2) 1
dt
t
t

  
  1
2 2
1
tan
dx x
using
x a a a




6. Physics Helpline
L K Satapathy
Indefinite Integrals - 13
Answer-3
2
2
.
2 1
x
I dx
x x


 

1 22 2
1 2 2
.
2 2 1 2 1
x dx
dx I I
x x x x

   
   
 
2
1 (2 2) 2
.
2 2 1
x
dx
x x
 

 

2
1 2 4
.
2 2 1
x
dx
x x


 

1 2
1 2 2
.
2 2 1
x
I dx
x x

 
 

2
2 1 (2 2).Put x x t x dx dt     
2
1 2 1
2
dt
I t x x
t
     

7. Physics Helpline
L K Satapathy
Indefinite Integrals - 13
2 2 2
2 1 2 1 2
dx dx
And I
x x x x
 
    
 
2 2
2 1 log ( 1) 2 [ ]1x x x x C Ansx        
 
2
2
log ( 1) ( 1) 2x x     2 2
2 2
log
dx
using x x a
x a
  


2
log ( 1) 2 1x x x    
1 2I I I  
 
2
2
( 1) 2
dx
x

 


8. Physics Helpline
L K Satapathy
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