Tích phân hàm lượng giác

Chương 5c Tích phân
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5/ Tích phân hàm lượng giác:.................................................................................................10
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1


dx

1/ I = ∫

x2 + a2
dx

a/ I=∫

⇒x=
=

x2 + a2
2

= ln x + x 2 + a 2

2

t −a
, dx =
2t

4t 2 − 2t 2 + 2a 2
4t 2

2

x 2 + a 2 = t − x ⇒ x 2 + a 2 = ( t − x ) = t 2 − 2t.x + x 2

, put

(

)

.dt =

t2 + a2
2t 2

t2 + a2
dx

⇒∫

x2 + a2

1

=∫

(

'
'
t 2 − a 2 .2t − ( 2t ) . t 2 − a 2

4t 2

) .dt

.dt

.dt

2

dt
2t
= ∫ = ln t + C = ln x + x 2 + a 2 + C
t
t2 − a2
t−
2t
1



⇒∫
=  ln x + x 2 + a 2  = ln 1 + 1 + a 2 − ln a
2
2

0
0 x +a
1b /

b/

dx

1
 i.a  
= ln tan  .arcsin   
 x 
2
x2 + a2
dx

∫
∫

dx
x2 + a2

dx

=∫

x2 −
'

⇒ dx =

−i.a. ( sin t ) .dt

=

(

−1.a

)

2

=∫

−i.a.cos t.dt

dx
x 2 − ( i.a )

2

 i.a 
, t = arcsin  
 x 

i.a
( i2 = −1) put x = sin t

sin 2 t
sin 2 t
 i.a.cos t.dt 
 i.a.cos t.dt 
1
1
⇒ I = −∫ 
= −∫ 


 sin 2 t   i.a 2
 sin 2 t  −a 2
2
+ a2

 − ( i.a )
2
 sin t 
sin t

2

 −i.a.cos t.dt 
= ∫

 sin 2 t 

 −i.a.cos t.dt 
= ∫

2
 sin 2 t 
−1 + sin t
1

a

sin 2 t

1
− cos 2 t

a

sin 2 t

 −i.a.cos t.dt  sin t
−dt
t
1
 i.a  
= ∫

=∫
= − ln tan = − ln tan  .arcsin   

sin t
2
 x 
2
 sin 2 t  i.a.cos t
1
 a 
= ln tan  .arcsin    put a = i.b
 x 
2
x2 − a2
dx

*∫

1
 i.b  
= ln tan  .arcsin   
 x 
2
x 2 + b2
dx

⇒∫

1c /

∫

c/

∫

dx
x2 + a2
dx
x2 + a2

= −i arcsin

i.x
+C
a
dx

=∫

−

(

−1.x

)

2

+ a2

=∫

dx
a 2 − ( i.x )

2

=∫

dx
 i.x 
a 1−  
 a 

i.x
i.dx
a.du i.a.du
= u ⇒ du =
⇒ dx =
=
= −i.a.du
2
a
a
i
i
dx
−1 i.a.du

i.x
⇒∫
= ∫
= −i arcsin
+C
2
2
2
a
a
a −x
1− u
Put

dx

*I=∫
⇒I=∫

Put x = a.sin t ⇒ dx = a.cos t.dt, t = arcsin

a2 − x2
a.cos t.dt

=∫

a.cos t.dt

= ∫ dt = t = arcsin

a 2 − a 2 .sin 2 t
a cos 2 t
i.dy
 i.y 
Put x = i.y ⇒ I = ∫
= arcsin  
 a 
a 2 + y2

⇒∫

 i.y 
= −i.arcsin  
 a 
a 2 + y2
dy

3

x
a

x
a

2

( i2 = −1)

x = i.y ⇒ when x = 1, y = −i and when x = 2, y = −2i
2

⇒∫

1

2

dx
x2 + a2

dx

=∫

−

1

(

−1.x

)

2

=
+ a2

2

−2i

∫

−i

dy
a 2 + y2

−2i

i.y
⇒ ln x + x + a
= −i.arcsin
a −i
1
i. ( −i )
1
⇔ ln 1 + 1 + a 2 = −i.arcsin
= −i.arcsin , arcsin z = −i ln  z ± z 2 + 1 


a
a


2

2

eiw − e −iw
w = arcsin z ⇒ z = sin w =
2
⇒ e2i.w − 2z.eiw − 1 = 0 ⇒ eiw = z ± z 2 + 1
⇒ i.w = ln  z ± z 2 + 1  ⇒ w = arcsin z = −i ln  z ± z 2 + 1 








x π
1
= ln tg  .arctan +  + C
a 4
2
x2 + a2
dx
a.dt
x
I=∫
Put x = a.tan t ⇒ dx =
, t = arctan
a
cos 2 t

x2 + a2
a.dt
a.dt
dt
⇒I=∫
=∫
=∫
1
cos 2 t a 2 .tan 2 t + a 2
a.cos 2 t tan 2 t + 1
cos 2 t
cos 2 t

*I=∫

dx



dt

t

π 

∫ cos t = ln  tan  2 + 4   ⇒ I = ∫



⇒∫

x π
1
= ln tg  .arctan + 
a 4
2
x2 + a2
dx

i.a 
1
= ln x + x 2 + a 2 = i.ln tg  .arcsin 
x 
2
x2 + a2
dx

= −i arcsin

i.x
x π
1
= ln tg  .arctan + 
a
a 4
2

4

2

so why

2

dx

∫

2

x +a

1

=∫

2

2

dx

(

x2 −

1

−1.a

)

dx

=∫

2

−

1

(

−1.x

)

