Transcript Document

Mathematics
Session
Indefinite Integrals -1
Session Objectives
 Primitive or Antiderivative
 Indefinite Integral
Standard Elementary Integrals
 Fundamental Rules of Integration
 Methods of Integration
1. Integration by Substitution, Integration
Using Trigonometric Identities
Primitive or Antiderivative
If
d
F  x  = ƒ  x  , then the function F(x) is
dx
called a primitive or an antiderivative of a function f(x).
For example :
x5
d  x5 
4
is a primitive of x because

 = x4
5
dx  5 
Cont.
If a function f(x) possesses a primitive, then it possesses
infinitely many primitives which can be expressed as F(x) + C,
where C is an arbitrary constant.
For example :
x5 x5
x5
,
-1,
+7 etc., are primitives of x4.
5 5
5
Indefinite Integral
Let f(x) be a function. Then collection of all its primitives
is called indefinite integral of f(x) and is denoted by
 f(x)dx.

d
F  x  + C  = f(x)   f(x)dx = F(x)+ C
dx
where F(x) + C is primitive of f(x) and C is an arbitrary
constant known as ‘constant of integration’.
Cont.
d
d
{F(x)} = f(x), then also
{F(x)+C} = f(x),
dx
dx
where C is an arbitrary constant.
If
If one integral of f(x) is F(x), then F(x) + C will be also
an integral of f(x), where C is a constant.
 f(x)dx will have infinite number of values and
hence it is called indefinite integral of f(x).
Standard Elementary Integrals
i
d  xn+1 
xn+1
n
n
+C, (n  -1)

 = x   x dx =


dx  n+1 
n+1
ii
d
1
1
log
x
=

dx =loge x +C, x  0

e 

dx
x
x
iii
d x
e = ex   exdx = ex +C
dx
 
d  ax 
ax
x
x
+C
 = a , a> 0, a  1   a dx =
iv  
dx  logea 
logea
Cont.
The following formulas hold in their domain
v
d
-cosx  = sinx   sinxdx =- cosx +C
dx
d
 vi
 sinx  = cosx   cosxdx =sinx +C
dx
 vii
d
 tanx = sec2x   sec2xdx =tanx+C
dx
 viii
d
-cotx  = cosec2x   cosec2xdx =- cotx +C
dx
ix
d
secx = secxtanx   secxtanxdx =secx+C
dx
Cont.
 x
d
-cosecx = cosecxcotx   cosecxcotxdx =- cosecx+C
dx
 xi
d
1
sin-1x =

dx
1- x2
 xii
 viii





d
-1
cos-1x =

dx
1- x2


dx
1- x2

=sin-1 x +C
-dx
1- x2
=cos-1 x +C
d
1
dx
tan-1x =

=tan-1 x+C
dx
1+x2
1+x2
Cont.
 xiv
 xv 
 xvi


d
-1
-dx
cot-1x =

=cot-1 x+C
dx
1+x2
1+x2


d
1
dx
sec-1x =
, x  1 
=s ec 1 x  C, x 2  1
dx
x x2 -1
x x2 -1


d
-1
-dx
cosec-1x =
, x  1 
= cosec 1 x  C, x2  1
dx
x x2 -1
x x2 -1
Fundamental Rules of Integration
  f  x  dx  = f  x 
1
d
dx
2 
 kf  x  dx = k  f x  dx
3  f(x)±g(x)dx =  f x dx±  g (x)dx
Example - 1
3
1 

