EX – Evaluate the following integrals:
1)

cosh(lnx)
dx
x
6)
 sec h
2
( 2 x  3 ) dx
2)
 sinh( 2 x  1 ) dx
e x  ex
7)  x
dx
e  ex
3)
sinhx
 cosh 4 x dx
8)
 e
4)
 x  cosh(3x
9)
 1  cosh x dx
5)
 sinh
4
2
) dx
x  cosh x dx
10)
ax

 e  ax dx
sinh x
 csch
2
Sol.-
 dx 
  sinh(ln x )  c
x 
1
1
2 )  sinh( 2 x  1 )  ( 2 dx )  cosh( 2 x  1 )  c
2
2
1
sinh x
3) 

dx   sec h 3 x  tanh x dx
3
cosh x cosh x
sec h 3 x
2
   sec h x   sec hx  tanh x dx   
c
3
1
1
4 )  cosh( 3 x 2 )  ( 6 x dx )  sinh( 3 x 2 )  c
6
6
sinh 5 x
4
5 )  sinh x  cosh x dx  
c
5
1
1
6 )  sec h 2 2 x  3   2 dx   tanh 2 x  3   c
2
2
x
x
e e
7)  x
dx   tanh x dx  ln(cosh x )  c
e  ex
e ax  e  ax
2
2
8 ) 2
dx   sinh ax (a dx)  cosh ax  c
2
a
a
sinh x dx
9) 
 ln1  cosh x   c
1  cosh x
csc h 2 x
10 )   csc hx   csc hx  coth x dx   
c
2
1)
 cosh(ln x )  
٦
x  cothx dx
Integrals of inverse hyperbolic functions:
The integration formulas for the inverse hyperbolic functions
are:
1 )

2 )

du
1 u
du
 sinh  1 u  c
2
u2  1
 cosh  1 u  c
 tanh 1 u  c if u  1 1 1  u
du

3 ) 
c
  ln
1  u 2  coth  1 u  c if u  1  2 1  u
 1
du
  sec h  1 u  c   cosh  1    c
4 ) 
u 1  u2
u
 1
du
5 ) 
  csc h  1 u  c   sinh  1    c
u 1  u2
u
EX – Evaluate the following integrals:
1)
4)

x
dx
1  4x2
dx
4  x2
2)
5)


dx
4  x2
sec 2  d
tan 2  1
Sol.-
1
2 dx
1
 sinh  1 2 x  c

2
2
1 4x2
1 dx
x
2
 sinh  1  c
2) 
2
2
1 x
2
dx
3) 
 tanh  1 x  c
if x  1
2
1 x
 coth  1 x  c
if x  1
1)
 
٧
3)
dx
 1  x2
6)
1
 tanh ln x 

 x1  dxln
2
x

4)
5)
6)


1
 
x 4  x2 2 x
1
dx
1
tan 2   1
let
sec
2
1
 tanh (ln x ) 
1
2
2
dx
 2
1 x
2
1
  csc h  1 x  c
2
2
 d   cosh  1 (tan  )  c
u  ln x 
tanh u
2
2
2
1
ln x
2
1
dx
2x
du 
dx
x ( 1  ln 2
x)

  tanh  1 u 

2
 c  tanh  1 (ln x )  c
٨
2 du
1  u2
Methods of integration
Integration by parts:
The formula for integration by parts comes from the product
rule:d ( u  v )  u  dv  v  du  u  dv  d ( u  v )  v  du
and integrated to give:
 u dv   d ( u  v )   v du
then the integration by parts formula is:-

u dv  u  v   v du
Rule for choosing u and dv is:
For u: choose something that becomes simpler when
differentiated.
For dv: choose something whose integral is simple.
It is not always possible to follow this rule, but when we can.
EX – Evaluate the following integrals:
 lnx 
7)  sin ax
 xe dx
2 )  x  cosx dx
x
1)
3)

x
x1
6)
-1
dx
8)
 x  ln x dx
5 )  x  sec x dx
4)
e
ax
x
10)  x
2
9)
2

1  x 2 dx
dx
 sinbx dx
3
 e x dx
3
 e x dx
2
Sol. –
1)
let
u x 
du  dx 

dv  e x dx  v  e x 
 xe
x
 udv  u  v   vdu
dx  x  e x   e x dx  x  e x  e x  c
١
2)
u  x  du  dx


dv  cosx dx  v  sin x 
let
 udv  u  v   vdu
 x  cosx dx  x  sin x   sin x dx  x  sin x  cos x  c
u x
3 ) let
dv 