2

but
+ a2

2

2

i.a  


1
2
2 
ln x + x + a  ≠ i.ln tg  .arcsin  

x  1

1 
2
2

2

i.x 
x π

1
≠  −i arcsin  ≠  ln tg  .arctg +  
a 1 
a 4  1

2

dx

2a / I = ∫
a/ I=∫

a2 − x2

a.cos t.dt

⇒I=∫
*

x
a

dat x = a.sin t ⇒ dx = a.cos t.dt, t = arcsin
=∫

a.cos t.dt

= ∫ dt = t = arcsin

x
a

x
a

a 2 − a 2 .sin 2 t
a cos 2 t
dx
dx
x
dx
=∫
Dat
= u ⇒ du =

⇒ dx = a.du
2
2
2
a
a
a −x
x
a 1−  
a

∫

dx

⇒∫

2b /
b/

a2 − x2
dx

= arcsin

a2 − x2

∫

∫

=

dx

1
a.du
x
= arcsin
∫
a 1 − u2
a
= −i.ln i.y + a 2 − y 2 + C

a2 − x2
dx
=∫
2
2
a −x

dx

( i.x )

2

+a

Dat

2

( i.x ) 2 + a 2

= t − i.x

2

⇒ − x 2 + a 2 = ( t − i.x ) = t 2 − 2t.ix − x 2
⇒x=

=

2

2

t −a
=
2t.i

(

)

t2 − a2 i
2t.i 2

−i4t 2 + 2t 2 .i − 2i.a 2

4t

2

dt =

=

(

)

− t2 − a2 i
2t

(

−i 2t 2 + 2a 2
4t 2

, dx =

(

) = −i ( t 2 + a 2 ) dt
2t 2

5

)

(

'
'
−i t 2 − a 2 .2t + ( 2t ) i t 2 − a 2

( 2t ) 2

) dt


(

−i t 2 + a 2
dx

⇒∫

(

2t
t − i.x

=∫

a2 − x2

2

)

−i. t 2 + a 2 dt

=∫



(

)

=∫

) dt

=∫

(

(



(

dx

*∫

)

−i. t 2 + a 2 dt

(

)

=∫

a 2 − y2

dy

⇒∫

∫

i.dy

∫

a 2 − y2
dx

a2 − x2

( )
t. ( t 2 + a 2 )

−i. t 2 + a 2 dt

= ln x + a 2 + x 2 + C put x = i.y, we have :

x2 + a2

dx = i.dy,

)

− t2 − a2 i 

2t t − i.


2t




2


i2 t 2 − a 2 
t2 − a2 
2


2t t +
2t t −





2t
2t








−i.dt
=∫
= −i.ln t + C = −i.ln i.x + a 2 − x 2 + C
t
2

)

−i. t 2 + a 2 dt

= ln i.y + a 2 − y 2 + C

= −i.ln i.y + a 2 − y 2 + C

= arcsin

x
i.y
= arcsin
= −i.ln i.y + a 2 − y2
a
a
2

i.y
i.y
 i.y 
arcsin z = −i ln  z ± z 2 + 1  , ⇒ arcsin
= −i.ln
+   +1


a
a


 a 
i.y + a 2 − y 2


= −i.ln
= −i.  ln i.y + a 2 − y 2 − ln a  = −i.ln i.y + a 2 − y 2 + C
a


2c / ∫

a
1
= i.ln tg  .arcsin  + C
x
2
a2 − x2
dx

6


dx

2c / ∫
=∫

a2 − x2
i.dx

i2

dx

=∫

( x2 − a2 )

=∫

( −1) ( x 2 − a 2 )

( x2 − a2 )
( i2 = −1)

−i.dx

=∫

dx

( x2 − a2 )

i

'

−a. ( sin t ) dt −a.cos t.dt
a
a
dat x =
⇒ dx =
=
t = arcsin
sin t
x
sin 2 t
sin 2 t
⇒I=∫

− ( −i ) a.cos t.dt
sin 2 t

1

.

a2

− a2

2

=∫

i.a.cos t.dt
sin 2 t

= i.ln tg

.
a

sin t
1
i.cos t.dt sin t
dt
=∫
.
= i.∫

sin t
sin 2 t cos t
1 − sin 2 t
sin 2 t

t
a
1
= i.ln tg  .arcsin  + C
2
x
2
2

Vay tai sao :

∫

1

dx
2

a −x

=∫

2

dx

( i.x )

2

+a

2

dx

=∫

( −1) ( x

2

−a

2

2

x
a


1
2
2

 arcsin a  ≠ i.ln tg  2 .arcsin x   ≠ ln i.x + a − x

1 

 1 

x2 − a2
−a.cos t.dt
sin 2 t

'

−a. ( sin t ) dt −a.cos t.dt
a
a
dat x =
⇒ dx =
=
t = arcsin
sin t
x
sin 2 t
sin 2 t
.

1
a2
2

sin t

=∫

2



1

a
1
= − ln tg  .arcsin 
x
2
x2 − a2

dx

⇒I=∫

)

nhung

dx

3a / I = ∫
a/∫

2

−a

2

=∫

−a.cos t.dt
sin 2 t

1

.
a

1 − sin 2 t
sin 2 t

− cos t.dt sin t
dt
t
a
1
.
= −∫
= − ln tg = − ln tg  .arcsin 
sin t
2
x
2
sin 2 t cos t

7


3b /
3b /

dx

∫

2

2

x −a
dx

∫

= ln x + x 2 + ( i.a )
dx

=∫

x2 − a2

2

x2 +

(

−1.a

2

)

+C

=∫

2

dx
x 2 + ( i.a )

Dat x + ( i.a ) = t − x ⇒ x − a = ( t − x )
2

(
dx =

)

2

'

'

