INTEGRATION(‫)التكامل‬
DEFINITE INTEGRAL)‫(التكامل المحدد‬:
ƒ(x) ‫[تعرف بالتكامل المحدود للداله‬a,b] ‫ في الفتره‬ƒ(x) ‫العمليه العكسيه لتفاضل الداله‬
b
x ‫ هى المكامل و‬ƒ ‫ هى عالمة التكامل والداله‬ ‫ و العالمة‬ f
( x )dx
‫ ويكتب‬x ‫بالنسبة ل‬
a
‫هو متغير التكامل‬
The set of all anti derivatives of ƒ(x) in [a,b]is the definite integral
b
of ƒ(x) with respect to x , denoted by  f (x )dx
a
The symbol  is an integral sign. The function ƒ is the integrand of
the integral, and x is the variable of integration.
PROPERTIES OF DEFINITE INTEGRALS)‫(خواص التكامل المحدد‬:
If
f (x )
and
g (x )
are integrable in [a,b] then:
: ‫[ عليه‬a,b] ‫دوال قابله للتكامل في الفتره‬
b
b
b
 f (x )  g (x )  dx  f (x )dx  g (x )dx




a
a
a
b
b
(2)  Af (x )dx  A  f (x )dx
a
a
(1)
b
c
b
(3)  f (x )dx   f (x )dx   f (x )dx
a
a
c
b
a
(4)  f (x )dx    f (x )dx
a
b
1
g (x )
‫و‬
f (x )
‫اذا كان‬
a
(5)   f (x )  dx  0
a
b
b
(6)  f (x )dx   f (x ) dx
a
a
if a  b
INTEGRAL OF ELEMENTARY FUNCTION)‫(تكامل الدوال االساسيه‬:
‫ وفي كل حاله يجب اضافة‬.‫النتائج االتيه يمكن التوصل اليها باجراء التفاضل للطرفين‬
‫ثابت اختياري‬
The following results can be demonstrated by differentiating both
sides. In each case an arbitrary constant c should be added.
(1)
POLYNOMIALS(‫)كثيرات الحدود‬:
(1)
(i) ∫ 0 dx = c
(ii) ∫ 1dx = x + c
(iii) ∫ Adx
n
= Ax + c
(iv) ∫ x dx
(iv )
n
 x dx 
(v )
x n 1
c
n 1
 ax  b 
x n+1
=
+c
(n + 1)
 n  1
ax  b 

n 1
n
a (n  1)
c
2
  g (x ) 
(vi )
n
g (x )dx
 g (x ) 

n 1
n 1
c
dx
f (x )dx
 ln x  c  (ii ) 
 ln f (x )  c
x
f (x )
If the numerator is the derivative of denominator then the integration is equal to

(2) (i )
the logarithmic of denominator
‫اذا كان البسط تفاضل المقام فان ناتج التكامل هو لوغاريثم المقام‬
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
3
(1)
4
  2x  4  dx
 x
(2)
1
 6x  5  dx
2
(3)
0
 x
2
 x  dx
(4)
  4x  7 
Solution:
3
(1)
  2x  4  dx  x
2
 4x    (3) 2  4(3)    (1) 2  4(1)   39  5  34
3
1
1
4
x 3
  (4)3

140
(2)   x  6x  5  dx    3x 2  5x   
 3(4) 2  5(4)    0   
3
 3
0  3

0
x 1 x 2
(3)   x 2  x  dx 

c
1 2
4
2
(4)
  4x  7 
21
dx
 4x  7 

22
4(22)
c
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
1
 4x dx
(2)
dx
x6
(3)
2xdx
 x 2  10
(4)
Solution:
dx
 4 ln x  c
x
dx
x 5
(2)  6   x 6dx 
c
x
5
(1)
 4x
dx  4 
1
3
3x 2  5
 x 3  5x
21
dx
2xdx
2
 10
Since the numerator is the derivative of denominator then we find
(3)
x
2xdx
 ln  x 2  10   c
2
x  10
3x 2  5
(4)  3
x  5x
Since the numerator is the derivative of denominator then we find
3x 2  5
(4)  3
 ln x 3  5x
c
x  5x



Examples:
Find the following integrals )‫(احسب التكامالت االتيه‬
3x 5  2x 3
(1) 
dx
x8
xdx
(2)  2
x  4x  5
5
(3)
2xdx
2
5
x
2
Solution:
3x 5  2x 3
2 
3x 2 2x 4
 3
3
5
dx


dx

3
x

2
x
dx


c


 x8
  x 3 x 5  
2
4
xdx
(2)  2
By completing the square
x  4x  5
xdx
xdx
 x 2  4x  5   (x  2)2  1 let u  x  2  du  dx and x  u  2
xdx
u 2
udu
2du
1

  2 du   2
 2
 ln u 2  1  2 tan 1 u
2
(x  2)  1
u 1
u 1 u 1 2
xdx
1
  2
 ln  (x  2) 2  1  2 tan 1 ( x  2)  c
x  4x  5 2
5
5
2xdx
(3)  2
 ln x 2  5   ln 20  ln 1  ln 20
2
x 5
2
(1)
5
another solution
20
2xdx
du
2 x 2  5  1 u  ln u  ln 20  ln 1  ln 20
(2)EXPONENTIAL AND LOGARITHMIC FUNCTIONS)‫(الدوال االسيه واللوغاريثيميه‬:
With the Fundamental Theorems of Calculus it is possible to rigorously
develop the logarithmic, exponential
4
ab x
(i )  a dx 
c
b ln a
1
(ii )  e a x dx  e a x  c
a
a  0, a  1
bx
(3)
 ln x dx  x ln x  x  c
(4)
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
5x
5x
 3 dx
(2)
1
  7  dx
e
(3)
8x
dx
(4)

log 3 xdx
x
(5)
 xe
Solution:
(1)
5x
 3 dx 
35x
c
5ln 3
7 9 x
9 x
1
dx

7
dx

  7 
 
9 ln 7
1
(3)  e 8 x dx  e 8 x  c
8
log 3 xdx
(4) 
x
9x
(2)
c
 ln x   c
log3 xdx
ln x
1 ln x
log 3 x 


dx

ln 3
x
ln 3  x
2 ln 3
2
2
2
1
1
(5)  xe x dx   e x d (x 2 )  e x  c
2
2
2
Exercise(1):
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
1
(1)
2
x
dx
(2)
0
 3x
3
dx
(3)

2
(4)
 x  2
x
dx
(5)
1
(7)

7
1
(6)   
3
0 
4
cos t
sin tdt
0

2  1 x 2 dx
e
(8)
2 1
dx

2
2
x
2
 ln 2  1
 x dx
(9)
1
5
2ln x
1 x
tan x
sec 2 xdx
x2
dx
4
(10)
4
(11)
log xdx
 10x
(14)
log xdx
1 5x
(15)
ln 2 log 2 xdx
x
1
(17)
log 2 (x  2)
dx
(x  2)
0
(18)
dx
 x log10 x
(20)
2
(13)
(19)

ln 4
(22)

1
(25)
x
5 x

log10 xdx
x
e
2 ln10 log10 xdx
x

1
2

2 log 2  x  1 dx
2  x  1
dx
 x  log x 
(23)
x
9
(21)
e dx
(26)

0
1
2 log10 (x  1)dx
 x  1
log 7 xdx
x
 t dt
(24)

t
(27)
 x x
1
6
2
8
3
e
1
dt
t
(12)
4
4
(16)
2 x dx
1 x
 3 1  ln x  dx
2 t3
e dt
2
 3 dx
4
‫سانبئك عن تفصيلها ببيان‬
‫اخي لن تنال العلم اال بستة‬
‫ذكاء وحرص واجتهاد وبلغة وصحبة استاذ وطول زمان‬
(3)TRIGONOMETRIC FUNCTIONS)‫(الدوال المثلثيه‬:
(5)
(6)
(7)
(8)
(9)
 sin xdx   cos x  C
 cos xdx  sin x
 tan xdx  ln sec x   ln cos x
 cot xdx  ln sin x  C
 sec xdx  ln sec x  tan x  C
(10)
(11)
C
(12)
(13)
(14)
 csc xdx  ln csc x  cot x  C
 sec xdx  tan x  C
 csc xdx   cot x  C
 sec x tan xdx  sec x  C
 csc x cot xdx   csc x  C
2
2
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
6
(1)
 sin 2xdx
(2)
 cos 6xdx
 tan xdx
(3)
(4)
 cot 3xdx
Solution:
1
cos 2x  c
2
1
(2)  cos 6xdx  sin 6x  c
6
(1)
 sin 2xdx

(3)
 tan xdx
(4)
 cot 3xdx
  ln cos x
c
1
 ln sin 3x  c
3
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬

2
(1)
4 cos xdx
 3  2sin x
(2)

2
 (tan 2x  sec x tan x )dx
Solution:

2
(1)



4 cos xdx

 
 2 ln 3  2sin x  2  2 ln 3  2sin
 2 ln 3  2sin  
3  2sin x
2
2
2
2
 2 ln 1  2 ln 5  2 ln 5
1
(2)  (tan 2x  sec x tan x )dx   tan 2xdx    sec x tan x  dx  (ln sec 2x )  (sec x )  c
2
Examples:
Calculate the following integrals )‫(احسب التكامالت االتيه‬
(1)
 sec 2xdx
(2)
 csc
2
7 xdx
(3)
 csc 9xdx
Solution:
1
ln sec 2x  tan 2x  c
2
1
(2)  csc 2 7 xdx  cot 7 x  c
7
1
(3)  csc9xdx  ln csc9x  cot 9x  c
9
(1)
 sec 2xdx

Examples:
prove the following integrals )‫(اثبت التكامالت االتيه‬
7
(i )
 tan xdx
  ln cos x
(ii )
 sec xdx
 ln sec x  tan x
Solution:
sin xdx
 sin xdx
 
  ln cos x  c
cos x
cos x
sec x  sec x  tan x  dx
(ii )  sec xdx  
 ln sec x  tan x  c
 sec x  tan x 
(i )
 tan xdx  
Exercise(2):
(A) Find the simplest antiderivatives you can for each part and add the
arbitrary constant of integration at the end.
‫احسب ابسط صوره ممكنه للتكامالت االتيه ثم اضف ثابت التكامل لكل حاله‬
(1)
(4)
(7)
 (3x
 5
 sec x )dx
(2)
 e  x  x 1  dx
(5)
2
3x
 x
2
  2  x


 dx

 sec2 x 
  4  dx
1 
1

(13)   x   x   dx
x 
x

 2 1 
(16)    2  dx
x x 
(10)
(43)
  x  4  x  2  dx
  x  5x  dx
 (e  cos x )dx
 sec 4xdx
 x tan  x  1 dx
 tan x sec xdx
 cot x csc xdx
(46)
 x sinh  x  dx
(49)
x
 xe dx
(19)
(28)
(31)
(34)
(37)
(40)
2
x
2
3
(8)
13
 2  13 1 
  3t  t  t 5  dt
2
(6)
 (tan 6x  sec x )dx
 x  x  1 dx
(9)
 
2
(3)

3
t t  t 
 dt
t2


  5 sec tan   d 
(14)
7 211
 5 x dx
(15)
2
(17)
 3e
(18)
  x
(20)
(29)
(32)
(35)
(38)
2
(41)
3
2
(44)
2
2x )dx
(11)
5
2
 (2x  cos 7x 
  4  8x  dx
dx
  2x  3dx
  r  r  1 dr
 e sin 2xdx
 pe dp
 sec (3x  1)dx
 sec x tan xdx
 sin x cos xdx
(47)
(50)
x
15
x dx
 2
 x
(21)
5
3

1 
 dx
x5
 2x 1  7 x  1 dx
2
(39)
2
(42)
3
(45)
 e sin  d 
 e cos 3xdx
 x ln xdx
 x coth x dx
 csc x cot xdx
 x tanh(x  1)dx
(48)
 cos
2
2
p2

 6cos3d 
(12)
 sin x 
4
dx
1  cos x
2
ex 2
 e x dx
8
4
(30)
2
(33)
1
(36)
2
3
2
2
3
3x sin 3xdx

(51)

0
1  cos 2 x dx
(52)



3
2
 (cos 3x  sin 4x )dx
4
 6 cot xdx
(53)
 cot

(54)


4
xdx
6


4
(55)
2
4
 3 tan xdx
(56)
0
4
x
2 x
 e sec e dx
(57)
 tan
2
xdx
0


0
12
x
(58)  3csc dx
2

2
(59)
 6sec
2
2xdx
(60)
 (sin 7x  cos 3x )dx

0
(B) prove that)‫(اثبت االتي‬:
(1)  cot xdx  ln sin x
(2)
 cot xdx
 ln sin x
(3)
 csc xdx
 ln csc x  cot x  c
ٍ ‫و َما ِمن َكات‬
ُ‫ّهر ما كتَبت يداه‬
ُ ‫ و يُب ِقي الد‬،،، ‫س َيـــفنَى‬
َ ّ‫ب إال‬
‫غير‬
ّ ،،، ‫شيئ‬
ُ‫يسر َك في ال ِقيا َم ِة أَن تـراه‬
َ ‫فالَ تكتُب ِبكفّ َك‬
ٍ
(4)HYPERBOLIC FUNCTIONS)‫(الدوال الزائديه‬:
(15)
(16)
(17)
(18)
(19)
(20)
 sec hx tanh xdx   sec hx  c
(22)  csc hx coth xdx   csc hx  c
(23)(i )  sec hxdx  tanh  sinh x   c
(ii )  sec hxdx  ln sec hx  tanh x  c
(24)(i )  csc hx   coth  cosh x   c
(ii )  csc hxdx  ln csc x  cot x  c
 sinh xdx  cosh x  c
 cosh x  sinh x  c
 tanh xdx  ln cosh x  c
 coth x  ln sinh x  c
 csc h x   coth x  c
 sec h xdx  tanh x  c
(21)
1
1
2
2
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
 sinh 8xdx
(2)
 sec h 2xdx
(3)
 cosh
9
2
5xdx
(4)
 coth 3xdx
Solution:
(1)
 sinh 8xdx
1
 cosh 8x  c
8
1
ln sec h 2x  tanh 2x  c
2
1
(3)  cosh 2 5xdx   coth 5x  c
5
1
(4)  coth 3xdx  ln sinh 3x  c
3
(2)
 sec h 2xdx

Exercise(3):
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
x
(1)
 sinh 2xdx
(2)
 sinh 5 dx
(4)
 4 cosh  3x  ln 8 dx
(5)
 tanh 7 dx
(7)
 sec h
(8)
 csc h  3  x  dx
2
1

 x   dx
2

x
ln 4
(10)
x
 coth
 ln 2
ln 2
 coth xdx
(11)
ln 2

d
5
csc h  ln x  coth  ln x  dx
(9) 
x
(6)
2

 6 cosh  2  ln 4  dx
(3)
 tanh 2xdx

(12)
2e x cosh e x dx
 ln 4
0

ln 2
(13)

