The document discusses techniques for solving indefinite integrals, focusing on integration by parts and substitution methods. It includes step-by-step instructions and examples using various functions to demonstrate the integration process. Additionally, it provides resources for further support and learning in physics and mathematics.

1. Physics Helpline
L K Satapathy
Indefinite Integrals 3

2. Physics Helpline
L K Satapathy
Indefinite Integrals - 3
Integration by Parts : The products of two functions may be integrated as follows
( . ) ( ) . .
d dv du
u v u v d uv u dv v du
dx dx dx
    
. ( ) . . ( ) .u dv d uv v du u dv d uv v du       
. .u dv uv v du   
The two parts of the function on LHS are (u) and (dv)
On the RHS , we need (u) , (v) and (du)
We get (v) by integrating (dv) and (du) by differentiating (u)
 Choose (u) which is differentiable and (dv) which is integrable.
We differentiate the product (u.v) , rearrange and then integrate as follows

3. Physics Helpline
L K Satapathy
Indefinite Integrals - 3
Answer-1
Substitution :
2 2
.I a x dx 
2
2 2
2 2
. .
x
I u dv uv v du x a x dx
a x
      

  
2 2 2 2
1 2 2 2 2
. .
x x a a
I dx dx
a x a x
 
 
 
 
2 2
u a x 
2 2 2 2
2 . .
2
x dx x dx
du
a x a x
  
 
Now
2 2
1 ( ) . . . (1)I x a x I say   
2 2 2
1 2 2 2 2
. .
a x a
I dx dx
a x a x

  
 
 
and dv dx
dv dx
v x

 
 
 
Separating the terms

4. Physics Helpline
L K Satapathy
Indefinite Integrals - 3
 2 2 2 2 2
(1) logI x a x I a x a x      
2 2 2 2 2
2 logI x a x a x a x     
2 2 2 2 2
logx a x I a x a x     
2
2 2 2 2
log
2 2
[ ]
x a
I a x x a x C Ans      
2 2 2
1 2 2
.
dx
I a x dx a
a x
   

 
2 2 2
1 logI I a x a x    
2 2
2 2
log
dx
x a x
a x
  



5. Physics Helpline
L K Satapathy
Indefinite Integrals - 3
Answer-2 2 2
.I a x dx 
2 2 2 2
2 . .
2
x dx x dx
du
a x a x
 
  
 
2
2 2
2 2
. .
x
I u dv uv v du x a x dx
a x

      

  
2 2 2 2
1 2 2 2 2
. .
x x a a
I dx dx
a x a x
   
 
 
 
2 2
u a x Substitution :
Now
2 2
1 ( ) . . . (1)I x a x I say   
2 2 2
1 2 2 2 2
. .
a x a
I dx dx
a x a x

  
 
  Separating the terms
and dv dx
dv dx
v x

 
 
 

6. Physics Helpline
L K Satapathy
Indefinite Integrals - 3
2 2 2 1
(1) sin
x
I x a x I a
a
  
      
  
2 2 2 1
2 sin
x
I x a x a
a
  
     
 
2 2 2 1
sin
x
I x a x I a
a
  
      
 
2
2 2 1
sin [ ]
2 2
x a
Anx s
x
I a C
a
  
     
 
2 2 2
1 2 2
.
dx
I a x dx a
a x
   

 
2 1
1 sin
x
I I a
a
  
    
 
1
2 2
sin
dx x
aa x
  
  
 


7. Physics Helpline
L K Satapathy
Indefinite Integrals - 3
Answer-3
2 2
.I x a dx 
2 2 2 2
2 . .
2
x dx x dx
du
x a x a
  
 
2
2 2
2 2
. .
x
I u dv uv v du x x a dx
x a
      

  
2 2 2 2
1 2 2 2 2
. .
x x a a
I dx dx
x a x a
 
 
 
 
Substitution :
Now
2 2
u x a 
2 2
1 ( ) ... (1)I x x a I say   
2 2 2
1 2 2 2 2
. .
x a a
I dx dx
x a x a

  
 
 
and dv dx
dv dx
v x

 
 
 
Separating the terms

8. Physics Helpline
L K Satapathy
Indefinite Integrals - 3
 2 2 2 2 2
(1) logI x x a I a x x a      
2 2 2 2 2
2 logI x x a a x x a     
2 2 2 2 2
logI x x a I a x x a      
2
2 2 2 2
log
2 2
[ ]
x a
I x a x x a C Ans      
2 2 2
1 2 2
.
dx
I x a dx a
x a
   

 
2 2 2
1 logI I a x x a    
2 2
2 2
log
dx
x x a
x a
  



9. Physics Helpline
L K Satapathy
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