(

2

t 2 + a 2 .2t − ( 2t ) t 2 + a 2

( 2t ) 2

2

t2 + a2
= t − 2t.x + x ⇒ x =
,
2t

2

2

2

) dt = 4t 2 − 2t 2 − 2a 2 dt = t 2 − a 2 dt
4t 2

2t 2

t2 − a2

dt
dt
2
2t 2
=∫
= ∫ = ln t + C = ln x + x 2 + ( i.a ) + C
2
2
t
t +a
x2 + a2
t−
2t
dx

⇒∫

3c / ∫
3c / ∫
=∫

2

dx
x2 − a 2
dx
x2 − a2
i.dx
2

i . a −x

2

= −i.arcsin

Vay tai sao :

∫

1

dx

=∫
=∫

2

x
a
=∫

dx

−1 a 2 − x 2
i. a 2 − x 2
−i.dx
x
= −i.arcsin

a
a2 − x2

dx
x2 − a2
2

2

=∫

1

2

dx
x2 +

(

−1.a

)

2

2

=∫

1

dx
−1 a 2 − x 2
2

a
x
1
− ln tg  .arcsin  ≠ ln x + x 2 − a 2 ≠ −i.arcsin
x1
a1
2
1
4/ I = ∫

dx
−x 2 − a 2

8


dx

I=∫

−x 2 − a 2

=∫

dx

−i.dx

=∫

i x2 + a2

x2 + a2

= −i.ln x + x 2 + a 2

i.a 
i.x
x π
1
1
= ln tg  .arcsin  = − arcsin
= −i.ln tg  .arctg + 
x 
a
a 4
2
2
i.a 
1
= ln x + x 2 + a 2 = i.ln tg  .arcsin 
x 
2
x2 + a2

dx

∫

= −i arcsin

i.x
x π
1
= ln tg  .arctg + 
a
a 4
2
2

∫

Vay tai sao :

−x 2 − a 2

1
2

2

dx

2

−idx

=∫

x2 + a2

1

dx

=∫

− x 2 + ( i.a )

1


=∫
nhung  −i.ln x + x 2 + a 2
2
2

1 ( i.x ) − a
dx

2
2

2


i.a  

1
 ≠ ln tg  2 .arcsin x  

 1
1 
2

2

i.x 
x π

1
≠  − arcsin  ≠  −i.ln tg  .arctg +  
a 1 
a 4  1

2

1
x
= .arctan + C
a
x2 + a 2 a
dx
x
* I=∫
Dat x = a.tgt ⇒ dx = a tg 2 t + 1 dt, t = arctan

a
x2 + a2
5/ I=∫

dx

(

dx

Vay :

∫ x2 + a2

*I= ∫

dx

*I= ∫

x2 + a2
dx
2

x +a
1

2

=

=∫

(

)

a tan 2 t + 1 dt

(

i
x + i.a
ln
2a x − i.a
dx

=∫

=

)

a 2 tan 2 t + 1

x2 −

(

−1.a

)

2

=

=∫

)

1
t 1
x
∫ dt = a = a .arctan a + C
a

dx
x − ( i.a )
2

2

=∫

dx
( x − i.a ) ( x + i.a )

A
B

+
⇒ A ( x + i.a ) + B ( x − i.a ) = 1
( x − i.a ) ( x + i.a )

( x − i.a ) ( x + i.a )
⇔ x ( A + B ) + i.a ( A − B ) = 1

9

( i2 = −1)


A + B = 0 ⇔ A = −B

⇔
1
i
−i
i
i.a ( A − B ) = 1 ⇒ A =
=
=
, B=

2i.a 2i 2 .a 2.a
2.a

d ( x − i.a )  i ( ln x + i.a − ln x − i.a )
i  d ( x + i.a )
i

x + i.a
⇒ I2 =  ∫
−∫
= ln
=
2a  x + i.a
x − i.a 
2a
2a x − i.a
2

2

2

2

x
 i
x + i.a 
1
so why : I = ∫
=∫
but  .arctg  ≠  ln
a 1  2a x − i.a 1
a

x 2 + a 2 1 x 2 − ( i.a ) 2
1
6/ I=∫

dx

dx

dx

x2 − a2
dx
dx
* I=∫
=∫
( x − a) ( x + a)
x2 − a2
1
A
B
=
+
⇒ A ( x + a ) + B( x − a ) = 1
( x − a) ( x + a) ( x − a) ( x + a)
A + B = 0 ⇔ A = −B

⇔ x ( A + B) + a ( A − B) = 1 ⇔ 
1
1
a ( A − B ) = 1 ⇒ A = 2a , B = − 2a

d( x + a)
1  d( x − a)

−∫
∫
2a  x − a
x+a

⇒I=

*I= ∫

dx
x2 − a2

=∫

dx
x2 +

(

−1.a

)

2

 ln x − a − ln x + a
1
x−a
= ln
=

2a
2a x + a

=∫

dx
x 2 + ( i.a )

2

(
)
i.a ( tg 2 t + 1) dt
dx
1
i.t
−i.t −i
x
Vay : ∫
=∫
= ∫ dt =
=
= .arctg + C
2
a
a
i.a
i 2 .a
x 2 + ( i.a )
( i.a ) 2 ( tg 2 t + 1) i.a

Dat x = i.a.tgt ⇒ dx = i.a tg 2 t + 1 dt, t = arctg

2

2

x
i.a

2

2

1
x −a 
x
 −i
Vay tai sao : ∫
=∫
nhung  ln
≠  .arctg 

2
2
2
2
i.a 1
 2a x + a 1  a
1x −a
1 x + ( i.a )

dx

dx

5/ Tích phân hàm lượng giác:
* I=∫

dx
a sin x + b cos x + c

10


Put t = tg
=

x
2dt
x
x
⇒ x = 2arctgt, dx =
, sin x = 2sin .cos
2
2
2
1 + t2

2sin ( x/2 )
2tg ( x/2 )
1

2t
/
=
=
cos ( x/2 ) cos2 ( x/2 ) tg 2 ( x/2 ) + 1 1 + t 2


x
1
cos x = 2 cos2 − 1 =  2 −

2
cos2 ( x/2 )