Evaluate :   x2 +
dx
2

x 
3
1
3 
 6
1 

2
=
x
+
+3x
+
dx
Solution: Let I =   x2 +
dx


6
2
2

x
x 

x 
  x6dx  
1
x
6
dx   3x2dx  
3
x
2
dx
=  x6dx+  x-6dx+3 x2dx+3 x-2dx
x7 x-5 3x3
x-1
=
+
+
+3×
+C
7
-5
3
-1
x7
1
3
=
+ x3 - +C
7 5x5
x
Example - 2
Evaluate: 
1
3x +1 - 3x -1
Solution: Let I = 
=
dx
1
3x+1 - 3x -1
1
3x +1 - 3x -1
×
dx
3x +1 + 3x -1
3x +1 + 3x -1
=
3x +1 + 3x -1
dx
3x +1 - 3x -1
=
3x+1+ 3x -1
dx
2
dx
Cont.
1
1
=  3x+1 dx+  3x -1 dx
2
2
3
2
=
3
2
1 3x +1
1 3x -1
+
+C
2 3
2 3
×3
×3
2
2
3
3
1
1
= 3x +1 2 + 3x -1 2 + C
9
9
Example - 3
Evaluate:  cos-1  sinx  dx, 


x
2
2
Solution:Let I=  cos-1 sinx dx
 
 
=  cos-1cos  - x  dx =   - x  dx
2 
2 


x2
=  dx -  x dx = x +C
2
2
2
Note : cos1 cos     0,  
Example - 4
Evaluate: 
x
x +2
dx
x
Solution: Let I = 
x+2
=
x+2
x+2
dx = 
x+2-2
dx -2
x+2
1
dx
x+2
1
2
-
=   x+2 dx - 2  x+2
=
=
 x +2 
3
2
-2
3
2
3
2
 x +2 
1
2
1
2
+C
2
 x +2 - 4 x +2 +C
3
dx
1
2 dx
Integration by Substitution
If g(x) is a differentiable function, then to
evaluate integrals of the form
 ƒ g x  g  x  dx
We substitute g(x) = t and g’(x) dx = dt,
then the given integral reduced to
 ƒ  t  dt
After evaluating this integral, we substitute back the value of t.
Cont.

Integrals of the form of ƒ  ax +b  dx

Let I = ƒ  ax +b  dx
Putting ax +b = t  adx = dt  dx =
I =
1
dt
a
1
1
ƒ  t  dt  I = F  t  +C
a
a

1
 I = F  ax +b  +C, where f  x  dx = F  x  +C
a

Example - 5
Evaluate:  (3x+5)5 dx
Solution :
Let I=  (3x+5)5dx
1
Putting 3x+5= t 3dx = dt  dx = dt
3
1 5
1 t5+1
I=  t dt = ×
+C
3
3 5+1
1 6
1
=
t +C =
3x+56 +C
18
18
Integration Using Trigonometric Identities


Integrals of the form sinmnxdx, cosmnxdx, wherem
is small positive integer can be evaluated using the following
identities.
1 sin2x =
1- cos2x
2
1+cos2x
2 cos x =
2
2
3 sin3x = 3sinx - 4sin3x
 4 cos3x = 4cos3x -3cosx
Example - 6
Evaluate:  cos4x dx
Solution:Let I=  cos4x dx
2
2
 2cos2 x 
 1+ cos2x 
= 
 dx =  
 dx


2
2




 1 cos2x cos2 2x 
x sin2x
 1+cos4x 
=  +
+
+
 dx = +
 dx
4

2
4
4
4
8




=
x sin2x 1
1
x sin2x x sin4x
+
+  dx+  cos4x dx = +
+ +
+C
4
4
8
8
4
4
8
32
=
3x sin2x sin4x
+
+
+C
8
4
32
Integration Using Trigonometric Identities


 cosmxcosnxdx and  cosmxsinnxdx
Integrals of the form sinmxcosnxdx, sinmxsinnxdx,
can be evaluated using the following identities.
1
2sinAcosB = sin A+B +sin A -B
2
2cosAsinB = sin A+B - sin A -B
3
 4
2cosAcosB = cos  A+B +cos  A -B
2sinAsinB = cos  A -B - cos  A+B
Example - 7
Evaluate:  sin3xcos2x dx
Solution:Let I=  sin3xcos2x dx
=
=
1
2sin3xcos2x dx

2
1
sin5x+sinx dx
2
[Using 2sinAcosB = sin (A + B) + sin (A – B)]

