Karnataka State Open University Study Material for MCA Mathematics - Code - MCA 11 by K.S. Srinivasa Retd. Principal & Professor of Mathematics Bangalore Published by Sharada Vikas Trust (R) Bangalore
Karnataka State Open University
Study Material for MCA
Mathematics - Code - MCA 11
by
K.S. SrinivasaRetd. Principal &
Professor of MathematicsBangalore
Published by
Sharada Vikas Trust (R)Bangalore
MCA 11 MATHEMATICS
Syllabus
1. Complex Trigonometry
Revision of Plane Trigonometry - trigonometric ratios, expressions for relation between allied angles and trigonometricalratios. Addition formulae for trigonometrical ratios and simple problems. Complex numbers and functions, definition,properties, De Moivre's Theorem (without proof), Roots of a complex number, expansions of sin (nθ), cos (nθ) inpowers of sin θ & cos θ, addition formulae for any number of angles, simple problems.
2. Matrix Theory :
Review of the fundamentals. Solution of linear equations by Cramers' Rule and by Matrix method, Eigen values andEigen vectors, Cayley Hamilton's Theorem, Diagonalization of matrices, simple problems.
3. Algebraic Structures
Definition of a group, properties of groups, sub groups, permutation groups, simple problems, scalars & vectors, algebraof vectors, scalar & vector products, scalar triple product, simple problems.
4. Differential Calculus
Limits, continuity and differentiability (definition only), standard derivatives, rules for differentiation, derivatives offunction of a function and parametric functions, problems. Successive differentiation, nth derivative of standard functions,statement of Leibnitz's Theorem, problems, polar forms, angle between the radius vector and the tangent to a polarcurve, (no derivation) angle between curves, pedal equation, simple problems, indeterminate forms, L' Hospital's rule,partial derivatives, definition and simple problems.
5. Integral Calculus
Introduction, standard integrals, integration by substitution and by parts, integration of rational, irrational and trigonometricfunctions, definite integrals, properties (no proof), simple problems, reduction formulae and simple problems.
6. Differential Equations of first order
Introduction, solution by separation of variables, homogeneous equations, reducible to homogeneous linear equation,Bernoulli's equation, exact differential equations and simple problems.
Text Books
1. Elementary Engineering Mathematics by Dr. B.S. Grewal, Khanna Publications
2. Higher Engineering Mathematics by B.S. Grewal, Khanna Publications
Reference Books
1. Differential Calculus by Shanti Narayan, Publishers S. Chand & Co.
2. Integral Calculus by Shanti Narayan, Publishers S. Chand & Co.
3. Modern Abstract Algebra by Shanti Narayan, Publishers S. Chand & Co.
CONTENTS
Page Nos.
1. Complex Trigonometry 01
2. Matrix Theory 27
3. Algebraic Structures 47
4. Differential Calculus 69
5. Integral Calculus 101
6. Differential Equations 123
COMPLEX TRIGONOMETRY
Trigonometric ratios of acute angles
Consider a right-angled triangle ABC, right angled at C. Let .� èBCA = The side ACopposite to angle θ is called ‘opposite side’. The side BC is called ‘adjacent side’. Theside AB is hypotenuse.
The ratio AB
ACie
hypotenuse
side opposite is defined as ‘sine of θθθθθ ’ & written as sinθ.
The ratio AB
BCie
hypotenuse
sideadjacent is defined as ‘cosine of θθθθθ ’ & written as cosθ.
The ratio BC
ACie
sideadjacent
side opposite is defined as ‘tangent of θθθθθ ’ & written as tanθ.
It can be seen by the above definition that èè
ètan
cos
sin =
The reciprocal of sinθ ie AC
ABie
èsin
1 is defined as ‘cosecant of θθθθθ ’ and written as cosecθ.
The reciprocal of cosθ ie BC
ABie
ècos
1 is defined as ‘secant of θθθθθ ’ and written as secθ.
The Reciprocal of tanθ ie AC
BC is defined as ‘cotangent of θθθθθ ’ and written as cotθ.
Identities:- (1) sin2θθθθθ + cos2θθθθθ = 1
(2) sec2θθθθθ = 1 + tan2θθθθθ
(3) cosec2θθθθθ = 1 + cot2θθθθθ
Proof:- From the right angled triangle ABC
AC2 + BC2 = AB2
1 have weby divide2
2
2
22 =+
AB
BC
AB
ACAB , ie sin2θ + cos2θ = 1
2
2
2
22 1 have weby divide
BC
AB
BC
ACBC =+, ie tan2θ + 1 = sec2θ
2
2
2
22 1 then by divide
AC
AB
AC
BCAC =+, ie 1 + cot2θ = cosec2θ
Trigonometric ratios of 30° & 60°
Consider an equilateral triangle of side 2 units AB = BC = AC = 2. BCAD r to Draw ⊥ , then BD = DC = 1. Then in the
triangle ABD, °=°=°= 30 & 6090 ADBBDADBA ��,� , further AD2 = AB2 – BD2 = 4 – 1 = 3, ∴ 3=AD
θB C
A
90º
In the triangle ABD
BDA� take2
360 ==°
AB
ADsin
2
160 ==°
AB
BDcos
1
360 ==°
BD
ADtan
3
160260
3
260cosec also =°=°=° cot&sec,
ADB� take2
130 ==°
AB
BDsin
2
330 ==°
AB
ADcos
3
130 ==°
AD
BDtan
3033
203230cosec also =°=°=° cot&sec,
From the above results, it can seen that sin60º = cos30º, cos60º = sin30º and tan60º = cot30º.
Trigonometric ratios of 45 º
Consider a right-angled isosceles triangle ABC where °= 90BCA �
ACBCBA �� =°= 45
Let AC = BC = 1 unit then 2=AB units
\2
145 ==°
AB
ACsin
2
145 ==°
AB
BCcos
11
145 ===°
BC
ACtan
145 and 2452cosec45 also, =°=°=° cotsec,
Note :- Trigonometric ratios of 30º, 45º and 60º are called ‘Standard Trigonometric’ ratios which are always useful,hence these values have to be always remembered.
Trigonometric ratios of any angle (from 0 º to 360 º)
Let XOX' & YOY' be co-ordinate axes where O is the origin. Consider a circle of radius r with centre O. Let P be any pointon the cirlce whose co-ordinates are (x, y).
Draw PM perpendicular to OX.
3
90º60º
30º
B D C
A
2 2
1 1
21
A
CB 1
2 KSOU Complex Trigonometry
Then OM = x & PM = y
èPOM =�et L
x
yè
r
xè
r
yè === tan,cos,sin
y
xè
x
rè
y
rè === cot,sec,cosec also
When ,�POM satisfy 0º < θ < 90º the print P will be in first quadrantof the circle, when it satisfy 90º < θ < 180º, P will in second quadrant,when 180º < θ < 270º, P will be in third quadrant & finally when 270º< θ < 360º the point P will be in fourth quadrant, because of thesepositions, signs of the Trigonometric ratios changes.
In the I quadrant both x & y are +ve and r is always +ve.
Therefore sinθ, cosθ & tanθ are +ve, their reciprocals are also +ve.
In the II quadrant x is –ve, y is +ve
Therefore sinθ & cosecθ are +ve cosθ, tanθ, secθ & cotθ are –ve.
In the III quadrant both x & y are –ve
Therefore, tanθ & cotθ are +ve and sinθ, cosθ, cosecθ & secθ are –ve.
In the IV quadrant x is +ve & y is –ve
Therefore, cosθ & secθ are +ve and sinθ, tanθ, cosecθ & cotθ are –ve.
Note :- The signs of the trigonometric ratios can be easily remembered with the help of the following diagram
Trigonometric ratios of angles 0 º, 90º, 180º, 270º and 360º(can be called border angles).
Let POM � be an angle whose measure is very close to zero (as in fig. 1)
rxy →→→ ,00, As θ
\ 00
0 ==°r
sin
Y
O
Y'
X' XM
P (x, y)
yxθ
(–, +) (+, +)
(–, –) (+, –)
S A
T C
Sine is +ve All are +ve
tan is +ve cos is +ve
in short
O M
P
yxθ
fig.1
MCA 11 - Mathematics SVT 3
10 ==°r
rcos
00
0 ==°r
tan
If ,�POM is very close to 90º as in fig. 2.
ryx →→°→ ,0,09 As θ
\ 190 ==°r
rsin
00
90 ==°r
cos
∞==°0
90r
tan
If θ is very close to 180º as in fig. 3
0081 As →−→°→ yrx ,,θ
\ 00
180 ==°r
sin
1180 −=−=°r
rcos
00
180 ==°r
tan
If θ is very close to 270º as in fig. 4
ryx −→→°→ ,, 0270 Asθ
\ 1270 −=−=°r
rsin
00
270 ==°r
cos
−∞=−=°0
270r
tan
If θ is very close to 360º as in fig. 5
0360When →→°→ yrx ,,θ
\ 00
360 ==°r
sin
1360 ==°r
rcos
00
360 ==°r
tan
O M
P
y
xθ
fig.2
OM
P
y
xθ
fig.3
OM
P
y
xθ
fig.4
r
M
P
y
xθ
fig.5
r
4 KSOU Complex Trigonometry
Thus we have the following
036013600360
27002701270
018011800180
90090190
001000
=°=°=°−∞=°=°−=°
=°−=°=°∞=°=°=°
=°=°=°
tan,cos,sin
tan,cos,sin
tan,cos,sin
tan,cos,sin
tan,cos,sin
Note :- From the above derivations it can be seen that the values of sinθ & cosθ always be between –1 & 1. Whereas thevalue of tan θ will be between −∞ & ∞ and hence the graphs of the trigonometric functions
xyxyxy tancos,sin === & are as follows.
fig.6
xy sin=
Ox
90° 270°180° 360°
xy cos=
Ox
90° 270°180° 360°
fig.7
fig.8
xy tan=
Ox
90° 270°180° 360°
MCA 11 - Mathematics SVT 5
Rules for allied angles
θθ cos)sin( =−°90
θθ sin)cos( =−°90
θθ cot)tan( =−°90
θθ cos)sin( =+°90
θθ sin)cos( −=+°90
θθ cot)tan( −=+°90
θθ sin)sin( =−°180
θθ cos)cos( −=−°180
θθ tan)tan( −=−°180
θθ sin)sin( −=+°180
θθ cos)cos( −=+°180
θθ tan)tan( =+°180
O
B
Ar
r r
1 radian
θθ cos)sin( −=−°270
θθ sin)cos( −=−°270
θθ cot)tan( =−°270
θθ cos)sin( −=+°270
θθ sin)cos( =+°270
θθ cot)tan( −=+°270
θθ sin)sin( −=−°360
θθ cos)cos( =−°360
θθ tan)tan( −=−°360
θθ sin)sin( =+°360
θθ cos)cos( =+°360
θθ tan)tan( =+°360
Using the Trigonometric ratios of standard angles 30°, 45° & 60° and using the above rules for allied angles, we canfind the trigonometric ratios of other angles as follows.
Eg.12
3303090120 =°=°+°=° cos)sin(sin
or 2
36060180120 =°=°−°=° sin)sin(sin
Eg.22
3606090150 −=°−=°+°=° sin)cos(cos
or 2
33030180150 −=°−=°−°=° cos)cos(cos
Eg.32
14545360315 −=°−=°−°=° sin)sin(sin
or 2
14545270315 −=°−=°+°=° cos)sin(sin
Radian Measure
Consider a circle of radius r with centre O. Let AB be a arc such that arc AB = r.
Measure of the BOA � is called a 'radian' denoted as 1C.
To prove that radian is a Constant angle
Consider a circle of radius r with centre O. Let arc AB = r so that BOA � is 1 radian.Let C be a point on the circle such that .� °= 90COA
Since the angle subtended at the centre of a circle is proportional to the corresponding arc.
6 KSOU Complex Trigonometry
AC
AB
COA
BOA
arc
arc=�
�
ππ
2
24190
radian 1 ==° )( r
r
== )( rAC π2
4
1 nceCircumfere
4
1Q
°=∴ 180radians πresult. imp.180 ie °=π
,, °=°= 454
902
ππ., °=°= 30
660
3
ππ
Using the relation π = 180° measurement of an angle in degrees can be converted to radians and vice-versa.
Eg. (1) Qn. : Convert 40° to radians
radius 9
2
1804040 : Solution
ππ =×=°
Eg. (2) Qn. : Convert 3
2π radians in to degrees
°=× 120180
3
2 : Solution
ππ
Some Problems
1. Show that
AAA 222 cosec211 =−++ )cot()cot(
AAAA cotcotcotcot 2121LHS :Solution 22 −++++=
A222 cot+=
)cot( A212 +=
.RHScosec2 2 == A
2. Show that
AAA
A
A
Atansec
sin
sin
sin
sin4
1
1
1
1 =+−−
−+
)sin)(sin(
)sin()sin
AA
AA
+−−−+=
11
11( LHS : Solution
22
(taking LCM)
A
AAAA2
22
1
2121
sin
)sinsin()sinsin(
−−+−++=
A
AAAA2
22 2121
cos
sinsinsinsin +−−++=
A
A
AA
A
cos
sin
coscos
sin ⋅== 14
42
RHS4 =⋅= AA tansec
O
B
Ar
rr
C
MCA 11 - Mathematics SVT 7
3. If sec A + tan A = a, then prove that Aa
asin=
+−
1
12
2
12
12
1
1 LHS :Solution
22
22
2
2
+++−++=
+−=
AAAA
AAAA
a
a
tansectansec
tansectansec
AAAA
AAAA
tansectansec
tansectansec
21
2122
22
+++++−=
identity) the(using 22
222
2
AAA
AAA
tansecsec
tansectan
++=
)tan(secsec
sectantan
AAA
A)AA(
++=
2
2
RHS1
==×== AA
A
A
A
Asin
cos
cos
sin
sec
tan
4. If ,sin5
4=θ find the value of θθθθ
cottan
cossin
−+
5
4 If :Solution =θsin then opposite side is 4 hypotenusis 5.
391625 sideadjacent ==−=∴3
4 &
5
3 ==∴ θθ tancos
5
12
1275
7
43
34
5
3
5
4
==−
+=
−+∴
θθθθ
cottan
cossin
5. If θθ
θθπθπθcosec38
85 find
24
3
−+<<−=
sec
tancos,tan
Solution : Since θ lies in II Quadrant sine is +ve cosine & tangant are –ve.
5
4 and
5
3 −==∴ θθ cossin
−
−
−+
−
=−+∴
3
53
4
58
4
38
5
45
cosec38
85
θθθθ
sec
tancos
3
2
15
10
510
64 ==−−−−=
6. Prove that
θθθθ
θθθ 2
9090360
90cosec180180sec
)sin()cot()sec(
)()sec()tan( =−°+°−°
+°−°+°
8 KSOU Complex Trigonometry
θθθθθθθ 2LHS :Solution sec
cos)tan(sec
sec)sec(tan =−−=
7. Prove that
θθθπθπ
θπθπθπsin
)sin(cottan
cot)cos()sin(
=−
+
+
−−+
2
3
2
22
)sin)(tan(cot(
tancos)sin(
θθθθθθ
−−−−=LHS :Solution
θθ tancos=
RHS==×= θθθθ sin
cos
sincos
Addition Formulae
BABABA sincoscossin)sin( +=+
BABABA sinsincoscos)cos( −=+
BA
BABA
tantan
tantan)tan(
−+=+
1
replacing B by –B
BABABA sincoscossin)sin( −=−
BABABA sinsincoscos)cos( +=−
BA
BABA
tantan
tantan)tan(
+−=−
1
Using the above formulae we can find the trigonometric ratios of 15°, 75°, 105° etc.
)sin(sin °+°=° 304575
°°+°°= 30453045 sincoscossin
22
13
2
1
2
1
2
3
2
1 +=×+×=
)cos(cos °+°=° 304575
°°−°°= 30453045 sinsincoscos
22
13
2
1
2
1
2
3
2
1 −=×−×=
13
13
75
7575
−+=
°°=°
cos
sintan
)sin(sin °−°=° 304515
°°−°°= 30453045 sincoscossin
MCA 11 - Mathematics SVT 9
2
1
2
1
2
3
2
1 ×−×= 22
13 −=
)cos(cos °−°=° 304515
°°+°°= 30453045 sinsincoscos
2
1
2
1
2
3
2
1 ×+×= 22
13 +=
13
13
15
1515
+−=
°°=°
cos
sintan
)sin(sin °+°=° 4560105
°°+°°= 45604560 sincoscossin
2
1
2
1
2
1
2
3 ×+×= 22
13 +=
)cos(cos °+°=° 4560105
°°−°°= 45604560 sinsinsincos
2
1
2
3
2
1
2
1 ×−×= 22
31−=
31
13105
−+=°tan
Alternate method
°=°−°=° 75759015 cos)sin(sin
22
13 −=
°=°−°=° 75759015 sin)cos(cos
22
13 +=
°=°−°=° 7575180105 sin)sin(sin
22
13 +=
°−=°−°=° 7575180105 cos)cos(cos
( )22
31
22
13 −=−−=
10 KSOU Complex Trigonometry
To find sin2 θθθθθ, cos2 θθθθθ & tan2 θθθθθ.
BABABA sincoscossin)sin( +=+
BA == θput
θθθθθ sincoscossinsin +=2
θθ cossin2=BABABA sinsincoscos)cos( −=+
θ== BAput
θθθθθ sinsincoscoscos −=2
θθ 22 sincos −=
θθ 22 1 using cossin −=
122 2 −= θθ coscos
θθ 22 1 using also sincos −=
θθ 2212 sincos −=
BA
BABA
tantan
tantan)tan(
−+=+
1
BA ==θput
θθθθθ
tantan
tantantan
−+=
12
θ
θ21
2
tan
tan
−=
To find sin3 θθθθθ, cos3 θθθθθ & tan3 θθθθθ.
)sin(sin θθθ += 23
θθθθ sincoscossin 22 +=
θθθθθ sin)sin(coscossin 2212 −+=
θθθθ 32 212 sinsin)sin(sin −+−=
θθθθ 33 222 sinsinsinsin −+−=
θθ 343 sinsin −=
)cos(cos θθθ += 23
θθθθ sinsincoscos 22 −=
θθθθ cossincos)cos( 22 212 −−=
θθθθ cos)cos(coscos 23 122 −−−=
θθθθ 33 222 coscoscoscos +−−=
θθ coscos 34 3 −=
MCA 11 - Mathematics SVT 11
)tan(tan θθθ += 23
θθθθ
tantan
tantan
21
2
−+=
θθ
θ
θθ
θ
tantan
tan
tantan
tan
×−
−
+−=
2
2
1
21
1
2
)tan(
tantan
tan
)tan(tantan
θθθ
θθθθ
2
22
2
2
1
211
12
−−−
−−+
=
θ
θθθ2
3
31
2
tan
tantantan
−−+=
θ
θθ2
3
31
3
tan
tantan
−−=
Thus we have,
θθθ cossinsin 22 =
θθθ 222 sincoscos −= θθ 22 2112 sincos −=−=
θθθ21
22
tan
tantan
−=
θθθ 3433 sinsinsin −=
θθθ coscoscos 343 3 −=
θθθθ
2
3
31
33
tan
tantantan
−−=
Problems
1. Show that 233 =−A
A
A
A
cos
cos
sin
sin
A
AA
A
AA
cos
coscos
sin
sinsin 3443 LHS :Solution
33 −−−=
3443 22 +−−= AA cossin
)cos(sin AA 2246 +−= RHS246 ==−=
2. If 3
2=Acos find sin2A & cos3A.
3
5
3
49
3
2Given :Solution =−== AA sin,cos
12 KSOU Complex Trigonometry
AAA cossinsin 22 = 9
54
3
2
3
52 =××=
AAA coscoscos 343 3 −=
3
23
27
84 ×−×= 2
27
32 −= 27
22
27
5432 −=−=
9
542 =∴ Asin &
27
223
−=Acos
3. Prove that
)(tansin
sin θθθ +°=
−+
4521
21 2
θθθθ
cossin
cossin
21
21 LHS :Solution
−+=
2
2
)sin(cos
)sin(cos
θθθθ
−+=
)cossincos
sincos θθθθθ
by bracket theinsideDr &Nr (dividing 2
−+=
2
1
1
−+=
θθ
tan
tan
RHS45 =+°= )tan( θ
4. Prove that 21
1 2 θθθ
tancos
cos =+−
and hence prove that 3215 −=°tan
−+
−−
=+
12
21
2211
1
-1 :Solution
2
2
θ
θ
θθ
cos
sin
cos
cosusing cos2 A formula
22
22
12
21
2211
2
2
2
2
θ
θ
θ
θ
cos
sin
cos
sin=
−+
+−=
22 θ
tan=
21
1 2 θθθ
tancos
cos =+−∴
°+°−=°°=
301
3011530put 2
cos
costan.θ
32
32
23
1
2
31
+−=
+
−=
( )( )( )( )3232
3232
−+−−=
( ) ( )22
3234
32 −=−
−=
322
−=∴ θtan
MCA 11 - Mathematics SVT 13
Complex Number
Definition : A number of the form 1 where −=∈∈+ iRyRxiyx &, is defined as a Complex Number and usually denoted
as Z. x is called Real part & y is called Imaginary part. x – iy is called Conjugate. Complex number denoted as .zZ or Acomplex number can be represented by a point on a plane by taking real part on x-axis & imaginary part on y-axis. The planeon which complex numbers are represented is called a Complex Plane. For every point in a plane there is a complex number& for every complex number there is a point in the plane. In a complex x-axis is called Real axis & y-axis Imaginary axis.
Properties
(1) Equality. Two complex numbers 222111 iyxziyxz +=+= , are said to be equal if 2121 yyxx == ,
(2) Addition. 222111 If iyxziyxz +=+= ,
)()( 212121 yyixxzz +++=+
(3) Subtraction. )()( 212121 yyixxzz −+−=−
(4) Multiplication. ))(( 221121 iyxiyxzz ++=
212
211221 yyiyixyixxx +++=
1212212121 −=++−= iyxyxiyyxx Q)(
etc111 : Note 432 =−=−=−= iiiii ,,,
(5) Division. )(
)(
22
11
2
1
iyx
iyx
z
z
++
=
multiply numerator & denominator by the conjugate of x2 + iy2 ie x2 – iy2, then
))((
))((
2222
2211
2
1
iyxiyx
iyxiyx
z
z
−+−+
=
22
22
21122121
yx
yxyxiyyxx
+−++= )(
22
22
211222
22
2121
yx
yxyxi
yx
yyxx
+−+
++=
which is a complex number.
Note :- Product of a complex number with its conjugate is always a positive real number.
iyzzxzzyxiyxiyxzz 22ie 22 =−=++=−+= ,&))((
i
zzy
zzx
22
−=+=∴ &
Polar form the complex number
z = x + iy is called the Cartesian form.
Let P(x, y) be any point in the plane which represents a complex number.
Draw rPM ⊥ to x-axis & join PM. Then OM = x, MP = y. Let θ=OPM� &OP = r.
From the triangle OPM,
O x M
ry
P (x, y)
x
y
θ
14 KSOU Complex Trigonometry
θθ coscos rxy
x
OP
OM =⇒==
θθ sinsin ryr
y
OP
PM =⇒==
θθ sincos irriyxz +=+=∴ )sin(cos θθ ir +=
This form of the complex number is called 'Polar form'. Where r is called Modulus & θ is called argument which aregiven by
x
yyxr 122 & −=+= tanθ
r is always positive and argument θ varies from 0° to 360°. The value of argument satisfying –π < θ ≤ π is definedas amplitude which is unique for a complex number.
Thus we have
22 Mod. yxrzz +===
x
yz 1 arg −= tan and amp. z = θ where –π < θ ≤ π
x
y
y
x == θθθ sincos and ie satisfying find tohave wenumber,complex a of amplitude thefinding while
Examples
1. Express i
i
++
1
32 in the form x + iy
))((
))((
ii
ii
i
i
−+−+=
++
11
132
1
32 : Solution
11
3232 2
+−−+= iii
2
5
2
32 ii +=++=2
1
2
5i+=
2. Express i
i
−+
3
1 2)(in the form x + iy
)(
)(
i
ii
i
i
−++=
−+
3
21
3
1 : Solution
22
)))((
)(1 (using
33
32 2 −=+−
+= iii
ii
10
62
19
26 2 iii +−=+
+=5
3
5
1
10
6
10
2 ii +−=+−=
3. Find the modulus and amplitude of 1 + i.
211 : Solution 22 =+=+== yxrz
2
1 and
2
1 ====r
y
r
x θθ sincos4
45πθ =°=∴
2 is Modulus∴4
is amplitude andπ
MCA 11 - Mathematics SVT 15
4. Find the modulus & amplitude of i+− 3
2413 : Solution ==+=z
6
5
2
12
3πθ
θ
θ=→
=
−=
sin
cos
6
5 is amplitude & 2 is Modulus
π∴
5. Find the modulus & amplitude of 31 i−−
2431 : Solution ==+=z
3
2
2
3
2
1
πθθ
θ−=→
−=
−=
sin
cos
3
2 is amplitude & 2 is Modulus
π−∴
6. Find the modulus & amplitude of 1 – i.
211 :Solution =+=z
4
2
12
1
πθθ
θ−=→
−==
==
r
y
r
x
sin
cos
4 is amplitude & 2 is Modulus
π−∴
7. 1( that prove then 1
If 2222 =+++
=+ ))(, baiba
i βαβα
))((
)(
ibaiba
iba
ibai
−+−=
+=+ 1
:Solution βα
222222 ba
bi
ba
a
ba
iba
+−
+=
+−=
Equating real and imaginary parts separately
2222 and
ba
b
ba
a
+−=
+= βα
16 KSOU Complex Trigonometry
Squaring and adding
( ) ( )222
2
222
222
ba
b
ba
a
++
+=+ βα ( ) 22222
22 1
baba
ba
+=
+
+= )(
gmultiplyin-cross ∴
12222 =++ ))(( baβα
DeMoivre's Theorem
Statement : If n is a +ve or –ve integer, then ni )sin(cos θθ + θθ nin sincos +=
If n is a +ve or –ve fraction, one of the values of ni )sin(cos θθ + is .sincos θθ nin +
Proof : Case (i) when n is a +ve integer proof by Mathematical Induction.
When n = 1,1)sin(cos θθ i+ θθ sincos i+= θθ .sin.cos 11 i+=
∴ result is true for n = 1.
Let us assume that the result is true for n = m
θθθθ mimi m sincos)sin(cos +=+ie
multiply both sides by θθ sincos i+
)sin(cos)sin(cos θθθθ ii m ++ )sin)(cossin(cos θθθθ imim ++=
1ie ++ mi )sin(cos θθ
θθθθθθθθ sinsinsincoscossincoscos mmimim −++=]sincoscos[sinsinsincoscos θθθθθθθθ mmimm ++−=
θθ )sin()cos( 11 +++= mim
∴ the result is true for n = m + 1. Thus if the result is true for n = m then it is true for n = m + 1.
ie If it is true for one integer it is true for next integer, hence by Induction the result is true for all +ve integers.
Case (ii) When q
pn = where p & q are +ve integers.
q
q
pi
q
p
+ θθ sincosConsider
θθ qq
piq
q
p ⋅+⋅= sincos θθ piq sincos += pi )sin(cos θθ +=
qp
q
pi
q
pi
+=+ θθθθ sincos)sincos( ie
taking qth roots on both sides. One of the values of θθθθq
pi
q
pi q
p
sincos)sin(cos +=+
Case (iii) when n is –ve integer or –ve fraction. Let n = –m where m is a +ve integer or +ve fractionmn ii −+=+ )sin(cos)sin(cos θθθθ
MCA 11 - Mathematics SVT 17
mi )sin(cos θθ += 1
θθ mim sincos += 1
)sin)(cossin(cos
)sin(cos
θθθθθθ
mimmim
mim
−+−=
θθθθ
mm
mim22 sincos
sincos
+−=
1
θθ mim sincos −= θθ )sin()cos( mim −+−= .sincos θθ nin +=
Important Results
(i) If θθ sincos ix += then θθ sincos ix
−=1
.sincos θθ ix
xx
x 21
and 21 =−=+∴
(ii) If θθ sincos ix +=
nn ix )sin(cos θθ += θθ nin sincos +=
θθ sincos ix
−=1 & n
ni
x)sin(cos θθ −=1
θθ nin sincos −=
θnx
xn
n cos21 =+∴ & θni
xx
nn sin2
1 =−
Note :- For convenience θθ sincos i+ can be written as .θcis
Roots of a complex number
Let z = x + iy, express the complex number in the polar form.