(

)

2 − tg2 ( x/2 ) + 1 1 − t 2

1
/
=
=
2
 cos2 ( x/2 )
tg ( x/2 ) + 1
1 + t2


2dt

* I=∫

dx
=
a sin x + b cos x + c ∫

1 + t2
b 1 − t2

) + c ( 1 + t2 )

(

a.t

+
1 + t2
1 + t2
1 + t2
2dt
2
dt
=∫
=
∫ 2 a.t c + b
t 2 ( c − b ) + a.t + b + c c − b t +
+
c−b c−b

2
dt
=
2
c−b∫

a
c+b
a2
−
t +
 +
2 ( c − b) 
c − b 4 ( c − b) 2

=

2
c−b∫

dt
2

(

2

)

4 c −b −a



a
t+
+


2
4 ( c − b)
 2 ( c − b) 

)

(

If 4 c2 − b2 − a 2 > 0 ⇒ put
t+

2

)

(

2

,

4 c 2 − b2 − a 2
4 ( c − b)

2

= q2 ,

a
x
a
= tan +
= y ⇒ dt = dy
2 ( c − b)
2 2 ( c − b)

⇒I=∫

(

2

dx
2
dy
2
y
=
=
arctan
a sin x + b cos x + c c − b ∫ y 2 + q 2 ( c − b ) .q
q

If 4 c − b
⇒I=∫

2

)

2

− a < 0 ⇒ put

(

)

4 c 2 − b2 − a 2
4 ( c − b)

2

= −q 2

dx
2
dy
2
dy
2
q+y
=

∫ y2 − q 2 = ( b − c ) ∫ q2 − y2 = ( b − c ) 2q ln q − y
a sin x + b cos x + c c − b

11


π

*

∫

−π

dx
=0
a sin x + b cos x + c

( With 4 ( c

2

)

− b2 − a 2 < 0

)

a
⇒ when y → +∞ ⇒ t → +∞, x = 2arctan t → π

2( c − b)

y=t+

when y → 0 ⇒ t →

−a
,
2 ( c − b)

 −a
x = 2 arctan t → 2 arctan 
 2 ( c − b)




a
= −2arctan 


 2( c − b)



( with 4 ( c

when y → −∞ ⇒ t → −∞, x = 2 arctan t → −π
π

2

)

− b2 − a 2 < 0

+∞

dx
dy
 2 
=
 ∫ 2
a sin x + b cos x + c  b − c  y − q 2
0


∫

⇒ J1 =






 a
−2arctan 

 2( c − b ) 

y=k

 q+y
2
=
. lim ln
=0
( b − c ) 2q k →+∞  q − y  y =0

π

+∞

dx
dy
 2 
J2 = ∫
=
 ∫ 2
a sin x + b cos x + c  b − c  y − q 2
−π
−∞
y =+∞

( With 4 ( c

 q+y 
2
=

ln
=0
( b − c ) 2q  q − y  y=−∞



(

2

)

− b2 − a 2 < 0

)

)

If 4 c2 − b 2 − a 2 > 0
⇒I=

π

∫

−π

=

dx

2
=
a sin x + b cos x + c c − b

+∞

2
( arctan ( +∞ ) − arctan ( −∞ ) )
( c − b ) .q

=

( c − b) .

(

2π

)

4 c2 − b2 − a 2
4( c − b)

2

=

(

+∞


2
y
∫ y2 + q 2 = ( c − b ) .q arctg q 

 −∞
−∞
dy

4π

)

4 c2 − b2 − a 2

12

)


π

sin n x.dx

∫

*

−π ( a sin x + b cos x + c )

( with 4 ( c
I( a) =

π

∫

−π

2

n +1

)

− b2 − a 2 < 0

)

=

π

cos n x.dx

∫

−π ( a sin x + b cos x + c )

dx
=0
a sin x + b cos x + c

( with 4 ( c

)

2

n +1

=0

− b2 − a 2 < 0

)

'

π

−1
'
⇒ I ( a ) =  ∫ ( a sin x + b cos x + c ) dx 
 −π


a
=

π

'

∫ − ( a sin x + b cos x + c ) a .( a sin x + b cos x + c )

−2

dx

−π

π

− sin x.dx
−1  '
−2 
'
=0
   u ( a )   = −u ( a ) .  u ( a )   = ∫

 

 
2


 −π ( a sin x + b cos x + c )
'

π
π

( −1) 2 2sin 2 x.dx
− sin x.dx
''
 = ∫
I ( a) =  ∫
=0
2
3
 −π ( a sin x + b cos x + c )
a sin x + b cos x + c )

 a −π (

( n) =
I ( a ) 


( n) =
I ( b ) 


π

*

∫

π

π
( −1) n n!.sin n x.dx
sin n x.dx
=0⇒ ∫
=0
∫
n +1
n +1
a sin x + b cos x + c )
a sin x + b cos x + c )
−π (
−π (
π

π
( −1) n n!.cosn x.dx
cos n x.dx
=0⇒ ∫
=0
∫
n +1
n +1
−π ( a sin x + b cos x + c )
−π ( a sin x + b cos x + c )

sin x.dx

−π ( a sin x + b cos x + c )

2

=

−4π.a
3
2
2
2 2
4 c −b −a

((

)

)

13

( with 4 ( c

2

)

− b2 − a 2 > 0

)

