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T m tài li u To n ? Chuy n nh - www.toanmath.com TÍCH PHÂN I Khái ni m tích phân Di n tích hình thang cong • Gi i thi u cho h c sinh v cách tính di n tích c a m t hình thang cong • T suy cơng th c : xlim →x S ( x ) − S ( x0 ) = f ( x0 ) x − x0 nh ngh a tích phân • Cho hàm f liên túc m t kho ng K a, b hai s b t k thu c K N u F m t nguyên hàm c a f K hi u s : F(b)-F(a) đ c g i tích phân c a f t a đ n b , ký hi u : b ∫ f ( x)dx a • Có ngh a : b F (b) − F ( a ) )dx ∫ f ( x= a b a • G i F(x) m t nguyên hàm c a f(x) F ( = x) F ( b ) − F ( a ) : b )dx ∫ f ( x= a b F ( x= ) F (b) − F ( a ) a • Trong : - a : c n , b c n d i - f(x) g i hàm s d i d u tích phân - dx : g i vi phân c a đ i s -f(x)dx : G i bi u th c d i d u tích phân II Tính ch t c a tích phân Gi s cho hai hàm s f g liên t c K , a,b,c ba s b t k thu c K Khi ta có : a ∫ f ( x)dx = a b ∫ f ( x)dx = −∫ f ( x)dx ( G a b ∫= f ( x)dx a b a i tích ch t đ i c n ) b c ∫ a b f ( x)dx + ∫ f ( x)dx c b b a a ∫ [ f ( x) ± g ( x)] dx = ∫ f ( x)dx ± ∫ g ( x)dx ( Tích phân c m t t ng ho c hi u hai tích a phân b ng t ng ho c hi u hai tích phân ) b b a a ∫ kf ( x)dx = k.∫ f ( x)dx ( H ng s k d u tích phân , có th đ a ngồi d u tích phân đ c ) Ngồi tính ch t , ng i ta ch ng minh đ N u f(x) ≥ 0∀x ∈ [ a; b ] : c m t s tính ch t khác nh : b ∫ f ( x)dx ≥ 0∀x ∈ [ a; b] a b b a a N u : ∀x ∈ [ a; b ] : f ( x) ≥ g ( x) ⇒ ∫ f ( x)dx ≥ ∫ g ( x)dx ( B t đ ng th c tích phân ) T m tài li u To n ? Chuy n nh - www.toanmath.com N u : ∀x ∈ [ a; b ] v i hai s M,N ta ln có : M ≤ f ( x) ≤ N Thì : b M ( b − a ) ≤ ∫ f ( x)dx ≤ N ( b − a ) ( Tính ch t giá tr trung bình c a tích phân ) III CÁC PH a NG PHÁP TÍNH TÍCH PHÂN A PH NG PHÁP PHÂN TÍCH 1.Trong ph ng pháp , c n : • K n ng : C n bi t phân tích f(x) thành t ng , hi u , tích , th ng c a nhi u hàm s khác , mà ta có th s d ng đ c tr c ti p b ng nguyên hàm c b n tìm nguyên hàm c a chúng • Ki n th c : Nh trình bày ph n " Nguyên hàm " , c n ph i n m tr c ki n th c v Vi phân , công th c v phép toán l y th a , phép toán c n b c n c a m t s bi u di n chúng d i d ng l y th a v i s m h u t Ví d áp d ng Ví d 1: Tính tích phân sau a/ ∫ x2 + 1 c/ ∫ ( ( x 1+ x ) ( x x4 −1 + a/ ∫ dx = + x 2 ⇒ ∫ x − 1d 1 ) x2 ∫ ( x + 1) b/ x x − x + ln + x Gi i ) dx ( x x4 −1 + dx ) dx ∫ d/ x3 + x − x + dx x4 − x2 +  2x x2 −1 x2 +  x  x  dx + = −2 x+ x     dx ∫1  ∫ 2  + + + 1 x x x  1   2 2 x2 −1 + ∫ d x2 + = x2 −1 + x2 + = + − 1 2 ) ( ) ) ( ( b/ ( x + − 1) ∫0 ( x + 1)3 dx =∫0 ( x + 1)3 1 x2   ( x + 1)2 x +1  1  − + = − + dx =∫  dx 2   3 ∫0  x + ( x + 1)2 ( x + 1)3  dx + + + 1 x x x ( ) ( ) ( )       1 1 d ( x + 1) d ( x + 1) d ( x + 1) 1 1 ln ⇒ I= ∫ − 2∫ + = + + − = ln + x ∫ x +1 x + ( x + 1) 0 ( x + 1) ( x + 1) c/ ∫ ( x x − x + ln + x ( x 1+ x ∫( ⇒= I d/ ∫ ) ( ln (1 + x ) +d (1= x ) ) 1+ x (1 + ) − ln x3 + x − x + dx = x4 − x2 + Trang ( ( )  ln + x   x −1 + =  dx− ∫1  + x 1+ x x    x − dx + ∫ = − + ln ) )= dx 2 −3 ) ( x )+   + ∫1   ( 3 x  + ln 2= 1  ( x3 − x ) dx  ∫  x − x +  + 2  ∫ ( x − 1) dx + 2 ∫ (x 2dx − 1) ) x 2 ( x Gv Ph m Minh T - 0968.469.299 )1 ( )   dx 1+ x x   ln + x ( ) T m tài li u To n ? Chuy n nh - www.toanmath.com = ∫ d ( x − x + 1) + ( x − x + 1) = ln ( x − 1) 2  1 1   ∫  x − − x +  dx + ∫  x − − x +  dx 2 2 x −1  x −1  + ln + − − − ln = x +  2 x +1 2  x −1 x +1 Ví d Tính tích phân sau π a/ ∫ 2sin x ( sin x − 1) π + cosx b/ dx ∫ 2sin sin x dx x + 3cos x π  2+ x c/ ∫ ln   dx 4− x  2− x  −1 ∫ b/ x2 −1 ∫1 x ( x + 1) dx Ví d Tính tích phân sau e2 a/ ln x + ∫e x ln x dx s inx+ 1+tanx dx cos x d/ π π + sin x c/ ∫ dx sin 2 x π d/ ∫ sin 3x.cosxdx B PH NG PHÁP I BI N S I Ph ng pháp đ i bi n s d ng tính tích phân d ng , ta c n th c hi n theo b c sau 1/ Quy t c : • B c 1: t x=v(t) • B c 2: Tính vi phân hai v đ i c n • B c 3: Phân tích f(x)dx=f(v(t))v'(t)dt • B • B b v (b ) a v(a) c 4: Tính ∫= f ( x)dx = ∫ g (t )dt G(t ) c 5: K t lu n : I= G (t ) v(b) v(a) v(b) v(a) 2/ Nh n d ng : ( Xem l i ph n nguyên hàm ) * Chú ý : a Các d u hi u d n t i vi c l a ch n n ph ki u thông th D u hi u a2 − x2 x2 − a2 ng : Cách ch n π π  =  x a sin t ↔ − ≤ t ≤  =  x a cost ↔ ≤ t ≤ π a   π π x ↔ t ∈ − ;  = sin t  2   a π  x ↔ t ∈ [ 0; π ] \   = cost 2  Gv Ph m Minh T - 0968.469.299 Trang T m tài li u To n ? Chuy n nh - www.toanmath.com   π π =  x a tan t ↔ t ∈  − ;      x a cot t ↔ t ∈ ( 0; π ) = a2 + x2 a+x a−x ∨ a−x a+x x=a.cos2t x=a+ ( b − a ) sin t ( x − a )( b − x ) b Quan tr ng nh t em ph i nh n d ng : - Ví d : Trong d ng phân th c h u t : β β β 1 *∫ dx ( ∆ < ) ∫ dx = = 2     α ax + bx + c α b  −∆ a  x+  +     2a   2a     b −∆ V= i :  u x+ ,k , du dx  = = 2a 2a   * áp d ng đ gi i toán t ng quát : β * β ∫ α + 2x − x dx = ∫ α ( 3) − ( x − 1) β ∫ α 1 du ∫ aα u +k dx (a +x ) 2 k +1 (k ∈ Z ) dx T suy cách đ t : x − = sin t 3/ M t s ví d áp d ng : Ví d 1: Tính tích phân sau a/ ∫ − x dx b/ 0 Gi i a/ ∫ 1− 2x c/ dx ∫ π π t x=sint v i : t ∈  − ;   2  x = ↔ sin t = → t = • Suy : dx=costdt :  π  x = ↔ sin t = → t = • Do : f(x)dx= − x dx =1 − sin tcostdt=cos 2tdt π • V y: b/ t:x= 0 ∫ f ( x)dx = ∫ (1 + cos2t ) dt = =+ (1 cos2t ) dt π 1 1  π  π −1   t + sin 2t  =  −  = 2  2 2  π π sin t t ∈  − ;   2  x=0 ↔ sint=0 → t=0 • Suy : dx = costdt ⇒  1 π t →t sin =  x= ↔= 2  Trang Gv Ph m Minh T - 0968.469.299 + x − x2 dx T m tài li u To n ? Chuy n nh - www.toanmath.com • Do : ∫ = dx − x2 π 1 = dx    − x  2 ∫ π ∫ π 1 = = costdt = ∫ dt 20 − sin t 2 t 2 c/ Vì : + x − x =4 − ( x − 1) Cho nên : π π x −1 ( *) 1−1   x = ↔ sin t = = → t =  π • Suy : dx= costdt :  ⇒ t ∈ 0;  → cost>0  6  x = ↔ sin t = − = → t = π  2 1 • Do : f(x)dx= = cos tdt dt dx dx = = 2 + 2x − x (1 − sin t ) − ( x − 1)   t:= sin t x − 2sin t t ∈  − ;  ↔ =  2 • π • V y: dx ∫= dt ∫ f ( x)= π π t= 6 Ví d 2: Tính tích phân sau a/ c/ ∫ 1 dx + x +1 ∫x b a − x2 12 x − x − 5dx b/ ∫2 x − x + dx * Chú ý : d/ ∫ tính tích phân d ng có ch a ( ( a + x2 ) dx ) x + a , a − x , ta s d ng ph ng pháp đ i bi n s : u(x)=g(x,t) Ví d : Tính tích phân sau 1 ∫ x2 + dx Gi i : • t: x2 + = x − t ⇒ x = t −1 2t  x =0 → t =−1; x =1 → t =1 − • Khi :  t2 +1 dx = 2t  • Do v y : ∫ 1− 1− −2t t + dt 1− dt − = − = ln t = ln dx = ∫ ∫ t + 2t t −1 x +1 −1 −1 Ví d 2: Tính tích phân : I = ( ) ∫x − x dx Gi i • t : t=sinx , suy dt=cosxdx x=0,t=0 ; Khi x=1 , t= Gv Ph m Minh T - 0968.469.299 π Trang π 2 T m tài li u To n ? Chuy n nh - www.toanmath.com • Do : f(x)dx= x − x dx= sin t − sin tcostdt=sin 2t cos=2 tdt π π 12 • V y : I= ∫ f ( x)dx = −∫ (1 cos4t= )−dt  t 80 8  sin = = 4t   II 1π 82  − cos4t    dt 4  π 16 i bi n s d ng Quy t c : ( Ta tính tích phân b ng ph ng pháp đ i bi n s d ng theo b sau : ) • B c 1: Khéo léo ch n m t hàm s u(x) đ t b ng t : t=u(x) • B c 2: Tính vi phân hai v đ i c n : dt=u'(x)dx • B c 3: Ta phân tích f(x)dx = g[u(x)]u'(x)dx = g(t)dt • B b u (b ) a u (a) c 4: Tính ∫= f ( x)dx = ∫ g (t )dt G(t ) • K t lu n : I= G (t ) Nh n d ng : u (b) u (a) u (b) u (a) TÍCH PHÂN HÀM PHÂN TH C H U T β P( x) dx α ax+b A D NG : I= ∫ * Chú ý đ n công th c : ( a ≠ 0) β m β β m dx = ln ax+b Và n ∫ α a α ax+b b ng ta chia t cho m u d n đ n β c β u b c c a P(x) cao h n ho c β P( x) m ∫α ax+b dx =α∫ Q( x) + ax+b dx =α∫ Q( x)dx + mα∫ ax+b dx Ví d : Tính tích phân : I= x3 ∫1 x + dx Gi i Ta có : f ( x=) Do : 3 27 x x − x+ − = 2x + 8 2x + 27  27 13 27 x3 1 1 3 2 ln = − + − = x − x + x − x + − − ln 35 dx x x dx    1= ∫1 x + ∫1  8 x +   8 16 16  Ví d 2: Tính tích phân : I= ∫ x2 − dx x +1 Gi i x −5 = x −1 − x +1 x +1  x2 −  1  dx= ∫  x − −  dx=  x − x − ln x +  = x +1 x +1 2  5 Ta có : f(x)= Do : ∫ Trang Gv Ph m Minh T - 0968.469.299  +1 − + ln     T m tài li u To n ? Chuy n nh - www.toanmath.com B D NG : β ∫ α ax P( x) dx + bx + c Tam th c : f ( x) = ax + bx + c có hai nghi m phân bi t β Cơng th c c n l u ý : β u '( x) dx = ln u ( x) ∫ α α u ( x) Ta có hai cách Cách 1: ( H s b t đ nh ) Cách 2: ( Nh y t ng l u ) Ví d 3: Tính tích phân : I= ∫x x + 11 dx + 5x + Gi i Cách 1: ( H s b t đ nh ) A ( x + 3) + B ( x + ) A B x + 11 x + 11 = = + = x + x + ( x + 2)( x + 3) x + x + ( x + 2)( x + 3) Ta có : f(x)= Thay x=-2 vào hai t s : 3=A thay x=-3 vào hai t s : -1= -B suy B=1 + x+2 x+3 1 x + 11   V y: ∫ = ∫ + = dx  dx x + 5x + x+2 x+3 0 Do : f(x)= ( 3ln x + + ln x + ) =0 ln − ln Cách 2: ( Nh y t ng l u ) ( x + 5) + 2x + 2x + 1 = 2 + = 2 + − x + 5x + x + x + ( x + )( x + 3) x + 5x + x + x + Ta có : f(x)= Do : I= ∫ f ( x)dx =  ∫  x 2x + + + 5x +   = dx  ln +x +5 x +6 x+3  − x+2 ln x+2  = x +  ln − ln 2 Tam th c : f ( x) = ax + bx + c có hai nghi m kép Công th c c n ý : Thông th β ∫ α β u '( x)dx = ln ( u ( x) ) α u ( x) ng ta đ t (x+b/2a)=t x3 Ví d : Tính tích phân sau : I= ∫ dx x + 2x +1 Gi i Ta có : ∫x 3 x x dx = ∫ dx + 2x +1 + x ( ) t : t=x+1 suy : dx=dt ; x=t-1 : x=0 t=1 ; x=3 t=4 Do : ∫ ( x + 1) x3 dx= ∫ ( t − 1) t dt=  Ví d 5: Tính tích phân sau : I= 1 1 1 dt=  t − 3t + ln t +  = ln −  t1  2 ∫  t − + t − t ∫ 4x 4x dx − 4x +1 Gv Ph m Minh T - 0968.469.299 Trang T m tài li u To n ? Chuy n nh - www.toanmath.com Gi i 4x 4x Ta có : = x − x + ( x − 1)2  x =0 ↔ t =−1 dt ;  x = ↔ t = 1 1 ( t + 1) 4x 4x 1 1   Do : ∫ dx = dx dt = = − +  dt =−  ln t  = 2 ∫ ∫ ∫ t t t  t  −1 4x − 4x +1  −1 −1  0 ( x − 1) t : t= 2x-1 suy : dt = 2dx → dx = Tam th c : f ( x) = ax + bx + c vô nghi m : b  u= x +  P( x) P( x) 2a  Ta vi t : f(x)= = ; 2 2  b   −∆   a ( u + k ) k = −∆ a  x +  +     2a 2a   2a     Khi : t u= ktant Ví d 6: Tính tích phân : I= ∫x x dx + 4x + Gi i • Ta có : ∫x 2 • x x dx = ∫ dx + 4x + ( x + 2) + t : x+2=tant , suy : dx= ↔ tan t = x = dt ; ⇒  2 ↔ tan t = cos t x = tan t − dt  sin t  • Do : ∫ dx = = ( ln cost  − −  dt−= 2 ∫ ∫ + tan t cos t t1  cost  t1 ( x + 2) + t2 x t  2  tan t = ↔ + tan t = ↔ cos t = → cost1 = T :  2 = ↔ + = ↔ = → cost tan tan 17 os t t c t  17  t ( ln cost • V y : ( − ln cost − 2t ) −2 = − 2−t2 ) t1 2t ) t2 (1) t1 = 17 ( ln cos− t cost cost1 −2= t1 )  ln+ 17 cost − ( t2 t1 ) cost1 • ⇔ − ln = = 2−( arctan4-arctan2 ) + ( t2 − t1 ) 2−( arctan4-arctan2 ) ln Ví d 7: Tính tích phân sau : I= ln 17 x3 + x + x + dx ∫0 x2 + Gi i x + 2x + 4x + = x+2+ 2 x +4 x +4 2  x + 2x + 4x + dx  1 2 • Do : ∫ = + J (1) dx = ∫  x + +  dx =  x + x  + ∫ 2 x +4 x +4 x +4 2  0 • Ta có : 2 Tính tích phân J= ∫x Trang dx +4 Gv Ph m Minh T - 0968.469.299 T m tài li u To n ? Chuy n nh - www.toanmath.com • t : x=2tant suy : dx = x = → t =   π dt ;  π ↔ t ∈ 0;  → cost>0 cos t  x = → t =  4  π • Khi : π 1 14 dt t = = = dx dt 2 ∫0 x += ∫ ∫ 4 + tan t cos t 20 • Thay vào (1) : I= + C D NG : π β ∫ α ax π π P( x) dx + bx + cx + d a th c : f(x)= ax + bx + cx + d ( a ≠ ) có m t nghi m b i ba Công th c c n ý : β ∫ α x m dx = Ví d 8: Tính tích phân : I= 1 β m −1 1− m x α x ∫ ( x + 1) dx Gi i Cách 1: • t : x+1=t , suy x=t-1 : x=0 t=1 ; x=1 t=2 t −1 1 1  1 12 • Do : ∫ dx = ∫ dt = ∫  −  dt =  − +  = t t t   t 2t 1 ( x + 1) 1 x Cách 2: 1) − ( x += 1 − 3 ( x + 1) ( x + 1) ( x + 1) x • Ta có : = ( x + 1)   1  1 1 = − − = + dx dx    2 ∫0 ( x + 1)3 ∫0  ( x + 1)2 ( x + 1)3   x + ( x += )     x4 Ví d : Tính tích phân : I= ∫ dx − x ( ) −1 • Do : • x Gi i t : x-1=t , suy : x=t+1 : x=-1 t=-2 x=0 t=-1 • Do : x4 ∫ ( x − 1) −1 dx= −1 −1 ∫ −2 ( t + 1) t dt= −1 −1 t + 4t + 6t + 4t +  = dt + ∫ +t + + ∫−2 t t t −2  −1 1  dt t3  • ⇔ ∫  t + + + +  dt =  t + 4t + ln t − −  = − ln t t t  t t  −2 2 −2  a th c : f(x)= ax + bx + cx + d ( a ≠ ) có hai nghi m : Có hai cách gi i : H s b t đ nh ph ng pháp nh y t ng l u 1 Ví d 10 : Tính tích phân sau : I= ∫ ( x − 1)( x + 1) 11 33 dx Gi i Cách ( Ph ng pháp h s b t đ nh ) Gv Ph m Minh T - 0968.469.299 Trang T m tài li u To n ? Chuy n nh - www.toanmath.com • Ta có : A ( x + 1) + B ( x − 1)( x + 1) + C ( x − 1) A B C = + + = 2 x − ( x + 1) ( x + 1) ( x − 1)( x + 1) ( x − 1)( x + 1)  A=  1 = A  Khi (1) • Thay hai nghi m m u s vào hai t s :  ⇔ = − C  C = −  2 ( A + B ) x + ( A + C ) x + A − B − C ⇒ A − B − C =1 ⇔ B =A − C − =1 + − =− ⇔ 4 ( x − 1)( x + 1) 1 1 1  + −  ∫2 ( x − 1)( x + 1) ∫2  x − ( x + 1) ( x + 1)2  dx   1 1 3 ln= ln I  ln ( x − 1)( x + 1) + ⇔= = ( x + 1)  4 4 • Do : 3 dx = Cách 2: • t : t=x+1, suy : x=t-1 x=2 t=3 ; x=3 t=4 4 dt 1 1 t − (t − 2) = = − dt dt  ∫3 t ( t − ) ∫3 t ( t − ) ∫  t ( t − ) 4 11  1  1 t −2 4 ln t  ln = ⇔I − − = − = dt   ln dt  ∫  ∫ t  4 t 22 2t −2 t  3 3 = dx 2 ( x − 1)( x + 1) • Khi : I= ∫ ( 3t − 4t ) 2 1  t − (t − 4)   Ho c : =  = t (t − 2)  t (t − 2)     ∫ t dt    3t − 4t ( 3t + )  3t − 4t   − = −  +    t  t − 2t  t t   t − 2t 1    ln t − 2t −  3ln t − t   3= ln     3t − 4t −  − =   t − 2t  t − 2t   3t − 4t    • Do : I= ∫  −  +   dt= t − 2t  t t   3 Ho c : = t − 2t t+2 1 − =  4t −2 t  1 1 2 − −   t − t t2  • Do : I=  1 2 1 t −2 2 1 1 2 1 1 ln ln + − ln −= − − 2= + = ln − ln −    dt      ∫ 3 4 6 t t  4 2 3t −2 t t  4 Ví d 11: Tính tích phân sau : I= x2 ∫ ( x − 1) ( x + ) dx 2 Gi i t : x-1=t , suy : x=t+1 , dx=dt : x=2 t=1 ; x=3 t=2 ( t + 1) dt x2 Do : ∫ = = dx ∫ 2 t ( t + 3) ( x − 1) ( x + ) Cách 1; ( H s b t đ nh ) t + 2t + At + B C Ta có : = + = t ( t + 3) t t +3 Trang 10 t + 2t + ∫1 t ( t + 3) dt ( At + B )( t + 3) += Ct ( A + C ) t + ( A + B ) t + 3B t ( t + 3) t ( t + 3) Gv Ph m Minh T - 0968.469.299 T m tài li u To n ? Chuy n nh - www.toanmath.com π I=∫ b C Y T – 2006 π 2 sin xdx sinx − cosx = = dx b I ∫ + sin2x π π KQ: ln Gi i π π s inx osx = − dx + ln 1= c= ∫0 1+cosx ln π sinx − cosx ∫ π dx sin xdx = ∫0 sin x + cos x (1 + cosx ) x sin x + cos x.cos 2 π + sin2x π π a I = ∫ sinx − cosx dx ( sinx+cosx ) π π π sinx − cosx ∫π sinx+cosx dx (1) π π π π Vì : s inx+cosx= sin  x +  ; ≤ x ≤ ⇒ ≤ x + ≤ ⇔ sin  x +  > 4 2 4 4   Do : s inx+cosx = s inx+cosx M t khác : d ( s inx+cosx ) = ( cosx-sinx ) dx π π d ( s inx+cosx ) ln s inx+cosx − = − = − π sinx+cosx Cho nên : I = ∫− π Ví d Tính tích phân sau a C S Ph m H i D ln1 ln  =  ln 2 π ng – 2006 I=∫ cos2x dx KQ: KQ: ln3 ( sin x − cosx + 3) π b C KTKT ông Du – 2006 cos2x dx 2sin2x + Gi i π a I = ∫ I=∫ cos2x ( sin x − cosx + 3) Cho nên : f ( x)dx = = dx Vì : cos x = cos x − sin x = ( cosx+sinx )( cosx-sinx ) cos2x ( sinx-cosx+3) dx ( cosx-sinx ) cosx+sinx dx ) ( ( sinx-cosx+3) π  dt= ( cosx+sinx ) dx; x = → t = 2, x = → t =  t −3 1 1  f ( x)= = dx dt  − 3  dt t t  t  t: t ⇒ = s inx-cosx+3 π V y : I = ∫ f ( x)dx = ∫  1  14 − = dt  − +  = t2 t3  32  t 4t  2 π cos2x dx + 2sin2x b I = ∫ Trang 28  = dt cos xdx → cos2xdx= dt t +: t = 2sin ⇒ 2x   x = → t = 1; x = π → t =  Gv Ph m Minh T - 0968.469.299 32 T m tài li u To n ? Chuy n nh - www.toanmath.com π V = y: I 3 cos2x dt ln3 = = = ln t dx ∫0 1+ 2sin2x ∫ 41 t 4 Ví d Tính tích phân sau : π 4sin3 x I=∫ dx + cosx a C S Ph m Qu ng Ngãi – 2006 KQ: π sin3x − sin3 3x dx B n Tre – 2006 I = ∫ + cos3x b C π π π Gi i π − cos2 x 4sin3 x − ∫ (1 cosx ) sinxdx=4 − (1 cosx = ) ∫0 1+ cosx dx= 4∫0 1+ cosx sinxdx=4 0 a =I ( ) π sin3x − sin3 3x dx cos3x + b I = ∫ Ta có : sin 3x − sin 3x = sin 3x (1 − sin 3x ) = sin 3x.cos 3x t : +t = ⇒ cos3x π V y: ∫  dt=-3sin3xdx → sin3xdx=- dt   x = → t = 2; x = π → t =  ( t − 1)  1 11 f ( x)dx =− ∫ dt = ∫  t − +  dt =  t − 2t + ln t 32 t 31 3 2 1 2 t  =− + ln 1 Ví d Tính tích phân sau π a I = ∫ π sin x − sin x cot gx dx sin x ∫ −π π − x) dx π + x) sin( sin( π π c I = b I = π 2 4 d I = ∫ cos x( sin x + cos x)dx ∫ sin x dx 0 Gi i a I = π ∫ π −   s inx  −  sin x − sin x sin x   cot gx dx = ∫ cot xdx sin x s inx π π 3 π π 3   = = ∫ 1 sin  cot xdx ∫ x π  π − cot x cot xdx Gv Ph m Minh T - 0968.469.299 Trang 29 T m tài li u To n ? Chuy n nh - www.toanmath.com π π − x) cosx-sinx b I = = dx dx ∫ ∫ − π sin( π + x) π cosx+sinx − 2 π sin( π ln cosx+sinx π − π d ( cosx+sinx ) ∫ cosx+sinx π − π π π 2 − cos2x   + cos4x   dx = ∫ 1 − 2cos 2x +  dx   0 0  ∫ sin x dx = ∫ c I = 0 π π 1 3π 3  3  = ∫− cos2x+ cos4x  dx =  x − sin 2x + sin 4x  = 32  8  16 0 π 4 d I = ∫ cos x( sin x + cos x)dx Vì : sin x + cos x− = sin x Cho nên : π π π π π 12 1   sin os2xdx= os2xdxsin x cos xdx = sin x − sin x = −I = x c c  ∫0  ∫ ∫ 2  0 0 2 Ví d Tính tích phân sau π π a I = ∫ sin xdx b I = ∫ π sin x cot gx dx π π d */I = ∫ ( cos x − sin x )dx c I = ∫ tg x + cot g x − 2dx π π π 0 Gi i ( a I = ∫ sin xdx = −∫ cos x − ) π sinxdx=- ∫ 1 − 2cos x + cos x  d ( cosx ) π 2   = − cos3 x = cos5 x   cosx+   15 Trang 30 Gv Ph m Minh T - 0968.469.299 T m tài li u To n ? Chuy n nh - www.toanmath.com π b I = ∫ π sin x cot gx dx  −2tdt = → −dx  sin x cot x ⇒ t 2= cot x ↔   x = π → t = 3; x =  t : =t 2tdt dx = sin x π →t = ∫ dt =− 2t ( 1) V y−: I = ∫== 2tdt t 3 1 π π π π π c I = ∫ tg x + cot = g x= − 2dx ∫ ( t anx-cotx ) dx Vì : tanx-cotx= sinx cosx sin x − cos x cos2x − =− − = = cosx sinx s inxcosx sin2x ∫ t anx-cotx dx π cot x  π π  t anx-cotx0;x ∈  ;  4 3  π π π 4 π = ) dx+ ∫ ∫ ( t−anx-cotx π π 6 V y−: I = + ) dx ∫ ( t anx-cotx π π cos2x cos2x dx ∫ dx = sin2x π sin2x π ln ( ln sin x ) π4 − 12 ( ln sin x ) π3 = π d I = ∫ ( cos x − sin x )dx (1) π π π 2 t : x = − t → dx =−dt , x =0 → t = ; x = → t =0 Do :  π  π  I ∫  cos  − t  − sin  t   ( − dt ) = =  2     π  π ∫( π ) sin t − cost= dt ∫( ) sin x − cosx dx ( 2) L y (1) +(2) v v i v : I = ⇒ I = Ví d Tính tích phân sau π π π cos x ∫ dx (NNI-2001) π sin x a ∫ tan xdx (Y-HN-2000) b π cos2x ∫0 ( sinx+cosx+2 ) dx (NT-2000) c 4 Gv Ph m Minh T - 0968.469.299 Trang 31 T m tài li u To n ? Chuy n nh - www.toanmath.com π d π π sin x ∫0 cos6 x dx ( GTVT-2000) e − 2sin x ∫0 + sin x dx (KB-03) sin x ∫0 − cos2 x dx f Gi i π sin x (1 − cos x ) 4 a ∫ tan xdx Ta có : f ( x) =tan x = = = −2 +1 cos x cos x cos x cos x π π π π π Do : I =∫ f ( x)dx =∫  1 dx  −2 + 1 dx =∫ (1 + tan x ) − [ tan x + x ] 2 π cos x  cos x π  cos x π 4 3 π π π   4  π  π    =  t anx+ tan x  −  3− 2+ =  3− −  3− 2+ = + 12   3  12  12  π  * Chú ý : Ta cịn có cách phân tích khác : f ( x) = tan x = tan x ( tan x + 1−) tan = x (1 tan + x ) tan − 2x π π π 3 2 + 1 dx ∫  tan x (1 + tan x ) − ( tan x + 1)= V= y: I tan = x (1 tan + x) ∫ tan x π π 4 (−tan x 1+) + π dx dx dx − + cos x π∫ cos x π∫ 4 π π  1 π π 1  1 I =  tan x − t anx+x  =  3 − +  −  − +  = +  3  12 3  π 3 π b cos2x ∫ ( sinx+cosx+2 ) dx Ta có : f ( x) = = = ( cos x − sin x ) ( cosx-sinx )( cosx+sinx ) cos2x ( sinx+cosx+9 ) π π 4 ( sinx+cosx+9 ) ( sinx+cosx+9 ) 3  ( cosx+sinx )  ∫0  ( sinx+cosx+2 )3  ( cosx-sinx ) dx (1)   π  cosx+sinx=t-2.x=0 → t=3;x= → t= + 2, t: t ⇒ = s inx+cosx+2  1 t −2 1 dt = −3 dt =  dt ( cosx-sinx ) dx ⇒ f ( x)dx =   t t  t Do= : I ∫= f ( x)dx V y:   1 1 1  1  2+2   −  − +  = − 1+ I = ∫  −  dt =  − +  = − +     t t   t t  3   2+ 2+   9 2+   t + cost ) ( sin t + cost ) cost sin t dt f ( x) ( sin = − − = ( sin t −cost )( dt ) ( ) ( sin t + cost+9 ) ( sin t + cost+9 ) +2 ( Trang 32 ) Gv Ph m Minh T - 0968.469.299 ( ) T m tài li u To n ? Chuy n nh - www.toanmath.com π cos x ∫ dx = π sin x c cos x Ta có : f = ( x) = sin x (1 − sin x= ) − 3sin x + 3sin x − sin x = − sin x sin x π π π 1 + 4− 3 sin x sin x sin x π 2 − cos2x  dx dx V y : I =∫ (1 + cot x ) − 3∫ + 3∫ dx − ∫   dx sin x π sin x π  π π  4 4 π 