PHƯƠNG PHÁP TÍCH PHÂN TỪNG PHẦN

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Chương II: Nguyên hàm và tích phân − Tr n Phương BÀI 8. PHƯƠNG PHÁP TÍCH PHÂN T NG PH N I. CÔNG TH C TÍCH PHÂN T N G PH N u = u ( x ) ; v = v(x) có Gi s o hàm liên t c trong mi n D, khi ó ta có: ∫ ∫ ∫ ∫ ∫ • d ( uv ) = udv + vdu ⇔ d ( uv ) = udv + vdu ⇔ uv = udv + vdu b b b ∫ udv = uv − ∫ vdu ⇒ ∫ udv = ( uv ) ∫ ⇒ − vdu a a a Nh n d n g: Hàm s dư i d u tích phân thư ng là tích 2 lo i hàm s khác nhau Ý nghĩa: ưa 1 tích phân ph c t p v tích phân ơ n gi n hơ n (trong nhi u trư ng h p vi c s d ng tích phân t ng ph n s kh b t hàm s dư i d u tích phân và cu i cùng ch còn l i 1 lo i hàm s dư i d u tích phân) Chú ý: C n ph i ch n u, dv sao cho du ơn gi n và d tính ư c v ng th i ∫ vdu ∫ udv ơ n gi n hơn tích phân tích phân II. CÁC D N G TÍCH PHÂN T N G PH N C Ơ B N VÀ CÁCH CH N u, dv 1 . D ng 1: u = P ( x )  sin ( ax + b ) dx   sin ( ax + b ) dx  cos ( ax + b ) dx  cos ( ax + b ) dx ∫ P (x)  ⇒ (trong ó P(x) là a th c) dv =  ax + b  e dx  ax + b  e dx  ax + b dx   m  ax + b m dx   2 . D ng 2: dv = P ( x ) dx  arcsin ( ax + b ) dx  arcsin ( ax + b )  arccos ( ax + b ) dx  arccos ( ax + b )  arctg ( ax + b ) dx    P (x)  ∫ arctg ( ax + b ) ⇒ ( ax + b ) dx u =  (trong ó P(x) là a th c)  arc cotg  arc cotg ( ax + b )  ln ( ax + b ) dx  ln ( ax + b )   log m ( ax + b ) dx    log m ( ax + b )    3 . D ng 3:  sin ( ln x )  eax+b sin ( αx + β) dx  eax+b sin ( ln x ) dx  cos ( ln x )  ax+b u =  ax+b ()  e cos ( αx + β) dx ⇒  m u = sin ( log x ) k cos ln x dx ∫ ∫ x ⇒ sin ( loga x) dx  cos ( log x)  max+b sin ( αx + β) dx  ; a dv = sin ( αx + β) dx   cos ( loga x ) dx   k  ax+b cos ( αx + β) dx a m cos ( αx + β) dx      dv = x dx 2 10
Bài 8. Phương pháp tích phân t ng ph n III. CÁC BÀI T P M U MINH H A : ∫ P ( x ) {sin ( ax + b ) ; cos ( ax + b ) ; e ; m ax+b } dx ax+b 1 . D ng 1: ∫ • A1 = x 3 cos x dx . u = x 3 du = 3x 2 dx   ⇒ Cách làm ch m: t . Khi ó ta có: dv = cos x dx  v = sin x   u = x 2  du = 2x dx  ∫ 3 2 ⇒ t A1 = x sin x − 3 x sin x dx . . Khi ó ta có: dv = sin x dx  v = −cosx   u = x du = dx   A1 = x sin x − 3  − x cos x + 2 x cos x dx  . ∫ 3 2 ⇒ t .   