1   cos5x


cos
x
C


2
5

cos5x cosx
=+C
10
2
Integration by Substitution
Integrals of the form
Let I =
ƒ  x 
 ƒ x
ƒ  x 
 ƒ x dx
dx
Putting ƒ  x  = t  ƒ'  x  dx = dt
I = 
dt
= loge t +C = loge f(x) +C
t
Example - 8
Evaluate : 
1- tanx
dx
1+ tanx
Solution: Let I = 
1- tanx
dx
1+tanx
=
cosx - sinx
dx
cosx+sinx
Putting cosx + sinx = t  -sinx + cosx  dx = dt
dt
 I =  =loge t +C
t
=loge cosx+sinx +C
Solution Cont.
Method - 2


I =  tan  - x  dx
4



= loge cos  - x   C
4

Example - 9
Evaluate : 
 x +1 ex
2
cos
 xe 
Solution: Let I = 
x
dx
 x +1 ex
2
cos

 xe 
x
dx

Putting xex = t  xex + ex dx = dt   x +1 exdx = dt
 
 I =  sec2t dt = tant +C = tan xex +C
Some Standard Results
1
 tanx dx = -log
2 
 cotxdx =loge sinx +C
e
3

 4

cosx +C = loge secx +C
  x
secx dx = loge secx + tanx +C = loge tan  +  +C
4 2
cosecx dx =loge cosecx - cotx +C =loge tan
x
+C
2
Integration by Substitution
Integrals of the form
Let I =

ƒ x f'x dx
n
n
ƒ x f' x dx
Puttingƒ  x  = t  ƒ'  x  dx = dt
 I =  tndt  I =
n+1
t
+C  I =
n+1
n+1
f  x 
n+1
+C, n  -1
Example - 10
Evaluate :

logex 3
x
Solution: Let I =
dx

3
logex 
x
dx
1
Putting logex = t  dx = dt
x
I =

t3dt
4
logex 
t4
=
+C =
+C
4
4
Integration by Substitution

Integrals of the form sinmxcosnxdx
Use the following substitutions.
(i) When power of sinx i.e. m is odd, put cos x = t,
(ii) When power of cosx i.e. n is odd, put sinx = t,
(iii) When m and n are both odd, put either sinx = t or cosx = t,
(iv) When both m and n are even, use De’ Moivre’s theorem.
Example - 11

Evaluate : sin3xcos5xdx

Solution: Let I = sin3xcos5xdx
Powers of sin x and cos x are odd.
Therefore, substitute sinx = t or cosx = t
We should put cosx = t, because power of cosx is heigher
Substituting cosx = t  -sinxdx = dt  dx =I =

 1 
sin3x t5  dt

 sinx 
1
dt
sinx
Cont.

= - t 1- t  dt

= - t5sin2xdt
5
=-

2

t5 - t7 dt
t6 t8
cos6 x cos8 x
= - + +C = +
+C
6
8
6
8
Example - 12
Evaluate : 

sin 2tan-1x
1+ x2
Solution: Let I = 

sin 2tan-1x
1+x2
Putting tan-1x = t 
1
cos2t
=+C
2
=-
cos2 tan-1x
2
2
1+x
 I =  sin2t dt

 dx
 +C
 dx
dx = dt
Example - 13
Evaluate :  ex -1 dx
Solution: Let I =  ex -1 dx
Putting ex - 1 = t2  exdt = 2tdt  dx =
2
I =  t .
= 2
2t dt
1+ t
t2 +1-1
1+ t2
2
= 2
2t dt
2t dt

dx
=
ex
1 + t2
t2dt
1+ t2


1
dt = 2   dt - 
dt 
1+ t2 

= 2t - 2tan-1t + C = 2 ex - 1 - 2tan-1  ex - 1  + C


Example - 14
x
x
Evaluate :  5 5 5x dx
55
5
x
x
Solution : Let I =  5 5 5x dx
55
x
55
Substituting 5
5
x
x
= t  5 5 5x loge5 dx = dt
55
3
5
x
55
x
 5 5 5x dx =
I =
1
log 5 
3
dt =
e
=
x
55
5
loge 5
3
+C
1
loge5
3
5
t +C
1
loge5
3
dt
Thank you