Basic Mathematical Knowledge

7/31/2019 Basic Mathematical Knowledge

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Chapter 0. Basic Mathematical Knowledge/P.1

Chapter 0. Basic Mathematical Knowledge

Section 0. Trigonometric Formulas

Section 1. Complex Numbers

Section 2. Exponential Functions, Logarithmic Functions and Their Limits

Section 3. Differentiation

Section 4. Integration


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Section 0. Trigonometric Formulas

0.0.0 Compound angle

1. BABABA sincoscossin)sin( = .

2. BABABA sinsincoscos)cos( m= .

3.BA

BABA

tantan1

tantan)tan(

m

= .

0.0.1 Double angle

1. AAA cossin22sin = .

2. AAAAA 2222 sin211cos2sincos2cos === .

0.0.2 Sum to Product

1.2

cos2

sin2sinsinBABA

BA+

=+ .

2.2

sin2

cos2sinsinBABA

BA+

= .

3.2

cos2

cos2coscosBABA

BA+

=+ .

4.

2

sin

2

sin2coscosBABA

BA+

= .

0.0.3 Product to Sum

1. )sin()sin(cossin2 BABABA ++= .

2. )cos()cos(coscos2 BABABA ++= .

3. )cos()cos(sinsin2 BABABA += .

0.0.4 General solution

1. If k=sin , then nn )1(+= ;

2. If k=cos , then = n2 ;

3. If k=tan , then += n , where Nn and is one of the roots.


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Section 1. Complex Numbers

Definition 0.1.0 A complex numberz is defined to be a number of the form a + bi,

where a b, R and i 2 1= .

a is called the real part ofz and is denoted by Re(z), b is calledthe imaginary

part ofz and is denoted by Im(z).

If Im(z)=0,z is real. If Re(z)=0 and Im( )z 0, z is called apurely imaginary

number.

Definition 0.1.1 Let z a bi z c di1 2= + = +, , where a b c d , , , R , be two

complex numbers, then z1 is equalto z2 if and only ifa = c and b = d, i.e. z z1 2=

iff Re( ) Re( )z z1 2= and Im( ) Im( )z z1 2= .

Definition 0.1.2 (Arithmetic operations of complex numbers)

Let z a bi z c di1 2= + = +, where a b c d , , , R , be two complex numbers, then

the rules of arithmetic operations are defined as

i) z z a c b d i1 2+ = + + +( ) ( )

ii) z z a c b d i1 2 = + ( ) ( )

iii) z z ac bd ad bc i1 2 = + +( ) ( )

iv)z

z

ac bd

c d

bc ad

c d

i1

2

2 2 2 2=

+

+

+

+

if z2 0 .

e.g.0.1.0 Evaluate i) )53)(21( ii + ; ii)i

i

67

32

+

.


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e.g.0.1.1 Find Cz such that iz 432 += .

e.g.0.1.2 Solve the following equations:

i) 0222 =++ xx ; ii) 01062 =+ xx ;

iii) 013 =x ; iv) 0124 =++xx .


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Section 2. Exponential Functions and Logarithmic Functions

0.2.0 Revision --Indices Notation and indices laws

For any real number a and any positive integer n, an is defined to be the product ofa

by itselfn times. That is,

an: = aa ... a (n times).

For any positive real number a, we define

am

:=

integer.negativeaisif

1

zero,isif1

m

a

m

m

For any non-negative real number a and any positive integer n, we define a1/n

to be

the unique non-negative real root of the equation

xn= a.

Sometimes, we write 21

: aa = and naan1

:= for n 3.

For any non-zero rational number r, there is a unique pair of non-zero integers m andn such that n is positive, m and n are relative prime (that is, they have not common

prime factor) and r=n

m, we define

ar

=

>=

>

.0and0if0

,0if)(1

a

aa nm

It is not difficult to prove that the following Indice laws.:

For any real numbers a, b > 0 and for any rational numbersx andy, we have

(2.0) a0= 1, (2.1) a

1= a,

(2.2) axa

y= a

x + y,

(2.3)

y

x

a

a= a

xy,

(2.4) (ab)x

= axb

x, (2.5) x

b

a)( =

x

x

b

a

(2.6) (ax)y

= axy

. (2.7) Ifax

= ay

thenx =y.

