Các phương pháp tính tích phân - Vi tích phân 1 | Trường Đại học Khoa học Tự nhiên, Đại học Quốc gia Thành phố Hồ Chí Minh
Các phương pháp tính tích phân - Vi tích phân 1 | Trường Đại học Khoa học Tự nhiên, Đại học Quốc gia Thành phố Hồ Chí Minh. Tài liệu được sưu tầm giúp bạn tham khảo, ôn tập và đạt kết quả cao trong kì thi sắp tới. Mời bạn đọc đón xem !
Môn: Vi tích phân 1 (MTH00005) 15 tài liệu
Trường: Trường Đại học Khoa học tự nhiên, Đại học Quốc gia Thành phố Hồ Chí Minh 1 K tài liệu
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN Chuyªn ®Ò 1: C¸ C c ¸ ph p − h ¬ng n p h p ¸ h p p t Ý t nh n h tÝ t ch h p h p © h n
Th«ng thường ta gÆp c¸c lo¹i tÝch ph©n sau ®©y:
+) Loại 1: TÝch ph©n cña h m sè ®a thøc ph©n thøc h÷u tû.
+) Loại 2: TÝch ph©n cña h m sè chøa c¨n thøc
+) Loại 3: TÝch ph©n cña h m sè l−îng gi¸c
+) Loại 4: TÝch ph©n cña h m sè mò v logarit
§èi víi c¸c tÝch ph©n ®ã cã thÓ tÝch theo c¸c ph−¬ng ph¸p sau:
I) Ph−¬ng ph¸p biÕn ®æi trùc tiÕp b
Dïng c¸c c«ng thøc biÕn ®æi vÒ c¸c tÝch ph©n ®¬n gi¶n v ¸p dông ®−îc b f (x d ) x ∫ = ∫ F(x)
= F(b) − F(a) a a
+) BiÕn ®æi ph©n thøc vÒ tæng hiÖu c¸c ph©n thøc ®¬n gi¶n VÝ dô 1. TÝnh: 2 2 2 2 1. 1 2 2 I ∫ x − =
2x dx ta cã I = ( − )dx = ln x + = (ln2 + )1 − (ln1+ ) 2 = ln 2 −1 ∫ 3 x x x 2 x 1 1 1 2 e 2 e 2 x − 4 2 e 2. = ∫ −1/2 2x
− 3+ dx = (4 x −3x + 4ln x ) J = ∫ 3x + 4 dx = − 2 e 3 + e 4 + 7 x x 1 1 1 8 3 5 8 8 4 x − 4 1 4 3 3. 207 K = ∫
3x − 1 dx = ∫ x − 1/3 x − −2 / 3 x dx = 2 x − 3 4 x − 3 x = 3 2 3 3 3 4 4 1 3 x 1 1
+) BiÕn ®æi nhê c¸c c«ng thøc l−îng gi¸c VÝ dô 2. TÝnh: π / 2 π / 2 π / 2 1. 1 1 sin 2x sin 8x
I = ∫ cos 3xcos5xdx = ∫ (cos2x +cos8x)dx = + = 0 2 2 2 8 −π / 2 −π / 2 −π / 2 π / 2 π / 2 π / 2 2. 1 1 sin 5x sin 9x 4
J = ∫ sin2xsin7xdx = ∫ (cos( 5 − x) − cos9x)dx = − = 2 2 5 9 45 −π / 2 −π / 2 −π / 2 π / 2 π / 2 π / 2 π / 2 3. 1 1 cos 4x cos10x
K = ∫ cos 3xsin7xdx = sin 7x cos x 3 dx = ∫
∫ (sin4x +sin10x)dx = − + = 0 2 2 4 10 −π / 2 −π / 2 −π / 2 −π / 2 π π 1 + π cos 2x 1 1 4. H = ∫ 2
sin 2x cos xdx = ∫sin 2x dx = − cos 2x − cos 4x = 0 hoÆc biÕn ®æi 0 2 4 16 0 0 0 π π 1 + π cos 2x 1 1 H = ∫ 2
sin 2x cos xdx = ∫sin 2x dx = − cos 2x − cos 4x = 0 2 4 16 0 0 0 π / 2 π / 2 π / 2 1 + sin 2x + 2 2 2 + + − 5. (sin x cos x) cos x sin x π / 2 G = ∫ cos 2x = dx = 2 cos xdx = ∫ ∫ (−2sin x) = 1 − sin x + dx + cosx sin x + cos x π / 6 π / 6 π / 6 π / 6 π / 2 π / 2 2 π / 2 π / 2 − π 6. 1 cos 2x 1 1 sin 4x 3 E = ∫ 4 sin xdx = dx = ∫
∫(3+cos4x −4cos2x)dx = 3x + − sin 2x = 2 8 8 4 16 0 0 0 0 π / 4 π / 4 π / 2 π −π 7. 1 / 4 4 F = ∫ 2 tan xdx = −1dx = − = ∫ . §Ò xuÊt: F cot xdx v 1 = ∫ 2 2 (tan x x) cos x 0 4 0 0 π / 4 π / 4 F = ∫ 4 tan xdx 2 0
+) BiÕn ®æi biÓu thøc ë ngo i vi ph©n v o trong vi ph©n VÝ dô 3. TÝnh: 1 1 1 4 + 1. 1 3 1 (2x ) 1
I = ∫ (2x + 3 ) 1 dx = (2x + ) 1 d(2x + ) 1 = = 10 ∫ 2 2 4 0 0 0 2 2 1 1 −2 − + 2. = ∫ 1 1 3 1 (2x ) 1 1 1 J dx = (2x − ) 1 d(2x − ) 1 = = − = 0 ∫ (2x − 3 ) 1 2 2 − 2 4 (2x − ) 1 2 1 1 0 0
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN 7 / 3 7 / 3 7 / 3 3. 1 1/ 2 2 3 16
K = ∫ 3x − 3dx = x 3 ( − ) 3 d 3 ( x − ) 3 = 3 ( x − ) 3 = ∫ 3 9 9 1 1 1 4 4 1 − − 4. = ∫ dx 1 1/ 2 2 1/ 2 10 2 13 H = − (25 − x 3 ) d(25 − 3x) = (25 − 3x) = ∫ − 25 − 3x 3 3 3 0 − 0 0 2 2 2 + − − − − 5. = ∫ 1 x 1 x 1 1 3 2 1 G dx = dx = ( x +1 − x −1)dx = ∫ ∫ x + 1 x 1 (x − ) 1 − (x + ) 1 2 3 1 + + − 1 1
(Nh©n c¶ tö v mÉu víi bt liªn hîp cña mÉu sè) §Ò xuÊt 1 1 G = dx ∫ = ∫
(ax + b)3 + (ax + c)3 + C víi a ≠ ; 0 b ≠ c 1
ax + b + ax + c a(b − c) 1 1 1 1 6. 4
P = ∫ x 1 − xdx = ∫ (x −1+ ) 1 1− xdx = −∫ (x − ) 1 1− xd 1 ( − x) + ∫ 1− xdx = 5 0 0 0 0 1 1 7. 2 1 1 2 2 Q ∫ 1− = x e xdx = − e1−x d 1 ( − x 2 ) = −e1−x = e −1 ∫ 2 0 0 0 1 − §Ò xuÊt 4 6 2 Q = x3 1 ∫ + ∫ x 2 dx =
HD ®−a x v o trong vi ph©n v thªm bít (x2 + 1 H 1). 1 15 0 VÝ dô 4. TÝnh: π π / 2 π / 2 1. 1 I = (2 cos 3x ∫ + ∫ 3 sin 2x d
