JEE Mathematics / Lakshmikanta Satapathy / Problems on Indefinite Integrals Trigonometric functions solved by the method of substitution

1. Physics Helpline
L K Satapathy
Indefinite Integrals 1

2. Physics Helpline
L K Satapathy
Indefinite Integrals - 1
4
cos 2 .I x dx Answer-1
2
4 2 2 1 cos4
cos 2 (cos 2 )
2
x
x x
 
   
 
1 1 1 1
cos4 cos8
4 2 8 8
x x   
21
(1 2cos4 cos 4 )
4
x x  
4 3 1 1
cos 2 cos4 cos8
8 2 8
x x x   
1 1 1 1 cos8
cos4
4 2 4 2
x
x
 
    
 
Power of cos x has to be removed by converting to multiple angle
2
[2cos 1 cos2 ]  

3. Physics Helpline
L K Satapathy
Indefinite Integrals - 1
3 1 1
cos4 cos8 .
8 2 8
x x dx
 
   
 

4
cos 2 .x dx 
3 1 1
cos4 . cos8 .
8 2 8
dx x dx x dx    
 
3 1 sin4 1 sin8
8 2 4 8 8
x x
x C
   
      
   
     
3 1 1
sin4 sin8
8 8 6
]
4
[x x x AnC s   

4. Physics Helpline
L K Satapathy
Indefinite Integrals - 1
Answer-2
4 2 2 2 2 2
tan tan (sec 1) tan .sec tanx x x x x x   
4
tan .I x dx 
2 2 2
tan .sec (sec 1)x x x  
2 2 2
tan .sec . sec .x x dx x dx dx    
4
tan .x dx 
1 2 3 [ ]I I I say  
2 2 2
tan .sec sec 1x x x  
We simplify the integrand as follows:

5. Physics Helpline
L K Satapathy
Indefinite Integrals - 1
Answer-2
4 3
1 2 3
1
tan . tan tan
3
[ ]x dx I I I x x x C Ans       
2 2
1 tan .sec .I x x dx 
2
tan sec .x t x dx dt  
3
2 3
1
1
t . tan
3 3
t
I dt x   
2
2 sec . tanI x dx x 
3I dx x 
We evaluate the integrals separately.
Substitution :

6. Physics Helpline
L K Satapathy
Indefinite Integrals - 1
Answer-3
2
cos2
.
(cos sin )
x
I dx
x x


2 2
2 2
cos2 cos sin
(cos sin ) (cos sin )
x x x
x x x x


 
2
(cos sin )(cos sin )
(cos sin )
x x x x
x x
 


(cos sin )
(cos sin )
x x
x x



We simplify the integrand as follows:

7. Physics Helpline
L K Satapathy
Indefinite Integrals - 1
Answer-3
(cos sin )
.
(sin cos )
x x
dx
x x


2
cos2
.
(cos sin )
x
I dx
x x


As the derivative of sinx + cosx = cosx – sinx
sin cosx x t 
cos sin
. log( )
sin cos
x x dt
dx t C
x x t

   
 
(cos sin ).x x dx dt 
2
cos2
. log(sin cos )
(cos sin
[ ]
)
x
I dx x x AnsC
x x
   

We substitute :
Putting back the value of t , we get
Differentiating we get

8. Physics Helpline
L K Satapathy
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