Table 0


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e.g.0.2.0

Solve the following equations.

a) 493x

= 343; b) 51

125

23

+

=

x

x

;

c) 22x 5(2x) + 4 = 0; d) 25x= 23(5x)+ 50.

e.g.0.2.1 Solve the following system of simultaneous equations:

=+=

.954

,75421 yx

yx


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0.2.1 Exponential functions and their graphs

We ploty = 2x against all rational numbersx. The following graph is obtained.

Since the symbol 2x

is not defined for all irrational numbers, the above graph is not a

continuous curve.

However, it is easily to see that there is one and only one continuous curvey = g(x)

satisfying g(x) = 2

x

for all rational numbersx. Then we can define

)(:2 xgx = for any irrational numberx.

In general, we can prove that

i) Iff(x +y) =f(x)f(y) for all rational numbersx andy,

thenf(r) =f(1)r

for all rational numbers r.

ii) For any real positive number a, there is one and only one continuous

functionfa defined on the real line such thatfa(r) = arfor all rational numbers

r.

We define

ax : =fa(x) for irrational numberx.

wherefa is the unique continuous function satisfyingfa(x) = ax for all rational number

x.

0

1

2

3

4

3 2 1 1 2x

y


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Furthermore, we can show that all indices laws ( (2.0) - (2.7) ) in table 0 hold even

the powers are irrational.

Definition 0.2.0

For any a > 0, the continuous function f(x) = axdefined on the real line is called an

Exponential Function with Base a.

Since a0

= 1 for any positive real number a, the exponential curvey = ax

must passes

through the point (0, 1). When a = 1, the graphy = 1x

is just a horizontal straight line.

Ifa > 1 then

ax < a

y wheneverx ay

wheneverx 1 and 0 < a < 1,x-axis is a horizontal asymptote to the curvey = ax.

Besides the graph of the curvey = )1

()1

(x

x

aa

= is the mirror image of that of the

curve xay = about they-axis.

y = ax , a > 1

y =x

a

)1

(

y = 1x

y

x

O

The graphs of exponential functionsy = ax

andy = ax

1

a

1

a

1

1

a

21


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e.g.0.2.2 Sketch the following curves in one graph.

a) y = 3x

b) y = )3

1()

3

1(

x

x = ;

c) y = 4(3x);

d) y = 4(3x) 1 ;

e) y =2

1 (3x);

f) y = 22

1 (3x).

Class Practice 0.2.0

Sketch the following curves in one graph.

a) y =x2.0 ;

b) y =

x5 ; c) y = 2(0.2x); d) y =5

2.0 x;

e) y = 32(0.2x) ; f) y =5

2.03

x

+ .

0.2.2 The limits of exponential functions

Theorem 0.2.0

(I) ifa >1,

(i) x

xa

+lim = + ,

(ii) x

xa

lim = 0,

(II) if 0< a


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d)x

x

x 3

3

31

1)3(4lim

+

; e)

)5.0(43

)5.0(2lim

x

x

x +; f)

x

x

x )7.0(45

)7.0(2lim

1

+

;

g)x

xx

x )1.0(27)1.0(4)1.0(3lim

32

+

+

.

e.g.0.2.4 Sketch the following graphs

a) y =13

1

+x; b) y =

13

2

+

x

; c) y =13

31

+

x

x

.


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1. Evaluate the following limit.

a) x

x4.0lim

+; b)

2

31lim

x

x

.

c)5

4lim

3x

x +;

d) kx

x5lim

where kis a positive constant;

e) bx

xa

+lim where a and b are positive constants but a


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Let a > 0 and a 1.

(2.8) logaax

=x for any real numberx;

(2.9) y = a aylog

for anyy >0;

(2.10)

loga 1 = 0, a

0= 1.

(2.11) logaa = 1. a1

= a .

For anyx,y > 0, we have

(2.12) log axy

=ylogax,

.