) x = 0 ; I = sin3 x cos xdx ∫ = ∫ v I = ecosx sin xdx ∫ = ∫ e − 1 1 2 4 3 0 0 0 π / 4 π / 2 π / 4 2. sin x 2 J = tan xdx ∫ = ∫ ln 2 ; J = cot xdx ∫ = ∫ ln 2 v J = dx ∫ = ∫ ln 2 1 2 3 1 + 3 cos x 3 0 π / 6 0
(®−a sinx, cosx v o trong vi ph©n) e 2 e 3 e 3. sin(ln x) cos(ln x) 1 K = dx ∫ = ∫
1 − cos1 ; K = dx ∫ = ∫
1 − cos 2 v K = dx ∫ = ∫ 2 1 x 2 x 3 x 1 + ln x 1 1 1
{®−a 1/x v o trong vi ph©n ®Ó ®−îc d(lnx)} ln 3 x ln 3 4. e x 5 H dx = ln 2 + e = ln
1 = ∫ 2 + x e 1 2 + e 1 ln 2 ln 2 ln 2 ln 2 1 − x 1+ x e − x x e 2e e H dx = ∫ dx = dx 2 dx 3ln 2 2 ln 3 x ∫ − ∫ = −
2 = ∫ 1 + x e 1+ e 1 + x e 0 0 0 0 ln 2 ln 2 ln 2 ln 2 ln 2 x x x + − dx 1 (e 5 e )dx 1 1 e dx 1 1 x 1 12 H = = dx − = x − ln e + 5 = ln ∫ ∫ ∫ 3 = ∫ x e + 5 5 ex + 5 5 5 ex + 5 5 5 5 7 0 0 0 0 0 1 x 1 1 e dx 2x 2 e dx 1 1 e 1 H = ∫ 2x + = ln e +1 = ln 4 = ∫ x − e + x e 2x e +1 2 2 2 0 0 0 b
+) BiÕn ®æi nhê viÖc xÐt dÊu c¸c biÓu thøc trong gi¸ trÞ tuyÖt ®èi ®Ó tÝnh I = ∫ f(x,m)dx a
H XÐt dÊu h m sè f(x,m) trong ®o¹n [a; b] v chia [a;b]= [a;c ]∪[c ;c ]∪...∪[c ;b] trªn mçi ®o¹n h m sè f(x,m) 1 1 2 n gi÷ mét dÊu c c 1 2 b
H TÝnh I = ∫ f(x,m)dx + ∫ f(x,m)dx +...+ ∫ f(x,m)dx a c c 1 n VÝ dô 5. TÝnh: 2 1. I = ∫ 2
x + 2x − 3 dx Ta xÐt pt: x 2 + 2 x 3 = 0
⇔ x = 1∨ x = 3 . B¶ng xÐt dÊu f(x) 0
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN 1 2 1 2
Suy ra I = x2 + 2x − 3dx + x2 + 2x − 3dx = − (x2 + 2x − ) 3 dx + (x 2 + 2x − ) 3 dx = 4 ∫ ∫ ∫ ∫ 0 1 0 1 1 −2 0 1
2. J = ∫ 4x − 3
x dx tÝnh t−¬ng tù ta cã J = 4x − x3 dx + 4x − x3 dx + 4x − x3 dx = 16 ∫ ∫ ∫ −3 −3 −2 0 3 3. 1 K = 2x ∫ − ∫ 4 dx = 4 + ln 2 0 2π 2π π 2π 4. H 1 cos 2xdx = 2 sin x dx = 2 sin x dx + 2 sin x dx = 2 2 ∫ ∫ ∫ 1 = ∫ − 0 0 0 π π H 1
sin 2xdx {ViÕt (1 – sin2x) vÒ b×nh ph−¬ng cña mét biÓu thøc råi khai c¨n} 2 = ∫ − 0 π π / 4 3π / 4 3π / 2
= sin x + cos x dx = sin x + cos x dx + sin x + cos x dx + sin x + cos x dx = 2 2 ∫ ∫ ∫ ∫ 0 0 π / 4 3π / 4
II) Ph−¬ng ph¸p ®æi biÕn sè
A ' Ph−¬ng ph¸p ®æi biÕn sè d¹ng 1: b
Gi¶ sö cÇn tÝnh tÝch ph©n I = ∫ f(x)dx ta thùc hiÖn c¸c b−íc sau: a H B−íc 1. §Æt x = u(t)
H B−íc 2. LÊy vi ph©n dx = u’(t)dt v biÓu thÞ f(x)dx theo t v dt. Ch¼ng h¹n f(x)dx = g(t)dt
H B−íc 3. §æi cËn khi x = a th× u(t) = a øng víi t = α ; khi x = b th× u(t) = b øng víi t = β β
H B−íc 4. BiÕn ®æi I = ∫g(t)dt (tÝch ph©n n y dÔ tÝnh h¬n th× phÐp ®æi biÕn míi cã ý nghÜa) α
C¸ch ®Æt ®æi biÕn d¹ng 1.
C¸ch ®Æt 1. NÕu h m sè chøa 2
1 − x th× ®Æt x = sin ; t t ∈[ π − / ; 2 π / ]
2 hoÆc ®Æt x = cos ; t t ∈[ ; 0 π ] VÝ dô 1. TÝnh: 1 2 1. A ∫ 1− =
x dx ta ®Æt x = sin t;t ∈[ π − / ;
2 π / 2] ⇒ dx = cost.dt; ®æi cËn khi x = 2 /2 th× t = π / 4 ; khi x 2 x 2 / 2 π / 2 π / 2 π / 2 2 2 2 − − − π = 1 th× t = π 1 sin t cos t 1 sin t 4 / 2 . Khi ®ã A = cos t d . t = d . t = d . t = ∫ ∫ ∫ sin 2 t sin 2 t sin 2 t 4 π / 4 π / 4 π / 4 1 2 1 2 2. = ∫ x x B dx ta viÕt B = ∫ dx . 2 2 2 1 (x / 2) 0 − 0 4 − x §Æt (x / 2) = cos t; t ∈ ; 0
[ π ] ⇒ x = 2 cos t ⇒ dx = 2 − sin tdt π / 3 π / 2 π / 2 2 π §æi cËn suy ra (2 cos t) B = ( 2 − sin tdt) = 4cos2 tdt = 2 ∫ ∫ ∫(1+ cos2t) 3 dt = − 2 3 2 π − / 2 2 1 cos t π / 3 π / 3 1 1 2 3. 3 x . C = ∫ 2 x 4 − 2
3x dx Tr−íc hÕt ta viÕt C = 2∫ 2 x 1 − dx . 2 0 0 §Æt 3 x = sin t; t ∈[ π − / ;
2 π / 2] ®−a tÝch ph©n vÒ d¹ng: 2 π / 3 π / 3 16 − π 2 2 4 1 cos 4t 2 3 1 C = sin t cos tdt = dt = + ∫ ∫ 3 3 3 3 2 27 12 0 0 Chó ý:
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN x = sin t;t ∈[−π π 2 / ; 2 / ] 2 H NÕu h m sè chøa x a
a − x2 ,a > 0 th× ta viÕt 2 a − x = a 1 − v ®Æt a x = cos t; t ∈[ ; 0 π ] a b π π 2 x = sin t; t ∈[− / ; 2 / ] 2 H NÕu h m sè chøa b a
a − bx2 ,a,b > 0 th× ta viÕt 2 a − bx = a 1 − x v ®Æt a b x = cost;t ∈[ ; 0 π ] a VÝ dô 2. TÝnh: 2 1. = ∫ 1 E
dx {ViÕt tÝch ph©n vÒ d¹ng 2 1 − X } 2 x x 1 2 / 3 − 2 π / 3 π ta viÕt = ∫ 1 1 E dx v ®Æt = sin t; t ∈ [− π / ; 2 π / 2] suy ra E = dt = ∫ 2 2 x 1 1/ x x 12 2 / 3 − ( ) π / 4 2 2 / 3 2 2. G ∫ 3x − =
4 dx {ViÕt tÝch ph©n vÒ d¹ng 2 1 − X } 3 x 2 / 3 2 2 / 3 3 . x . 1 (2 / 3x)2 ta viÕt 2 G ∫ − = dx v ®Æt = sin t; t ∈ [− π / ;