2
2
4sinh xdx
(14)
0
 2sinh(sin t ) cos tdt
cosh(ln t )dt
t
1

(15)
0

4
4
(16)
 cosh(tan x ) sec
2
xdx

4
(19)
x 
cosh 2   dx
2
 ln 2
(22)
 2  x 
8cosh x
(17) 
dx
x
1
0
(25)

dx
2
2
 x x  7dx
sds
(20)

(23)
 sin 3x cos 2xdx
(26)

s2 2
3x 2dx
x 3 2
General Exercise(1):
10
ln10
 4sinh
(18)
2
0
x 
  dx
2
cos 3xdx
(21)
 4  sin 3x
(24)

(27)
 sin 2x sin 5xdx
2xdx
x 2 4
(A) Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
  3sec x tan x  2sec x  dx
(3)
  sin 3x  csc x  dx
2
(10)
 (2  tan
2
x  csc x cot x  dx
 1  tan   d 
2
(4)

2
1
 Hint :1  tan   sec 2  
csc x
(6)  cos   tan   sec  d  (7) 
dx
csc x  sin x
(9)  1  cot 2 x  dx   h int :1  cot 2 x  csc 2 x  
2
1  cos 5x
dx
7
(8)  cot 2 xdx
(5)
 2  csc
(2)

dx
(11) 
2 x  2x
x )dx
x
 2 cot 3 dx

(12)
2
2
(13)
4
2ln x
x

1
dx
 x  ln x 
(14)

sin udu
 2  cosu
(15)
2
2
0

16
(16)
(19)
4sin xdx
(17) 
1  4 cos x
0
sec xdx
ln(sec x  tan x )

4
3
dx
2 2x ln x
(20)
0
sec y tan ydy
 2  sec y
8rdr
2
5
r
(18)
3sec 2 xdx
 6  3 tan x
(21)

e 2x
2
(22)
e
sin x
cos xdx

(23)
0
(25)
e4
  2e
x
 3e 2 x 
(26)
ln xdx

(24)

(27)
e
1
x
e
(28)  3
x
r
r
ln16
e
(29)  2 dx
x

e  x dx
 ln 2
x
(2 x 1)
 2e dx
1
x2
e
(30)
dr
x
4
0

 1  e  csc
2
(31)
0
cot x
 1  e  sec
4
2
xdx

(32)
tan x
2
xdx
(33)
e
sec  x
sec  x tan  xdx
0
4
(34)
csc(  x )
csc(  x ) cot(  x )dx
e
ln 
(37)

0
2
2
2xe x cos(e x )dx
(38)
ln 3
(35)
e
x
ln 2
dx
 1 e x
(39)
11
(36)
 e
e x dx
 1 e x
3x
 5e  x  dx
(40)
(43)
(46)
 2 cos xdx
 x ln e dx
(44)
 e sec xdx
 x sin x dx
(45)
3
x
(47)
 sin
(48)
 cos
sin x
(41)
x2
cos xdx
 2  3sin x
tan x
2
2
4
xdx
(42)
2cos x
sin xdx
x 2  4dx
3
xdx
(B) Right, or wrong? Say which for each formula and give a brief reason
For each answer.
‫وضح اي االجابات االتيه صحيحه ومن ثم علل لذلك في كل اجابه‬
(1)
2
 tan x sec xdx 
(i )
sec3 x
c
3
1
tan 2 x  c
2
1
(iii )  tan x sec 2 xdx  sec 2  c
2
(ii )
(2)
 tan x sec
2
xdx 
x2
sin x  c
2
(ii )  x sin x 2dx  x cos x 2  c
2
 x sin x dx 
(i )
 x sin x
(iii )
1
dx   cos x 2  c
2
2
(5)INVERSE TRIGONOMETRIVC FUNCTIONS)‫(الدوال المثلثيه العكسيه‬:
a0
‫ناتج التكامالت االتيه دوال مثلثليه عكسيه وجميع النتائج صحيحه عند‬
Integrals evaluated with inverse trigonometric functions
The following formulas hold for any constant a  0
(25)

dx
 sin 1
x
x
  cos 1
a
a
a x
dx
1
x
1
x
(26)  2
 tan 1   cot 1
2
a x
a
a
a
a
(27)
x
2
2
dx
 Valid for x
2
 a2 
(Valid for all x)
1
x
1
x
 sec 1   csc 1
a
a
a
a
x a
2
2
 Valid for
Examples:
12
x > a >0 
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
dx
 16
x
3
2
(5)
(2)
2
1 x
2
2
9x
(3)
2
1
dx

dx

(6)
2
dx
0 1  x 2
(7)
x
dx

dx
x 4
2
(4)

 sin 3x  x
2
3  4x
2
(8)
x
2
3
6 
dx
 5 
2
dx
x 2 1
Solution:
dx
1
x 
 tan 1    c
 16 4
4
(1)
x
(2)

(3)
x
(4)
 sin 3x  x
2
9x
1
x 
  sec1    c
2
2
x 4
2

dx

1
2
0
(7)


(8)
dx
3  4x
dx
3  4x 2
2
x
2
3
3
2
2
2
 3
 2   
1
 sin 1 
  sin 
   
 2 
 2  3 4 12
1

 
  tan 1 x    tan 1 1   tan 1 0       0  
0
4
4
dx
 1 x
6 
dx
1
6
 x 
dx   sin 3xdx  6 2
 cos 3x 
tan 1 
 c

 5
x 5 3
5
 5
2
 sin 1  x  
1 x 2
2
2
(6)
2
dx
3
2
(5)
x 
 sin 1    c
3
dx
a  3 and
2

1
2
dx
x 1
2
du
 3
2

u 2
 sec 1 x 
2
2
3
u  2x  du  2dx  dx 
du
2
1  1  u   1 1  2x 
sin 
   sin 
 c
2
 3  2
 3
    
    
 4   6  12
Example:
13
By completing the Square evaluate)‫(باكمال المربع‬
dx

(1)
4x  x
2
(2)
x
2
dx
 4x  20

(3)
dx
e 2x  6
Solution:
dx

(1)
4x  x 2
The expression
4x  x 2 does not match any of the formulas of integtation so we first rewrite
4x  x 2 by completing the square: 4x  x 2    x 2  4x     x 2  4x  4   4
   x  2    4  4   x  2 


dx
dx


2
2
4x  x
4   x  2
2
2
Then we substitute a = 2 and u = (x-2)  du = dx

dx

4x  x 2
x
(2)

dx
4   x  2
2

  x  2 
u 
 sin 1    c  sin 1 
 c
2
4 u 2
 2 
du
dx
 4x  20
2
We complete the square on the x 2  4x  20 

dx
  x  2

x
(3)


2
2
 16

e 2x  6
dx
e 2x  6

2
  x  2
 4  20 
2
 16

du
1
1
u 
 x 2
 tan 1    c  tan 1 
 c
u  16 4
4
4
 4 
2
dx
dx

2
 4x  20
 x  2   16
dx
 x  2
let
u  x  2  du  dx
Using Substitution u = e x  du  e x dx  dx 
du
ex
du
x

du
du
1
 u  1
1  e
ex



sec 1 

sec

 c

6
6
 6
u2 6
ex u2 6
u u2 6
 6
Exercise(4):
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
14
(1)
dx

9x 2
dx
(4) 
9  3x 2
(7)
4dt

4 t

(5)
x
 2
2

(13)
1
25x 2  2
y 4 y 2 1
x
2
dx
5x 2  4
dt

9  4t 2
0
 2
3

(12)
2
2
3
ds
 8s
(14)
(6)
(9)
2
4   r  1
2
dy
 17  x
6dr

(11)
dx
(3)
3 2
4
1  4  r  1
2
dt
(10) 
4  3t
2
1  4x 2
dx
3dr

(8)
2
dx
(2)
y 9 y 2 1
dx
 2   x  1
(15)
2
dy
0

(16)


2 cos  d 
2
dx
 2x  1  2x  1
2
 1   sin  
(17)
4
2

2
4
2
csc 2 xdx
 1   cot x 
(18)

2
6

ln 3

e dx
1 e x

sec 2 ydy
(19)
0
(22)

6dt
3  2t  t
1
(28)

1
2
2
2
6dt
3  4t  4t
2
8dx
(31)  2
x  2x  2
1
dx
(23)

(26)
x
 x  3  x  3
(32)

1 x 2
(35)

2
 25
dx
 2x  5
2
ds
 6s  10
  x  1
1
(34)
2
s
(29)
2
e cos x dx
(21)

(24)

(27)

1
1  tan y
1
4dt
 t 1  ln t 
(20)
2
0
(25)
e4
x
 sin
1
(30)
xdx
1 x 2
dx
 x  4x  3
2
dx
2x  x 2
  x  2
dx
x 2  4x  3
1
dx
x 2  2x
x  dx
(33)

e sin x dx
2
(36)
1 x 2
15

1 x 2
tan 1 x dx
1 x 2
ds
  tan 1 s 1  s 2 
(37)
2
(38)

sec 2  sec 1 s  ds
2
s s 2 1
2
(39)

cos  sec 1 x  dx
2
3
x x 2 1
(6)INVERSE HYPERBOLIC FUNCTIONS)‫(الدوال الزائديه العكسيه‬:
x
(i )
(28) (i )

(ii )

(iii )

(iv )

(30)
x
(31)
x
dx
1
x
1
ax
 tanh 1 
ln
2
x
a
a 2a a  x
a  x 
dx
1
x
1
ax
  coth 1 
ln
2
a
a
a 2a a  x
a  x 
a
(29) (i )
2
2
dx
a x
dx
2
x a
2
2
2
xdx
a x
2
2
xdx
a x
2
dx
a2  x 2
dx
a2  x 2
2
 sinh 1
x
 c  ln  x  x 2  a 2   c


a
 cosh 1
x
 c  ln  x  x 2  a 2   c


a
 a2  x 2  c
  a2  x 2  c

1
x
sec h 1
a
a

1
x
csc h 1
a
a
‫ومن هدم دينه كان لمجده أهدم‬
‫من ظلم نفسه كان لغيره أظلم‬
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)

dx
4x
2
(2)
x
3dx
(3)
2
 16
x
dx
x  25
2
Solution:
16
(4)
3
  x 

x  dx

dx
x 
 sinh 1    c
2
(1)

(3)
x
(4)
1
dx
2 23
3

2

x
dx

3

x
dx

3ln
x

x c
  x
x 

3
4x
3dx
3
x 
(2)  2
 tanh 1    c
x  16 4
4
2
dx
1
x
 sec 1  c
5
x  25 5
2
Exercise (5):
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
2 3
(1)

(4)
x
1
5

(7)

0
0
2
3dx
1  9x 2
e
cos xdx
(8)
1  sin 2 x

1
2
(6)
x
1
dx
x 1   ln x 
2
5
4
dt
(5) 
1t 2
0
2
dx
 1 x
(3)
1
2
dx
1  16x

(2)
4x 2
0
3
13
1
3
dx
2
(9)
dx
4x 2
1  sin 3 x
 sin 2 x dx
(B) Right, or wrong? Say which for each formula and give a brief reason
For each answer.
‫وضح اي االجابات االتيه صحيحه ومن ثم علل لذلك في كل اجابه‬
(1)
sec3 x
(i )  tan x sec xdx 
c
3
1
(ii )  tan x sec 2 xdx  tan 2 x  c
2
1
(iii )  tan x sec 2 xdx  sec 2  c
2
2
17
(2)
x2
(i )  x sin xdx 
sin x  c
2
(ii )  x sin xdx  x cos x  c
 x sin xdx  x cos x  sin x  c
(iii )
(3)
dx
(i )

(ii )

(iii )

1 x
dx
1 x
dx
 sin 1 x  c
2
2
1 x
2
 sinh 1 x  c
 cosh 1 x  c
(C) Verify the integration formulas in Exercises 1-10
4 ‫ الى‬1 ‫حقق التكامالت االتيه في التمارين من‬
(1)
tan 1 xdx
 x2
1
tan 1 x
 ln x  ln 1  x 2  
c
2
x
x4
5
x 4dx
(2)  x 3 cos 1 5xdx 
cos 1 5x  
4
4 1  25x 2
(3)
  sin
(4)
 ln a
1
2
x  dx  x  sin 1 x   2x  2 1  x 2 sin 1 x  c
2
2
 x 2  dx  x ln  a 2  x 2   2x  2a tan 1
x
c
a
‫ شريف السمع كريم النظر‬..... ‫الطريق عفيف الخطى‬
‫كن في‬
ِ
‫مــــــــــــــر و هذا األثر‬
‫ يقولون‬..... ‫و كن رجال إن أتو بعده‬
َ
SPECIAL METHODS OF INTEGRATION ‫طرق خاصه للتكامل‬
(1)INTEGRATION BY SUSTITUTION )‫(التكامل بالتعويض‬:
‫اذا كانت الداله المعنيه وتفاضلها موجدين داخل عالمة التكامل فيمكن استبدال‬
u  g (x )  du  g (x )dx ‫ ومن المناسب استخدام‬.‫المتغيرات‬
18
Many functions are formed by using compositions. In dealing with
composite function it is useful to change variables of integration. It is
convenient to use the following differential notation:
If
u  g (x )  du  g (x )dx
  f  g (x )   g (x )dx   f (u )du
Example:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
 x cos x
2
dx
(2)
 3x
2 x3
e dx
 sin x cos xdx
(3)
(4)
Solution:
(1)
 x cos x
let
u  x 2  du  2xdx  dx 

2
dx
du
du
  (x cos u )( )
2x
2x
1
1
1
cos udu  sin u  c  sin x 2  c

2
2
2
(2)
 3x
2 x3
e dx
let
u  x 3  du  3x 2dx  dx 
du
3x 2
3
3
 du 
  3x 2e x dx   3x 2  e u
  e u du  e u  c  e x  c
2 
 3x 
(3)
 sin x cos xdx
let
u  sin x  du  cos xdx  dx 
 du 
1
 sin x cos xdx   (u cos x )  cos x    udu  2 u
2
19
1
 sin 2 x  c
2
du
cos x
dx
 1  x  tan
2
1
x
dx
 1  x  tan
(4)
2
1
x
dx
 dx  1  x 2  du
1 x 2
1  x 2  du

dx
du



 ln u  ln  tan 1   c


2
1
2
1  x  tan x 1  x u u
u  tan 1 x  du 
let
Example:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬

2
6
 tan 2xdx
(1)
(2)
0

1
dx
x
2
 2x  4 
3
2
Solution:

6

(1)
tan 2xdx
u  2x  dx 
let
x 0

du
2


1 3
1
1
1
  tan 2xdx   tan udu   ln sec u 03   ln 2  ln1  ln 2
2 u 0
2
2
2
0
6
2

(2)
1
2
dx
x
2
 2x  4 
3
2

1
dx
 (x  1)
3
2
 3 2
x  1  3 tan u  dx  3 sec udu

2


1
 (x  1)2  3

6
dx
3
2

0
6
3 sec udu
(
3 tan u )  3
2

3
2

0

6
3 sec udu
 3 tan 2 u  3
3
2

0
3 sec udu



16
1
1
6 
cos
udu

sin
u



0
30
3
6
Example:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
20
3
3sec2 u  2
(1)
 (x  2)sin(x
(4)

2
 4x  6)dx
cot(ln x )dx
x
2
1 x
(5)
(2)
dx
(x  2)(3  x )

tanh 21x dx
(6)
(3)

x sin 1 x 2dx
1 x 4
xdx

x  x 1
2
Solution:
 (x  2) sin(x
(1)
2
 4x  6)dx
1
u  (x 2  4x  6)  du  (2x  4)dx  du  (x  2)dx
2
1
1
1
  (x  2) sin(x 2  4x  6)dx   sin udu   cos u   sin(x 2  4x  6)  c
2
2
2
dx
(x  2)(3  x )

(2)
u x 

25
1
 (x  ) 2
4
2


x sin 1 x 2dx

(3)
6x x
2

dx
6  (x  x )
2

dx
25
1
 (x  ) 2
4
2
1
 du  dx
2
dx

dx

1 x
1 x 4
let
4
x sin 1 x 2dx


 
 
du
 2x  1 
1 u
 sin    sin 1 
 c
5
25
 5 
2


u
2
4
u  sin 1 x 2  du 
2xdx
1 x 4
1
1 2 1
1
2 2
udu

u

sin
x

 c
2
4
4
cot(ln x )dx
dx
let
u  ln x  du 
x
x
cot(ln x )dx
 
  cot udu  ln sin u  ln sin(ln x )  c
x
(4)

(5)
2

1 x
2
1 x
tanh 21 x dx
tanh 21 x dx 
let
u  21 x  du  21 x (ln 2)dx  21 x dx 
1
1
1
tanh udu 
ln cosh u 
ln cosh 21 x  c

ln 2
ln 2
ln 2
21
 du
ln 2

1  2x  1  1  1  (2x  1)dx
dx
 
 



x 2  x 1 2  x 2  x 1  2  x 2  x 1
x 2  x 1 
1
dx
1
dx
 x 2  x 1  
 x 2  x 1  
2
2
1
3
x 2  x 1
(x  ) 2 
2
4
1 
1
1
3
 x 2  x  1   ln  x    (x  ) 2    c
2 
2
2
4
(6)
xdx

Example:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
(4)
 sin x dx
 cot xdx
2
(2)
2
(5)
 cos
 sec
2
xdx
(3)
2
xdx
(6)
 tan
 csc
2
xdx
2
xdx
Solution:
(1)
 sin
(2)
 cos
(3)
 tan
 cot
(4)
(5)
(6)
x dx 
2
1
1
cos 2x 
C
1  cos 2x  dx  x 

2
2
2 
1
1
cos 2x
1  cos 2x  dx   x 

2
2
2
2
xdx 
2
xdx    sec 2 x  1 dx  tan x  x  C
2
xdx    csc 2 x  1 dx   cot x  1  C
2
xdx  tan x  C
2
xdx   cot x  C
 sec
 csc

  C
Example:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
 sinh
2
x dx
(2)
 cosh
2
xdx
(3)
 tanh
2
Solution
22
xdx
(4)
 coth
2
xdx
since (i ) cosh 2 x  sinh 2 x  1
 (ii ) 1  tanh 2 x  sec h 2 x (ii ) coth 2 x  1  csc h 2 x
1
1
(iii ) cosh 2 x  1  cosh 2x  (iv ) sinh 2 x   1  cosh 2x 
2
2
(1)
 sinh
(2)
 cosh
(3)
 tanh
 coth
(4)
xdx 
2
1
1
sinh 2x
 1  cosh 2x  dx  x 

2
2
2
1
1
sinh 2x
1  cosh 2x  dx  x 

2
2
2
2
xdx 
2
xdx   1  sec h 2 x  dx  x  tanh x  C
2

  C

  C
xdx   1  csc h 2 x  dx  x  coth x  C
Exercise(5):
(1)
e
(4)
 csc(5x  2)dx
 cot x csc xdx
(7)
x3
x 2dx
3
2
sin
1
x
(2)
3
sinh(x 2  6)dx
(8)
x  dx

2
 sin x cos xdx
 sin 9xdx
4
(6)
(9)
(13)
  3x  2 
(16)
x
(19)
2
dx
x 2 3
dx
3
 sec x tan xdx
(11)
1 x 2
x 2dx
(21) 
1 x 6
2
 sec(3x  2)dx
4
(10)
2 x 1
 tan x sec xdx
 x tan(x  1)dx
5
(5)
(3)
(14)
(17)
 Hint

e
x
2
x
sin e x dx
2x 3
y  2x  1 
 Hint u  x 3  (22)

(12)
 csc
(15)
(18)
(20)
e x dx
 1  e 2x
 arctan x 
1 x
3
3
dx
2
x cot xdx
 x tanh(2  x
x
3
2
)
ln x dx
u  e x 
‫ارض بما قسم هللا لك تكن اغنى الناس‬
(2)INTEGRATION BY PARTS)‫(التكامل بالتجزئيه‬:
23
d (uv )  udv  vdu
‫ اعتمادا على قانون تفاضل ضرب دالتين‬. ‫ دوال قابله للتفاضل‬v‫ و‬u ‫اذا كانت‬
uv   udv  vdu
 udv
 uv  vdu
:‫وباخذ التكامل لطرفي المعادله نجد ان‬
: ‫وصيغة التكامل بالتجزئيه تكتب على الصوره‬
Let u and v be differentiable functions. According to the product rule for
differentials d (uv )  udv  vdu
Upon taking the antiderivatives of both sides of the equation, we obtain
uv   udv  vdu
This is the formula for integration by parts when written in the form
 udv
Or if
f (x ), g (x ), f (x ) and g  ( x )
 uv  vdu
be continuous on an interval [a,b] , then
‫[ عليه نجد ان‬a,b] ‫دوال مستمره في الفتره‬
 f (x ) g (x )dx
(i )
‫او اذا كانت‬
 f (x ) g (x )   g (x )f (x )dx
b
(ii )
f (x ), g (x ), f (x ) and g (x )
b
 f (x ) g (x )dx  f (x ) g (x )a   g (x )f (x )dx
b
a
Where u  f
a
(x ) , dv  g (x ) are
parts of the integration .
‫تمثل اجزاء من التكامل‬
u  f (x ) , dv  g (x )
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
 x sin xdx
(2)
 xe
4x
dx
(3)
x
2
cos xdx
Solution:
24
(4)
 ln xdx
(5)
x
3 x2
e dx
‫بحيث ان‬
(1)
 x sin xdx
let
u  x  du  dx
and dv  sin xdx  v   cos x
  x sin xdx  x cos x    cos xdx  x cos x  sin x  c
(2)
 xe
let
u  x  du  dx
4x
dx
dv  e 4 x dx   dv   e 4 x dx
and
x 4x
1
x
1
e   e 4 x dx  e 4 x  e 4 x  c
4
4
4
16

 xe
(3)
x
let
u  x 2  du  2xdx and dv  cos xdx  v  sin x
4x
dx 
2
cos xdx
 x cos xdx  x
2  x sin xdx
2
1
v  e 4 x
4
2
sin x   2x sin xdx
let again u  x  du  dx and dv  sin xdx  v   cos x
 2  x sin xdx  2x cos x  2   cos xdx  2  x cos x  sin x 

(4)
x
2
 ln xdx
 ln xdx
(5)
cos xdx  x 2 sin x  2  x cos x  sin x   c
x
u  ln x  du 
let
dx
x
and dv  dx  v  x
 dx 
 x ln x   x 
  x ln x   dx  x ln x  x  c
 x 
3 x2
e dx
  x 3e x dx 
2
let
u  x 2  du  2xdx
and
2
1 2
v  xe x  dv  e x
2
1 2 x2 1 x2
x e  e 
2
2
Examples:
Find the following integrals )‫(احسب التكامالت االتيه‬
(1)
 cos
1
xdx
(2)
 tan
1
xdx
(3)
 x sin
1
25
xdx
(4)
 x sec
1
xdx
Solution:
(1)
 cos
1
xdx
  u sin udu    u cos u  sin u 
  cos 1 xdx
 cos

(2)
1
 tan
xdx
 tan
(3)
xdx
 x sin
 x sin

1
1
1
u  tan 1 x  x  tan u  dx  sec2 udu
let
  u sec 2 udu  u tan u  ln cos u   c
  tan 1 xdx

 x cos 1 x  sin  cos 1 x   c
xdx
1
u  cos 1 x  x  cos u  dx   sin udu
let
 tan 1 x tan(tan 1 x )  ln cos(tan 1 x )  c
xdx
u  sin 1 x  x  sin u  dx  cos udu
let
xdx   u  sin u  cos udu 
1
u sin 2udu
2
1  1
1
 1
u cos 2u  sin 2u    sin 1 x cos(2sin 1 x )  sin(2sin 1 x )   c

2 2
4
 4
(4)
 x sec
 x sec
1
1
xdx
let
u  sec 1 x  x  sec u  dx  sec u tan udu
xdx   u sec u  sec u tan udu    u sec 2 u tan udu
1
1
1
1
 u tan 2 u  u  tan u   sec 1 x tan 2 (sec 1 x )   sec 1 x  tan(sec 1 x )   c
2
2
2
2
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
 ln(x
2
 2)dx
(2)
 sec
3
xdx
Solution:
2xdx
and dv  dx  v  x
x 2 2


x 2dx
2
  ln(x 2  2)dx  x ln(x 2  2)  2  2
 x ln(x 2  2)  2  1  2
 dx
x 2
 (x  2) 
4
 x 
  ln(x 2  2)dx  x ln(x 2  2)  2x 
tan 1 
 c
2
 2
(1)
 ln(x
(2)
 sec
3
2
 2)dx
xdx
let
let
u  ln( x 2  2)  du 
u  sec x  du  sec x tan xdx
26
and dv  sec 2 xdx  v  tan x
I   sec3 xdx  sec x tan x   sec x tan 2 xdx
  sec x tan 2 xdx   sec x (sec 2 x  1)dx   sec3 xdx   sec xdx  I  ln sec x  tan x
 I   sec3 xdx  sec x tan x   I  ln sec x  tan x   sec x tan x  I  ln sec x  tan x
 I   sec3 xdx  sec x tan x  I  ln sec x  tan x
 I  I  2I  sec x tan x  ln sec x  tan x
 I   sec3 xdx 
1
 sec x tan x  ln sec x  tan x   C
2
Exercise(6):
‫ باستخدام تعويض مناسب‬12 ‫ الى‬1 ‫احسب التكامالت غير المحدوده في التمارين من‬
‫الختزال التكامالت الى الصورالقانونيه‬
(A)Evaluate the indefinite integrals in Exercises 1–12 by using the
suitable substitutions to reduce the integrals to standard form.
(2)
 x sin(2x
z 
z

(4)  1  cos  sin dz
2
2

(5)
 32  3x  5
sec x tan x
dx
sec x
(8)
 3x
(1)
 sin 3xdx
2
(7)
(10)
(13)

 csc
 sec
1
5
xdx
xdx
(11)
(14)
5
2
)dx
6
dx
x 3  1 dx
 x csc xdx
 csc xdx
1
5
(3)
 sec 2x tan 2xdx
(6)

(9)
(12)
(15)
cot y csc 2 ydy
1
x
2
1
cos 2   dx
x 
 x cos x
 cot xdx
2
3
dx
4
(B) prove that)‫(اثبت االتي‬:
(1)
 cot xdx
(4)
 csc xdx
(2)
sec2 xdx
1 1
 1  4 tan 2 x  2 sin  2 tan x 
(5)
 cot xdx  ln sin x
(3)
sec2 xdx
1 1
 1  4 tan 2 x  2 sin  2 tan x 
(6)
 sec xdx  ln sec x  tan x
 ln sin x
27
 ln csc x  cot x  c
c
General Exercise:
(1)
 sin
1
3xdx
 x sin
(2)
2
z 
z

(4)  1  sin  cos dz
2
2

(5)
csc x cot x
dx
scsx
(8)
(7)

(13)
 csc
 sec
(16)
 x tan
(19)
x 
cosh 2   dx
2
 ln 2
(22)
 2  x 
(10)
1
7
xdx
2
x
1
x
0
(25)
dx
2
2
 x x  7dx
(43)
  x  5x e dx
 e cos xdx
 x sec xdx
 x tan  x  1 dx
 tan x sec xdx
 cot x csc xdx
(46)
 x sinh  x  dx
(49)
x
 xe dx
(28)
(31)
(34)
(37)
(40)
2
x
x
1
2
3
5
2
3
2
2
2
2
(15)
sin 1 xdx
sds
(20)

(23)
 sin 3x cos 2xdx
(26)

s2 2
3x 2dx
x 3 2
  r  r  1e dr
(32)  e sin 2xdx
(35)  p e dp
(38)  sec (3x  1)dx
(41)  sec x tan xdx
(44)  sin x cos xdx
(29)
2
r
2 x
4 p
(47)
(50)

1
x
(12)
6
5
2x tan 2xdx

(6)
(9)
 x sec xdx
 csc xdx
(17)

x 5  1 dx
4
 sec
(3)
6
(14)
xdx
xdx
2
 x  3x  5 dx
(11)
xdx
1
2
tan x sec 2 xdx
1
sin 2   dx
x 
 x cos(x
 cot xdx
2
7
(18)
 x sin
(21)
 4  5sin 2x
(24)

(27)
 sin 2x sin 5xdx
3
(45)
ex 2
 e x dx
28
 e sin  d 
 e cos 3x dx
 x ln xdx

(36)
(42)
1 x
x 2 4
(33)
3
x  dx
xdx
2xdx
(30)
(39)
1
2
dx
2
 sin
 10)dx
3
2x
3
 x coth x dx
 csc x cot xdx
 x tanh(x  1)dx
2
3
3
2
4
2
(48)
 cos
3
3x sin 3xdx

(51)

0
1  cos 2x dx
(52)



3
2
 cos 3x sin 4xdx
5
 6 cot xdx
(53)
 cot

(54)