)(

log

log

xy

yxy

a

a

a

ax

=

=

(2.13) log a(xy) = logax + logay,

.

))((

loglog

loglog

yx

yx

aa

aa

a

aaxy

+=

=

(2.14) logax

1= log ax ,

.

11

log

log

x

x

a

a

a

ax

=

=

(2.15) loga(y

x) = logax logay,

.loglog

log

log

yx

y

x

aa

a

a

a

a

a

y

x

=

=

(2.16)If logax = logay thenx =

y.

(2.17) logax =a

x

b

b

log

log

(change of base)

whenever b > 0 and b1.

).)(log(log

)(logloglog

ax

ax

ba

xbb

a

=

=

Table 1

Definition 0.2.2

The inverse function of the exponential function with base a is called theLogarithmic

Function withBasea and is denoted by loga. That is, forx > 0,

logax =y if y is the unique number satisfying ay=x.


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Ifa > 1, then logax is strictly increasing.

If 0 < a < 1, then logax is strictly decreasing.

The graph of logarithmic functiony = logax

(a) passesthrough the point (1, 0).

(b) has a vertical asymptotex = 0 (the y-axis)

(c) is the mirror image of that of its inverse functiony = axabout the liney =x

e.g.0.2.5 Solve the following equations.

a) log8(x 3) + log8(x 5) = 1; b) log10(x2

+ 4) 2 log10x = 1;c) 3

2x+1= 5

x;d) 2

32x7

x+1=56.

y

= lo ax

= ax

=x

O x

y

= lo ax

= ax =x

O1 1


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e.g.0.2.6 Solve the following simultaneous equations.

=+

=+

.0)33(log

,log22log)2(log

10

101010

yx

yx

Definition 0.2.3

Thebase of Natural Logarithm is denoted by e.This number e be defined by

e = nn n

)11(lim +

2.7182818284

The exponential function with base e is simply calledthe Exponential function and

denoted by exp. That is,

exp(x) = ex

for any real number x.

The Natural Logarithmic Function and denoted by ln or log is the logarithmic

Function with base e. That is, xx elogln = .

The importance ofe will be shown in next sections.


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Section 3. Differentiation

You should learn the following formulas/theorems in CE Additional Mathematics:

0.3.0 Let u and v be differentiable functions ofx and c is a constant, then

i)d

dxc = 0 ;

ii)d

dxu v

du

dx

dv

dx( ) = ;

iii)d

dxuv u

dv

dxv

du

dx( ) = + (Product rule);

iv)d

dx

u

v

vdu

dxu

dv

dx

vv( ) ,=

20 (Quotient rule).

0.3.1 (Chain Rule)

If y is a differentiable function ofu and u is a differentiable function ofx,

thendx

du

du

dy

dx

dy= .

0.3.2 (Inverse Function Theorem)

Ify is a differentiable function ofx given by y=f(x), and if x y= ( ) is the

inverse function ofy=f(x), then

'( )'( ( ))

yf y

=1

, or

dy

dxdx

dy 1= .


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e.g.0.3.0 Prove or disprove:2

2

2

2

2

2

dx

ud

du

yd

dx

yd= .

e.g.0.3.1 sin: ( . ) ( , )

2 211 is differentiable and bijective. Let y x= sin 1 be

its inverse function, finddy

dx.

e.g.0.3.2 It is known thatd

dxx

x aa(log )

ln=

1(where a > 0, 1 ). Find

d

dxa

x .


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Theorem 0.3.0

i)d

dxx x

= 1 , is a real constant;

ii)d

dxx xsin cos= ;

iii)d

dxx xcos sin=

iv)d

dxx xtan sec= 2 ;

v)d

dxx xcot csc= 2 ;

vi) ddx

x x xsec sec tan= ;

vii)d

dxx x xcsc csc cot= ;

viii) 1)d

dxe ex x= ; 2)

d

dxa a a

x x= ln , a>0 is a constant;

ix) 1)d

dxx

xln =

1; 2)

d

dxx

x aalog

ln,=

1 a > 0 1, is a constant.

x) ddx

xx

sin =

12

11

;

xi)d

dxx

xcos =

1

2

1

1;

xii)d

dxx

xtan

=+

1

2

1

1;

xiii)d

dxx

xcot

=

+1

2

1

1.

e.g.0.3.3 Find the derivatives of the following functions

a) e7x

; b) 4 22

x

e

; c)xe4 ; d) xex

2; e)

x

x

e

e

+

1

1; f)

21 )(tan xe

.