1   5π 23 = − cot x + 3cot x + x − x + sin x  = + 12  π π sin x d ∫ = dx os c x π π − cos x dx ∫0 cos6 x =  1  − = − dx ∫0  cos6 x cos4 x  ∫0 cos4+x cos2 x dx π = π π ∫ (1 + tan x ) 2 π ∫ (1 tan x ) π 1 dx − ∫ (1+ tan x ) dx = cos x cos x dx cos x π ∫ (1+ tan x + tan x ) d ( tan x ) − ∫ (1+ tan x ) d ( t anx ) 0 π π 1   1  =  t anx+ tan x + tan x − t anx- tan x  =  tan x + tan x  = 5   3  15 π π π 2 π π d ( − cos2x ) 2sin x = − = −= − ln cos2x =2 dx ∫0 − cos2x ∫0 − cos2x sin x sin x e ∫ dx = dx ∫ + cos2x − cos x 0 4− π π π π − 2sin x d (1 + sin x ) cos2 x f ∫ ln = sin x = dx = dx ∫0 + sin x ∫0 + sin x+ = + sin x 0 4 Ví d Tính tích phân sau : 2 a ∫ sin x cos xdx b π sin x ∫ + 2cos3x dx π J cos x ∫0 s inx+ 3cosx dx K π cos2x dx s inx ∫π cosx- Gi i π π a ∫ sin x cos xdx = − ∫ (1 cos x ) cos x.s −inxdx ln 2 π sin x c I = dx = = ∫0 s∨inx+ 3⇒ cosx π π ln ∫ ( cos x cos x= ) d ( cosx ) π 1  =  cos x − cos x  = 7  35 Gv Ph m Minh T - 0968.469.299 Trang 33 T m tài li u To n ? Chuy n nh - www.toanmath.com π π 2 π π sin x −3sin x d (1 + cos x ) b ∫ − dx = − = − = + dx + 2cos3x ∫0 + cos x ∫0 + cos x π sin x + cos x dx s inx+ 3cosx c = Ta = có : I + = J ∫ π π 16 ∫0 1 dx ∫ 201 s inx+ cosx 2 π π sin x − 3cos x dx s inx+ 3cosx - M t= khác= : I − 3J ∫ ln dx π  sin  x +  3    x π  d  tan  +   1    x π x π x π tan  +  2cos  +  tan  +  2 6 2 6 2 6 1 Do : = = = π  x π  π sin  x +  2sin  +  cos  x+  3  2 6  6   x π  π d  tan  +   π x π    V y: I= ∫ = ln tan  +  6= 20 x π 2 6 tan  +  2 6 = ( ln cos 3x ) 6 ∫ (sin x − ln 3= ln (1) )( 3cosx sin x + 3cosx s inx+ 3cosx π ) dx π Do : I − 3J = (2) ∫ ( s inx- 3cosx ) dx (=c−osx- s inx ) =− 0  3 −1  ln − = I + = I J ln  16 4 ⇔ T (1) (2) ta có h :  −1   =  I − J =−  J 16 ln + π ( 3) π π π tính K ta đ t t = x − → dt = dx ⇔ x = ; t = 0.x = → t = π cos ( 2t+3π ) V y : K =∫ π π π   cos  t+3  − sin  t+3  2 2   dt cos2t −1 = I J = ln − − ∫0 sint+ 3cost dt =− Ví d 10 Tính tích phân sau π π a ∫0 + sin x dx ( C -99) b ∫ ( sin H-LN-2000) π π c dx ∫ + s inx+cosx ( 10 x + cos10 x − sin x cos x ) dx (SPII-2000) d π π 4 a ∫ dx = x + sin Trang 34 π dx (M C-2000)  π s inxsin  x+   6 ∫ π Gi i π π  dx = ∫ dx = tan  x −  = π 4  cos  x −   4  ∫ ( s inx+cosx ) Gv Ph m Minh T - 0968.469.299 T m tài li u To n ? Chuy n nh - www.toanmath.com π b dx ∫ + s inx+cosx x t : t = tan ⇔ dt = cos x dx = 1 2dt π x ; x = → t = 0, x = → t = 1 + tan  dx; ⇔ dx = 2 2 1+ t 1 1 2dt 2dt dt ∫= =∫ ( 2) 2 ∫0 = 2t − t (1 + t ) t + 2t + ( t + 1)2 + 2+ + 1+ t2 1+ t2  du →; t = →u= = 0= tan ; t tan u dt = cos u  t +1 tan u ⇒  t := 2dt 2  f (t )dt = du = = 2du 2  cos 2u tan + u + + t ( ) ( )  u2   u2 V y := ( u2 − u1= = 2u = I ∫ 2du )  arxtan − arctan  u1 u1   V y: I = π ∫ ( sin c 10 x + cos10 x − sin x cos x ) dx Ta có : sin10 x + cos10 x − sin x cos x ( sin x + cos x ) = ( cos x − sin x )( cos6 x − sin x ) = (−cos2 x sin x )(−cos2 x sin x )(+cos4 x +sin x cos2 x sin x ) 1 + cos4x − cos8x 15 1   = + − = cos4x+ cos8x cos 2 x 1 − sin 2 x  = cos 2 x − sin x = 16 32 32 32   π  15 V y: I= ∫0+ 32 1 15 π  + cos4x+ cos8x=  dx 32 32  π +sin x π = sin x 32.8 15π 64 π dx  π s inxsin  x+   6 π π  π  π π Ta có :  x +  − x = ⇒ sin  x +  − x  =sin  x +  cosx-sinxco  x +  = (*) 6 6  6 6     π π   sin  x +  cosx-sinxco  x +  6 6  Do : f ( x) 2  = = = π π π       s inxsin  x+  s inxsin  x+  s inxsin  x+   6  6  6 π π   π  π  cos  x+  cos  x+   3 cosx cosx   ⇒=    dx  ln s inx − ln sin  x+ π   = − = 2∫  − = I ∫ f ( x)dx    π π  sinx    6 π π  sinx  + sin  x +  sin x   6 6      d ∫ π π π π I= ln s inx 3 = − ln = ln ln 2  π π sin  x+   6 Gv Ph m Minh T - 0968.469.299 Trang 35 T m tài li u To n ? Chuy n nh - www.toanmath.com * Chú ý : Ta có cách khác 1 = = f(x)=  π   s inxsin  x+  s inx  s inx+ cosx   6   π π dx + cot x sin x V y: I= ∫ π ( sin x ( ) s inxcos x ∫0 + cos2 x dx (HVBCVT-99) b ∫ cos x cos 2 xdx ( HVNHTPHCM-98) π 4 sin x ∫0 cos6 x + sin x dx ( HNT-01) π d dx ∫ cos x ( HTM-95) Gi i π s inxcos3 x cos x dx (sin x)dx = ∫0 + cos2 x ∫0 + cos x a π π π c ) ) π a + cot x 2d + cot x 3 −∫ = − = = 2+ln cot ln x π + cot x π 6 Ví d 11 Tính tích phân sau ( (1) 2sin x cos xdx − sin xdx = − dt =  t : +t = ⇒x  cos π cos x = t − 1; x = → t = 2; x = → t = 1 2 ln − 1 ( t − 1) 1  V y: = ln − = − = − = t t I dt dt ( ) ( )   ∫2 t ∫1  t  2 π b ∫ cos x cos 2 xdx + cos2x + cos4x + = (1 cos2x+cos4x+cos4x.cos2x ) 2 1 1  = cos2x+ cos4x+ cos6x= 1 cos2x+cos4x+ ( cos6x+cos2x+)  4  Ta có : f (= cos 2 x x) cos x = π π 1 1 π  1  cos2x+ cos4x+ cos6x  dx =  x + sin x + sin x + sin x  = 16 16 48 8  4  1 V y: I= ∫0+ π c sin x ∫ cos x + sin 6 x dx Vì : d ( sin x + cos = x ) ( 6sin x cos x − 6cos5 x sin = x ) dx 6sin x cos x ( sin x − cos x ) ⇔ d ( sin x + cos x ) = 3sin x ( sin x − cos x )( sin x −+ cos x ) dx = 3sin x cos xdx − sin xdx = ⇒ − π sin xdx = + d ( sin x cos x ) π π 6 sin x d ( sin x + cos x ) 6 V y: ∫ − dx = − = + ln= ( sin x cos x ) cos x + sin x ∫0 ( sin x + cos x ) 0 Trang 36 Gv Ph m Minh T - 0968.469.299 ln + T m tài li u To n ? Chuy n nh - www.toanmath.com π π π π 1 dx dx   d ∫ = tan x ) d ( t anx )  t anx+ = + tan x  = = ( 2 ∫ ∫ cos x cos x cos x   4 Ví d 12 Tính tích phân sau π π a ∫ sin11 xdx ( HVQHQT-96) b ∫ sin x cos xdx (NNI-96) 0 π π c ∫ cos x cos xdx (NNI-98 ) d ∫ + cos2x dx ( HTL-97 ) 0 Gi i π a ∫ sin11 xdx Ta có : sin11 x sin10 x.s+inx= (1-cos−2 x ) s inx=+(1-5cos x− 10 cos3 x 10 cos x 5cos5 x cos x ) s inx π Cho nên : = I ∫ (1-5cos x + 10 cos x − 10 cos x + 5cos5 x − cos x ) s inxdx 5 1  π −118 =  cos x − cos x + cos5 x − cos x + cos3 x − cosx  = 21 7 0 π b ∫ sin x cos xdx H b c:  − cos2x   + cos2x  sin x cos x = cos2x )−(1 cos 2+x cos 2 x+) (=    2    =(1 + cos x + cos 2 x − cos2x-2cos 2 x − cos3 x ) 1 1+cos4x  1+cos4x   x+) cos2x- − = + (1 cos2x-cos−2 x cos3= cos2x    8 2   2 1 cos6x+cos2x  = + (1 cos2x-cos4x+cos4x.cos2x ) 1 cos2x-cos4x+=  16 16   ( + 3cos x + cos6x-cos4x ) 32 π + π 1 1  V y I =∫ ( + 3cos x + cos6x-cos4x ) dx = x + sin x + sin x − sin x  = 64 32.6 32.4 32  32  0 π d ∫ = π + cos2x= dx ∫ π cos xdx =  π π    s inx − s inx π =     ∫ cosx = dx  π2  π    ∫ cosxdx − ∫ cosxdx  π 0    (1 + 1= ) 2 Gv Ph m Minh T - 0968.469.299 Trang 37 T m tài li u To n ? Chuy n nh - www.toanmath.com III M T S CHÚ Ý QUAN TR NG ng pháp đ i bi n s d ng Trong ph * S d ng công th c : b b 0 )dx ∫ f (b − x)dx ∫ f ( x= Ch ng minh : x = → t = b x = b → t = t : b-x=t , suy x=b-t dx=-dt , ⇒  • • Do : b ∫ f ( x)dx = ∫ b f (b − t )(−dt= ) b ∫ b f (b − t )dt = ph thu c vào bi n s Ví d : Tính tích phân sau ∫ f (b − x)dx Vì tích phân khơng π a/ π 4sin xdx ∫ ( s inx+cosx ) 5cos x − 4sin x ∫ ( s inx+cosx ) π π b/ 3 dx sin x ∫0 sin x + cos6 x dx c/ ∫ log (1 + t anx ) dx d/ π sin x cos x f/ ∫ dx sin x + cos3 x e/ ∫ x (1 − x ) dx n m Gi i π 4sin xdx a/ I = ∫ ( s inx+cosx ) (1) t: π π  dt =−dx, x =0 → t =2 ; x =2 → t =0  π  π π  4sin  − t  t= −x⇒ x= −t ↔    2  f ( x)dx = dt ) (   π  π   sin  − t  + cos  − t         cos t ( cost+sint ) dt f (t )dt = − Nh ng tích phân không ph thu c vào bi n s , : π I = f (t )dt ∫= π 4cosx ∫ ( sinx+cosx ) dx ( 2) π ( s inx+cosx ) L y (1) +(2) v v i= v ta có : 2I ∫ ⇒ = π ( s inx+cosx ) π dx I 2∫ ( s inx+cosx ) π ⇔= = dx I 2∫ π cos  x −   4  π  tan − x =  4  Trang 38 Gv Ph m Minh T - 0968.469.299 2 dx T m tài li u To n ? Chuy n nh - www.toanmath.com π b/ I = ∫ 5cos x − 4sin x ( s inx+cosx ) dx T ng t nh ví d a/ ta có k t qu sau : π π 5cos x − 4sin x 5sin t − cos t I= ∫0 ( s inx+cosx )3 dx = ∫π ( cost+sint )3 dt 5sin x − 4cosx ∫ ( s inx+cosx )− dx ( ) 2 = π π V y : 2I =∫ π 1 π  dx =∫ dx = tan  x −  =1 ⇒ I = π 4  