dv = cos x dx  v = sin x   ( ) ∫ 3 2 3 2 A1 = x sin x + 3x cos x − 6 xsin x − sin xdx = x sin x + 3x cos x − 6 ( xsin x + cos x ) + c ∫ P ( x ) L ( x ) dx = ∫ P ( x ) du Cách làm nhanh: Bi n i v d ng A1 = x3 cos x dx = x 3 d ( sin x ) = x 3 sin x − sin x d ( x3 ) = x3 sin x − 3 x 2 sin x dx ∫ ∫ ∫ ∫ = x sin x + 3 x d ( cos x ) = x sin x + 3  x cos x − cos x d ( x )  ∫ ∫ 3 2 3 2 2   ∫ ∫ 3 2 3 2 = x sin x + 3x cos x − 6 x cos x dx = x sin x + 3x cos x − 6 x d ( sin x ) ( ) ∫ 3 2 3 2 = x sin x + 3x cos x − 6 xsin x − sin xdx = x sin x + 3x cos x − 6 ( xsin x + cos x) + c 1 3 ( 5 x −1 ) 1  3 5 x −1 − e5 x −1 d ( x3 )  ∫ ∫ ∫ • A2 = x 3 e 5x − 1 dx = = xe xd e 5  5 1 3 2 ( 5x −1 )  1 =  x 3 e5x −1 − 3 x 2 e5x −1 dx  =  x 3 e5x −1 − ∫ ∫ xde    5  5 5 1 3 1 3 6 = x 3 e5x −1 −  x 2 e5x −1 − e5x −1 d ( x 2 )  = x 3 e5x −1 − x 2 e5x −1 + ∫ ∫ xe5x −1 dx  5 5 25 25 25 1 3 6 1 3 ∫ x d ( e5x −1 ) = x 3 e5x −1 − x 2 e5x −1 + = x 3 e5x −1 − x 2 e5x −1 + 5 25 125 5 25 6  5x −1 1 3 6 6 5x −1 − e5x −1 dx  = x 3 e5x −1 − x 2 e5x −1 + ∫ xe5x −1 − + +c xe e 125  5 25 125 625 Nh n xét: N u P(x) có b c n thì ph i n l n s d ng tích phân t ng ph n. 2 11
Chương II: Nguyên hàm và tích phân − Tr n Phương π2/4 x 0 π2 /4 ∫ t t = x ⇒t = x ⇒ • A3 = x sin x dx . 2 t 0 π/2 0 dx 2tdt π2 π2 π2 π2 π2 + 2 ∫ cos td ( t ) = 6 ∫ t cos t dt ∫ ∫ 3 3 3 3 2 A3 = 2 t sin t dt = −2 t d ( cos t ) = −2t cos t 0 0 0 0 0 π2 π2 π2 π2 3π2 3π2 π2 = 6 t 2 d ( sin t ) = 6t 2 sin t 0 − 6 sin td ( t 2 ) = ∫ ∫ ∫ ∫ + 12 td ( cos t ) −12 t sin t dt = 2 2 0 0 0 0 π2 3π 2 3π2 3π2 π2 π2 ∫ = + 12t cos t 0 − 12 cos t dt = − 12 sin t 0 = − 12 2 2 2 0 π6 π6 π6 π6 x cos3 x 1 1 ∫ ∫ ∫ cos x d ( cos3 x ) = − 2 • A4 = x sin x cos xdx = − 3 + x dx 3 3 3 0 0 0 0 π6 π6 (1 − sin x ) d ( sin x ) = − π 3 + 1  sin x − sin x  = 11π − π 3  3 π3 1 ∫ 2  =− + 48 3  3 0 48 3 72 48 0 u = x 2 e x du = x ( x + 2 ) e x dx 1 2 x  x e dx  ∫ ( x + 2) dx ⇒  • A5 = t . 1 2 dv = v = − 0 ( x + 2 )2   x+2 2 x1 1 1 1 xe e e + xe dx = − + xe dx = − + xd ( e ) ∫ ∫ ∫ x x x A5 = − x+20 0 30 30 1 1 1 e e e ∫ x x x = − + xe − e dx = − + e − e =1− 0 0 3 3 3 0 ∫ P ( x ) {arcsin u; arccos u; arctg u; arc cotg u ; ln u ; log u u = ax + b} dx 2 . D ng 2: m   e e e e 1 ( ln x ) 2 d ( x 3 ) = 1  x3 ( ln x )2 1 − x3 d ( ln x ) 2  2 ∫ ∫ ∫ • B1 = x 2 ( ln x ) dx = 3  31   1 1 1 dx  1  3  1 3 2 e  e e ln x d ( x 3 )  ∫ ∫ ∫ =  e3 − 2x 3 ln x 2  =  e − 2x ln x dx  =  e − 3 x  3  3  31       1 1 e 2 ( 3 3 e e e 3 3 3 e ) 1 − x3d ( ln x ) = e − 2 e3 + 2 x2 dx = e + 2 x3 = 5e − 2 ∫ ∫ = −  x ln x 3 9 39 91 9 27 1 27 27   1 2 12