)sin(cos θθ irz +=ie
[ ])sin()cos( θπθπ +++= kikr 22 Ik ∈where
[ ] nnn kikrz111
22 )sin()cos( θπθπ +++=
+++=
n
ki
n
kr n
)(sincos
θπθπ 221
where k = 0, 1, 2, ......... n – 1. Let us denote the nth roots of the complex number by z0, z1, z2, ..............zn – 1
Then,
for nr
ni
nrzk nn
θθθcis0
11
0 =
+== sincos,
nr
ni
nrzk nn
θπθπθπ +=
+++== 2
cis22
111
1 sincos,
nr
ni
nrzk nn
θπθπθπ +=
+++== 4
cis44
211
2 sincos,
n
nr
n
ni
n
nrznk nn
nθπθπθπ +−=
+−++−=−= −
)()(sin
)(cos,
12cis
12121
11
1
the above n values gives nth roots of z = x + iy
18 KSOU Complex Trigonometry
Note :- If k = n, n + 1, n + 2 etc. The values will repeat. Hence these will be only n values of nz1
which are distinct.Using the polar form of the complex number we can plot the nth roots of the complex in the following way.
Draw a circle of radius nr1
whose centre is O. Mark a point on the circle and take
OA as intial line. Take a point B such that n
OBAθ=� . Then B represent z0. Take a
point C such that n
OCAθπ += 2� then C represent z1, like this all the nth roots can
be represented. This diagram is called 'Argand Diagram'.
Problems :
59
32
44
7755Simplify 1
)sin(cos)sin(cos
)sin(cos)sincos)(
θθθθθθθθ
ii
ii
+−+− −
Solution : G.E. (given expression)
536
2110
)sin(cos)sin(cos
)sin(cos)sin(cos
θθθθθθθθ
ii
ii
++++= −
−−
5362110 −+−−+= )sin(cos θθ i 10 =+= )sin(cos θθ i
104
35
44
3322Simplify 2
)sin(cos)sin(cos
)sin(cos)sincos()(
θθθθθθθθ
ii
ii
+−−+ −−
1016
910
G.E. :Solution )sin(cos)sin(cos
)sin(cos)sin(cos
θθθθθθθθ
ii
ii
++++= −
−
1016910 −++−+= )sin(cos θθ i 5)sin(cos θθ i+= θθ 55 sincos i+=
yy
xx
12 &
12 If 3 +=+= φθ coscos)(
Show that
)cos(( φθ nmyx
yxmm
mm +=+ 21
i) & )cos(( φθ nmx
y
y
xm
n
n
m
−=+ 2 ii)
x
x
xx
112 : Solution
2 +=+=θcos
xx ⋅=+∴ θcos212 012ie 2 =+− xx θcos
2
442 2 −±=⇒ θθ coscosx
2
142 2 )cos)((cos θθ −−±=
2
42 2θθ sincos −±= 2
22 θθ sincos i±= θθ sincos i±=
θθθθ sincos,sincos ix
ix −=+=∴ 1 then If
φφφφφ sincos&sincos,cos iy
iyy
y −=+=+= 112 ifSimilarly
AO
C
B
z2
z1
z0
zn
nr1
n/θ
MCA 11 - Mathematics SVT 19
(i) mm ix )sin(cos θθ += θθ mim sincos +=
nn iy )sin(cos φφ += φφ nin sincos +=
)sin)(cossin(cos φφθθ ninmimyx nm ++=
)sin()cos( φθφθ nminm +++= (1)
)sin()cos( φθφθ nminmyx nm
+−+= 21
(2)
adding (1) & (2)
)cos( φθ nmyx
yxnm
nm +=+ 21
(ii) θθ mimxm sincos +=
φφ ninyn
sincos ==1
)sin()cos( φθφθ nminmy
xn
m
−+−=∴ & )sin()cos( φθφθ nminmx
ym
n
−−−=
adding
)cos( φθ nmx
y
y
xm
n
n
m
−=+ 2
(4) Prove that
+=−++ −
a
b
n
mbaibaiba n
mn
mn
m 122 22 tancos)()()(
)sin(cos θθ iriba +=+Let :Solution a
bbar 122 & where −=+= tanθ
nm
nm
nm
iriba )sin(cos)( θθ +=+∴
+= θθ
n
mi
n
mr n
m
sincos (1)
)sin(cos θθ iriba −=−
−=− θθ
n
mi
n
mriba n
mn
m
sincos)( (2)
adding (1) & (2)
=−++ θ
n
mribaiba n
mn
mn
m
cos)()( 2
a
b
n
mbaibaiba
nm
nm
nm 122 2
2 −
+=−++∴ tancos)()(
a
b
a
bbar 122 −=⇒=+= tantan& θθQ
( )
+=−++∴ −
a
b
n
mbaibaiba n
mn
mn
m 122 22 2tancos)()(
20 KSOU Complex Trigonometry
(5) Find the cube roots if 1 + i and represent them on Argand diagram.
say 1Let :Solution )sin(cos θθ iriz +=+=
11 == θθ sin,cos rr
11 adding & Squaring 222 +=+ )sin(cos θθr
222 =⇒= rr
4
2
12
1
πθθ
θ=→
=
=
sin
cos
+=∴
442
ππsincos iz
++
+=
42
422
ππππ kik sincos Ik ∈for
+=+=∴
42cis21
ππkiz
( ) 31
31
31
31
4
8cis21
+=+= ππk
iz )( 2 1, 0,for 12
8cis2 6
1=+= k
k ππ
°=== cis15212
cis20when 61
61
0π
zk ,
°==== 135cis24
3cis2
12
9cis21 6
16
16
1
1ππ
zk ,
°===== cis255212
17cis22
12
17cis22 6
16
16
1
22ππ
zkzk ,,
To represent them on Argand diagram.
Draw a circle of radius 61
2 with centre O. Let OA be the initial line.
Take a point B on the circle such that ,� °= 15OBA take a point C on the circle
such that °= 135OCA� & take a point D on the circle such that °= 255ODA� thenthe points B, C, D represent z0, z1, z2.
(6) Find all the values of 4
1
2
3
2
1
− i & find their continued product.
2
3
2
1Let :Solution iz −= )sin(cos θθ ir +=
2
3
2
1 −==∴ θθ sin,cos rr
Squaring & adding
14
4
4
3
4
12 ==+=r
AO
C
B
61
2
D
15°
135°
255°
MCA 11 - Mathematics SVT 21
3
2
3
2
1
πθθ
θ−=→
−=
=
sin
cos
−+
−=∴
331
ππsincos iz
−=
32cis
ππkz Ik ∈for
41
41
2
3
2
1
−= iz
41
32cis
−= ππk
41
3
6cis
−= ππk
3210for 12
6cis ,,,=−= k
k ππ
−==
12cis0 for 0
πzk ,
==
12
5cis1 1
πzk ,
==
12
11cis2 2
πzk ,
==
12
17cis3 3
πzk ,
Their continued product
−=
12
17cis
12
11cis
12
5cis
12sci
ππππ
+++−=
12
17
12
11
12
5
12cis
ππππ3
8cis
12
32cis
ππ =
=
3
2cis
3
22cis
πππ =
+=
3
2
3
2 ππsincos i+=
2
3
2
1i+−=
Expansion of sin ( nθθθθθ) and cos ( nθθθθθ) in powers of sin θθθθθ & cos θθθθθni )sin(cos θθ +Consider θθ nin sincos +=
nnin )sincos θθ +( Expand
Using Binomial Theoremn
nnnnn anCaxnCaxnCxax ++++=+ −− ........)( 22
21
1
Equating real and imaginary parts separately we get the expressions of θθ nn sin&cos
Eg. (1) Express θθ 55 sincos i+ in powers of θsin & θcos .
555 :Solution )sin(cossincos θθθθ ii +=+5
54
432
323
24
15 55555 )sin()sin(cos)sin(cos)sin(cossincoscos θθθθθθθθθθ iCiCiCiCiC ++++⋅+=
θθθθθθθθθθ 54322345 510105 sinsincossincossincossincoscos iii ++−−+=
]sinsincossincos[sincossincoscos θθθθθθθθθθ 53244235 105510 +−++−= i
22 KSOU Complex Trigonometry
Equating real & imaginary parts separately
θθθθθθ 4235 5105 sincossincoscoscos +−= and
θθθθθθ 5324 1055 sinsincossincossin +−=
Eq. (2) θθθθ sincossin&cos i+ of powersin 66 Express
666 :Solution )sin(cossincos θθθθ ii +=+42
433
324
25
16 6666 )sin(cos)sin(cos)sin(cos)sin(coscos θθθθθθθθθ iCiCiCiC ++++=
66
55 66 )sin()sin(cos θθθ iCiC ++
θθθθθθθθθθθθ 6542332456 61520156 sinsincossincossincossincossincoscos −++−−+= iii
]sincossincossincos[sinsincossincoscos θθθθθθθθθθθθ 5335642246 62061515 +−+−+−= i
Equating real and imaginary parts separately
and15156 642246 θθθθθθθ sinsincossincoscoscos −+−=
θθθθθθθ 5335 62066 sincossincossincossin +−=
Addtion formulae for any number of angles
We have,
)...................... nn θθθθθθθθ ++++= 321321 cis(ciscisciscis
nθθθθ ciscisciscis Now 321 .........
)tan(cos).........tan(cos)tan(cos nn iii θθθθθθ +++= 111 2211
)tan.....().........tan)(tan(cos.........coscos nn iii θθθθθθ +++= 111 2121
)............( nθθθθ ++++ 321cis ie
....)..........(cos..........coscoscos ++++= 33
22
1321 1 SiSiiSnθθθθ
where 11 θtan∑=S
212 θθ tantan∑=S
3213 tantantan θθθ∑=S
.........................................
)(cis 21 nè............èè +++∴
....)..........(cos..........coscos 432121 1 SiSSiSn +−−+= θθθ
..)..........(cos..........coscos....)..........(cos..........coscos 31214221 1 SSiSS nn −++−= θθθθθθ
Equating real and imaginary parts separately
.......)(cos..........coscos).............cos( 4221321 1 SSnn +−=++++ θθθθθθθ
and
.......)(cos..........coscos).............sin( 3121321 SSnn −=++++ θθθθθθθ
.................SS
.................SSè.............èèè n
42
31321
1)tan(
+−−
=++++∴
MCA 11 - Mathematics SVT 23
Exercise
unity. of rootsfourth ofproduct continued theFind1.
.)( 32334 of value thefind then unity, ofroot cube theis '' If2. ωωω −−
8
88
88 of value theFind3
−+
)/cos()/sin(
)/cos()/sin(.
ππππ
i
i
).cossin. θθ i+(number complex theof inverse tivemultiplica theof conjugate theWrite4
?represents 11 doest then wha If5 =++= ziyxz.
Theorem. sMoivre' De using 22Simplify 6. 2)cos(sin xix +
.)()( 01232equation esatisfy th which and of valuesreal theFind7. =+++− yxiyxyx
.)()( 0432equation esatisfy th which and of valuesreal theFind8. =−−++ yixiyx
.. 321 find integer,any is '' If9 +−+−+−− +++ nnnn iiiin
?. 101 of inverse tivemultiplica theisWhat 10 i
.)1()1( of value thefind ,'1' ofroot cube in the '' If.11 323 ωωω +−+
..432
4242Simplify 12
iiii+++
form.Cartesian in the 4
3
4
32 Express13
+ ππ
sincos. i
.sincos
1
551 ofpart real theFind14.
−
++ ππ
i
.)(,.22
22222 that prove If15
dc
bayx
idc
ibaiyx
++=+
++=+
.. 11
1 for which ''integer positiveleast theFind16 =
−+ n
i
in
.,. 9 that show326 and If17 22 =++=++= yxzziyxz
.,. 1 that show122 and If18 22 =+−=−+= bazzibaz
.,. 12 that show41
1amp such that be If19 22 =−+=
+−+= yyx
z
ziyxz
π
form.polar in the1 Express20 i−.
form.polar in the2
3 Express21
i−.
.
2
3
2 of amplitude and modulus theFind22.
−+
i
i
2
3
Simplify 23)sin(cos
)cos(sin.
ααθθ
i
i
++
24 KSOU Complex Trigonometry
4
42 3377Simplify 24 −
−
+++
)sin(cos
)sin(cos)sin(cos.
θθθθθθ
i
ii
.))()()((. 161111 that show unity, ofroot cube a is If25 16884422 =+−+−+−+− ωωωωωωωωω
.,5
5 1 find then cisIf26.
xxx += θ
terms.2 find then 1 If27 6422 )(,. niiii ++++−= L
−=
+−==
2 that prove cis and cis If28.
βαβα tan, iyx
yxyx
.cos,.3
2 that prove then 042 of roots theare & If29 12 πβαβα nxx nnn +=+=+−
( ) ( ) .)()(
.
462121 ii)411 i)
following theProve306633 =++−−=−++ iii
.sincos,sincos. θθθθ niZ
ZnZ
ZiZn
nn
n 21
and 21
that show If31 =−=++=
).sin()cos(
,,.
βαβα
βα
+=−+=+
= =
ixy
xyxy
xy
yx
21
ii)21
i)
thatproveciscisIf32
).32sin(21
ii))32cos(21
i)
thatprove,cis,cisIf.33
3232
3232 βαβα
βα
+=−+=+
= =
iyx
yxyx
yx
yx
.tan,. θθ nix
xx
n
n
=+− =
1
1 that provecisIf34
2
2
).2/(cos)4cos(2)sincos1()sincos1( that Prove.35 8988 θθθθθθ ⋅⋅=−++++ ii
).sin(sinsinsin))cos(coscoscos)
,sinsinsincoscoscos.
γβαγβαγβαγβαγβαγβα
++=++++=++++==++
18327383ii18327383i
thatprove 32032 If36
).sin())cos()
,coscos,cos.
γβαγβα
γβα
++=−++=+
+=1+=+=
ixyz
xyzxyz
xyz
zz
yy
xx
21
ii21
i
thatprove1
2 and 21
2 If37
[ ])/(tansin)()()(. abnbaiibaiba nnn 12222 22 that Show38 −⋅⋅+=−−+
−+
−=
−+++ θπθπ
θθθθ
nn
inn
i
in
221
1 that Prove39 sincos
cossin
cossin.
.tansincos
sincos. θθ
θθθθ
7cis21
1 that Prove40 7
7
⋅
=
−+−−
ii
i
.tan
tan
tan
tan.
αα
αα
i
i
i
in
−+=
−+
1
1
1
1 that Prove41
MCA 11 - Mathematics SVT 25
.,,);/(. iZZZrZ rr =∞⋅⋅== LLL 321 that prove 3213 cis If42 π
.31 of roots cube theFind.43 i+−
.)(. /321 of values theall Find44 i+
.)(. /321 of values theall Find45 i−
diagram. Argand in the themrepresent and 31 of rootsfourth theFind46 i+.
.0 Solve47. 7 =− xx
.. 01 Solve48 5 =+x
.cos&sincos&sin θθθθ of powersin 77 Express49.
.sin&cossin&cos. θθθθ of powersin 88 Express50
–––––––––––––
26 KSOU Complex Trigonometry
MATRIX THEORY
Review of the fundamentals
A rectangular array of mn elements arranged in m rows & n columns is called a 'Matrix' of a order m × n matrices are denotedby capital letters of The English Alphabet.
Examples
Matrix of order 3 × 2 is
33
22
11
ba
ba
ba
Matrix of order 4 × 3 is
321
321
321
321
ddd
ccc
bbb
aaa
Matrix of order 3 × 3 is
333
222
111
cba
cba
cba
Note :- Elements of Matrices are written in rows and columns with in the bracket ( ) or [ ].
Types of Matrices
(1) Equivalent Matrices : Two matrices are said to be equivalent if the order is the same.
(2) Equal Matrices : Two matrices are said to be equal if the corresponding elements are equal.
(3) Rectangular & Square Matrices : A matrix of order m × n is said to be rectangular if m ≠ n, square if m = n.
(4) Row Matrix : A matrix having only one row is called Row Matrix.
(5) Column Matrix : A matrix having only one column is called Column Matrix.
(6) Null Matrix or Zero Matrix : A matrix in which all the elements are zeros is called Null Matrix or Zero Matrixdenoted as O. [English alphabet O not zero where as elements are zeros]
(7) Diagonal Matrix : A diagonal matrix is a square matrix in which all elements except the elements in the principaldiagonal are zeros.
400
010
002
60
04 Example
are diagonal matrices of order 2 & 3.
(8) Scalar Matrix : A diagonal matrix in which all the elements in the principal diagonal are same.
800
080
008
40
04 Example
are Scalar Matrices of order 2 & 3.
(9) Unit Matrix or Identity Matrix : A diagonal matrix in which all the elements in the principal diagonal is 1 iscalled Unit Matrix or Identity Matrix denoted by I.
1000
0100
0010
0001
10
01: Example ,
are unit matrices of order 2 & 4.
(10) Transpose of a Matrix : If A is any matrix then the matrix obtained by interchanging the rows & columns of A iscalled 'Transpose of A and it is written as A' or AT.
′
=
fdb
ecaA
fe
dc
ba
A is then If : Example
A is of order 3 × 2 but A' is of order 2 × 3.
Matrix addition
Two matrices can be added or subtracted if their orders are same.
=
=
222
111
222
111 If : Exampleedc
edcB
cba
cbaA &
++++++
=+222222
111111
ecdbca
ecdbcaBA
−−−−−−
=−222222
111111
ecdbca
ecdbcaBA
Matrix Multiplication
If A is a matrix of order m × p and B is matrix of order p × n, then the product AB is defined and its order is m × n. (ie. for ABto be defined number of columns of A must be same as number of rows of B)
=
=
33
22
11
222
111Let : Example
βαβαβα
Bcba
cbaA &
++++++++
=322212322212
312111312111then βββαααβββααα
cbacba
cbacbaAB
which is of order 2 × 2.
Note :- If A is multiplied by A then AA is denoted as A2, AAA.... as A3 etc.
28 KSOU Matrix Theory
Scalar Multiplication of a Matrix
If A is a matrix of any order and K is a scalar (a constant), then KA represent a matrix in which every element of A is multipliedby K.
=
=
333
222
111
333
222
111
then If : Example
KcKbKa
KcKbKa
KcKbKa
KA
cba
cba
cba
A
Symmetric and Skew Symmetric Matrices
Let A be a matrix of order n × n an element in ith row and j th column can be denoted as aij. Hence a matrix of order n × n canbe denoted as (aij) or [aij] where i = 1, 2, .......n, j = 1, 2, .......n
A matrix of order n × n is said to be Symmetric if aij = aji and Skew Symmetric if aij = –aji or A is symmetric if A = AT
or A = A', skew symmetric if A = –AT or A = –A' also A + A' is symmetric & A – A' is skew symmetric.
Note :- In a skew symmetric matrix the elements in principal diagonal are all zeros.
AAA ′=
= wheresymmetric is
865
673
532
: Example
BBB ′−=
−−
−= wheresymmetric skew is
067
602
720
Determinant
A determinant is defined as a mapping (function) from the set of square matrices to the set of real numbers.
If A is a square matrix its determinant is denoted as .A
333
222
111
333
222
111
det. then Let : Example
cba
cba
cba
AorA
cba
cba
cba
A =
=
Minors and Co-factors
321321Let ,,,,)( === jiaijA
=
333231
232221
131211
ie
aaa
aaa
aaa
A
333231
232221
131211
aaa
aaa
aaa
A =
3332
2322Consider aa
aa which is a determinant formed by leaning all the elements of row and column in which all lies. This
determinant is called Minor of a11. Thus we can form nine minors. In general if A is matrix of order n × n then minor of aij is
MCA 11 - Mathematics SVT 29
obtained by leaning all the elements in the row and column in which aij lies in .A The order of this minor is n – 1where as theorder of given determinant is n if this minor is multiplied by (–1)i + j then it is called Co-factors of aij.
=
333231
232221
131211
Let : Example
aaa
aaa
aaa
A
3332
232211 ofMinor
aa
aaa =
3332
2322
3332
23221111 1 offactor -Co
aa
aa
aa
aaa =−= +)(
3332
131221 is ofMinor
aa
aaa
3332
1312
3332
13121221 1 is offactor -Co
aa
aa
aa
aaa −=− +)(
Value of a determinant
Consider a matrix A of order n × n. Consider all the elements of any row or column and multiply each element by its correspondingco-factor. Then the algebraic sum of the product is the value of the determinant.
22
11Let : Exampleba
baA =
Co-factor of a1 is b2
Co-factor of b1 is –a2
2121 abbaA −=∴
333
222
111
Let
cba
cba
cba
A =
33
22
33
22111 1 is offactor -Co
cb
cb
cb
cba =− +)(
33
22
33
22211 1 is offactor -Co
ca
ca
ca
cab −=− +)(
33
22
33
22311 1 is offactor -Co
ba
ba
ba
bac =− +)(
33
221
33
221
33
221 ba
bac
ca
cab
cb
cbaA +−=∴
)()()( 233212332123321 babaccacabcbcba −+−−−=
123132213312231321 cbacbacbacbacbacba −+−−−= )
30 KSOU Matrix Theory
Properties of determinants
(1) If the elements of any two rows or columns are interchanged then value of the determinant changes only in sign.
(2) If the elements of two rows or columns are identical then the value of the determinant is zero.
(3) If all the elements of any row or column is multipled by a constant K, then the value of the determinant is multipledby K.
(4) If all the elements of any row or column are written as sum of two elements then the determinant can be written assum of two determinants.
(5) If all the elements of any row or column are multiplied by a constant and added to the corresponding elements ofany other row or column then the value of the determinant donot alter.
Adjoint of a Matrix
=
333
222
111
Let
cba
cba
cba
A
Let us denoted the co-factors of a1, b1, c1, a2, b2, c2, a3, b3, c3 as A1, B1, C1, A2, B2, C2, A3, B3, C3 transpose of matrix ofco-factors is called Adjoint of the Matrix.
=
333
222
111
factors-Co ofMatrix
CBA
CBA
CBA
=
321
321
321
ofAdjoint
CCC
BBB
AAA
A
AAIAAA .... adjadj ==Theorem
=
321
321
321
333
222
111
adj
CCC
BBB
AAA
cba
cba
cba
AA ..
++++++++++++++++++
=
333333232323131313
323232222222121212
313131212121111111
CcBbAaCcBbAaCcBbAa
CcBbAaCcBbAaCcBbAa
CcBbAaCcBbAaCcBbAa
∆=+−=++33
221
33
221
33
221111111 Now
ba
bac
ca
cab
cb
cbaCcBbAa The value of the det. A.
∆=++ 222222 Similarly CcBbAa
∆=++ 333333 CcBbAa
33
111
33
111
33
111212121 ba
bac
ca
cab
cb
cbaCcBbAa −+−=++
)()()( 133111331113311 babaccacabcbcba −−−+−−=
0113131113311131311 =+−−++−= cbacbacbacbacbacba
MCA 11 - Mathematics SVT 31
Similarly the other five elements of A adj.A is zero.
AAA =∆
∆∆
∆=∴ where
00
00
00
adj..
I.∆=
∆=
100
010
001
AIAAA ... adjadj =⋅=∴
Singular and Non-singular Matrices
A square matrix A is said to be singular if 0=A and non-singular if .0≠A
Inverse of a Matrix
Two non-singular matrices A & B of the same order is said to be inverse of each other if AB = I = BA. Inverse of A is denotedas A–1. Inverse of B is denoted as B–1 and further (AB)–1 = B–1A–1.
To find the inverse of A
IAAA ⋅=..adj
111 adjby multiply −−− = AAAAAA ..,
A
AAAAA
..
adjadjie 11 =⇒= −−
Example : Find the inverse of
−−−
−
121
452
241
−−−
−=
121
452
241
Let A
−−−−
−−
−−
−−
−−
−
−−−−
−−−
=
52
41
42
21
45
2421
41
11
21
12
2421
52
11
42
12
45
factors-Co ofMatrix
+−−−−−−−+−−
+−−−+−=
)()()(
)()()(
)()()(
85441016
422144
544285
=
306
630
963
32 KSOU Matrix Theory
=∴
369
036
603
adj.A
121
452
241
−−−
−=A )()()( 542424851 +−−−−+−= 918243 =−+=
AA
A adj.11 =−
=
369
036
603
9
1
=
93
96
99
93
96
96
93
0
0
=−
31
32
31
32
32
31
1
1
0
0
A
Solutions of Linear equations
Cramer's Rule
To solve the equations
1111 dzcybxa =++
2222 dzcybxa =++
3333 dzcybxa =++
333
222
111
Consider
cba
cba
cba
=∆ (1)
first evaluate & if it is not zero then multiply both sides of (1) by x.
333
222
111
333
222
111
cba
cbxa
cbxa
cba
cba
cba
xx
x
==∆
multiply the elements of columns 2 & 3 by y & z and add to elements of column 1.
33333
22222
11111
then
cbzcybxa
cbzcybxa
cbzcybxa
x
++++++
=∆
(say)1
333
222
111
∆==cbd
cbd
cbd
(2)
multiply both sides of (1) by y
333
222
111
333
222
111
cyba
cyba
cyba
cba
cba
cba
yy ==∆
MCA 11 - Mathematics SVT 33
multiply the elements of columns 1 & 3 by x & z and add to the elements of column 2.
33333
22222
11111
czcybxaa
czcybxaa
czcybxaa
++++++
= (say)2
333
222
111
∆==cda
cda
cda
(3)
multiply both sides of (1) by z
zcba
zcba
zcba
cba
cba
cba
zz
333
222
111
333
222
111
==∆
multiply the elements of columns 1 & 2 by x & y and add to the elements of column 3.
zcybxaba
zcybxaba
zcybxaba
33333
22222
11111
++++++
=
(say)3
333
222
111
∆==dba
dba
dba
(4)
then (2) from 1
∆∆
=x
(3) from 2
∆∆
=y
(4) from 3
∆∆
=z
Note :- Verification of values of x, y, z can be done by substituting in the given equations.
Example - 1
32 Solve =−+ zyx
1=++ zyx
432 =−− yyx
321
111
112
Let
−−=∆
-
(1)
)()()( 121131232 −−−−−−+−= 5342 =++−=
multiply both sides of (1) by x
321
111
112
then
−−=∆
-
xx 32
11
112
−−=
x
x
-x
multiply the elements of columns 2 & 3 by y and z and add to the elements of column 1.
3232
11
112
−−−−++−+
=∆yyx
zyx
-zyx
x
34 KSOU Matrix Theory
324
111
113
−−=
-
)()()( 421431233 −−−−−−+−= 10673 =++−=
25
1010 ==∆
=∴ x
multiply both sides of (1) by y
321
11
12
341
111
132
−−==
−=∆
y
y
-y-
yy
multiply the elements of column 1 by x & 3 by z and to the corresponding elements of column 2.
3321
11
122
then
−−−++−+
=∆zyx
zyx
-zyx
y
341
111
132
−=
-
)()()( 141133432 −−−−−−−= 531214 −=−+−=
15
55 −=−=∆
−=∴ y
multiply both sides of (1) by z
z
z
-z-
zz
341
11
32
341
111
132
−=
−=∆
multiply the elements of column 1 by x & column 2 by y and to the corresponding elements of column 3.
zyx
zyx
zyx
z
3221
11
212
then
−−−++−+
=∆
421
111
312
−= )()()( 123141242 −−+−−+= 09312 =−−=
05
00 ==∆
=∴ z
Thus solution is x = 2, y = –1 & z = 0 which can be verified by substituting in the given equations.
Example - 2
74 Solve =+ yx
543 =+ zy
235 =+ zx
302
435
014
=∆ )()()( 15002001094 −+−−−= 562036 =+=
MCA 11 - Mathematics SVT 35
302
435
017
1 =∆ )()()( 6008151097 −+−−−= 56763 =−=
325
450
074
2 =∆ 16814028250020078154 =+=−+−−−= )()()(
205
530
714
3 =∆ )()()( 15072501064 −+−−−= 561052524 −=−+=
156
561 ==∆∆=∴ x
356
1682 ==∆
∆=y
156
563 −=−=∆∆=z
Solution of Linear equations by Matrix Method
Given 1111 dzcybxa =++
2222 dzcybxa =++
3333 dzcybxa =++
=
333
222
111
Consider
cba
cba
cba
A
=
=
3
2
1
d
d
d
B
z
y
x
X &
then given equations can be written in Matrix form as AX = B. If 0≠A solution exists multiply both sides by A–1
BAAXA 11 −− =)(
BAAXA 11 −− =
BAIX 1 ie −=
BAX 1−=∴
Example
1323 Solve =+− zyx
32 =−+ zyx
853 −=−+ zyx
−=
=
−−
−=
8
3
13
531
112
213
Let B
z
y
x
XA ,
then given equations can be written as AX = B
36 KSOU Matrix Theory
BAX 1−=∴ (1)
To find A–1
531
112
213
−−
−=A 510961621101353 −=+−−=−++−++−= )()()(
−−
−−
−
−−
−−−
−
−−
−−−
=
12
13
12
23
11
2131
13
51
23
53
2131
12
51
12
53
11
factors-Co ofMatrix
+−−−−+−−−−−
−+−−+−=
)()(
)()()(
)()()(
234321
1921565
1611035
−−−−
−=
571
10171
592
−−
−−=
5105
7179
112
adj.A
−−
−−−==∴ −
5105
7179
112
5
1adj.
