1. introduction - ahmad...

UNIVERSITI KUALA LUMPUR INTEGRATION – F2-B

1. INTRODUCTION When x2 is differentiated with respect to x, the derived function is 2x. Conversely, given that an unknown function has a derived function of 2x, it is clear that the unknown function could be x2. This process of finding a function from its derived function is called integration and it reverses the operation of differentiation. 1.1 DEFINITIONS : THE INDEFINITE INTEGRAL Let the function f be defined on an interval I. A function F is called an antiderivative for f on the interval I if F is differentiable on I and F’(x) = f(x) for all x ∈ I The indefinite integral of the function f on the interval I is the family of all antiderivatives for f on I and is denoted by

∫ = +f x dx F x C( ) ( ) where • the symbol ∫ is called an integral sign • ∫…dx means ‘ the integral of … w.r.t x ‘ • f(x) is called the integrand • C is called the arbitrary constant of integration

F2-B - 1 - MATHEMATICS UNIT


2. BASIC RULES 2.1 BASIC RULES These rules are used to simplify the process of finding the integration of a function. (a)

Integration of a

constant

∫ a dx = ax + C

a = constant,

C = real number.

(b)

Integration of

STANDARD FORM of a polynomial

Remark :

∫ nx dx = Cnxn

++

+

1

1

C and n = constant except for n = -1.

∫ nax dx = a dx ∫ nx

= Cnaxn

++

+

1

1

C , a and n = constant except for n = -1.

(c)

Integration of

GENERAL FORM of a polynomial

∫ + nbax ][ dx = ))(1(][ 1

anbax n

++ +

+ C

C and n = constant except for n = -1.

2.2 PROPERTIES OF INTEGRALS (a) Sum and difference

∫ ∫ ±=± dxxgdxxfdxxgxf )()()]()([ ∫

(b) Constant multiple of a function

[ ( )] ( )cf x dx c f x dx= ∫∫



EXAMPLE 1 : Integrate the following function .

(a) 4dx∫ cxdx +=∫ 44

(b) x dx5∫ cxdxx +=∫ 6

65

(c)

dtt∫ ∫∫ +== ctdttdtt

32 2

3

21

(d)

dx

x∫ 6

3 ∫∫ −= dxxdx

x6

6 33

c

x

cx

+−=

+−

=−

5

5

535

3

(e)

dxx

x∫

+− 4

512 3

4

( )

cxx

x

cxxx

dxdxxdxx

dxx

x

+++=

++−

−=

+−=

+−

−

−∫ ∫ ∫

∫

410

15

2

425

15

2

4512

4512

2

5

25

34

34

(f)

( ) dxx∫ + 432

( ) ( ) cxdxx ++

=+∫ )2(53232

54

( ) cx

++

=10

32 5

(g)

dxx∫ −1

1

( ) dxxdxx ∫∫ −−=

−21

111

( )( )

( ) cx

cx

+−−=

+−

−

=

21

21

12

1211



3. THE INTEGRAL OF CERTAIN FUNCTIONS 3.1 TRIGONOMETRIC FUNCTIONS Whenever a function f ’(x) is recognised as the derivative of a function f(x), then d

dxf x f x f x dx f x C( ) ' ( ) ' ( ) ( )= ⇔ = +∫

Thus, referring to the derivatives of trigonometric functions in Differentiation module, the integration of STANDARD FORM of trigonometric functions are as follow :

DIFFERENTIATION ( standard form )

INTEGRATION

( standard form )

( ) xxdxd

cossin = ∫ += Cxxdx sincos

( ) xxdxd sincos −= ∫ +−= Cxxdx cossin

( ) xxdxd 2sectan = ∫ += Cxxdx tansec 2

( ) xxxdxd tansecsec = ∫ += Cxxdxx sectansec

( ) xxxdxd cotcsccsc −= ∫ +−= Cxxdxx csccotcsc

( ) xxdxd 2csccot −= ∫ +−= Cxxdx cotcsc 2



The integration of GENERAL FORM of trigonometric functions are as follow : Remark : f(x) must be a LINEAR FUNCTION which means f(x) = ax+b

DIFFERENTIATION

( general form )

( ) )(').(cos)(sin xfxfxfdxd

=

( ) )(').(sin)(cos xfxfxf

dxd

−=

( ) )(').(sec)(tan 2 xfxfxf

dxd

=

( ) )(').(tan)(sec)(sec xfxfxfxf

dxd

=

( ) )(').(cot)(csc)(csc xfxfxfxf

dxd

−=

( ) )(').(csc)(cot 2 xfxfxf

dxd

−=

INTEGRATION ( general form )

∫ += Cxfxfdxxf)('