4
xdx
6


4
(55)
5
 3 tan
3
xdx
(56)
0
e
4
x
3
x
sec e dx
 sec
(57)
4
xdx
0


0
12
x
(58)  3csc 4 dx
2

(59)
4
 6sec 2xdx
 sin 7x cos 3xdx
(60)

0
(3)Integration by partial fractions)‫(التكامل بالكسور الجزيئيه‬:
‫كثيرات حدود ودرجة البسط تكون اقل من درجة‬
q (x )
‫و‬
p (x )
‫بحيث‬
p (x )
q (x )
‫كل داله كسريه‬
‫ فانه يمكن ان يكتب الكسر في صورة كسور جزئيه على الصوره‬.‫المقام‬
‫والتي يمكن ايجاد تكاملها باستحدام قوانين‬
n  1, 2,3,....
‫بحيث ان‬
A
ax  b 
n
,
Ax  B
(ax  bx  c ) n
2
‫التكامل االوليه‬
Any rational function
the degree of
p (x )
p (x )
q (x )
where
less than that of
rational functions having the form
n  1, 2,3,....
p (x )
and
q (x ) ,
q (x ) are
polynomials, with
can be written as the sum of
A
ax  b 
n
,
Ax  B
(ax  bx  c ) n
2
which can always be integrated in term of elementary
functions
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
dx
  x  2 x  3
(2)
dx
  x  1 x  6 x  5
Solution:
(1)
, where
dx
  x  2 x  3
29
(3)
x
3
dx
 3x 2  2x
1
let
 x  2 x  3
A
B

 1  A  x  3  B  x  2 
 x  2   x  3
x  3  1  5B  B 
put

dx
 x  2  x  3

dx

  x  2  x  3


1
and
5
x  2  1  5A  A 
put
1
5
Adx
Bdx
dx
dx

 A
B
 x  2   x  3
 x  2
 x  3
1
dx
1
dx
1
1
 
 ln  x  2   ln  x  3  c

5  x  2  5  x  3 5
5
xdx
(2)
  x  1 x  6  x  5
let
x
A
B
C



 x  1 x  6  x  5   x  1  x  6   x  5 
 x  A  x  6  x  5   B  x  1 x  5   C  x  1 x  6 
6
5
and let x  5  5  44C  C 
77
44
1
let x  1  1  77 A  A 
77
xdx
dx
dx
dx
  x  1 x  6  x  5  A   x  1  B   x  6   C   x  5
let

x  6  6  77 B  B 
1
dx
6
dx
5
dx
1
6
5
 
 

ln  x  1  ln  x  6   ln  x  5   c

77  x  1 77  x  6  44  x  5  77
77
44
(3)
x
3
dx
dx
dx


2
2
 3x  2x
x  x  1 x  2 
x  x  3x  2 
A
1
B
C 
 

  1  A  x  1 x  2   Bx  x  2   Cx  x  1
x  x  1 x  2   x  x  1  x  2  
put
put
1
and
2
1
x  2  1  2C  C 
2
x  0  1  2A  A 
put
x  1  1   B  B   1
30
 1 
1 






dx
A
B
C
 2   1  2  dx

  3



dx



  x  x  1  x  2  
x  3x 2  2x   x  x  1  x  2  



1 dx
dx
1
dx
1
1
 
 
 ln x  ln  x  1  ln  x  2   c

2 x
2
 x  1 2  x  2 2
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
(x  1)dx
2
 1
  x  4  x
(2)
 x
2
dx
 9  x 2  5 
(3)
dx
  x  1  x  2 
3
Solution:
(1)
(x  1)dx
2
 1
  x  4  x
 x  1
A
Bx  C

 2
  x  1  A  x 2  1   Bx  C  x  4 
2
 x  4   x  1  x  4   x  1
put
x  4  3  17A  A 
3
17
3
17
12 5
and by equating x coefficients 1  4B  C  C  1  
17 17
by equating x 2 coefficients
0  A  B  B  A 
3
5
x
(x  1)dx
3
dx

 
  17 2 17 dx
2
 x  4   x  1 17  x  4 
 x  1

3
5
dx
 3  1  2xdx
ln  x  4       2
  2
17
 17  2   x  1 17  x  1

3
3
5
ln  x  4   ln  x 2  1  tan 1 x  c
17
34
17
31
 x
(2)
2
dx
 9  x 2  5 

dx
 x  3 x  3  x 2  5 
1
A
B
Cx  D


 2
2
 x  3 x  3  x  5   x  3  x  3  x  5 
 1  A  x  3  x 2  5   B  x  3  x 2  5   Cx  D  x  3 x  3
1
84
put
x  3  1  84A  A 
put
x  3  1  84B  B 
1
84
by equating x 3 coefficients in both sides :  0  A  B  C  C  0
by equating x 2 coefficients in both sides :  0  3A  3B  D  D 
 x

2
6
84
dx
1
dx
1
dx
6
dx
 
 
  2
2
 9  x  5  84  x  3 84  x  3 84  x  5 
1
1
6
x
ln  x  3  ln  x  3 
tan 1
c
84
84
84 5
5
(3)
dx
  x  2  x  1
1
 x  2  x  1

3
3
A
B
C
D



2
 x  2   x  1  x  1  x  13
 1  A  x  1  B  x  2  x  1  C  x  2  x  1  D  x  2 
3
put x  1  1  3D  D 
2
1
3
put x  2  1  27A  A 
1
27
by equating x 3 coefficients in both sides :  0  A  B  B  A 
1
27
by equating x 2 coefficients in both sides :  0  2A  C  C  2A 
2
27
by equating x coefficients in both sides :  0  3A  B  C  D  D 
6
27
32
dx
  x  2  x  1
3
1
dx
1
dx
2
dx
6
dx
 
 
 
2

27  x  2  27  x  1 27  x  1 27  x  13

1
1
2  x  1
6  x  1

ln  x  2   ln  x  1 

c
27
27
27  1
27  2 
1
2
Reduction formulas)‫(صيغ االختزال‬:
Evaluate the given integrals in terms of integrals of the same kind but
with a lower power of the integrand. Such formulas are called the
reduction formulas. In this way, we have a reduction formula by which
we can compute the integral of any positive integral power of:
(1)Sine)‫(الجيب‬
 cos x  sin x 
I n   sin xdx 
n
n 1

n
(n  1)
n 2
 sin x  dx

n
Or
 cos x  sin x 
In 
n
n 1

(n  1)
I n 2
n
Proof:
I n   sin n xdx
u   sin x 
by part
n 1
 du  (n  1)  sin x 
  sin n xdx   cos x  sin x 
  cos x  sin x 
n 1
n 1
n 2
 (n  1)   sin x 
n 2
cos 2 xdx
 (n  1)  sin n xdx  (n  1)   sin x 
  sin n xdx  (n  1)  sin n xdx   cos x  sin x 
 n  sin n xdx   cos x  sin x 
 cos x  sin x 
  sin xdx 
n
n
 cos x  sin x 
In 
n
n 1

n 1
n 1
 (n  1)   sin x 
n 1

and dv  sin xdx  v   cos x
cos xdx
n 2
33
dx
 (n  1)   sin x 
n 2
dx
(n  1)
n 2
 sin x  dx

n
(n  1)
I n 2
n
 let cos
n 2
dx
2
x  1  sin 2 x 
Example
If I n   sin n xdx Apply the reduction formulas for n = 3 and n = 4.
 cos x  sin x  3
(1) I 4 
 I2
4
4
3
I2 
and
 cos x sin x 1
 I0
2
2
and
I0  x
 cos x  sin x  3   cos x sin x 1 
 I4 
 
 x
4
4
2
2 
3
Another solution:
 cos x  sin x  3
2
   sin x  dx
 sin xdx 
4
4
 cos x sin x 1
2
   sin x  dx 
  dx
2
2
3
4
 cos x  sin x  3   cos x sin x 1 
  sin xdx 
 
 x  c
4
4
2
2 
3
4
 cos x  sin x 
2
(2)  sin xdx 
   sin x dx
3
3
2
3
 cos x  sin x  2
  sin xdx 
 cos x  c
3
3
2
3
(2)Tangent)‫(الظل‬:
 tan
n
xdx   tan n  2 xd (tan 2 x )   tan n  2 x (sec 2 x  1)dx   tan n  2 x sec 2 xdx  tan n  2 xdx
  tan
n 2
xd  tan x    tan
Or
In
 tan x 

n 2
xdx
 tan x 

n 1
n 1
  tan n  2 xdx
n 1
n 1
 I n 2
Example:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
 tan
3
xdx
(2)
 tan
4
xdx
34
Solution:
(1)
3
 tan xdx 
(2)
 tan
4
xdx
 tan x 
2
  tan xdx 
2
 tan x 

3
3
  tan xdx
2
 tan x 
2
2
 tan x 

 ln cos x
3
3
c
  x  tan x   c
(3) lnx)‫(اللوغريثم الطبيعي‬
(A)
  ln x 
n
dx  x  ln x   n   ln x 
n
u  (ln x ) n  du  n (ln x ) n 1
dx
x
n 1
dx
and dv  dx  v  x
I n  x (ln x ) n  n  (ln x ) n 1dx
 I n  x  ln x   n I n 1
n
(B)
n
 x ln xdx
let
u  ln x  du 
dx
x
dv  x n  v 

n
 x ln xdx 
x n 1 ln x
1
x n 1 ln x
x n 1
n

x
dx


(n  1)
(n  1) 
(n  1)
(n  1) 2

n
 x ln xdx 
x n 1 ln x
x n 1

(n  1)
(n  1) 2
(4)Exponential function:
I n   x ne a x 
1 n ax n
x e  I n 1
a
a
proof :
let
u  x n  du  nx n 1 and
1
dv  e a x  v  e a x
a
1 n ax n
x e   x n 1e a x dx
a
a
n
 I n 1
a
 I n   x ne a x 
 In 
1 n ax
x e
a
35
x n 1
(n  1)
Exercise(7):
(A) Evaluate the following integrals )‫(احسب التكامالت االتيه‬
dx
(1)
  x  1 x  2  x  4 
(4)
 x  x  1  x
(7)
  x  3  x
(10)
dx
dx
2
2
 x  12
2
x
 2  dx
2
 2
dx
  x  510  x 
xdx
  x  2  x
(5)
(8)
 2x  2 
 3x  2  dx
x
(2)
x
2
 1
dx
2
6
 x  1 dx
  x  2  x
(11)
(3)
2
(6)
x
2
(9)
x
2
(12)
 x  3
xdx
(14) 
 x  1  x 2  1
 x
 x  1 dx
(16)
 x ln(x  1)dx (17)  x cos 3xx (18)  e sin 3xdx
 sin (2x ) (20)  arctan 2xdx (21)  arc cos 2xdx
 x sin xdx (23)  x cos xdx (24)  tan xdx
(19)
(22)
2
2x
1
3
8
7
(B)prove that:
  sin x 

n
n 1
(1)
 sin
(2)
 cos
(3)
 tan
n
xdx
n
n
xdx
 cos x 

xdx
 tan x 

n 1
sin x
n
 sinh xdx
 n  1
n
 n  1
n
n 1
  cot n 2 xdx
 sinh x 

n 1
n
cosh x

 sin
 cos
  tan n  2 xdx
n 1
  cot x 
(4)  cot xdx 
n 1
(5)


n 1
n
n
cos x
n 2
n 2
xdx
xdx
 n  1
 n  1
n 1
sinh n 2 xdx

n
36
2dx
 5x  6
  x  2

x 3  6x
dx
 4  x 2  1
dx
 81
(13)
(15)
2
4
x
2
 1 dx
  x  1 x  2 x  3
(6)
 cosh xdx
(7)
 csc
n
 cosh x 

n 1
sinh x
n

n 1
cosh n 2 xdx

n
  csc x  cot x
m
m
xdx 

 csc x  dx

m 1
m 1
m
m 2
 sec x 

m
tan x
m
m

 sec x  dx

m 1
m 1
n x
n x
n 1 x
(9)  x e dx x  x e  n  x e dx
(8)
 sec
m 2
xdx
e ax
(10)  e sin bxdx  2
a sin bx  b cos bx   c
a b 2
e ax
(11)  e ax cos bxdx  2
b sin bx  a cos bx   c
a b 2
ax
sin m 1 x cos n 1 x
n 1

I m ,n  2
m n
m n
x m 1
n
n
n
m
  x  ln x  dx 
I m ,n 1
 ln x  
m 1
m 1
(12) I m ,n   sin m x cos n xdx 
(13) I m ,n
‫فصرت باذيالها متمسك‬
‫رايت القناعة راس الغنى‬
‫وال ذا يراني به منهمك‬
‫فال ذا يراني على بابه‬
‫امر على الناس شبه الملك‬
‫فصرت غنيا بال درهم‬
37
TRIGONOMETRIC SUBSTITUTION)‫(تعويضات مثلثيه‬:
Three Basic Substitutions
The most common substitutions are x  a sin t ,
and .They come from the reference right triangles .
x  a tan t
(1) If we find the form  a 2  x 2  .Substitute that x  a sin t
x  sec t
or x  a cos t
(2) If we find the form  a 2  x 2  . Substitute that x  a tan t
(3) If we find the form  x 2  a 2  . Substitute that x  a sec t
and
or x  a sinh t
or x  a cosh t
Example:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)

(4)
x
x 2 4
dx
x
dx
2
9x
2
x2
 1  x 2 dx (3)
xdx
(5) 
(6)
1 x 4
(2)

3  x 2 dx

16  25x 2 dx
Solution:
(1)


x 2 4
dx
x
let
x  2sec  dx  2sec  tan  dt
x 2 4
4sec2   4
4 tan 2 
dx  
2sec

tan

d



 
 2sec tan  d  
x
2sec
2sec 
 x 2 4
x
 4 tan 2  d   4  sec2   1 d   4  tan      4 
 sec 1 
2 
 2
38
x2
(2) 
dx
1 x 2
let x  tan   dx  sec2  d 

x2
tan 2 
dx

sec2  d     tan 2  d    tan       x  tan 1 x 

2
2

1 x
1  tan 
--------------------------------------------------------------------------------------(3)

let
x  5sin   dx  5cos  d 
25  x 2
  25  x 2   25  25sin   5cos  d     5 1  sin 2   5cos  d  
 25 cos   cos  d    25 cos 2  d   25