11 AA
A
Using this in (1)
−
−−
−−−=
8
3
13
5105
7179
112
5
1X
−−−−
−−=
403065
5651117
8326
5
1
−=
−
−−=
1
2
3
5
10
15
5
1
∴ x = 3, y = –2, z = 1 is the solution
Verification : Consider the first equation
1322923 =++=+− zyx
Characteristic equation, Eigen Values & Eigen Vectors
Let A & I be square matrices of same order and λ a scalar then 0=− IA λ is called Characteristic equation and the roots of
this equation ie values of λ are called Eigen Values or Characteristic roots. The matrix X satisfying AX = λX is called EigenVector.
Example - 1
Find the eigen roots and eigen vectors of the matrix
32
41
=
32
41Let A Characteristic equation is 0
10
01
32
41=
−
λ
034
41 ie =
−−
λλ
0831 =−−−⇒ ))(( λλ
0833 2 =−+−−⇒ λλλ 0542 =−−⇒ λλ
MCA 11 - Mathematics SVT 37
51015 ,))(( −=⇒=+−⇒ λλλ
∴ Eigen roots are –1 & 5.
To find eigen vector X, corresponding to –1,
AX = –1
−−
=
y
x
y
x
32
41 ie
−−
=
+
+⇒
y
x
yx
yx
32
4
yx
yix
yyx
xyx
2 ie
4
32
4
−=−=
⇒−=+−=+
⇒ 12
yx =−
∴⇒
∴ Eigen vector corresponding to eigen value –1 is (–2, 1)
To find the eigen vector corresponding to 5.
=
2
1
2
1 532
41 ie
x
x
x
x
=
+
+⇒
2
1
21
21
5
5
32
4
x
x
xx
xx
=+=+⇒
221
121
532
54
xxx
xxx
11 ie ie 21
21xx
xx ==
∴ Eigen vector is (1, 1)
∴ Eigen vector corresponding to eigen root 5 is (1, 1)
Example - 2
Find the eigen roots and eigen vectors of the matrix
−−−
−
312
132
226
−−−
−=
312
132
226
Let A Characteristic equation is 0
312
132
226
=−−−−−
−−
λλ
λ
[ ] [ ] [ ] 032222322136 ie 2 =−−++−−+−−− )()()()( λλλλ
[ ] [ ] [ ] 0262222621696 ie 2 =+−+++−+−−+− ))( λλλλλ
0422422866 ie 2 =−+−++−− )()())(( λλλλλ
084848648366 ie 232 =−+−+−+−+− λλλλλλλ
0323612 ie 23 =+−+− λλλ 0323612 ie 23 =−+− λλλ
which is the characteristic equation, by inspection 2 is a root
have we2by 323612 dividing 23 ,−−+−∴ λλλλ
016102)- 2 =+− )(( λλλ
0822( ie =−−− ))()( λλλλ 822 ,,=∴ λ
38 KSOU Matrix Theory
To find eigen vector or λ = 2
Consider AX = 2X
=
=
−−−
−
3
2
1
3
2
1
3
2
1
where2
312
132
226
ie
x
x
x
X
x
x
x
x
x
x
1321 2226 xxxx =+−⇒ 0224 321 =+−⇒ xxx
2321 232 xxxx =−+− 02 321 =−+− xxx
3321 232 xxxx =+− 02 321 =+− xxx
the above three equations represent one equation 2x1 – x2 + x3 = 0.
21 ie 2 then 0Let 21
213xx
xxx === , 021
ie 321 xxx ==
∴ Eigen Vector is (1, 2, 0)
To find the eigen vector for λ = 8
Consider AX = 8X
=
−−−
−
3
2
1
3
2
1
8
8
8
312
132
226
ie
x
x
x
x
x
x
1321 8226 ie xxxx =+− 0222 ie 321 =+−− xxx
2321 832 xxxx =−+− 052 321 =−−− xxx
3321 832 xxxx =+− 052 321 =−− xxx
0 ie 321 =−+ xxx (1)
052 321 =++ xxx (2)
052 321 =−− xxx (3)
adding (1) & (2) we get 063 21 =+ xx
02 ie 21 =+ xx 21 2 ie xx −= (Say) 1221 K
xx ==−
∴
KxKx =−= 21 2then ,
subsitituting in (1)
KxxKK −=⇒=−+− 33 02
KxKxKx −==−=∴ 321 2 &,
1) 1, 2,( isor Eigen vect −−∴ 1) 1, (2,or −
8 2, 2, are rootsEigen ∴
),,(&),,(& 112021 are orsEigen vect −
Properties of Eigen values
(1) The sum of the eigen values of a matrix is the sum of the elements of the principal diagonal.
(2) The product of the eigen values of a matrix is equal to the value of its determinant.
(3) If λ is an eigen value of A then λ1
is the eigen value of A–1.
MCA 11 - Mathematics SVT 39
Cayley - Hamilton Theorem
Every square matrix satisfies its characteristic equation.
=
dc
baALet Characteristic equation is 0=
−−
λλ
dc
ba
which on simplification becomes a quadric equation in λ in the form 0212 =++ aa λλ where a1, a2 are constants.
Cayley Hamilton Theorem states that 0212 =++ IaAaA where I is a unit matrix of order 2 & 0 is a null matrix of order 2.
=
333
222
111
If
cba
cba
cba
A
then characteristic equation is 0
333
222
111
=−
−−
λλ
λ
cba
cba
cba
which on simplification becomes 0322
13 =+++− aaa λλλ which is a cubic equation.
Then as per Cayley Hamilton Theorem 0322
13 =+++− IaAaAaA where I is a unit matrix of order 3 & 0 is a null
matrix of order 3.
In general if A is a square matrix of order n then characteristic equation will be of the form
01 21 =++++− −− Iaaa n
znnnn ........)( λλλ
and by Cayley Hamilton Theorem
01 21 =++++− −− IaAaaAA n
znnnn ........)(
where I is a unit matrix of order n & 0 is a null matrix of order n.
Note :- If we put λ = 0 in the characteristic equation then Aan =
∴ If an = 0, matrix A is singular & 0≠na the matrix A is non-singular & hence inverse exists and we can find the
inverse of A using Cayley Hamilton Theorem.
Example - 1
=
dc
baALet
Characteristic equation is 0=−
−λ
λdc
ba
0 ie 212 =++ aa λλ where a1, a2 are constants.
By Cayley Hamilton Theorem
0212 =++ IaAaA
multiply both sides by A–1
0then 12
11
12 =++ −−− AaAAaAA
0 ie 121 =++ −AaIaA
)( IaAAa 11
2 +−=∴ −
)( IaAa
A 12
1 1 +−=∴ −
40 KSOU Matrix Theory
Example - 2
The characteristic equation of a matrix A of order 2 is .A find 01052 =+− λλ
Solution : put λ = 0 in C.E. then the constant 10 is .A
Example - 3
Find the inverse of
−
−43
12 using Cayley Hamilton Theorem.
−
−=
43
12Let :Solution A 0
43
12 is C.E. =
−−−−
λλ
0342 ie =−−− ))(( λλ 03248 ie 2 =−+−− λλλ
056 ie 2 =+− λλ
by Cayley Hamilton Theorem
0562 =+− IAA
multiply both sides by A–1
056 1 =+− −AIA
IAA 65 1 +−=∴ −
+
−
−−=
10
016
43
12
+−+
++−=
6403
0162
=
23
14
1=∴ −
23
14
51A
Diagonalisation of Matrices
If A is a square matrix of order n where all the eigen values are linearly independent then a matrix P can be found such thatP–1AP is a Diagonal Matrix.
Let A be a square matrix of order 3 and let λ1, λ2, λ3 be the eigen values, corresponding to these. Let X1, X2, X3 be threevectors where
=
=
=
3
2
1
3
3
2
1
2
3
2
1
1
z
z
z
X
y
y
y
X
x
x
x
X ,,
=
333
222
111
Let
zyx
zyx
zyx
P
=−
3
2
11
00
00
00
Then
λλ
λAPP
Example - 1
−
−=
45
21Let A
01041045
21 is C.E. =−−−⇒=
−−−−
))(( λλλ
λ
MCA 11 - Mathematics SVT 41
061065 ie 2 =−+⇒=−− ))(( λλλλ eseigen valu are 61,−=⇒ λ
XAX −=−= let 1For ,λ
−−
=
−
−
2
1
2
1
45
21 ie
x
x
x
x
21221
121
45
2 ie xx
xxx
xxx=⇒
−=+−−=−
=∴
1
1 isor eigen vect
1121 ....
xx
XAX 66For == ,λ
=
−
−
2
1
2
1
6
6
45
21 ie
x
x
x
x
21221
121 2545
2xx
xxx
xxx−=⇒
−=+−−=−
52 ie 21 xx =
−
−∴
5
2 isor eigen vect
−=
51
21Let P
−
=−
11
25
7
1Then 1P
−
−
−
−
=∴ −
51
21
45
21
11
25
7
11APP
−
+−−
+−−=
51
21
4251
810105
7
1
−
−
−−=
51
21
66
25
7
1
−=
++−−−−
=420
07
7
1
301266
101025
7
1
−=
60
01
−
−
−=
45
21matrix theediagonaliz
51
21 Thus P
Example - 2
=
113
151
311
Let A
Characteristic equation is 0
113
151
311
=−
−−
λλ
λ
[ ] [ ] [ ] 05313311115(1 ie =−−+−−−−−−− )())(() λλλλλ
03143246(1 ie 2 =+−+−−−+−− )()())( λλλλλ
094224646 ie 232 =+−++−+−+− λλλλλλλ
0367 ie 23 =−+− λλ 0367 ie 23 =+− λλby inspection –2 is a root ∴ λ + 2 is a factor. ∴ equation becomes
01892 2 =+−+ ))(( λλλ
42 KSOU Matrix Theory
0632 ie =−−+ )()(( λλλ 632 ,,−=∴ λ
ie characteristic roots are –2, 3, 6.
To find the eigen vector for λ = –2 Consider 11 2XAX −=
=
3
2
1
1 where
x
x
x
X
−−−
=
3
2
1
3
2
1
2
2
2
113
151
311
ie
x
x
x
x
x
x
1321 23ie xxxx −=++ 033 ie 321 =++ xxx (1)
2321 25 xxxx −=++ 07 321 =++ xxx (2)
3321 23 xxxx −=++ 033 321 =++ xxx (3)
(1) & (3) are same.
0 then (2),or (1)in 0Put 312 =+= xxx
11 ie 31
31xx
xx =−
⇒−=
−=∴
1
0
1
or eigen vect 1X
Let X2 be the eigen vector for λ = 3.
22 3ie XAX =
=
=
3
2
1
2
3
2
1
3
2
1
where
3
3
3
113
151
311
ie
y
y
y
X
y
y
y
y
y
y
1321 33ie yyyy =++ 032ie 321 =++− yyy (1)
2321 335 yyyy =++ 02 321 =++ yyy (2)
3321 333 yyyy =++ 023 321 =−+ yyy (3)
Let us eliminate y1 from (1) & (2)
032is 11 321 =++−× yyy)(
0242 is 22 321 =++× yyy)(
055adding 32 =+ yy
032 =+⇒ yy11
ie 3232
yyyy =
−⇒−= (4)
Let us eliminate y2 from (2) & (3)
055gsubtractin
0426 is 23
02is 12
32
321
321
=+−
=−+×=++×
yy
yyy
yyy
)(
)(
3131 55 yyyy =⇒=⇒11
31 yy =∴ (5)
MCA 11 - Mathematics SVT 43
111 (5) & (4) From 321 yyy =
−=
−=∴1
1
1
2X
Next, let X3 be the eigen vector for λ = 6
33 6 ie XAX =
=
=
3
2
1
3
3
2
1
3
2
1
where
6
6
6
113
151
311
ie
z
z
z
X
z
z
z
z
z
z
1321 63 ie zzzz =++ 035 ie 321 =++− zzz (1)
2321 65 zzzz =++ 0321 =+− zzz (2)
3321 63 zzzz =++ 053 321 =−+ zzz (3)
044 (2), & (1) adding 31 =+− zz
11 ie 31
31zz
zz =∴= (4)
Let us eliminate z3 from (2) & (3)
048adding
053is 13
0555 is 52
21
321
321
=−
=−++×=+−×
zz
yzz
zzz
)(
)(
212 ie 21
21zz
zz =⇒= (5)
121 (5), & (4) From 321 zzz ==
=∴
1
2
1
3X
−
−=
111
210
111
Let P
−=−
600
030
002
Then 1APP
−
−=∴
113
151
311
ediagonaliz
111
210
111
P
Exercise
..
132
321
321
Evaluate1 −
..
cbbaac
baaccb
accbba
−−−−−−−−−
Evaluate2
44 KSOU Matrix Theory
..
443
432
321
Evaluate3
.,. xx find then 0
131
56
402
If4 =−−
..
1
1
1
Evaluate5
rq
rp
qp
−−
−
..
32
02ofadjoint theFind6
1. offactor -co thefind
243
432
321
If7
=A.
.&. AAAAA ′′
= find
730
512If8
.sectan
tansec.
θθθθ
of inverse theFind9
.. ABBABA 76and35 find 240
321 and
501
432If10 −−
−=
−
=
.. BAABBA and find 321
4022 and
673
5243If11
−
=+
−=+
.. yxxy
yx
x
xxand find
412
54
75
1If12
−=
−
+
+
( ) .. AA =′′
= Aat verify th
1098
765If13
.. BABABA −′′+
−−=
−
−= and find
23
34
42
and 126
543If14
.. 010 that prove 75
43If15 2 =+−
= IAAA
..
23
56 of inverse theFind16
..
40
21 ofequation sticcharacteri theFind17
MCA 11 - Mathematics SVT 45
..
40
32 of eseigen valu theFind18
Rule. sCramer'by 25212 Solve19 −=−=+−=+ yzxyzx ,,.
Solve20. 545 =+− zyx
2532 =++ zyx
5627 =+− zyx
by matrix method.
.10
12 of roots sticcharacteri theFind 21.
−
.
322
121
101
of roots sticcharacteri theFind 22.
−
.43
21matrix for the TheoremHamilton -CayleyVerify 23.
.11
02matrix for the TheoremHamilton -CayleyVerify 24.
−
.12
21matrix for the orseigen vect theFind 25.
.
312
132
226
matrix for the orseigen vect and eseigen valu theFind 26.
−−−
−
46 KSOU Matrix Theory
BCA 21 / IMCA 21 / Mathematics SVT 21
ALGEBRAIC STRUCTURES
Abreviations used
N : represent set of natural numbers.
Z or I : represent set of +ve and –ve integers including zero.
Z+ : represent non-negative integers ie. +ve integers including zero.
Q : represent set of rational numbers.
R : represent set of real numbers.
C : represent set of complex numbers.
Zn = {0, 1, 2, 3, .............. n – 1} ie. Zn represent set of integers modulo n.
Q+ : represent set of +ve rational numbers.
z - {o} : represent set of integers except 0.
Q - {o} : represent set of rational numbers except zero.
R - {o} : set of real numbers except zero.
A set in general is denoted by S.
∀ : for all
∈ : belongs to
Binary Operation
If S is a non-empty set then a mapping (function) from S × S to S is defined as Binary Operation (in short B.O.) and denotedby ∗ (read as star). ie. : S × S → S (Star maps S cross S to S)
Another Definition
If S is non-empty set then ∗ (star) is said to be a Binary operation if ∀ a, b ∈ S, a ∗ b ∈ S.
Examples
(1) on N + and × (ie addition & multiplication) are B.O.
N∈=+ 532 N∈=× 632
(2) on Z, +, – & × are B.O.
,, ZZ ∈=−∈+ 14354 ZZ ∈=×∈=− 3065134 ,
(3) On Q & R +, – & × are B.O. but ÷ is not a B.O. on Q & R Q for 0, RQa
&∉0
but on Q - {o} & R - {o} ÷ is a B.O.
(4) on C, + and × are B.O.
Cyyixxiyxiyx ∈+++=+++ )()()()( 21212211
Cyxyxiyyxxiyxiyx ∈++−=+++ )()()()( 122121212211
48 KSOU Algebraic Structures
Definitions
(1) A non-empty set S with one or more binary operations is called an 'Algebraic Structure'.(N, +), (Z, +, ×), (Q, +, ×) are all algebraic structures.
(2) Closure Law : A set S is said to be closed under a B.O. ∗ if ∀ a, b ∈ S, a ∗ b ∈ S.
(3) Associative Law : A B.O. ∗ is said to be associative on S if ∀ a, b, c ∈ S(a ∗ b) ∗ c = a ∗ (b ∗ c)
(4) Commutative Law : A B.O. ∗ is said to be commutative on S if ∀ a, b ∈ S, a ∗ b = b ∗ a.
(5) Identity Law : An element e ∈ S satisfyinga ∗ e = a = e ∗ a. ∀ a ∈ S is called an identity element for the B.O. ∗ on S.
Examples
(i) + and × (addition and multiplication) are associative and commutative on N, Z, Q & R.
(ii) B.O. – (subtraction) is not associative & commutative.
(iii) 1 is an identity for B.O. × on N but + has no identity on N. Where as O is an identity on Z, Q and R for the B.O. +.
(iv) If S is a set of 2 × 2 matrices and B.O. is matrix multiplication then
=
10
01I is an identity element.
Group
A non-empty set G together with a B.O. ∗ ie (G, ∗) is said to form a group if the following axioms are satisfied.
G1. Closure Law : ∀ a, b ∈ G, a ∗ b ∈ G
G2. Associative Law : ∀ a, b, c ∈ G, (a ∗ b) ∗ c = a ∗ (b ∗ c)
G3. Identity Law : There exists an element e ∈ G such that ∀ a ∈ G, a ∗ e = a = e ∗ a.
G4. Inverse : ∀ a ∈ G, there exists an element b such that a ∗ b = e = b ∗ a. This b is called inverse of a and usually denotedas a–1
ie. a ∗ a–1 = e = a–1 ∗ a.
In addition to the above four axions if ∀ a, b ∈ G, a ∗ b = b ∗ a. Then (G, ∗) is called an 'abelian group' or 'commutativegroup'.
Note (1) If for (G, ∗) only G1 is satisfied it is called a 'groupoid'.
(2) If for (G, ∗) G1 & G2 are satisfied it is called a 'semi-group'.
(3) If for (G, ∗) G1, G2 & G3 are satisfied it is called 'Monoid'.
Examples
(i) (N, +) is a groupoid and semigroup.
(ii) (N, ×) is a groupoid, semigroup and Monoid (identity for × is 1)
(iii) ( Z, +) is a group (identity is O and a–1 = –a) ie. a + (–a) = 0 = –a + a.
Note :- Every group is a monoid but the converse is not true, (Z, +) is a group and also a monoid but (N, ×) is a monoid butnot a group.
MCA 11 - Mathematics SVT 49
Properties of Groups
1. Cancellation laws are valid in a group
ie if ),( ∗G is a group then ,,, Gcba ∈∀
law)on cancellatileft i ()( cbcaba =⇒∗=∗
law)on cancellatiright ii ()( cbacab =⇒∗=∗
Proof :- GaGacaba ∈∈∗=∗ −1as ,,
)()( caabaa ∗∗=∗∗∴ −− 11
caabaa ∗∗=∗∗ −− )()( 11ie
identity. theis whereie ecebe ∗=∗
cb =⇒
Similarly by considering11 −− ∗∗=∗∗ aa)(caa)(b
cb =get we
2. In a group G, the equations ., Gbabaybxa ∈∀=∗=∗ solutions, unique have and
Proof :- bxa =∗
Operating on both sides by a–1
baxaa ∗=∗∗ −− 11 )(
baxaa ∗=∗∗ −− 11ie )(
baxe ∗=∗ −1ie
bax ∗=∴ −1
To prove that the solution is unique, let x1 & x2 be two solutions of .bxa =∗bxabxa =∗=∗ 21ie &
21 xaxa ∗=∗⇒
Operating on both sides by a–1
)()( 21
11get We xaaxaa ∗∗=∗∗ −−
21
11ie xaaxaa ∗∗=∗∗ −− )()(
21ie xexe ∗=∗
21 xx =⇒∴ solution is unique.
3. In a group the identity element and inverse of an element are unique.
Proof :- To prove identity is unique. If possible let e1 & e2 are two identities then
aeaeaGa ∗==∗∈∀ 11, (1)
aeaea ∗==∗ 22& (2)
From LHS of (1), aeaea ∗==∗ 21 (using (2))
221ie eaaeea ∗=∗=∗ (using LHS of (2))
50 KSOU Algebraic Structures
by left cancellation, .21 ee = Thus the identity is unique.
To prove that inverse of an element is unique.
Let ,Ga∈ if possible let b & c are inverses of acaba == −− 11 and ie
ecaeba =∗=∗ & Now,
where e is the identity of the group
caba ∗=∗∴
by left cancellation law b = c. Thus inverse of an element is unique.
4. In a group ( ) GaaaG ∈∀=−− 11,
Proof :- As a–1 is the inverse of a
aaeaa ∗==∗ −− 11 have We
it can be easily seen from above relation that inverse of a–1 is a ie ( ) aa =−− 11 .
Note :- If b & c are elements of G, such that bcecb ∗==∗ then each is the inverse of the other.
5. In a group ),( ∗G
111)(,, −−− ∗=∗∈∀ abbaGba
Proof :- )()(Consider 11 −− ∗∗∗ abba
eaaaeaabba =∗=∗∗=∗∗∗= −−−− 1111)(
111 −−− ∗=∗∴ abba )(
6. A group of order 3 is abelian.
Proof :- Order of group means the number of elements in a group. If a group G has n elements. The order of G is n, whichis denoted as O(G) = n.
If n is finite it is called finite group and n is Infinite then it is called Infinite group.
},,{ baeG =Let be a group with a binary operation ∗.
e is the identity by definition of identity it commutes with every element aeaea ∗==∗Q
So we have to prove that abba ∗=∗
aabaaaba ∗=∗∗⇒=∗ −− 11 )(Let ebe =∗⇒ eb =⇒
which is not possible.
bba =∗Let 11 −− ∗=∗∗⇒ bbbba )( eea =∗⇒ ea =⇒
which is not possible.
ba∗ cannote be equal to a or b eba =∗∴
Similarly we can prove that
eab =∗abba ∗=∗∴ abelian. is ),( ∗∴ G
MCA 11 - Mathematics SVT 51
Subgroups
A non-empty subset H of a group G is said to form a subgroup with respect to the same binary operation ∗ if ),( ∗H is agroup.
Eg. (1) (z, +) is a subgroup of (Q, +)
(2) },,{ 420=H is a subgroup of },,,,,{ 543210=G with B.O. 6ie6 ⊕+ mod
(3) },{ 11−=H is a subgroup of },,,{ iiG −−= 11 with respect to the B.O. multiplication.
Theorem
A non-empty subset H of a group G is a subgroup of G if and only if .,, HabHba ∈∈∀ −1
Proof :- case (i) Let H be a subgroup of G then H is a group
law) closure & axion) inverse 11 ((,, GabGbGba ∈∈∈∀∴ −−
∴ condition is satisfied.
case (ii) Let H be a non-empty subset of G with the property .,, HabHba ∈∈∀ −1
We have to prove that H is a subgroup.
HaHeaHaeHeaaHaaHa ∈∈⇒∈∈=⇒∈∈ −−−− 1111 ieLet ,&,,
Since all elements of H are elements of G, associative property is satisfied.
( ) HabbaHaHbHb ∈∈∈∈−−− iefor Let 111 ,&, ie closure property is satisfied.
Hence H is a group and hence a subgroup.
Permutation group
}..........,,,{ naaaaS 321Let =
Then a one-one and onto mapping or function from S onto itself is called a Permutation.
Permutation is denoted as
)(..........)()()(
..........
n
n
afafafaf
aaaa
321
321
There will be n ! ie n∠ permutations the set of permutations is denoted by Sn.
., nSgf ∈Let There is a composite mapping for f & g denoted as ,gf o this can be taken as binary operation. Then the
set Sn with binary operation 'O' (ie composite mapping) will form a group. For convenience gf o is denoted as gf.
41432
4321
2413
4321Let Sgf ∈
=
= & the B.O. composite function is given by
=
1432
4321
2413
4321ogf o
=
????
4321
to fill up the second row, following is the procedure.
14433221in ==== )(,)(,)(,)(: ggggg
24431231in ==== )(,)(,)(,)(:& fffff
1211Now === )()]([)( fgfgf o
4322 === )()]([)( fgfgf o
2433 === )()]([)( fgfgf o
3144 === )()]([)( fgfgf o
52 KSOU Algebraic Structures
=∴
3241
4321gf o
for convenience gf o is written as gf
=
1432
4321
2413
4321ogfie o
=
3241
4321
=
2413
4321
1432
4321gf&
=
3241
4321
the composite function is also called product function.
Eg. (1) },{ 21Let =S
=
12
21
21
21then 2 ,S
Let B.O. be product permutation
=
12
21
12
21
21
21 then
=
21
21
21
21
21
21
=
21
21
12
21
12
21
closure law is satisfied, associative law can be easily verified. inverse
=
21
21e
=
−
21
21
21
211
=
−
12
21
12
211
∴ S2 forms a group.
Eg. (2) },,{ 321Let =S
=
123
321
213
321
132
321
123
321
231
321
321
3213S
Let us denote the elements as 654321 ffffff &,,,, respectively.
{ }6543213ie ffffffS ,,,,,=
The following is the multiplication table.
f1
f2
f3
f4
f5
f6
f1
f1
f2
f3
f4
f5
f6
f2
f2
f1
f5
f6
f3
f4
f3
f3
f4
f1
f2
f6
f5
f4
f4
f3
f6
f5
f1
f2
f5
f5
f6
f2
f1
f4
f3
f6
f6
f5
f4
f3
f2
f1
MCA 11 - Mathematics SVT 53
It can be seen from the table that closure law is satisfied.
For associative law
356543Consider ffffff ==)(
313543 ffffff ==)(
)()( 543543 ffffff =∴ hence associative law is satisfied.
1identity fe=
21
211
1 ffff == −− , 51
431
3 ffff == −− , 61
641
5 ffff == −− &
inverse of all elements exists.
∴ S3 forms a group, but it is not an abelian group
234643 but ffffff ==Q .3443 ffff ≠∴
Examples
(1) Show that the set R - {o} with B.O. × forms a group.
Solution : For any elements a, b ∈ R - {o}. a ∗ b ∈ R - {o}
2, 3 ∈ R - {o}, 2 × 3 = 6 ∈ R - {o}
∴ closure law is satisfied.
For any three elements a, b, c ∈ R - {o}
(a × b)× c = a × (b × c)
(–3 × 4) × 5 = –12 × 5 = –60.
–3 × (4 × 5) = –3 × 20 = –60.
∴ associative law is satisfied.
Identity element is 1,
ie. ∀ a ∈ R - {o}, a × 1 = a = 1 × a.
Let a ∈ R - {o} then there exists
aaa
aRa
×==×∈ 11
1 such that o-
1}{
∴ inverse exists for all elements R - {o}.
∴ (R - {o}, ×) forms an abelian group.
(2) Show that ),( 55 ⊕Z forms an abelian group.
Solution : Let us construct table for ),( 55 ⊕Z
5⊕
0
1
2
3
4
0
0
1
2
3
4
1
1
2
3
4
0
2
2
3
4
0
1
3
3
4
0
1
2
4
4
0
1
2
3
54 KSOU Algebraic Structures
From the above table it can be easily seen that closure law is satisfied.
422432 555 =⊕=⊕⊕ )(
440432 555 =⊕=⊕⊕ )(
432432ie 5555 ⊕⊕=⊕⊕ )()(
∴ associative law can be versified.
identity element is 0.
inverse of 0 is 0
inverse of 1 is 4
inverse of 2 is 3
inverse of 3 is 2
inverse of 4 is 1
∴ ),( 55 ⊕Z form a group, further it can be seen from the table that for any two element a, b ∈ Z5
abba 55 ⊕=⊕
∴ ),( 55 ⊕Z is an abelian group.
(3) Show that G = {1, 2, 3, 4} with B.O. multiplication mod 5 ie 5⊗ is an abelian group.
Solution : The following in the relevant table for elements
From the above table it can be seen that closure law is satisfied.
441432 555 =⊗=⊗⊗ )(
422432 555 =⊗=⊗⊗ )(
∴ associate law is satisfied.
identity is 1.
inverse of 1 is 1
inverse of 2 is 3
inverse of 3 is 2
inverse of 4 is 4
∴ ),( 5⊗G forms a group and it can be seen from the table that it forms an abelian group.
4. If R is the set of real numbers and ∗ is a binary operation defined on .,, RyxxyyxR ∈∀+=∗ 1as
Show that ∗ is commutative but not associative.
Rbaabba ∈∀∗=∗ ,,property eCommutativ
xyyx +=∗ 1
5⊗
1
2
3
4
1
1
2
3
4
2
2
4
1
3
3
3
1
4
2
4
4
3
2
1
MCA 11 - Mathematics SVT 55
xyyxyxxy ∗=∗∴+=∗ 1
)()( zyxzyx ∗∗=∗∗property eAssociativ
xyzzzxypzzPzxy ++=++=+=∗=∗+= 11111LHS )()(
xyzxyzxxQQxyzx ++=++=+=∗=+∗= 11111RHS )()(
eassociativ anot isRHSLHS ∗∴≠∴
5. Show that set of integers with group anot iswhere Ibababa ∈−=∗ ,
satisfied is axiom closure∴∈−=∗ Ibaba
.,,),()( IbaIbabacbacba ∈∀∈−=∗∗∗≠∗∗
cbacba −−=∗−= )(LHS
satisfiednot is axiom eassociativRHS ∴+−=−∗= cbacba )(
),( ∗∴ I is not a group.