Bài 8. Phương pháp tích phân t ng ph n  1 + x  ( 2 ) 1  2 1 + x  1 + x  12 12 12 12  1+ x  2 1 ∫ ∫ ∫  d x =  x ln  • B2 = x ln   dx =  − x d  ln ln   1− x   1− x  2  1− x  0  1 − x  2   0 0 0 12 12 2 x 1 − x dx 1 1 ∫ ∫ 2   dx = ln 3 − x⋅ ⋅ = ln 3 − 1+ x  2 1+ x 1− x 8 8 0 0 12 12 2  1  2 1 1 1 ∫ ∫ 1 + (1 + x ) 1 −  dx = ln 3 − = ln 3 − − dx 1+ x   1+ x  2 8 8   0 0 12   1 1 ln 3 35 = ln 3 −  x − − 2 ln 1 + x  = + 2 ln −  0 1+ x 8 8 26 1 1 1 ) dx =  x ln ( x + ) 0 − ∫ xd ln ( x + 1 + x2 ) • B3 = ∫ ln ( x + 2 1+ x 2 1+ x     0 0 1 1  x dx x dx = ln (1 + 2 ) − x  1 + = ln (1 + 2 ) − ∫ ∫ 2 2 2 0 1+ x  x + 1+ x 0 1+ x 1 d (1 + x 2 ) 1 1 = ln (1 + 2 ) − = ln (1 + 2 ) − 1 + x = ln (1 + 2 ) + 2 − 1 ∫ 2 0 2 0 1+ x 2 x ln ( x + 1 + x 2 ) dx = 1 ln ( x + 1 1 + x2 ) d ( 1 + x2 ) ∫ ∫ • B4 = 2 1+ x 0 0 1 1 1 + x d  ln x + 1 + x  ln ( x + ) ( ) ∫ 2 2 2 2 = 1+ x 1+ x −    0 0 1 dx 1 + x 1 +  = 2 ln (1 + 2 ) − x ∫ 2  2 2  1+ x  x + 1+ x 0 1 = 2 ln (1 + 2 ) − dx = 2 ln (1 + 2 ) − 1 ∫ 0 u = ln x + 1 + x 2 ( ) x ln ( x + 1 + x 2 ) dx . 1  ∫ • B5 = t ( ) x dx 2 dv = =x 1 + x − x dx x + 1 + x2 0 2  x + 1+ x   (x + ) x dx 1 + x2 = ⇒ du =  1 +  dx 2 2   1+ x 1+ x 2 13
Chương II: Nguyên hàm và tích phân − Tr n Phương (1 + x 2 )1 2 d (1 + x 2 ) − ∫ x 2 dx = 1 (1 + x 2 )3 2 − x 3  1 ∫   v= 2 3 1 1 1 2 B5 =  (1 + x ) − x  ln x + 1 + x  − 1 ( 3 ( ) dx 2 32 2 32  1+ x ) − x  ∫ 3   3 0 3 0 2 1+ x (2 2 − 1) ln (1 + 2 ) 1 1 1 1 x 3 dx dx ∫ ∫ = − + 3 0 1 + x2 3 0 1 + x2 3 1 (1 + x ) − 1 ( (2 2 − 1) ln (1 + 2 ) 1 1 1 2 d 1+ x ) ∫ 2 = − arctg x + 3 3 60 2 1+ x 0 (2 2 − 1) ln (1 + 2 ) π 1 ( 1 2 −1 2  2 12  1 + x ) − (1 + x )  d (1 + x ) ∫ 2 = − + 3 12 6 0 (2 2 − 1) ln (1 + 2 ) π 1  2 ( 1 2 12 2 32 +  1 + x ) − 2 (1 + x )  = − 12 6  3 0 3 (2 2 − 1) ln (1 + 2 ) π 2 − 2 = − + 3 12 9 1 1 • B6 = ∫ x ln ( x + ) dx = 1 ∫ ln ( x + 1 + x2 ) d ( x2 ) 2 1+ x 2 0 0 1  x 2 ln x + 1 + x 2 ) ( 1 1 2 x d  ln x + 1 + x  ( ) ∫ =  2 −   0 2   2 0 1 1 2 1 2  1 x dx 1 1 x