Tich phan 210_ham_lg_co_ban_doi_bien_129

Bài 5. Các phép i bi n s cơ b n và nâng cao tích phân hàm lư ng giác

BÀI 5. CÁC PHÉP I BI N S CƠ B N VÀ NÂNG CAO
TÍCH PHÂN HÀM LƯ NG GIÁC

I. CÁC D NG TÍCH PHÂN VÀ PHÉP BI N I CƠ B N

• tv n :

∫
Xét tích phân d ng I = R ( sin x,cos x ) dx

1. i bi n s t ng quát:

x 2 dt 2t 1− t2
t t = tg ⇒ x = 2 arctg t ;dx = ; sin x = ; cos x =
2 1+ t2 1 + t2 1 + t2

 2
 2 dt
Khi ó: I = R ( sin x,cos x ) dx = R  2t 2 , 1 − t 2 
∫ ∫
 1 + t 1 + t  1 + t2
Ta xét 3 trư ng h p c bi t thư ng g p sau ây mà có th i bi n s b ng
cách khác hàm s dư i d u tích phân nh n ư c ơn gi n hơn.

2. N u R ( sinx, cosx ) là hàm l theo sin: R ( −sinx, cosx ) = −R ( sinx, cosx )

thì c n bi n i hàm s và vi phân th c hi n phép i bi n t = cosx.

3. N u R ( sinx, cosx ) là hàm l theo cosin: R ( sinx, − cosx ) = −R ( sinx, cosx )

thì c n bi n i hàm s và vi phân th c hi n phép i bi n t = sinx.

4. N u R ( sinx, cosx ) tho mãn i u ki n: R ( −sinx, − cosx ) = R ( sinx, cosx )

thì c n bi n i hàm s và vi phân th c hi n phép i bi n t = tgx.
II. CÁC BÀI T P M U MINH H A

1. D ng 1: i bi n s t ng quát

3sin2x − 2cos2x − 1
I= ∫ 3cos2x + 4sin2x + 5 dx
2
dt 2t 1− t
t t = tg x ⇒ x = arctg t ; dx = 2
; sin 2x = 2
; cos 2x = 2
1+ t 1+ t 1+ t

3.2t − 2 (1 − t ) − (1 + t ) dt 1 ( t + 6t − 3) dt
2 2 2 2
1 t + 6t − 3 dt
⇒ I= ∫ 3 (1 − t 2 ) + 4.2t + 5 (1+ t2 ) ⋅ 1+ t2 = ∫2
⋅
2 t + 4t + 4 1 + t 2
= ∫
2 ( t + 2)2 (1 + t 2 )

169

Chương II: Nguyên hàm và tích phân − Tr n Phương

t 2 + 6t − 3 A B Ct + D
Gi s = + + , ∀t
( t + 2 ) (1 + t 2 )
2
t + 2 (t + 2) 2
1 + t2

⇔ t 2 + 6t − 3 = A ( t + 2 ) (1 + t 2 ) + B (1 + t 2 ) + ( Ct + D ) ( t + 2 ) , ∀t (*)
2

⇔ t 2 + 6t − 3 = ( A + C) t 3 + ( 2A + B + 4C + D) t 2 + ( A + 4C + 4D) t + ( 2A + B + 4D)

Thay t = −2 vào (*) thì −11 = 5B ⇒ B = −11/5
A + C = 0 A + C = 0 A = −34 25
  
2A + B + 4C + D = 1 2A + 4C + D = 16 5 B = −11 5
(*) ⇔  ⇔ ⇔
A + 4C + 4D = 6 A + 4C + 4D = 6 C = 34 25
2A + B + 4D = −3 2A + 4D = −4 5 D = 12 25
  
2
1 t + 6t − 3 34 dt 11 dt 1 24t + 12
I= ∫
2 ( t + 2 ) (1 + t )
2 2
dt = − −
25 t + 2 5 ( t + 2 ) 2
+ ∫
25 1 + t 2
dt ∫ ∫
12 d ( t ) 12
2
34 dt 11 dt dt
=− ∫ −
25 t + 2 5 ( t + 2 ) 2
+ ∫
25 1 + t 2
+
25 1 + t 2 ∫ ∫
34 11 12 ( 12
ln 1 + t ) +
2
=− ln t + 2 + + arctg t + c
25 5 ( t + 2 ) 25 25
34 11 12 12
ln (1 + tg x ) +
2
=− ln tg x + 2 + + x+c
25 5 ( tg x + 2 ) 25 25

2. D ng 2: R ( −sinx, cosx ) = −R ( sinx, cosx )

sin2xdx 2 sin x cos xdx
• J1 = ∫ cos 3 2
x − sin x − 1
= ∫
cos 3 x + cos 2 x − 2
2 sin x cos x
R ( sin x, cos x ) = 3 2
⇒ R ( − sin x, cos x ) = −R ( sin x, cos x )
cos x + cos x − 2
−2t dt −2t dt  A Bt + C 
t t = cos x ⇒ J1 = ∫t 3 2
+t −2
= ∫ ( t − 1) ( t 2
+ 2t + 2)
∫
= −2  + 2  dt
 t − 1 t + 2t + 2 
t A Bt + C
⇔ t = A ( t + 2t + 2) + ( Bt + C) ( t − 1)
2
Ta có: = + 2
( t − 1) ( t + 2t + 2)
2 t − 1 t + 2t + 2