When one of the axiom is not satisfied, it is not a group. Hence, we need not have to check the rest of the axioms.
6. In a group .),,(2
abbaG =∗∗ Find the identity element, inverse of 4 and solve 54 =∗ x
aeaeae ∗==∗: find To
2. iselement identity i.e.22
=∴==∗ eaae
ea
aaeaa ∗==∗ −− 11
aa
aaaa
42
22 1
11 =∴=⇒=∗ −
−−
1. is 4 of inverse14
44 1 ∴==− .
2
5
4
105
2
454 ==∴=⇒=∗ x
xx
7. Prove that { }real is θθθ sincos iG += is an abelian group under multiplication
{ }real is θθθ sincos iG +=
Let x, y, z be any three elements of G.
.sincos,sincos,sincos γγββαα iziyix +=+=+= Take
numbers. real are , , where γβα
i) real is)()sin()cos()sin)(cossin(cos βαβαβαββαα +∈+++=++= QGiiixy
∴ Closure axiom is satisfied
ii) )sin)](cossin)(cossin[(cos) γγββαα iiiz(xy +++=
)sin)}(cossin(){cos( γγβαβα ii ++++=
})sin{(})cos{( γβαγβα +++++= i
)}(sin{)}(cos{ γβαγβα +++++= i
= x (yz) Q multiplication is associative on R ∴ Associative axiom is satisfied.
iii) elementidentity theis 001 Gi ∈+= sincos
56 KSOU Algebraic Structures
iv) .sincossincos)sin)(cossincos θθθθθθθθ iiii +−⇒=−+ of inverse tivemultiplica theis 1(
v) )sin)(cossin(cos ββαθ iixy ++=
)sin()cos( βαβα +++= i
Ri on ecommutativ isQ)sin()cos( αβαβ +++= yx=∴ Commutative law is satisfied.
So all the axioms are satisfied. Hence, G is an abelian group under multiplication.
8. Show that the cube roots of unity form an abelian group under multiplication
We know that the cube roots of unity are },,{., 22 1Let and1 ωωωω =G
1 here 3 =ω
.& ωωωω =⋅= 34
i) All the entries in the table are the same as the elements of the set. This means the closure law is satisfied.
ii) 1111Consider 2 =⋅=⋅⋅ )( ωω
11 22 =⋅=⋅⋅ ωωωω)(
satisfied. is law eAssociativ11 22 .)()( ωωωω ⋅⋅=⋅⋅∴
iii) The row heading 1 is the same as the topmost row.
∴ 1 is the identity element.
iv) Inverse of 1 is 1, inverse of ω is ω2 and inverse of ω2 is ω.
Every element has a inverse.
v) The table is symmetrical about the principal diagonal
∴ commutative law holds good.
So G is an abelian group under multiplication.
9.
−
−
−
−
10
01 and
10
01
10
01
10
01 matricesfour that theShow ,, form an abelian group under matrix
multiplication.
},,,{;,,, CBAGCBAI 110
01
10
01
10
01
10
01 Take ==
−
−=
−
=
−=
CCIICBBIIBAAIIA ====== ,,
CAB =
−
−=
−+++−
=
−
−=
10
01
1000
0001
10
01
10
01
etc. shown that becan it Similarly, IAAACBBCBCAACCBA =⋅===== ,,,
.
1
ω
ω2
1
1
ω
ω2
ω
ω
ω2
1
ω2
ω2
1
ω
MCA 11 - Mathematics SVT 57
The composition table is
i) The entries in the table are the same as the elements of the set G. ∴ Closure law is satisfied.
ii) ICCCABIAABCA ==== )()(;)()( ∴ Associative law is satisfied.
iii) I is the identity element.
iv) Inverses of BAI ,, are respectively GCBAI ∴.,,, is a group under matrix multiplication
),( ⋅∴ G is a group.
v) Since the entries on either side of the leading diagonal are symmetric, ),( ⋅G is an abelian group.
10. If every element of a group G has its own inverse, show that G is abelian
Gbaaa ∈∀=− ,,1Given (1)
111 that,know we −−− =∈∀ ababGba )(,, (2)
.)(, ababbbaa === −−− 111 and (1), from Using these in (2), Gbabaab ∈∀= ,,
∴ the commutative law is satisfied, so G is abelian.
11. In a group .,,)(),( GbabaabG ∈∀=⋅ 222if Prove that ),( ⋅G is abelian and conversely.
))(())(()( bbaaababab ==2
L.C.L. using)]([)]([ bbaaabba =⇒
RCL using babbbabbaabb )()()()( =⇒=⇒ abba =⇒ ∴ it is abelian
Conversely
If ab = ba pre operating by a we get, aabbaabaaaba )()()()( =⋅⇒=
baabbaab ])[(])[( =⋅⋅getweby operatingpost
222 )())(())(( abbaababbbaa =⇒=⋅⋅⇒
12. Given Q0, the set of non zero rational numbers is a multiplicative group and { },ZnH n ∈= 2 show that H is a subgroup
of Q0 under multiplication.
{ } { }.....22,2,2...2 1,012 −−=∈= ZnH n
i) satisfied is law closure22222 ∴∈=⋅∈ + HH nmnmnm ,,
ii) zrnmrnmrnm ∈∀⋅⋅=⋅ ,,),()( 222222 ie )()( rnmrnm ++ = 2222
satisfied. is law eAssociativ22i.e.22 ∴== ++++++++ rnmrnmrnmrnm )()(
iii) 20 is the identity element
iv) .; mmmmmm −−− ∴=⋅∀ 2is2 of Inverse222such that 2exist there2 0
∴ H is a group under multiplication and 0QH ⊂ ie H is a subgroup of Q0 under multiplication.
.
I
A
B
C
I
I
A
B
C
A
A
I
C
B
B
B
C
I
A
C
C
B
A
I
58 KSOU Algebraic Structures
Exercise
1. If N = {1, 2, 3, ....}, which of the following are binary operation of N.
b
abababababa =∗−=∗+=∗ )()()( 543221
2. Which of the following operations on the given set are binary
(1) on I, the set of integers, baba 43 −=∗
(2) 22on babaR −=*,
(3) abbaR =∗,on
3. If ∗ is given by ,baba =∗ show that ∗ is not a binary operation of Z.
4. Why the set of rationals does not form a group w.r.t. multiplication ?
5. If x, y, z are any three elements of a group G, find (xyz)–1.
6. In a group ( ) .,,,111find
−−−∈∀ baGbaG
7. If the binary operation ∗ on the set Z is defined by ,5++=∗ baba find the identity element.
8. In the group of non zero integers mod 5. Find the multiplicative inverse of 4.
9. ., gfSgf ofindin nspermutatio are132
321and
321
321If 3
=
=
10. If S = {1, 2, 3, 4, 5, 6} w.r.t. multiplication (mod 7), solve the equation 3x = 5 in S.
11. Show that S = {1, 2, 3} under multiplication (mod 4) is not a group.
12. .gfgf find4132
4321and
2143
4321If
=
=
13. The binary operation ∗ is defined by ,7
abba =∗ on set of rational numbers, show that ∗ is associative.
14. If ∗ is defined by ,2
abba =∗ on the set of real numbers, show that ∗ is both commutative and associative.
15. If ∗ is defined by ,22 baba +=∗ show that ∗ is associative.
16. On the set of real numbers, .,,, RbababaR ∈∀−+=∗ 1 Show that ∗ is associative.
17. On the set of real numbers, R, ∗ is defined by ,abbaba +−=∗ 32 examine whether ∗ is commutative and associative.
18. In the set of rationals except 1, binary operation ∗ is defined by .abbaba −+=∗ Find the identity and inverse of 2.
19. On the set of positive rational numbers .,,, ++ ∈∀=∗ Qbaab
baQ4
Find the identity element and the inverse of 8.
20. In a group of integers, an operation ∗ is defined by .1−+=∗ baba Find the identity element.
MCA 11 - Mathematics SVT 59
Vectors and Scalars
Vector : A physical quantity which has both direction and magnitude is called a 'Vector'.
Eg. Velocity, acceleration, force, weight etc.
Scalar : A physical quantity which has only magnitude and no direction is called a 'Scalar'.
Eg. speed, volume, mass, density, temperature etc.
Vectors are represented by directed line segments. Let AB be the line segment. Vector from A to B is denoted by AB and
vector from B to A is denoted by ABBA. can also be represented by .a The length of AB is magnitude of the vector denoted
as AB or a or simply a. For ,AB A is the initial point and B the terminal point.
Scalar multiplication of a vector
Let λ be a scalar and a a vector then aλ represents a vector whose magnitude is λ times the magnitude of a and the direction
is same as that of a if λ is positive but opposite to that of a if λ is negative. If 0=λ it represents a null vector denoted by
''0 ie a null vector is a vector of magnitude zero but its direction is arbitrary.
A vector whose magnitude is 1 is called a unit vector and a unit vector in the direction of a is written as
cap) as (readoror aaa
a
a
a�
Like and unlike vectors
Vectors having same direction are called 'like vectors' and those having the opposite direction are called 'unlike vectors'.
Co-initial vectors : Vectors having the same initial point are called co-initial vectors.
Coplanar vectors : Vectors in the same plane are called 'coplanar vectors'.
Parallel vectors : Vectors having same direction but different initialpoints are called 'parallel vectors'.
Triangle Law for addition of vectors
Let AB & BC represents two vectors then AC represents
baBCAB ++ ie
Parallelogram Law
bOBaOA == &Let
Complete the parallelogram OACB.
law eby trianglthen OCACOA =+
vectors)(parallelbut OBAC =
baOBOAOC +=+=∴
OAOBAB −=-: Note
OBOABA −=&
A
B
C
AO
b
a
B
C
BA
b
a
60 KSOU Algebraic Structures
Properties
(i) Vector addition is commutative ie .abba +=+
(ii) Vector addition is associative ie ( ) ( ) .cbacba ++=++
(iii) Set of vectors V, with binary operation vector addition will form a 'Group' . The identity being 0 (null vector) and
inverse of .aa − is
Position vectors
(i) Let P be a point in a plane where O is the origin and OX & OY are co-
ordinate axes. OP is called position vectors of P.
yQPxOQOXPQ r ==⊥ &, thento Draw
ji �&�Let represents unit vectors in the direction of OX & OY.
jyQPixOQ �,� ==Then
jyixQPOQOP �� +=+=∴
22 yxOP +=
Note :- A plane vector is an ordered pair of real numbers and the distance between O & P is the magnitude of .OP
(ii) Let P be a point in three dimensional space where OX, OY & OZ are co-ordinate axes. Let (x, y, z) be the co-ordinates of
P. Draw rPQ⊥ to the plane XOY & QA & QB parallel to OY & OX respectives to meet OX at A & OY at B.
kji �&�,�Let be unit vectors in the direction of OX, OY & OZ
kzQPjyOBixOA �,�,� ===Then
law) ramparallelog(by OBOAOQ +=
jyix �� +=
law) le(by triangQPOQOP +=
kzjyix ��� ++=
222 zyxOP ++=
This position vector OP of the point P is usually
denoted by rOPr =ie
222by given is ofdirection in ther unit vecto
zyx
kzjyixOP
++
++ ���
O Q
P (x, y)
X
Y
xi�
y
j�
Z
Q
P (x, y, z)
X
Yi�
j�O B
k�
A
MCA 11 - Mathematics SVT 61
),,(&),,(Let 222111 zyxPzyxP be any two points in 3-space
OPQPQ triangle thefrom find to ,
OQPQOP =+
kzjyixkzjyixOPOQPQ ������111222 −−−++=−=∴
kzzjyyixx �)(�)(�)( 121212 −+−+−=
and unit vector in the direction of
212
212
212
121212is)()()(
�)(�)(�)(
zzyyxx
kzzjyyixxPQ
−+−+−
−+−+−
Scalar product of two vectors
kbjbibbkajaiaa ������321321 and If ++=++= are two non-zero vectors, then 332211 bababa ++ is defined as 'scalar product'
of two vectors baba ⋅ as denoted ,& also known as 'dot product'.
Vector product of two vectors
kbjbibbkajaiaa ���,���321321 If ++=++= are two non-zero vectors, then a vector
kbabajbabaibaba �)(�)(�)( 122131132332 −+−+−
product. cross asknown also as denoted vector twoof product'vector ' as defined isie
321
321 baba
bbb
aaa
kji
×,&
���
ibaba
bbb
aaa
kji
ba �)(
���
∑ −==× 2332
321
321 ie
Geometrical Meaning of baba ×⋅ and
θ=== BOAbOBaOA �& let andLet
θcosabba =⋅then
bbaa == & where
ababba =°=⋅°= 00when cos,θ
09090when =°=⋅°= cos, abbaθhence two vectors are said to be 'orthogonal' if .0=⋅ba
Let ON represent a line which is r⊥ to both OA & OB. ie ON is r⊥ to the plane OAB. Let n� represent a unit vector inthe direction of ON.
0then 0 if is and is Then =°=×−× θθθθ sin,�sin�sin abnabbanab
abba ×−×∴ or is a null vector, hence two vectors are said to be parallel or coincident if vector)(null0=× ba
O Y
Z
X
),( , 111 zyxP
),( , 222 zyxQ
b
aO A
B
θ
62 KSOU Algebraic Structures
Note :-
rotation. clockwise-antirepresent nabba �sinθ=×
rotation. clockwiserepresent nabab �sinθ−=×
also for unit vectors kj,i �&��
ikjjikkji ���,���,��� =×=×=× .���,���,���& ijkjkikij −=×−=×−=×
Projection of b upon a
..�&, abODOABDOBAbOBaOA r upon of projection theis then to DrawLet ⊥=== θ
ab
baabba
⋅==⋅ θθ cos,cos Since
b
OD
OB
ODOBDle ==∆ θcos thefrom
ab
babbOD
⋅⋅==∴ θcos
baba
a ⋅=⋅= �
further area of parallelogram whose adjacent sides are OA & OB is given by
.sinsinsin θθθ abbaOB
BDabBDOA =×==⋅ butQ
∴ Area of parallelogram whose co-terminus edges are baba ×by given is&
.baOABle ×=∆2
1 of area thereforeand
Also area of the parallelogram whose diagonals are 21212
1 is dddd ×&
A
B
O
n�−
N
θa
b
N
O A
B
θ
n�
a
b
b
aO D
B
θA
MCA 11 - Mathematics SVT 63
Scalar Triple Product
kcjcicckbjbibbkajaiaa ���&���,���321321321Let ++=++=++=
cbacba ⋅××⋅ )()( or Then is called 'Scalar Triple Product'
both on computation gives 123213312132131321 cbacbacbacbacbacba −+−+−
321
321
321
iswhich
ccc
bbb
aaa
cbacba ⋅××⋅ )()( or Thus
which is 'Scalar Triple Product' which is usually denoted as [ cba ] also called 'Box-Product'.
Geometrical Meaning of [ cba ]Consider a parallelopiped whose co-terminus edges are
cbaOCOBOA ,,,, ie
Area of parallelogram cbOBDC × is
to is which upon ie of projection thebeLet rONaOAOP ⊥the plane OBDC.
( )cb
cbaa
cb
cbOP
×
×⋅=⋅×
×=∴
Volume of parallelopiped = area of parallelogram × OP
( )cb
cbacb
×
×⋅×=
( )=×⋅= cba [ cba ]∴ Volume of parallelopiped whose coterminus edges are given by cba ,, is
[ cba ]321
321
321
ccc
bbb
aaa
=
If [ cba ] = 0 coplanar. are vectorsthen the cba &,
Vector Triple Product
If cba ,, are three non-zero vectors then
( ) ( ) ( )cbabcacba ⋅−⋅=××
also ( ) ( ) ( )acbbcacba ⋅−⋅=××
b
a
O
D
B
A
C
P
N
c
64 KSOU Algebraic Structures
Examples
1. .),(),,( baba ⋅−== find 3252 If
Solution : 11154 =+−=⋅ba
2. ( ) ( )βαβαβα 22 find2332 If +⋅+−−=−+= ,���,��� kjikji
Solution : ( ) ( ) 91560357783522 =++=−⋅−+=+⋅+ )��()���( kikjiβαβα
3. other.each lar toperpendicu are 4223 vectors that theProve kjibkjia ���,��� −+=−−=
Solution : other.each lar toperpendicu are0426 baba &⇒=−−=⋅
4. .orthogonal are 82and3such that Find kjikjmim ������ −−++
20860823:Solution −=−⇒=−−⇒=−−⋅++ mmkjikjmi )���()���(
5. other.each toorthogonal are and that Show If bababa .−=+
baba −=+Given :Solution
( ) ( ) ( ) ( )babababa −⋅−=+⋅+ sides, both the Squaring
( ) 02 i.e. =⋅⋅+⋅−⋅−⋅=⋅+⋅+⋅+⋅ babbabbaaabbabbaaa
.baba lar toperpendicu is 0 ∴=⋅∴
6. .������ kjibkjia ++−=++= 3on 53 of projection theFind
.11
5
119
533on of Projection:Solution =
++++−=⋅=
b
baba
7. .,��������� bcakjickjibkjia on of projection thefind243and 22and 2 If ++−=+−=++=
( ) ( ) ( ) ( )3
17
3
665
441
22335on of Projection:Solution =++=
+++−⋅+−=⋅+=+ kjikji
b
bcabca
������
8. kjibkjia ������ −+=+−= 2and 334 ctorsbetween ve angle theof cosine theFind
634
2
1149916
338:Solution =
++++−−=⋅=
ba
baθcos
9. kjibkjia ���,��� +−=+−= 8 49 vectorsofproduct cross theFind
kjikji
kji
ba ���)(�)(�)(�
���
−+=+−+−−+−=−−=× 2338932941
118
419:Solution
10. bakjibkjia ×−+=++= findthen 22 and 22 If ,������
MCA 11 - Mathematics SVT 65
kjikji
kji
ba ˆ3ˆ6ˆ6)41(ˆ)42(ˆ)24(ˆ
212
221
ˆˆˆ
:Solution −+−=−+−−−−−=−
=×
98193636 ==++=×ba
11. kjibkjia ������ 23 and 26 vectorsofpair thelar toperpendicur unit vecto theFind −+=+−=
kjikji
kji
baba ���)(�)(�)(�
���
& 121536631214
213
126is lar toperpendicuVector :Solution ++=++−−−−=−
−=×
378
12153
1442259
12153 kjikji
ba
ban
�������
++=++++=
×
×=
12. ( ) ( ) ( )bababa ×=+×+ 322 that Prove
( ) ( ) ( ) ( ) ( ) ( ) ( ).baabbabbabbaaababa ×=×+×=×+×+×+×=+×+ 3424222:Solution
13. ( ) ( ) ( ) 0 that prove vectorszeronon are If =+×++×++× bacacbcbacba ,,
.0:Solution =×−×−×−×+×+×=×+×+×+×+×+× cbcabacbcababcacabcbcaba
14. kjibkjia ������ 232 and vectorsebetween th angle theof sine theFind +−=++=
,,��)(�)(�)(�
���
22 5555232232
232
111:Solution +=×−=−−+−−+=−
=× bakikji
kji
ba
173
25
494111
55 22
=++++
+=∴ θsin
15. accbbacba ×=×=×=++ that showthen ,0 If
0Given :Solution =++ cba
( ) ( ) [ ] acbaaacabacabaaacbaa ×=×∴=××−=×∴=×+×+×=++× 00 Q
cbbababcbcab ×=××−=××=× oror Similarly
accbbacbac ×=×=×∴×=× Similarly
16. kjibkjia ������ 53 and 23 are sides whose triangle theof area theFind +−=+−=
baA ×=2
1:Solution
kjikji
kji
ba ���)(�)(�)(�
���
714729115310
531
123 −−−=+−+−−+−=−−=×
66 KSOU Algebraic Structures
sq.units2
294
2
4919649area =++=
17. .������,���, kjikjikjiCBA +−−++− 23 and 2ly respective are and points theof ectorsPosition v Find the area of triangle
ABC.
.���)���()���( kjikjikjiOAOBAB 222:Solution −+=+−−−+=−=
.��)���(��� jikjikjiOAOCAC −=+−−+−=−= 223
.���)(�)(�(�
���
kjikji
kji
ACAB 542414120
012
221 −−−=−−+−−=−
−=×
sq.units2
4525164
2
1area =++=
18. .������ kjibkjia 32 and 23 are sidesadjacent whoseramparallelog a of area theFind ++=−+=
kjikji
kji
ba ���)(�)(�)(�
���
4108261926
321
123:Solution +−=−++−+=−=×
sq.units1801610064 =++=A
19. .������ kjidkjid 43 and 23 are diagonals whoseramparallelog a of area theFind 21 +−=++=
212
1:Solution ddA ×=
kjikji
kji
dd ���)(�)(�)(�
���
1010101921264
431
21321 −−=−−+−−+=−
=×
sq.units.35sq.units3002
1
2
1100100100 2121 ==×++=× dddd ;
20. .������,��� kjickjibkjia −+=++=+−= 3 and 3232 vectorsofproduct plescalar tri theFind
( ) .)()()( 35151010613911322
113
321
312
:Solution −=−−−=−+−−+−−=−
−=×⋅ cba
21. [ ]ikkjji ��,��,�� −−− Evaluate
001111
101
110
011
:Solution =+−+=−
−−
)()(
22. Find the volume of parallelopiped whose coterminus edges are
.���,���,��� kjickjibkjia +==+−=+−= 53542
MCA 11 - Mathematics SVT 67
( ) units cubic 10213211210115212541
153
542
111
:Solution =+−=+−+−++−=−−−
=×⋅ )()()(cba
23. coplanar. are 21and 121432 vectors theif Find ),,(),,(),,( −=−=−= λλ cba
[ ] 0
21
121
432
:Solution =−
−−
=λ
cba
02142312(4 i.e., =−−+++− )()() λλ
.;5
885084366 ==⇒=−−++ λλλλ
24. coplanar. are 444 and493110154 points that theShow ),,(),,(),,,(),,,( −−− DCBA
kjiODkjiOCkjOBkjiOA ���,���,��,��� 44449354:Solution ++−=++=−−=++=
kjiOAODADkjiOAOCACkjiOAOBAB ���,���,��� 3834264 +−−=−=++−=−=−−−=−=
Consider
( ) 06612660321224363124
318
341
264
=−+−=+−+−++−=−−
−−−−
=×⋅ )()()(DACAAB
coplanar. are andDCBA ,,∴
Exercise
1. The position vectors of the points of in terms vectorsExpress 32 and ly respective are and ABACBCbabaCBA, ,,., −
.ba and
2. ly.respective 2 and 32433542 are and are ectorsPosition v cacbcbacaDCBA ++−+++ ,,,,
CDABCDAB2
3 and that Show =||
3. bcaacbjicjibjia 232ii)22(i) find3243 If +−+−+=−=−= (,��,��,��
4. .������,��� kjickjibkjiacba 622 and23324 where ofdirection in ther unit vecto theFind ++−=−−=++=++
5. .,���� bajibjia ⋅+−=+= find2 and 32 If
6. Show that the vectors (–1, 2, 3) and (2, –5, 4) are orthogonal.
7. .orthogonal bemay 2 and 3such that of values theFind kjikji ������ +++− λλλλ
8. .orthogonal are and vectors that theshowr unit vecto are and If bababa −+
9. .),,,(),,( baba ×=−= find112 and 311 If
10. .����� jibkjiaba 32and 32 whereon of projection theFind +−=−+=
11. ..������ abkjibkjia on of projection theFind 2and 32 If +−=−+=
68 KSOU Algebraic Structures
12. .),(),( mbamba find vector,null a is and8 and 32 If ×==
13. .),,,(),,(),,,( cbacba ++−≡−=−= ofdirection in ther unit vecto thefind712 and412312 If
14. r.unit vecto a is that Show kji �cos�sin�sincos φθφθ ++
15. Show that points A (3, –2, 4), B (1, 1, 1) and C (–1, 4, –2) are collinear.
16. Show that points A (1, 1, 1), B (7, 2, 3), C (2, –1, 1) form a triangle.
17. kjikjikjiCBA ������,���, 443 and 532 are ectorsposition v whose and points that Show −−−+++ respectively form a
right-angled triangle.
18. .������ kjibkjia 232 and vectorsebetween th angle theof cosine theFind −+=++=
19. .������ kjikji 23 and 32 vectorsebetween th angle theof sine theFind −++−
20. .������ kjikji 422 and 23 vectors thelar toperpendicur unit vecto theFind +−++
21. Find the volume of the parallelopiped whose co-terminal edges are represented by and 2432 kjibkjia ���;��� −+=+−=
.��� kjic 23 ++=
22. coplanar. are543 and 322 ors that vectShow kjikjikji ������,��� −−−++−
23. . find coplanar. are293 and 10252 vectors theIf xkjikjxikji ������,��� −+−+++
24. Show that points A (2, 3, –1), B (1, –2, 3), C (3, 4, –2) and D (1, –6, 6) are coplanar.
25. .���,���,��� kjickjibkjia ++=++=+−= 2 22 of,product plescalar tri theFind
26. ( ) ( ) ( )bababa ×=+×+ 5432 that Prove
27. .���,��� kjibkjia ++=−−= 223by drepresente sides 2 whose triangle theof area theFind
28. .����,��� kjkjkji 3 and 22 are vertices whose triangle theof area theFind +++−
29. .������ kjibkjia 43and23 are diagonals whoseramparallelog of area theFind +−=−+=
30. ( ) ( ) .).,,(&),,(),,,( cbacbacba ××××−==−= and Find423321211 If
BCA 21 / IMCA 21 / Mathematics SVT 65
DIFFERENTIAL CALCULUS
Limits of functions
1
1Consider
2
−−==
x
xxfy )( the function is defined for all values of x except for x = 1.
0
0
11
111for =
−−==∴ )(, xfx which is indeterminate.
Let us consider the values of f (x) as x approaches 1
x1
12
−−=
x
xxf )(
.9 1.9
.99 1.99
.999 1.999
1.01 2.01
1.001 2.001
1.0001 2.0001
It can be seen from the above values that as x approaches 1, 1
12
−−
x
x approaches 2.
.lim 21
1or
1
1 as written becan which 1" approaches as
1
1 ofLimit " called is 2 valueThis
2
1
2
1
2
=−−
−−
−−
→→ x
xLt
x
xx
x
xxx
In general the limit of a function f (x) as x approaches a is denoted as l and which is written as
lxfLtlxfaxax
==→→
)()(lim or
Properties
(1) [ ] )(lim)(lim)()(lim xgxfxgxfaxaxax →→→
±=±
(2) [ ] )(lim)(lim)(lim)(lim)()(lim xfkxkfxgxfxgxfaxaxaxaxax →→→→→
⋅== particularin where k is a contant
(3) .)(lim)(lim
)(lim
)(
)(lim 0 provided ≠=
→→
→→
xgxg
xf
xg
xfax
ax
ax
ax
Standard Limits
(1) 1−
→=
−− n
nn
axna
ax
axlim
70 KSOU Differential Calculus
(2) 1 also radians)in 100
==→→ θ
θθθ
θθθ
tanlim(
sinlim
(3) exeen
x
x
n
n=+<<=
+
→∞→
1
01or 32
11 )(limlim
(4) .lim)(loglim 11
particularin 01
00=−>=−
→→ x
eaa
x
a x
xe
x
x
Examples
(1)4
3
400
300
45
342
2
0=
+−++=
+−++
→ xx
xxxlim
(2) )lim 22
2
by Dr&Nr (dividing123
432x
xx
xxx ++
+−∞→
3
2
003
00212
3
432
2
2=
+++−=
++
+−=
∞→
xx
xxxlim
(3) 108343
3
3
81 344
3
4
3==
−−=
−−
→→)(limlim
x
x
x
xxx
(4) 24
6
5
77
55
77
55
77
5
7
5
7a
a
a
ax
ax
ax
ax
ax
axax
ax
ax
axaxaxax
=−−
−=
−−−−−−−−
=
++++
=++
5−→−→−→ )(
)(
)(
)(
)(
)(
limlimlim
(5) 771777
00=×=×=
→→ x
x
x
xxx
sinlim
sinlim
(6) 2221
2
2
020==−
→→ θθ
θθ
θθ
sinlim
coslim
(7) )sin
tanlim x
xx
xxx
by Dr&Nr (dividing3
30 −
−→
113
13
3
13
0=
−−=
−
−=
→
x
xx
x
x sin
tan
lim
(8) ab
ab
ax
x
xb
xeaxax =
+=+
→→
1
0011 )(lim)(lim
(9) 3
333
13
1 enn
n
n
n
n=
+=
+
∞→∞→limlim
(10) 2
2
21
0
1
02121 −
−−
→→=
−=− exx x
x
x
x)(lim)(lim
(11)x
xbxa
x
ba xx
x
xx
x
)()(limlim
−−−=−→→ 00 b
aba
x
xb
x
xaeee
x
x
x
xloglogloglimlim =−=−−−=
→→ 00
MCA 11 - Mathematics SVT 71
(12) 21
212
1200
ee
x
x
x
x
x
xx
xlog
log
sinlim
sinlim ==
−
=−→→
Continuity of a function
)()(lim)( afxfaxxfax
==→
ifat continuous be tosaid isfunction A
equal. are they and exists exists, if continuous be tosaid isfunction A 0
)()(lim)( afxfxfx→
∴
If these donot happen then the function is said to be not continuous or discontinuous.