dx ln (1 + 2 ) − = ln (1 + 2 ) − ∫ ∫ = x 1+ 20  2 2 2 2 2 2  1+ x  x + 1+ x 1+ x 0 0 1 x 1 x 2 dx ) t x = tg t ; t ∈  0, π ⇒ ∫ Xét I = 0 t . π/4 2  1 + x2 0 dx dt cos 2 t π4 π4 π4 1 2 2 2 2 x dx tg t dt sin t sin t ∫ ∫ ∫ cos ∫ cos ⇒I= d ( sin t ) = ⋅ = dt = 2 3 4 2 2 1 + tg t cos t t t 1+ x 0 0 0 0 π4 2 22 22 2 2  (1 + u ) − (1 − u )  sin t d ( sin t ) u du 1 ∫ ∫ ∫ = = =  (1 + u ) (1 − u )  du   (1 − sin 2 t )2 (1 − u 2 )2 4 0 0 0 2 14
Bài 8. Phương pháp tích phân t ng ph n 22 22 2 1 1 1 2 1 1 1 ∫ ∫ =  −  du = + − du ( (1 + u ) 1 − u 2  1− u 1+ u  2 2  1 − u) 4 4  0 0 22 1 1 1+ u  1 2 − ln (1 + 2 ) = − − 2 ln =  4 1− u 1+ u 1− u  0 2 1 2  1 1 1 2 ⇒ B6 = ln (1 + 2 ) − I = ln (1 + 2 ) −  − ln (1 + 2 )  = − + ln (1 + 2 ) 2 2  2 2 2 4 0 0 0 1 1 1 0 x 2 d ( ln 1 − x ) ∫ ∫ ∫ ln 1 − x d ( x2 ) = x2 ln 1 − x −8 − • B7 = x ln 1 − xdx = 2 −8 2 2 −8 −8 0 0 2 −1 1 dx 1 x dx ∫ ∫ x2 ⋅ = −32ln 3 − ⋅ = −32 ln 3 + 4 −8 1 − x 2 −8 2 1− x 1− x 1 1 − (1 − x ) 0 0 2 1 1  ∫ ∫ − (1 + x )  dx = −32ln 3 + dx = −32 ln 3 + 4 −8 1 − x   4 −8 1 − x 0 1 1 2 l 63 = −32ln 3 +  − ln 1 − x − x − 2 x  = −32 ln 3 + 6 + 2 ln 3 = 6 − 2 ln 3 4  −8 x 0 −3 0 ln 1 − x ∫ (1 − x ) t t = 1− x ⇒ • B8 = dx . t 2 1 1− x −3 dx −2tdt 1 2 2 (t) ln t ( −2t dt ) = 2 ln t dt = 2 ln t d −1 ∫ ∫2∫ Khi ó ta có: B8 = 3 t t 2 1 1 2 2 2 2 −2 ln t −1 dt 2 ∫ ∫ d ( ln t ) = − ln 2 + 2 2 = − ln 2 − = −2 = 1 − ln 2 t1 t t1 1t 1 3 1 ln x d ( x 2 + 1) 1 3 3 x ln x dx  −1  ∫ (x ∫ ∫ • B9 = = = ln x d  2  2  x + 1 2 + 1) 2 1 ( x 2 + 1) 21 2 1 3 3 3 − ln x − ln 3 1 1 1 dx ∫ ∫ d ( ln x ) = = + + 2 ( x + 1) 1 2 1 x ( x 2 + 1) 2 2 2 1 x +1 20 − ln 3 1 ( x + 1) − x 3 3 2 2 − ln 3 1  1 x ∫ ∫ = + dx = + −2  dx 2 1 x ( x + 1) 2 1  x x +1 2 20 20 3 − ln 3 1   12 9 ln 3 + ln x − ( x + 1)  = = −2 2 1 20 2 20 2 15
Chương II: Nguyên hàm và tích phân − Tr n Phương 3 . D ng 3: Tích phân t ng ph n luân h i 1 1 13 ∫ ∫ ∫ sin ( ln x ) d ( x3 ) = x 3 sin ( ln x ) − • C 1 = x 2 sin ( ln x ) dx = x d ( sin ( ln x ) ) 3 3 3 1 1 dx 1 1 ∫ ∫ = x3 sin ( ln x ) − x3 cos ( ln x ) = x3 sin ( ln x ) − x2 cos ( ln x ) dx 3 3 x3 3 13 1 13 13 13 = x sin ( ln x ) − cos ( ln x ) d ( x ) = x sin ( ln x ) − x cos ( ln x ) + x d ( cos ( ln x ) ) ∫ ∫ 3 3 9 3 9 9 13 13 12 13 13 1 ∫ = x sin ( ln x ) − x cos ( ln x ) − x sin ( ln x) dx = x sin ( ln x ) − x cos ( ln x ) − C1 3 9 9 3 9 9 10 1 1 1 C1 = x3 sin ( ln x ) − x3 cos ( ln x ) ⇒ C1 = 3x3 sin ( ln x ) − x3 cos ( ln x )  + c ⇒   9 3 9 10 π π π π e2π −1 1 e2 x 1 2x 1 2x ∫ ∫ ∫ 2x 2 • C2 = e sin x dx = e (1 − cos 2x ) dx = − e cos 2 x dx = −J 20 4 20 4 2 0 0 π π π π 1 2x 1 2x 1 sin 2x d ( e ) ∫ ∫ ∫ 2x e d ( sin 2x ) = e sin 2x − J = e 2 x cos 2 x dx = 20 2 20 0 0 π π π π 1 2x 1 1 cos 2x d ( e 2x ) ∫ ∫ ∫ = − e 2x sin 2x dx = e d ( cos 2x ) = e 2x cos 2x − 20 2 20 0 0 π 2π 2π 2π 2π −1 e −1 e −1 e −1 e ∫ 2x − J ⇒ 2J = ⇒J= = − e cos 2x dx = 2 2 2 4 0 e2 π − 1 1 e2π − 1 e2 π − 1 e2 π − 1 ⇒ C2 = − J= − = 4 2 4 8 8 eπ eπ eπ eπ ∫ cos ( ln x ) dx = ∫ xd ( cos ( ln x )) = − ( e ∫ sin ( ln x ) dx + 1) + π • C3 = x cos ( ln x ) 1 − 1 1 1 π π e e eπ = − ( e + 1) + ∫ sin ( ln x ) dx = − ( e + 1) + ∫ − xd ( sin ( ln x ) ) π π x sin ( ln x ) 1 1 1 π e −1 ( π ) = − ( e + 1) − cos ( ln x ) dx = − ( e + 1) − C3 ⇒ 2C3 = − ( e + 1) ⇒ C3 = ∫ π π π e +1 2 1 eπ eπ eπ eπ eπ −1 1 1 1 1 • C4 = ∫ cos ( ln x ) dx = ∫ [1 + cos ( 2ln x)] dx = x ∫ 2 cos ( 2ln x) dx = − −I 2 2 21 2 2 1 1 1 2 16
Bài 8. Phương pháp tích phân t ng ph n eπ eπ eπ 2sin ( 2lnx) eπ ∫ ∫ ∫ Xét I = cos ( 2 ln x ) dx = xcos( 2lnx) 1 − xd( cos( 2lnx) ) = e −1+ x π dx x 1 1 1 π π e e π e = e − 1 + 2 ∫ sin ( 2 ln x ) dx = e − 1 + 2x sin ( 2 ln x ) 1 − 2 ∫ xd ( sin ( 2 ln x ) ) π π 1 1 π π e e 2 cos ( 2 ln x ) ∫ ∫ π π π dx = e − 1 − 4 cos ( 2 ln x ) dx = e − 1 − 4I = e −1− 2 x x 1 1 eπ − 1 eπ − 1 6 ( π = e − 1) ⇒ 5I = eπ − 1 ⇒ I = ⇒ C4 = e π − 1 + I = e π − 1 + 5 5 5 1 + sin x ) ( 1 + sin x ( x ) x 1 + sin x − e x d 1 + sin x ∫ ∫ ∫ • C5 = e x dx = d e =e 1 + cos x 1 + cos x 1 + cos x 1 + cos x x x 1 + sin x x 1 + cos x + sin x x 1 + sin x e dx e sin x dx ∫ ∫ ∫ x =e −e dx = e − − 2 (1 + cos x )2 1 + cos x 1 + cos x 1 + cos x (1 + cos x ) x x 1 + sin x e dx e sin x dx ∫ ∫ x − I − J (1) ; I = =e ;J= (1 + cos x ) 2 1 + cos x 1 + cos x du = e x dx u = e x   e x sin x dx ∫ (1 + cos x ) sin x dx ⇒  t Xét J = . −d (1 + cos x ) 1 ∫ 2 dv = v = = 2 (1 + cos x ) 2  1 + cos x (1 + cos x )  x x x e e dx e ∫ ⇒ J= (2). Thay (2) vào (1) ta có: − = −I 1 + cos x 1 + cos x 1 + cos x  ex  x 1 + sin x x 1 + sin x e ⇒ C5 = e x −I− − I + c = e − +c  1 + cos x  1 + cos x 1 + cos x 1 + cos x π π π sin 2 x π 1 −x 1 −x 1 −x ∫ ∫ ∫ ∫ • C6 = dx = e (1 − cos 2 x ) dx = e dx − e cos 2 x dx x e 20 20 20 0 −x π π π −π −π −e 1− e 1− e 1 −x 1 −x 1 ∫ ∫ = − e cos 2 x dx = − e cos 2 x dx = − J 2 20 2 20 2 2 0 π π π π −x 1 −x e sin 2x 1 sin 2x d ( e ) ∫ ∫ ∫ −x −x e d ( sin 2x ) = J= e cos 2 x dx = − 20 2 20 0 0 π π π π e− x cos 2 x −1 − x 1 −x 1 ∫ ∫ ∫ cos 2 x d ( e − x ) e d ( cos 2 x ) = − = e sin 2 x dx = + 20 40 4 40 0 2 17
Chương II: Nguyên hàm và tích phân − Tr n Phương π 1 − e −π 1 − x 1 − e −π 1 1 − e −π 1 − e −π 5 ∫ − J ⇒ J= ⇒J= = − e cos 2 x dx = 4 40 4 4 4 4 5 1 − e −π 1 1 − e −π 1 − e −π 2 ( = 1 − e −π ) ⇒ C6 = − J= − 2 2 2 10 5 a ∫ a 2 − x 2 dx ; ( a > 0 ) • C7 = 0 a2 − ( a2 − x2 ) a a a x 2 dx a − x d ( a2 − x2 ) = ∫ ∫ ∫ C7 = x a 2 − x 2 = dx 0 a2 − x2 a2 − x2 0 0 0 a a a a πa 2 dx x ∫ ∫ ∫ = a2 a 2 − x 2 dx = a 2 arcsin a 2 − x 2 dx = − − − C7 2 a a2 − x2 0 0 0 0 2 2 πa πa ⇒ 2C7 = ⇒ C7 = 2 4 a ∫ a 2 + x 2 dx ; ( a > 0 ) • C8 = 0 a a x2 a ( ) = a2 0 − ∫ xd ∫ 2 2 2 2 C8 = x a + x a +x 2− dx a2 + x2 0 0 ( a2 + x2 ) − a2 a a a dx ∫ ∫ ∫ 2 2 2 2 2 =a 2− dx = a 2− a + x dx + a 2 2 a + x2 2 a +x 0 0 0 a a a 2 + x 2 dx = a 2 2 + a 2 ln (1 + 2 ) − C8 ∫ = a 2 2 + a 2 ln x + a 2 + x 2 − 0 0 2 + ln (1 + 2 ) 2 ⇒ 2C8 = a 2 2 + a 2 ln (1 + 2 ) ⇒ C8 = a 2 du = dx u = x a   ∫ 2 2 2 ⇒ • C9 = x a + x dx ; ( a > 0 ) . t 3 dv = x a 2 + x 2 dx  v = 1 ( a + x ) 2 2 2  0  3 a a 3 3 x2 1 (2 C9 = ( a + x ) 2 a + x ) 2 dx ∫ 2 2 − 3 30 0 a a 2 2 4 a2 2 2 4 a2 122 1 ∫ ∫ 2 2 2 = a− a + x dx − x a + x dx = a− C8 − C9 3 3 30 3 3 3 0 2 18
Bài 8. Phương pháp tích phân t ng ph n 2 + ln (1+ 2 ) 3 2 − ln (1+ 2 ) 3 2 − ln (1+ 2 ) 2 4 224 a ⇒ C13 = ⇒ C9 = a− ⋅ = 3 3 3 2 6 8 du = dx u = x a   ∫ 2 2 2 ⇒ • C 10 = x a − x dx ; ( a > 0 ) . t 3 dv = x a 2 − x 2 dx  v = −1 ( a − x ) 2 2 2  0  3 a 1  a a a 3 3 −x ( 2 1 (2 a − x2 ) 2 a − x 2 ) 2 dx =  a 2 a 2 − x 2 dx + x 2 a 2 − x 2 dx  ∫ ∫ ∫ C10 = + 30  3 30   0 0 2 2 4 4 πa πa a 1 2 a ⇒ C10 = C7 = ⇒ C10 = = C7 + C10 3 3 3 3 12 8 2a 2a 2a x d ( x2 − a2 ) ∫ ∫ x 2 − a 2 dx = x x 2 − a 2 • C 11 = − a2 a2 a2 a + (x − a ) 2a 2a 2 2 2 x = (2 3 − 2 ) a − dx = ( 2 3 − 2 ) a − ∫ ∫ 2 2 x dx 2 2 