)(sin)(cos

∫ +

−= C

xfxfdxxf

)(')(cos)(sin

∫ += C

xfxfdxxf)('

)(tan)(sec 2

∫ += C

xfxfdxxfxf)('

)(sec)(tan)(sec

∫ +

−= C

xfxfdxxfxf

)(')(csc)(cot)(csc

∫ +

−= C

xfxfdxxf

)(')(cot)(csc 2



EXAMPLES : Evaluate the following :

(a)

∫ xdxcos12

cxxdx +=∫ sin12cos12

(b)

( )∫ ++ dxxxx cossin3

( ) cxxxdxxxx ++−=++∫ sincos4

cossin4

3

(c)

∫ − tdt2csc5

( ) cttdt +−−=−∫ cot5csc5 2

ct += cot5

(d)

∫ θθd4sin

cd +−=∫ 44cos4sin θθθ

(e)

dxx∫

+

432csc2 π

cx

dxx +

+−

=

+∫ 2

432cot

432csc2

ππ

cx +

+−=

432cot

21 π

(f)

∫ θθθ d5cot5csc

cd +−=∫ 55csc5cot5csc θθθθ

(g)

dxxx∫

−

− 6

2tan6

2sec ππ

cx

dxxx

+−

−

=

−

−∫

6

62

sec

62

tan62

sec

π

ππ



3.2 EXPONENTIAL FUNCTION

It is known that xx ee

dxd

=)(

So , the integration of :

(a)

STANDARD FORM

of exponential function

∫ += Cedxe xx

(b)

GENERAL FORM of exponential function

Cxf

edxexf

xf +=∫ )('

)()(

where f(x) must be a LINEAR FUNCTION

EXAMPLE Integrate the following functions with respect to x . (a) xexf 3)( = cedxe xx +=∫ 33

(b) xexf 5

2)( =

c

ecedxe

exf

xxx

x

+−=+−=

=

−−

−

∫ 555

5

52

522

2)(

(c) xexf 83)( =

cedxe xx +=∫ 88

833



3.3 INTEGRATION OF THE FORM

x1

It is already known that for x > 0, xxdx

dxd 1ln =

So, for all values of x, the integration of :

(a)

STANDARD FORM of x1

Cxdxx

+=∫ ln1

(b)

GENERAL FORM of )(1xf C

xfxf

dxxf

+=∫ )()(ln

)(1

'

where f(x) must be a LINEAR FUNCTION

Remark :

We have absolute value of x that is x because the logarithm of a negative number does not exist. EXAMPLE Integrate the following functions with respect to x

(a) ∫∫ +== Cxdxx

dxx

ln4144

(b) ( ) Cxdx

x+

+=

+∫ 252ln

521

(c) ( ) ( ) CxCxdx

xdxx

+−−=+−−

=−

=− ∫∫ 24ln

23

224ln3

2413

243



4. INTEGRATING PRODUCTS

(a)

METHOD I BY RECOGNITION

( only for function ( )xfe )

( ) ( ) ( ) cedxxfe xfxf +=∫ '

(b)

METHOD II

BY SUBSTITUTION

( for products with the same type of functions )

STEPS TO EVALUATE THE INTEGRAL

( )[ ] ( )∫ dxxfxfa n '

where a and n are constants 1. Substitute u = f(x) and du = f ‘ (x) dx to

obtain the integral ( )∫ duug

2. Evaluate by integrating with

respect to u

( )∫ duug

3. Replace u by f(x) in the resulting expression

(c)

METHOD III

INTEGRATION BY PARTS

( for products with different types of functions )

Many products are not in the form of

( )[ ] ( )∫ dxxfxf n '

and cannot be expressed in the form ( )ufdxdu '

. Therefore we cannot integrate by using substitution method .To solve this problem , we have to use integration by parts formula as follows :

∫ ∫−= vduuvudv

where u and v are both functions of x .



4.1 BY RECOGNITION

(a) Integrate 2 : 2xxe Cedxxe xx +=∫

22

2

(b) Integrate cos : xxesin Cedxxe xx +=∫ sinsincos 4.2 BY SUBSTITUTION

(a) Determine ( )∫ x + dxx 532

STEP 1 : Let then 53 += xu 23 xdxdu

= ,

therefore : dxxdu 2

3=

STEP 2 : Hence ,

( )3

5 21

32 duudxxx∫ ∫=+

CuCu

duu

+=+

=

= ∫

232

3

21

92

32

31

31

STEP 3 : but =u 53 +x

Therefore ( ) ( ) Cxdxx ++=+ 23

33 5925x∫ 2

Find :

(b) (c) ( ) ( dxxxx∫ +++ 2362 243 ) dxxx

+sin1cos

∫

(d) (e) dxxx∫ cossin3dx

xx

∫ln



4.3 BY INTEGRATION BY PART

(a) Integrate with respect to x . xxe

SOLUTION :

Let xu = then 1=dxdu

, therefore : dxdu =

Let then dxedv x= xedvv ∫ == By using the formula of integration by parts , :

∫ ∫−= vduuvudv

∫∫ −= dxexedxxe xxx

Cexe xx +−=

NOTE : 1.The choice must be such that the u part becomes a constant after successive differentiation .