cos  x −   4  ( s inx+cosx ) π c/ ∫ log (1 + t anx ) dx t: π π  dx =−dt , x =0 → t = ; x = → t =0  4 π π  −x→x= −t ⇔  t= 4  f ( x)dx= log (1 + t anx ) dx= log 1 + tan  π − t   ( −dt )     − tan t  Hay: f (t=) log 1 + ) log ( −dt=) log 2 − log t  ( −dt= + tan t  + tan t  V y: I= π π 4 ∫ f (t )dt = ∫ dt − ∫ log π π π π ⇔I= tdt ⇒ I = t = π sin x d/ I = ∫ dx (1) sin x + cos x π π  sin  − t  cos x  = d ( −t ) ∫ dx ∫ π  π = cos x + sin x  π sin  − t  + cos  − t  2  2  π π cos x + sin x C ng (1) (2) ta có : I = ∫ dx= cos x + sin x e/ ∫ x m (1 − x ) dx I (2) ∫ dx = π π π x = ⇒I= t : t=1-x suy x=1-t Khi x=0,t=1;x=1,t=0; dt=-dx n 0 1 Do : I = ∫ (1 − t ) t (−dt ) = ∫ t (1 − t ) dt = ∫ x n (1 − x)m dx m n n M TS m BÀI T P T π LUY N π 4sin x ∫0 + cosx dx cosx+2sinx ∫ cos x + 3sin x dx (XD-98 ) Gv Ph m Minh T - 0968.469.299 Trang 39 T m tài li u To n ? Chuy n nh - www.toanmath.com π π 3 s inxcos x ∫0 + cos2 x dx x + s inx dx ( HVNHTPHCM-2000 ) cos x π ∫ x5 (1 − x3 ) dx ( HKT-97 ) x sin x ∫ + cos x dx π π 6 ( AN-97 ) + s inx  ∫ ln   dx ( C SPKT-2000 )  1+cosx  ∫ s inx+2cosx ∫0 3sin x + cosx dx ( C SPHN-2000) π π x sin x ∫ dx ( HYDTPHCM-2000 ) x cos + β * D ng : I = ∫ α asinx+bcosx+c dx a 's inx+b'cosx+c' β Ta phân tích : sin x cos x ∫0 sin x + cos3 x dx 10 Cách gi i : asinx+bcosx+c ∫ α a 's inx+b'cosx+c' B ( a ' cosx-b'sinx ) dx = A + a 's inx+b'cosx+c' C + a 's inx+b'cosx+c' - Sau : Quy đ ng m u s - ng nh t hai t s , đ tìm A,B,C - Tính I : β B ( a ' cosx-b'sinx )  C + I= + ∫α  A a 's inx+b'cosx+c' a 's inx+b'cosx+c' Ví d Tính tích phân sau : β β  dx Ax+Bln 's inx+b'cosx+c' a C = + dx ( )  ∫ α  α a 's inx+b'cosx+c' VÍ D ÁP D NG π π s inx-cosx+1 a ∫ dx ( B đ ) s inx+2cosx+3 b cosx+2sinx ∫ cos x + 3sin x dx ( XD-98 ) π π c s inx+7cosx+6 ∫0 4sin x + 3cos x + dx d I = Gi i π a cos x − 3sin x + dx ∫ sin x + 3cos x + s inx-cosx+1 ∫ s inx+2cosx+3 dx Ta có : B ( cosx-2sinx ) s inx-cosx+1 f ( x) = = A s inx+2cosx+3 s inx+2cosx+3 Quy đ ng m u s đ ng nh t h s hai t s : C + s inx+2cosx+3  A = −  A − 2B =  A − B ) s inx+ ( 2A+B ) cosx+3A+C  (  ⇔ f ( x) = ⇒2 A +B =−1 ⇔ B =− Thay vào (1) s inx+2cosx+3 3 A + C =    C =  Trang 40 Gv Ph m Minh T - 0968.469.299 (1) + T m tài li u To n ? Chuy n nh - www.toanmath.com π π 2 d ( s inx+2cosx+3)  1 I= ∫0  −  dx − ∫0 s inx+2cosx+3 π 4 I= − − ln − J 10 5 π 42 +∫ dx s inx+2cosx+3 π π −=− ln s inx+2cosx+3 − J 10 5 ( 2) - Tính tích phân J : dx π  dt x t x ; 0, = = → = = →t =  cos x  x t : =t tan ⇒  ⇔= J 2dt 2dt  f ( x)dx = =  2t 1− t2 + t t + 2t + 3 + +  1+ t2 1+ t2 Tính (3) : t +1 =  du t = 0= tan → →= u == = u1 ; t dt = os c u  tan u ⇒  2du du = =  f (t )dt 2 cos u  cos 2u  +2 (3) tan u u2  2 π 4  tan u1 = du = ( u2 − u1 ) ⇒ I =I =− − ln − ( u2 − u1 )  2 10 5  tan u =  V y : j= ∫ u π 2dt ∫ ( t + 1) t: u2 b cosx+2sinx ∫ cos x + 3sin x dx; B ( 3cos x − 4sin x ) cosx+2sinx C + + → (1) f ( x) = = A cos x + 3sin x cos x + 3sin x cos x + 3sin x Gi ng nh phàn a Ta có : A = ; B = − ;C=0 5 π π  ( 3cos x − 4sin x )  2  V y : I−= ln 4+cos= x 3sin x   dx  +x ∫0  5 cos x + 3sin−= x  5  Gv Ph m Minh T - 0968.469.299 π 10 ln Trang 41 T m tài li u To n ? Chuy n nh - www.toanmath.com Trang 42 Gv Ph m Minh T - 0968.469.299 ... tích phân ) III CÁC PH a NG PHÁP TÍNH TÍCH PHÂN A PH NG PHÁP PHÂN TÍCH 1.Trong ph ng pháp , c n : • K n ng : C n bi t phân tích f(x) thành t ng , hi u , tích , th ng c a nhi u hàm s khác , mà... ng khéo léo h n cách gi i s hay h n Sau đay minh h a b ng m t s ví d Ví d Tính tích phân sau 2 a dx ∫1 x ( x + 1) b x2 + ∫ ( x − 1) ( x + 3) dx Gi i a dx N u theo cách phân tích b ng đ ng nh... 3  =  t − sin 2t + sin 4t  = 32 4  Gv Ph m Minh T - 0968.469.299 Trang 25 T m tài li u To n ? Chuy n nh - www .toanmath. com TÍCH PHÂN CH A CÁC HÀM S I KI N TH C Thu c nguyên hàm : β ∫α sin