1 
1
dx  v  2  x  1 2 
x1

x

du  dx
x1
dx  2 x   x  1
2  x  1
 2x x  1 
3
2
5)
u x 
2
 2   x  1 2 dx
1
 c  2x x  1 
 x  1 3
4
3
c
du 
du  dx


dv  sec 2 x dx  v  tan x 
let
 x  sec
6)
3
2
1

dx 

x
let
   udv  u  v   vdu
3
x
2

dv  x dx  v 

3
x3
1
x3
1 3
2
2
x

l
nx
dx


ln
x

x
dx


ln
x

x c

3
3
3
9
u  ln x 
4)
1
 udv  u  v   vdu
2
x dx  x  tan x   tan x dx  x  tan x  ln cos x  c

let u  ln x  1  x
dv  dx

 udv  u  v   vdu

2

vx

2x
1
du 

2 1  x2
x  1  x2


dx
2
2
2
 ln x  1  x dx  x  ln x  1  x   x 1  x

 x  ln x  1  x 2


1 1  x2
 
1
2
2
٢

1
2


1
2
dx

 c  x  ln x  1  x 2  1  x 2  c
let u  sin  1 ax  du 
7)
 sin
1
a dx
ax dx  x  sin  1 ax  
 x  sin  1 ax 
1a x


8)

1
2
dx
2
1
 2a 2 x 1  a 2 x 2

2a
1 1  a2 x2
 x  sin ax 

1
2a
2
1
2
2
dv  dx  v  x
&
1a x
ax
2

1
2
dx
1
 c  x  sin ax 
1  a2 x2
c
a
1
& dv  sin bx dx  v   cos bx
b
1 ax
a ax
ax
 e  sinbx dx   b e  cos bx  b  e  cos bx dx ...........(1)
1
let u  e ax  du  a  e ax dx & dv  cos bx dx  v  sin bx
b
1 ax
a ax
ax
e
cosbx
dx
e
sin
bx
e  sin bx dx
.......... .(2)





b
b
sub. (2) in (1) 
let u  e ax  du  a  e ax dx
1 ax
a ax
a2
ax
 e  sinbx dx   b e  cos bx  b 2 e  sin bx dx  b 2  e  sin bx dx
a2
1 ax
a ax
ax
ax
e
sinbx
dx
e
sin
bx
dx
e
cos
bx
e  sin bx dx  c








b
b2 
b2

a2 
e ax
 1  2   e ax  sinbx dx  2 a sin bx  b cos bx   c
b 
b

ax

e ax
 e  sinbx dx  a 2  b 2 a sin bx  b cos bx   c
ax
9) derivative of u
x3
3x 2
6x
6
0
integration of dv
+
+
-
ex
ex
ex
ex
ex
  x 3 e ax dx  x 3 e x  3 x 2 e x
 6 xe x  6 e x  c


 ex x3  3x2  6 x  6  c
٣
10 )
u  x2
let

3
x
 x  e dx 
2
du  2 x dx
dv  x  e x dx  v 
2
&
1 x2
e
2
2
2
1 2 x2 1
1
1 2
x  e   2 x  e x dx  x 2  e x  e x  c
2
2
2
2
Odd and even powers of sine and cosine:
To integrate an odd positive power of sinx (say sin2n+1x ) we
split off a factor of sinx and rewrite the remaining even power in
terms of the cosine. We write:-
and
 sin
 cos
2 n 1
2 n 1


x  dx   1  sin x 
x  dx   1  cos 2 x
2
n
 sin x dx
n
 cos x dx
EX - Evaluate:
1)
 sin
3
x dx
2)
 cos
5
x dx
Sol.-
1)
 sin
3


x dx   sin 2 x  sinx dx   1  cos 2 x  sinx dx
  sinx dx   cos 2 x   sinx  dx   cos x 
2)

