Transcript Document 7898194

Mathematics
Session
Indefinite Integrals - 3
Session Objectives
 Three Standard Integrals

a2 - x2 dx,

x2 +a2 dx,

x2 - a2 dx
 Integrals of the form

 Integrals of the form
 px +q
 Integrals of the form

ax2 +bx +c dx
ax2 +bx +c dx
1
dx,
a+bsinx
 Integration Through Partial Fractions
 Class Exercise

1
dx
a+bcosx
Three Standard Integrals
x
a2
2
2
-1 x
1
a
x
dx
=
a
x
+
sin
+C
 
2
2
a
Put x = asin  or acos

2
2
x
a2
2
2
x + a dx =
x +a +
loge x + x2 + a2 + C
2 
2
2
Put x = atan  or acot

3
2
2
2
x
2
2 a
x - a dx =
x -a loge x + x2 - a2 +C
2
2
Put x = asec  or acosec

2
2
Integrals of the Form

ax2 +bx +c dx
Reduce the given integral to one of
the following forms:

a2 - x2 dx or

x2 + a2 dx or

x2 - a2 dx
Example-1
Evaluate:

16x2 +25 dx
Solution: Let I =
=4


16x2 +25 dx
2
5
x +   dx
 4
2
2

5

2
2
 4
x 2 5


2 5

=4
x +  +
loge x + x +  
2
2
 4
 4


= 2x x2 +
25 25
25
+
loge x + x2 +
+C
16
8
16


 +C



Example - 2
Evaluate:

x2 +8x + 4 dx

Solution: Let I =
=
 

x2 +8x+16 -16+4 dx

 x + 42 -12 dx

 x + 4 2 - 2
=
=
x2 +8x + 4 dx
3

2
dx
Solution Cont.
x + 4

=
2




2
 x + 4

- 2 3

2

-
2 3
2

2
loge  x + 4  +
x + 4

x 2 2 a2
x - a dx =
x - a - loge x + x2 - a2 +C 
2
2

2
 x + 4
=
2
x2 +8x + 4
2
2
- 6loge  x + 4  + x2 +8x + 4 +C

- 2 3

2
+C
Example - 3
Evaluate :

7x -10 - x2 dx
Solution: Let I =



-10 - x2 - 7x dx

7
3 
x
dx
2 

2
  
=
=

7x -10 - x2 dx
2
7

x

2 

=
2
=
=

-10+
49  2
49 
-  x - 7x +
dx
4 
4 
2
2
2
2
7
3 
x
2 
2 
  
3
2
+   sin-1
2
7

x

2 +C


3


 2 
1
9
2x - 7 
2x - 7 7x -10- x2 + sin-1 
+C
4
8
3


Integrals of the Form
 px +q
ax2 +bx +c dx
We use the following method:
i 
Express px + q = A


d
ax2 +bx + c +B
dx
Identity
 px+q= A 2ax+b +B
(ii) Obtain the values of A and B by comparing the coefficients of
like powers of x. Then the integral reduces to

A 2ax+b  ax2 +bx+c dx+B
iii

ax2 +bx+c dx
To evaluate first integral, put ax2 +bx + c = t
 2ax +b  dx = dt and second integral by the
method discussed earlier.
Example - 4
Evaluate :  x x + x2 dx
Solution: Let I =  x x + x2 dx
Putting x = A

 x = A 1+2x  +B
 A=
I =

d
x + x2 + B
dx
Identity
1
1
, B=2
2
1
1
2
2
1+2x
x
+
x
dx
x
+
x
dx


2
2
Solution Cont.
Let I1 =  1+2x  x + x2 dx and I2 =  x + x2 dx
I1 =  1+2x  x + x2 dx
Putting x + x2 = t  1 + 2x  dx = dt
3
 I1 = 

2
2
t dt= t 2 = x + x2
3
3

3
2
1 1

I2 =  x + x2 dx =   x2 + x +  - dx
4 4

=
2
2
1 1

x
+

 -  2  dx
2

  
Solution Cont.
2
2
2
2
1
1 
1 1 1 1
=  x +   x +  -   - × loge
2
2 
2 2 2 4
1
1 
1 1 1
=  x +   x +  -   - loge
2
2 
2 2 8
2
2