2 π / 2] suy ra tÝch ph©n cã d¹ng 3 x 3x 2 / 3 π / 3 3 3 + − 2 3( 6 3 3) G = cos tdt = ∫ π
{NÕu tÝch ph©n cã d¹ng ax 2 − b th× viÕt vÒ d¹ng 2 1 − X } 2 16 π / 4
C¸ch ®Æt 2. NÕu tÝch ph©n cã chøa 2 1 + x hoÆc ( 2
1 + x ) th× ta ®Æt x = tan t;t ∈ (− π / ; 2 π / 2) hoÆc x = cot t; t ∈ ( ; 0 π ) VÝ dô 3. TÝnh: 3 π / 3 π 1. = ∫ 1 M
dx ta ®Æt x = tan t; t ∈ (− π / ; 2 π / 2) suy ra M = dt = ∫ 1 + 2 x 6 1 / 3 π / 6 3 π / 3 − 2. = ∫ 1 cos t 18 2 3 N
dx ta ®Æt x = tan t; t ∈ (− π / ; 2 π / 2) suy ra N = dt = ∫ 2 2 2 x . 1 x sin t. 3 1 + π / 4 a 3. = ∫ 1 P d ; x a ≠ 0 2 (a + 2 2 x ) 0 a π / 4 + ta viÕt = ∫ 1 x 1 2 P dx v ®Æt = tan t; ⇒ 2 P = cos tdt = ∫ π 3 3 x 2 a a 4a 0 4 2 a 1 + ( ) 0 a 1 4. = ∫ 1 Q dx 2 x + x + 1 0 1 1 π ta viÕt = 4 2 1 4 3 3 Q ∫ 1 dx v ®Æt
x + = tan t; t ∈ (− π / ; 2 π / 2) ⇒ Q = dt = ∫ 3 2 3 2 3 2 9 0 1 + 2 1 0 (x + ) 3 2
Chó ý: NÕu gÆp tÝch ph©n chøa 2 a + bx hoÆc 2
a + bx th× ta viÕt: 2 2 b 2 b b a + bx = a 1 + x hoÆc 2 a + bx = a 1 + x v ta ®Æt x = tan t; t ∈ (− π / ; 2 π / 2) a a a
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN − +
C¸ch ®Æt 3. NÕu tÝch ph©n cã chøa a x hoÆc a x th× ta ®Æt ta ®Æt x = acos t 2 ;t ∈ [ ;
0 π / 2] v l−u ý vËn dông a + x a − x 1 − cos 2t = 2sin2 t 1 + cos 2t = 2cos2 t VÝ dô 4. TÝnh: 0 π / 4 a + 1 + − π 1. cos 2t 4 I = ∫ x d ;
x a > ta ®Æt x = a cos 2t; t ∈ ; 0 [ π / 2] suy ra I = ∫ (−2a sin 2t)dt = a a − 0 − x 1 − cos 2t 4 −a π / 2 2 / 2 π / 8 1 + 1 + 2. cos 2t J = ∫
x dx ta ®Æt x = cos2t;t ∈[ ;0π / 2] suy ra J = ∫ (−2 sin 2t d ) t = 1 − x 1 − cos 2t 0 π / 4 π / 4 + − 1 + x 2 4 2 2 J = 4 cos tdt = ∫ π {cã thÓ ®Æt
= t suy rra tÝch ph©n J vÒ d¹ng tÝch ph©n cña h m sè h÷u tû} 4 1 − x π / 8
B ' Ph−¬ng ph¸p ®æi biÕn sè d¹ng 2: b
Gi¶ sö cÇn tÝnh tÝch ph©n I = ∫ f(x d
) x ta thùc hiÖn c¸c b−íc sau: a H B−íc 1. §Æt t = v(x)
H B−íc 2. LÊy vi ph©n dx = u’(t)dt v biÓu thÞ f(x)dx theo t v dt. Ch¼ng h¹n f(x)dx = g(t)dt
H B−íc 3. §æi cËn khi x = a th× u(t) = a øng víi t = α ; khi x = b th× u(t) = b øng víi t = β β
H B−íc 4. BiÕn ®æi I = ∫g(t)dt (tÝch ph©n n y dÔ tÝnh h¬n th× phÐp ®æi biÕn míi cã ý nghÜa) α
C¸ch ®Æt ®æi biÕn d¹ng 2.
C¸ch ®Æt 1. NÕu h m sè chøa Èn ë mÉu th× ®Æt t = mÉu sè. VÝ dô 1. TÝnh: π / 2 4 1. dt 4
I = ∫ sin 2x dx ta cã thÓ ®Æt t = 4 H cos2x suy ra I = = ln ∫ 4 − 2 cos x t 3 0 3 π / 4 2 2. dt 3 J = ∫ sin 2x
dx ®Æt t = sin 2 x + 2 cos2 x = 1 + cos2 x suy ra J = ∫ = ln 2 sin x + 2 2cos x t 4 0 3 / 2
{cã thÓ h¹ bËc ®Ó biÕn ®æi tiÕp mÉu sè vÒ cos2x sau ®ã ®−a sin2x v o trong vi ph©n} π / 2 §Ò xuÊt: sin x cos x J
dx víi a 2 + b2 > 0 1 = ∫ 2 2 a sin x + 2 2 b cos x 0 ln 2 3. = ∫ 1 K
dx ta ®Æt t = ex + 5 ⇒ ex = t − 5 ⇒ exdx = dt sau ®ã l m xuÊt hiÖn trong tÝch ph©n biÓu x e + 5 0 ln 2 7 7 x − thøc e dx dt 1 t 5 1 12 ex dx ⇒ K = = = ln = ln ∫ ∫ ex (ex + ) 5 t(t − ) 5 5 t 5 7 6 0 6 ln 2 ln 2 ln 2 x x x x + − +
{Cã thÓ biÕn ®æi trùc tiÕp 1 e 5 e 1 e 5 1 e 1 12 K = dx = dx − dx = ln ∫ ∫ ∫ } 5 e x + 5 5 e x + 5 5 e x + 5 5 7 0 0 0 π / 2 7 sin 2x + 4. 1 1 2 H = ∫ cos x
dx ta ®Æt t = 2sin x − cos 2x + 4 ⇒ H = dt = ∫
(2sin x − cos 2x + 2 4) 2 t 2 21 0 3
{®«I khi kh«ng ®Æt c¶ MS} π / 2 3
5. G = ∫ sinxcos x dx chó ý r»ng t¸ch mò 3 = 2 +1 ®Æt 1 + 2 cos x 0 2 2 1 (t − ) 1 1 1 − ln 2 t = 1 cos2 +
x ⇒ cos2 x = t −1 ⇒ 2sin x cos xdx = −dt khi ®ã: G = dt = (t − ln t ) = ∫ 2 t 2 2 1 1
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN π / 4 6. M = ∫ cos 2x dx
sin x + cos x + 2 0
ta ®Æt t = sin x + cos x + 2 ⇒ dt = (cos x − sin x)dx l−u ý cos2x = (cosx+sinx)(cosxHsinx) π / 4 2+ 2
(cos x + sin x)(cos x − sin x) (t − 2)dt 2+ 2 + M = dx = = ∫ ∫ (t − 2lnt ) 2 2 = 2 −1 + ln sin x + cos x + 2 t 3 3 0 3 π / 4 7. N = ∫ cos 2x
dx ®Æt t = sin x + cos x + 2 suy ra (sin x + cos x + 3 2) 0 2+ 2 2+ 2 (t − ) 2 dt 1 1 1 1 1 1 2 1 N = = − = − − + = − ∫ t 3 t 2 t (2 + 2)2 2 + 2 9 3 9 2 1 ( + 2 ) 3 3 π / 4 π / 4 §Ò xuÊt: cos 2x cos 2x M dx v N dx 1 = ∫
1 = ∫ sin x − cos x + 2
(sin x − cos x + 3 2) 0 0 8.