1
cos 3 x  c
3

5
4
2
 cos x dx   cos x  cosx dx   1  sin x  cos x dx
2
  cosx dx  2  sin 2 x  cos x dx   sin 4 x  cos x dx
 sinx 
2
1
sin 3 x  sin 5 x  c
3
5
To integrate an even positive power of sine (say sin2nx ) we use
the relations:-
cos 2 
1  cos 2
2
or
sin 2 
then we can write:-
٤
1  cos 2
2
 sin
and
 cos
n
2n
 1  cos 2 x 
x  dx   
 dx
2


n
2n
 1  cos 2 x 
x  dx   
 dx
2


EX - Evaluate:
1)
 cos
2
 d
2)
 sin
4
 d
Sol.1)
1  cos 2
1
1

d    d   2 cos 2 d 
2
2
2

1
1

   sin 2   c
2
2

 cos  d  
2


1
 1  cos 2 
2
2 )  sin  d   
 d   d   cos 2 ( 2 d )   cos 2 d
2
4


1
1
1  cos 4
1
 1

   sin 2  
d     sin 2  (   sin 4 )  c
4
2
2
4
 4

1
3
1

  sin 2 
sin 4  c
32
8
4
2
4
To integrate the following identities:-
 sin mx  sin nx
dx ,
 sin mx  cos nx
dx
, and
we use the following formulas:-
cos( m  n ) x  cos( m  n ) x
2
sin( m  n ) x  sin( m  n ) x
sin mx  cos nx 
2
cos( m  n ) x  cos( m  n ) x
cos mx  cos nx 
2
sin mx  sin nx 
٥
 cos mx  cos nx
dx
EX - Evaluate:
1)
 sin 3 x  cos 5 x dx
2)
 cos x  cos 7 x dx
3) sin x  sin 2 x dx
Sol.1)
 sin 3 x  cos 5 x dx  2  sin( 3 x  5 x )  sin( 3 x  5 x ) dx
1

1 1
1
1
 1

sin
2
x
(
2
dx
)

sin
8
x
(
8
dx
)

cos
2
x

cos 8 x  c
 4
2  2 
8
16

2)
 cos x  cos 7 x dx 
1
1
1


cos(
6
x
)

cos(
8
x
)
dx

sin
6
x

sin 8 x  c
2
12
16
3)
 sin x  sin 2 x dx 
1
1
1


cos
x

cos
3
x
dx

sin
x

sin 3 x  c
2
2
6
Trigonometric substitutions:
Trigonometric substitutions enable us to replace the binomials
a  u 2 , a 2  u 2 , and u 2  a 2 be single square terms. We can
use:2
u  a sin 
for a 2  u 2  a 2  a 2 sin 2   a 2 ( 1  sin 2  )  a 2 cos 2 
u  a tan 
for a 2  u 2  a 2  a 2 tan 2   a 2 ( 1  tan 2  )  a 2 sec 2 
u  a sec 
for u 2  a 2  a 2 sec 2   a 2  a 2 (sec 2   1 )  a 2 tan 2 
EX - Evaluate the following integrals:
1)

2)

z 5 dz
1  z2
dx
4  x2
dx
3) 
4  x2
٦
4)

5)

6)

x2
9  x2
dt
dx
25t 2  9
dy
25  9y 2
Sol.-
z  tan 
1 ) let
z 5 dz

1  z2

dz  sec 2  d

tan 5   sec 2  d

1  tan 2 
tan  
z
1
  tan 5  sec  d

  tan   sec  sec 2  1 d
2
  sec 4  tan   sec  d   2  sec 2  tan   sec  d    tan   sec  d
1
2
sec 5   sec 3   sec   c
5
3
1
2
 ( 1  z 2 )5  ( 1  z 2 )3  1  z 2  c
5
3

x  2 tan 
2 ) let

dx
4  x2


dx  2 sec 2  d

2 sec 2  d
4  4 tan 2 
 ln
1  z2
1
tan  
4  x2
4 x
x
 c
2
2
θ
 ln 4  x 2  x  c 
x
4  x2
c
1
2 x
1
ln
 c  ln
2
2
( 2  x )( 2  x )
٧
x
2
wh ere c   c  ln 2
3 ) let x  2 sin  
dx  2 cos  d
dx
2 cos  d
1
d
1
 4  x 2   4  4 sin 2   2  cos   2  sec  d
1
2
 ln sec   tan   c
2
θ