2.The dv part can be integrated from standard integrals

3.Normally , we give priority to the following functions as u part by following this order :

i . ( L ) logarithm function

ii ( I ) inverse function



(b) Evaluate ∫ xdxx ln4

SOLUTION : Let u then xln=xdx

du 1= , therefore : dx

x1

=du

Let then dxxdv 4=5

5xdvv ∫ ==

By using the formula of integration by parts , :

∫ ∫−= dxx

xxxxdxx 1.5

ln5

ln55

4

Cxxx

dxxxx

+−=

−= ∫

255ln

51

5ln

55

45

(c) Evaluate ∫ xdxx sin2

(repeating process of integration by parts ) SOLUTION : Let then 2xu = x

dxdu 2= , therefore : xdxdu 2=

Let then xdxdv sin= xdvv cos−== ∫ By using the formula of integration by parts , : ∫ ∫ −−−= xdxxxxxdxx cos2cossin 22

∫+−= xdxxxx cos2cos2 Let xu then = 1=

dxdu , therefore : dxdu =

Let then xdxdv cos= xdvv sin∫ == ∫ ∫ −−−= xdxxxxxdxx cos2cossin 22

[ ]

CxxxxxCxxxxx

xdxxxxx

+++−=

+−−+−=

−+−= ∫

cos2sin2cos)cos(2sin2cos

sinsin2cos

2

2

2



(d) Evaluate ∫ xdxe x cos

( special case of integration by parts ) SOLUTION : Let then xeu = xe

dxdu

= , therefore : du dxex=

Let then xdxdv cos= xdvv sin∫ == By using the formula of integration by parts , : ∫ ∫−= xdxexexdxe xxx sinsincos

Let then xeu = xedxdu

= , therefore : du dxex=

Let then xdxdv sin= xdvv cos−== ∫ [ ]∫ ∫ −−−−= xdxexexexdxe xxxx coscossincos

∫−+= xdxexexe xxx coscossin

[ ]

[ ] Cxxexdxe

Cxxexdxex

x

xx

++=

++=

∫

∫cossin

2cos

cossincos2

(e) Evaluate ∫ xdxln

( integration of ln x ) SOLUTION :

Let then xu ln=xdx

du 1= , therefore : dx

x1

=du

Let then dxdv = xdvv ∫ == By using the formula of integration by parts , :

∫ ∫−= dxx

xxxxdx 1lnln

Cxxx

dxxx

+−=

−= ∫ln

ln



5. INTEGRATING QUOTIENT

(a)

METHOD I

BY RECOGNITION

( only for dxxfxf

∫ )()('

)

Cxfdxxfxf

+=∫ )(ln)()('

Integrating fractions by recognition method will be used only when the numerator is basically the derivative of the complete denominator.

(b)

METHOD II

BY SUBSTITUTION

(only for dxxfxfn∫ )]([

)(' )

Integrating fractions by substitution will be used when an integral whose numerator is the derivative, not of the complete denominator, but of a function within the denominator.

(c)

METHOD III

BY PARTIALFRACTIONS

Integrating fractions by partial fractions method will be used

only for integral other then dxxfxf

∫ )()('

and dxxfxfn∫ )]([

)('

form.

5.1 BY RECOGNITION

(a) Evaluate dxxx

∫ + sin1cos

: ( ) Cxdxxx

++=+∫ sin1ln

sin1cos

(b) Evaluate dxeex

x

∫ + 4 : ( ) Cedx

ee xx

x

++=+∫ 4ln

4

(c) Evaluate dxxx

∫ + 3

2

1 : ( ) Cxdxxx

++=+∫

33

2

1ln31

1



5.2 BY SUBSTITUTION

(a) Determine dxxx

∫+ )1(

22

SOLUTION :

STEP 1 : Let then 12 += xu xdu 2dx

=

rearranging , gives xdxdu 2= STEP 2 : Hence,

∫∫ =+ u

dudxx

x

)1(

22

Cu

duu

+=

= ∫−

21

21

2

STEP 3: but 12 += xu

therefore, ( ) Cxdxxx

++=+

∫ 21

2

212

)1(

2

Find :