C¸ch ®Æt 2. NÕu h m sè chøa c¨n thøc n ϕ(x) th× ®Æt n
t = ϕ(x) sau ®ã luü thõa 2 vÕ v lÊy vi ph©n 2 vÕ. VÝ dô 1. TÝnh: 1 4x − 1. 1 2 I = ∫ 3 dx ta ®Æt t = x 3 + 1 ⇒ x =
(t2 − )1⇒ dx = tdt khi ®ã ®−a tÝch ph©n vÒ d¹ng:
2 + 3x + 1 3 3 0 2 2 2 3 − 2 4t 13t 2 I = dt = ∫ ∫(4t2 −8t + 3) 2 6 2 4 4 dt − dt = − ln ∫ 9 2 + t 9 9 2 + t 27 3 3 1 1 1 7 3 2 2. = ∫ x 3 4 141 J dx ta ®Æt 3 2 t = 1 + x
⇒ x 2 = t3 −1 ⇒ 2xdx 3t 2 = dt ⇒ J = (t − t)dt = ∫ 3 2 2 20 0 1 + x 1 2 5 5 − 3. = ∫ 1 tdt 1 t 1 K dx ta ®Æt 2 t = 1 + x
⇒ x 2 = t2 −1 ⇒ xdx = tdt ⇒ J = = ln ∫ 2 2 (t − ) 1 t 2 t + 1 1 x 1 + x 2 2 2 4. = ∫ 1 H dx ta ®Æt 3
t = 1 + x ⇒ x 3 = t 2 − 1 ⇒ 3x2dx = t
2 dt nh©n c¶ tö v mÉu sè víi x2 ta ®−îc: 3 1 x 1 + x 2 3 3 xdx 2 dt 1 t −1 2 2 + 1 H = = = ln = ln ∫ ∫ 2 3 3 t 2 + −1 3 t +1 3 1 x 1 x 2 2 2 3 5 x + 3 5. G = ∫ 2x dx ta ®Æt 2 t = 1 + x
⇒ x 2 = t2 −1 ⇒ xdx = tdt nhãm x2.x.(x2 +2) ta ®−îc: 2 0 x + 1 2 3 2 (x 2 + ) 2 x 2 x . (t 2 + ) 1 (t 2 − ) 1 tdt t5 26 G = dx = = − t = ∫ ∫ 2 t + 5 5 0 x 1 1 1 6 6. = ∫ 1 1 2 M dx ta ®Æt 6 t = 9x + 1 ⇒ x = (t6 − )1 ⇒ dx t5 =
dt luü thõa bËc hai v bËc ba
9x + 1 + 3 9x + 9 3 − 1 1 2 2 2 2 t 5dt 2 t 3dt 2 2 1 2 11 2 ta cã: M = = = (t − t + 1 − )dt = + ln ∫ 3 2 ∫ ∫ 3 t + t 3 t + 1 3 t +1 3 6 3 1 1 1 VÝ dô 2. TÝnh: π / 2 sin 2x + 1. [§H.2005.A] 1 2 P = ∫
sin x dx ta ®Æt t = 1+ 3cosx ⇒ cosx = (t2 − )1 ⇒ sin xdx = − tdt nhãm 1 + 3 cos x 3 3 0 π / 2 2 2 (2 cos x + 3 nh©n tö sinx ta cã: 2 2 2 2t 34 P = ∫ )
1 sin xdx = ∫(2t + )1dx = + t = 1 + 3cos x 9 9 3 27 0 1 1
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN π 2 + 2. cos 3x sin 2x 1 2 Q = d . x ∫
ta ®Æt t = 1+ 3sin x ⇒ sin x = (t 2 − ) 1 ⇒ cos xdx = tdt ¸p dông c«ng thøc 1 + 3 sin x 3 3 0 π π 2 3 − + 2 2 − − +
nh©n ®«i v nh©n 3 ta viÕt: 4 cos x 3 cos x 2 sin x cos x 4 4 sin x 3 2 sin x Q = d . x ∫ = .cos xdx ∫ 1 + 3sin x 1 + 3sin x 0 0 2 2 VËy = 2 2 4 5 14 3 206 Q ∫(− 4 4t + 2 14t − d ) 1 t = − t + t − t = 27 27 5 3 405 1 1 π 2 3. [§H.2006.A] 2 1 R = ∫ sin 2x
dx ta ®Æt t = 1 3sin 2 + x ⇒ sin x = (t 2 − ) 1 ⇒ 2 2 3 0 cos x + 4 sin x 2 2 2 2 tdt 2 2 sin 2xdx = tdt . khi ®ã: R = = t = ∫ 3 3 t 3 3 1 1 4. VÝ dô 3. TÝnh: e 1. P ∫ lnx 1 + = 3 ln x dx x 1 2 Ta ®Æt 1 dx 2 2 4 116 t = 1 + 3ln x ⇒ ln x = (t 2 − ) 1 ⇒
= tdt khi ®ã: P = ∫(t − t)dx = 3 x 3 9 135 1 e 3 − 2. Q = ∫ 2 ln x dx x 1 + 2 ln x 1 Ta ®Æt 1 dx t = 1 + 2 ln x ⇒ ln x = (t 2 − ) 1 ⇒ = tdt . Khi ®ã: 2 x 3 2 2 3 − (t 2 − ) 1 tdt − 2 t 3 9 3 11 Q = = (4 − t )dt = 4t − = ∫ ∫ t 3 3 1 1 1 2 ln 2 5 2dt 5 − 1 3 + 3. = ∫ dx 1 R
. Ta ®Æt t = ex + 1 suy ra exdx = 2tdt ⇒ R = ∫ = ln . x 2 t − 1 5 + 1 3 − 1 ln 2 e + 1 3 3 e 4. = ∫ dx d 3 x e 2 S . Ta ®Æt 3 x t = e suy ra S = = 3ln ∫ 3 x t(t + ) 1 e + 1 0 1 + e 1 ln 5 x x e e − 5. X = ∫ d 1 x x e + 3 0
C¸ch ®Æt 3. NÕu h m sè chøa c¸c ®¹i l−îng x x
sin x , cos x v tan
th× ta ®Æt t = tan khi ®ã 2 2 t 2 2 1 − t sin x = , cos x = 2 1 + t 2 1 + t VÝ dô 4. TÝnh: π / 2 1. 1 Q = d . x ∫ 5 sin x + 3cos x + 5 0 1 1 + Ta ®Æt x 2dt 1 1 t 1 1 8 t = tan ⇒ dx = v Q = dt = ln = ln ∫ 2 2 1 + t t 2 + 5t + 4 3 t + 4 3 5 0 0 x π / 3 tan 1 / 3 1 / 3 2. 2 x d 2 t 2tdt 2 10 L = d . x ∫ ta ®Æt t = tan ⇒ dx = v L = = ln t + 3 = ln ∫ cos x + 2 2 2 1 + t t 2 + 3 0 9 0 0
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN π 4 1 1 ( + 1 1 3. dt t)dt dt tdt V = ∫ cos 2x dx ta ®Æt t = tan x ⇒ dx = v V = ∫ = 2 ∫ + 2 ∫ cos 2x + sin 2x + 1 2 1 + t 2 1 ( + t ) 2 1 ( + t ) 1 ( 2 + 2 t ) 0 0 0 0 1 π 4 1 1 t =tan y π + 2 = dt cos 2x 2 ln 2 V + ln t + 1 ta tÝnh V = = ∫ suy ra V = dx = ∫ π 1 1 2 0 4 1 ( 2 + t ) 8 cos 2x + sin 2x + 1 8 0 0 π 4 π 4 1 + 2 1 + 2 4. tan x N = ∫ tan x dx ta viÕt N = ∫ dx v ®Æt 2 cos x + sin 2x − 2 sin x + 1 cos 2x + sin 2x + 1 0 0 1 1 dt 1 1 + t 2 1 t 2 3 + 2 ln 2 t = tan x ⇒ dx = suy ra N = dt = + t + ln t +1 = ∫ 2 1 + t 2 t + 1 2 2 4 0 0 π π 4 sin x − π 4 1 (sin x − cosx) 5. [§H.2008.B] F = ∫ 4 dx ta viÕt F = ∫ dx dùa sin 2x + 2 1 ( + sin x + cos x) 2 2 sin x cos x + 1 ( 2 + sin x + cos x) 0 0
v o mèi quan hÖ gi÷a sin x + cos x v sin x cos x ta ®Æt t = sin x + cos x ⇒ dt = (cos x − sin x)dx v 2 2 2 t 2 − 1 1 − dt 1 dt 1 1 1 1 sin x cos x = khi ®ã F = ∫ = − 2 ∫ = = − 2 2 t − 1 + 1 ( 2 + 2 t) 2 t + 2t + 1 2 t + 1 2 2 2 2 1 1 + 1
C¸ch ®Æt 4. Dùa v o ®Æc ®iÓm hai cËn cña tÝch ph©n. a 0 a 0
NÕu tÝch ph©n cã d¹ng I = ∫f(x)dx th× ta cã thÓ viÕt I = ∫f(x)dx + ∫f(x)dx ®Æt t = H x ®Ó biÕn ®æi I f (x d ) x 1 = ∫ −a −a 0 −a π