A + B = 0 A = 1 5
2  
⇔ t = ( A + B ) t + ( 2A − B + C ) t + ( 2A − C ) ⇔ 2A − B + C = 1 ⇔ B = −1 5
 
2A − C = 0 C = 2 5

170


2  1 t−2  2 dt 1 2t + 2 − 6
J1 = −
5  ∫
 t −1 − 2  dt = − 5 t − 1 + 5 2
t + 2t + 2  t + 2t + 2
dt ∫ ∫
2 dt 1 d ( t 2 + 2t + 2 ) 6 dt
=− ∫
+
5 t −1 5 2
t + 2t + 2
− ∫
5 ( t + 1) 2 + 1 ∫
2 1 2 6
= − ln t − 1 + ln t + 2t + 2 − arctg ( t + 1) + c
5 5 5
2 1 2 6
= − ln (1 − cos x ) + ln cos x + 2 cos x + 2 − arctg (1 + cos x ) + c
5 5 5
dx sin x dx −d ( cos x ) dt
• J2 = ∫ 6
sinxcos x
=
sin 2 x cos 6 x
= ∫ ∫ (1 − cos 2
x ) cos x6
= ∫t 6 ( t 2 − 1)
t − ( t − 1)  1 t + t +1
6 6 4 2
t −1 1 1 1
= ∫ t ( t − 1)
6 2
dt =  2
 t −1
−
t
6

∫
 dt = ln + + 3 + 5 +c
t + 1 t 3t 5t

1 − cos x 1 1 1
= ln + + + +c
1 + cos x cos x 3 cos3 x 5 cos5 x
sinx + sin3x 2 sin 2 x cos x 4 sin x cos 2 x
• J3 = ∫ cos2x
dx =
cos 2 x
dx = ∫
2 cos 2 x − 1
dx ∫
2 2
4 cos xd ( cos x ) 4t dt  2  dt
= ∫ 1 − 2 cos x
2
= ∫ 1 − 2t = ∫  1 − 2t
2 2
− 2  dt =

∫ 1 −t 2 ∫
− 2 dt
2

1 1 + 2t 1 1 + 2 cos x
= ln − 2t + c = ln − 2 cos x + c
2 1 − 2t 2 1 − 2 cos x
π 2 π 2 π 2
4sin 3 x 4 sin 2 x 4 (1 − cos 2 x )
• J4 = ∫
0
1 + cosx
dx = ∫0
1 + cos x
sin x dx = − ∫0
1 + cos x
d ( cos x )

4 (1 − t )
0 2 1
1

∫ dt = 4 (1 − t ) dt = ( 4t − 2t
∫ )0 =4−2=2
2
=−
1
1+ t 0

π 2 π 2 π 2 π 2
sin 2 x sin 2 x dx sin x dx sin x dx
• J5 = ∫
π6
sin3x
dx =
π 6
∫ = =
3 sin x − 4 sin x π 6 3 − 4 sin x π 6 4 cos 2 x − 1
3 2 ∫ ∫
π6 3 2 3 2 3 2
d ( cos x ) dt 1 d ( 2t )
1 2t − 1 1
= ∫ 2
= ∫ 2
= ∫ 2
= ln = ln ( 2 − 3 )
π2
4cos x − 1 0
4t − 1 2 0
( 2t ) − 1 4 2t + 1 0 4

171


3. D ng 3: R ( sinx, − cosx ) = −R ( sinx, cosx )

cos 9 x cos8 x (1 − sin 2 x )4 (1 − t 2 )4
• K1 = ∫ sin 20 x
dx =
sin 20 x ∫
cos x dx = ∫ sin 20 x
d ( sin x ) = ∫ t 20
dt

1 − 4t 2 + 6t 4 − 4t 6 + t 8 −1 4 6 4 1
= ∫ t 20
dt =
19t19
+
17t 17
− 15 + 13 − 11 + c
15t 13t 11t
−1 4 6 4 1
= 19
+ 17
− 15
+ 13
− 11
+c
19 ( sin x ) 17 ( sin x ) 15 ( sin x ) 13 ( sin x ) 11 ( sin x )

cos 3 x + cos5 x ( cos2 x + cos4 x ) ( cos2 x + cos4 x )
• K2 = ∫ sin2 x + sin4 x
dx = ∫ sin2 x + sin4 x
cos x dx = ∫ sin2 x + sin4 x
d ( sin x )

2
1 − t 2 + (1 − t 2 ) t 4 − 3t 2 + 2  2 6 
= ∫ 2
t +t 4
dt = ∫ t 2
(1 + t ) 2

∫
dt =  1 + 2 −
t 1 + t2
 dt

2 2
=t− − 6 arctg t + c = sin x − − 6 arctg ( sin x ) + c
t sin x

4. D ng 4: R ( −sinx, − cosx ) = R ( sinx, cosx )

π6 π6 π6
dx dx d ( tg x) 3− 3
∫ ∫ cos ∫
π6
• L1 = = = = ln tg x −1 0 = ln
0
cosx ( sinx − cosx ) 0
2
x ( tg x −1) 0
tg x −1 3

π 3 π 3 π 3 π 3
dx dx dx d ( tg x )
• L2 = ∫ 4 3 5
= ∫ 4 3 8
= ∫ 2
cos x . 4 tg x 3
= ∫ 3
π 4 sin xcos x π 4 tg x cos x π 4 π 4 ( tg x ) 4
π3 −3 1 π3
= 4 ( 3 ) − 1 = 4 ( 8 3 − 1)
14
=
π4
∫ ( tg x ) 4 d ( tg x ) = 4 ( tg x ) 4 π4  