π

I( a) =

∫

−π

dx
=
a sin x + b cos x + c

(

4π
2

4 c −b

) −a

2

( with 4 ( c

2



π

( −1) sin x.dx
4π
'
I ( a )  = ∫
=


2
 4 c2 − b2 − a 2
−π ( a sin x + b cos x + c )


(

((

)

)

'
 1
4π.  −  . 4 c2 − b 2 − a 2
a
 2
=
=

( 4( c

⇒

π

−b

2

)

3
2 2
−a

)

sin x.dx

∫

2
−π ( a sin x + b cos x + c )

π

*

2

cos x.dx

∫

−π ( a sin x + b cos x + c )
π

2

=

=

)

− b2 − a 2 > 0

)

'





a

4π.a

( 4( c

2

−b

2

)

3
2 2
−a

)

−4π.a
3
2
2
2 2
4 c −b −a

((

)

)

3

4 c2 − b2 − a 2 2

((

)

)

(

( with 4 ( c

)

'

π

dx

I ( b)  =  ∫


a sin x + bcos x + c 
 −π

b
'




( −1) cos x.dx
4π
= ∫
=
2
 4 c2 − b2 − a 2
−π ( a sin x + bcos x + c )

π

(

)

14

( with 4 ( c
( with 4 ( c

−16π.b

dx
4π
=
∫ a sin x + bcos x + c
4 c 2 − b2 − a 2
−π

I ( b) =

)

2

'





b

2

2

2

)

− b2 − a 2 > 0

)

− b2 − a 2 > 0

)

− b2 − a 2 > 0

)

)

)


((

)

)

,
 1
4π.  −  4 c2 − b 2 − a 2
b
 2
=
=
3
4 c2 − b2 − a 2 2

((

⇒

π

cos x.dx

∫

−π ( a sin x + b cos x + c )

π

*

)

)

dx

∫

2

−π ( a sin x + b cos x + c )

I ( c) =

2

=

=

π

−2π. ( −8b )
3
4 c2 − b2 − a 2 2

((

)

)

−16π.b

( 4( c

2

−b

2

3
2 2
−a

)

)

( with 4 ( c

16π.c
3
4 c2 − b2 − a 2 2

((

)

)

dx
4π
=
a sin x + b cos x + c
4 c 2 − b2 − a 2
−π

∫

( with 4 ( c

2

)

− b2 − a 2 > 0

)

(

)
'

π

dx

I ( c )  =  ∫


a sin x + b cos x + c 
 −π

c
'



( −1) dx
4π
= ∫
=
2
 4 c2 − b2 − a 2
−π ( a sin x + b cos x + c )

π

)

(

((

)

)

,
 1
4π.  −  4 c2 − b2 − a 2
c
 2
=
=

( 4( c

⇒

π

∫

2

−b

2

)

3
2 2
−a

)

dx

−π ( a sin x + b cos x + c )

2

=

'





c

−2π.8c

( 4( c

2

−b

2

)

3
2 2
−a

)

16π.c

( 4( c

2

−b

2

)

3
2 2
−a

)

15

2

)

− b2 − a 2 > 0

)


2π n
* lim
.∑
n →+∞ n i =0

1
 −n.π + 2i.π 
 − n.π + 2i.π 
a sin 
 + b cos 
+c
n
n






(

)

0
with 4 c2 − b 2 − a 2 < 0


4π
=
=
with 4 c2 − b2 − a 2 > 0

4 c2 − b2 − a 2



(

)

(

π

)

+∞

dx
dy
 2 
J2 = ∫
=
 ∫ 2
a sin x + b cos x + c  b − c  y − q 2
−π
−∞
y =+∞

 q+y 
2
=
ln
=0
( b − c ) 2q  q − y  y =−∞


1
2π
, x i ∈ [ −π, π] , ∆x =
,
a sin x + b cos x + c
n
2i.π − n.π + 2i.π
x i = −π +
=
n

n
Dat f ( x i ) =

2π n
.∑
n →+∞ n i = 0

⇒ lim I = lim
n →+∞

n

π

i =0

−π

∑ f ( xi ) .∆x = ∫
n →+∞

= lim

(

1
 − n.π + 2i.π 
 − n.π + 2i.π 
a sin 
 + b cos 

+c
n
n




dx
a sin x + b cos x + c

)

0
with 4 c2 − b2 − a 2 < 0


=  2π
with 4 c2 − b2 − a 2 > 0

 ( c − b ) .q


(

)

2dt
dx
dt
1 + t2

=∫
= 2∫
∫ 3cos x + 4sin x + 5
2t 2 + 8t + 8
3 1 − t2
8t
+
+5
1 + t2
1 + t2
dt
1
1
=∫
=−
=−
+C
2
x
t+2
t + 2)
(
tg + 2
2

(

)

16

* I=∫

dx
=
a cos x + b

(

 x −a + b 
.arctg  tg .
+C
2 a+b 
2
2

b −a
2

)

dx
x
1 − t2
2dt
* I=∫
dat t = tg ⇒ cos x =
, x = 2arctan t ⇒ dx =
a cos x + b

2
1 + t2
1 + t2
2dt
2dt
1 + t2
⇒I=∫
=∫
t 2 ( −a + b ) + ( a + b )
a 1 − t2 + b 1 + t2

(

) (

)

1 + t2

If ( a + b ) ( −a + b ) = b 2 − a 2 > 0 ⇔ b > a
⇒I=

2
( −a + b )

dt

∫

2

=

2
( −a + b )

 a+b 
t2 + 

−a + b 

dx
2
⇒I=∫
=
.arctan
a cos x + b
( −a + b ) ( a + b )
=

(

−a + b
.arctan
a+b
t
a+b
−a + b

t

a+b
−a + b

+C

+C

 x −a + b 
.arctan  tg .
+C
2 a+b 
2
2

b −a
2

)


a+b
If ( a + b ) ( −a + b ) = b2 − a 2 < 0 ⇔ b < a ⇒
= −

−a + b



b − a = −  b2 − a 2 



2

⇒I=

2

2
( −a + b )