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e.g.0.3.4 Find the derivatives of the following functions

a) f(x) = 3lnx; b) g(x) = )1ln(2 +x ; c) h(x) =

x

xln; d)

y(x)= xx ln2

.

e.g.0.3.5 a) Let y =1

32)54(

2

2

+

+

x

xxforx >

3

2. Find )(lny

dx

dand

dx

dy.

b) Lety =xx

forx > 0. Find )(lnydx

dand

dx

dy.


1. Find the derivatives of the following functions.

a)y = ekx

(kis a constant.); b) y = 4e8x

+ 15 3x

e ;

c)y = 4ex( )2 5 6+ ; d) y =x3e2x ;


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e)y =e

x

x2 1

7 4

+

; f) y = log3x;

g)y = log2 (x2 +x 1); h) y = ln lnx;

i)y = 3x; j) y = 5

lnx

k)y =54 23x x ; l) y = ee

x

.

2. (Logarithmic differentiation) Find ddx

(lny) and dydx

of the following functions.

a)y = (x2

+ 3)4(2x

3 1)5; b) yx x

x x=

+ +

( )( )

( )( )

1 3

1 3;

c)y =xlnx

d) y = (log2x)x.


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Section 4. Integration

0.4.0 Method of Substitution

Definition 0.4.0

Let )(xfy = be a function ofx, and let x be a small increment ofx. The

product xxf )(' is defined to be thedifferential of y and denoted by dy.

Since xxdx

dxdx == 1 , we have dxxfxxfdy )(')(' == .

e.g.0.4.0 Find the differential of xyi cos) = and xxyii 2sin) = .

Theorem 0.4.0 (Method of Substitution/ Change of Variable)

Ifd

dug u f u( ) ( )= and u h x= ( ) , then

f h x h x dx f u du g h x C( ( )) ' ( ) ( ) ( ( ))= = + ,

where Cis a constant.

e.g.0.4.1 Evaluate i) dxx 2)72(1

by the substitution 72 = xu ;

ii) dxx)37sin( by the substitution xu 37 = ;

iii) + dxxx20052

)1( by the substitution 12

+=xu ;

iv) + dxxx 1032 )12( by the substitution 12 3 += xu ;

v) xdxx cossin by the substitution xu sin= or xu cos= .


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Remark: i) If 1n , Cna

baxdxbax

nn +

++

=++

)1()(

)(1

.

ii) ++

=+ Ca

baxdxbax

)cos()sin( .

iii) ++

=+ Ca

baxdxbax

)sin()cos( .

e.g.0.4.2 Evaluate the following integrals by a suitable substitution:

i) dx

x

x

cos1

sin; ii) +

dxx

x

tan20071

sec2; iii) dxxxxx )43()2( 22

1

23 ++

;

iv) d6csc ; v) dxxx

6)1(; vi) +

dx

xx

x22 )12(

1.


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e.g.0.4.3 Evaluate the integrals in e.g.0.4.1 and 0.4.2 directly.


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e.g.0.4.4 (Trigonometric Substitution)

Case 1. If the integrand contains )( 22 xa , make the substitution x a= sin .

Case 2. If the integrand contains )( 22 xa + , make the substitution tanax = .

Case 3. If the integrand contains )( 22 ax , make the substitution secax = .

Evaluate i) dxx 21 ; ii) dxxx

2

24; iii)

+ 23

2 )5( x

dx; iv)

+dx

x

x

92

3

.

0.4.1 Formula of Integration

e.g.0.4.5 i) Find the derivative of lnx for x 0.

ii) Find the indefinite integral1

x

dx

.