π 4 π 4
sin 2 xdx cos 4 x sin 2 x dx
• L3 = ∫ cosx ( 2sin
0
3
x + 3cos 3 x )
= ∫0 cos x ( 2 sin3 x + 3 cos 3 x ) cos x
4

d ( 3 + 2 tg 3 x )
π4 π4 π4
tg 2 x tg 2 x 1
= ∫ ⋅ dx = ∫ d ( tg x ) = ∫
0
3 + 2 tg x cos 2 x
3
0
3
3 + 2 tg x 6 0
3 + 2 tg 3 x
π4
1 1 1 5
= ln ( 3 + 2 tg 3 x ) = ( ln 5 − ln 3) = ln
6 0 6 6 3

172


II. BI N I VÀ I BI N NÂNG CAO TÍCH PHÂN HÀM S LƯ NG GIÁC

dx
1. D NG 1: M U S LÀ BI U TH C THU N NH T C A SIN
∫ ( sinx ) n

2

dx dx dx 1 (
1 + tg 2 x
2 ) d (tg 2 )
x
• A1 = ∫
sin3 x
= ∫ 3
= 3 6 ∫ 3
= ∫
2 sin x cos x
2 2 ( 8 tg x cos x
2 2 ) ( )( ) 4
(tg 2 )
x

2 x 4 x
1 1 + 2 tg 2 + tg 2 1  −1 2

= ∫ 3
d tg x = 
2 2 ( ) x 1
+ 2 ln tg + tg x  + c
2 ( )
4
tg x
2 ( ) 4


2 tg x
2
2 2 

 ( )
dx sin x d x d ( cos x ) d ( cos x )
Cách 2: A1 = ∫ 3
=∫ 4
= −∫ 2
= −∫
sin x sin x (1 − cos 2 x ) [(1 + cos x ) (1 − cos x )] 2
2 2
−1  (1 + cos x ) + (1 − cos x )  1  1 1 
=
4  ∫
 (1 + cos x ) (1 − cos x )  d ( cos x ) = 4  1 − cos x + 1 + cos x  d ( cos x )
   ∫
−1  1 1 2  ( − cos x 1 1 + cos x
=  ∫ 2
+ 2
+ 2 
4  (1 − cos x ) (1 + cos x ) 1 − cos x 
d cos x ) = − ln
2sin 2 x 2 1 − cos x
+c

dx dx dx
• A2 = ∫ sin 5
= ∫ 5
= ∫ 5 10
x
( 2 sin x cos x
2 2 ) ( )(
32 tg x
2
cos x
2 )
4
1 + tg x ) d ( tg x )
1 (
2 2 x + 6 tg 4 x + 4 tg 6 x + tg8 x
1 1 + 4 tg
= ∫
2 2
= ∫ 5
2 2
5
2 2 d tg x
2 ( )
16
( tg x )
2
16
( ) tg x
2
1  −1  2 4
+ 6 ln tg + 2 ( tg x ) + ( tg x )  + c
2 x 1
=  4
− 2 2 2
16 


4 tg x
2( ) ( tg x )
2
2 4 


dx sin x dx d ( cos x ) d ( cos x )
Cách 2: A2 = ∫ sin 5
x
= ∫ 6
sin x
=− ∫
(1 − cos2 x )
3
=−
(1 + cos x ) (1 − cos x )
3∫
 
3 3
−1  (1 + cos x ) + (1 − cos x )  1  1 1 
=
8  ∫
 (1 + cos x ) (1 − cos x )  d ( cos x ) = 8  1 − cos x + 1 + cos x  d ( cos x )
   ∫
−1  1 1 3 d ( cos x )  − cos x 3
= 
8  2 (1 − cos x ) 2
−
2 (1 + cos x )
2
+
2 ∫ (1 − cos 2
x)
2 = − A
 4 sin 4 x 4 1
 

173


− cos x 3  − cos x 1 1 + cos x  − cos x 3cos x 3 1 + cos x
= 4
−  − ln = + + ln +c
4 sin x 4  2 sin x 2 1 − cos x  4 sin x 8sin 2 x 8 1 − cos x
2 4

dx dx
• A3 = ∫ ( sinx ) 2n+1
= ∫ 2 n +1

( 2 sin x cos x
2 2 )
2n

=∫
dx 1 (1 + tg x ) d ( tg x )
2 2
2

2 n +1
= ∫ 4n+2 2n 2 n +1
2 ( tg x ) ( cos x ) 2
2 n +1
2 2 ( tg x )
2
n 2n
C + C tg x + ... + C ( tg x ) + ... + C ( tg x )
0 1 2 n 2 2n 2

d ( tg x )
1 2n 2n 2n 2n
2 2 2
2 ∫
= 2n 2 n +1
2
( tg x )
2
1  −C 
0 n −1 n +1 2 2n 2n
= 2n
2 

2n ( tg x )
2n
− ... −
C
2n
2 ( tg x )
+ C ln tg +
x C
2 2 ( tg x ) + ... + C2n ( tg x )  + c
2n
2 2
n
2n
2
2n 2n


 2 2 

dx n
• A10 = ∫ sin 2n+ 2
x ∫
= − (1 + cotg 2 x ) d ( cotg x ) =

= − C0 + C1 cotg 2 x + ... + Cn ( cotg2 x ) + ... + Cn ( cotg 2 x )  d ( cotg x )
 k n
∫
k
n n n 
 0 Cn
1
Cn
k
2k +1 Cn
n