A function f (x) is said to be continuous in an interval if it is continuous at all points in the interval.
Examples
(1) exists.donot 0
00but exists 2
1
11at continuousnot is
1
1 2
1
2
==−−=
−−=
→)(lim)( f
x
xx
x
xxf
xQ
where as it is continuous at all other values of x.
(2)
<+≥+
=4for 73
4for 34function of continuity theDiscuss
x
xxxf )( at x = 4.
Solution : While finding the limit of a function f (x) as x approaches a, if we consider the limit of the function as x approachesa from left hand side, the limit is called 'Left Hand Limit' (LHL) and if x approaches a from right hand side thelimit is called 'Right Hand Limit' (RHL) and the limit of the function is said to exists if both LHL & RHL existsand are equal, for convenience LHL & RHL are denoted as )(lim)(lim xfxf
axax +− →→ and and further
)(lim)(lim&)(lim)(lim hafxfhafxfhaxhax
+==−==→→→→ +− 00
RHLLHL
For the given problem
197434 is4at LHL004
=+−=−==→→→ −
)(lim)(lim)(lim hhfxfxhhx
193444 is4at RHL004
=++=+==→→→ +
)(lim)(lim)(lim hhfxfxhhx
194 and =)(f
4at continuous isfunction The =∴ x
(3)
=
≠=
0for 2
0for of continuity theExamine
x
xx
xxf
sin
)( at x = 0.
20but 1 :Solution 0
==→
)(,sin
lim fx
xx
.0at ousdiscontinu isfunction The =∴ x
(4)
≥+<<−
≤<−
=2for 44
21for 24
10for 45
function theof continuity theExamine 2
xx
xxx
xx
xf )(
.2 and1at == xx
72 KSOU Differential Calculus
,1at =x 14151LHL001
=−−=−==→→→ −
)(lim)(lim)(lim hhfxfhhx
22412141RHL 2
00=−=+−+=+=
→→)()(lim)(lim hhhf
hh
1at RHLLHL =≠ x
.1at ousdiscontinu isfunction The =∴ x
,2at =x 1241622242LHL 2
00=−=−−−=−=
→→)()(lim)(lim hhhf
hh
124242RHL00
=++=+=→→
)(lim)(lim hhfhh
124242 and =+×=)(f
)()(lim 22
fxfx
=∴→
.2at continuous isfunction The =∴ x
Differentiability of a function
is derivative theand exists if point aat abledifferenti be tosaid is function A 0 h
afhafaxf
h
)()(lim)(
−+→
).(' af as denoted
).(')()(
lim)( xfx
xfxxfxxf
x as denoted is derivative theand exists if at abledifferenti be tosaid is function A
0 δδ
δ
−+→
[δx is called the increment in x which is very very small].
)('),( xfdx
dyxfy or by denoted is derivative the If =
Note :- A function which is differentiable is always continuous but the converse is not always true.
≥<−
===0for
0for eg.
xx
xxxxfy )( is continuous for all x but not differentiable at x = 0.
To find the derivates of xn, loge x, ax, sin x, cos x and a constant C with respect to x.
(1) 1
0Let −
→=
−+−+== n
nn
x
n nxxxx
xxx
dx
dyxy
δδ
δ
)(lim,
( ) ( ) 5467 4ii7 (i) Eg. −− −== xxdx
dxx
dx
d)(, ( ) 4
54
94
9
4
9
4
9 (iii) 1 xxx
dx
d == −
( ) 38
35
35
3
5
3
5(iv) 1 −−−− −=−= xxx
dx
d ( )x
xdx
d
2
1 v) =(
(2) xy elog=Let x
xxx
dx
dy ee
x δδ
δ
log)(loglim
−+=→0
then
.logloglimloglimx
exx
x
xx
xx
x
x
x ex
x
xx
111
1100
==
+⋅=
+⋅=
→→
δ
δδ
δδδ
(3) xay =Let
)(log)(
limlim 01
00>=−=−=
→
+
→aaa
x
aa
x
aa
dx
dye
xx
x
x
xxx
x δδ
δ
δ
δ
δ in particular xx ee
dx
d =)(
MCA 11 - Mathematics SVT 73
(4) xy sin=et L
x
xxx
dx
dyx δ
δδ
sin)sin(lim
−+=→0
.sin)cos(sin
coslimsincos
lim xxx
xx
xx
xxxxxx
xx=⋅+=⋅
+=
−+++
=→→
10
2
22
222
00 δ
δδ
δ
δδ
δδ
(5) xy cos=Let
x
xxx
dx
dyx δ
δδ
cos)cos(lim
−+=→0
.sin)sin(sin
sinlim xxx
xx
xx
−=⋅+−=⋅
+−=
→10
2
220 δ
δδ
δ
(6) constant) aLet (cy = 00
=−=→ x
cc
dx
dyx δδlim
Thus derivative of a constant is zero.
Thus we have the following standard derivatives
Function y = f (x) Derivativenx 1−nnx
xlogx
1
xa aa ex log
xsin xcos
xcos xsin−constant zero
Rules for differentiation
I Sum Rule
dx
dyxvuvuy find then to of functions are whose If &+=
ly.respective bein increments ingcorrespond let the toincrement small a Give yvuyvuxx δδδδ &,&,,
vvuuyy δδδ +++=+Then
Subtracting
vuvvuuyyy −−+++=−+ δδδ vuy δδδ += ie
x
v
x
u
x
yx
δδ
δδ
δδδ +=thenby out through divide
.0 as sidesboth on limits take →xδ
x
v
x
u
x
yxxx δ
δδδ
δδ
δδδ 000Then
→→→+= limlimlim
74 KSOU Differential Calculus
dx
dv
dx
du
dx
dy += ie
dx
dw
dx
dv
dx
du
dx
dywvuy −+=−+= then If -: Note
)(cos)(sin)(cossin xdx
dx
dx
dx
dx
d
dx
dyxxxy −+=−+= 66 then If (1) Eg.
.sincos)sin(cos xxxxxx ++=−−+= 55 66
.loglog)( 01
33then 3 If2 +−=+−=xdx
dycxy e
xe
x
II Product Rule
.&dx
dyxvuuvy find then to of functions are where If =
ly.respective be in increments ingcorrespond let the toincrement small a Give yvuyvuxx δδδδ &,&,,
vuuvvuuvvvuuyy δδδδδδδ ⋅+++=++=+ ))((then
vuuvvuyyyy δδδδδδ ⋅++=−+=∴
,xδby out through divide
x
vu
x
uv
x
vu
x
y
δδδ
δδ
δδ
δδ ⋅++=
,0 as sidesboth on limit take →xδ
x
vu
x
uv
x
vu
x
yxxxx δ
δδδδ
δδ
δδ
δδδδ⋅++=
→→→→ 0000then limlimlimlim
dx
dvo
dx
duv
dx
dvu
dx
dy ++= ie
dx
duv
dx
dvu
dx
dy += ie
constant a is where If (1) Note kkvy = dx
dvk
dx
dy =then
dx
duvw
dx
dvuw
dx
dwuv
dx
dyuvwy ++== then If2 ,)(
+== 2
32
32
3then If (1) Eg. x
dx
dxx
dx
dx
dx
dyxxy sin)(sinsin
21
23
2
3xxxx ⋅+= sincos
xdx
dyxy sincos)( 8then8If2 −==
)(logsin)(sinlog)(logsinlogsin)( xxxx edx
dxxx
dx
dxex
dx
dxe
dx
dyxxey ⋅+⋅+=⋅= then If3
.logsincoslogsin xxx exxxxex
xe ⋅⋅+⋅+⋅= 1
MCA 11 - Mathematics SVT 75
III Quotient Rule
.&dx
dyxvu
v
uy find then to of functions are where If =
ly.respectivebein increments ingcorrespond let the toincrement small a Give yvuyvuxx δδδδ &,&,,
vv
uuyy
δδδ
++=+then
)()(
)()(
vvv
vuuvuvvu
vvv
vvuuuv
v
u
vv
uuy
δδδ
δδδ
δδδ
+−−+=
++−+=−
++=∴
xδby out through divide
)( vvvx
vu
x
uv
x
y
δδδ
δδ
δδ
+
−=
0 as sidesboth on limit take →xδ
)(limlim
vvvx
vu
x
uv
x
yxx δ
δδ
δδ
δδ
δδ +
−=
→→ 00 ie
2 ie
vdx
dvu
dx
duv
dx
dy −=
This rule can be easily remembered in the following manner
(say) IfDr
Nr
v
uy ==
2
of derivative of derivativeThen
)(
)()(
Dr
DrNrNrDr
dx
dy −=
constant a is where If -: Note kv
ky =
dx
dv
v
k
dx
dy2
then −=
x
xxy
cos
sintan == If (1) Eg.
x
xdx
dxx
dx
dx
dx
dy2
then cos
)(cossin)(sincos −=
.seccoscos
sincos
cos
)sin(sincoscosx
xx
xx
x
xxxx 222
22
2
1 ==+=−−⋅=
.sec,tan xdx
dyxy 2then If ==∴
x
xxy
sin
coscot)( == If2
x
xxxx
dx
dy2
then sin
)(coscos)sin(sin −−=
.sinsin
cossinx
xx
xx 222
22
cosec1 −=−=−−=
76 KSOU Differential Calculus
.,cot xdx
dyxy 2cosecthen If −=∴
xxy
cossec)(
1 If 3 ==
)(coscos
xdx
d
xdx
dy ⋅−=2
1then
.tanseccos
sin
cos)(sin
cos)sin(
cosxx
x
x
xx
xx
x=⋅==−−= 111
2
.tansec,sec xxdx
dyxy ==∴ then If
xxy
sin)(
1cosec If 4 ==
xxdx
dycos
sin⋅−=
2
1then
.cotsin
cos
sinxx
x
x
x⋅−=⋅−= cosec
1
.cot, xxdx
dyxy ⋅−==∴ cosecthen cosec If
IV Chain Rule or function of a function rule
x
u
u
y
x
y
dx
dyxguufy
δδ
δδ
δδ ×=== consider find to where If ,)()(
x
u
u
y
x
yxx δ
δδδ
δδ
δδ×=∴
→→ 00limlim
dx
dv
dv
du
du
dy
dx
dyxhvvguxfy
dx
du
du
dy
dx
dy ⋅⋅====⋅= then if also ie )(&)(),(
cbxaxucbxaxy n ++=++= 22 put then If (1) Eg. )(
1−=∴= nn nudu
dyuy
cbxaxu ++= 2
baxdx
du += 2
).()( baxcbxaxdx
du
du
dy
dx
dy n +++=⋅=∴ 22
)log()( 72 If2 23 +−= xxy
72
4372
72
1then
23
223
23 +−−=+−×
+−=
xx
xxxx
dx
d
xxdx
dy)(
)(
)sin()( 342 If3 2 −+= xxy
)cos()())(cos( 3421444342then 22 −++=+−+= xxxxxxdx
dy
MCA 11 - Mathematics SVT 77
+−=
1
1 If4
2
2
x
xy tan)(
+−
+−=
1
1
1
1then
2
2
2
22
x
x
dx
d
x
x
dx
dysec
22
22
2
22
22
22
2
22
1
112
1
1
1
2121
1
1
)(
)(sec
)(
)())((sec
++−+×
+−=
+−−+×
+−=
x
xxx
x
x
x
xxxx
x
x
222
22
1
4
1
1
)(sec
+×
+−=
x
x
x
x
Derivative of Hyperbolic functions
,sinh)( xy = If1
.cosh)(sinh xeeee
dx
dx
dx
d
dx
dy xxxx
=−=
+==−−
22then
,cosh)( xy = If2
.sinh)(cosh xeeee
dx
dx
dx
d
dx
dy xxxx
=+=
−==−−
22then
,cosh
sinhtanh)(
x
xxy == If3
=
x
x
dx
d
dx
dy
cosh
sinhthen
.coshcosh
sinhcosh
cosh
sinhsinhcoshcoshx
xx
xx
x
xxxx 222
22
2sech
1 ==−=−=
,sinh
coshcoth)(
x
xxy == If4
=
x
x
dx
d
dx
dy
sinh
coshthen
.sinhsinh
coshsinh
sinh
coshcoshsinhsinhx
xx
xx
x
xxxx 222
22
2cosech
1 −=−=−=−=
,cosh
)(x
xy1
sech If5 ==
)(coshcoshcosh
xdx
d
xxdx
d
dx
dy2
11then −=
=
.tanhcosh
sinhsinh
coshxx
x
x
xx
xsech
sech
112
−=⋅−=⋅−=
78 KSOU Differential Calculus
,sinh
)(x
xy1
cosech If6 ==
)(sinhsinh
xdx
d
xdx
dy2
1then
−−=
.cothsinh
cosh
sinhcosh
sinhxx
x
x
xx
x⋅−=⋅−=⋅−= cosech
112
Implicit Functions
Function of the type 0=),( yxf is called Implicit function.
rule.chain use offunction a astreat find To &xydx
dy
02 If (1) Eg. 22 =++ byhxyax
0222then =+
++
dx
dybyy
dx
dyxhax
hyaxdx
dybyhx 2222( ie −−=+ )
byhx
hyax
byhx
hyax
dx
dy
++−=
++−=∴ )(
)(
)(
2
2
10 If (2) Eg. =+ xyyx sinsin
differentiating w.r.t. x
0=+++dx
dyxxyy
dx
dyyx sincossincos
xyydx
dyxyx cossin)sincos( −−=+ ie
)sincos(
)cos(sin
xyx
xyy
dx
dy
++−=∴
Parametric functions
)(),( tgytfx == type theof Functions taken together is called Parametric function, where t is the paramter. Parametric
functions are also denoted as parameter. theis whereθθθ )(),( fyfx ==
θθ d
dy
d
dx
dt
dy
dt
dx
dx
dy&or consider find To &,
θθ
ddx
ddy
dx
dy
dtdx
dtdy
dx
dy == or then
dx
dytaytax find If (1) Eg. 33 ,sin,cos ==
ttadt
dxtax sincos,cos 23 3 :Solution −== , tta
dt
dytay cossin,sin 23 3==
ttta
tta
dtdx
dtdy
dx
dytan
sincos
cossin −=−
==∴2
2
3
3
MCA 11 - Mathematics SVT 79
dx
dyayax find4343 If 2 33 ),cossin(&)sincos()( θθθθ −=−=
)cossin12sin3(
)sincos12cos3( :Solution
2
2
θθθθθ
θθ
−−+==
a
a
ddx
ddy
dx
dy
θθθθ
θθθcot
)cossin41(sin3
)cossin41(cos3 −=+−
+=
Differentiation using Logarithms
)()( xgxfy =If
use logarithms on both sides
)(log)()(loglog )( xfxgxfy xg ==
differentiating w.r.t. x
)(log)(')(
)(')()(log)(')('
)()( xfxg
xf
xfxgxfxgxf
xfxg
dx
dy
y+=+= 11
+=∴ )(log)('
)(
)(')()( )( xfxg
xf
xfxgxf
dx
dy xg
xxydx
dy = if Find (1) Eg.
xxy loglog = :Solution
111 ⋅+⋅=∴ xx
xdx
dy
ylog
]log[ xxdx
dy x += 1 ie
.,)(cos)( tansin
dx
dyxxy xx find If 2 +=
xx xvxuvuy tansin )(cos& ==+= whereLet :Solution
xxu logsinlog =
xxx
x
dx
du
ulogcos
sin +=∴ 1
+= xx
x
xx
dx
du x logcossinsin ie
xxv tan)(cos=
xxv coslogtanlog =
xxx
xx
dx
dv
vcoslogsec
cos
)sin(tan 211 +−=∴
]coslogsectan[)(cos tan xxxxdx
dv x 22 +−=∴
]coslogsectan[)(coslogcossin tansin xxxxxx
x
xx
dx
dv
dx
du
dx
dy xx 22 +−+
+=+=∴
80 KSOU Differential Calculus
Derivatives of inverse Trigonometric functions
1for 1
11
2
1 <−
=− xx
xdx
d)(sin)(
1for 1
12
2
1 <−
−=− xx
xdx
d)(cos)(
21
1
13
xx
dx
d
+=− )(tan)(
21
1
14
xx
dx
d
+−=− )(cot)(
1for 1
15
2
1 >−
=− ||)(sec)( xxx
xdx
d
1for 1
1cosec6
2
1 >−
−=− xxx
xdx
d)()(
Derivatives of inverse hyperbolic functions
xx
xdx
d ∀+
=−2
1
1
11 )(sinh)(
1for 1
12
2
1 >−
=− xx
xdx
d)(cosh)(
1for 1
13
21 <
−=− x
xx
dx
d)(tanh)(
1for 1
14
21 >
−−=− x
xx
dx
d)(coth)(
1for 1
1sech5
2
1 <−
−=− xxx
xdx
d)()(
2
1
1
1cosech6
xxx
dx
d
+
−=− )()(
Exercise
following the ofFinddx
dy
xyxyaxy. ee coslog.)(log. =+=+= 32321 22
22 32651
14 )(... +==
−+= xyexy
x
xy x
312 53987 /)(... +==++= xyeycbxaxy x
xyxxxyexxxy xe
152 12cosec1110 −=+−=+++= sinh.tan.logsin.
xxyxxyxxy 1111 cosec151413 −−−− +=⋅== sec.sinsin.sinhsinh.
MCA 11 - Mathematics SVT 81
x
xxyxxyxxy
118117116
2212 ++=+=+= − .sin)(.tan)(.
)cos(sin.log.)()(. xxeyxeyxxy xe
x +=+=++= 2145202119 2
xxyee
yx
ey
xxx
2242
23222
sinsin... ⋅=+==−
−+=
+=+++= −−
ax
axy
x
xycxbxaxy
127
1
22625 1
21 tan.sin.))()((.
xyeyx
xy e
x sinlog..cos
cos. sinh ==
+−= 3029
1
128
−+=−== −−
x
xyxxyxxy e 1
13313231 1122 tan.cos)(.)tan(log.
036351
134 1121 =+=
+−= −−− yxcxy
x
xy sinsin..tan.
atyatxayxayx 2393837 2323232222 ===+=+ ,... ///
tbytaxtytxt
ytx sin,cos.sinh,cosh.,. ====== 4244411
40
xyyxyxytytx e sin.cos.tan,seclog. ==== 454443 2
xxyee
eey
xx
exy
xx
xx
e
x
tan)(..log
)(. 2
2
14847cosec
146 +=
+−=
⋅+= −
−
x
xy
xx
exy
x
xey
ee
xx
log
cos.
log.
cos.
12
511
501
49−
=−
+=+
=
−−=
−+== −−
− 2
31
2
21
1
3
31
354
1
15352
x
xxy
x
xy
x
exy
x
tan.sec.cos
.
xx
xy
x
xy
x
xxy sec.
sin
sin.
cosh.
1
357
1
156
cosech
555
3
−+=
−+==
060591cosec158 12 =++=+−−= − yxxyyxyxxxy e sinsin.)(log)sin(.)()(.
taytaxtteyttex tt 446261 sin,cos.)sin(cos),sin(cos. ==−=+=
nxxyt
ay
t
axtytx n
e sinsin.cos
,cos
.tan,seclog. ⋅=−
=+
=== 6511
6463 2
xxx xyxyxy1
686766 1 −=== − tansintan )(sin.)(sin.)(sin.
yxyxxx eeexy e +== +.)(log. log 7069
82 KSOU Differential Calculus
Successive Differentiation
obtained. becan sderivativefurther hence offunction a also is then If ,)()( xxfdx
dyxfy ′==
.)( yDyxfdx
yd 222
2
or or or by denoted is derivative second The ′′
.)( yDyxfdx
yd 333
3
or or or by denoted is derivative thirdThe ′′′
.)()( yDyxfdx
ydn n
nn
n
n
or or or as denoted is derivative generalIn th
Examples
1at find1 If12
212 =+= − x
dx
ydxxy tan)(.
xxy 121 :Solution −+= tan)(
.tan)(tan)( xxxxx
xdx
dy 112
2 2121
11 −− +=⋅+
+×+=
xx
xdx
yd 122
2
21
120 −+
+×+= tan
.,2
14
211
21at
12
2 ππ +=×++
=
=
=xdx
ydx
32
22 4
that show4 If2y
a
dx
ydaxy
−== ,.
axy 4 :Solution 2 =
x w.r.t.atingdifferenti
y
a
dx
dya
dx
dyy
242 =⇒=
x w.r.t.atingdifferentiagain
3
2
222
2 4222
y
a
y
a
y
a
dx
dy
y
a
dx
yd −=
−=−=
ta
t
dx
ydtay
ttax
42
2
that showthen 2
If3cos
sin,sin&tanlogcos. ==
+=
+=
2 :Solution
ttx tanlogcos
+−=∴22
1
21
1 2 tta
dt
dxsec
tan
sin
MCA 11 - Mathematics SVT 83
t
ta
t
ta
tta
ttta
tta
sin
cos
sin
sin
sinsin
cossin
sin
costan
sin22
2
11
222
1
221
2
1 =
+−=
+−=
+−=
+−=
tadt
dytay cos,sin =∴=
ttta
ta
dtdx
dtdy
dx
dytansin
cos
cos =×==∴2
t
ta
t
dtdxt
dx
dt
dx
dy
dt
d
dx
dy
dx
d
dx
yd
sin
cos
secsec
2
22
2
2 1 =×=
=
=∴
ta
t
dx
yd42
2
cos
sin=∴
2
22 find If4
dx
ydxy ,sin. =
xxxdx
dy22 :Solution sincossin ==
xxdx
yd2222
2
2
coscos =⋅=
2
22 find If5
dx
ydxxy elog. =
)log(log 1221
:Solution 2 +=⋅+⋅= xxxxx
xdx
dyee
xxx
xdx
ydee log)log( 2312
22
2
+=++
=
2
2
find If6dx
ydcbxey ax )sin(. +⋅=
axax aecbxbcbxedx
dy ⋅++⋅+⋅= )sin()cos( :Solution
{ })sin()cos( cbxacbxbedx
dy ax +++=
{ } { })sin()cos()cos()sin( cbxacbxbaecbxabcbxbedx
yd axax +++++++−= 22
2
{ })sin()()cos( cbxbacbxabeax +−++= 222
2
2
find1 If7dx
ydbyax )cos(),sin(. θθθ −=−=
θθ
θθ
sin),cos( bd
dya
d
dx =−= 1 :Solution
)cot()(sin
)cos()sin(
)cos(
sin2
22
222
1 2θ
θθθ
θθ
a
b
a
b
a
b
dx
dy =⋅
⋅=−
=
84 KSOU Differential Calculus
).()cos(
2cosec41
1
2cosec
22
1
2cosec 4
222
2
2
θθ
θθθa
b
aa
b
dx
d
a
b
dx
yd −=−
⋅−=⋅⋅−=
01 that provethen If8 212
21
=−−−=−
ymxyyxey xm )(,. sin
21 :Solution
1
x
me
dx
dy xm
−⋅=
−sin
cross multiplying & squaring, we have
2221
21 ymyx =− )(
differentiating w.r.t. x
122
1212 2221 yymxyyyx ⋅=−+⋅− )()(
01get we2by t throughouDividing 212
21 =−−− ymxyyxy )(,
01 that provethen If9 212
21 =+−−= − ymxyyxxmy )(),sinsin(.
2
11
1 :Solution
x
mxmy
−⋅= − )sincos(
)()( 2221
2
2
2
1 111
1ymyx
x
ymy −=−⇒
−
−=
differentiating w.r.t. x
)()()( 122
1212 2221 yymxyyyx −=−+−
.)( 01get we2by ut throughtoDividing 212
21 =+−− ymxyyxy
01121 that prove If10 21
22
21 =−++−−+= − yxyxxyxxey x )()()(,tan.
xex
edx
dy xx 121
1 :Solution −+
+⋅= tan
xeyyx =−+ ])[( 121
differentiating w.r.t. x
xeyyxyyx =−+−+ )())(( 1122 21
yxyxxyyxxyx )()()()( 21
21
22
2 112211 +−+=−−+−+
01121 21
22
2 =−+−+−+⇒ yxyxxyx )()()(
Exercise
22
222
9
16 that show3694 If1
ydx
ydyx −==+)(
32
222
3
2 that prove132 If2
)()(
yxdx
ydyxyx
+−==++
2
233 find If3
dx
ydayax θθ sin,cos)( ==
2
2
find22
1 If4
dx
ydayax θθ sin,tan)( ==
MCA 11 - Mathematics SVT 85
θθθθθθθθ
adx
ydayax
3
2
2
that show If5sec
)cos(sin),sin(cos)( =−=+=
2
21 findthen2 If6
dx
ydxy ,sin)( −=
2
2
find then If7dx
ydxxy e ,log)( =
2
24 find then3 If8
dx
ydxey x ,sec)( ⋅=
2
2
find then If9dx
yday x ,)( =
2
22 find then2 If10
dx
ydatyatx ,,)( ==
2
233 findthen If11
dx
ydayx x ,)( =
0 that prove If12 22
2
=++= ymdx
ydmxbmxay ,sincos)(
01 that prove1 If13 212
22 =−++
++= ymxyyxxxym
)(,)(
01 that prove If14 221 =+−+= −+ ynnyxbxaxy nn )(,)(
01 that prove If15 212
2 =+−−== ypxyyxptytx )(,sin,sin)(
062 that provethen 1 If16 2322 =++=++ yyxyxyx )(,)(
02 that prove If17 2212 =++−= ybaayybxey ax )(sin)(
02 that show then If18 122
2=−++= )()( yxyyx
x
baxy
ynnyxx
baxy
nn )()( 1 that show then If19 2
21 +=+= +
0 that show then If20 22
2
=++= xndt
xdntbntax sincos)(
nth derivative of Standard functions
nm ybaxy find to If1 )(. +=
222
11 1 :Solution abaxmmyabaxmy mm −− +−=+= ))((;)(
have we times, atingdifferenti n
nnmn abaxnmmmy −++−−= ))(()( 11 L
baxym
+=−= 1
ie1 if particularin ,
nnn abaxny −−+−−−−= 1321 ))(())()(( L
1
1 ie ++
−=n
nn
nbax
any
)(
!)(
nybaxy find to If2 )log(. +=
abax
y ⋅+
= 1 :Solution 1
86 KSOU Differential Calculus
times1 atingdifferenti )( −n
n
nn
n
nn
nbax
aa
bax
ay
)(
)(
)(
)(
+−=⋅
+−=
−−− 111 11
nmx yay find to If3 =.
2221 :Solution )(log;)(log amayamay mxmx ⋅=⋅=
mxnnn aamy )(log=∴ general,In
nybaxy find to If4 )sin(. +=
++=+=
2 :Solution 1
πbaxabaxay sin)cos(
x w.r.t.atingdifferentiagain
⋅++=
++=
22
222
2ππ
baxabaxay sincos
⋅++=
⋅++=
23
22 33
3ππ
baxabaxay sincos
⋅++=∴
2 general,In
πnbaxay n
n sin
nybaxy find to If5 )cos(. +=
++=+−=
2 :Solution 1
πbaxabaxay cos)sin(
x w.r.t.atingdifferentiagain
⋅++=
++−=
22
222
2ππ
baxabaxay cossin
⋅++=∴
2 general,In
πnbaxay n
n cos
nax ycbxey find to If6 )cos(. +=
x w.r.t.atingdifferenti :Solution
)sin()cos( cbxbecbxaey axax +−+=1
αα sin,cos rbra ==put
αα sin)sin(cos)cos( cbxrecbxrey axax +−+=1then
)cos(]sin)sin(cos)[cos( ααα ++=+−+= cbxrecbxcbxre axax
get weg,simplifyin w.r.t.atingdifferentiagain &x
)cos( α222 ++= cbxery ax
a
bbarncbxery axn
n122 where general,In −=+=++= tan&)cos( αα
MCA 11 - Mathematics SVT 87
nax ycbxey find to If7 )sin(. +=
x w.r.t.atingdifferenti :Solution
)cos()sin( cbxbecbxaey axax +++=1
αα sin,cos rbra ==put
)sin(]sin)cos(cos)[sin( ααα ++=+++= cbxrecbxcbxrey axax1then
get weg,simplifyin w.r.t.atingdifferentiagain &x
)sin( α222 ++= cbxery ax
a
bbarncbxery axn
n122 where general,In −=+=++= tan&)sin( αα
8. Statement of Leibmitz's Theorem on nth derivative of a product
bygiven is product theof derivative the of functions are If th uvnxvu ,&
nnnnnn uvncvuncvuncvuuv ++++= −− LL222211)(
.&& vuvu of sderivative theoforder represent of suffixes where
Examples
86
1 of derivative theFind1
2th
+− xxn.