x
2
  sec  d  ln sec   tan   c
2
2
1
 ln

2
4  x2
z
Ө
x
4  x
2
2 x
1 2 x
 c  ln
c
2 x
4 2 x
4 ) let

x  3 sin
x 2 dx

dx  3cos  d

9 sin 2 

3cos  d  9  sin 2  d
9  9 sin 2 
3
1  cos 2
9
1

d     sin 2   c
 9
x
2
2
2

θ
9
9  x2
   sin   cos    c
2
9 
x
9  x 2 
9
x
x
-1 x

 
sin
 c  sin -1
  9  x2  c

2 
3
3
3
2
3
2

9  x2
5t  3 sec  
5dt  3 sec   tan  d
3 sec   tan  d
dt
1
 5
  sec  d
5
25 t 2  9
9 sec 2   9
1
 ln sec   tan   c
5t
5
5 ) let

6 ) let

25 t  9
c
3

1 5t
ln 
5
3

1
ln 5 t  25 t 2  9  c 
5
3y  5 tan 
dy
25  9 y

2



25 t 2  9
3
where c   c 
1
ln3
5
3dy  5 sec 2  d
5 sec 2  d
1
3
  sec  d
3
25  25 tan 2 
1
ln sec   tan   c
3
25  9 y 2 3 y
1
 ln

c
3
5
5

θ
2
1
ln 25  9 y 2  3 y  c 
3
٨
25  9 y 2
θ
where c   c 
3y
5
1
ln5
3
EX - Prove the following formulas:
1)
du

 sin  1
a 2  u2
u
c
a
2)
du
1
1 u

tan
c
 a 2  u2 a
a
Proof.-
u  a sin 
1 ) let
du

a u
2
2 ) let

2


du  a cos  d
u
a cos  d
  d    c  sin  1  c
a
a 2  a 2 sin 2 
u  a tan 
du  a sec 2  d

du
a sec 2  d
1
1
1
1 u
 a 2  u 2   a 2  a 2 tan 2   a  d  a   c  a tan a  c
Integral involving a x 2 + b x + c :
By using the algebraic process called completing the square, we
can convert any quadratic: a x 2 + b x + c , a ≠ 0 to the form:
a( u 2  A 2 )
we can then use one of the trigonometric
substitutions to write the expression as a times a single square
term.
EX – Evaluate:
dx
1) 
2x  x2
dx
2) 
2x2  2x  1
3)

4)

5)

dx
1  x  x2
dx
x2  2x  8
dx
x2  2x  2
Sol.
1)

dx
2x  x2

let
x  1  sin 

dx
2x  x
2

dx
1 ( x2  2x  1 )



dx
1  ( x  1 )2
dx  cos d
cos d
1  sin 
2
  d    c  sin  1 ( x  1 )  c
٩
2)
3)
1
dx
dx
1
dx
  2
 
2 x1 2  1
 2x  1 2 x  x  1
2
2
4
1 1
1
let x   tan   dx  sec 2 d
2 2
2
1 sec 2  d
dx
1
1
 2 x 2  2 x  1  2  1 2tan 2  1   d    c  tan ( 2 x  1 )  c
4
4
 2x


2
dx
x2  2x  2
dx

x  2x  2
2
dx

x  1  tan 
let

( x  1 )2  1

dx  sec 2 d
sec 2  d

x2  2x  2
tan   1
2
θ
  sec  d
x+1
1
 ln sec  tan   c  ln x 2  2 x  2  x  1  c
4)

dx
1 x  x2
5)

dx

5  x1
2
4

2
1
5
5
sin   dx 
cos d

2
2
2
5 cos  d
 2x  1
2
  c
  d    c  sin  1 
5  5 sin 2 
5 

4
4
x
let



dx
x2  2x  8

dx
( x  1 )2  9
let x  1  3 sec   dx  3 sec   tan  d


3 sec  tan  d
9 sec 2   9
  sec  d
x1
 ln sec  tan   c  ln

3
 ln x  1 
x 2  2 x  8  c
١٠
x1
θ
x2  2x  8
3
x2  2x  8
c
3
w here c   c  ln3
Partial fractions:
Success in separating
f(x)
g( x )
into a sum of partial fractions
hinges on two things:1- The degree of f ( x ) must be less than the degree of g ( x ) .
(If this is not case, we first perform a long division, and then
work with the remainder term).
2- The factors of g ( x ) must be known. If these two conditions
are met we can carry out the following steps:
Step I - let x  r be a linear factor of g ( x ) . Suppose ( x  r ) m
is the highest power of ( x  r ) that divides g ( x ) . Then assign
the sum of m partial factors to this factor, as follows:
Am
A1
A2
.........