= − Cn ( cotg x ) +
3
cotg x + ... + ( cotg x ) + ... + ( cotg x )2n +1  + c
 3 2k + 1 2n + 1 
dx
2. D NG 2: M U S LÀ BI U TH C THU N NH T C A COSIN
∫ ( cos x ) n

B1 = ∫ dx =
d x+ π
2 = du = ( du ) du
cos 3 x ∫ sin 3 x + π ∫ sin 3 u ∫ =∫
2 (
2 sin u cos u
2 2
) ( )
3

( )(
8 tg u
2
3
cos u
2 )
6

2

(1 + tg 2 u ) d ( tg u ) = 1  −1
( )  + c ; (u = x + π )
2
=1∫ 2 2 + 2 ln tg u + 1 tg u
4 3 4 2 2 2 2 2
tg u )
( 2  2 ( tg u )
 2 
dx cos x d x d ( sin x ) d ( sin x )
Cách 2: B 1 = ∫ =∫ =∫ =∫
cos 3 x cos 4 x (1 − sin 2 x ) 2 [(1 + sin x ) (1 − sin x )] 2
2 2
1  (1 + sin x ) + (1 − sin x )  1  1 1 
= ∫  (1 + sin x ) (1 − sin x )  d ( sin x ) = 4 ∫  1 − sin x + 1 + sin x  d ( sin x )
4    

174


1  1 1 2  ( sin x 1 1 + sin x
=
4 ∫  (1 − sin x ) 2 + (1 + sin x ) 2 + 1 − sin 2

)
 d sin x = 2 cos 2 x + 2 ln 1 − sin x + c
x

dx
i B 2 = ∫ 2n+1 = ∫
d x+π
2 =∫
du ( =∫
) du
cos x sin 2 n +1 x + π
2
( sin u ) 2 n +1
( 2 sin u cos u
2 2
) ( )
2 n +1

2n

=∫
du 1 (1 + tg u ) d ( tg u )
2
2
2
2 n +1 4n + 2
=
2 2n ∫ 2 n +1

( ) (cos u )
2 2 n +1 tg u
2 2 ( tg u )
2

1  −C2n 2n 
0 n− n+1
C 2n 1 2 2n
= 
2 2n  2n
− ... − 2
n
+ C2n ln tg
u C 2n
+ tg u
2 ( ) + ... +
C 2n
( )
tg u
2
+c

 ( )
2n tg u
2 ( )
2 tg u
2
2 2 2n 


dx n
iB3 = ∫ 2n+ 2
= ∫ (1 + tg 2 x ) d ( tg x ) =
cos x

= ∫ C n + C n tg 2 x + ... + C n ( tg 2 x ) + ... + C n ( tg 2 x )  d ( tg x )
0 1 k k n n
 

 0 C1 Cnk
2 k +1 Cnn

= C n ( tg x ) + n tg 3 x + ... + ( tg x ) + ... + ( tg x ) 2 n+1  + c
 3 2k + 1 2n + 1 
dx
3. D NG 3: C= ∫ a ( sinx ) 2
+ bsinxcosx + c ( cosx )
2

dx dx
•C =∫ =∫
cos 2 3x ( 5 tg 3x + 2 ) − 21(1 + tg 2 3x ) 
2 2
( 5sin3x + 2cos3x ) - 21  

1 d ( tg 3x ) 1 d ( tg 3x ) 1 2 tg 3x + 5
= ∫ 4 tg 2 3x + 20 tg 3x − 17 = 12 ∫ 2
= arc tg +c
3
( tg 3x + 5
2 ) +
42
4
6 42 42

dx
4. D NG 4: D= ∫ a sin x + b cos x + c
dx dx
• D1 = ∫ 2sinx + 5cosx + 3 = ∫ 4 sin x cos x + 5
2 ( cos x − sin x ) + 3 ( cos x + sin x )
2 2 2
2
2 2
2 2 2

d ( tg x − 1) tg x −1 − 5
dx 2 −1 2
= ∫ cos = −∫ = ln +c
( 2)
2
2 5 tg x − 1 + 5
2 x 4 tg x + 8 − 2 tg 2
2 2
x
( tg 2 )
x −1 − ( 5 )
2
2

175


5. D NG 5: TÍCH PHÂN LIÊN K T

cosxdx sin x dx
∫ sinx + cosx . Xét tích phân liên k ∫ sin x + cos x
*
• E1 = t v i E1 là: E1 =

 cos x + sin x
∫ ∫
*
E1 + E1 = sin x + cos x dx = dx = x + ( c1 )

Ta có: 
E − E* = cos x − sin x dx = d ( sin x + cos x ) = ln sin x + cos x + ( c )
 1

1 ∫
sin x + cos x ∫
sin x + cos x
2

E = 1 ( x + ln sin x + cos x ) + c
 1 2
Gi i h phương trình suy ra: 
E1 = 1 ( x − ln sin x + cos x ) + c
*
 2

sin3xdx cos 3 x dx
• E2 = ∫ 2cos3x − 5sin3x . Xét tích phân liên k t là: E* =
2 ∫ 2 cos 3x − 5 sin 3x
Ta có:
 * 2cos3x − 5sin3x
2E2 − 5E2 =

