mathematics mca11

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


ryx −→→°→ ,, 0270 Asθ

\ 1270 −=−=°r

rsin

00

270 ==°r

cos

−∞=−=°0

270r

tan


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


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°


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.


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


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( =−°+°−°

+°−°+°


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


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 −=−−=


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 −=


)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


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


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


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


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


π−∴

6. Find the modulus & amplitude of 1 – i.

211 :Solution =+=z

4

2

12

1

πθθ

θ−=→

−==

==

r

y

r

x

sin

cos


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


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


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


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


(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)()(


(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°


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


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


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


.......)(cos..........coscos).............cos( 4221321 1 SSnn +−=++++ θθθθθθθ

and

.......)(cos..........coscos).............sin( 3121321 SSnn −=++++ θθθθθθθ

.................SS

.................SSè.............èèè n

42

31321

1)tan(

+−−

=++++∴


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

++


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


.,,);/(. 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

–––––––––––––


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


321

321

321

321

ddd

ccc

bbb

aaa


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


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


cb

cb

cb

cba =− +)(

33

22

33


ca

ca

ca

cab −=− +)(

33

22

33


ba

ba

ba

bac =− +)(

33

221

33

221

33

221 ba

bac

ca

cab

cb

cbaA +−=∴

)()()( 233212332123321 babaccacabcbcba −+−−−=

123132213312231321 cbacbacbacbacbacba −+−−−= )


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


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


=∴

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 ==∆


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


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 =+=


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


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


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


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.


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 +−=∴ −


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


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


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)


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


..

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


..

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.

−−−

−


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.


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))


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


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


=∴

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


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


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


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


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


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


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.


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


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


),,(&),,(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


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


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


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 ,��


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 −−−=+−+−−+−=−−=×


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


( ) 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 +−=−+=


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


(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


,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 =)(


(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


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


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 −=−=−−=


.,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


+−=

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

−=⋅−=⋅−=


,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


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 ==


)(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 +−+

+=+=∴


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.


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


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


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 =⋅

⋅=−

=


).()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 =− )(


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 −=−⇒

−

−=


)()()( 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


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)( ==


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 ,)( =


2

=++= ymdx

ydmxbmxay ,sincos)(

01 that prove1 If13 212

22 =−++

++= ymxyyxxxym

)(,)(

01 that prove If14 221 =+−+= −+ ynnyxbxaxy nn )(,)(


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


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(


⋅++=

++=

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(


⋅++=

++−=

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


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


−−=∴ 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


xxb

xxay

111 )cos(log)sin(log +−=


)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 )()(


21

=+−+−−= ++−

nnnxm ymnxynyxey )()()(,. cos

xmey1

:Solution −

= cos


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


12

442in equation quadratic a iswhich 2

211 −±=−±=∴ xx

xxyy mm

mm xxyxxy

−+=∴−+= 11 Consider, 221


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 =− )(


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)(.


21 =−−+−−= ++−

nnn ymnxynyxxmy )()()(),sinsin(.


21

=+−+−−= ++−

nnnxa yanxynyxey )()()(,. sin

0121 that prove1 If6 1222 =+−+−−= ++ nnn

n ynnxyyxxy )()(,)(.


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


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


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)


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 =+ )(


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 =+−=


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

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


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→


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


2020

2

01

1 x

xxx

x

xxx

x

xx

xxx

x

x

xxx

sincoslim

sincos

sinlim

sincos

sinlim−×=−×=

−×=

→→→

rule sHospital'L' theapplyingagain 0


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


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−


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

∂∂∂=

∂∂∂

= −



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


∫∫ ′−=′∴ 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


(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


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)(


∫∫∫∫ +−=

++−=

+−=

−− 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)(


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


2 xx

dx)(

cx

x

dx

x

dx

xx

dx +−=−+

=+−

=+−

−∫∫∫ 2

1

122152 :Solution 1

22222sinh

)()(


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 ,


∫∫∫∫+−

++−=+−

++−

−=+−

+∴

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


∫∫∫∫ +++−=

+++−=

++

++−

+=++

∴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


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


∫∫∫∫∫+−

+

=++

=++

=

++

+=+

∴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


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,,, −==′=′=


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 )()(


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, )()(


∫ +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


[ ])(

)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)(


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


( )∞

−∞∞∞

=

+=

−++=

+−+

+=+ ∫∫∫∫

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


∫ +−=

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


∫∫∫ +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


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


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


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)( ,


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


( )∫∫∫ −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)(,)(,)(



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


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


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)()(


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 )(


∫ =−

++ 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 ==∫


∫ +=∴ ∫∫ 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


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 &


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


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


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


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