JEE Mathematics/ Lakshmikanta Satapathy/ Questions on Indefinite Integration Part 14 taken from previous Board papers are solved using integration by parts with alternate methods

1. Physics Helpline
L K Satapathy
Indefinite Integrals 14

2. Physics Helpline
L K Satapathy
Indefinite Integrals - 14
Answer-1 log .I x dx 
. . .u dv u v v du  We use
log
dx
Put u x du
x
  
2
. .
2
x
And dv x dx v x dx   
2 2
(log )
2 2
x x dx
I x
x
 
    
 

21 1
(log ) .
2 2
x x x dx  
2
21 1
(log )
2 2 2
x
x x C   
2 2
(log )
2 4
[ ]
x x
x sC An  

3. Physics Helpline
L K Satapathy
Indefinite Integrals - 14
Answer-2 2
sin .I x x dx 
2
2 .Put u x du x dx  
sin . sin . cosAnd dv x dx v x dx x    
We use
2 2
( cos ) ( cos ).2 . cos 2 . cos .I x x x x dx x x x x dx        
. cos . , cos . sinFor x x dx put u x du dx and dv x dx v x     
. cos . sin sin . sin ( cos ) sin cosx x dx x x x dx x x x x x x        
2
cos 2[ sin cos ]I x x x x x C     
. . .u dv u v v du  
2
(2 )cos 2 sin [ ]x x x x nsC A   

4. Physics Helpline
L K Satapathy
Indefinite Integrals - 14
Answer-3 1
sin .I x dx
 
1
2
sin
1
dx
Put u x du and dv dx v x
x

     

Method 1 :
We use
1
2
(sin ). .
1
x
I x x dx
x

  


2 2 2
2
. , 1 1 2 . 2 . . .
1
x
For dx put x t x t x dx t dt x dx t dt
x
          


1 1 1.
sin sin sin
t dt
I x x x x dt x x t C
t
  
        
1 2
sin 1 [ ]x Anx x sC
   
. . .u dv u v v du  

5. Physics Helpline
L K Satapathy
Indefinite Integrals - 14
1
sin sin cos .Put x t x t dx t dt
    Method 2 :
1 2
sin 1 [ ]x Anx x sC
   
1
sin . cos .I x dx t t dt
   
cos . sinPut u t du dt and dv t dt v t     
sin sin .I t t t dt   
We use
2
sin ( cos ) sin 1 sint t t C t t t C       
. . .u dv u v v du  

6. 10/11/2015
Physics Helpline
L K Satapathy
Indefinite Integrals - 14
Answer-4 1
tan .I x dx
 
1
2
tan
1
dx
Put u x du and dv dx v x
x

     

Method 1 :
We use
1
2
(tan ). .
1
x
I x x dx
x

  

2
2
. , 1 2 . .
1 2
x dt
For dx put x t x dx dt x dx
x
     

1 11 1
tan tan log
2 2
dt
I x x x x t C
t
 
     
1 21
tan log 1
2
[ ]x x x AnsC
   
. . .u dv u v v du  

7. Physics Helpline
L K Satapathy
Indefinite Integrals - 14
1 2
tan tan sec .Put x t x t dx t dt
    Method 2 :
1 2
tan . sec .I x dx t t dt
   
2
sec . tanPut u t du dt and dv t dt v t     
sin .
tan tan . tan tan log(cos )
cos
t dt
I t t t dt t t t t t C
t

        
We use
1 21
tan log 1
2
[ ]x x x AnsC
   
21
tan log(sec ) tan logsec
2
t t t C t t t C     
21
tan log 1 tan
2
t t t C   
. . .u dv u v v du  

8. Physics Helpline
L K Satapathy
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