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Physics Helpline
L K Satapathy
Indefinite Integrals 18
Physics Helpline
L K Satapathy
Indefinite Integrals - 18
Method : [ ( ) ( )]. . ( )x x
e f x f x dx e f x C  
Proof : [ ( ) ( )]. ( ). ( ).x x x
I e f x f x dx e f x dx e f x dx      
( ). , ( ) ( ).x x x
For e f x dx put u f x du f x dx and dv e dx v e     
, . .we use u dv uv v du  
( ). . ( ). ( ).x x x
I f x e e f x dx e f x dx     
. ( ) . ( ). ( ).x x x
e f x e f x dx e f x dx    
. ( )x
e f x C 
Physics Helpline
L K Satapathy
Indefinite Integrals - 18
Answer-1
2
1
.x x
I e dx
x
 
  
 

2 2
1 1 1 1
. . . . .x x x
e dx e dx e dx
x x x x
 
    
 
  
2
1 1 1
. . ,x x x
For e dx put u du dx and dv e dx v e
x x x

     
2 2
1 1 1
. . . .x x x
I e e dx e dx
x x x

    
2 2
1 1 1
. . . .x x x
e e dx e dx
x x x
   
[
1
]x
e C s
x
An 
Physics Helpline
L K Satapathy
Indefinite Integrals - 18
Answer-2 2
.
( 1)
x x
I e dx
x


2 2 2
1 1 1 1
. . .
( 1) ( 1) ( 1)
x x xx x
e dx e dx e dx
x x x
  
  
    
2
1 1
. .
( 1) ( 1)
x x
e dx e dx
x x
 
  
2
1 1 1
. ,
( 1) ( 1) ( 1)
x x x
For e dx putu du dx and dv e dx v e
x x x

     
  
2 2
1 1 1
. .
1 ( 1) ( 1)
x x x
I e e dx e dx
x x x

   
   
2 2
1 1 1
. .
1 ( 1) ( 1)
x x x
e e dx e dx
x x x
  
   
[
1
1
]x
e nsC
x
A 

Physics Helpline
L K Satapathy
Indefinite Integrals - 18
Answer-3 2 sin2
.
1 cos2
x x
I e dx
x
 
   

2 2
2 2sin cos 1 sin cos
. .
2cos cos
x xx x x x
e dx e dx
x x
    
    
   
 
2
(sec tan ).x
e x x dx 
2
, tan . , tan sec ,x x x
For e x dx put u x du xdx dv e dx v e     
2 2
tan sec . sec .x x x
I e x e x dx e x dx    
[ ]tanx
e x C Ans 
2
tan . sec .x x
e x dx e x dx  
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L K Satapathy
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Indefinite Integral 18

  • 1. Physics Helpline L K Satapathy Indefinite Integrals 18
  • 2. Physics Helpline L K Satapathy Indefinite Integrals - 18 Method : [ ( ) ( )]. . ( )x x e f x f x dx e f x C   Proof : [ ( ) ( )]. ( ). ( ).x x x I e f x f x dx e f x dx e f x dx       ( ). , ( ) ( ).x x x For e f x dx put u f x du f x dx and dv e dx v e      , . .we use u dv uv v du   ( ). . ( ). ( ).x x x I f x e e f x dx e f x dx      . ( ) . ( ). ( ).x x x e f x e f x dx e f x dx     . ( )x e f x C 
  • 3. Physics Helpline L K Satapathy Indefinite Integrals - 18 Answer-1 2 1 .x x I e dx x         2 2 1 1 1 1 . . . . .x x x e dx e dx e dx x x x x             2 1 1 1 . . ,x x x For e dx put u du dx and dv e dx v e x x x        2 2 1 1 1 . . . .x x x I e e dx e dx x x x       2 2 1 1 1 . . . .x x x e e dx e dx x x x     [ 1 ]x e C s x An 
  • 4. Physics Helpline L K Satapathy Indefinite Integrals - 18 Answer-2 2 . ( 1) x x I e dx x   2 2 2 1 1 1 1 . . . ( 1) ( 1) ( 1) x x xx x e dx e dx e dx x x x            2 1 1 . . ( 1) ( 1) x x e dx e dx x x      2 1 1 1 . , ( 1) ( 1) ( 1) x x x For e dx putu du dx and dv e dx v e x x x           2 2 1 1 1 . . 1 ( 1) ( 1) x x x I e e dx e dx x x x          2 2 1 1 1 . . 1 ( 1) ( 1) x x x e e dx e dx x x x        [ 1 1 ]x e nsC x A  
  • 5. Physics Helpline L K Satapathy Indefinite Integrals - 18 Answer-3 2 sin2 . 1 cos2 x x I e dx x        2 2 2 2sin cos 1 sin cos . . 2cos cos x xx x x x e dx e dx x x                 2 (sec tan ).x e x x dx  2 , tan . , tan sec ,x x x For e x dx put u x du xdx dv e dx v e      2 2 tan sec . sec .x x x I e x e x dx e x dx     [ ]tanx e x C Ans  2 tan . sec .x x e x dx e x dx  
  • 6. Physics Helpline L K Satapathy For More details: www.physics-helpline.com Subscribe our channel: youtube.com/physics-helpline Follow us on Facebook and Twitter: facebook.com/physics-helpline twitter.com/physics-helpline