2 2 x −a x −a a2 a2 2a 2a dx = (2 3 − 2 ) a − a ∫ ∫ 2 2 2 2 − x − a dx 2 2 x −a a2 a2 2a 2a = ( 2 3 − 2 ) a − a ln x + x − a ∫ 2 2 2 2 2 2 − x − a dx a2 a2 2+ 3 = ( 2 3 − 2 ) a − a ln 2 2 − C11 1+ 2 a 2+ 3 2 2+ 3 ⇒ 2C11 = ( 2 3 − 2 ) a − a ln ( 2 3 − 2 ) − ln 2 2 ⇒ C11 =  2 1+ 2  1+ 2 π2 π2 π2 π2 dx ∫ cotg x d ( sin x ) 1 cotg x 1 ∫ ∫d ( cotg x ) = − • C 12 = =− + 3 sin x sin x sin x π 4 π4 π4 π 4 π2 π2 cos x 1 ∫ sin x  sin  1 ∫ cotg x sin =− 2 − dx = − 2 − − 1 dx  2 2   x x π4 π4 π2 π2 π2 dx dx sin x dx ∫ ∫ ∫ =− 2 + − =− 2 + − C12 sin x π 4 sin 3 x 2 1 − cos x π4 π4 π2 − 2 + ln (1 + 2 ) 1 1 + cos x = − 2 + ln (1 + 2 ) ⇒ C12 = ⇒ 2C12 = − 2 − ln 2 1 − cos x 2 π4 2 19
Chương II: Nguyên hàm và tích phân − Tr n Phương 4 . D ng 4: Các bài toán t ng h p 3 x3 ( x 2 + 2 ) x 3 ( x 2 + 1) 3 3 3 x 5 + 2x 3 x3 ∫ ∫ ∫ ∫ • D1 = dx = dx = dx + dx x2 + 1 x2 + 1 x2 + 1 x2 + 1 0 0 0 0 3 3 x dx ∫ ∫x 2 2 2 = x .x x + 1 dx + =I+J 2 x +1 0 0 du = 2x dx u = x 2 3   ∫x ⇒ 2 2 t Xét I = .x x + 1 dx . 12 32 dv = x x 2 + 1 dx  v = ( x + 1)  0  3 3 3 3 12 2 2 1 32 32 32 I = x ( x + 1) x ( x + 1) dx = 8 − ∫ ( x + 1) d ( x + 1) ∫ 2 2 2 − 3 3 3 0 0 0 3 2( 2 2 ( 32 − 1) = 58 52 x + 1) =8− =8− 15 15 15 0 u = x 2 du = 2x dx  3   x dx ∫x x dx ⇒  2 t Xét J = . dv = x2 + 1 v = x 2 + 1  0  x2 + 1  3 3 3 3 22 4 32 + 1 d ( x + 1) = 6 − ( x + 1) ∫ 2x ∫ 2 2 2 2 2 J=x x +1 0 − x + 1 dx = 6 − = x 3 3 0 0 0 58 4 26 ⇒ D1 = I + J = += 15 3 5 2 1 d ( 1 + x3 ) 2 2 2 1 + x3 3  = − 1+ x 1 1 + x3 d  13 ∫ ∫ ∫ • D2 = dx = − +   4 x x  3 x3 x3 31 31 1 1 2 1 1 d (1 + x ) 2 2 2 3 211 x dx ∫ ∫ = −+ = −+ 3 8 2 1 x3 1 + x3 3 8 6 1 x3 1 + x3 d (u2 ) 3 3 211 211 du ∫ (u ∫u = −+ = −+ − 1) u 2 2 386 3 83 −1 2 2 3 2 1 1 1  2 1 1 u −1 − +  ln + 2 ln (1 + 2 )  = − + ln = 8 3 2  3 8 3 u +1 3 2 2 20
Bài 8. Phương pháp tích phân t ng ph n π2 π2 1 sin 2 x ∫e ∫ 2 cos 2 3 x ( 2 sin x cos x ) esin • D3 = sin x cos x dx = 2 x dx 4 0 0 π2 π2 π2 1 ) = 1 (1 + cos 2x ) esin 1 (1 + cos 2x ) d ( esin 2 2 sin 2 x ∫ ∫e x x d (1 + cos 2x ) = − 4 4 4 0 0 0 π2 π2 π2 −1 1 11 1 1 sin e ∫ d (e ) = − + e sin 2 x 2 2 ∫e sin x x = + sin 2x dx = − + = −1 22 22 2 2 2 0 0 0 π2 π2 ∫ ∫ ln (1 − cos x ) d ( sin x ) • D4 = cos x ln ( 1 − cos x ) dx = π3 π 3 π2 π2 3 sin x dx π2 ∫ ∫ sin xd ( ln (1 − cos x ) ) = = sin x ln (1 − cos x ) − ln 2 − sin x 1 − cos x 2 π3 π3 π3 π2 π2 2 1 − cos x 3 3 ∫ ∫ ln 2 − (1 + cos x ) dx = ln 2 − dx = 1 − cos x 2 2 π3 π3 π 3 3 3 π2 ln 2 − ( x + sin x ) = = ln 2 − − 1 + 2 2 6 2 π3 π3 π3 π3 π3 ∫ sin x ln( tg x) dx = −π∫ ln ( tg x) d ( cos x) = − cos x ln ( tg x) π ∫ cos x d ( ln ( tg x)) • D5 = + 4 π4 π4 4 π3 π3 π3 1 cos x dx 1 dx 1 sin x dx ∫ ∫ ∫ = − ln 3 + = − ln 3 + = − ln 3 + 2 2 4 4 sin x 4 π 4 cos x tg x π 4 sin x π4 π3 π3 d ( cos x ) 1 1 + cos x 1 1 3 = ln (1 + 2 ) − ∫ = − ln 3 − = − ln 3 − ln ln 3 2 2 1 − cos x 4 4 4 π 4 1 − cos x π4 π4 π4 π4 π4 π4 x + sin x d (1 + cos x ) ∫ xd ( tg 2 ) sin x dx x dx x ∫ ∫ ∫ 2 cos ∫ • D6 = dx = + =− + 1 + cos x x 1 + cos x 1 + cos x 2 0 0 0 0 0 2 π4 π4 π4  x ππ x 4 x ∫ =  − ln 1 + cos x + x tg  − dx = ln + tg + 2 ln cos tg  20 2 2+ 2 4 8 2 0 0 π ( 2 − 1) + ln 2 + 2 = π ( 2 − 1) + ln1 = π ( 2 − 1) 4 = ln + 4 4 4 4 2+ 2 2 21
Chương II: Nguyên hàm và tích phân − Tr n Phương π2 π2 π2 • D7 = ∫ sin 2x cos ( sin x ) dx = 2 ∫ sin x cos x cos4 ( sin x) dx = 2 ∫ sin x cos4 ( sin x) d( sin x) 4 0 0 0 1 1 1 1 1( t 1 + 2 cos 2t + cos 2t ) dt ∫ ∫ ∫ 4 2 2 = 2 t ( cos t ) dt = t (1 + cos 2t ) dt = 20 20 0 1 1 1 1 + cos 4t  1 ∫ ∫ t ( 2 + 4 cos 2t + cos 4t ) dt t  1 + 2 cos 2t +  dt = = 20   2 40 21 1 1 1   1 1 t 1 1 ∫ ∫ ∫ t ( 4 cos 2t + cos 4t ) dt = t d  2 sin 2t + sin 4t  = 2t dt + + 40   40 40 4 4 0 1 1 1 1  1  1 1 ∫ = + t  2 sin 2t + sin 4t  −  2 sin 2t + sin 4t  dt 4 4 0 4 0  4 4 1 1 1  1  1 1 +  2 sin 2 + sin 4  −  − cos 2t − cos 4t  = 4 4  4 0 4 16 1 1 1 1 31 = sin 2 + cos 2 + sin 4 + cos 4 + 2 4 16 16 64 π4 π4 π4 tg x 2 − cos2 x sin x ∫ ∫ ∫ 1 + sin2 x dx = d ( cos x ) • D8 = 2 − cos2 x dx = − cos x 2 cos2 x cos x 0 0 0 1 2 1 2 1 2 1 2 2 2 () 2−u 2−u ( ) 1 2−u d 1 = ∫ ∫ ∫ 2 2 =− du = − 2−u d 2 u u u u 1 1 1 1 1 2 1 2 π du u ∫ = 3 −1− = 3 − 1 − arcsin = 3 −1− 12 2 2 2−u 1 1 π3 π3 x 2 dx x cos x x ∫ ( x sin x + cos x ) ∫ cos x ⋅ ( x sin x + cos x ) • D9 = = dx 2 2 0 0 π3 π3 π3 x ( ) x x 1 1 1 ∫ ∫ x sin x + cos x d  cos x  =− =− ⋅ + d   cos x x sin x + cos x cos x x sin x + cos x 0 0 0 π3 −4π cos x + x sin x  dx = −4π + tg x π 3 = 3 3 − π ∫ x sin x + cos x  1 = +   0 2   3+ π 3 3+ π 3 3+ π 3 cos x 0 2 22