⇒I=∫

∫


t −








2

2

dt
2

a+b
−a + b

a+b
−a + b

dx
1
=
a cos x + b ( a − b )






2

=

2
.
( a − b) 
2



−a + b
.ln

a+b

1
a+b
−a + b

tg

x
+
2

a+b
−a + b

tg

x
−
2

a+b
−a + b

17

t+






a+b
−a + b

t−

a+b
−a + b

.ln


π

π
if b > a
 2
π
dx
2
* lim ∑
=
=  b −a
 ∫ a cos x + b 
n →+∞ i =1 
 i.π 
n  a cos   + b  0
if b < a
0

 n 


n

π

* I=∫

0

dx
if b > a
a cos x + b

π

dx
⇒I=∫
=
a cos x + b

(

0

2

=

( b2 − a 2 )

if b > a ⇒ I = ∫

0

1
( a − b)

)

. ( arctan ( +∞ ) − arctan 0 ) =

π

=

π


 x −a + b  
 arctg  tg .

2 a + b 
2
2 

0
b −a 
2

)

dx
a cos x + b


x
 tan +
2
−a + b 
. ln
a+b 
x
 tan −
2



Dat f ( x i ) =

(

π
π
. =
2
b2 − a 2
b2 − a 2
2

a+b
−a + b
a+b
−a + b

π



 = 0  tan π = +∞ 



2



0


1
π
i.π
, x i ∈ [ 0, π] , ∆x = , x i =
a cos x i + b
n
n
π

n

n
π
dx
⇒ lim In = lim ∑
= lim ∑ f ( x i ) .∆x = ∫
a cos x + b
 n →+∞ i =0
n →+∞
n →+∞ i =1 
 i.π 
0
n  a cos   + b 
 n 


π

if b > a
 2
2
= b −a
0
if b < a

π

*∫

cos x.dx

0 ( a cos x + b )

2

=

−a.π

(

3
2
2 2
b −a

)

with sign ( a + b ) = sign ( −a + b )

18


π

* I( a) = ∫

0

dx
if b > a
a cos x + b

π

dx
⇒I=∫
=
a cos x + b

(

0

2

=

( b2 − a 2 )

π


 x −a + b  
arctg  tg .


2
2 

 2 a + b 0


b −a
2

)

. ( arctan ( +∞ ) − arctan 0 ) =

(

π
π
. =
2
b2 − a 2
b2 − a 2
2

)

'

π ( −1) ( a cos x + b ) '
π

−1
'
a .dx

⇒ I ( a ) =  ∫ ( a cos x + b ) .dx  = ∫
2
0


a 0 ( a cos x + b )
π

( −1) cos x.dx
2
0 ( a cos x + b )

=∫

π

⇒∫

cos x.dx

0 ( a cos x + b )

2


=


=

π

if b < a ⇒ I ( a ) = ∫

0

)

(

−a.π

(

3
b2 − a 2 2

( cos x ) n .dx
*∫
=0
n +1
0 ( a cos x + b )
π

)

(

'
 1

π.  −  b2 − a 2

π
a.π
2
a
=
 = 
3
3
b2 − a 2 a

b2 − a 2 2
b2 − a 2 2
'

)

(

with b > a

with b < a

dx
1
=
a cos x + b ( a − b )



x
 tan +
2
−a + b 
. ln
a+b 
x
 tan −
2



'

π
π

( −1) cos x.dx
π
dx


'
 =∫
tan = +∞  ⇒ I ( a ) =  ∫
= 0,

2
2



 0 a cos x + b 

 a 0 ( a cos x + b )
n
n
n
π
π
( n ) = ( −1) ( cos x ) .dx = 0 ⇒ ( cos x ) .dx = 0
⇒ I ( a ) 
∫
∫


n +1
n +1
0

)

( a cos x + b )

0 ( a cos x + b )

19

π

a+b 


−a + b 
=0

a+b

−a + b 
0




2
b2
 x a

.arctg  tg + 
 b2 − a 2
 2 b  b 2 − a 2



dx

x a
a 2 − b2
* I=∫
=
tan + +
a sin x + b 

2 b
1
b2
.ln

 a 2 − b2
x a
a 2 − b2
tan + −

2 b
b2








with b2 − a 2 > 0

with b2 − a 2 < 0

dx
x
2t
2dt
Put t = tg ⇒ sin x =
, x = 2arctgt ⇒ dx =

a sin x + b
2
1 + t2
1 + t2
2dt
2dt
2
dt
1 + t2
⇒I=∫
=∫
= ∫
bt 2 + 2at + b b t 2 + 2at + 1
2at + b 1 + t 2
b
2
1+ t
2
dt
2
dt
= ∫
= ∫
where b2 − a 2 > 0
2
2
2
2
2
b  a

b  a
b −a
a
t +  +1−  

t +  +
 b
b
 b
b2

* I=∫

(

)


 t+a

2
1
b
⇒I= .
.arctg 
b b2 − a 2
 b2 − a 2

b2
b2






 2
b2
b2
 t + a 
.arctg 

 = b. 2
b  b2 − a 2

b − a2





dx
2
b2 
 x a

⇒I=∫
=
.arctg  tg + 
2
2

a sin x + b
 2 b  b 2 − a 2 
b −a


2
If b2 − a 2 < 0 ⇒ a 2 − b 2 > 0 ⇒ I = ∫
b

dt
a   a 2 − b2

t +  −
b 

b2

2






2

a
a 2 − b2
x a
a 2 − b2

t+ +
tan + +
b
2 b
2
1
1
b2
b2
= .
.ln
=
.ln
b
a 2 − b2
a
a 2 − b2
a 2 − b2
x a
a 2 − b2
2
t+ −
tan + −
b
2 b
b2
b2
b2