1
= x + x2
3


3
2
-
2
1
1 1


x
+
+
x
+
- 




2
2 2


3

2
2
1 2
1
1
1 
1 1 1
2 2

I = × x+ x
x +   x +  -   - loge
2 3
2  2 
2 
2 2 8


2
1
1 1


x
+
+
x
+
- 




2
2 2


1 
1
2 1
x
+
x
+
x
- loge


4 
2
4
2
2 
1
1 1 


 x + 2  +  x + 2  -  2   +C



  

1

x
+
+ x + x2


2


 +C

Integrals of the Form

1
dx,
a+bsinx

1
dx
a+bcosx
We use the following method:
i 
ii
x
x
1- tan2
2 , cosx =
2
Write sinx =
x
x
1+ tan2
1+ tan2
2
2
2tan
x
1
x
Putting tan = t  sec2 dx = dt, we get
2
2
2
1
the integral in the form
dx
2
at +bt +c

(iii) Now, we evaluate the integral by the method discussed earlier.
Example - 5
Evaluate : 
1
dx
4 cosx -1
Solution: Let I = 
1
dx
4 cosx -1
x
2 , we get
Putting cosx =
x
1+ tan2
2
1- tan2
I= 
1

2x 
1tan

2  -1
4

x
2
 1+ tan


2
dx = 
sec2
x
2
x
x
4 - 4tan2 -1- tan2
2
2
dx
Solution Cont.
=
sec2
x
2
x
3 - 5tan
2
2
Putting tan
I =
dx
x
1
x
x
= t  sec2 dx = dt  sec2 dx = 2dt
2
2
2
2
2
dt
2
dt
=
2
53 2 5

-t
3
2

 - t
5
 5
Solution Cont.
2
1
= ×
loge
5
3
2
5
=
1
15
loge
3
+t
5
+C
3
-t
5
x
2 +C
x
3 - 5tan
2
3 + 5tan
Integration Through Partial
Fractions (Type – 1)
When denominator is non-repeated linear factors
Let
f(x)
A
B
C
=
+
+
 x - a  x - b   x - c  x - a x - b x - c
where A, B, C are constants and can be calculated by equating the
numerator on RHS to numerator on LHS and then substituting
x = a, b, c, ... or by comparing the coefficients of like powers of x.
Example - 6
Evaluate : 
2x +1
dx
x
+1
x
+2



Solution: Let I = 
Let

2x +1
dx
 x +1 x +2 
2x +1
A
B
=
+
 x +1 x +2   x +1  x +2 
 x +2 A +  x +1B
2x +1
=
 x +1 x +2
 x +1 x +2 
 2x +1 = A  x +2 +B  x +1
Identity
Solution Cont.
Putting x = -1, - 2, we get
A = -1 and B = 3
 I = -
dx
dx
+3
x +1
x +2
= -loge x +1 +3loge x +2 + C
= loge
 x +23
 x +1
+C
Type - 2
When denominator is repeated linear factors
Let
f(x)
(x - a) (x - b)2 (x - c)3
=
A
B
C
D
E
F
+
+
+
+
+
x - a x - b (x - b)2 x - c (x - c)2  x - c 3
where A, B, C, D, E and F are constants and value of the constants are
calculated by substitution as in method (1) and remaining are obtained
by comparing coefficients of equal powers of x on both sides.
Example - 7
Evaluate :

3x +1
x - 2
 x +2 

3x +1
Solution: Let I =
Let
2
3x +1
 x - 22  x +2 
2
x - 2
=
dx
 x +2 
dx
A
B
C
+
+
x - 2  x - 2 2 x +2
2
 3x+1= A  x -2 x+2 +B  x+2 +C x -2


2
 3x+1= A x2 - 4 +B  x+2 +C  x -2
Putting x = 2, we get 7 = 4B  B =
7
4
Identity
Solution Cont.
Putting x = -2, we get - 5 =16C  C =
-5
16
Comparing coefficients of x2 on both sides, we get
A +C = 0  A = -C  A =