25  1  x 
25  x 2
 sin   
2 
5
5




1
25
sin 2 
1  cos 2  d    

2
2 
2 
 hint : sin2 = 2sin cos 
--------------------------------------------------------------------------------------(4)

x
dx
2
9x 2
dx
x 2 9x 2

let
x  3sec   dx  3sec  tan  d 
3sec  tan  d 
9sec2  9  9sec2 

3sec  tan  d 

9sec2  3 1  sec2 
39


1 tan  d 
1 d
1
1
1 
x
 
  cos  d   sin   sin  sec1   c

1
9 sec  tan  9
9
9
9 
3
cos 
(5)
xdx
 1 x
let
4
1
x 2  tan t  2xdx  sec 2 tdt  xdx  sec 2 tdt
2
1
sec 2 tdt
xdx
1 sec 2 tdt 1
1
1
2



  dt  t  tan 1 x 2
4
2
2


1 x
1  tan t
2 sec t
2
2
2
(6)
let


16  25x 2 dx   16  (5x ) 2 dx
4
4
5x  4 cos   x  cos   dx 
sin  d 
5
5
 4
 16
16  25x 2 dx   16  16 cos 2   sin  d   
sin   sin  d  
5 
 5


16
16 1
8
sin 2 
sin 2  d  
1  cos 2  d    


5
5 2
5 
2 

8  1  5x
 sin 
5 
 4
2
 5x 16  25x


16





 hint : sin2 = 2sin cos 
Exercise(8):
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)

(5)

(9)

3
2 2
1  x 
x
dx
6
(2)
2
v dv
(6)
5
1 v 2  2
dy
y 1   ln y 
2


(10)
1
2 2
1  x  dx
x
4
(3)
5
2 2
1  r 
r8
dr
dx
 1 x
2
(7)

8dx
 4x

(11)
40
2
 1
2
e t dt
(8)
e 2t  9
x
dx
x 1
2

(4)

 9t
6dt
2
 1
2
e t dt
3
1  e 2t  2
(12)

dx
9x 2
(13)

(17)

3
2
(21)

0
(25)

3dx
1  9x
2
25  t 2 dt
dx
3
2 2
4  x 
x 3dx
x 4
2
(14)
(18)
(22)
(26)
dx
 8  2x
(15)
5dx

8dw
4 w
2
2dx
3
x 1
2
dx

9x
2

(23)
(27)

2
(16)
y 2  49
dy
y

(19)
25x 2  9
w
x
2
9 w 2
dw
w2
dx
4x  9
2
(28)

1  4t 2 dt
(20)
(24)

y 2  25dy
y3

x
dx
2
x 2 1
xdx
3
6 2
1  x 
‫ ما كان هذا من صنيع محمـد‬--- ‫يا راقصــا او زاحفــا لتَعَبّـــــُـ ٍد‬
‫ او كان يزحف للقبور بمسجد‬--- ‫ما كان يرقص بالدفوف عبــادة‬
41
Products of Powers of Sines and Cosines
We begin with integrals of the form:
 sin
m
x cos n xdx
Where m and n are nonnegative integers (positive or zero). We can
divide the work into three cases.
‫ اعداد صحيحه غير سالبه فيمكننا ان نقسم الحل الى ثالثة حاالت وهى‬n‫ و‬m ‫اذا كان‬
Case 1:
2
2
If m is odd, we write m  2k  1 and use the identity sin x  1  cos x to
m
2 k 1
2
2
(1)
obtain sin x  cos x   cos x  cos x  1  cos x  sin x
Then we combine the single sin x with dx in the integral and set sin xdx
equal to d (cos x ) .
‫ومن ثم نستخدم المتطابقه‬
m  2k  1 ‫الصوره‬
‫ عدد فردي يكتب على‬m ‫اذا كان‬
2
2
‫ لنحصل على عليه‬sin x  1  cos x
sin m x  cos 2 k 1 x   cos 2 x  cos x  1  cos 2 x  sin x
(1)
sin xdx
Case 2:
m
n
If m is even and n is odd in  sin x cos xdx we write
2
2
identity cos x  1  sin x to obtain
cos n x  cos 2 k 1 x   cos 2 x  cos x  1  sin 2 x  cos x
k
k
‫بدال عن‬
n  2k  1
n  2k  1 ‫الصوره‬
(2)
k
k
cos xdx
‫بدال عن‬
Case 3:
m
n
If both m and n are even in  sin x cos xdx we substitute
1  cos 2x
2
, sin 2 x 
1  cos 2x
2
(3)
42
d (sin x )
‫ عدد فردي يكتب على‬n‫ عدد زوجي و‬m ‫اذا كان‬
2
2
‫ لنحصل على‬cos x  1  sin x
cosn x  cos2 k 1 x   cos2 x  cos x  1  sin 2 x  cos x
cos 2 x 
‫نضع‬
and use the
We then combine the single cos x with dx and set cos xdx equal to
‫ومن ثم نستخدم المتطابقه‬
d (cos x )
(2)
d (sin x )
‫عليه نضع‬
to reduce the integrand to one in lower powers of cos 2x .
Here are some examples illustrating each case.
cos 2 x 
1  cos 2x
2
, sin 2 x 
1  cos 2x
2
(3)
‫ اعداد موجبه نعوض‬n‫ و‬m ‫اذا كان‬
cos 2x ‫فالتكامل يختزل الى‬
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
 sin
3
x cos 2 xdx
(2)
 cos
5
xdx
(3)
 sin
2
x cos 4 xdx
Solution:
(1)
 sin
x cos 2 xdx   sin 2 x cos 2 x sin xdx   1  cos 2 x  cos 2 x  d (cos x ) 
3
  1  u 2 u 2   du    u 4  u 2  du 
(2)
 cos
5
xdx
u5 u3 1
1
   cos5 x    cos3 x   c
5 3 5
3
  cos 4 x cos xdx   1  sin 2 x  d (sin x )
2
2
1
1
1
1
  1  u 2  du   1  2u 2  u 4  du  u  u 3  u 5  sin x  sin 3 x  sin 5 x  c
3
5
3
5
1
 1  cos 2x   1  cos 2x 
2
(3)  sin x cos xdx   

 dx   (1  cos 2x )(1  2 cos 2x  cos 2x )dx
2
2
8



2
2
4
1
1
1

(1  cos 2x  cos 2 2x  cos3 2x )dx   x  sin 2x   (cos 2 2x  cos 3 2x )dx 

8
8
2

1
1
1

(1  cos 4x )dx   x  sin 4x 

2
2
4

1
1
1

and  cos3 2xdx   1  sin 2 2x  cos 2xdx   1  u 2  du   sin 2x  sin 3 2x 
2
2
3

1
1
1
1
1
 1
 
  sin 2 x cos 4 xdx   x  sin 2x    x  sin 4x    sin 2x  sin 3 2x   
8
2
4
3
 2
 
2
but
 cos
2
2xdx 
Combining everything and simplifying we get(‫)بعد التجميع والتبسيط نجد‬
43
 sin

2
x cos 4 xdx 
1
1
1 3 
 x  sin 4x  sin c   c
16 
4
3

Eliminating Square Roots)‫(حذف الجذر التربيعي‬
In the next example, we use the identity
square root.
‫لحذف الجذر التربيعي‬
cos 2 x 
cos 2 x 
1  cos 2x
2
1  cos 2x
2
to eliminate a
‫في المثال التالي سوف نستخدم المتطابقة‬
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬

4

1  cos 4x dx
0
Solution:
To eliminate the square root we use the identity
‫لحذف الجذر التربيعي نستخدم المتطابقة االتية‬
cos 2  
1  cos 2
2
or 1  cos 2  2 cos 2  , with   2x  1  cos 4x  2 cos 2 2x



4
4
4
0
0
0
  1  cos 4x dx   2 cos 2 2x dx  

2
2
 sin 2x  4
2 cos 2xdx  2 

1  0 

2
2
 2 0
Integrals of Powers of tan x and sec x
We know how to integrate the tangent and secant and their squares. To
integrate higher powers we use the identities tan 2 x  sec2 x  1 and
sec2 x  tan 2 x  1 and integrate by parts when necessary to reduce the
higher powers to lower powers.
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
 tan
4
xdx
(2)
 sec
3
xdx
solution:
44
(1)
 tan xdx    tan x  tan x  dx    tan x  sec x  1 dx
  tan x sec xdx   tan xdx   tan x sec xdx   (sec x  1)dx
4
2
2
2
2
2
2
2
2
2
2
1
  tan 2 x (d tan x )   sec 2 xdx   dx  tan 3 x  tan x  x  c
3
3
(2)  sec xdx
let u  sec x
dv  sec2 xdx  du  sec x tan xdx
To integrate by
part
and
and v  tan x
  sec3 xdx  sec x tan x    tan x  sec x tan x  dx
 sec x tan x    sec 2 x  1 sec xdx  sec x tan x   sec xdx   sec3 xdx
 sec xdx  sec x tan x   sec xdx   sec xdx
 2 sec xdx  sec x tan x  ln sec x  tan x
1
  sec xdx   sec x tan x  ln sec x  tan x   c
2
3
3
3
3
Products of Sines and Cosines
The integrals
(1)
 sin(mx ) sin(nx )dx
(2)
 sin(mx ) cos(nx )dx
(3)
 cos(mx ) sin(nx )dx
Can be evaluate through integration by parts, but two such integrations
are required in each case. It is simpler to use the identities
1
cos(m  n )x  cos(m  n )x 
2
1
sin(mx ) cos(nx )  sin(m  n )x  sin(m  n )x 
2
1
cos(mx ) cos(nx )   cos(m  n )x  cos(m  n )x 
2
sin(mx ) sin(nx ) 
These come from the angle sum formulas for the sine and cosine
functions. They give functions whose antiderivatives are easily found.
Examples:
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
(1)
 sin 3x cos 5xdx
(2)
 sin 3x sin xdx
45
Solution:
(1)
1
1
 sin 3x cos 5xdxdx  2  sin(2x )  sin(8x )dx  2  sin 8x  sin 2x  dx
1   cos8x sin 2x
 

2
8
2

 c

1
1  sin 2x cos 4x 
(2)  sin 3x sin xdx    cos 2x  cos 4x  dx  

 c
2
2 2
4 
Exercise(9):
Evaluate the following integrals )‫(احسب التكامالت االتيه‬


(1)
4
 7 sin xdx
(10)
(13)
(16)


6
2
5
 3cos 3xdx
7
d 
2
 sin
(6)
0
 sin 3x cos 3xdx (8)  sin x cos xdx (9)  cos 3x cos 4xdx
 cos x cos 7xdx (11)  8cos xdx (12)  8sin xdx
 16sin x cos xdx (14)  8sin y cos ydy (15)  sin 2x cos 2xdx
 35sin x cos xdx (17)  8cos 2x sin 2xdx (18)  sin 3x cos 3xdx
4

0
2
2
4
3

4
4
2
2
3
2

1  cos x
dx
2
1  cos 2x dx

(21)
0

4

(23)
sec x  1dx
2
3
2 2
 1  cos x 
(24)

4
0
1  sin 2 y dy
0

1  cos x dx
3


(20)

(22)
xdx

7
 7 cos xdx
(5)
3

2
0
2
(19)
 cos
(3)
0
0
(7)
2
5
 sin xdx
(2)
0
(4)

2
dx


0
(25)
 2sec
4
3
xdx
(26)

3
 sec
xdx
(27)
e
3xdx
(30)
 3csc
4
0
sec3 xdx
x
dx
2
(28)
 csc
(31)
 6 tan xdx (32)  4 tan xdx (33)  cot xdx
 sin 3x cos 6xdx (35)  sin 2x cos8xdx (36)  sin x cos 3xdx
(34)
4
xdx
4
(29)
 3sec
x
4
3
3
46
4
Tangent x/2 Substitution :
If the integrand contains an expression of the form a  b sin x or a  b cos x ,
then the following theorem may be helpful in evaluating the integral.
‫ عليه فان الفرضية االتيه سوف‬a  b sin x or a  b cos x ‫اذا كان التكامل يحتوي على الصيغ‬
‫تخدمنا في حساب التكامل‬
x
2u
1u 2
2
Suppose that : u  tan  sin x 
, cos x 
and dx 
du
2
2
2
1u
1u
1u 2
2
2
2
1u 2
x 
cos x  2 cos 2    1 
1 
1 

1

1u 2
1u 2
x 
x 
2
sec 2  
1  tan 2  
2
2
x 
x 
x 
sin    cos 2  
2 tan  
1
2
x 
x 
 2   2 tan  x  
2
sin x  2sin   cos    2  



x 
2
2
 2  sec 2  x  1  tan 2  x 
cos  
 
 
2
2
2
2u
sin x 
1u 2
2du
x  2 tan 1 u  dx 
1u 2
For example:
 2 

 du
dx
2du
1u 2 

(1) 


a  b sin x
a (1  u 2 )  2bu
 2 
a b 
2 
 1u 
 2 

 du
dx
2du
1u 2 

(2) 


2
2
a  b cos x
a (1  u )  b (1  u 2 )
 1u 
a b 
2 
 1u 
Example:
(1)
dx
 1  cos x
(2)
dx
 2  sin s
Solution:
2du
2
dx
x 
(1) 
  1  u 2   du  u  c  tan    c
1u
1  cos x
2
1
2
1u
47
2du
2
dx
du
du
(2) 
  1u
 2

2
2u
2  sin x
u u 1
1 3

1
u

2

 
1u
2 4

u
du
1
u 
 tan 1  
2
a
a
a 
2

x 
1  2 tan   

du
2
 2u  1  2
 2   c
 

tan 1 
tan 1 

2
3
3
3


1 3
 3 



u   


2 4

Exercise(10):
Evaluate the following integrals )‫(احسب التكامالت االتيه‬
dx
2  sin x
dx
(4) 
2sin x  3cos x
(1) 
(7) 
sin xdx
sin x  cos x
2  sin x
dx
3  cos x
dx
(13) 
1  sin x  cos x
(10) 
dx
sin x  cos x
dx
(5) 
3  cos x
dx
sin x  cos x
dx
(6) 
2  sin x
(2) 
(8) 
(3) 
dx
3  cos x
(9) 
1  sin x  dx
1  sin x 
1  cos x
2  cos x
dx
(12) 
dx
1  cos x
2  sin x
dx
(14) 
2  cos x  2sin x
(11) 
GENERAL EXERCISE(2):
‫ واختبر اجابتك بالتفاضل‬100 ‫ الى‬1 ‫احسب التكامالت غير المحدوده في التمارين من‬
(A)Find the indefinite integrals (most general antiderivatives) in Exercises
1-100. Check your answers by differentiation.
(1)
 x
(4)
  r  9
3
 5x  4  dx
dr
2
(2)
(5)
 3 t2

  5t  2  t  dt
7dr

r  2 
(3)
(6)
3
48

  3
 3
u
6 
 du
u2 
 2  1d 
(7)
t

dt
5 t 2
v
(10)  sec 2 dv
3
(13)

x
(8)

 sec 3 tan 3 d 
3
1  x 
4
(11)
 csc
(14)
 sin
2
1
4
3
dx
 xdx
2
  2  x  5 dx
(9)
 csc
(12)
x
dx
4
2x cot 2x dx
 cos
(15)
x
dx
2
2
1
1