20







n

2π
=0


 −n.π + 2i.π 
i =1 n. a sin

 + b

n





∑
n →+∞

* lim

* I=

+π

∫

−π

dx
= 0 ( sin x là hàm le ) ,
a sin x + b
1
2π
, x i ∈ [ −π, π] , ∆x =
,
a sin x i + b
n

dat f ( x i ) =

n
2i.π − n.π + 2i.π
2π
x i = −π +
=
, ⇒ lim I = lim ∑
n
n

n →+∞
n →+∞ i =1 

 − n.π + 2i.π 
n.  a sin 
 + b
n




n

+π

i =1

−π

∑ f ( x i ) .∆x = ∫
n →+∞

= lim

* I=∫

dx
sin x

Put t = tan

dx
=0

a sin x + b

x
2dt
⇒ x = 2 arctan t, dx =
,
2
1 + t2

sin ( x/2 ) 
x
x
1
sin x = 2sin .cos = 2
.
2
2
cos ( x/2 )  cos2 ( x/2 )


x
1
cos x = 2 cos2 − 1 =  2 −

2
cos2 ( x/2 )

=

(

) = 1 − t2






−1

2 tan ( x/2 )

=


1
.
  cos2 ( x/2 )


tan 2 ( x/2 ) + 1






=

2t

1 + t2

−1

2 − tan 2 ( x/2 ) + 1
tan 2 ( x/2 ) + 1

⇒I = ∫

1 + t2

dx
2dt
2t
dt
x
=∫
÷
= ∫ = ln t = ln tan + C
sin x
t
2
1 + t2 1 + t2

 at 2 + 2t + a 
sin x.dx
2t
2dt  2t
4t.dt


* I=∫
=
.
÷
+a =
÷

2 
2

sin x + a ∫ t 2 + 1 t 2 + 1  t 2 + 1  ∫ 2


 t +1 
t +1

)

(

=∫

4t.dt

, Put

4t

=

A

+

B

( t2 + 1) ( at2 + 2t + a ) ( t2 + 1) ( at2 + 2t + a ) t2 + 1 at2 + 2t + a
⇒ A ( at 2 + 2t + a ) + B ( 1 + t 2 ) = 4t ⇒ ( a.A + B ) .t 2 = 0, 2A.t = 4t, a.A + B = 0
⇒ A = 2, a.A + B = 2a + B = 0 ⇒ B = −2a

21


⇒I=∫

4t.dt

( t2 + 1) ( at2 + 2t + a )

= 2.arctan t − 2 ∫

* I=∫

=∫

2dt
t2 + 1

− 2a ∫

dt
at 2 + 2t + a

dt
2t
t2 + + 1
a

 tan ( x/2 ) .a + 1 
sin x.dx
a
=x−
.arctan 



2
2
sin x + a
a −1
a −1 


I1 = −2 ∫

( where

a > 1)

dt

dt
dt
= −2 ∫
= −2 ∫
2
2
2t
1
a2 − 1
 1
 t.a + 1 
t2 + + 1
 t +  +1− 2

 + 2
a
 a
 a 
a
a
a2 − 1

1
a
=
M
a2
a2 − 1
dt
1

 t.a + 1 
⇒ I1 = −2 ∫
= − .arctan 

2
M
 a.M 
 t.a + 1 
2

 +M
 a 
2

If a > 1 or a < −1 ⇒ a − 1 > 0 ⇒

= M2 ⇒

 t.a + 1
 t.a + 1 
a 
a
=−
.arctan 
.
.arctan 
=−

 a


 2

2
2
2
a −1
a −1 
a −1

 a −1 
a

 t.a + 1 
1
dt
a
= 2.arctan t −
.arctan 

 2

2
2 ∫ t 2 + 2t + 1
a −1
 a −1 
a
 tan ( x/2 ) .a + 1 
x
a


= 2.arctan  tan  −
.arctan 



2
2
2

a −1
a −1 


⇒ I = 2.arctan t −

⇒I=∫

 tan ( x/2 ) .a + 1 
sin x.dx
a
=x−
.arctan 



2
2
sin x + a
a −1
a −1 



π

( where

a > 1)


 tan ( π /2 ) .a + 1 
 tan 0.a + 1  
sin x.dx
a
= π−
.  arctan 
 − arctan 





2
2
2
sin x + a
a −1 
a −1 

 a − 1 
0




⇒J=∫

π
 1 
= π−
.  − arctan 

 2

2
2
a −1 
 a − 1 

a

22


sin x.dx
a
* I=∫
=x−
sin x + a
1 − a2
I1 = −2 ∫






( where

− 1 < a < 1)

