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 : cos1 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+56 +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