Hint : cos 2   1  cos 2  , sin 2   1  cos 2  
2
2


(16)
 cos(7x  3)dx
(19)

(22)

(25)
x
e
(28)
(31)
2zdz
3
z 1
2
x 1
dx
x5
2
 sin
(46)
 csc
(49)
 cos
3
5
 x (13) dx
 x sin 5xdx
x cos xdx
(35)
(38)
(41)
(44)
 xe
x
x
x
2 4x
e dx
(36)
2
ln xdx
(39)
3
cosh xdx
(42)
 ln x dx
 x sec xdx
 x ln(2x )
 (1  sin x )
10
(50)

(55)
(58)

dx
(53)
(56)
(59)
dx
x
x sin xdx
 x 4  x 
 x 4 dx
2 x
(33)
 sin
1
2 2
xdx
 (3x  1)e 2 dx
(47)
cos xdx
sin 5 x
(24)
2
dx
3x
(30)
2
xdx
x
 cos
2
 cot xdx
 x sec xdx
2x
(32)
2
4
(21)
(27)
x
3 x
 x 7 dx
21
xdx
1
(29)
(43)
(52)
 x cos  ln x  dx
2
1
 cos
(18)
(23)
cos xdx
(40)
(37)
sin x 3dx
 sin
(26)
3x
2
(20)
sin(x 3  1)dx
  x  ln x  dx
 x ln xdx
 x sinh(3x )dx
(34)
x
(17)
3
cos xdx
xdx
cos xdx
sin 2 x
  x  3 dx
 tan x sec xdx
10
3
 tan
2
2
xdx
2
2
ln xdx
2
3
sin(ln x )dx
x
3
(48)  cos xdx
(45)

(51)
 tan
(54)
(57)
(60)
49
1
4
xdx
 x ln xdx
 csc x (csc x  cot x )dx
dx
  x  a  x  b 
(61)
t
2
t dt
 1  t   2t  1 dt
2
(62)
1 5
(64)   2  3  dt
t 
t
 3x
(65)
dx
2
3
(67)
x
(70)
(76)
 x tan x dx
 x sin dx
 x sec h xdx
(79)
 x sec(x  1)dx
(82)
 x ln x
e  x dx
(80)   x
e 1
5  ln x
(83) 
dx
x ln x
(85)
3
(86)
(88)
 3 7
(91)

(94)

(97)
1
  sin x  dx
(73)
4
dx
(68)
1
x
2
3dx
dx
4
 1 dx
(89)
2
1 3 4 x
x e dx
2
(92)
 3xe
(95)
x 
x
x
2e 2 x dx
(98)

x
3x
(69)


x  8 dx
x sin 1 xdx
2
2
4
dx
2
dx
x
1
2
2x
 ln x 
(81)

(84)
 25
7
dx
x
1
2
ln x
x
dx
dx
x
(90)
 3x
dx
(93)
e
dx
(96)
 3x
(99)
3
 x cos xdx
 x sinh(7x )dx
 e cos(3x )dx
(87)
dx
1 x
x
(78)
dx
e 3x dx
x2
(66)
(75)
2
(77)
 1  x  dx
(72)
1
(74)
2
dx
 e cos xdx
 csc xdx
 x sinh xdx
(71)
1
2x 7
x
5
2
(63)
5 x 8
dx
 1 x
1
x
dx
(100)
  x log x 
(B) calculate the following integrals:
(1)
 x tan
(4)
  x  1 x  5
(7)
 cos
2
1
xdx
xdx
6
dx
(2)
 x sin
(5)
 x sin 3x cos xdx
(8)
 sin
1
3
xdx
xdx
(3)
 x cos
(6)
x
(9)
50
2
xdx
dx
2
1
 x sec
1
xdx
2
dx
(10)
(13)
(16)
2x  1
  x  1 4x  1dx

3x 2  3x  2

1 x
dx
x
x
2
 x  1
(19)

(22)
 3  5cos x

(14)
2x 3  x 2  2
 x 4  4 dx
(17)

dx
3x 2  5x dx
dx
1  tan x
 25 
2
(12)
  2x
dx
2
 1 3x 2  2 
(15)
 9x
6x  5
dx
 18x  25
dx
  3x  2 
x 2  5x  6
2

 H int
sin 2xdx
 3  5cos
2
x
put
u
(27)
 sin
dx
(29)  cos3 x csc 2 xdx (30)
2
x cos x
dx
xdx
xdx
(31) 
(32) 
(33) 
3
x x 4
x 2
 2x  3 2
(28)
 sin
(34)
 x 2x  3 (35)
4
3
 x  7x  5 2
1
x 3dx
(37)  2
(38)
x  2x  1
0
r 2dr
(40)

(43)
 4  5sin 2x
(46)

(49)
x
(52)
 csc
4r2
dx
ds
9  s 
2 2
2
xdx
dx
x 2 2
(44)
(47)
tan 1 xdx
5

(41)
x
(50)
(53)
9  4x
dx
x2

e
3 x
2
1 
x  1 
5
x cos 2 xdx
dx
x 3
x
x  x 2 dx
x
 x tan
(45)
 sin 3x cos 5xdx
3
(2x  3) x 2  3x  1
dx
(54)
1
xdx

3t  4
dt
t
(51)
 sec
(48)
u
u
5
xdx
 sin 3 sin 6 du
51
dx
2
dx
 5  4sin 2x
sin 4xdx
 16x  ln x 
6  x  x 
dx
4x  9
2

(39)
(42)
4x  x 2 dx
x
(36)
2x  7
dx
(24(i )) 
 16
2
 sin x  cos x  1
(21)
dx

(18)
x  6x  13
(23)
(26)
2
2

1  2x  x 2
 1  tan x dx
x
(20)
dx
  x  1
(24(ii ))
(25)
2
dx
(11)
t
(55)
 8sin 4t sin 2 dt
(58)

25  x 2 dx
dx
(57)
x 2 4
dx
x
x
x
(64)
 tan x
(67)
2
 cot x 1  sin x dx
(70)

dx
 cos
(65)
4  sin x
2

(68)
2x
dx
2
(79)
 tan
(82)
 4v
2
(85)
e
csc e x  1 cot e x  1 dx
(88)
 1  sin
(91)
 
(94)

(97)
 tan  3  dx
(100)
x
4
x tanh xdx
x 
  dx
2
(80)
 4 cot
2dv
 4v  4
(83)
v
cos xdx
2
x
(89)
 8ln 3log3 x
x
dx
2x  x
2
x 

 ln x 
3

 dx

2
3
 tan
x
dx
1 x
(69)
cos 1 x
 x dx
5  sin x
3
dv
 v lnv
x
(95)

(98)

 csc
2xdx
e
2
dx
2
dx
 x  4x  1
2
cos(1  lnv )dv
v
dx
x
52
cosh 3xdx
x cos 5 xdx
xe cot x dx
(87)

ln x
dx
x
tan(lnv )dv
v
(93)
5x 2  3
2
sec 2 e x  7 dx
x

(90)
x
(78)
 csc
(81)
2dv
(84)
 4v  5
(86)
5
y dy
 x sinh 5xdx
(75)
2x coth 2xdx
2xdx
1
 sin
(72)
2
(92)
dx
(66)
 csc h
(77)
25 v 2 dv
2
x dx
1
cosh 4 x
(74) 
dx
x
 sec h
v
7x 2
x 2  6x
(63) 
dx
2
2
 x  3
cos xdx

(71)
(76)
7
(60)
x 3  x 1
 x 2  1 2 dx


(62)
1
(73)  sinh 5 xdx
8
x
 cos 3 cos 4 dx

(59)
(61)
7x 2
x
(56)
3dx
 4  3x
1
2
(96)
 r csc (1  ln r )dr
(99)

2
ln  x  5  dx
x 5
(C) Evaluate the following integrals )‫(احسب التكامالت االتيه‬
3
(1)
(3)


4x 2 dx
1 x
dx
1
2
1 e
2x
xdx
x 2
(5)

(8)
1 x1
 x 2 e dx
x  u 
6
e
0
2x
dx

(u 2  1  e 2 x ) (4)
2
x dx
x 4
(6)

(9)
 x log
ln 3
(11)
(2)
(12)
dx

3x
ln x dx
x
1
2
(7)
x  u 
dx
1 x
dx
x

3
1
3
x 8
3
u
2
 x 3  8
1  ln x
dx
x
(10)
x
(13)
 x ln xdx
2
cos 3xdx
‫ و المستقبل حصاد غرسك‬..‫والحاضر غرس‬... ‫الماضي درس‬
‫باألمس‬
53
Vector Integral)‫(تكامل المتجه‬
The indefinite integral of a vector A  A (t ) with respect to t is the set of
all antiderivatives of A.
 A (t ) dt
 i  A1 (t ) dt  j  A 2 (t ) dt  k  A 3 (t ) dt
The usual arithmetic rules for indefinite integrals apply.
Example:
Finding the vectors Integral
(1)

4

(cos t ) i  j  2t k dt
(2)
 [3t
2
i  2t j  5t 4 k ]dt
1
Solution
(1)
 ((cos t ) i  j  2tk )dt  (sin t ) i  t j  t
4
(2)
4
4
2
k C
Where C  c1i  c 2 j  c 3k
4
2
4
 A (t ) dt  i  3t dt  j  2tdt  k  5t dt
1
1
4
1
4
1
4
  t  i   t  j   t  k  63 i  255 j  1055 k
1
1
1
3
2
5
Problems
Find the vectors Integral
1
(1)
 t
3

i  7 j  (t  1)k dt
2
(2)
0
(3)

 

  (6  6t ) i  3
tj
 t

1
2
 1t
2
i 
3 
k  dt
1  t 2 
(4)
54
3
i  j  ln tk dt
4 
k  dt
t2 

1
1 
1
1  t i  5  t j  2t k  dt
4
(5)
  (sec t tan t )i  (tan t ) j  (2sin t cos t )k  dt
3
(6)
0

  (sin t ) i  (1  cos t ) j  (sec
4
(7)
2

4

t )k dt
(8)

 

i   25t  j  e sin t cos t  dt
t2 3

t
Line Integral)‫(تكامل الخط‬
Work over a Smooth Curve)‫(الشغل المبذول على منحنى املس‬
Case(1):
The work done by a force F  F1i  F2 j  F3k over a smooth curve C from
a to b is
b
W   F dr 
a
 F dx
1
 F2 dy  F3dz
C
Is called the line integral. If c is closed path then
W 
 F  dr   F dx
1
C
 F2dy  F3 dz
C
Example :
Evaluate
y
dx  x 2dx
2
C
Where C is the triangle with vertices (0,0) , (1, 0) , (1, 1)
Solution:
55
One has to compute three integrals. The first is the integral from (0, 0)
to (1, 0); along this path
the first integral is :
and x is the parameter,
y 0
0  0
dy  0
. Hence
       (1)
The second integral is that from (1,0) to (1, 1); if y is used as
parameter, this reduces to x  1 since dx  0 . Hence the first integral is:
1
 dy   y 
1
0
1
         (2)
0
For the third integral, from (1, 1 ) to (0, 0), x can be used as parameter,
since x  y  dx  dy . Thus the last integral is:
0
 2x 3 
2
2
x
dx

 3   3
1

1
0
         (3)
2
Thus finally
y
dx  x 2dx  (1)  (2)  (3)  0  1 
2
C
Example:
2 1

3 3
Find the integral of vector
A (x , y , z )
if
A  ( x 2  y ) i  (3x  y) j Where C is
1)
y = x2
3)
2
from (0,0) to (1,1)
2)
y=x
from (0,0) to (1,1)
x = y from (0,0) to (1,1)
Solution:
 A  dr   A dx
1
 A 2 dx  A3 dz   (x 2  y )dx  (3x  y ) dy
56
(1) y  x 2  dy  2 x dx
 (x

2
 x 2 ) dx  (3x  x 2 )(2xdx ) 
 (2x
2
 6x 2  2x 3 ) dx
1
 8x 3 2x 4 
8
1 16  3
13
  (8x  2x ) dx  


 


4 0
3
2
6
3
 3
0
1
2
3
(2) y  x  dy  dx 
1
(y
2
 y ) dy  (3 y  y )dy
0
1
 y3
29
3y 2 
1
3
11
 




 
2 0
3
2
6
6
 3
1
  ( y 2  3 y) dy
0
x  y  dx  2 y dy 
2
1
(y
4
 y )(2 ydy)  (3 y 2  y ) dy
0
(3)
1
 2y6
5y3
y2 
1 5 1 9
  (2 y  5 y  y )dy  


    
3
2 0
3 3 2 6
0
 6
1
5
2
Example:
Find  x dy  y dx
where C is the vertices of triangle (0,0), (1,0) , (0,1)
c
Solution:
From
B (0,1)
O (0, 0)  A (1, 0)
0 (0,0)
The equation of a straight line is
y  0  dy  0
57
A (1,0)
  (00) 0
From
 0      (1)
A (1, 0)  B (0,1)
 the equation of a straight line is
y 0
1 0
y


  1  y   x  1  dy   dx
x 1
0 1
x 1

0
 x ( dx )  (x  1) dx   (x  x  1) dx 
1
From
1
 1    (2)
1
B (0,1)  O (0, 0) 
  0 0  0

  x 0
 A. dr
the equation
x  0  dx  0
       (3)
 (1)  ( 2)  ( 3)  0  1  0  1
c
Example:
Find the integral  (x  y ) dx  (x  y ) dy
c
if c is the first quarter of a circle x 2  y 2  a 2 .
Solution:
From O ( 0 , 0 ) to A (a , 0)

y  0  dy  0
58
From A(a,0) to ( 0 , a )  the curve equation
is x 2  y 2  a 2 x  a cos , y  a sin 
 :0 

dx   a sin  d  ,
2
dy  a cos  d 

2
  (a cos   a sin  )(a sin  d )  (a cos   a sin  )(a cos  d  )
0

2
  a 2 sin  cos   a 2 sin 2   a 2 cos 2   a 2 sin  cos   d 
0


2
a 2
  a 2 d    a 2   2 
0
2
0
Case(2):
To evaluate the work done by a force F  F1i  F2 j  F3k along a smooth
curve
r (t ) ,
we take these steps:
(1) Evaluate F on the curve as a function of the parameter t.
(2) Find
dr
dt
(3) Integrate
F
dr
dt
from t = a to t = b.
Example:
Find the work done by
r (t )  ti  t 2 j  t 3k
F  ( y  x 2 ) i  ( z  y 2 ) j  ( x  z 2 )k
from (0, 0, 0) to (1, 1, 1)
59
over the curve
Solution:
First we evaluate F on the curve:
F  ( y  x 2 )i  (z  y 2 ) j  (x  z 2 )k  (t 2  t 2 )i  (t 3  t 4 ) j  (t  t 6 )k
Then we find
dr
 i  2tj  3t 2 k
dt
Finally we find
F
F
dr
dt
, and integrate from t = 0 to t =1
dr
 (t 3  t 4 )(2t )  (t  t 6 )(3t 2 )  2t 4  2t 5  3t 3  3t 8
dt
 dr
W  F 
dt
0
1
1
29