(Say) 4242
1
86
1Let :Solution
2 )()())(( −+
−=
−−=
+−=
x
B
x
A
xxxxy
))(( 42by t throughougmultiplyin −− xx
)()( 241 −+−= xBxA
2
1212put −=⇒−== AAx )(,
2
1214put =⇒== BBx ,
)()( 421
221
−+
−
−=∴
xxy
11 4
12
1
2
12
1
have we times, atingdifferenti ++ −
−+
−
−−=
n
n
n
n
nx
n
x
nyn
)(
!)(
)(
!)(
xxn 32th of derivative theFind2. cossin
4
33
2
21Let :Solution 32 xxx
xxycoscos)cos(
cossin+×−==
]coscoscoscoscos[cos xxxxxxy 2323338
1 ie −−+=
+−+−+= )cos(cos)cos(coscoscos xxxxxx 3
2
35
2
133
8
1
−−−−+= )coscoscoscoscoscos xxxxxx
2
33
2
3
2
15
2
133
8
1
88 KSOU Differential Calculus
−−=∴ xxxy 5
2
13
2
1
8
1coscoscos
+⋅−
+⋅−
+=
255
2
1
233
2
1
28
1have we times, atingdifferenti
πππnxnxxyn nn
n coscoscos
nx yxey find to If3 22 sin. =
)cos( xey x 212
1 :Solution 2 −= xeey xx 2
2
1
2
1 22 cos−=
)cos( α−+⋅−= nxereyn xnxnn 2
2
12
2
1have we times, atingdifferenti 22
41
2
2844 where 11 πα ====+= −− tantan,r
( )
⋅+−=∴ −
228
2
12 21 π
nxeynxn
n cos
nx yxxey find 35 If4 4 cossin. =
]sin[sin xxey x 282
1 :Solution 4 += xexey xx 2
2
18
2
1 ie 44 sinsin +=
( ) ( )
+++
++= −−
4
22416
2
1
4
886416
2
1have we times, atingdifferenti 1414 tansintansin nxenxeyn xnxn
n
( ) ( )
+++= −−
2
1220
2
12880
2
1 ie 1414 tansin)tansin( nxenxey xnxn
n
nyxxy find3 If5 2 log. =
xxy 3 :Solution 2 log=
xxu logloglog +== 33Let
211
xvx
un
n
n =−=∴−,
)(
Theorem, sLeibnitz' using times, atingdifferenti n
21
211
2
3
21
2
12
1
⋅−+−+⋅−== −
−
−
−−
n
n
n
n
n
n
nnx
nCxx
nCxx
uvy)()()(
)(
[ ] [ ]nnnx
nnnx n
n
n
n
−+−−=−+−+−−= −
−
−
−2
2
312
2
3
211
12111 )(
)()()()(
[ ]131
ie 22
3
+−−= −
−nn
xy
n
n
n)(
0112 that show If6 212
2 =+++++= ++ nnx
n ynynyxxbxay )()()sin(log)cos(log.
)sin(log)cos(log xbxay +=Let :Solution
x w.r.t.atingdifferenti
xxb
xxay
111 )cos(log)sin(log +−=
MCA 11 - Mathematics SVT 89
)cos(log)sin(log xbxaxy +−=1 ie
again w.r.t.atingdifferenti x
xxb
xxayxy
1112 )sin(log)cos(log −−=+
yxbxaxyyx −=−−=+ )sin(log)cos(log122 ie
0 ie 122 =++ yxyyx
have weTheorem, sLeibnitz' using times, atingdifferenti n
01112 adding,
0
1
22
122
11
21122
=++−+++
=+⋅++⋅++
++
+
++
nnn
n
nn
nnn
ynnnxynyx
y
ynCxy
ynCxynCyx
])([)(
0112 ie 212
2 =++++ ++ nnn ynxynyx )()(
0121 that prove If7 2212
21
=+−+−−= ++−
nnnxm ymnxynyxey )()()(,. cos
xmey1
:Solution −
= cos
x w.r.t.atingdifferenti
21
1 x
mey xm
−
−⋅= cosmyyx −=− 1
21 ie
squaring both sides 2221
21 ymyx =− )(
,x t.again w.r. atingdifferenti
122
1212 2221 yymxyyyx =−+− )()(
have we2by dividing 1,y
01 212
2 =−−− ymxyyx )(
Theorem, sLeibnitz' using times w.r.t.atingdifferenti nx,
0121 adding,
0
1
221
2212
2
2
11
21122
=++−−+−−
=−
−+−−+−+−
++
+
++
nnn
n
nn
nnn
ymnnnxynyx
ym
ynCxy
ynCxynCyx
)()()(
)(
)()()(
0121 ie 2212
2 =+−+−− ++ nnn ymnxynyx )()()(
0121 that prove2 If8 2212
211 =−+++−=+ ++−
nnnmm ymnxynyxxyy )()()(,.
xyy mm 2 :Solution 11 =+ −
( ) mm xyy 121 21 ie =+
( ) 012 121 =+−∴ mm xyy
90 KSOU Differential Calculus
12
442in equation quadratic a iswhich 2
211 −±=−±=∴ xx
xxyy mm
mm xxyxxy
−+=∴−+= 11 Consider, 221
x w.r.t.atingdifferenti
1
11
12
211
2
21
2
2
12
1−
+−
−+=
−+
−+=
−−
x
xxxxm
x
xxxmy
mm
myyx =− 12 1 ie
2221
2 1 sidesboth Squaring ymyx =− )(
x w.r.t.atingdifferenti
122
1212 2221 yymxyyyx =+− )()(
have we2by t throughoudividing 1,y
01 212
2 =−+− ymxyyx )(
Theorem sLeibnitz' using times atingdifferenti n
0121 adding,
0
1
221
2212
2
2
11
21122
=−+−+++−
=−
++++−
++
+
++
nnn
n
nn
nnn
ymnnnxynyx
ym
ynCxy
ynCxynCyx
)()()(
)(
0121 ie 2212
2 =−+++− ++ nnn ymnxynyx )()()(
mm xxyxxy
−−=
−−= 1 ie1 ifresult same obtain the We 221
Exercise
65
1 of derivative theFind1
2th
+− xxn.
.coscos)(sinsin)(cossin)(cos)(sin)(. xxvxxivxxiiixiixin 54834 of derivative theFind2 33th
xxeiiixeiixein xxx 25 of derivative theFind3 2223th cossin)(cos)(sin)(.
0121 that prove If4 2212
21 =−−+−−= ++−
nnn ymnxynyxxmy )()()(),sinsin(.
0121 that prove If5 2212
21
=+−+−−= ++−
nnnxa yanxynyxey )()()(,. sin
0121 that prove1 If6 1222 =+−+−−= ++ nnn
n ynnxyyxxy )()(,)(.
MCA 11 - Mathematics SVT 91
Polar Co-ordinates
curve. on thepoint any be Let line. initial the andpoint fixed a be Let POAO
rOPPOA == &� θLet
ordinates.-coPolar called arewhich are of ordinates-cothen ),( θrP
direction.anticlock in measured theis and
from of distance is which vector,radius called is
POAO
Pr
�θ
)(θfr = as taken is curve theoEquation t
Let PT be the tangent to the curve r = f (θ) at P, then angle made by
the tangent with radius vector OP is denoted as ϕ ϕ=PTO� ie and anglemade by the tangent with initial line OA is denoted by ψ. It can be seenfrom the figure that
ϕθψ += (1)
Let OQ represent perpendicular from the pole O upon the tangent at P, it is denoted as p.
r
p
OP
OQOPQ ==φsin that triangle thefromseen becan It
φsinrp =∴ (2)
result) (importantthat prove Todr
rdθφ =tan
θcos),( rxyxP =thenare of ordinates-coCartesian theIf dx
dyry == ψθ tansin and
θθ
θ
θθθ
θθψ
sincos
sincostan
rd
drd
drr
ddx
ddy
−
+==∴
have weby RHS ofDr &Nr dividing & (1) using ,cosθ
θd
dr
θθ
θθ
ϕθtan
tan)tan(
dr
dr
dr
dr
−
+=+
1dr
dr
dr
dr
θθ
θθ
φθφθ
⋅−
+=
−+
tan
tan
tantan
tantan
11 ie
Comparing LHS & RHS, we have dr
dr
θφ =tan (3)
φsinrp = (2) from [ ] (3) using 11
11
cosec11
2
222
22
22
+=+==∴
θφφ
rd
dr
rrrrpcot
2
422
111
+=
θd
dr
rrp(4)
curve, for thebetween relation aobtain and eliminatecan we4 using rpfr &)(&)( θθ= which is called Pedal
Equation or (p, r) equation of the curve.
P r( , )q
r = f ( )q
j
q yA
Q
T
pO
r
92 KSOU Differential Calculus
To find the angle between two curves )(&)( θθ grfr ==
.)(&)( Pgrfr at intersect curves twoLet the θθ ==
21Let PTPT & be the tangents to the two curves and let 21 φφ & bet the angles
made by the tangents with the radius vector OP.
21by given is curves obetween tw Angle φφ −∴
( )21
2121
1 Now
φφφφφφ
tantan
tantantan
+−
=− (5)
21 findcan we(3)in result theusing φφ tan&tan and hence angle between two
curves at the point of intersection can be found out.
( ) 0 then If 2121 =−= φφφφ tantantan
.at other each touch curves twothe ie 21 P∴= φφ
2 then 1 If 2121
πφφφφ =−−=tantan
ly.orthogonalintersect tosaid are curves twothe∴
Examples
vector.radius the toangleconstant aat inclined is tangent the spiralr equiangula in the that Show(1) αθ cotaer =αθ cotaer = :Solution
ααθ
θ αθ cotcot, w.r.t.atingdifferenti cot raed
dr ==
αφαα
θ
θφ =⇒==== tancot
tanr
r
d
drr
dr
dr hence the result.
line. initial the toparallel is 3
point at the 1 Cardiod theo tangent t that theShow2πθθ =+= )cos()( ar
)cos( θ+= 1 :Solution ar
)sin0(, w.r.t.atingdifferenti θθ
θ −= ad
dr
+=−=
−=
−+===
22222
221
2
θπθθθ
θ
θθ
θ
θφ tancot
cossin
cos
sin
)cos(tan
a
a
d
drr
dr
dr
22
θπφ +=∴ ππππφθψππφπθ =++=+=+==623623
when &,
line. initial the toparallel is tangent the∴
2cosec( and
2( that show1
2 curve For the(3)
θθπφθ apiiir
a =−=−= ))cos
θcos−= 12
:Solution r
a
r = f ( )qr = g ( )q
P
T2
T1
f2
f1
f1-f2
MCA 11 - Mathematics SVT 93
a
r
d
dr
d
dr
r
a
2
sinsin
2, w.r.t.atingdifferenti
2
2
θθ
θθ
θ −=∴=−
222
2212
2
Now
2
2
θθθ
θ
θθ
θθθ
θφ tan
cossin
sin
sin
cos
sinsintan −=
−=
−−=−=
−===
r
a
a
r
r
d
drr
dr
dr
−=−=
22 ie
θπθφ tantantan222
θθπφθπφ sinsinsin& rrrp =
−==−=∴
222
2
21
2
2 ie
2 θθθ
θθ
sinsin
sincos
sinaaa
p =⋅=−
⋅=2
cosecθ
ap =∴
.cossin&sin)( θθθ +== rr 2 curves ebetween th angle theFind4
θθ
θ cos,sin 22 :Solution ==d
drr
vectorradiuson with intersecti ofpoint at the tangent by the made angle thebe Let 1φ
θθθ
θ
θφ tancos
sintan ====∴
2
21
d
drr
dr
dr
θφ =∴ 1 (1)
θθ cossin +=r curve for the
θθθ
sincos −=d
dr
curve this w.r.t.angle thebeLet 2φ
+=
−+=
−+===∴
41
12
πθθ
θθθθθ
θ
θφ tantan
tan
sincos
cossintan
d
drr
dr
dr
42πθφ +=∴ (2)
θπθφφ −+=−4
(2) & (1) from 21
4 is curves obetween tw angle
π∴
angles.right at intersect 11 curves that theProve5 )cos(&)cos()( θθ −=+= brar
)cos( θ+= 1Consider :Solution ar
θθ
sinad
dr −=
θθ
θθ
θ
θφsin
)cos(
sin
)cos(tan
−+=
−+===∴ 11
1a
a
d
drr
dr
dr
(1)
94 KSOU Differential Calculus
θθ
θ sin),cos( bd
drbr =−= 1Consider
θθ
θθ
θ
θφsin
)cos(
sin
)cos(tan
−=−=== 112
b
b
d
drr
dr
dr
(2)
(2) & (1) from
1111
2
2
2
2
21 −=−
=−−=−×
−+=
θθ
θθ
θθ
θθφφ
sin
sin
sin
cos
sin
)cos(
sin
)cos(tantan
ly.orthogonalintersect curves twothe∴
θcos) += 12
ofequation pedal theFind(6r
a
θcos+= 12
:Solution r
a
θ w.r.t.atingdifferenti
θθ
sin−=−d
dr
r
a2
2
ad
dr
r 2
1 ie
2
θθ
sin= (1)
(1) using 4
111111 Now
2
2
2
2
22
2
422 ard
dr
rrd
dr
rrp
θθθ
sin+=
+=
+=
[ ]
+−−+=
−−+=−+=
r
a
r
a
arr
a
arar
41
41
4
111
21
4
111
4
112
2
22
2
222
22θcos
ararrrr
a
r
a
arp
111144
4
111 ie
222
2
222=+−=
+−+=
equation. pedal required theiswhich 2 arp =∴
Exercise
)cos1( curve. theo tangent ton the pole thefromlar perpendicu theoflength theFind1. θ−= ar
2sin2: Ans 3 θ
a
.3 that show2 curve the2 22 θψθ == sinFor. ar
θθ cos,sin 22curves theofon intersecti of angle theFind3. == rr
2 :Answer
π
ly.orthogonalintersect 11 curves that theShow4 )sin(&)sin(. θθ −=+= arar
θmar mm cos= curve theofequation pedal theFind5. 1 :Answer += mm rpa
θar = ofequation pedal theFind6. 4222 :Answer rarp =+ )(
MCA 11 - Mathematics SVT 95
Indeterminate Forms
[ ] °∞°∞−∞∞∞− ∞
→→→01
0
0 forms theit takeswhen or or evaluating While ,,,,,)(lim)()(lim
)(
)(lim )(xg
axaxaxxfxgxf
xg
xf they
are called Indeterminate forms and to evaluate such forms the following rule known as L' Hospital's Rule is used.
L' Hospital's Rule
then or 0
0 form theof is thisifagain then or
0
0 form theof is If
∞∞
′′
=∞∞
→→→ )(
)(lim
)(
)(lim
)(
)(lim
xg
xf
xg
xf
xg
xfaxaxax
applied. becan rule Thisor 0
0 theof isit whenever
∞∞
′′′′
=→→ )(
)(lim
)(
)(lim
xg
xf
xg
xfaxax
[ ] then form theof isit when evaluate To ,)()(lim ∞−∞−→
xgxfax
[ ] applied. becan Rule sHospital' L' hence 0
0 form theof is which
1
11
Consider &
)()(
)()(lim)()(lim
xfxg
xfxgxgxf
axax
−=−
→→
say)Consider ()(lim )( yxf xg
ax=
→
)(
)(loglim)(log)(lim)(loglimlog )(
xg
xfxfxgxfy
axax
xg
ax 1then
→→→===
applied. becan rule sHospital' L' hence and or 0
0 form theof iswhich
∞∞
Examples
23 Evaluate(1)
21 +−→ xx
xx
loglim
0
0 form theof is this
23 :Solution
21 +−→ xx
xx
loglim rule sHospital' L' using∴
132
1
32
1
1−=
−=
−=
→ xx
xlim
30
22 Evaluate(2)
x
xxx
sinsinlim
−→
30
22 :Solution
x
xxx
sinsinlim
−→
rule sHospital' L' applying0
0 form theof is This ∴
20 3
222
x
xxx
coscoslim
−=→
again rule sHospital' L' applying0
0 form theof isit again ∴
113
41
6
2
2
2
3
4
6
2
6
242000
=×+×−=
+⋅−=+−=
→→→ θθ
θ
sinlim
sinsinlim
sinsinlim Q
x
x
x
x
x
xxxx
13
4
3
1 =+−=
96 KSOU Differential Calculus
xx
xxx tan
tanlim)(
20 Evaluate3
−→
1 :Solution 0303020
=−=×−=−→→→→ x
x
x
xx
x
x
x
xx
xx
xxxxxx tanlim
tanlim
tan
tanlim
tan
tanlim Q
rule sHospital' L' using0
0 form theof is This ∴
0
0
3
12
2
0=−=
→ x
xx
seclim again rule theusing∴
3
11
6
2
6
2
6
2 2
00=×=×=⋅=
→→ x
xx
x
xxxxx
tanseclim
tansecseclim
−−
→ 1
11 Evaluate(4)
0 xx exlim
∞−∞ form theof is This :Solution
0
0 form theof iswhich
1
1
1
1100 xe
xe
ex x
x
xxx )(
)(limlim
−−−=
−−∴
→→ rule sHospital' L' using∴
xx
x
x xee
e
+−−=
→ 1
10
lim
again rule theusing0
0 form theofagain iswhich ∴
2
1
101
10
=++
=++
=→ xxx
x
x exee
elim
−
−→ xx
xx loglim)(
1
1 Evaluate5
1
∞−∞ form theof This :Solution
−
−−=
−
−∴
→→
x
xx
x
xx
xx
xxx 1
11
1 11log
loglim
loglim
rule sHospital' L' using0
0 form theof is This ∴
rule sHospital' L' theapplyingagain form 0
0 is
1
1
111
11
1
2
2
1 −+−=
×−+
×
−=
→→ xx
x
xx
x
xx
xxxx loglim
log
lim
2
1
11
11
=+
=→
xxlim
x
x
x cot)(tanlim)(
2
Evaluate6π→
MCA 11 - Mathematics SVT 97
0 form theof is This :Solution ∞x
x
xy cot)(tanlim
2
Let π→
=
x
xxxxy
xx
x
x tan
tanloglimtanlogcotlim)log(tanlimlog cot
222
πππ →→→=== rule sHospital' L' using
01
1
22
2
2
2
===⋅
=→→→
xxx
xx
xxx
cotlimtan
limsec
sectanlim
πππ
10 ==∴ ey
x
x x
x1
0 Evaluate7
→
sinlim)(
∞1 form theof is This :Solution
x
x x
xy
1
0Let
=
→
sinlim
x
x
x
x
x
xy
xx
=
=
→→
sinlog
limsin
loglimlog00
1
rule sHospital'L' theapplying0
0 form theof is This ∴
2020
2
01
1 x
xxx
x
xxx
x
xx
xxx
x
x
xxx
sincoslim
sincos
sinlim
sincos
sinlim−×=−×=
−×=
→→→
rule sHospital'L' theapplyingagain 0
0 form theof is This ∴
0222 000
=−=−=−−=→→→
x
x
xx
x
xxxxxxx
sinlim
sinlim
cossincoslim
10 ie =∴= yylog
limit. theand of value thefind finite, is 2
If830
ax
xxax tan
sinsinlim)(
−→
rule sHospital'L' theapplying0
0 form theof islimit given The :Solution ∴
xx
xxax 220 3
22
sectan
coscoslim
⋅−=
→
0
0 form theof is thisif existslimit
20for 022 =∴==−∴ axxxa coscos
0
0 form of is 11
3
2231
3
2222022
2
20××−=××−=
→→ x
xx
xx
x
x
xxxx
coscoslim
sectan
coscoslim
98 KSOU Differential Calculus
rule sHospital'L' using∴
13
4
3
1
3
4
6
2
2
2
3
4
6
2
6
24200
=+−=+−=×+−=+−=→→ x
x
x
x
x
xxxx
sinsinlim
sinsinlim
1. islimit theand2=∴ a
Exercise
Evaluate the following
xx
ee
xx
xx xx
xx sinlim)(
sin
coscoshlim
sin
−−−
→→ 002(1)
200
14
1
13
x
xxe
x
xe x
x
x
x
)log(lim)(
)log(
sinlim)(
+−+
−−→→
−
−
→→ xxxx xx
116
115
0220 sinlim)(
sinlim)(
x
x
x
xxx −
→→1
1
1
1
087
2)(lim)()(coslim)(
xxxx
x
x
x
cbax
1
0
1
0 3109
++→→lim)()(cotlim)( log
31
101
91
81
7063
15
2
3 (4) 2 (3) 1 (2) 1 (1) : Answers )()()()()()()( abc
eee−
MCA 11 - Mathematics SVT 99
Partial Derivatives
A function of two independent variables and a dependent variable is denoted as ),( yxfz = which is explicit function where
x & y are independent variables and z a dependent variable. Implicit function is denoted by Czyx =),,(φ
x
f
x
zxfz
x
yxfyxxfx ∂
∂∂∂−+
→or by denoted and w.r.t.or of derivative Partial called isit then exists If
0 δδ
δ
),(),(lim
y
f
y
zyfz
y
yxfyyxfy ∂
∂∂∂−+
→or by denoted and w.r.t.or of derivative Partial called isit then exists If
0 δδ
δ
),(),(lim
y
zyx
x
z
∂∂
∂∂
finding whileandconstant a astreating.r.t.function wgiven theatedifferenti derivative theobtaining while
differentiate the given function with respect to y, treating x as a constant.
yxy
zyxyx
x
zyxyxz 2202 then If (1) Eg. 22 −=
∂∂+=++=
∂∂−+= &
)cos(cos&sincossin)( xyxxxyxxy
zxyyxyxxy
x
zxyxyxz −=⋅−=
∂∂−⋅−=
∂∂−= 12then If2 222
( )( )222
22
222
22221 222
41
1 then
2If3
yx
xxyyyx
yx
yxx
z
yx
xyz
−
⋅−−×
−+
=∂∂
−= − )(tan)(
( )( ) ( ) ( ) ( ) 22222
22
222
23
222
232
22222
222 2222422
4 yx
y
yx
xyy
yx
yxy
yx
yxyyx
yxyx
yx
+−=
+
+−=+
−−=−
−−×+−
−= )(
( )( )222
22
222
22
222
41
1
yx
yxyxyx
yx
yxy
z
−
−−−×
−+
=∂∂ )())((
&
( )( ) ( ) ( ) ( ) 22222
22
222
23
222
223
22222
222 2222422
4 yx
x
yx
yxx
yx
xyx
yx
xyxyyx
yxyx
yx
+=
+
+=+
+=−
+−×+−
−= )(
Successive derivatives
are sderivative partialorder second thes,derivative partialorder first are &),(function For they
z
x
zyxfz
∂∂
∂∂=
. generalin but ,,, as denoted are which ,,,22
2
222
2
2
xy
z
yx
z
y
z
xy
z
yx
z
x
z
y
z
yx
z
yy
z
xx
z
x ∂∂∂=
∂∂∂
∂∂
∂∂∂
∂∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
1&1,2,2 (1) exampleIn 22
2
2
2
2
=∂∂
∂=∂∂
∂−=∂∂=
∂∂
xy
z
yx
z
y
z
x
z
xyxxyyxxyx
zxyxxyxxyyxx
xy
zcos2sin2&coscossin2 (2) exampleIn 2
22
2
−+=∂∂
∂−−+=∂∂
∂
xy
z
yx
z
∂∂∂=
∂∂∂∴
22
( ) ( ) ( )222
22
222
222
222
22
22
2 )(242222)2)((2 (3) exampleIn
yx
yx
yx
yyx
yx
yyyx
yx
y
yxy
z
+
−−=+
+−−=+
⋅+−+=
+−
∂∂=
∂∂∂
( ) ( ) ( )222
22
222
22
222
22
22
2 )(222222)(2 and
yx
yx
yx
xy
yx
xxyx
yx
x
xyx
z
+
−−=+
−=+
⋅−+=
+∂∂=
∂∂∂
xy
z
yx
z
∂∂∂=
∂∂∂ 22
always general,in Thus
Exercise
xy
z
yx
zyxz
y
z
x
z
∂∂∂=
∂∂∂+=
∂∂
∂∂ 22
22 that show and )log(for , Find)1(
constant. a is where that show )()(If)2(2
22
2
2
cx
zc
t
zctxctxfx
∂∂=
∂∂−++= φ
abzy
za
x
zbbyaxfez byax 2 that show then )( If(3) =
∂∂+
∂∂−= +
∂∂−
∂∂−=
∂∂−
∂∂
++=
y
u
x
u
y
u
x
u
yx
yxu 14 that show then If)4(
222
. that show then sin If)5(22
1
xy
u
yx
u
x
yu
∂∂∂=
∂∂∂
= −
100 KSOU Differential Calculus
105 KSOU Matrix Theory
INTEGRAL CALCULUS
called is finding of process theGiven yxfdx
dy,)(= 'Integration' and the resulting function is called 'Integral'. If g (x) is
the integral then )()( xgdxxf =∫ is the notation used to represent the process.
In the above notation f (x) is called 'Integrand' and further [ ] .)()( xfxgdx
d =
[ ] tconstan a is when But cxgcxgdx
d)()( ′=+ ∫ +=∴ cxgdxxf )()(
Thus integral of a function is not unique and two integrals always differ by a constant.
Properties
[ ] ∫∫∫ ±=± dxxdxxfdxxxf )()()()()( ϕφ1
constant a is where2 KdxxfKdxxKf∫ ∫= )()()(
∫ = constant) (a0(3) cdx
Standard Integrals
nnn
n xcn
x
dx
dnc
n
xdxx =
+
+−≠+
+=
++
∫ 11
11
11
Q)(.
xcx
dx
dcxdx
x e11
2 =++=∫ )(loglog. Q
∫∫ +==
++= cedxeac
a
a
dx
dc
a
adxa xxx
xxx particularin 3
loglog. Q
∫ +−= cxdxx cossin.4
∫ += cxdxx sincos.5
∫ += 16 xdxxx sectansec.
∫ +−= cxdxxx coseccosec7 cot.
∫ += cxdxx coshsinh.8
∫ += cxdxx sinhcosh.9
∫ +−= cxdxxx sechsech10 tanh.
∫ +−= cxdxxx coseshcosech11 coth.
cxcxdxx
+−+=+
−−∫ 112
or 1
112 cottan.
cxcxdxx
+−+=−
−−∫ 11
2or
1
113 cossin.
cxdxx
+=+
−∫ 1
21
114 sinh.
cxdxx
+=−
−∫ 1
2 1
115 cosh.
cxcxdxxx
+−+=−
−−∫ 11
2cosecor
1
116 sec.
cxdxxx
+−=−
−∫ 1
2sech
1
117.
cxdxxx
+−=+
−∫ 1
2cosech
1
118.
Methods of Integration
There are two methods (1) Integration by substitution & (2) Integration by parts.
1. Integration by substitution
∫ dxxf )(Consider dttdxtdt
dxtx )()()( φφφ ′=′== iethen put
[ ] dtttfdxxf )()()( φφ ′=∴ ∫∫now for the new integrand, we can use the standard forms, ie. we have to make a proper substitution so that the given
integrand reduced to a standard one.