NÕu tÝch ph©n cã d¹ng I = ∫f(x)dx th× ta cã thÓ ®Æt t = π H x 0 π 2
NÕu tÝch ph©n cã d¹ng I = ∫f(x)dx th× ta cã thÓ ®Æt t = 2π H x 0 π / 2 π
NÕu tÝch ph©n cã d¹ng I = ∫f(x)dx th× ta cã thÓ ®Æt t = H x 2 0 b
NÕu tÝch ph©n cã d¹ng I = ∫f(x)dx th× ta cã thÓ ®Æt t = (a + b) H x a VÝ dô 4. TÝnh: 1 0 1 1. I = ∫ 2008 x sin xdx ta viÕt I = 2008 x sin xdx + ∫ x 2008 sin xdx = A + B ∫
. Ta ®Æt t = Hx th× A = H B. vËy I = 0. −1 −1 0 π π π π 2. sin t t sin t
J = ∫ x sin x dx ta ®Æt t = π − x khi ®ã J = ∫ dt − dt ta ®æi biÕn tiÕp: 2 ∫ 1 + 2 cos x 1 + cos t 1 + 2 cos t 0 0 0 π π π sin t 2 cos t =tan u π t sin t t =−x 2 2 π π J = dt ==== ∫ v J = dt === J ∫ .VËy J = − J ⇒ J = 1 2 1 + cos2 t 2 1 + cos2 t 2 4 0 0
C¸ch ®Æt 4. NÕu tÝch ph©n cã chøa ax 2 + bx + c; a > 0 th× ta cã thÓ ®Æt t − ax = ax 2 + bx + c sau ®ã tÝnh x theo t
v tÝnh dx theo t v dt.{PhÐp thÕ ¬le} VÝ dô 5. TÝnh: 1 2 − 2 1. = ∫ dx 1 t 2dt I
ta ®Æt t − x = x 2 − x + 1 ⇒ x = ⇒ I = = ln 3 ∫ 2 2t + 1 2t − 1 0 x − x + 1 1 1 2 − 2 2 − 2. = ∫ dx t 1 dt 1 6 2 1 J
ta ®Æt t − 3x = 9x 2 − 2x + 1 ⇒ x = ⇒ J = = ln ∫ 2 ( 2 3t − ) 1 t 3 − 1 3 2 0 9x − 2x + 1 1
II )Ph−¬ng ph¸p tÝch ph©n tõng phÇn
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN b
HGi¶ sö cÇn tÝnh tÝch ph©n I = ∫f(x)dx . Khi ®ã ta thùc hiÖn c¸c b−íc t×nh: a b b
B−íc 1. ViÕt tÝch ph©n d−íi d¹ng: I = ∫f(x)dx = ∫g(x) h.(x)dx a a du = u = g(x) g' (x)dx B−íc 2. §Æt ⇒ dv = h(x) d . x v = ∫h(x) d.x b b
B−íc 3. ¸p dông c«ng thøc: hay ∫ u d.v = b v . u − d . v u a ∫ a a
C¸c c¸ch ®Æt ®Ó tÝch ph©n tõng phÇn: b du = P' (x)dx u = P(x)
+C¸ch ®Æt 1. NÕu tÝch ph©n cã d¹ng I = ∫ P(x).sin ax d.x th× ta sÏ ®Æt ⇒ cos ax dv = sin ax d . x v = − a a b du = P' (x)dx u = P(x)
NÕu tÝch ph©n cã d¹ng ∫ P(x).cosax d.x th× ta ®Æt ⇒ sin ax dv = cos ax d . x v = a a du = b P' (x)dx u = P(x) NÕu tÝch ph©n cã d¹ng ∫ ax P(x) e . d . x th× ta ®Æt ⇒ eax dv = eax d . x v = a a VÝ dô 5. TÝnh: π 1. I = ∫( x 3 − ) 1 .sin 2 d . x x ta ®Æt 0 du = d 3 x π π u = x 3 − 1 cos 2x 3 3π ⇒ cos 2x ⇒ I = − ( x 3 − ) 1 + cos 2x d . x = − ∫ dv = sin 2x d . x v = − 2 2 2 2 0 0 π / 2 u = x2 + du = 2xdx 2. 1 J = ∫ 2 (x + ) 1 .cos x d . x ta ®Æt ⇒ dv = cos d . x x v = sin x 0 π π 2 / 2 π + ⇒ / 2 4 2 J = (x + ) 1 sin x − 2 . x .sin x d . x = − 2J ∫ π ta tÝnh J . x sin d . x x b»ng c¸ch ®Æt 1 = ∫ 1 0 4 0 0 π / 2 u = x π 2 2 / 2 π + 4 π − 4
sau ®ã suy ra J = − x cos x + cos xdx =1 ∫ .VËy J = − 2 = dv = sin d . x x 1 0 4 4 0 1 3. L = ∫ 2 (x − x + 3x ) 1 e . d . x ta ®Æt 0 1 1 u = x 2 − x + 1 3 1 1 e − 1 1 ⇒ 2 3x 3x L = (x − x + ) 1 e − (2x − ) 1 e . d . x = − L ∫ 1 dv = e3x d . x 3 3 3 3 0 0 1 u = 2x − 1 3 − 3 − TÝnh tiÕp 4e 4 e 5 5 L (2x ) 1 e . d . x ®Æt ⇒ L = suy ra L = 1 = ∫ − 3x 1 dv = e3x d . x 9 27 0 π π π 1 cos 2x x 1 2 − π π 2 4. M = ∫ 2 ( x sin x) d . x ta viÕt M = ∫ x sin d . x x = ∫ x d . x = − ∫ x cos2xdx 2 4 2 0 0 0 0 0 π u =x 2 π xÐt M = x cos 2x d . x 0 ∫ === . vËy ta cã M = 1 4 dv=cos 2xdx 0 2 π / 4 π / 2 u = 2t
5. M = ∫sin x d.x ta ®æi biÕn t = x ®Ó ®−a M = ∫ 2tsin tdt b»ng c¸ch ®Æt ⇒ M = 2 dv = sin t d . t 0 0
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN du = b b cos bxdx u = sin bx
+C¸ch ®Æt 2. NÕu tÝch ph©n cã d¹ng I = ∫ ax e sin bx d . x th× ta ®Æt ⇒ eax dv = eax d . x v = a a du = − b b sin bxdx u = cos bx
NÕu tÝch ph©n cã d¹ng I = ∫ ax e cos bx d . x th× ta ®Æt ⇒ eax dv = eax d . x v = a a VÝ dô 6. TÝnh: π du = / 2 3 cos x 3 dx u = sin x 3 1. I = ∫ 2x e .sin 3 d . x x ta ®Æt ⇒ e 2x dv = e2x dx v = 0 2 π / 2 π π 2x π u = cos3x ⇒ e 3 e 3 2 x I = sin x 3 − e cos x 3 d . x = − − I ∫ (*). Ta xÐt I e cos 3 d . x x v ®Æt ⇒ 1 = ∫ 2x 1 2 2 2 2 dv = e2x dx 0 0 0 π / 2 π e2x 3 eπ 3 1 3 π 2e + 3 2 x 1 3 I = −cos 3x + e sin 3 d . x x = + I ∫ thay v o (*) ta cã: I = − − + I ⇒ I = − 1 2 2 2 2 2 2 2 2 13 0 0 π π 1 cos 2x 1 1 2x − π π 2. F = ∫ x 2 (e .sin x) d . x ta viÕt F = ∫ e d . x = ∫ 2x e d . x − ∫ 2x e cos 2 d . x x 2 2 2 0 0 0 0 π π 2π − 2π − Ta xÐt 1 1 2x e 1 2x e 1 F = e d . x = ∫
. Sau hai lÇn tÝch ph©n tõng phÇn ta tÝnh ®−îc F = e cos 2x d . x = ∫ . 1 2 2 2 2 4 0 0 π 2π − VËy ta cã: x 2 e 1 F = (e .sin x) d . x = ∫ 8 0 P' (x) b du = dx u = ln[P(x)]