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Theorem 0.4.1

1. x dx n x C

n n

= + ++1

1

1

for 1n .

2.1

xdx x C = + ln .

3. i) sin cosxdx x C= + ; ii) cos sinxdx x C= + ;

iii) sec tan2xdx x C= + ; iv) csc cot2xdx x C= + ;

v) sec tan secx xdx x C= +

; vi) csc cot cscx xdx x C= +

.

4. tan ln secxdx x C= + .

5. cot ln sinxdx x C= + .

6. sec ln sec tan ln tan( )xdx x x Cx

C= + + = + +

4 2.

7. csc ln csc cot ln tanxdx x x Cx

C= + = +2

.

8. i) Cedxe xx += ; ii) Cekdxekxkx +=

1, where kis a constant;

iii) a dxa

aC

xx

= + ln , where a > 0 1, is a constant.

9. i)1

1 21

+= + x dx x C tan ;

ii)1 1

2 2

1

x adx

a

x

aC

+= + tan , where a is a constant.

10. i)1

1

1

2

1

12=

+

+ x dx

x

xCln ;

ii)1 1

22 2a xdx

a

a x

a xC

=

+

+ ln , where a is a constant.

11. i)1

1 21

= +

xdx x C sin ;

ii)1

2 2

1

a xdx

x

aC

= + sin , where a is a constant.


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12. i)1

11

2

2

xdx x x C

= + + ln ;

ii)1

2 2

2 2

x adx x x a C

= + + ln , where a is a constant.

e.g.0.4.6 Evaluate the following indefinite integrals by suitable substitutions:

i)1

2 1xdx

+ ; ii)1

x xdx

ln ,x > 0; iii) x e dxx2 3 ;

iv)e

edx

x

x + 1 ; v)e x

edx

x

x

sinsin

2 2

2 .

e.g.0.4.7 (Integrals of the form1 1

2 2ax bx cdx

ax bx cdx

+ + + + , )

Find i)1

12x xdx

+ + ; ii)1

4 42 x x dx ;

iii)1

5 42

x x

dx ; iv)1

6 12

x xdx

+ + .


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e.g.0.4.8 (Integrals of the formex f

ax bx cdx

ex f

ax bx cdx

+

+ +

+

+ + 2 2, )

Find i)8 3

2 2 12t

t tdt

+ + ; ii)x

x xdx

+

+ +5

2 82;

iii)2 1

3 8 12

x

x xdx+

+ .


Find the following integrals by the substitution.

a) dxx)25( ; b) dxxxx + 21

)13)(32( 2 ;

c) dxxx

xx

++

23

42

23

2

; d) dxxx + )56()12( ; e) dxxx ln1

.


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0.4.2 Integration by Partial Fractions

Definition 0.4.1

An algebraic fraction or a rational function is an algebraic expression of the

formp x

q x

( )

( ), where p(x) and q(x) are polynomials in x and q x( ) 0 .

Definition 0.4.2

A rational functionp x

q x

( )

( )

is called

i) aproper fraction if deg( ( )) deg( ( ))p x q x< ,

ii) an improper fraction if deg( ( )) deg( ( ))p x q x , where deg(p(x)) is the degree of

the polynomialp(x).

For instance,1

1

2 3 1

1 2 3

2

+

+

+ + +x

x x

x x x

,

( )( )( )

are proper fractions while

x

x

+

1

1,x x

x x

3

2

4 1

2 1

+ +

+ are improper fractions.

Ifp x

q x

( )

( )is an improper fraction, then by dividing p(x) by q(x), we get

p x q x Q x R x( ) ( ) ( ) ( ) + , where Q(x), R(x) are polynomials and

deg( ( )) deg( ( ))R x q x< . It follows thatp x

q xQ x

R x

q x

( )

( )( )

( )

( ) + , which is the sum of a

polynomial and a proper fraction.

Definition 0.4.3

Partial fractions is the process which breaks a given fraction (proper or

improper) into an algebraic sum of several proper fractions and a polynomial.

