Physics Helpline
L K Satapathy
Indefinite Integrals 15
Physics Helpline
L K Satapathy
Indefinite Integrals - 15
Answer-1
. . .u dv v u v du  We use
1
tan .I x x dx
 
2
1
2
tan
1 2
dx x
Put u x du and dv xdx v
x

     

2 2
1
2
1
(tan ) . .
2 2 1
x x
I x dx
x

  

2 2
1
2
1 1 1
(tan ) .
2 2 1
x x
x dx
x
  
 

2 2
1
2 2
1 1 1
(tan ) . .
2 2 1 1
x x
x dx dx
x x
  
   
  
 
2
1
2
1
(tan )
2 2 1
x dx
x dx
x
  
    
 
 
2
1 11
(tan ) tan [ ]
2 2
x
x x x C Ans 
   
Physics Helpline
L K Satapathy
Indefinite Integrals - 15
Answer-2
sin
.
1 cos
x x
I dx
x



sin
. .
1 cos 1 cos
x x
dx dx
x x
 
  
2 2
2sin cos
2 2. .
2sin 2sin
2 2
x x
x
dx dx
x x
  
2
cosec . cot .
2 2 2
x x x
dx dx  
1 cot . [ ]
2
x
I dx say  
Physics Helpline
L K Satapathy
Indefinite Integrals - 15
2 2
1
1
cosec . cosec .
2 2 2 2
x x x
For I dx x dx  
2
cosec 2cot
2 2
x x
and dv dx v   
1
1
2cot 2cot .
2 2 2
x x
I x dx
    
        
    

Put u x du dx  
cot cot . cot .
2 2 2
x x x
I x dx dx     
cot
2
[ ]
x
x C Ans  
cot cot .
2 2
x x
x dx   
Physics Helpline
L K Satapathy
Indefinite Integrals - 15
Answer-3 3 2
sec . sec .sec .I x dx x x dx  
2
sec sec tan & sec tanPut u x du x xdx dv xdx v x     
sec tan tan .sec tan .I x x x x x dx   
2
sec tan sec (sec 1).x x x x dx  
3
sec tan sec . sec .x x x dx x dx   
sec tan log sec tanx x I x x   
2 sec tan log sec tanI x x x x   
1
sec tan log sec tan [
2
]I x sC nx x Ax      
Physics Helpline
L K Satapathy
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Indefinite Integral 15

  • 1.
    Physics Helpline L KSatapathy Indefinite Integrals 15
  • 2.
    Physics Helpline L KSatapathy Indefinite Integrals - 15 Answer-1 . . .u dv v u v du  We use 1 tan .I x x dx   2 1 2 tan 1 2 dx x Put u x du and dv xdx v x         2 2 1 2 1 (tan ) . . 2 2 1 x x I x dx x      2 2 1 2 1 1 1 (tan ) . 2 2 1 x x x dx x       2 2 1 2 2 1 1 1 (tan ) . . 2 2 1 1 x x x dx dx x x             2 1 2 1 (tan ) 2 2 1 x dx x dx x             2 1 11 (tan ) tan [ ] 2 2 x x x x C Ans     
  • 3.
    Physics Helpline L KSatapathy Indefinite Integrals - 15 Answer-2 sin . 1 cos x x I dx x    sin . . 1 cos 1 cos x x dx dx x x      2 2 2sin cos 2 2. . 2sin 2sin 2 2 x x x dx dx x x    2 cosec . cot . 2 2 2 x x x dx dx   1 cot . [ ] 2 x I dx say  
  • 4.
    Physics Helpline L KSatapathy Indefinite Integrals - 15 2 2 1 1 cosec . cosec . 2 2 2 2 x x x For I dx x dx   2 cosec 2cot 2 2 x x and dv dx v    1 1 2cot 2cot . 2 2 2 x x I x dx                     Put u x du dx   cot cot . cot . 2 2 2 x x x I x dx dx      cot 2 [ ] x x C Ans   cot cot . 2 2 x x x dx   
  • 5.
    Physics Helpline L KSatapathy Indefinite Integrals - 15 Answer-3 3 2 sec . sec .sec .I x dx x x dx   2 sec sec tan & sec tanPut u x du x xdx dv xdx v x      sec tan tan .sec tan .I x x x x x dx    2 sec tan sec (sec 1).x x x x dx   3 sec tan sec . sec .x x x dx x dx    sec tan log sec tanx x I x x    2 sec tan log sec tanI x x x x    1 sec tan log sec tan [ 2 ]I x sC nx x Ax      
  • 6.
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