+C¸ch ®Æt 3. NÕu tÝch ph©n cã d¹ng I = ∫ ln[P(x)]Q . (x)dx th× ta ®Æt ⇒ P(x) dv = Q(x) d . x a v = ∫ Q(x)dx VÝ dô 7. TÝnh: 1 5 du = dx 5 5 u = ln[x − ] 1 2 2 1. x − 1 x x I = ∫ . x ln(x − ) 1 dx ta ®Æt ⇒ ⇒ I = ln(x − ) 1 − ∫ dx dv = x d . x x 2 2 2x − 2 2 2 v = 2 2 48 ln 2 + 27 = 4 1 3 2 u = ln x + 1 + x du = dx 2. J = ∫ln(x + 1+ 2 x )dx ta ®Æt ⇒ 1 + x 2 ⇒ J = 3 ln( 3 + 2) −1 0 dv = dx v = x e e e e u = 2 2 3. ln x x K = ∫ 2 . x ln xdx ta ®Æt suy ra K = 2 ln x − ∫ x.ln xdx . XÐt K x.ln xdx v ®Æt 1 = ∫ dv = xdx 2 1 1 1 1 u = ln x e2 + 1 e 2 − 1 th× K = ⇒ K = . dv = xdx 1 4 4 2 u = ln x 2 e − 4. 1 1 5 15 4 ln 2 H = ∫ ln x dx ta ®Æt suy ra H = − ln x + x dx = ∫ − . 5 x dv = − x 5dx 4x 4 4 256 1 1 1 π / 3 5. G = ∫ ln(sin x) dx ®Æt 2 cos x π / 6 u = ln(sin x) π / 3 du = cot xdx π / 3 3 3l 3 n −4 3l 2 n −π 1 ⇒ ⇒ I = tan x ln(sin x) − dx = π / 6 ∫ dv = dx v = tan x 6 cos2 x π / 6
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN π π e sin(ln x) e u = cos(ln x) du = − π 6. dx e F = o c s(ln x)dx ∫ ®Æt ⇒ x ⇒ I = x cos(ln x) + sin(ln x)dx (*). Ta xÐt 1 ∫ dv = dx 1 v = x 1 π π e cos(ln x) e u = sin(ln x) du = dx π e F sin(ln x)dx ®Æt ⇒ ⇒ F = x sin(ln x) − cos(ln x)dx = −F ∫ thay 1 = ∫ x dv = dx 1 1 1 v = x 1 π π + v o (*) ta cã: e 1
F = −e − 1 − F ⇒ F = − . 2
II )Ph−¬ng ph¸p t×m hÖ sè bÊt ®Þnh AH Khi gÆp tÝch ph©n:
I = ∫ P(x) dx víi P(x), Q(x) l c¸c ®a thøc cña x. Q(x)
B−íc 1: NÕu bËc cña P(x) ≥ bËc cña Q(x) th× ta lÊy P(x) chia cho Q(x) ®−îc th−¬ng A(x) v d− R(x),
tøc l P(x) = Q(x).A(x) + R(x), víi bËc R(x) < bËc Q(x). Suy ra : P(x) R(x) = P(x) R(x) A(x) + ⇒ ∫ dx = ∫ A(x d ) x + ∫ dx ( Q x) ( Q x) ( Q x) ( Q x)
B−íc 2: Ta ®i tÝnh : I = ∫ R(x) dx , víi bËc R(x) < bËc Q(x). Q(x)
Cã thÓ x¶y ra c¸c kh¶ n¨ng sau : + +Kh¶ n¨ng 1: Víi R(x) M x . N
Q(x) = ax2 + bx + c ,( a ≠ 0 ) th× bËc R(x) < 2 ⇒ R(x) = M.x+N v = ( Q x)
ax2 + bx + c
TH1 : Q(x) cã 2 nghiÖm x , x , tøc l : Q(x) = a(x – x )(x – x ). 1 2 1 2 +
Chän h»ng sè A, B sao cho: R(x) M x . N A B = = + Q(x)
a(x − x )(x − x ) x − x x − x 1 2 1 2
TH2 : Q(x) cã nghiÖm kÐp x , tøc l : 2 = − . 0 Q(x) a(x x ) 0 +
Chän h»ng sè A, B sao cho: R(x) M x . N A B = = + 2 2 Q(x) a(x − x ) x − x (x − x ) 0 0 0
TH3 : Q(x) v« nghiÖm. Chän h»ng sè A, B sao cho: R(x) A ' Q . (x) B R(x) = A ' Q . (x) + B và = + ( Q x) Q(x) Q(x)
+Kh¶ n¨ng 2: Víi Q(x) = ax3 + bx2 + cx + d ,( a ≠ 0 ) th× bËc R(x) < 3
TH1: Q(x) cã 3 nghiÖm x ,x ,x . tøc l : Q(x) = a(x − x )(x − x )(x − x ) 1 2 3 1 2 3
Chän h»ng sè A, B, C sao cho: R(x) R(x) A B C = = + + Q(x)
a(x − x )(x − x )(x − x ) x − x x − x x − x 1 2 3 1 2 3
TH2: Q(x) cã 1 n ®¬n kÐp = − − 0 x , 1 n x , tøc l : 2 Q(x) ( a x x )(x x ) 1 0 0 1 0
Chän h»ng sè A, B, C sao cho: R(x) R(x) A B C = = + + 2 2 ( Q x) (
a x − x )(x − x ) x − x x − x (x − x ) 1 0 1 0 0
TH3: Q(x) cã mét nghiÖm x (béi 3), tøc l : 3
Q(x) = a(x − x ) 0 0
Chän h»ng sè A, B, C sao cho: R(x) R(x) A B C = = + + 3 2 3 Q(x) ( a x − x ) x − x (x − x ) (x − x ) 0 0 0 0
TH4: Q(x) cã ®óng mét nghiÖm ®¬n x , tøc l : Q(x) = (x − x ) a
( x2 + βx + γ ) (trong ®ã 2 ∆ = β − a 4 γ < 0 ). 1 1 R(x) R(x) A Bx +
Chän h»ng sè A, B, C sao cho: = = + C ( Q x) (x − x ) a ( x2 1
+ βx + γ ) x − x ax2 1 + βx + γ
+Kh¶ n¨ng 3: Víi bËc Q(x) >3 th× th«ng th−êng ta gÆp Q(x) l c¸c biÓu thøc ®¬n gi¶n nh−: x4 + 1 ; x4 ± x2 + 1 ; x6 + 1
VÝ dô 1. TÝnh c¸c tÝch ph©n:
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN 0 2 0 x + x + 4x − 1 − 1. 4x 1 A B I = ∫
1 dx ta viÕt I = ∫1+ dx v viÕt = + Sau ®ã chän ®−îc 2
x − 3x + 2 2 x − x 3 + 2 x 2 − x 3 + 2 x − 1 x − 2 −1 −1 A = H3; B = 7. Khi ®ã: 0
I = (x − 3ln x − 1 + 7 ln x − 2 ) = 10 ln 2 − 7 ln 3 . 1 − 1 2. = ∫ x J
dx ta viÕt x = A(x2 + x + 1)’ + B suy ra A = 1/2; B = H 1/2. VËy J = J + J víi 2 1 2 x + x + 1 0 1 1 1 1 d(x 2 + x + ) 1 1 1 dx 1 4 dx J = = ln 3 ∫ v J . 2 = − ∫ = − 2 ∫ 1 2 x 2 + x + 1 2 2 x + x + 1 2 3 2 0 0 2 1 0 x + + 1 3 2 π / 3 π Ta ®Æt 2 1 2 3 3 x + = tan u suy ra J = − du = ∫ . 2 3 2 3 9 π / 6 3 + 3. = ∫ 1 1 A Bx c K dx ta viÕt = +
sau ®ã chän ®−îc A = 1/3, B = H 1/3, C = 0. V× thÕ viÕt ®−îc 3 x + x 3 x 3 + 3x x x 2 + 3 1 3 3 1 x 1 K = dx − dx = ln 3 ∫ ∫
{V× ®−a ®−îc x v o trong vi ph©n}. 3x ( 3 x 2 + ) 3 6 1 1 4. β a sin x +
B – Khi gÆp tÝch ph©n I = ∫
b cos x dx (c, d ≠ 0) th× ta viÕt TS = A.(MS) + B.(MS)’ tøc l chän A, B sao cho: c sin x + d cos x α x 2t 2 1 − t asinx + b cosx = A
(csinx + dcosx) + B(csinx + dcosx)' hoÆc ®Æt t = tan ⇒ sin x = cos x = 2 2 1 + t 2 1 + t VÝ dô 1. TÝnh: π / 2 3sin x + 1. I = ∫ 5 cos x dx ta viÕt 3sinx + c 5 osx = A
(sinx + cosx) + B(cosx - sinx) suy ra A = 4; B = 1. sin x + cos x 0 π / 2 π / 2 + Khi ®ã: d(sin x cos x) / 2 I = d 4 x + = ∫ ∫ (4x + lnsinx + cosx )π = π2 sin x + cos x 0 0 0 π / 2 3sin x + 2. J = ∫