Process of resolving a rational fraction in partial fractions:

1. Express the given rational function as a sum of a polynomial (it may be a zero


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polynomial if the given rational function is proper) and aPROPERfraction.

2. Factorize the denominator completely into product of linear and quadratic factors.

3. To a factor of the form ( ) ,ax b nn

+ 1, there corresponds a group of fractions

A

ax b

A

ax b

A

ax b

nn

1 22+

++

+ ++( ) ( )

L

where A A An1 2, , ,L are constants.

4. To a factor of the form ( ) ,ax bx c nn2 1+ + , there corresponds a group of

fractions

A x Bax bx c

A x Bax bx c

A x Bax bx c

n nn

1 12

2 22 2 2

++ +

+ ++ +

+ + ++ +( ) ( )

L

where A A An1 2, , ,L , B B Bn1 2, , ,L are constants.

5. Find the constants A A An1 2, , ,L , B B Bn1 2, , ,L .

e.g.0.4.9 Resolve the following fractions in partial frations:

i) x xx x

3

23 42 1

; ii) 443x x+

; iii) 113x +

;

iv)1

15 4 3 2x x x x x + + ; v)

x x x

x

3 2

4

6 4 8

3

+ +

( ).

e.g.0.4.10 Evaluate

i)1

2 2

a x

dx

; ii)

1

1

5 4 3 2

x x x x x

dx

+ + ;


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iii)2 3 3

1

4 3 2

3

x x x x

xdx

+ +

; iv)2 6 8

1 1

2

4

x x

x xdx

+ +

+ ( )( ) .

e.g.0.4.11 Using the substitution t ex= , find

i)

4 6

9 4

e e

e e dx

x x

x x

+

; ii)e

e dx

x

x

2

1 + .


By the method of partial fractions, find the following indefinite integrals.

a)1

1x xdx

( )+ ; b)x

x xdx

+

+ 1

4 52;

c)

2 3

6 3 2

x

x dx

+

+ ( ) ; d)x

x x dx2 4 4 + .


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0.4.3 Integration by Parts

Theorem 0.4.2 (Integration by Parts)

Let u=f(x) and v=g(x) be two real-valued functions with continuous first

derivatives. Then udv uv vdu= , i.e.

f x g x dx f x g x g x f x dx( ) ' ( ) ( ) ( ) ( ) ' ( )= .

Integration by parts is particularly useful when the integrand contains

i) a product of two factors;

ii) an inverse trigonometric functions;

iii) a logarithmic functions.

e.g.0.4.12 Evaluate

i) lnxdx ; ii) sin 1xdx .

In the integral f x g x dx( ) ( ) , if f xdu

dx( ) = for some function u(x) and g(x) cant be

integrated by ordinary means, then try

f x g x dx g x du u x g x u x dg x( ) ( ) ( ) ( ) ( ) ( ) ( )= = .

e.g.0.4.13 Evaluate

i) x xdxtan 1 ; ii) sin ln(tan )x x dx ; iii) x x dx(tan ) 1 2 .


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In the integral x g x dxn ( ) , if g xdv

dx( ) = for some function v(x), then try

x g x dx x dv x v x v x dxn n n n( ) ( ) ( )= = .

e.g.0.4.14 Evaluate

i) xe dxx3

; ii) x xdx2

cos .


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If there are sinnx or cosnx in the integrand, we should reduce them to

compound angle.

e.g.0.4.15 Evaluate x xdxcos2

.

Sometimes, after integrating by parts, the new integral on the right hand side

may be expressible as a sum of a multiple of the original integral and an integral

which can be readily integrated. This technique is illustrated in the following

examples.

e.g.0.4.16 Evaluate

i) sec3xdx ; ii) e xdxx sin ; iii) cos(ln )x dx .


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1. Using the method of integration by parts. Find

a) x xdxk ln where k 1; b) ln( )

x

xdx

12

and

c) sec tanx xdx2 .

2. a) Derive the following reduction formula

x e dxa

x en

ax e dx

n ax n ax n ax = 1 1 where a 0.

b) Hence evaluate x e dxx3 2 .

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Basic Mathematical Knowledge