∫ 2cos3x − 5sin3x dx = ∫dx = x + ( c )1


 * 5cos3x + 2sin3x 1 d( 2cos3x − 5sin3x) ln 2cos3x − 5sin3x
5E2 + 2E2 =
 ∫ 2cos3x − 5sin3x dx = − 3 ∫ 2cos3x − 5sin3x
=−
3
+ ( c2 )

Gi i h phương trình suy ra:

 2 x
E = 1 ⋅ −1  2 ln 2 cos 3x − 5sin 3x 
 2
29 5 − ln 2 cos 3x − 5 sin 3x + c = 29 
 3
+ 5x  + c

 3

 x −5
1  5 ln 2 cos 3x − 5sin 3x 
 E* = 1 ⋅ ln 2 cos 3x − 5sin 3x
2 +c=  2x − +c
 29 − 2 29  3 
 3

( sin x)4 ( cos x)4
• E3 = ∫ ( sin x)4 + ( cos x)4 dx . Xét tích phân liên k t là: E* = ∫
3 dx
( sinx)4 + ( cos x)4

( sin x ) 4 + ( cos x )4
Ta có: E* + E 3 =
3 ∫ ( sin x )4 + ( cos x )4 dx = ∫ dx = x + ( c1 ) (1). M t khác:
( cosx )4 − ( sin x )4 ( cos 2 x + sin 2 x )( cos 2 x − sin 2 x )
E* − E 3 =
3 ∫ ( sin x )4 + ( cos x )4 dx = ∫
( cos 2 x + sin 2 x )2 − 2 cos 2 x sin 2 x
dx

176


cos 2x d ( sin 2x ) 1 2 + sin 2x
= ∫ 1 2
dx = ∫ ( 2) 2
2
− sin 2x
=
2 2
ln
2 − sin 2x
+ c ( 2)
1 − sin 2x
2
T (1) và (2) suy ra:

1 1 2 + sin 2x  * 1 1 2 + sin 2x 
E3 = x −
 ln  + c ; E3 =  x +
  ln +c
2 2 2 2 − sin 2x  2 2 2 2 − sin 2x 

π 2 π 2
( cosx )99 ( sin x )99
• E4 = ∫ ( sinx )
0
99
+ ( cosx )
99
dx . Xét tích phân: E* =
4 ∫ ( sin x )
0
99
+ ( cos x )
99
dx

π π π
t x= − u ⇒ dx = −du. V i x = thì u = 0 và x = 0 thì u = . Ta có:
2 2 2
99
π2
( sinx ) dx 99
sin π − u  ( −du)

 2
0

 ( )
π2
( cosu )99 du
∫ ∫ ∫
*
E4 = = = = E4
( sinx)99 + ( cosx )99 π 2  π 99 99
( )99 + ( sinu )99
0


sin − u
2
 + cos π − u 

 
 2 
 ( ) ( )
0 cosu

π2 π2 π2
( sin x )99 + ( cos x )99 π π
∫ dx = ∫ dx = x
* *
Ta có: E 4 + E 4 = 99 99
= ⇒ E4 = E4 =
0
( sin x ) + ( cos x ) 0
0 2 4

π 2 π 2

∫ ( cos3x ) 2 ( cos6x ) 2 dx . Xét tích phân: E5 =
∫ ( sin 3x )
2
• E5 = ∗
( cos 6 x )2 dx
0 0

π2 π2
( cos 3x )2 + ( sin 3x )2  ( cos 6x )2 dx =
∫ ∫ ( cos 6x )
2
Ta có: E 5 + E∗
5 =   dx
0 0

π2 π2
1
(1 + cos12x ) dx = 1  x + sin12x 
 π
=
2 ∫ 0

2 12 0
=
4
. M t khác:

π2 π2
( cos 3x )2 − ( sin 3x )2  ( cos 6x )2 dx =
∫ ∫ cos 6x ( cos 6x )
2
E 5 − E∗ =
5   dx
0 0

π2
1
π2
 ( )3  π
1 − ( sin 6x )2  d ( sin 6x ) = 1 sin 6x − sin 6x 
=
6 ∫
0
 
6 3 0
= 0 ⇒ E 6 = E* =
6
8

π 2 π 2
sinx dx cos x dx
∫ ( sinx + cosx ) ∫ ( sin x + cos x )
*
• E6 = 3
. Xét tích phân: E6 = 3
0 0

177


π2 π2
( cos x + sin x ) dx dx
Ta có: E∗
6 + E6 = ∫0
( sin x + cos x ) 3
= ∫ ( sin x + cos x )
0
2

π2 π2 π2
= ∫
dx
=
1
∫ sin
dx
=
−1
(
cotg x + π ) =
1 1
+ =1
0
(
 2 sin x + π 

 4 

)
2
2 0
2
( x+ π
4 ) 2 4 0 2 2

π2 π2
( cos x − sin x ) dx d ( sin x + cos x )
M t khác: E∗ − E 6 =
6 ∫
0
( sin x + cos x ) 3
= ∫
0
( sin x + cos x )3
π2
−1 1
= 2
= 0 ⇒ E 6 = E* =
6
2 ( sin x + cos x ) 0
2

a sin x + b cos x
6. D NG 6: F= ∫ m sin x + n cos x dx
a. Phương pháp:

Gi s : a sin x + b cos x = α ( m sin x + n cos x ) + β ( m cos x − n sin x ) , ∀x