dt
dt
dt
= −2 ∫
= −2 ∫
2
2
2t
1
a2 − 1
 1
 t.a + 1 
t2 + + 1
t +  +1−
+



a
 a
 a 

a2
a2
2

If a < 1 ⇒ a − 1 < 0 ⇒
⇒ I1 = −2 ∫

=


2
 ln tan ( x/2 ) .a + 1 + 1 − a

2
 tan ( x/2 ) .a + 1 − 1 − a

a2 − 1
a2

dt
2

 t.a + 1 
2

 −M
 a 

1 − a2
= −M ⇒ M =

a
2

=−


1   t.a + 1 
 t.a + 1 
.  ln 
 − M − ln 
+M 
M   a 
 a 




.  ln  t.a + 1 + 1 − a 2  .a −1 − ln  t.a + 1 − 1 − a 2  .a −1 








1 − a2  
a



2
 ln t.a + 1 + 1 − a
=
1 − a 2  t.a + 1 − 1 − a 2

a


,



1 

dx

dx 

dx

∫ x 2 − a 2 = 2a  ∫ x − a − ∫ x + a 




dt
a  t.a + 1 + 1 − a 2
⇒ I = 2.arctan t − 2 ∫
= 2.arctan t −
ln

2 
2 2t
1 − a  t.a + 1 − 1 − a 2
t + +1
a

2 
x
a  tan ( x/2 ) .a + 1 + 1 − a 

= 2.arctan  tan  −
ln

2
2

1 − a  tan ( x/2 ) .a + 1 − 1 − a 2 
sin x.dx
a
⇒I=∫
=x−
sin x + a
1 − a2

π

sin x.dx
a
⇒J=∫
= π−

sin x + a
1 − a2
0
= π+

a
1− a

2

.ln

1 + 1 − a2
1− 1− a

2


2
 ln tan ( x/2 ) .a + 1 + 1 − a

2
 tan ( x/2 ) .a + 1 − 1 − a

2
 ln1 − ln 1 + 1 − a

1 − 1 − a2


( where a < 1)

Check the result

23











( where






− 1 < a < 1)


'


 tan ( x/2 ) .a + 1  

x
a

 2.arctan  tan  −
.arctan 



2
2
2


a −1
a − 1  x



2


x
= 2.  tan  + 1
2





−1

x

.  tan 
2


'

2


tan ( x/2 ) .a + 1 
a

−
.
 + 1



a 2 − 1 
a2 − 1 




= 2. ( cos ( x/2 ) ) . ( cos ( x/2 ) )
2

−2

−1

 tan ( x/2 ) .a + 1 
.



2
a −1 


. ( ( x/2 ) )

'

  tan ( x/2 ) .a  2 + 2 tan ( x/2 ) .a + 1 + a 2 − 1 
a


−
. 
2

a −1
a2 − 1 


2

 

x
 x 
2  2 
= 1 − a . a  tan  + 1 + 2a. tan   
2
 2 
 


 


−1

x

.  cos 
2


 2
2
x 
2  a + 2a.sin ( x/2 ) .cos ( x/2 ) 
= 1− a .
.  cos 
2


2 

 cos ( x/2 ) 




= 1 − a 2 .  a 2 + a.sin x 


22 / I = ∫
=∫

=∫

= 1−

dx
dx
=∫
cos x
sinπ/2 −
( x

d ( xπ/2
+
sin ( xπ/2
+

* I=∫

−1

)

)

'

−1

−2


 x 
.
 tan  2  
 
a2 − 1 
a

x
. 
2

−1

a
a + sin x − a
sin x

=
=
( a + sin x ) ( a + sin x ) a + sin x

)

=∫

= ln tgπ/2 +
( π/4

)

dx
sinπ/2− (x −
(π

))

+
C

cos ( x/2 ) .d ( x/2 )
dx
dx
=∫
=∫
sin x
2sin ( x/2 ) .cos ( x/2 )
sin ( x/2 ) .cos2 ( x/2 )

d ( tan ( x/2 ) )
tan ( x/2 )

= ln tan

* I = ∫ tan x.dx = ∫

'

x
+C
2

−d ( cos x )
sin x.dx
=∫
= − ln ( cos x )
cos x
cos x

24

'


* I=∫

 tan ( x/2 ) a − 1 
cos x.dx

2a
=x−
.arctan 
 where a < −1 or a > 1


2
cos x + a
a +1
a −1



* I=∫

cos x.dx
cos x + a

Put t = tan

x
2dt
⇒ x = 2arctan t, dx =
,
2
2
1+ t


x

1
cos x = 2 cos2 − 1 =  2 −

2
cos2 ( x/2 )

=

(

) = 1 − t2 ⇒ I =

2 − tan 2 ( x/2 ) + 1
tan 2 ( x/2 ) + 1

1 + t2


1
.
  cos2 ( x/2 )


−1


cos x.dx
2dt  1 − t 2   1 − t 2
=∫
.

+a
. 2
∫ cos x + a t 2 + 1  t 2 + 1   t + 1 




)
)

(
(






(

)

−1

2 1 − t 2 dt  1 − t 2 + a.t 2 + a −1
2 1 − t 2 dt
=∫
.
 =∫
2 

2

2
t +1
t 2 + 1 t 2 ( a − 1) + a + 1


t +1
=∫

( t + 1) ( t
2

2dt
2

( a − 1) + a + 1)
2

Put

)(

(

−∫

=

( t + 1) ( t

2

A.t + B

+

2t 2 .dt
2

( a − 1) + a + 1)

)

,

C.t + D

( t2 + 1) ( t2 ( a − 1) + a − 1) t2 + 1 t2 ( a − 1) + a + 1
⇒ ( A.t + B) ( t 2 ( a − 1) + a + 1) + ( C.t + D ) ( t 2 + 1) = 2
⇒ t 3 ( a.A − A + C ) = 0, t ( a.A − A + C ) = 0, t 2 ( a.B − B + D ) = 0, a.B + B + D = 2
⇒ A = C = 0, 2B = 2 ⇒ B = 1, a + 1 + D = 2 ⇒ D = 1 − a
⇒ I1 = ∫
=∫

2.dt

( t2 + 1) ( t2 ( a − 1) + a + 1)

dt
t2 + 1

+∫

( 1 − a ) .dt = arctan t −
dt
∫ 2 a +1
t 2 ( a − 1) + a + 1
t +
a −1

25