x  r ( x  r )2
( x  r )m
Do this for each distinct linear factor of f ( x ) .
Step II - let x 2  p x  q be an irreducible quadratic factor of
g ( x ) . Suppose ( x 2  p x  q ) n is the highest power of this
factor that divides g ( x ) . Then, to this factor, assign the sum
of the n partial fractions:
Bn x  C n
B1 x  C 1
B2 x  C 2

 ......... 
2
2
2
2
x  p xq (x  p xq)
( x  p x  q )n
Do this for each distinct linear factor of g ( x ) .
Step III - set the original fraction
f(x)
g( x )
equal to the sum of
all these partial fractions. Clear the resulting equation of
fractions and arrange the sums in decreasing powers of x.
Step IV - equate the coefficients of corresponding powers of x
and solve the resulting equations for the undetermined
coefficients.
١١
EX – Evaluate the following integrals:
2x  5
 x 2  9 dx
x dx
2)  2
x  4x  3
sin x dx
 cos x  5 cos x  4
2x2  3x  2
5) 
dx
 x  1 2  x  2 
1)
3)
 x
4)
x3  x
2

 1   x  1
2
dx
6)
2

x3  4 x2
dx
x2  4x  3
Sol.-
1)
2x  5
2x  5
dx

 x2  9
  x  3    x  3  dx
2x  5
A
B



x  3 x  3 x  3 x  3

2)
2x  5  A x  3   B  x  3 
at
x3

6A  6  5

at
x  3

 6B  6  5

11
6
1
B
6
A
1 
 11
2x  5
11
1


6
dx   
ln x  3   ln x  3   c
 6  dx 
2
6
6
x 9
 x3 x 3


x dx
x dx

 x 2  4 x  3   x  3  x  1
x
A
B


 x  A x  1  B  x  3 
 x  3  x  1 x  3 x  1
3
1
at x  3  A 
and at x  1  B  
2
2
 3
 1 
x dx
3
1
 2
2
 x 2  4 x  3    x  3  x  1  dx  2 ln x  3   2 ln x  1  c


١٢
3)

 x
x3  x
2

 1  x  1
2
dx 
x  x  1 x  1
 x
2

 1  x  1
x2  x
Ax  B
C
 2

2
x  1  x  1 x  1 x  1

x3  x
2
4 ) let

dx 



x 2  x   Ax  B  x  1   C x 2  1
x 2  x   A  C  x 2   A  B  x   B  C 
A  C  1 ......( 1 ) 

 A  B  1 ......( 2 ) 
 B  C  0 ......( 3 )
 x

2
x2  x
dx
x 2  1  x  1

 1  x  1
2
A0 , B1 , C 1
1 
 1
1
dx    2

 dx  tan x  ln x  1  c
 x  1 x  1
y  cosx
dy   sinx dx

sin x dx
dy
dy




 y 2  5 y  4   y  4  y  1
cos 2 x  5 cos x  4
dy
A
B


 y  4  y  1 y  4 y  1
at
y4  A
1
3
and

1  A y  1  B  y  4 
at y  1 
B
1
3
 1
 1 
sin x dx

3
 cos 2 x  5 cos x  4     y  4  y  31  dy


1
1
1
1
  ln y  4   ln y  1  c   lncos x  4   lncos x  1   c
3
3
3
3
2
2x  3x  2
A
B
C



5)
 x  1 2  x  2  x  1  x  1 2 x  2
2 x 2  3 x  2  A x  1 x  2   B  x  2   C  x  1
A  C  2 .......... ..........( 1 ) 

 3 A  B  2C  3 ......( 2 ) 
2 A  2 B  C  2 ...........( 3 ) 
A  2 , B  1 , C  4
 2
1
4 


dx


  x  1 2  x  2 
  x  1 (x  1 )2 x  2  dx
1
 2 ln( x  1 ) 
 4 ln( x  2 )  C
x1
2x2  3x  2
١٣
2

x3  4 x2
3x
6) 2
 x
 x  3  x  1
x  4x  3
x  4x  3
2
x
x3  4x2
 x3  4 x2  3x
 3x
3x
A
B


