JEE Mathematics/ Lakshmikanta Satapathy/ Indefinite Integrals/ Questions on Integration by parts involving trigonometric, inverse trigonometric and logarithmic functions solved with substitution

1. Physics Helpline
L K Satapathy
Indefinite Integrals 17

2. Physics Helpline
L K Satapathy
Indefinite Integrals - 17
Answer-1 2
(log ) .I x x dx 
2
2
x
and dv xdx v  
2
(log ) 2log .
dx
Put u x du x
x
  
2 2
2 2log
(log ) .
2 2
x x x
I x dx
x
   
2
2
(log ) log .
2
x
x x x dx  
We use udv uv vdu  

3. Physics Helpline
L K Satapathy
Indefinite Integrals - 17
2
2
x
and dv xdx v  
log
dx
Put u x du
x
  
2 2 2
2
(log ) log .
2 2 2
x x x dx
I x x
x
 
    
 

2 2
2 1
(log ) log
2 2 2
x x
x x xdx   
2 2 2
2
(log ) log
2 2 4
x x x
x x C   
2
2 1
(log ) log
2 2
[ ]
x
Anx x C s
 
    
 
log .For x x dx

4. Physics Helpline
L K Satapathy
Indefinite Integrals - 17
Answer-2
2 1
2 3 2
sin
.
(1 )
x x
I dx
x



1
2
sin , sin
1
dx
Put x t dt x t
x

   

2 1
2 2
sin
.
(1 ) 1
x x
I dx
x x

 
 

2 2
tan . (sec 1).t t dt t t dt   
2
2 2
sec . . sec .
2
t
t t dt t dt t t dt     
2 2
2 2
(sin ). sin
. .
(1 sin ) cos
t t t t
dt dt
t t
 
 

5. Physics Helpline
L K Satapathy
Indefinite Integrals - 17
2
sec . tanand dv t dt v t  
Put u t du dt  
2 2
sin
tan tan . tan .
2 cos 2
t t t
I t t t dt t t dt
t

       
2
tan logcos
2
t
t t t C   
2
2
cos 1 , tan
1
x
Putting t x t
x
  

1 2
1 2
2
(sin )
sin log 1
21
x x
I x x C
x


     

1 2 1 2
2
1 1
sin log 1 (sin )
2 21
[ ]
x
x x Ansx C
x
 
    

2
sec .For t t dt

6. Physics Helpline
L K Satapathy
Indefinite Integrals - 17
Answer-3 1
tan .I x dx
 
2
2Put x t x t dx tdt    
1
(tan )2 .I t t dt
  
2
1
2
tan
1 2
dt t
Put u t du and dv tdt v
t

     

2 2
1
2
2 tan .
2 2 1
t t dt
I t
t
 
   
 

1
2 tan .t t dt
 
2
2 1
2
tan .
1
t
t t dt
t

 


7. Physics Helpline
L K Satapathy
Indefinite Integrals - 17
2
2 1
2
1 1
tan .
1
t
I t t dt
t
  
  

2 1 1
tan tant t t t C 
   
1 1
tan tanx x x x C 
   
1
( 1) [ ]tanx x A sx nC
   
2
2 1
2 2
1
tan .
1 1
t dt
t t dt
t t
 
  
  
2 1
2
tan
1
dt
t t dt
t

  
 
[ ]Putting t x

8. Physics Helpline
L K Satapathy
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