Examples
dxx
xdxx ∫∫ =
cos
sintan.1
dtdxxdx
dtxtx −==−= sinsin,cos ie then put
cxxtt
dtdxx +=−=−=−=∴ ∫ ∫ seclogcosloglogtan
∫ ∫= dxx
xxdxx
sec
sectantanor
xtx w.r.t.atingdifferentiput ,sec =
dx
dtxx =tansec dtdxxx =∴ tansec
102 KSOU Integral Calculus
cxtt
dtdxx +===∴ ∫ ∫ secloglogtan
∫ ∫= dxx
xdxx
sin
coscot.2
xtx w.r.t.atingdifferentiput ,sin = dtdxxdx
dtx == coscos ie
∫ ∫ +===∴ cxtt
dtdxx sinloglogcot
∫ += cxdxx coshlogtanh.3
∫ += cxdxx sinhlogcoth.4
∫ ++= cxxdxx )tanlog(secsec.5
∫ +−= cxxdxx )cotlog(. coseccosec6
[ ] [ ])(
)()()()(log
)(
)(1
1 also generalIn
1
−≠++
=′+=′ +
∫∫ ncn
xfdxxfxfcxfdx
xf
xf nn
2. Integration by parts
dx
duv
dx
dvuuv
dx
dxvu +=)(,& that,know we of functions are If
nIntegratio of definitionBy ∴
(1)property using ∫ ∫∫ +=
+= dx
dx
duvdx
dx
dvudx
dx
duv
dx
dvuuv
∫ ∫−=∴ dxdx
duvuvdx
dx
dvu
The result can be used as the standard result. Out of the two functions of the product, one has to be taken as u & another
dx
dv then the RHS after evaluation gives the integral or if both functions have taken as u & v then the result is as follows
dxdxvdx
dudxvudxuv ∫ ∫∫∫
⋅−=
∫∫ ′−=′ dxvuuvdxvu as written be alsocan onefirst The e.conveniencon depending used becan form oneany
Examples
xxx evuevvudxxe ==′=′=∫ ,,,. 1put 1
cexedxexedxvuuvdxxe xxxxx +−=⋅−=′−=∴ ∫∫∫ 1
cxxxdxxxxdxxxxdxxduxxdxxx ++−=+−=−−−=⋅−= ∫∫∫∫∫∫ sincoscoscoscoscossinsinsin. 12
∫ ==′=′= xvx
uvxudxx ,,,loglog.1
1put 3
MCA 11 - Mathematics SVT 103
∫∫ ′−=′∴ dxvuuvdxvu
cxxxdxxxdxxx
xxdxx +−=⋅−=⋅−= ∫∫ ∫ loglogloglog 11
ie
xvx
uvxudxx =−
=′=′= −−∫ ,,,sinsin.2
11
1
11put 4
dxxx
xxdxx ∫∫ −−=∴ −−
2
11
1
1sinsin
22
21put
1 evaluate to txdx
x
x =−−
−∫ x w.r.t.atingdifferenti
dttdxxdttdxx =−⇒=− 22
2
2211
1xtdt
t
dtt
x
dxx −==⋅==−
−∴ ∫∫∫
cxxxdxx +−−=∴ ∫ −− 211 1sinsin
Special Types of Integrals
Type I
∫∫∫∫ ++−−+ CBxAx
dx
xa
dx
ax
dx
xa
dx2222222 432 (1) )(&)(,)(,
dtadxatx == ,put (1) evaluate to
ca
x
at
at
dt
ataa
dta
xa
dx +==+
=+
=+
∴ −−∫∫∫ 11222222
11
1
1tantan
fractions partial use (3) & (2) evaluate to
(Say)11
22 ax
B
ax
A
axaxax −+
+=
−+=
− ))((
22by hroughout multiply t ax −
)()( axBaxA ++−=1
aBaBax
21
201put =∴⋅+== ,
aAaAax
2
1021put
−=∴+−=−= )(,
axa
axa
ax −+
+
−
=−
∴ 2
1
2
11
22
ax
ax
aax
aax
aax
dx
aax
dx
adx
ax +−=−++−=
−+
+−=
−∴ ∫∫∫ log)log()log(
2
1
2
1
2
1
2
1
2
1122
cax
ax
aax
dx ++−=
−∴ ∫ log
2
122
104 KSOU Integral Calculus
(Say)11
next,22 xa
B
xa
A
xaxaax −+
+=
−+=
− ))((
thenby t throughougmultiplyin 22 ,xa −
)()( xaBxaA ++−=1
aBaBax
2
1201put =⇒+== )(,
aAaAax
2
102put =⇒+−= )(,
xaa
xaa
xa −+
+=
−∴ 2
1
2
11
22
−+=−−+=
−+
+=
−∴ ∫∫∫ xa
xa
axa
axa
adx
xaadx
xaadx
xalog)log()log(
2
1
2
1
2
11
2
11
2
1122
cxa
xa
adx
xa+
−+=
−∴ ∫ log
2
1122
∫ ++ cBxAx
dx2
(4) evaluate to
∫∫∫ −−
+
=
+−
+
=++
=
2
22
2
222
4
42
1
42
11G.I.
A
ACB
A
Bx
dx
A
A
C
A
B
A
Bx
dx
AA
Cx
A
Bx
dx
A
This integral will take any one of (1), (2) or (3) and hence can be evaluated.
Examples
∫ +− 423 Evaluate1
2 xx
dx)(
∫∫∫∫ +−+
−
=
+−
−
=+−
=+−
9
124
3
13
1
3
4
9
4
3
13
1
3
4
3
23
1
423 :Solution
222
2
x
dx
x
dx
xx
dx
xx
dx
cxx
x
dx
x
dx +
−=−
×=
−+
=
−+
= −−∫∫ 22
13
22
1
32231
322
1
3
1
31
3223
1
31
383
1 11
2222tantan
∫ +− 2110 Evaluate2
2 xx
dx)(
cx
x
x
x
x
dx
x
dx
xx
dx +−−=
+−−−
×=
−−=
+−−=
+− ∫∫∫ 3
7
4
1
25
25
22
1
25212552110 :Solution
2222loglog
)()(
∫ 2−− xx
dx
246 Evaluate3)(
MCA 11 - Mathematics SVT 105
∫∫∫∫ +−=
++−=
+−=
−− 22222 122
1
1132
1
232
1
246 :Solution
)()()( x
dx
x
dx
xx
dx
xx
dx
cx
x
x
x +
−+=
+−++
××=
1
3
8
1
12
12
22
1
2
1log
)(
)(log
Type II
∫∫∫∫ ++−+− CBxAx
dx
ax
dx
xa
dx
xa
dx2222222
432(1) )(,)(,)(,
θθθ dadxaxxa
dxcos,sin ==
−∫ put evaluate to
22
a
xd
a
da
aa
da
xa
dx 1
222221 −==⋅==
−=
−∴ ∫∫∫∫ sin
cos
cos
sin
cos θθθ
θθ
θ
θθ
ca
x
xa
dx+=
−∴ −∫ 1
22sin
θθθ dadxaxxa
dxcosh,sinh ==
+∫ put evaluate to22
a
xd
a
da
aa
da
xa
dx 1
222221 −==⋅==
+=
+∴ ∫∫∫∫ sinh
cosh
cosh
sinh
cosh θθθ
θθ
θ
θθ
ca
x
xa
dx+=
+∴ −∫ 1
22sinh
θθθ dadxaxax
dxsinh,cosh ==
−∫ put evaluate to22
ca
x
ax
dx +=−
∴ −∫ 1
22cosh
a
xd
a
da
aa
da
ax
dx 1
222221 −==⋅==
−=
−∴ ∫∫∫∫ cosh
sinh
sinh
cosh
sinh θθθ
θθ
θ
θθ
∫ ++ CBxAx
dx2
evaluate to
∫∫∫−−
+
=
+−
+
=++
=
2
22
2
222
4
4
2
1
42
11G.I.
A
ACB
A
Bx
dx
A
A
C
A
B
A
Bx
dx
A
A
Cx
A
Bx
dx
A
This will reduce to any one of (1), (2) & (3) and hence can be evaluated.
Examples
∫ − 252 Evaluate1
xx
dx)(
106 KSOU Integral Calculus
cxx
x
dx
xx
dx
xx
dx +−×=−
=
−−
=
−−
=−
−−∫∫∫ )(sinsin 155
1
5
15
1
5
1
5
1
5
15
1
5
25
1
5
2
5
1 :Solution 11
2222
∫ +− 52 Evaluate2
2 xx
dx)(
cx
x
dx
x
dx
xx
dx +−=−+
=+−
=+−
−∫∫∫ 2
1
122152 :Solution 1
22222sinh
)()(
∫ +− 8124 Evaluate3
2 xx
dx)(
∫∫∫−−
−
=
+−
−
=+−
=
489
232
1
249
232
1
234
1G.I. :Solution
222
x
dx
x
dx
xx
dx
cxx
x
dx +−=−
=
−
−
= −−∫ )(coshcosh 322
1
2
12
3
2
1
2
1
2
32
1 11
22
Type III
∫∫ ++
+++
+
CBxAx
qpxdx
CBxAx
qpx22
and
mBAxlmCBxAxlqpx ++=+++=+ )()( 2 of derivativeput evaluate to 2
where l & m are the constants to be found out by equating the co-efficients of corresponding terms on both sides. ie. tosolve for m & n from the equations
qmlBpAl =+= and2
∫∫∫∫ +++++⋅=
+++
+++=
+++
CBxAx
dxmCBxAxl
CBxAx
dxmdx
CBxAx
BAxldx
cBxAx
qpx2
2222
2then )log(
the second integral in RHS is Type I and hence can be evaluated.
∫∫ ∫∫ +++++=
+++
++
+=
++
+
CBxAx
dlmCBxAxl
CBxAx
dxmdx
CBxAx
qpxldx
CBxAx
qpx2
2
2222
the second integral in RHS is Type II and hence can be evaluated.
Examples
dxxx
x∫ +−+
543
32 Evaluate1
2)(
mllxmxlx +−=+−=+ 464632Put :Solution )(
3
13
3
43 ie34
3
126 =+==+−=⇒=∴ mmlll ,
MCA 11 - Mathematics SVT 107
∫∫∫∫+−
++−=+−
++−
−=+−
+∴
3
5
3
49
13543
3
1
5433
13
543
46
3
1
543
32
2
2222
xx
dxxx
xx
dxdx
xx
xdx
xx
x)log(
∫∫
+
−
++−=
+−
−
++−=22
22
2
3
11
3
29
13543
3
1
3
5
9
4
3
29
13543
3
1
x
dxxx
x
dxxx )log()log(
cx
xxx
xx +
−++−=
−
×++−= −−
11
23
113
13543
3
1
311
32
11
3
9
13543
3
1 1212 tan)log(tan)log(
dxxx
x∫ −−
−
23
75 Evaluate2
2)(
mllxmxlx ++−=+−=− 322375Put :Solution )(
2
1
2
1573773
2
552 =+−=−−=⇒−=+−=⇒=−∴ lmmlll &
∫∫∫ −−−+
−−
−−=−−
−∴)( xx
dxdx
xx
xdx
xx
x
322
1
23
23
2
5
23
75222
∫∫
−−
1+−−−=
+
−−−
+−−⋅−=22
2
2
2
2
3
2
12235
4
9
2
32
2
1232
2
5
x
dxxx
x
dxxx
cxxxx
xx +−+−−−=
−×+−−−= −− )(sinsin 32
2
1235
21
23
2
1235 1212
Type IV
∫ ++ cxbxa
dx
sincos
dx
dtxxt
x ==22
1 w.r.t.atingdifferentithen
2put evalute to 2sectan
2
2
2
2
222
222 1
1
11
1
221
2
21
2
2
2 ie
t
t
t
t
t
xxx
t
dtx
dtx
dtdx
+−=
+−
+=−=
+=
+== sincoscos&
tansec
222 1
2
1
1
12
222
t
t
tt
txxx
+=
+×
+== cossinsin&
22
2
2 1
2
1
1
1
2
2when
t
tx
t
tx
t
dtdxt
x
+=
+−=
+==∴ sin&cos,,tan
108 KSOU Integral Calculus
∫∫∫∫ +++−=
+++−=
++
++−
+=++
∴cabttac
dt
tcbtta
dt
ct
tb
t
ta
t
dt
cxbxa
dx
2
2
121
2
1
2
1
1
1
2
222
22
2
2
)()()(
)()(
)(
)(
sincos
which is Type I and hence can be evaluated.
Examples
∫ +− 532 Evaluate1
xx
dx
sincos)(
22
2
2 1
2
1
1
1
2then
2Put :Solution
t
tx
t
tx
t
dtdxt
x
+=
+−=
+== sin&cos,,tan
∫∫∫∫+−
2=+−
=++−−
=+
+×−
+−
+=
3
723763
2
15612
2
51
23
1
121
2
G.I.2
222
22
2
2
tt
dt
tt
dt
ttt
dt
t
t
t
tt
dt
)()()(
)(tantan
)()(
123
3
1
3
21
3
21
32
3
21
32
37
1132 11
22
2−=
−×=
+−
=+−−
= −−∫∫ tt
t
dt
t
dt
cx
xx
dx +
−=
+−∴ −∫ 1
22
3
3
1
5321 tantan
sincos
∫ − x
dx
cos)(
53 Evaluate2
2
2
2 1
1
1
2then
2Put :Solution
t
tx
t
dtdxt
x
+−=
+== cos&,tan
∫∫∫∫∫∫
−
=−
=−
=−−+
=
+−−
+=−
∴2
22222
2
2
2
2
14
1
8
28
2
28
2
1513
2
1
153
1
2
53t
dt
t
dt
t
dt
tt
dt
t
t
t
dt
x
dx
)()()(
)(
cos
2
1
2
2
1
24
1
2
12
1
2
12
1
4
1
+
−=
+
−
××=
x
x
t
t
tan
tanloglog
cx
x
x
dx ++
−=
−∴ ∫
12
2
12
2
4
1
53 tan
tanlog
cos
∫ + x
dx
sin)(
23 Evaluate3
22 1
2 and
1
2then
2Put :Solution
t
tx
t
dtdxt
x
+=
+== sin,tan
MCA 11 - Mathematics SVT 109
∫∫∫∫∫+−
+
=++
=++
=
++
+=+
∴1
9
4
3
23
2
3
413
2
413
2
1
43
1
2
23 22
2
2
2
t
dt
tt
dt
tt
dt
t
tt
dt
x
dx
)(sin
5
23
5
2
3532
35
1
3
2
35
323
2 1122
+=+
×=
+
+
= −−∫ tt
t
dttantan
c
x
x
dx +
+=
+∴ −∫ 5
22
3
5
2
231
tantan
sin
Type V
dxxexc
xbxa∫ ++
cossin
sincos
r)denominato of e(derivativr)Denominatoput evaluate to:Solution mlxbxa +=+ (sincos
)sincos()cossin(sincos ie xexcmxexclxbxa −++=+
.separately of efficients-co theequatingby foundout be toconstants are where xxml cos&sin&
amclebmelc =+=− & equations thefrom ie
cxexcmlxdxxexc
xexcmdx
xexc
xexcldx
xexc
xbxa +++=+−+
++=
++ ∫∫∫ )cossinlog(
cossin
sincos
cossin
cossin
cossin
sincosthen
Examples
dxxx
xx∫ +−cossin
sincos
4
23 Evaluate
)sincos()cossin(sincos xxmxxlxx −++=− 4423Put :Solution
24 −=−∴ ml (1)
34 =+ ml (2)
17
5517adding
3412
841641
−=⇒−=
=+×−=−×
ll
ml
ml
)(
)(
17
14
17
34202
17
2024 (1), from =+−=+−=+= lm
)cossinlog(cossin
sincos
cossin
cossin
cossin
sincosxxdx
xx
xxdx
xx
xxdx
xx
xx ++⋅−=+−+
++−=
+−∴ ∫∫∫∫ 4
17
141
17
5
4
4
17
14
4
4
17
5
4
23
cxxx +++−= )cossinlog(417
14
17
5
110 KSOU Integral Calculus
Type VI
)()()()( xxxfdxexf x φφ ′+=∫ where
∫ ∫ ∫ ′+= dxexdxexdxexf xxx )()()( φφ :Solution (1)
∫ dxex x)(φConsider
xx evxuevxu ==′=′= ),(,),( φφput
∫∫ ′−=∴ dxexexdxex xxx )()()( φφφ
have we(1),in thisngsubstituti
∫ ∫ ∫ +=′+′−= cexdxexdxexexdxexf xxxxx )()()()()( φφφφ
Examples
dxx
xex
∫ + 21 Evaluate1
)()(
∫ ∫∫ ∫∫∫ +−
+=
+−
++
=+
−+=
+dx
x
edx
x
edx
x
edx
x
exdx
x
exdx
x
xe xxxxxx
22222 1111
1
1
11
1 :Solution
)()()(
)(
)(
)(
)((1)
dxx
ex
∫ + )(1Consider
xx evx
uevx
u =+−=′=′
+= ,
)(,,
21
1
1
1put
∫∫∫ ++
+=
+−
−+
=+
∴ dxx
e
x
edx
x
e
x
edx
x
e xxxxx
22 11111 )()()()()(
(1)in ngsubstituti
cx
edx
x
edx
x
e
x
edx
x
xe xxxx
++
=+
−+
++
=+ ∫∫∫ 11111 222
2
)()()()(
dxx
xx∫ −−cos
sin)(
1 Evaluate2
dxx
xx
dxx
xdx
x
xdx
x
xdx
x
xx ∫∫∫ ∫∫ −=−
−−
=−−
22
222
22111
:Solution 22 sin
cossin
sincos
sin
coscos
sin
∫ ∫−= dxx
dxx
x22
cosec2
1 2 cot (1)
∫ dxxx 2cosec2
1Consider
21
2cosec
2
1put 2 x
vux
vxu cot,,, −==′=′=
MCA 11 - Mathematics SVT 111
dxxx
xdxx
x ∫∫ +−=222
cosec2
1 2 cotcot
(1)in ngsubstituti
cx
xdxx
dxxx
xdxx
xx +−=−+−=−− ∫∫∫ 22221
cotcotcotcotcos
sin
Other examples
dxx
xx∫ + 41 Evaluate1
sin
cossin)(
dtdxxxtx == cossin,sin 2then Put :Solution 2
cxtt
dtdx
x
xx +==+
=+
∴ −−∫∫ )(sintantansin
cossin 21124 2
1
2
1
12
1
1
dxxx
x∫ +++
))(()(
21
1 Evaluate2
2
2
2121
1Let :Solution
22
2
+++
+=
+++
x
CBx
x
A
xx
x
))((
))(( 21by t throughougmultiplyin 2 ++ xx
))(()( 121then 22 ++++=+ xCBxxAx
3
202121put =⇒++=−= AAx )(,
31
34
1210put −=−=⇒+== CCAx ,
sidesboth on ofefficient -co Equating 2x
3
1
3
2111 =−=−=⇒=+ ABBA
23
1
3
1
13
2
21
122
2
+
−+
+=
+++∴
x
x
xxx
x
))((
∫∫∫∫∫∫ +−
++
+=
+−+
+=
+++∴
231
2
261
132
2
131
132
21
12222
2
x
dxdx
x
x
x
dxdx
x
x
x
dxdx
xx
x
))((
cx
xx +−+++= −
223
12
6
11
3
2 12 tan)log()log(
Type VII
∫ ∫∫∫ ++−+− dxCBxAxdxaxdxxadxxa 2222222 432 (1) )()()(
∫ +++ dxCBxAxqpx 25 )()(
112 KSOU Integral Calculus
θθθ dadxaxdxxa cos,sin ==−∫ put evaluate To(1) 22
∫∫ ∫∫ ∫ +==⋅=⋅−=− θθθθθθθθθθ da
dadaadaaadxxa )cos(coscoscoscossin 212
22222222
θθθθθθθθ 2222222
122222
2
22sinsincossin
sin −⋅⋅+=+=×+= aaaaaa
222
21
2
2
221
2
221
22xa
a
xa
a
xa
a
x
a
xa
a
xa −⋅+=−⋅+= −− sinsin
∫ ++−=−∴ − ca
xaxa
xdxxa 1
22222
22sin
θθθ dadxaxdxxa cosh,sinh)( ==+∫ then put evaluate To2 22
∫ ∫∫∫ +==⋅+=+ θθθθθθθ da
dadaaadxxa )cosh(coshcoshsinh 212
22222222
θθθθθθθθ coshsinhsinh
cosh ×+=⋅+=+⋅= ∫∫ 222
2
222
21
2
222222 aaaad
ad
a
2212
2
221
22
22
221
221
22xa
x
a
xa
a
x
a
xa
a
xaaa ++=+⋅+=++= −− sinhsinhsinhsinh θθθ
∫ +++=+∴ − ca
xaxa
xdxxa 1
22222
22sinh
θθ dadxaxdxax sinh,cosh)( ==−∫ thenput evaluate To3 22
∫∫∫∫∫ −==⋅=⋅−=− θθθθθθθθθθ da
dadaadaaadxax )(coshsinhsinhsinhsinhcosh 122
22222222
a
xaaaad
ad
a 1222222
2222
2
21
22
2−−=−⋅=⋅−= ∫∫ coshcoshsinh
sinhcosh θθθθθθθ
a
xaax
xxa
a
x
a
xa
a
xaa 12
2212
2
221
22
2
22221
221
2−−− −−=−⋅−=−−= coshcoshcoshcoshcosh θθ
ca
xaax
xdxax +−−=−∴ −∫ 1
22222
22cosh
∫∫ ++=++ (3)or (2) (1), form the take willThis evaluate To4 22 .)( dxA
Cx
A
BxAdxCBxAx and hence can be
evaluated.
∫ +++ dxCBxAxqpx 2 evaluate To5 )()(
mBAxlmCBxAxlqpx ++=+++=+ )()( 2 of derivativeput 2
out, found be toconstants are where ml &
∫ ∫∫ ++++++=+++ CBxAxmdxCBxAxBAxldxCBxAxqpx 222 2 then, )()(
MCA 11 - Mathematics SVT 113
∫ +Β++++= dxA
Cx
AxAmCBxAx
l 2232
3
2)(
evaluated. becan hence and (3)or (2) (1), toreduces integral second the
Type VIII
∫ +++ CBxAxqpx
dx2)(
−=−==+ q
tpxdt
tdxp
tqpx
111then
1put
2&
∫∫∫+−+−
−=
+
−Β+
−
−=
+++∴
2222
2
2
2
21111 Ctqtt
p
Btq
p
A
dt
Cqtp
qtp
A
t
t
dt
CBxAxqpx
dx
)()()(
This integral reduces to any one of Type II and hence can be solved.
Type IX
∫∫ ++ dxcbxedxcbxe axax )sin()cos( and
parts.by n integratio use tohave weevaluate To
∫∫ +=+= dxcbxeSdxcbxeC axax )sin(&)cos(Let
∫ += dxcbxeC ax )cos(Consider
b
cbxvcbxvaeueu axax )sin(
),cos(,,+=+=′=′=put
∫ +−+= dxcbxeb
a
b
cbxeC ax
ax
)sin()sin(
aScbxebC ax −+= )sin(
)sin( cbxebCaS ax +=+∴ (1)
∫ += dxcbxeS ax )sin(Consider
b
cbxvcbxvaeueu axax )cos(
),sin(,,+−=+=′=′=put
∫ +++−=∴ dxcbxeb
a
b
cbxeS ax
ax
)cos()cos(
aCcbxeabS ax ++−= )cos(
)cos( cbxeaCbS ax +−=− ie (2)
[ ])cos()sin()
)cos()(
)sin()(
cbxbcbxaeSb(a
cbxbeabCSbb
cbxaeabCSaa
ax
ax
ax
+−+=+
+−=−×
+=+×
22
2
2
adding
2
1
114 KSOU Integral Calculus
[ ])(
)cos()sin(22 ba
cbxbcbxaeS
ax
++−+=∴
[ ])cos()sin()(
)cos()(
)sin()(
cbxacbxbeCba
cbxaeCaabSa
cbxbeCbabSb
ax
ax
ax
+++=+
+−=−×
+=+×
22
2
2
gsubtractin
2
1
[ ])(
)sin()cos(22 ba
cbxbcbxaeC
ax
++++
=∴
Examples
∫ ∫∫∫ +=+= dxedxxedxxxedxxxe xxxx sinsin]sin[sincossin)( 2222
2
15
2
15
2
1231
cxx
exx
e xx +−+−=5
2
2
1
29
5552
2
1 22 )cossin()cossin(
49
2223
2
1
32
12
2
1
2
121
2
12 333323
+++×=+=+=
3
∫∫∫∫ )sincos(cos)cos(cos)(
xxe
edxxedxedxxedxxe x
xxxxx
cxxee xx
+++= )sincos( 2223266
33
Exercise
x w.r.t.following theIntegrate
222
1
13
12
11
1
x
e
xxx
x x
−−
−− sin
)()(logcos
)(sin
)(
))(()(
)()()(
)tan(tan
sec)(
416
31
235
24
22
2
+−++−
+ xx
x
xx
x
xx
x
186
149
222
548
1
27
222
3
+−
+++
+−
−−
xx
x
xx
x
x
xx)()()(
xxxe
x
x x
cos)(
sincos)(
)(
)()(
43
112
2
111
2
110
2 +−+++
xex
x
xxxx
xx
)cos(
)sin()(
sincos)(
cossin
cossin)(
++
+++
1
115
94
114
54
3213
22
235218246171
116 22 +−−−−
+xxxxx
xx n)()()(
)()(
xxexxxxx
x 242111
120
4321
119 2
22sinsin)(
)()(
)()(
−++++
xexexxe xxx 34332 2423322 sin)(cos)(coscos)(
MCA 11 - Mathematics SVT 115
Definite Integrals
∫ += cxgdxxfbaxf )()(),()( andinterval in the definedfunction a beLet
[ ] [ ]cagcbgaxbx +−+== )()( ieat integral theof value theminusat integral theof valueThe
∫−b
adxxfagbg )()()( as denoted and integral Definite as defined is ie
∫ ∫ −== )()()()()( agbgdxxfxgdxxfb
a then If ie
limit.lower called is andlimit upper called is ab
Examples
∫ +−2
1
23 32 Evaluate(1) dxxx )(
33
2
4
16
3
1643
3
2
4
123
3
22
4
23
32
432 :Solution
1342
1
342
1
23 −+−+==
+−−
⋅+×−=
+×−=+−∫ x
xxdxxx )(
12
25
12
836484
3
2
4
1
3
167 =+−−=+−−=
dxx
x∫ −
−1
0 2
21
1 Evaluate2
)(sin)(
dtdxx
tx =−
=−2
1
1
1then Put :Solution sin
211when 000when 11 π====== −− sin,sin, txtx
2423
1
31
332
0
32
0
21
0 2
21 πππ
π=
=
==
−∴ ∫∫
− tdttdx
x
x)(sin
∫− ++
1
1 2 52 Evaluate3 dx
xx
dx)(
+−−
+=
+=
++=
++−−
−
−
−− ∫∫ 2
11
2
1
2
11
2
1
2
1
2
1
2152 :Solution 11
1
1
11
1 22
1
1 2tantantan
)(
x
x
dxdx
xx
dx
8
042
10
2
11
2
1 11 ππ =−=−= −− tantan
∫ +
π
0 34 Evaluate4
x
dx
cos)(
2
2
2 1
1
1
2 then
2Put :Solution
t
tx
t
dtdxt
x
+−=
+== cos,tan
∞======2
when 000when ππ tan,tan, txtx
116 KSOU Integral Calculus
( )∞
−∞∞∞
=
+=
−++=
+−+
+=+ ∫∫∫∫
0
1
0 220 220
2
2
2
0 77
1
7
2
1314
2
1
134
1
2
34
t
t
dt
tt
dt
t
tt
dt
x
dxtan
)()(
)(
)(cos
π
7227
10
7
1
7
1 11 ππ =×=−∞= −− tantan
Properties of Definite Integrals
∫∫ =b
a
b
adyyfdxxf )()(.1
∫∫ −=a
b
b
adxxfdxxf )()(.2
bcadxxfdxxfdxxfb
c
c
a
b
a<<+= ∫∫∫ where3 )()()(.