cos x dx ta viÕt 3sinx + cosx = A
(sinx + cosx) + B(cosx - sinx) suy ra A = 2; B = H1. (sin x + 3 cos x) 0 π / 2 π / 2 π / 2 + π Khi ®ã: 2 d(sin x cos x) 1 I = dx − = − cot(x + ) + = 2 ∫ ∫ (sin x + cos x)2 (sin x + cos x)3 4 2(sin x + cos x)2 0 0 0 β a sin x + bcos x +
C – Khi gÆp tÝch ph©n I = ∫
m dx (c, d ≠ 0) th× ta viÕt TS = A.(MS) + B.(MS)’ + C. Chän A, B,C sao cho: c sin x + d cos x + n α asinx + b cosx + m = A
(csinx + dcosx + n) + B(csinx + dcosx + n)'+C hoÆc cã thÓ ®Æt x 2t 2 1 − t t = tan ⇒ sin x = cos x = 2 2 1 + t 2 1 + t VÝ dô 1. TÝnh: π / 2 7sin x − cos x + 1. I = ∫
7 dx ta viÕt 7sinx − cosx + 7= A (4sinx + c 3 osx + ) 5 + B(4cosx - 3sinx) + C 4 sin x + 3cos x + 5 0 π / 2 π / 2 π d(4 sin x + 3 cos x + / 2
Khi ®ã A = 1; B = H1; C = 2 v I = ∫dx − ∫ ) 5 + ∫ 2 dx dx 4 sin x + 3cos x + 5 4 sin x + 3 cos x + 5 0 0 0
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN π / 2 2 − XÐt 2 x 2t 1 t I dx ®Æt t = tan ⇒ sin x = cos x = suy ra 1 = ∫ 4 sin x + 3cos x + 5 2 2 1 + t 2 1 + t 0 1 1 1 π / 2 π 1 9 I = 2 dt = ∫
. VËy I = (x − ln 4sin x + 3cos x + 5 ) + I = + − ln 1 1 (t + ) 2 2 3 0 2 3 8 0
V)Ph−¬ng ph¸p dïng tÝch ph©n liªn kÕt VÝ dô 1. TÝnh: π 2 π 2 π 1. cos xdx
I = ∫ sin xdx ta xÐt thªm tÝch ph©n thø hai: J = ∫ Khi ®ã: I + J = (*). sin x + cos x sin x + cos x 2 0 0 π 2 π 2 − + π MÆt kh¸c (sin x cos x)dx d(sin x cos x) I − J = = − = 0 ∫ ∫
(**). Gi¶I hÖ (*) v (**) suy ra I = J = . sin x + cos x sin x + cos x 4 0 0 π 2 π n 2 n π 2. sin x cos x I = dx ∫ ta xÐt J = dx ∫ . Khi ®ã: I + J = (*) n n n n sin n x + cosn x sin n x + cosn x 2 0 0 π π 2 π 2 n n π MÆt kh¸c nÕu ®Æt x = H t th× cos t cos x I = dt = dx = J ∫ ∫
(**). Tõ (*), (**) ta cã I = 2 n n n n n n n sin t + cos t sin x + cos x 4 0 0 π 2 π 2 n n π 3. sin x cos x I = dx ∫ t−¬ng tù xÐt J = dx ∫ v suy ra I = J = n n n n n sin x n + cos x n sin x n + cos x 4 0 0 π 6 π π 2 6 2 6 4. sin x cos x 1 1 E = dx ∫ v F = dx ∫ ta cã E + F = dx = ln 3 ∫ (*) sin x + 3 cos x sin x + 3 cos x sin x + 3 cos x 4 0 0 0 π 6 − L¹i cã 1 1 3 E − F 3 = (sin x − 3 cos x d ) x = 1 − 3 ∫
(**). Gi¶I hÖ (*), (**) ta ®−îc: E = ln 3 − v 16 4 0 π 6 3 1 − 3 cos 2x 1 1 − 3 F = ln 3 + . Më réng tÝnh E = dx = F − = ∫ E ln 3 + 16 4 sin x + 3 cos x 8 2 0 π 6 §Ò xuÊt cos 2x L = dx ∫ sin x − 3 cos x 0 C¸c ¸ c b i t o¸n ¸ n t− t ¬ − ng n tù t . ù
A – Ph−¬ng ph¸p biÕn ®æi trùc tiÕp 1
+ B×nh ph−¬ng v ph©n tÝch th nh 2 ph©n sè ®¬n gi¶n. 1 ( + x 2
1. [§HNNI.98.A] M = ∫ e ) dx + BiÕt ®æi biÕn. 1 + 2x e 0 1 1 1 + 1 2x x x Gi¶i: e 2 dx M = ∫ e dx + e 2 dx ta tÝnh M
®Æt ex = tan t,t ∈ (− π / ;
2 π / 2) khi ®ã víi tan α =e v 1 = ∫ 2x ∫ 1 + e 1 + 2x e 1 + 2x e 0 0 0 α α α 2 tan tdt α 1 1 + e 2 M = 2 tan tdt = −2 ln cos t = 2 − ln = ln ∫ 1 = ∫ 1 ( + 2 2 tan t) cos t π / 1 + tan 2 4 t 2 π / 4 π / 4 π / 4 π 2 + 3
2. [§HTCKT.97] ∫ 3sin xdx 1 + cos x 0 Gi¶i: π 2 + 3
2. [§HTCKT.97] ∫ 3sin xdx 1 + cos x 0 Gi¶i:
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN π 2 + 3
2. [§HTCKT.97] ∫ 3sin xdx 1 + cos x 0 Gi¶i: π 2 + 3
2. [§HTCKT.97] ∫ 3sin xdx 1 + cos x 0 Gi¶i: 1 1 ( + x 2 1. [§HNNI.98.A] ∫ e ) dx 1 + 2x e 0 π 2 3
2. [§HTCKT.97] ∫ 3sin xdx 1 + cos x 0 π 2 3. [§HBK.98] ∫ 4 cos 2 ( x sin x + 4 cos x d ) x 0 2 4. [§HDL§.98] ∫ dx
x + 1 + x − 1 1 π 6 5. ∫ 1 dx π 0 cos x. cos(x + ) 6 2 e
6. ∫ ln x + ln(ln x) dx x e π 3 7. [§HMá.00] ∫ 1 dx π π 6sin x sin(x + ) 6 π 3 8. ∫ 4 cos x sin 2xdx 0 π 4 9. [§HNN.01] ∫ sin 4x dx 6 sin x + 6 cos x 0 π 2 6
10. [§HNNI.01] ∫ cos x dx 4 sin x π 4 π 3 11. ∫ 4 tg xdx π 4 3
12. [C§GTVT.01] ∫ 2 x + 3x d . x −2 3
13. [C§SPBN.00] ∫ 2
x − 4x + 4dx 0 π 14. ∫ cosx sinxdx 0 π 3 15. ∫ 2 tg x + 2
cot g x − 2dx π 6
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN 3 16. ∫ 3 x − 2 2x + xdx 0 1
17. ∫ 4 − x dx §¸p: ( 2 5 − 3 ) −1 1
18. ∫ x − xdx 2 2 − 3 1 5
19. ∫(x + 2 − x − 2 )dx 8 −3 3 2
x + x − 1 20. ∫ d . x 2
x + x − 2 0 π
21. ∫ 2 + 2cos2xdx 4 0 π
22. ∫ 1 − sin2xdx 2 2 0 π / 2
23. ∫ 1 + sinxdx 4 2 0 1
− m + 1 / 2 ~ m ≤ 0
24. ∫ x − ad ; x a ∈ R
m 2 − m + 1 / 2 ~ 0 < m ≤ 1 0 2
a ≥ 2 th× ®s: (3a – 5)/6; 25. ∫ 2
x − (a + ) 1 x + a d ; x a ∈ R
1 < a < 2 th× ®s: (aH1)3/3 – (3a H 5)/6 1
a ≤ 1 th× ®s: (5 – 3a)/6 π 2 3 26. ∫ cos x dx 1 + cos x 0 π 2
28. [§H.2005.D] ∫ sinx (e + cos x) cos xdx 0 2 28. [§H.2003.D] ∫ 2 x − x dx 0 π / 4 1 − 2 29. [§H.2003.B] ∫ 2 sin x dx 1 + sin 2x 0 π 2
