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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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Chapter 0. Basic Mathematical Knowledge/P.2
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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Chapter 0. Basic Mathematical Knowledge/P.3
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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Chapter 0. Basic Mathematical Knowledge/P.4
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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Chapter 0. Basic Mathematical Knowledge/P.5
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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Chapter 0. Basic Mathematical Knowledge/P.6
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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Chapter 0. Basic Mathematical Knowledge/P.7
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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Chapter 0. Basic Mathematical Knowledge/P.8
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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Chapter 0. Basic Mathematical Knowledge/P.9
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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Chapter 0. Basic Mathematical Knowledge/P.10
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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Chapter 0. Basic Mathematical Knowledge/P.11
Class Practice 0.2.1
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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Chapter 0. Basic Mathematical Knowledge/P.12
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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Chapter 0. Basic Mathematical Knowledge/P.13
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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Chapter 0. Basic Mathematical Knowledge/P.14
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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Chapter 0. Basic Mathematical Knowledge/P.15
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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Chapter 0. Basic Mathematical Knowledge/P.16
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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Chapter 0. Basic Mathematical Knowledge/P.17
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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Chapter 0. Basic Mathematical Knowledge/P.18
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.
Class Practice 0.3.0
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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Chapter 0. Basic Mathematical Knowledge/P.19
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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Chapter 0. Basic Mathematical Knowledge/P.20
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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Chapter 0. Basic Mathematical Knowledge/P.21
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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Chapter 0. Basic Mathematical Knowledge/P.22
e.g.0.4.3 Evaluate the integrals in e.g.0.4.1 and 0.4.2 directly.
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Chapter 0. Basic Mathematical Knowledge/P.23
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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Chapter 0. Basic Mathematical Knowledge/P.24
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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Chapter 0. Basic Mathematical Knowledge/P.25
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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Chapter 0. Basic Mathematical Knowledge/P.26
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+
+ .
Class Practice 0.4.0
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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Chapter 0. Basic Mathematical Knowledge/P.27
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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Chapter 0. Basic Mathematical Knowledge/P.28
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 + .
Class Practice 0.4.1
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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Class Practice 0.4.2
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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