2 5 2 6 3 4 3 9
4
5
3
8
dt    2t  2t  3t  3t dt   t  t  t  t  
6
4
9  0 60

5
0
1
Example:
If A  (3x 2  6 y)i  14 yz j  20 xz 2 k calculate  A. dr
c
from (0,0,0) to (1,1,1) . if the path C is:
1) x  t , y  t 2 , z  t 3
2) line from (0,0,0) to (1,0,0) and to (1,1,0) and to (1,1,1)

3) the straight line (0,0,0) 
(1,1,1)
Solution:
60
1)
x  t  dx  dt , y  t 2  dy  2t dt and z  t 3  dz  3t 2 dt
1
 (3t

2
 6t 2 ) dt  14(t 2 )(t 3 )(2tdt )  20(t )(t 6 )(3t 2dt )
0
1
   9t 2  28t 6  60t 9  dt  3t 3  4t 7  6t 10   3  4  6  5
0
1
0
2) From
(0, 0, 0)  (1, 0, 0)  z  0
1
 3x
x 0
, dz  0
and y  0  dy  0
1
dx   x 3   1        (1)
0
2
From (1,0,0) to (1,1,0)  x 1 
dx  0 , z  0 
1


000  0
     (2)
y 0
Last from
(1, 0, 0)  (1,1,1)
x  1 , dx  0 ,
1


z 0
y  1 , dy  0
1
 20 z 3 
20
0  0  20z dz  


3
 3 0
2
  A  dr  (1)  (2)  (3)  1  0 
       (3)
20
23

3
3
3) The equation of a straight line is
x  x1
y  y1
z  z1


 t
x 2  x1
y 2  y1
z 2  z1
61
dz  0
x 0
y  0
z 0


 t  x t , y  t
1 0
10
1 0
and z  t
 dx  dt , dy  dt and dz  dt
1


14 t 3
13
  (3t  6t  14t  20t )dt  t 3  3t 2 
 5t 4  
3
3

0
0
1
2
2
3
Example:
Find the work done by force
F  (2 x  y  z )i  ( x  y  z 2 ) j  (3x  2 y  4 z )k
To move a particle on a circle lie on x y- play which has center at origin
and radius of 3 unit.
Solution:
Work =  Force dr
c
w   F . dr   (2 x  y ) dx  ( x  y )dy
c
Because
z  0  dz  0 Let x  3cos  , y  3sin  ,  : 0 
 2 .
2
w   (6 cos   3sin  )(3sin  d  )  (3cos   3sin  )(3cos  d  )
0
2
  ( 9  9sin  cos  ) d   9  0
2
0
 (9) (2 ) 

2
0


2
0
sin cos  d 
2
sin 2
 cos 2 
1 1
d   18  

18


 4   18
2
 4  0
62
Flow Integrals and Circulation for Velocity Fields
Instead of being a force field, suppose that V represents the velocity
field of a fluid flowing through a region in space (a tidal basin or the
turbine chamber of a hydroelectric generator, for example). Under
these circumstances, the integral of along a curve in the region gives
the fluid’s flow along the curve.
The Flow Integral)‫(تكامل التدفق‬
If r(t) is a smooth curve in the domain of a continuous velocity field V,
the flow along the curve from t = a to t = b is:
 dr 
Flow   V  dt
dt 
a
b
Example:
A fluid velocity field is
V  xi  zj  yk
r (t )  cos ti  sin tj  3tk
0 t 
find the flow along the helix

2
Solution:
V  xi  zj  yk  cos ti  3tj  sin tk
V 



dr
 cos ti  3tj  sin tk   sin ti  cos tj  3k   sin t cos t  3t cos t  3sin t
dt

 dr
Flow   V 
dt
0
2


 cos 2 t
2

dt


sin
t
cos
t

3
t
cos
t

3sin
t
dt




 2  3t sin t  3cos t  3cos t 



0
0
2
63
2



cos

  cos 0 2


  3 0 sin 0  6 cos 0    3



2








 3   sin   6  cos    
   
 

2
2 
2    2
 2 
  2




 11

 2
Circulation Integral)‫(تكامل التدوير‬
If the curve is a closed loop, the flow is called the circulation around the
curve.
Circulation 

 V

dr
dt

 dt

We evaluate flow and circulation integrals the same way as we evaluate
work integrals.
Example:
Find the circulation of the field V
r (t )  cos ti  sin tj
 (x  y ) i  xj
around the circle
0  t  2
Solution:
V  (x  y ) i  xj  (cos t  sin t ) i  cos tj
V 

dr
 (cos t  sin t ) i  cos tj
dt
2
Circulation 
  sin ti  cos tj   cos
 1  sin t cos t  dt
0
 cos t 
t 
2
2
t  sin 2t  sin t cos t  1  sin t cos t
2
 2
‫ارض بما قسم هللا لك تكن اغنى الناس‬
64
PROBLEMS
1. Evaluate the following integrals along the straight line paths joining
the end points:
(1)

(1,2)
(2,3)
y 2dx
(0,0)

(2)
(2,1)
(2,1)
ydx
(3)

xdy
(1,1)
2. Evaluate the following line integrals:
(0,1)

(1)
y 2dx  x 2dy
Where C is the semicircle
x  1 x 2
(0, 1)
(2,4)

(2)
ydx  xdy
Where C is the parabola
y x2
(0,0)
ydx  xdy
x2y2
(1,0)
(0,1)
(3)

Where C is the curve
x  cos3 t
y  sin 3 t
0 t 

2
3. Evaluate the following line integrals:
(1)
y
dx  xydy
Where C is the square with vertices (1, I), (-1, I),
2
C
-I), (1, -1);
(2)
x
2
y 2dx  xy 3dy
Where C is the triangle with vertices (0, 0),
C
(1, 0), (1, 1).
(3)
y
dx  xdy
Where C is the circle
C
65
x2y2 4
(-1,
surface Integration)‫(تكامالت السطح‬
Let
S (x , y , z )
be a surface on xyz-plane , to find the area of S we find the
surface element ds which has projection on xy-plane or xz-plane or yzplane and then we find the unit tangent vector n to ds then:
Area   ds  
s
R
dxdy
dxdz
dydz


n  k R n  j R n i
The flux)‫(الفيض‬
Flux    A  n  ds    A  n 
s
   A  n 
R
R
dxdy
n k
dxdz
dydz
   A  n 
nj R
n i
Example:
Find the flux of
A  18i  12 j  3yk
over the surface
2x  3 y  6z  12
in the
first octant
Solution:
Flux    A  n  ds    A  n 
s
R
dxdy

n k
To obtain n (the perpendicular vector to the surface 2x  3 y  6z
given by
n
 2 i  3 j  6 k 2
3
6

 i  j k

7
7
7
4  9  36
66
 12 )
is
Then a unit normal to any point of S.
Thus
Also
3
6
2
n k   i  j  k
7
7
7
6

k 
7


dxdy
dxdy

6
n k
7
3
6 
36z  36  18 y 36  12x
2
A  n   i  j  k   18i  12 j  3 yk 

7
7 
7
7
7

Using the fact that z 
12  2 x  3 y
from the equation of S. Then
6
 A  n dS   A  n
R

R
dx dy

n .k
 36  12x  7
 dxdy 
7
6
 
R
 (6  2x ) dxdy
R
To evaluate this double integral over R. keep x fixed and integrate with
12  2x
3
0
from
. In this manner R is completely covered. The
x  0 to
x 6
to
y 
respect to y from y
then integrate with respect to x
integral becomes.
6 (12 2 x ) / 3

x 0
4x2
 (6  2 x)dydx   (24  12 x  3 )dx  24
 0
x 0
6
If we had chosen the positive unit normal n opposite to that above, we
would have obtained the result -24.
Example:
Evaluate
 A  n dS where
A  zi  xj  3y 2zk
S
67
and S is the surface of the
cylinder x 2  y 2  16 included in the first octant between
z  0 and
z 5
.
Solution:
Project of S on the xy plane as in the figure below and call the
projection R.
Note that the projection of S on the xy - plane cannot be used here.
Then:
A normal to x 2  y 2  16 is ( x 2  y 2 )  2 xi  2 yj . Thus the unit normal
to S as shown in the adjoining figure, is
n
2 xi  2 yj

(2 x) 2  ( 2 y ) 2
xi  yj
4
Since x 2  y 2  16 on S .
A  nˆ  (zi  xj  3 y 2 zk )  (
1
(xz  xy )
4
 x i  yj
nˆ  j  
 4
x i  yj
4
)


y
 j 
4

Then the surface integral equals
68


xz
   16  x 2  x  dxdz 

z 0 x 0 
xz  xy
 y dxdz 
R
5
4
5
 (4 z  8)dz  90
z 0
Example:
Evaluate    n dS
S
3
where   xyz
8
and S is the surface of problem above.
Solution:
We have    n dS     n dxdz
n. j
S
Using
nˆ 
R
xi  yj
,
4
nˆ  j 
y
4
As in example above, this last integral becomes.
3
3 5 4 2
2
 8 xz( xi  yj)dxdz  8   ( x zi  xz 16  x j )dxdz
R
z 0 x 0
3 4 64
64
  ( zi  zj )dz  100i  100 j
8 x 0 3
3
Example:
If
F  yi  (x  2xz ) j  xyk
evaluate
 (  F )  nˆ dS where S is the surface
S
of the sphere x 2  y 2  z 2  a 2 above the xy- plane.
Solution:
69
i

F 
x
y
j

y
x  2xz
k

 x i  y j  2z k
z
xy
A normal to x 2  y 2  z 2  a 2 is
(x 2  y 2  z 2 )  2x i  2 y j  2z k
Then the unit normal n of the figure above is given by
2x i  2 yj  2zk
n
4x 2  4 y 2  4z 2

x i  yj  zk
a
Since x 2  y 2  z 2  a 2
The projection of S on the xy- plane is the region R bounded by the
circle x 2  y 2  a 2 , z  0 then
 (  F )  n dS
 (  F )  n

S

 x i  yj  zk  dxdy
 z
a

a
 (x i  yj  2zk )  
R
a

dx dy
n k

a2 x 2

3(x 2  y 2 )  2a 2
x  a y  a 2  x 2
a2  x 2  y 2
dydx
Using the fact that z  a 2  x 2  y 2 . To evaluate the double integral,
transform
to
polar
x   cos  , y   sin  and dydx  d d 
coordinates
( , )
where
. The double integral becomes:
70
2
3 2  2a 2
a
 
a2   2
 0  0
d d  
2
3(  2  a 2 )  a 2
a
 
a2   2
 0  0
2

a
 
 0  0
d d 

 3 a 2   2 


2
a

 d d 
a 2   2 
a2
2
  (a 2   2 )3 2  a 2 a 2   2  d    (a 3  a 3 )d   0


 0
 0
0
Example:
Find the flux of
F  4zi  y 2 j  yzk
where S is the surface of the cube
0  x  1 , 0  y  1 , 0  z  1.
bounded by
Solution:
On the face DEFG :

ni
and
x  1.
Then
1 1
F  n dS 
DEFG
  (4z i
 y 2 j  yzk )  ( i )dydz
0 0
1 1

   4z  dydz
 2
0 0
On the face ABCO

ABCO
n   i and x  0 .Then
1 1
F  n dS 
  ( y
2
j  yzk )  (i ) dydz  0
0 0
On the face ABEF :
n  j , y 1
. Then
71

1 1
F  n dS 
ABCO
  (4xzi
0 0
1 1
 j  zk )  (  j ) dxdz    dxdz   1
0 0
On the face OGDC : n   j , y  0. then

1 1
F  n dS 
OGDC
On the face BCDE :

  (4xzi )  ( j )dydz
n  k , z 1
. Then .
1 1
F  n dS 
OCDE
2
  (4xi  y j  yk )  (k ) dxdy 
0 0
On the face AFGO :
n  k , z  0

0
0 0
1 1
  y dxdy
0 0

1
2
. Then
1 1
F  n dS 
AFGO
  ( y
2
j )  (k )dxdy  0
0 0
Lastly we find that:
Flux   F  n dS  2  0  (1) 
S
1
3
0 
2
2
Volume integral)‫(تكامل الحجم‬
Let S be a surface contained a volume V to evaluate the value of V we
take volume element
V
and integrating it as :
 V   dV   dxdydz   dzdydx
V
V
V
72
V
Example :
Let
F  2xzi  xj  y 2k
. Evaluate
 F dV where V is the region
V
bounded by the surfaces x
0, y 0,y 6,z x 2 ,z 4
.
Solution:
The region V is covered
(i) By keeping x and y fixed and integrating from
z x2 ,z 4
(base to
top of column PQ),
(ii) Then by keeping x fixed and integrating from
y 0, y 6
(R to S in
the slab),
(iii) Finally integrating from x
 0 to x  2
the required integral is
73
(where
z  x 2 meets z  4 ).
Then
2
6
4
   (2 xzi  xj 
y 2 k )dzdydx
x 0 y 0 z  x 2
26 4
   2 xzdzdydx 
 i
26 4
j
0 0 x2
26 4
   xdzdydx  k    y
0 0 x2
2
dzdydx
0 0 x2
 128 i  24 j  384 k
Example:
Find the volume of the region common to the intersecting cylinders
x2  y2  a2
x2  z2  a2
and
Required volume = 8 times volume of region shown in above figure.
8
a

x 0
8
a

x 0
a2  x2

y 0
a2  x2

y 0
a2  x2

dzdydx
z 0
16a 3
a  x dydx  8  (a  x ) dx 
3
x 0
a
2
2
2
74
2
Integral theorems)‫(نظريات التكامل‬
The following theorem and its generalizations are fundamental in the
theory of line, surface and volume integrals
(1)Green's theorem in the plane)‫(نظرية قرين على المستوى‬
Let D be a domain of the xy-plane and let C be a piecewise
smooth simple closed curve in D whose interior is also in D. Let P(x,y)
and Q(x,y) be functions defined and continuous and having continuous
first partial derivatives in D. Then:
 P dx
 Q dy 
C
 Q
  x
R

P 
 dxdy
y 
Where R is the closed region bounded by C. The theorem will be proved
first for the case in which R is representable in both of the forms:
Example:
Verify green’s theorem to   5  xy  y 2  dx   2xy  x 2  dx
c
Where C is the square (0,0),(1,0),(1,1),(0,1) (traveled counter clockwise)
Solution:
R.H.S:
P  5  xy  y 2 