⇔ a sin x + b cos x = ( mα − n β ) sin x + ( nα + mβ ) cos x , ∀x

 am + bn
 mα − n β = a α = m 2 + n 2
 
⇔  ⇔ . Khi ó ta có:
 nα + m β = b
  bm − an
 β = m2 + n2


am + bn m sin x + n cos x bm − an m cos x − n sin x
F= 2 2 ∫
m + n m sin x + n cos x
dx + 2
m + n 2 m sin x + n cos x
dx ∫
am + bn bm − an d ( m sin x + n cos x )
= 2
m +n 2 ∫
dx + 2
m + n2 ∫
m sin x + n cos x
am + bn bm − an
= 2 2
x+ 2 ln m sin x + n cos x + c
m +n m + n2
b. Các bài t p m u minh h a:

4sin2x − 7cos2x 1 4sin 2x − 7 cos 2x 1 4sin u − 7 cos u
• F1 = ∫ 5sin2x + 3cos2x dx = 2 ∫ 5sin 2x + 3cos 2x d ( 2x ) = 2 ∫ 5sin u + 3cos u du
Gi s 4 sin u − 7 cos u = α ( 5 sin u + 3 cos u ) + β ( 5 cos u − 3 sin u ) , ∀u

⇔ 4 sin u − 7 cos u = ( 5α − 3β ) sin u + ( 3α + 5β ) cos u , ∀u

178


5α − 3β = 4
 α = −1 34

⇔ ⇔ . Khi ó ta có:
3α + 5β = −7
 β = −47 34


1 4 sin u − 7 cos u −1 5sin u + 3 cos u 47 5 cos u − 3sin u
F1 = ∫
2 5 sin u + 3cos u
du = ∫
68 5sin u + 3 cos u
du − ∫
68 5 sin u + 3cos u
du

−1 47 d ( 5 sin u + 3cos u ) −1
=
68
du − ∫
68 ∫
5 sin u + 3 cos u
= ( u + 47 ln 5 sin u + 3cos u ) + c
68
−1
= ( 2x + 47 ln 5 sin 2x + 3 cos 2x ) + c
68
c. Các bài t p dành cho b n c t gi i:

4sin 3x + 5cos 3x 2sin 5x − 7 cos 5x 4sin 9x + 5cos 9x
F1 = ∫ 7 cos 3x − 8sin 3x dx ; F = ∫ 3sin 5x − 4 cos 5x dx ; F = ∫ 7 cos 9x − 3sin 9x dx
2 3

a sin x + b cos x + c
7. D NG 7: G= ∫ m sin x + n cos x + p dx
a. Phương pháp:

Gi s a sin x + b cos x + c = α ( m sin x + n cos x + p ) + β ( m cos x − n sin x ) + γ , ∀x

⇔ a sin x + b cos x + c = ( mα − n β ) sin x + ( nα + mβ ) cos x + pα + γ , ∀x

 mα − n β = a α = ( am + bn ) ( m 2 + n 2 )

 
⇔  nα + mβ = b ⇔  β = ( bm − an ) ( m2 + n 2 ) . Khi ó ta có:

 γ = c − am + bn p
 pα + γ = c
 
 m2 + n2

am + bn msin x + ncos x + p bm − an mcos x − nsin x
G= 2 2 ∫
m + n msin x + n cos x + p
dx + 2 2 ∫
m + n msin x + n cos x + p
dx +

 am + bn  dx
+ c − 2 2 p  ∫
 m + n  msin x + n cos x + p

am + bn bm − an d ( m sin x + n cos x + p )  am + bn  dx
= 2
m +n
2 ∫ dx + m 2
+n
2 ∫m sin x + n cos x + p
+ c − 2
 m +n
2
p ∫
 m sin x + n cos x + p
am + bn bm − an  am + bn  dx
= 2
m +n
2
x+ 2
m +n
ln m sin x + n cos x + p +  c − 2
2
 m +n
2
p ∫
 m sin x + n cos x + p

sinx + 2cosx − 3
• G1 = ∫ sinx − 2cosx + 3 dx .
179


Gi s sin x + 2 cos x − 3 = α ( sin x − 2 cos x + 3) + β ( cos x + 2 sin x ) + γ , ∀x

⇔ sin x + 2 cos x − 3 = (α + 2 β ) sin x + ( −2α + β ) cos x + ( 3α + γ ) , ∀x

α + 2β = 1 α = −3 5

 

⇔ −2α + β = 2 ⇔  β = 4 5 . Khi ó ta có:
 
3α + γ = −3
 γ = −6 5


−3 sin x − 2 cos x + 3 4 sin x − 2 cos x 6 dx
G1 = ∫
5 sin x − 2 cos x + 3
dx + ∫
5 sin x − 2 cos x + 3
dx − ∫
5 sin x − 2 cos x + 3
−3 4 d ( sin x − 2 cos x + 3) 6 dx
=
5
dx + ∫
5 ∫
sin x − 2 cos x + 3
dx − ∫
5 sin x − 2 cos x + 3
−3 4 6
= x + ln sin x − 2 cos x + 3 − J
5 5 5
dx dx
J= ∫ sin x − 2 cos x + 3 = ∫ 2sin x cos x − 2 cos =
2 2 ( 2
2 ) (
x − sin 2 x + 3 cos2 x + sin 2 x
2 2 2 )
dx 2 ( )
d tg x
2
∫
) ∫( ) ( )
= =
cos
2 x
2 2 (
2 tg x + 1 + 5 tg
2 x
2
5
2
2
tg x + 2 tg x + 1
5 2 5
2
d tg x
2 ( ) 2 5 1 + 5 tg x
2 + c = arctg
1 + 5 tg x
2 +c
= ∫ 2 2
= ⋅ arctg
5
(
tg x + 1 + 2
2 5 5 ) () 5 2 2 2

−3 4 6 5 tg x + 1
⇒ G1 = x + ln sin x − 2 cos x + 3 − arctg 2 +c
5 5 5 2
π 2
sinx − cosx + 1
• G2 = ∫
0
sinx + 2cosx + 3
dx .