∫∫∫∫ −+=−=b
a
b
a
aadxxbafdxxfdxxafdxxf )()()()(. also 4
00
=∫
∫−function oddan is if
functioneven an is if5
0
20
)(
)()(.)(
xf
xfdxxf
adxxfa
a
−=−
=−=∫
∫)()(
)()()(.)(
xfxaf
xfxafdxxf
adxxfa
2 if
2if6
0
22
0
0
Examples
∫ +2
0 Evaluate1
πdx
xx
xnn
n
sincos
sin)(
∫∫∫∫ −=
−+
−
−
=+
=aa
nn
n
nn
n
dxxafdxxfdxxx
xdx
xx
xI
00
2
0
2
0 using
22
2Let :Solution )()(
sincos
sin
sincos
sin ππ
ππ
π
∫ += 2
0
πdx
xx
xnn
n
cossin
cos
212 2
0
2
0
2
0
2
0
2
0
ππππππ==⋅=
++=
++
+=∴ ∫∫∫∫ ]
)sin(cos
)cos(sin
cossin
cos
sincos
sinxdxdx
xx
xxdx
xx
xdx
xx
xI
nn
nn
nn
n
nn
n
4
π=∴ I
∫ +
π
0 1 Evaluate2 dx
x
xx
sin
sin)(
∫∫∫∫ −=−+
−−=+
=aa
dxxafdxxfdxx
xxdx
x
xxI
0000 using
11Let :Solution )()(
)sin(
)sin()(
sin
sin ππ
πππ
MCA 11 - Mathematics SVT 117
∫ +−=
π π0 1
ie dxx
xxI
sin
sin)(
∫∫∫∫∫ +=
+=
+−+=
+−+
+=∴
πππππππππ
00000 111112
x
xdx
x
xdx
x
xxxxdx
x
xxdx
x
xxI
sin
sin
sin
sin
sin
sin)(sin
sin
sin)(
sin
sin
∫ ∫∫∫ −=−=−+
−=π πππ
ππππ0 0
2
0 2
2
0 11
1dxxdxxxdx
x
xxdx
xx
xxtantansec
cos
sinsin
)sin)(sin(
)sin(sin
[ ] [ ] [ ]0001 000
2
0+−−+−=+−=
+−= ∫∫∫ tansectansectansecsectansec πππππππ ππππ
xxxdxdxxxx
[ ] [ ] [ ]πππππ +−=−+−= 211
)( 22
−=∴ ππI
∫ +3
6 1 Evaluate3
π
π x
dx
tan)(
∫∫∫∫ −=
−+
−
−
=+
=b
a
b
adxxafdxxf
xx
dxx
xx
dxxI )()(
sincos
cos
sincos
cos using
22
2Let :Solution 3
6
3
6
π
π
π
π ππ
π
∫ +=∴ 3
6
π
πdx
xx
dxxI
cossin
sin
3
6
3
6
3
6
3
6
12
π
π
π
π
π
π
π
π
=⋅=
++=
++
+=∴ ∫∫∫ xdxdx
xx
xxdx
xx
dxx
xx
dxxI
sincos
sincos
cossin
sin
sincos
cos
663
πππ =−=
12
π=∴ I
Exercise
Evaluate the following
∫∫∫ +42
0 210 232
11
ππ
dxx
edxxdx
x
x xe
cos)(log)(
cos
sin)(
tan
∫∫∫ −
−−
− −
1
1
1
0
12
659
42
1
21
dxxedxxxx
dx x)(tan)()(
∫∫∫ ++++
∞ 42
00 20 222219
1187
ππ
θθ dxx
dxx
xbxa
dx)tanlog()(
))(()(
sincos)(
∫∫∫ ++
+−+
1
0 20
2
0 1
11211
210
2dx
x
x
xx
dxxdx
xafxf
xfa
)(
)log()(
cossin)(
)()(
)()(
π
,)(,)(,)(,)(,log)(,)(,)(,4
82
72
621
45
57
31
413124
(1) : Answersππππ
abee −−−
( ) 28
121222
111028
9 log)(log)()(log)(πππ +a
118 KSOU Integral Calculus
Reduction formulae
I. ∫ ∫=2
0 evaluate tohence andfor formulareduction obtain the To
π
dxxdxxI nnn sinsin
∫ ⋅= − dxxxI nn sinsin 1 :Solution
xvxxnuxvxu nn cos&cossin)(,sin&sin −=−=′=′= −− 21 1put
∫∫ −−+−=−+−=∴ −−−− dxxxnxdxxxnxxI nnnnn )sin(sin)(cossincossin)(cossin 221221 111
nnnnnn InInxdxxndxxnxx )()(cossinsin)(sin)(cossin 1111 2
121 −−−+−=−−−+−= −−−− ∫∫
21 11 −
− −+−=−+∴ nn
nn InxxInI )(cossin)(
21 11(1 ie −
− −+−=−+ nn
n InxxIn )(cossin)
2
1 1 ie −
− −+−= n
n
n In
n
n
xxI
cossin
oddor even is as accordingor is integral ultimate the 10 nII
∫∫ −==== xdxxInxdxIn cossin10 odd is If 1even is If
then If2
0∫=π
dxxI nn sin
22
2
0
1 10
1−−
− −+=−+
−= nn
n
n In
nI
n
n
n
xxI
πcossin
642 4
5
2
31
2
311−−− −
−×−−×−=
−−×−=−=∴ nnnn I
n
n
n
n
n
nI
n
n
n
nI
n
nI
generalin
×××−−×−
×××−−×−
=
odd. is if 132
231
even. is if 22
1231
nn
n
n
n
nn
n
n
n
I n
LL
LL
π
∫ =×××==2
0
66 32
5
22
1
4
3
6
5 (1) Eg.
π ππdxxI sin
∫ =××==2
0
55 15
81
3
2
5
42
π
dxxI sin)(
II. ∫∫=2
0 evaluate toandfor formulareduction obtain the To
π
dxxdxI nnn coscos
obtain weIin as partsby n integratio using :Solution
n
n
n In
n
n
xxI
11 −+=− sincos
∫∫∫∫ −=
−==
aann
n dxxafdxxfdxxdxxI0000
using 2
iffurther and22
)()(coscosππ π
MCA 11 - Mathematics SVT 119
Idxxn iswhich 2
0∫=π
sin
35
161
3
2
5
4
7
6 (1) Eg.
2
0
7 =×××=∴ ∫π
dxxcos
256
35
22
1
4
3
6
5
8
72
2
0
8 πππ
=××××=∫ dxxcos)(
III. ∫= dxxI nn tanfor formulareduction obtain the To
∫ ∫∫∫ −−−− −⋅=−⋅=⋅= dxxdxxxdxxxdxxxI nnnnn
2222222 1 :Solution tansectan)(sectantantan
formula. required theis which 1 2
1
−
−−
−= n
n
n In
xI
tan
IV. ∫= dxxI nn cotfor formulareduction obtain the To
∫ ∫ ∫∫ −−−− −=−⋅=⋅= dxxdxxxdxxxdxxxI nnnnn
2222222 cosec1cosec :Solution cotcot)(cotcotcot
2
1
1 −
−−
−−= n
n
n In
xI
cot
V. ∫= dxxI nn secfor formulareduction obtain the To
∫ ⋅= − dxxxI nn
22 :Solution secsec
xvxxxnuxvxu nn tan&tansecsec)(,sec&sec =⋅−=′=′= −− 322 2put
∫∫ −−−=⋅−−=∴ −−−− dxxnxxdxxxnxI nnnnn )1(secsec)2(tansectansec)2(tansec 222222
∫ ∫ −− −+−−= dxxndxxnxx nnn 22 22 sec)(sec)(tansec
22 22 −
− −+=−+∴ nn
nn InxxInI )(tansec)(
22 221 −
− −+=−+ nn
n InxxIn )(tansec)(
formula.reduction required theis which 1
2
1 2
2
−
−
−−+
−=∴ n
n
n In
n
n
xxI
tansec
VI. ∫= dxxI nn cosecfor formulareduction obtain the To
∫ ⋅= − dxxI nn
22 coseccosec :Solution
xvxxxnuxvxu nn cot&cot)(,& −=⋅−−=′=′= −− coseccosec2coseccosecput 322
dxxxnxxdxxxnxxI nnnnn )()(cotcot)(cot 1coseccosec2coseccosec2cosec 222222 −−−−=⋅−−−=∴ ∫∫ −−−−
2222 22coseccosec2cosec2cosec −
−−− −+−−−=−+−−−= ∫∫ nnnnnn InInxxdxxndxxnxx )()(cot)()(cot
120 KSOU Integral Calculus
22 2cosec2 ie −
− −+−=−+ nn
nn InxxInI )(cot)(
22 2cosec1 ie −
− −+−=− nn
n InxxIn )(cot)(
formula.reduction required theis which 1
2
1
cosec ie 2
2
−
−
−−+
−−= n
n
n In
n
n
xxI
)(
)(
)(
cot
VII. ∫∫=2
0 evaluate tohence and of formulareduction obtain the To
π
dxxxdxxxI nmnmnm cossincossin,
∫ ⋅= − dxxxxI nmnm coscossin,
1 :Solution
xvxxnxxmuxvxxu nmnmnm sin&cossin)(cossin,cos&cossin =⋅−−⋅=′=′= −+−− 2111 1put
( ) dxxxnxxmxxxI nmnmnmnm sincossin)(cossinsincossin, ∫ −+−− −−−⋅=∴ 2111 1
∫∫ −+−+ ⋅−+−= dxxxndxxxmxx nmnmnm 2211 1 cossin)(cossincossin
∫ −−+ −−+−= dxxxxnmIxx nmnm
nm 2211 11 cos)cos(sin)(cossin ,
∫ −−+−= −−+ dxxxxxnmIxx nmnmnm
nm )cossincos(sin)(cossin ,211 1
nmmmnmnm InInmIxx ,,, )()(cossin 11 2
11 −−−+−= −−+
211 11 ie −
−+ −+=−++ nmnm
nmnmnm InxxxInmII ,,,, )(cossin)(
2
11 1 ie −
−+
+−+
+= nm
nm
nm Inm
n
nm
xxI ,,
)(
)(
cossin
∫=2
0 if formula,reduction required theiswhich
π
dxxxI nmnm cossin,
22
2
0
11 10
1then −−
−+
+−+=
+−+
+
= nmnm
nm
nm Inm
nI
nm
n
nm
xxI ,,,
)(
cossinπ
21
−+−=∴ nmnm I
nm
nI ,,
have wely,continuous formulareduction thisapplying
×××−−×−×
+××
−+−×
+−
×××−−×−×
+××
−+−×
+−
+×
+××
−+−×
+−
=
even is &even is if 22
1
2
31
2
1
2
31
odd is &even is if 13
2
2
31
2
1
2
31
evenor odd & odd is if 1
1
3
2
2
31
mnm
m
m
m
mnm
n
nm
n
mnm
m
m
m
mnm
n
nm
n
mnmmnm
n
nm
n
I nm
πLLLL
LLLL
LL
,
Examples
60
1
6
1
8
2
10
41
2
0
5555 =××== ∫
π
dxxxI cossin)( ,
693
8
7
1
9
2
11
42
2
0
5656 =××== ∫
π
dxxxI cossin)( ,
MCA 11 - Mathematics SVT 121
3465
481
3
2
5
4
7
6
9
1
11
33
2
0
4747 =×××××== ∫
π
dxxxI cossin)( ,
2048
5
22
1
4
3
6
5
8
1
10
3
12
54
2
0
6666
πππ
=××××××== ∫ dxxxI cossin)( ,
∫ −
1
0 2
9
1 Evaluate5 dx
x
x)(
θθθ ddxx cos,sin ==put :Solution 2
1when 00when πθθ ==== ,, xx
315
1281
3
2
5
4
7
6
9
8
11
222
0
9
0
9
0 2
91
0 2
9
=××××==⋅=−
⋅=− ∫∫∫∫
πππ
θθθ
θθθ
θ
θθθd
dddx
x
xsin
cos
cossin
sin
cossin
∫ −
a
xax
dxx2
0 2
3
2 Evaluate6)(
∫∫ −−=
−
aa
axa
dxx
xax
dxx 2
0 22
32
0 2
3
2 :Solution
)( θθθ dadxaax cos,sin ==−put
212when
210when
πθθπθθ =⇒==−=⇒−== sin,sin, axx
∫∫∫ −−−+++=+=
−
⋅+=∴2
2
2
2
2
2
23333
222
3
3311
G.I.π
π
π
π
π
πθθθθ
θθθθ
θ
θθθda
a
daa
aa
daaa)sinsinsin(
cos
cos)sin(
sin
cos)sin(
+++=
+++= ∫ ∫∫ ∫∫∫ −− −−−
2
2
22
2
2
2
2
2
2
2 0
23233 600331π
π
ππ
π
π
π
π
π
π
πθθθθθθθθθθ dadddda sinsinsinsin
333
2
5
2
3
22
16
22aaa
ππππππ =
+=
××+
−=
Exercise
Evaluate the following
( )∫∫∫ −1
0
24
0
7
0
5 236
13231 dxxxdxxxd )(sin)(sin)(ππ
θθ
( ) ∫∫∫ −−+
∞ 2
0
1
0
26
0 22615
14 2
5
27 dxxxdxxx
x
dx)()()(
∫∫∫ +−π
θθθθ
π
π
π
π 0
254
1
19cosec87
2
6
2
4
ddxxdxxcos
cossin)()(cot)(
( ) ∫∫∫ −+−
a
xa
dxxdx
x
xdx
x
x0 22
71
0 42
31
0 4
7
121
111
10 )()()(
( ) ,)(,log)(,)(,)(,)(,)(,)(,)(,3
28932
8
3
4
3118
12
837
8
56
256
55
15
84
256
33
35
162
45
8 (1) : Answers ++−πππππ
35
1612
24
111
3
110
7a)(,)(,)(
122 KSOU Integral Calculus
127 KSOU Matrix Theory
DIFFERENTIAL EQUATIONS
An equation which consists of one dependent variable and its derivatives with respect to one or more independent variables iscalled a 'Differential Equation'. A differential equation of one dependent and one independent variable is called 'OrdinaryDifferential Equation'. A differential equation having one dependent and more than one independent variable is called'Partial Differential Equation'.
Examples of ordinary differential equation
0201 =−=− dxydxxx
y
dx
dy..
04032
2
==+dx
yddybydxax ..
164
4326
43
7325
+−+−=
+−−+=
yx
yx
dx
dy
yx
yx
dx
dy..
08172
32
2
2
=+
+= dyxdxy
dx
dy
dx
yda ..
0321003492
22
2
2
2
=+−=+
− y
dx
dyx
dx
ydxy
dx
dy
dx
yd..
Order and Degree of a differential equation
Order
The order of the highest derivative occurring in a differential equation is called 'Order' of the differential equation.
Degree
The highest degree of the highest order derivative occurring in a differential equation (after removing the radicals if any) iscalled 'Degree' of the differential equation.
In the examples given above (1), (2), (3), (5), (6) & (8) are of order one and degree one, (4), (9) & (10) are of order twoand degree one where as (7) is of order two and degree two after removing the radicals.
Formation of Differential Equation
Differential equations are formed by eliminating the arbitrary constants. Arbitrary constants are eliminated by differentiation.
Examples
222 from constant thegeliminatinby equation aldifferenti theForm1 ayxa =+)(
222 :Solution ayx =+
have we w.r.t.atingdifferenti ,x
equation. aldifferenti theiswhich 0022 =+⇒=+ dyydxxdx
dyyx
cmxycm += from ' geliminatinby equation aldifferenti theForm(2) ''&'
cmxy += :Solution
mdx
dyx = have we w.r.t.atingdifferenti ,
have we w.r.t.atingdifferentiagain ,x
equation. aldifferenti required theis 02
2
=dx
yd
xbxayba 33 from ' geliminatinby equation aldifferenti Obtain the(3) sincos''&' +=
xbxay 33 :Solution sincos +=
have we w.r.t.atingdifferenti ,x
xbxadx
dy3333 cossin +−=
have we w.r.t.atingdifferentiagain ,x
yxbxadx
yd93939
2
2
−=−−= sincos
equation. aldifferenti required theis which 09y ie2
2
=+dx
yd
Note :- It can be seen from the above examples that the order of the differential equation depends on the number ofarbitrary constants in the equation. ie. if arbitrary constant is one then order is one and if the arbitrary constants aretwo then the order is two.
Solution of equations of first order and first degree
I. Variable Separable
,separable' Variable' called is typethen this0 as written becan equation aldifferentigiven theIf =+ dyygdxxf )()( solution
is obtained by integration
∫ ∫ =+ Constant ie dyygdxxf )()(
011 Solve 1. Eg. 22 =−+− dxyxdyxy
22 11by t throughoudivide :Solution yx −−
011
becomesequation 22
=−
+− x
dxx
y
dyy
Constant11
gintegratin22
=−
+− ∫∫
x
dxx
y
dyy
Cxy −=−−−− 22 11 ie
solution. theis 11 ie 22 Cxy =−+−
013 Solve 2. Eg. 2 =−+ dyyedxye xx sec)(tan
have we1by t throughoudividing :Solution ),(tan xey −
124 KSOU Differential Equations
01
3 2
=+−
dyy
ydx
e
ex
x
tan
sec
∫∫ =+−
Constant1
3 g,integratin
2
dyy
ydx
e
ex
x
tan
sec
Cyex logtanlog)log( =+−− 13
Ce
yx
log)(
tanlog =
− 31 ie
solution. theis 1 3)(tan xeCy −=∴
xyyxdx
dyxy +++=1 Solve 3. Eg.
))(( yxdx
dyxy ++= 11 isequation given :Solution
dxyxdyxy ))(( ++= 11 ie
)( yx +1by t throughoudivide
x
dxx
y
dyy )( +=+
1
1then
dxx
xdy
y
y
+=
+
−+ 1
1
11 ie
dxx
dyy
+=
+
− 11
1
11 ie
soluton theis 1 g,integratin cxxyy ++=+− log)log(
22 Solve 4. Eg. adx
dyyx =− )(
uyx =−put :Solution
dx
du
dx
dy =−1then dx
dy
dx
du =−1 ie
22 1 becomesequation given adx
duu =
−
222 ie uadx
duu −=−
dxau
duu =− 22
2
ie
dxduau
aau =
−+−22
222
ie
dxduau
a =
−+
22
2
1 ie
cxau
au
a
au +=
+−+ log
2 gintegratin
2
MCA 11 - Mathematics SVT 125
cxayx
ayxayx +=
+−−−+− log
2 ie
solution. theis 2
ie cyayx
ayxa +=+−−−
log
0 when 0given Solve 5. Eg.2
=== − xyxedx
dy xy
2
isequation given :Solution xy exedx
dy −⋅=
dxxedye xy 2
ie −− =
cee xy +−=− −− 2
2
1 g,integratin
2
1
2
1100when −=⇒+−=−∴== ccyx ,
2
1
2
1 issolution
2
−−=−∴ −− xy ee 12 ie2
+= −− xy ee
II. Homogeneous Equation
sexpression shomogeneou are where0or type theofEquation ),(&),(),(),(),(
),(yxgyxfdyyxgdxyxf
yxf
yxf
dx
dy =+=
Equation'. sHomogeneou' a called is degree same of &in yx
vxy =put solve To
dvxdxvdydx
dvxv
dx
dy +=+= or then
solved. becan hence and form separable variable toreducesequation given the this,ngsubstitutiby
2232 Solve (1) Eg. xydx
dyxy +=
dx
dvxv
dx
dyvxy +== then put :Solution
2222 32 becomesequation given xvxdx
dvxvvx +=
+
2by t throughoudivide x
132then 2 +=
+ v
dx
dvxvv
1322 ie 22 +=+ vdx
dvvxv
12 ie 2 += vdx
dvvx
x
dx
v
dvv =+1
2 ie
2
cxv loglog)log( +=+1 have weg,integratin 2
126 KSOU Differential Equations
cxx
ylogloglog +=
+1 ie
2
2
cxx
xyloglog
)(log +=+
2
22
ie
cxxxy logloglog)(log +=−+ 222 ie
322 3 cxcxxy logloglog)(log =+=+∴322 issolution cxxy =+∴
011 Solve (2) Eg. =
−+
+ dy
y
xedxe y
xy
x
dvydyvdxvyx +== then put :Solution
01(1 becomesequation =−+++ dyvedvydyve vv )())(
0 ie =−++++ dyvedyedvyedyvedvydyv vvvv
01 ie =++−++ dveydyeevev vvvv )()(
01 ie =+++ dveydyev vv )()(
)( vevy +by t throughoudivide
01
=+
++∴
v
v
ev
dve
y
dy )(
cevy v log)log(log =++ g,integratin
cevy v log)(log =+ ie
cevy v =+ )( ie
cey
xy y
x
=
+
solution theis ie cyex yx
=+
Equations reducible to homogeneous equatins
0or Give 222111222
111 =+++++++++= dycybxadxcybxa
cybxa
cybxa
dx
dy)()(
solved. becan hence andequation shomogeneou toreducesit then put If(i) Case 112
1
2
1 tybxab
b
a
a =+=
kYyhXxb
b
a
a +=+=≠ &put If (ii) Case2
1
2
1
00 such that out found be toconstants are where 222111 =++=++ ckbhackbhakh &&
toreducesequation given then
solved. becan hence and shomogeneou is which 0or 221122
11 =+++++
= dYYbXadXYbXaYbXa
YbXa
dX
dY)()(
MCA 11 - Mathematics SVT 127
322
1 Solve (1) Eg.
++−+=yx
yx
dx
dy
dx
dt
dx
dytyx =+=+ 1 then put :Solution
32
11 becomesequation given
+−=−∴
t
t
dx
dt
32
23
32
3211
32
1
++=
+++−=+
+−=
t
t
t
tt
t
t
dx
dt
dxdtt
t =++
23
32 ie
∫ +=++
cxdtt
t
23
32 gintegratin
mlltmtlt ++=++=+ 332332put )(
3
223 =⇒=∴ ll
3
5
3
432332 =−=−=⇒=+ lmml
∫ ∫ ∫ ∫
++=
++=
++
++=+∴
3
2
9
5
3
2
329
51
3
2
233
5
23
23
3
2tt
t
dtdt
t
dtdt
t
tcx log
++++=+∴
3
2
9
5
3
2yxyxcx log)(
solution theis 3
2
9
5
3
2
3
1 ie
+++=+ yxycx log
dxyxdyyx )()( 3232 Solve (2) Eg. −+=−+
kYyhXx +=+= ,put :Solution
such that choose k h &
32 =+ kh (1)
32 =+ kh (2)
12323133gsubtractin
3212
62421
=−=−==⇒=
=+×=+×
hkhh
kh
kh
,
)(
)(
dXYXdYYX )()( 22 toreducesequation given The +=+
dvXdXvdYvXY +== then put
dXvXXdvXdXvvXX )())(( 22 +=++∴
Xby t throughoudivide
dXvdvXvdXvv )()()( 21222then +=+++
02212 2 =++−−+∴ dvXvdXvvv )()(
021 ie 2 =++− dvXvdXv )()(
01
2 ie
2=
−++v
dvv
X
dX )(
128 KSOU Differential Equations
∫ =−
++ Constant1
2 g,integratin
2v
dvvX
)(log
∫ ∫ =−
+−
+ Constant1
21
2
2
1 ie
22 v
dv
v
dvvXlog
cv
vvX loglog)log(log =
+−×+−+
1
1
2
121
2
1 ie 2
c
X
YX
Y
X
YX loglogloglog 2
1
1212 ie
2
2
=+
−+
−+
cXYXYXXYX log)log()log(log)log(log 22222 ie 22 =+−−+−−+
cXYXYXYXY log)log()log()log()log( 222 ie =+−−+−++
cXYXY log)log()log( 23 =+−−
23
ie cXY
XYlog
)(log =
+−
11but 23 −=−=+=−∴ yYxXXYcXY ,)()(
solution. theis 223 )()( −+=−∴ xycxy
III. Linear Equation (Leibnitz's Equation)
Equation'.Linear ' a called is of functions are where type theofEquation xQPQPydx
dy&=+
∫ dxPebyequation given theof sidesboth multiply solution thefind To
∫∫ =
+
dxPdxPQeePy
dx
dythen
∫∫∫ =+dxPdxPdxP
QeyPeedx
dy ie
∫∫ =
dxPdxPQeye
dx
d ie
∫ +=∴ ∫∫ solution. required theis cdxQeyedxPdxP
∫ +==+ ∫∫ cdyQexeQPQPxdy
dx dyPdyP issolution its andequation linear a also isy offunction are where -: Note &
xxydx
dycostan =+equation theSolve (1) Eg.
equation standard with theComparing :Solution
xQxP cos&tan ==
∫∫ == xdxxdxP seclogtan
xee xdxPsecseclog ==∫
MCA 11 - Mathematics SVT 129
∫ +=∴ ∫∫ cdxQeyedxPdxP
isSolution
∫∫ +=+=+⋅= cxcdxcdxxxxy 1 ie seccossec
0(1 (2) Eg.12 =−++
−− dyexdxy y )() tan
dyy )2(1by t throughoudivide :Solution +
01 2
1
=+
−+∴−−
)(
tan
y
ex
dy
dx y
11 ie
22
1
+=
++
−−
y
e
y
x
dy
dx ytan
equation standard with Comparing
11
122
1
+=
+=
−−
y
eQ
yP
ytan
∫ −= ydyP 1tan cdyeQxedyPdyP
+=∴ ∫∫ ∫ isSolution
cdyey
exe y
yy +
+=
−−
−
∫−
11
1 tan2
tantan
)1( ie
cycy
dyxe y +=+
+= −∫
− 12
tan tan1
ie1
cyxe y +=∴ −− 1tan tan isSolution 1
Bernoulli's Equation
Equation. sBernoulli' called is of functions are whereequation The xQPQyPydx
dy n &=+
Qy
P
dx
dy
yy
nnn =+ −1
1 then by dividesolution thefind To
zyzy
nn
== +−−
11
ie 1
put
x w.r.t.atingdifferenti
dx
dz
dx
dyyn n =+− −)( 1 )(
)(1
1
11 ie −≠
+−= n
dx
dz
ndx
dy
yn
QPzdx
dz
n=+
+−∴
)( 1
1 becomesequation
solved. becan hence andequation linear a is which 11 ie QnPzndx
dz)()( +−=+−+
QnPzndy
dzQxPx
dy
dx n )()( 11by given issolution whoseequation, sBernoulli' also is -: Note +−=+−+=+
yQP of functions are where &
130 KSOU Differential Equations
xyxydx
dysectan 2 Solve (1) Eg. =+
2by t throughoudivide :Solution y
xxydx
dy
ysectan =+ 11
then 2
dx
dz
dx
dy
yz
y=−=
2
1 then
1put
xzxdx
dzsectan =⋅+−∴ becomesequation
xzxdx
dzsectan −=⋅− ie
xQxP sec,tan −=−= relinear whe iswhich
∫ ∫ =−= xdxxdxP coslogtan
xee xdxp coscoslog ==∫
cxcdxcdxxxxz +−=+−=+⋅−=∴ ∫∫ 1 isSolution cosseccos
cxy
x +−=∴ cos isSolution
yexx
y
dx
dy x sec)()(
tan +=+
− 11
Solve (2) Eg.
xexx
y
dx
dyy )(
)(
sincos +=
+− 1
1 isequation given the:Solution
dx
dz
dx
dyyzy == cos,sin then put
hereequation wlinear a is which 11
becomesequation xexx
z
dx
dz)( +=
+−∴
xexQx
P )(, +=+
−= 11
1
)log()(
xdxx
dxP +−=+
−= ∫∫ 11
1
xeee xxdxP
+===∴ ++−∫
1
11
11
log)log(
∫∫ +==+
+=+
⋅∴ cedxedxx
exx
z xxx
)()(
1
11
1
1 isSolution
cex
y x +=+
∴1
isSolution sin
MCA 11 - Mathematics SVT 131
IV. Exact Differential Equation
if Exact'' tosaid isequation the of functions are hereequation w aldifferenti thebe 0Let ,&& yxNMdyNdxM =+
solution thefind toand x
N
y
M
∂∂=
∂∂
∫ dxMConsider (1)
∫ dyNyx takeandconstant a as treating w.r.t.done isn integratio thewhere (2)
Nxy incontaining terms theomitting w.r.t.done isn integratio thehere
Constant. is (2)(1) issoltuion then the +
02322 Solve (1) Eg. 223 =+−+−− dyxyxdxyxyx )()(
equation standard with Comparing :Solution
xyxNyxyxM 2,322 223 −−=+−−=
22,22 −−=∂∂−−=
∂∂
xyx
Nxy
y
M exact isequation hence
x
N
y
M
∂∂=
∂∂∴
xxyyxx
xxyx
yx
dxyxyxdxM 3222
3224
23222242
24
23 +−−=+−⋅−⋅=+−−= ∫∫ )( (1)
∫∫∫ ⋅=−−= dydyxyxdyN 022 )( (omitting the terms which contain x)
= Constant
232
22 issolution
224 cxxy
yxx =+−−∴
cxxyyxx =+−− 64 ie 224
0 Solve (2) Eg. 222222 =−−+−+ dyybyxdxxayx )()(
equation standard with Comparing :Solution
ybyyxNxaxyxM 232223 −−=−+= ,
xyx
Nxy
y
M22 =
∂∂=
∂∂
,
exact isequation ∴∂∂=
∂∂
x
N
y
M
constant) a as (treating 224
consider solution,for 22224
223 yxayxx
dxxaxyxdxM −+=−+= ∫∫ )(
xdyybydyybyyxdyN contain which terms theomitting 23232 ∫∫∫ −−=−−= )()(
24
224 yby −−=
424224 issolution
22422224 cybyxayxx =−−−−+∴
cybyxayxx =−−−+ 22422224 222 ie
132 KSOU Differential Equations
Exercise
Solve the following
yyx exedx
dyxy
dx
dyyx 323222 420111 −− +==++− )()()()(
dxdyyxyyy
xx
dx
dy =++
+= )cos()()cos(sin
)log()( 4
123
yedx
dyx
dx
dyya
dx
dyxy −=++
+=− 1165 2 )()()(
x
y
x
y
dx
dydxyxdxydyx sin)()( +=+=− 87 22
01251521009 332 =−+−++=+− dyyxdxyxdyyxdxyx )()()()()(
33
4212022316411
+−−−==−−+−−
xy
xy
dx
dydyxydxyx )()()()(
1 when 2 if 21413 32 ==−==+ xyxyxdx
dyxy
dx
dyxx )()(loglog)(
633 16215 yxydx
dyxxyxy
dx
dy =+=+ )(sincos)(
0218117 22212 =+−+−+−=+ − dyyyxxdxyyxydyxydxy )sectan()tan()()(tan)()(
01
120032
194
22
3=−++
+
+=−+ dyyxxxdyy
xydy
y
xydx
y
x)sinlog(cos)()(
Answers
cexcyx
ycxxyycxeecyxy
x yyx =−+++=+==+−=−− ))(()(tan)(logsin)()(log 2152
43823211
(1) 2332
37
31222
3
21039287116
−−=+=
==++=−+ − yxcyxcy
y
xcxxycxyxycex y )()(log)()(tan)()())(()(
( )( ) 32 where
2152
2152121287
4
5211
21
1
22 −=−=
−−++=+−=+−+− yYxX
XY
XYcYXYXcyxyx ,)()()log()()(
12
516121541214
3
113 25322123 2
=
+++==+−+= −− cxyxxceyexycxxy xx )(sin)()()(loglog)( sin
cyxxxycyyxcyyxxyyxceyx y =++=−=+−++−=−−− cos)log()()(tantan)(tan)( tan 201918117 32221 1
MCA 11 - Mathematics SVT 133