3x +1
 x - 2 2  x +2 
I =
5
16
=

=
5
16
5
7
5
+
16  x - 2  4  x - 2 2 16  x +2 
1
7
dx +
4
 x - 2

1
2
x - 2
dx -
5
16

1
dx
 x +2 
5
7
5
loge x - 2 loge x +2 +C
16
4  x - 2  16
Type - 3
When denominator is non-repeated quadratic factors
Let
f(x)
 x - a px2 +qx +r 
=
A
Bx +C
+
x-a
px2 +qx +r


where A, B, C are constants and are determined by
either comparing coefficients of similar powers of x
or as mentioned in method 1.
Example - 8
Evaluate :

8

x +2  x + 4
Solution: Let I =
Let

8
 x +2  x2 + 4

2

dx
8

x +2  x + 4
=
2

dx
A
Bx +C
+
x +2 x2 + 4

 8 = A x2 +4 + Bx+C  x+2
Putting x = -2 in i ,we get A = 1
Putting x = 0 and1 in i ,we get
...i
Identity
Solution Cont.
C = 2 and B = -1
8
 x +2  x2 + 4
I =
=


=
1
dx +
x +2
1
1
dx x +2
2

1
-x +2
+
x +2 x2 + 4

-x +2
2
x +4
2x
2
x +4
dx

dx +2
1
2
x +4
dx
1
1
x
= loge x +2 - loge x2 + 4 +2× tan-1 +C
2
2
2
1
x
= loge x +2 - loge x2 + 4 + tan-1 +C
2
2
Type - 4
When denominator is repeated quadratic factors
Let
f(x)
ax
2

2
+bx + c px + qx +r

2
=
Ax +B
ax2 +bx + c
+
Cx +D
px2 + qx +r
+
Ex +F
px
2
+ qx +r
where A, B, C, D, E and F are constants and are determined by equating
the like powers of x on both sides or giving values to x.
Note: If a rational function contains only even powers of x,
then we follow the following method:
(i)
(ii)
(iii)
Substitute x2 = t
Resolve into partial fractions
Replace t by x2

2
Example – 9
Evaluate :
 x
2
Solution :Let I =
Let

x2 +2
x2 +2

2
+1 x + 4
 x
2
 x2 +1 x2 + 4

dx
x2 +2

2
+1 x + 4
=

dx
t +2
A
B
=
+
 t +1 t + 4 t +1 t + 4
 t + 4 A +  t +1B
t +2
=
 t +1 t + 4
 t +1 t + 4
 t  2 =  t +4 A+  t +1B
Putting t = -1, - 4, we get
1
2
A= , B=
3
3
Indentity
Solution Cont.
t +2
1
2
x2 +2
1
2

=
+

=
+
 t +1 t + 4 3  t +1 3  t + 4 
x2 +1 x2 + 4
3 x2 +1 3 x2 + 4

I =
1
1
2
1
dx
+
dx
3  x2 +1
3  x2 + 4
=
1
2 1
x
tan-1x +  tan-1  C
3
3 2
2
=
1
1
x
tan-1x + tan-1  C
3
3
2

 
 

Example - 10
x3dx
Evaluate : 
 x -1  x - 2 
Solution: Here degree of Nr > degree of Dr.
x3
7x - 6

= x +3 +
 x -1  x - 2 
 x -1  x - 2 
Let
7x - 6
A
B
=
+
 x -1  x - 2  x -1 x - 2
 7x - 6 =  x -2 A+  x -1B
Putting x =1, we get A = -1
Identity
Solution Cont.
Putting x = 2, we get B = 8

7x - 6
-1
8
=
+
x -1
x-2
 x -1  x - 2 
x3
1
8 


=   x +3 +
 dx
x
-1
x
2
x
-1
x
2
 






dx
dx
+8
x -1
x-2

= xdx + 3 dx -


x2
=
+ 3x - loge | x -1|+ 8 loge | x - 2|+ C
2
Thank you