(b) ∫ dxxx

4sincos

(c) cos

sinxxdx

+1∫



5.3 BY PARTIAL FRACTION

(a) Evaluate dxxx∫ ++ )2)(1(

1

SOLUTION : Find the partial fractions :

( )( ) ( ) ( )21211

++

+=

++ xB

xA

xx

( ) ( )( )( )21

12++

+++=

xxxBxA

Compare the numerator of the left hand side to the right hand side : ( ) ( 121 )+++= xBxA

Let : 1−=x 2 ( )( ) ( )( )BA +−++−= 1111 ,

, Let : 2−=x Hence :

dxx∫ ++ )2)(1(

1

NOTE : Only PROPER FIMPROPER FRAproper

(b) Evaluate ∫ ++

xx

)(1(

2

A=1 ( )( ) ( )( )BA +−++−= 12221

B=−1

( ) ( )dxxdx

xx ∫∫ +

−+

=2

11

1

( ) ( ) Cxx ++−+= 2ln1ln

RACTIONS can be converted directly into partial fractions. An CTIONS must first be divided out until the remaining fraction is

−dx

x )11

(c) Evaluate ∫ ++ dxxx

12



6. INTEGRATION of TRIGONOMETRIC EXPRESSIONS

(a)

STANDARD RESULT

Cxxxdx

Cxxxdx

Cxxdx

Cxxdx

++−=

++=

+=

+=

∫∫∫∫

.cotcsclncsc

tanseclnsec

sinlncot

seclntan

(b)

INTEGRATION OF SOME

TRIGONOMETRIC EXPRESSIONS

EVEN POWERS OF SIN θ OR COS θ

The double angle trigonometric identities are useful here:

θθθ 22 sincos2cos −=

1cos2

sin212

2

−=

−=

θ

θ

where :

2

2cos12cos θθ +=

22cos1sin2 θθ −

=

ODD POWERS OF SIN θ OR COS θ In this case, we can use the identity:

1sincos 22 =+ θθNote :

cos sin sin

sin cos cos

θ θ θ θ

θ θ θ θ

n n

n n

dn

C

dn

C

=+

+

=−+

+

+

+

∫

∫

11

11

1

1

Can you prove these formulae? (use substitution method)

POWERS OF TAN θ

The identity : θθ 22 sec1tan =+

and the formula Cn

dn

n ++

=∫+

1tansectan

12 θθθθ is

helpful in integrating any power of tan θ.



(c)

MULTIPLE ANGLE

When integrating products such as sin nθ cos mθ, one of the factor formulae should be used :

( ) ([ ]BABABA −++= sinsin21cossin )

( ) ([ ]BABABA −−+= sinsin21sincos )

( ) ([ ]BABABA −++= coscos21coscos )

( ) ([ ]BABABA +−−= coscos21sinsin )

6.1 EVEN POWERS OF SIN θ OR COS θ

(a) Evaluate ∫ cos θθd2

SOLUTION :

∫∫

+= θθθθ dd

22cos

21cos 2

C

dd

++=

+= ∫ ∫

42sin

2

22cos

21

θθ

θθθ

(b) Evaluate ∫ cos (c) Evaluate θθd4 ∫ θθd2sin



6.2 ODD POWERS OF SIN θ OR COS θ (a) Find ∫ sin θθd3

SOLUTION :

( )( )∫∫ = θθθθθ dd 23 sinsinsin

( )

C

dd

d

++−=

−=

−=

∫ ∫∫

3coscos

cossinsin

cos1sin

3

2

2

θθ

θθθθθ

θθθ

(b) Find ∫ sin θθd5

6.3 ODD AND EVEN POWERS FOR TANGENT

(a) Evaluate ∫ θθd2tan

( )∫∫ −= θθθθ dd 1sectan 22 SOLUTION:

C

dd

+−=

−= ∫ ∫θθ

θθθ

tan

sec 2

(b) Evaluate ∫ θθd3tan

( )∫∫ −= θθθθθ dd 1sectantan 23 SOLUTION:

C

dd

+−=

−= ∫ ∫θθ

θθθθθ

secln2

tan

tansectan2

2

(c) Evaluate (d) Evaluate ∫ tan ∫ θθd4tan θθd5




6.4 MULTIPLE ANGLE

(a) Determine sin cos5 3θ θd∫ θ SOLUTION :

( )∫∫ += θθθθθθ dd 2sin8sin213cos5sin

C

dd

+−−=

+= ∫ ∫

42cos

168cos

2sin218sin

21

θθ

θθθθ

(b) Determine ∫ θθθ d2cos3sin CONCLUSION : The aims when dealing with trig integrals are usually :

a) to convert the integral to the form

∫ dxufdxdu )('

b) to reduce the trig expression to a number of single trig ratios

1. introduction - ahmad...

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