P
 x  2 y
y
and
Q  2xy  x 2 
Q P

 3x
x y
75
Q
 2  2x
x
1
1
  5  xy  y  dx   2xy  x  dx =    3x  dxdy  2
2
3
2
y 0 x 0
c
L.H.S: line integral
  5  xy  y  dx   2xy  x  dx
2
2
c
(i) From (0,0) to (1,0)
y  0  dy  0
1
  5dx  5x 0  5
1
0
1
(ii)From (1,0) to (1,1)
x  1  dx  0
  (1  2 y )dy   y  y 2   0
0
1
0
0

x2
1  7

  (4  x )dx   4x    0   4   
2 1
2 2


1
0
(iii)From (1,1) to (0,1)
y  1  dy  0
(iv)From (0,1) to (001)
x  0  dx  0
1
 0  0
0
7
3
   5  xy  y 2  dx   2xy  x 2  dx  (i )  (ii )  (iii )  (iv ) (5)  (0)  ( )  (0) 
2
2
c
R.H.S = L.H.S
Thus Green’s theorem is verified
Example:
Let C be the circle:
x 2  y 2 1
. Then evaluate
76
 4xy
dx  6x 2 y 2dy
3
Solution:
 4xy
C
dx  6x 2 y 2dy   (12xy 2  12xy 2 )dxdy  0
3
R
Example:
Verify green’s theorem to   x 2  y  1 dx   2xy  x  dx
c
Where C is the place which is bounded by
x2y2 4
Solution:
The R.H.S:
P  x 2  y 1 
 x
2
c
P
 1 and
y
Q  2xy  x 
Q
Q P
 2y 1 

 2y
x
x y
 y  1 dx   2xy  x  dx     2 y  dxdy
But x 2  y 2  4 x  r cos  , y  r sin   dxdy  rdrd 
   2 y  dxdy
2

2
2
   2r sin   rdrd   4  sin d   4 cos 0
 0 r 0
2
 zero
0
The L.H.S:
 x
2
 y  1 dx   2xy  x  dx
c
But x 2  y 2  4 x  2cos  , y  2sin   dx  2sin  d  , dy  2cos  d 
 x
2
 y  1 dx   2xy  x  dx
c
77
2

 (4 cos
2
  2sin   1)(2sin  d  )  (8sin  cos   2 cos  )(2 cos  d  )
0
2

 (8sin  cos
2
  4sin 2   2sin   16sin  cos 2   4 cos 2  )d 
0
2
2
2
1
 1

 8  sin  cos 2  d   2  sin  d   4   (1  cos 2 )  (1  cos 2 )  d 
2
2

0
0
0 
2
 cos3  
2
2
 8
 2  cos  0   cos 2 0  zero

 3 0
Example:
Let C be the ellipse
4x 2  y 2  4
. Then evaluate  (2x  y )dx  (x  3y )dy
C
Solution:
(Where A is the area of R. Since the ellipse has semi-axes
area is  ab  2 , and the value of the line integral is 4 )
a  2 ,b 1
 (2x  y )dx  (x  3y )dy   (1  1)dxdy  2A  4
C
R
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78
the
Problems
(A).Evaluate by Green's theorem
(1)
 aydx  bxdy on any path;
C
(2)
e
x
sin ydx  e x cos ydy
around the rectangle with vertices
C


(0, 0) , (1, 0) , (1, ) , (0, )
2
2
(3)
 (2x
2
 y 3 )dx  (x 3  y 3 )dy
around the circle
x 2  y 2 1
C
(B) Let C be the ellipse
4x 2  y 2  4
. Then prove that Green's theorem is
not applicable to the integral  2 y 2 dx  2 x 2 dy
x y
x y
C
(C) Let C be the square with vertices (1, 1), (1, -1), (- 1, -1), (- 1, 1). Then
prove Green's theorem to   x 2  2 y 2  dx
C
(2)The divergence theorem of Gauss)‫(نظرية التباعد لقاوس‬
State that if V is the volume bounded by a closed surface S and A
is a vector function of position with continuous derivates, then:
    A  dV
V

  A  n  dS
S

  A  n  dS
S
Where n is the positive (outward drawn) normal to S.
79
Example:
Verify the divergence theorem to a vector
region bounded by
A  4xi  2 y 2 j  z 2 k
over the
x 2  y 2  4 ,z  0 , z  3
L.H.S:
   A dV
V
  (4  4 y  2z ) dV 
V
2

4 x 2

3
 (4  4 y  2z ) dzdydx  84
x 2 y  4 x 2 z  0
R.H.S:
The surface S of the cylinder consists of abase
and the convex portion
S1 : z  0
,the top
S3 : x 2  y 2  4
Surface integral    A  n  ds    A  n  dS 1    A  n  dS 2    A  n  dS 3
S
(i)On
S1
S2
S3
S 1 : z  0 , n   k , A  4x i  2 y 2 j  A  n  0    A  n  dS 1  0
S1
(ii)On
S 2 : z  3 , n  k , A  4x i  2 y 2 j  9k  A  n  9    A  n  dS 2  36
S2
(iii)On
S 3 : x 2  y 2  4 , Aperpendicular vector to x 2  y 2  4 is
n
2x i  2 yj
x i  yj
(x 2  y 2  4)


2
2
2
(x  y  4)
4x 2  4 y 2
 x i  yj 
2
3
 A  n  (4x i  2 yj  z 2 k )  
  2x  y
 2 
80
S2 :z  3
Then the surface integral is
 0  36  48  84
Agreeing with the volume integral and verifying gauss (divergence)
theorem
Exercise:
(1)Verify the divergence theorem to a vector
the region bounded by
A  x 2 z 3 i  2xyz 3 j  xz 4k
over
(1, 2, 3)
(2) Use the divergence theorem to calculate  x 2 z 3dx  2xyz 3dy  xz 4dz over
C
the region bounded by
(1, 2, 3)
(3) Use the divergence theorem to calculate  x 2e 5z dx  x cos ydy  3ydz over
C
the region bounded by
x  0 , y  2  2 cost , z  2  2sin t
, 0  t  2
The parametric equations above describe a circle of radius 2 on the yzplane:
z
y
x
81
(3)Stokes' theorem)‫(نظرية ستوكس‬
States that if S is an open, two-side surface bounded by a closed, noninteger section curve C (simple closed curve) then if A has continuous
derivatives.
 A  dr

C
 (  A )  ndS
S
Where C is traversed in the positive direction. The direction of C is
called positive if an observer, walking on the boundary of S in this
direction, with his head pointing in the direction of the positive normal
to S. has the surface on his left.
Example:
Verify the Stokes’ theorem to a vector
the upper half surface of the sphere
A  (2x  y ) i  yz 2 j  zy 2k
x 2  y 2  z 2 1
Solution:
The L.H.S:
2
2
 A  dr   (2x  y )dx  yz dy  y zdz 
C
2
 (2 cos   sin  )( sin  )  
0
The R.H.S:
82
where S is
 i


A  
 x

 2x  y
j

y
 yz 2
k 

 
k
z 

zy 2 
     A   ndS   k  ndS   dxdy
S
S
R
And R is the projection of S on the xy-plane
1

1 x 2
1 1 x 2

x 1 y  1 x 2
dydx  4
0

0
n  kdS  dxdy
since
1
dydx  4 1  x 2 dx  
0
Stokes’ theorem is verified.
Example:
Use Stoke's Theorem to calculate the line integral  y 3dx  x 3dy  z 3dz .
C
Where the curve C is the intersection of the cylinder x 2  y 2  a 2 And the
plane
x  y z b
Solution:
We suppose that S is the part of the plane cut by the cylinder. The
curve C is oriented counterclockwise when viewed from the end of the
normal vector , which has coordinates
n
And
i  j k
3
 A  0i  0 j  (3x 2  3y 2 )k  3(x 2  y 2 )k
83
Applying Stoke's Theorem, we find:
y
dx  x dy  z dz      A   ndS  3
3
3
3
x
S
C
2
 y 2  dS
3
S
  3  (x 2  y 2 )dS   3a  dS
S
The projection of the surface S onto the xy-plane is the circle
S
x  y  a2
2
2
of radius a. Therefore, representing the equation of the plane in the
form
z b x  y
and using the formula
 3  (x 2  y 2 )dS   3a  dS  3a 2  dxdy  (3a 2 )  ( a 2 )  3 a 4
S
S
Exercise:
(A)Use Stoke's Theorem to calculate the line integral
(i) C (x  z )dx  (x  y )dy  xdz
Where the curve C is the ellipse
(ii)
x2 y2

1,
4
9
and
z  1 ( (Ans  6 ) )
 (z  y )dx  (x  z )dy  ( y  x )dz
C
The curve C is triangle with vertices

 Hint n  AB  BD

AB  BD





A (2, 0, 0) , B (0, 2, 0) , D (0, 0, 2)
(Ans=12)
84
(iii)  (z 2  y 2 )dx  (x 2  z 2 )dy  ( y 2  x 2 )dz
C
Where the curve C is formed by intersection of the paraboloid
z  5x 2  y 2
with the plane x  y  z
 1 .  Ans  18 
(iv) Use the stokes theorem to calculate  x 2e 5z dx  x cos ydy  3ydz over the
C
region bounded by
x  0 , y  2  2 cost , z  2  2sin t
, 0  t  2
(B) Verify stokes theorem to:
(1) A  (2x  y ) i  yz 2 j  zy 2 k
where S is the surface
(i) z  5  x 2  y 2
(ii ) x  y  z  1
(iii ) A (2, 0, 0) , B (0, 2, 0) , D (0, 0, 2)
(iv )
x2 y2

1,
4
9
(2) F  (x  z ) i  (x  y ) j  xk
where S is the surface
(i) x 2  y 2  16
(ii ) x 2  y 2  z 2  1
(iii ) x 2  y 2  4 , z  0 , z  3
(iv ) (1, 1), (1, -1), (- 1, -1), (- 1, 1)
(v) 4x 2  y 2  4
(3) F  (x  y 2 ) i  ( y  z 2 ) j  (z  x 2 )k
(i ) 2x  y  2z  2 (ii ) z  9  x 2  y 2
(iv ) (1, 2, 3)
where S is the surface
(iii ) x 2  y 2  4 ,  3  z  1
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85
Basic Algebra
(1)
Arithmetic Operations
a b ay  bx
 
x y
xy
a
a y ay
(4) x   
b x b bx
y
(1) a (x  y )  ax  by
(2)
(3)
a b ab
 
x y xy
(2) Laws of Signs
(1)  (x )x
(2)
x
x
x
 
y
y y
(3)
 x   y   xy
(3) Division by zero:
if
a0
then
(1)
0
0
x
(2) x 0  1
(4) Laws of Exponents;
m
(1) x m x n  x m  n
(4)  x  n
(2)
 x m   x mn
(3)
 xy 
n
n
 x ny n
n xm 
(3) 0a  0
 x
n
m
xm
 x m n
n
x
1
(6) x  n  n
x
(5)
(5) From The Binomial Theorem:
(1) (x  y )2  x 2  2xy  y 2
(3) (x  y )3  x 3  3x 2 y  3xy 2  y 3
(2) (x  y )2  x 2  2xy  y 2
(4) (x  y )3  x 3  3x 2 y  3xy 2  y 3
86
(6) Factoring the Difference of Powers:
(1) x 2  y 2  (x  y )(x  y )
(3) x 3  y 3  (x  y )  x 2  xy  y 2 
(2) x 3  y 3  (x  y )  x 2  xy  y 2 
(4) x 4  y 4  (x  y )  x 3  x 2 y  xy 2  y 3 
(7) Completing the Square:
b

ax 2  bx  c  a  x 2  x
a

2
2
 2 b

b  b  

c

a
x

x




      c
a

 2a   2a  

2
2
2
 2 b

b  
b2
b   b 
 a  x  x      a    c  a  x     c 

a
2a  
4a
 2a    2a 



b 

where u   x   and
2a 


 ax 2  bx  c  au 2  C
C c 
(8) The Quadratic Formula:
if
ax 2  bx  c  0
then  x 
 b  b 2  4ac
2a
87
a  0 
b2 

4a 
(9) Trigonometry Identities:
(A )
(1) sin( x )   sin x
(2) cos(  x )  cox
(3) sin 2 x  cos 2 x  1
(4) sec 2 x  1  tan 2 x
(5) csc x  1  cot x
1
(6) cos 2 x  (1  cos 2x )
2
1
(7) sin 2 x  (1  cos 2x )
2
2
2
(C )
(B )
(1) sin(A  B )  sin A cos B  cos A sin B
(2) sin(A  B )  sin A cos B  cos A sin B
(3) cos(A  B )  cos A cos B  sin A sin B
(4) cos(A  B )  cos A cos B  sin A sin B
tan A  tan B
(5) tan(A  B ) 
1  tan A tan B
tan A  tan B
(5) tan(A  B ) 
1  tan A tan B
(E )


(1) sin  A     cos A
2



(2) sin  A    cos A
2



(3) cos  A    sin A
2



(4) cos  A     sin A
2

(1) sin A  sin B  2sin
 A  B  cos  A  B 
2
2
 A  B  sin  A  B 
(2) sin A  sin B  2 cos
2
2
 A  B  cos  A  B 
(3) cos A  cos B  2 cos
2
2
 A  B  sin  A  B 
(4) cos A  cos B  2sin
2
2
(D )
1
sin(A  B )  sin(A  B ) 
2
1
(2) sin A sin B   cos(A  B )  cos( A  B ) 
2
1
(3) cos A cos B   cos(A  B )  cos(A  B ) 
2
(1) sin A cos B 
88
(10)Hyperbolic functions:
e x  e x
e x  e x
e x  e x
, cosh x 
, tanh x  x
2
2
e  e x
1
1
cosh x
(ii ) sec hx 
, csc hx 
, coth x 
cosh x
sinh x
sinh x
(iii ) cosh(x  y )  cosh x cosh y  sinh x sinh y
(iv ) sinh(x  y )  sinh x cosh y  cosh x sinh y
1
1
(v ) sinh x  sinh y  2sinh  x  y  cosh  x  y 
2
2

(i ) sinh x 
(1) cosh 2 x  sinh 2 x  1
(2) sinh 2x  2sinh x cosh x
(3) cosh 2x  cosh 2 x  sinh 2 x
1
(4) cosh x  1  cosh 2x 
2
2
1
 1  cosh 2x 
2
(6) tanh 2 x  1  sec h 2 x
(5) sinh 2 x 
(7) coth 2 x  1  csc h 2 x
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‫فكم من جاهل اردى حليما حين اخاه‬
‫يقاس المرء بالمرء اذا المرء ماشاه‬
‫وللشيء على الشيء مقاييس واشباه‬
89