 3x  A x  1  B  x  3 
 x  3  x  1 x  3 x  1
9
3
at x  3  A 
and at x  1  B  
2
2
9
3 

x3  4 x2


2
 x 2  4 x  3 dx    x  x  3  x 21  dx


x2 9
3

 ln x  3   ln x  1  c
2
2
2
Rational functions of sinx and cosx, and other trigonometric
integrals:
We assume that z  tan
Since
cos 2
x
2
dz
then x  2 tan 1 z and dx 
2
1  z2
x 1  cos x

2
2

 cosx  2 cos 2
2
2
1 2
1
z 1
2 x
tan
1
2
x
1
2
2
1
x
sec
2
1  z2
 cosx 
1 z2
2
Since
x
x
x
2  cos 2 x  2 tan x  1
sin x  2 sin  cos 
2
2 cos x
2
2 sec 2 x
2
2
x
1
2z
 2 tan 
 sinx 
2
1  z2
2 x
tan
1
2
2sin
١٤
EX – Evaluate:
dx
 1  sin x  cos x
dx
2) 
sin x  tan x
dx
3) 
2  sin x
Sol.1)
3 dx
2  4 sin x
4)

5)
 secx dx
6)
cos x dx
 1  cos x
2
dz
2
dx
1

z


1) 
1  sin x  cos x 
2z
1  z2

1
1 z2 1  z2
x
 ln 1  z  c  ln 1  tan  c
2
2)
3)

2
dz
2
dx
1 1


1
z

    z dz
2z
2z
sin x  tan x
2 z


2
2
1 z
1 z
1
z2 
1
x 1
x
  ln z    c   ln tan  tan 2   c
2
2
2
2 2
2
dx
 2  sin x  
2
dz
1 z2

2z
2
1 z2
z
2
dz

z1

dz
1
3
( z  )2 
2
4
1
3
3
tan   dz 
sec 2  d

2
2
2
3
sec 2  d
dx
2
2
d 
 c
 2


3
3
2  sin x
2
3
3
tan  
4
4
x


2 tan  1 



2
2
z
1
2

2
c
  c 
tan  1 
tan  1 



3
3 
3
3





let z 

dz
 1 z
١٥
2
dz
2
3 dx
3
dx
3
dz

z
1
 
 
 3 2
4) 
2z
2  4 sin x
2 1  2 sin x 2
z  4z  1
1 2 2
z 1
dz
dz
 3

 z  2 2  3  ( z  2 ) 2  1
3
z2
 sec   dz  3 sec   tan  d
let
3

3 dx

2  4 sin x

3 sec   tan  d
sec 

3
 tan  d
sec 2   1
 3  csc  d   3 ln csc   cot   c
  3 ln
z2
z2  4z  1

z2
θ
z2  4z  1
3
x
2 3
3
2
 c   3 ln
c
x
x
z2  4z  1
tan 2  4 tan  1
2
2
tan
1  z2
2
1


5 )  secx dx  
dz
2
 1  z 1  z  dz
1  z2 1  z2
1
A
B


 A1  z   B 1  z   1
1  z 1  z  1  z 1  z
1
1
and at z  1  B 
at z  1  A 
2
2
1 
 1

2
2 
 secx dx  2   1  z  1  z  dz   ln1  z   ln1  z   c


x
1  tan
x
x


2 c
 ln 1  tan   ln 1  tan   c  ln
x
2
2


1  tan
2
By substituti ng
tan
x

2
1  cos x
1  cos x
 secx dx  ln sec x  tan x  c
١٦
implies
1 z2
2
cos x dx
2
1 z2

1
z

6) 

dz  
dz
1  cos x 
1  z2 1 z2
( 1  z2 ) z2
1
1  z2
1 z2
A B C zD




z z2
(1 z2 ) z2
1 z2
Az  Az 3  B  Bz 2  Cz 3  Dz 2  1  z 2
A  C  0 ......( 1 )
B  D  1 ....( 2 ) 

A  0 .............( 3 ) 
B  1 .............( 4 ) 
cos x dx
 1
A  0 , B  1 , C  0 , D  2
2 
1
1
 dz    2 tan z  c
z
z  1
1
x
x

 2   c   cot
 xc
x
2
2
tan
2
 1  cos x    z
2

2
١٧