29. M = ∫ (sin6 x + cos6 x − sin2 x.cos2 x)d.x 0 π 4 2 30. sin x N = d . x ∫ cos8 x 0 31.
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN
B – Ph−¬ng ph¸p ®æi biÕn 1 3 1. [C§BN.01] ∫ x dx HD ®Æt fsf 1 ( + 2 3 x ) 0 1
2. [PVB.01] ∫ 3 x 1 − 2 x d . x 0 ln 3 3. ∫ dx x e + 2 0 π 2
4. [C§XD.01] ∫ sin2x dx 1 + 2 cos x 0 1 1 5. [§HKTQD.97] ∫ 5 x 1 ( − 3 6 x ) dx §Ò xuÊt: ∫ 2 x 1 ( − 7 x) dx 0 0 1
6. [§HQG.97.B] ∫ dx 1 + x 0 2
7. [§H.2004.A] ∫ xdx
1 + x − 1 1 2 3 ``8. [§H.2003.A] ∫ dx 2 x x 4 5 + π
9. [§HSPHN.00.B] ∫ 2 2 x a − 2 x dx 0 ln 2 2x
10. [§HBK.00] ∫ e dx x 0 e + 1 23 11. ∫ dx
x + 8 − 5 x + 2 14 π 2 12. ∫ sin2x dx 2 2 sin x 0 ( + ) π 4 13. ∫ dx 3 cos x 0 6 14. ∫ 1 dx
2x + 4x + 1 2 3 15. ∫ 5 x 1 + 2 x dx 0 2 e
16. ∫ ln x + ln(ln x) dx x e 4
17. ∫ x + 1 dx x 2 1 3 10x + 2 3 x + 1 + 18. ∫ 10x dx 2 2 0 1
( + x ) x + 1
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN e 19. ∫ 1 dx 2 1 x 1 − ln x e 20. ∫ lnx dx
x 1 + ln x 1 4 21. ∫ 1 dx
3 2x + 1 + 2x + 1 0 4 23. ∫ dx 2 x. x 9 7 + 7 24. ∫ 1 d . x x + x + 2 2 1 25. ∫ dx 2 2 1 3x 0 ( + ) π 2 x + 26. [GTVT.00] cos x ∫ dx 4 − 2 sin x −π 2 π
27. [§HAN.97] ∫ xsinxdx 1 + 2 cos x 0 π 2 28. [§HLN.00] ∫ 1 2 + dx + sinx + cosx 0 π 4 29. [§HH§.00] ∫ 1 1 + dx + tgx 0 π 4
30. [§HVH.01] ∫ sinxcosx sin 2x + dx + cos2x 0 π 2 3
31. [HVBCVT.98] ∫ sinxcos xdx 1 + 2 cos x 0 1 32. = ∫ 1 I dx 2 (x + 2 ) 1 −1 (1+ 5 ) 2 2 x + 33. [§HTN.01] 1 ∫ dx 4 x − 2 x + 1 1 1 34. [§HTCKT.00] ∫ x dx 4 x + 2 x + 1 0 1 35. [HVKTQS.98] ∫ dx 2 − 1 ( x 1 x ) 1 + + + 1
36. [PVB¸o.01] ∫ 3 x 1 − 2 x d . x 0 e
37. [§H.2004.B] ∫ 1+ 3ln x ln x dx x 1 π 2sin 2x + 38. [§H.2005.A] ∫ sin x dx 1 3 cos x 0 +
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN π 2 39. [§H.2006.A] ∫ sin 2x dx 2 2 cos x 4sin x 0 + π 2
40. [§H.2005.B] ∫ sin 2xcos x dx 1 + cos x 0 ln 5 41. . [§H.2005.B] ∫ dx x e + −x e 2 − 3 ln 3 2 3 42. [§H.2003.A] ∫ dx 2 x x 4 5 + 2 43. [§H.2004.A] ∫ x dx 1 x 1 1 + − π / 6 4
44. [§H.2008.A] ∫ tan x dx cos 2x 0 π / 4 45. [§Ò thi thö §H] ∫ sin 4x dx 6 sin x + 6 cos x 0 e
46. [§Ò thi thö §H] ∫ 1 2
+ x .ln x d.x x. 1 + ln x 1 + 4
47. [§Ò thi thö §H] ∫ 2x+1 e dx 0 π 2 2 ( 2 t − d ) 1 t 1
48. [§Ò thi thö §H] ∫ sin2x = + ⇒ ∫ = ∫ dx HD: §Æt t 1 sin x 2t 3 8 1 ( 2 + 3 sin x) 1 0 8 2 49. I ∫ x − = 16 dx x 4 4 x − 1 + 50. J = ∫ x + 1 dx x 2 1 3 10x + 2 3 x + 1 + 51. K = ∫ 10x dx 2 2 0 1
( + x ) x + 1 − ln 2 x 52. = ∫ e H dx 2x −ln 2 1 − e ln 3 53. = ∫ dx G 2x 0 e + 1 π 2 54. = ∫ dx F
3 + 5 sin x + 3 cos x 0 π 3 55. D = ∫ cos x 2 dx 4 2 0
cos x − 3 cos x + 3 ln 5 x x e e − 56. S = ∫ d 1 x x e + 3 0 π 57. = ∫ dx T π
2 + sin x − cos x 2
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN 4 58. = ∫ dx R 2 x. x 9 7 + 7 59. = ∫ 1 E d . x x + x + 2 2 4 2x + 60. W = ∫ 1 d . x 1 + 2x 1 0 + + π / 2 1 t = 3 tan u π 61. = ∫ dx x dt 3 Q
§Æt t = tan th× Q = ∫ = ∫ === 2 + cos x 2 ( 3 )2 + t 2 9 0 0 2 62. = ∫ dx M 2 0
− 3x + 6x + 1
C – Ph−¬ng ph¸p tÝch ph©n tõng phÇn 1 1. [§HC§.97] ∫ 1 ( + 2 2x x) e dx 0 π 4 2. [§HTCKT.98] ∫ 2 ( x 2 cos x − d ) 1 x 0 2 10 3. ∫ 3 2 x ln(x + d ) 1 x v ∫ 2 xlg xdx 0 0 e 4. [PVB¸o.98] ∫ 2 (x ln x) dx 1 π 5. [HVNH.98] ∫ 2 xsin xcos xdx 0 2
6. [§HC§.00] ∫ ln(x + d ) 1 x 2 x 1 π 4
7. [§HTL.01] ∫ ln 1 ( + tgx d ) x 0 π 2 8. ∫ 2 xtg xdx 0 3
9. [§HYHN.01] ∫ 2 x − 1 d . x 2 2
10. [§Ò thi thö] ∫ xln(3x − 2 x d ) x 1 e
11. [§H.2007.D] ∫ 2 2 x ln xdx 1 1
12. [§H.2006.D] ∫ (x − 2x 2 e ) dx 0 0 13. I = ∫ 2x x e ( + 3 x + 1 d ) x −1 2 e 14. J ∫ lnx + = ln(ln x) dx x e
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CÁC PHƯƠNG PHÁP TÍNH TÍCH PHÂN π 15. K = ∫ 2 (x sin x) dx 0 1 16. = ∫ x H dx 2 sin (2x + ) 1 0 π 4
17. G = ∫ x.sin x dx 3 cos x 0 ln 2 18. 2 F = ∫ 5 x x e . dx 0 4 ln( x + 19. D = ∫ ) 1 dx x + x 1 e
20. S = ∫ lnx 2 + ln xdx x 1 + ln x 1 2 π 21. A = ∫ 2 cos x d . x 2 π / 4 π
22. P = ∫ (e .cosx)2 x dx 0 2 π 23. U = x . sin x d . x ∫ 0 π 24. Y = x.si x n . cos 2 x d . x ∫ 0 π / 3 1 + §Æt sin x u = sin x ;dv = ..XÐt T
e dx v ®Æt tptp suy ra 1 = ∫ x π / 3 sin x + 1 + 1 + cos x cos x 25. 0 T = ∫ sin x x sin x + e dx + cos x π / 3 x 3 π 0 e sin x e T = = 1 + cos x 3 0 1 26. 2 + ln 1 ( + 2 )
R = ∫ 1 + 2 x dx §s: 2 0 1 27. 1 E = ∫ 2 x ln(x + d ) 1 x §S: ln 2 − 2 0 π / 2 28. π
W = ∫ cos x ln 1 ( + cos x d ) x §s: − 1 2 0 e 29. e 2
Q = ∫ ln x dx §s: (x + 2 ) 1 e + 1 1 / e π / 2 π / 3 ViÕt M = M sin x 3 x 1 + M2. Víi M = dx ∫ = ∫ ln & M dx . 2 = ∫ 1 1 + cos x 2 1 + cos x π / 3 π / 6 π / 2 u = du = x + = x dx 30. M = ∫ sin x − π §Æt ⇒ ⇒ (3 2 3 ) = − 1 + dx + cos x 1 x M ln 4 2 π dv = v = / 3 = dx 2 cot 3 1 + cos x 2 − π VËy (3 2 3 ) 3 M = + ln 3 8
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