Gi s sin x − cos x + 1 = α ( sin x + 2 cos x + 3) + β ( cos x − 2 sin x ) + γ , ∀x

⇔ sin x − cos x + 1 = (α − 2 β ) sin x + ( 2α + β ) cos x + ( 3α + γ ) , ∀x

α − 2 β = 1 α = −1 5
 
⇔ 2α + β = −1 ⇔  β = −3 5 . Khi ó ta có:
3α + γ = 1 γ = 8 5
 

180


π2 π2 π2
1 sin x + 2 cos x + 3 3 cos x − 2 sin x 8 dx
G2 = −
5 0
∫ sin x + 2 cos x + 3
dx −
5 ∫
0
sin x + 2 cos x + 3
dx +
5 ∫ sin x + 2 cos x + 3
0
π2 π2 π2
1 3 d ( sin x + 2 cos x + 3) 8 dx
=−
5 ∫ dx − 5 ∫
0 0
sin x + 2 cos x + 3
+
5 ∫ sin x + 2 cos x + 3
0
π2
 −1 3  8 −π 3 5 8
=  x − ln sin x + 2 cos x + 3  + J = + ln + J
 5 5 0 5 10 5 4 5

π2 π2
dx dx
J= ∫ = ∫ 2sin x cos x + 2 cos
0
sin x + 2 cos x + 3 0
22 22 2 (
x − sin 2 x + 3 cos2 x + sin 2 x
2
2 ) ( )
π2
dx
π2 d tg x
2 ( )
= ∫ =2 ∫
0 cos
2 x
2 2 (
2 tg x + 2 − 2 tg x + 3 + 3 tg x
2
2
π2
2
2
0 tg
2 x
2
+ 2 tg x + 5
2 )
π2 d 1 + tg x
2 ( )
1 + tg x
2 π 1 3π 3 5 8 1
=2 ∫ 2
= arctg = − arctg ⇒ G 2 = + ln − arctg
(
0 1 + tg x
2
+2
2
) 2 0 4 2 10 5 4 5 2

a sin x + b cos x
8. D NG 8: H= ∫ ( m sin x + n cos x ) 2
dx

a. Phương pháp:

Gi s a sin x + b cos x = α ( m sin x + n cos x ) + β ( m cos x − n sin x ) , ∀x

⇔ a sin x + b cos x = ( mα − n β ) sin x + ( nα + m β ) cos x , ∀x

 am + bn
mα − n β = a α = m 2 + n 2
 
⇔  ⇔ . Khi ó ta có:
nα + m β = b
  bm − an
 β = m2 + n2


am + bn m sin x + n cos x bm − an m cos x − n sin x
H= 2 2 ∫
m + n ( m sin x + n cos x ) 2
dx + 2
m + n ( m sin x + n cos x )2
2
dx ∫
am + bn dx bm − an d ( m sin x + n cos x )
= ∫ + 2
m + n m sin x + n cos x m + n 2
2 2 ∫ ( m sin x + n cos x ) 2

am + bn dx bm − an 1
= 2 2 ∫ − 2 2
⋅
m + n m sin x + n cos x m + n m sin x + n cos x
+c

181


2. Các bài t p m u minh h a:

7 sin x − 5 cos x
• H1 = ∫ ( 3 sin x + 4 cos x ) 2
dx .

Gi s 7 sin x − 5 cos x = α ( 3 sin x + 4 cos x ) + β ( 3 cos x − 4 sin x ) ; ∀x

⇔ 7 sin x − 5 cos x = ( 3α − 4 β ) sin x + ( 4α + 3β ) cos x; ∀x

α = 1
3α − 4β = 7
 
⇔ ⇔ 5 . Khi ó ta có:
4α + 3β = −5  β = −43

 5
7 sin x − 5cos x 1 3sin x + 4 cos x 43 3cos x − 4sin x
H1 = ∫ (3sin x + 4 cos x ) 2
dx = ∫
5 ( 3sin x + 4 cos x ) 2
dx − ∫
5 ( 3sin x + 4 cos x )2
dx

1 dx 43 d ( 3sin x + 4 cos x ) 1 43
= ∫ − ∫
5 3sin x + 4 cos x 5 ( 3sin x + 4 cos x ) 2
= J+
5 5 ( 3sin x + 4 cos x )

dx dx
d tg x
2 ( )
J= ∫ = ∫ =2 ∫
3 sin x + 4 cos x cos
2 x
2
6 tg
2 (
x + 4 − 4 tg 2 x
2
6 tg
2 )
x + 4 − 4 tg 2 x
2

x x
−2 2 tg 2 − 4 −2 2 tg 2 − 4 43
= ln + c ⇒ H1 = ln + +c
5 2 tg x + 1 25 2 tg x + 1 5 ( 3sin x + 4 cos x )
2 2
3. Các bài t p dành cho b n c t gi i:

2 sin 5x − 3cos 5x 5 sin 7x + 4 cos 7x
H1 = ∫ ( 4 cos 5x + 9 cos 5x ) 2
dx ; H 2 = ∫ ( 2 sin 7x − 3cos 7x ) 2
dx

2 2
a ( sin x ) + b sin x cos x + c ( cos x )
9. D NG 9: I= ∫ m sin x + n cos x
dx

a. Phương pháp:

2 2
Gi s : a ( sin x ) + b sin x cos x + c ( cos x ) =

= ( p sin x + q cos x ) ( m sin x + n cos x ) + r ( sin 2 x + cos 2 x ) , ∀x

2 2
⇔ a ( sin x ) + b sin x cos x + c ( cos x ) =
2 2
= ( mp + r ) ( sin x ) + ( np + mq ) sin x cos x + ( nq + r ) ( cos x ) ; ∀x

182


 ( a − c ) m + bn
p =
 m2 + n2
 mp + r = a  mp + r = a
   ( a − c ) n − bm
  
⇔ np + mq = b ⇔  np + mq = b ⇔ q = . Khi ó ta có:
   m2 + n2
 nq + r = c
  mp − nq = a − c
  2 2
 r = an + cm − bmn

 m2 + n2
2 2
 ( a − c) m + bn ( a − c) n − bm  an + cm − bmn dx
I=  ∫
 m +n
2 2
sin x + 2
m +n
2
cos x  dx +

2
m +n
2
msin x + n cos x ∫
2 2
( a − c) n − bm ( a − c) m + bn an + cm − bmn dx
=
m +n2 2
sin x − 2
m +n 2
cos x + 2
m +n 2 ∫ msin x + n cos x
π 3
( cos x ) 2 dx
• I1 = ∫
0 sin x + 3cos x
.

Gi s ( cos x )2 = ( a sin x + b cos x ) ( sin x + 3 cos x ) + c ( sin 2 x + cos 2 x ) ; ∀x

⇔ ( cos x ) = ( a + c ) ( sin x ) + ( a 3 + b ) sin x cos x + ( b 3 + c ) ( cos x ) ; ∀x
2 2 2

a + c = 0 a = −1 4
  π3 π3
  1  3 1  1 dx
⇔ a 3 + b = 0 ⇔ b = 3 4 ⇒ I =
 
 cos x − sin x  dx +
2 0 2 2  ∫
4 0 sin x + 3 cos x ∫
b 3 + c = 1  c = 1 4

π3 π3
1  π π  1 dx
=
2 ∫  cos cos x − sin sin x  dx +
 6 6  8 ∫ π π
0 0 cos sin x + sin cos x
3 3
π3 π3 π3
1  π 1 dx 1  π 1 x π
= ∫ cos  x +  dx +
 6 ∫  π
=  sin  x +  + ln tg  +  
 6 8 2 6
sin  x +  
2 0
8 0
2
0
 3

1 1  1 1  1 1 1
=  + ln 3  −  − ln 3  = + ln 3 = (1 + ln 3 )
2 8  4 8  4 4 4

183


m sin x + n cos x
10. D NG 10: J= ∫ a ( sin x ) 2
+ 2b sin x cos x + c ( cos x )
2
dx

a. Phương pháp:

a−λ b
•G i λ1 , λ 2 là nghi m c a phương trình =0
b c−λ

2
a + c ± ( a − c ) + 4b 2
⇔ λ 2 − ( a + c ) λ + ac − b 2 = 0 ⇔ λ1,2 =
2
2 2
Bi n i a ( sin x ) + 2b sin x cos x + c ( cos x ) = λ1 A12 + λ 2 A2 =
2

2 2
λ1  b  λ2  b 
=  cos x − a − λ sin x  +  cos x − a − λ sin x 
b2  1  1+ b2  2 
1+ 2 2
( a − λ1 ) ( a − λ2 )
b b 1 1
t u1 = cos x − sin x ;u2 = cos x − sin x ; k1 = ; k2 =
a − λ1 a − λ2 a − λ1 a − λ2

1 1
A1 =
2
( cos x − bk1 sin x ) ; A2 = ( cos x − bk2 sin x )
1+ b k12 1 + b 2 k2
2

ý r ng A12 + A2 = 1 ⇒ λ1 A12 + λ 2 A2 = ( λ1 − λ 2 ) A12 + λ 2 = ( λ 2 − λ1 ) A2 + λ1
2 2 2

 b   b 
•Gi s m sin x + n cos x = p  sin x + cos x  + q  sin x + cos x  , ∀x
 a − λ1   a − λ2 

p + q = m
 bm − n ( a − λ2 ) bm − n ( a − λ1 )
⇔  p q n ⇔ p= b λ −λ ( a − λ1 ) ;q = ( a − λ2 )
 + = ( 2 1) b ( λ1 − λ2 )
 a − λ1 a − λ2 b

m sin x + n cos x − pdu1 −qdu2
J= ∫ a ( sin x) 2
+ 2b sin x cos x + c ( cos x)
2
dx = ∫ (λ − λ ) A
1 2
2
1 + λ2
+ ∫ (λ
2 − λ1 ) A2 + λ1
2

dA1 dA2
= − p 1 + b2 k12 ∫ ( λ −λ ) A − q 1 + b2 k2 ∫ (λ
2

1 2
2
1 + λ2 2 − λ1 ) A2 + λ1
2

184

Tich phan 210_ham_lg_co_ban_doi_bien_129

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