Tài liệu Phân dạng và phương pháp giải trắc nghiệm tích phân

GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN Chủ đề 1 : NGUYÊN HÀM ( TÍCH PHÂN BẤT ĐỊNH ) 1) Định nghĩa : F(x) là nguyên hàm của hàm số f(x) trên (a; b) ⇔ ………………. 2) Họ nguyên hàm : , với C là hằng số 3) Bảng nguyên hàm : Hàm cơ bản : ∫ dx= Hàm chứa (ax + b) x+C x α +1 +C .dx ∫ x= α +1 α dx = ∫ x ln x + C dx 1 = − ∫ x2 x + C dx = ∫ x 2 x +C ax = +C ∫ a dx lna x x x e dx = e +C ∫ −cosx + C ∫ sinx.dx = = ∫ cosx.dx sinx + C dx tanx + C 2 ∫ cos= x dx −cotx + C ∫ sin 2 x = dx 1 x −a = ln ∫ x 2 − a 2 2a x + a + C dx −1 = ∫ x n (n − 1)x n −1 + C Đt : 0914.449.230 (Zalo – Facebook) 1 (ax + b)α +1 +C ) dx ∫ ( ax + b= a α +1 dx 1 = ∫ ax + b a ln ax + b + C α dx 1 1 = − ∫ (ax + b)2 a . ax + b + C ∫ dx 2 = ax + b + C ax + b a 1 a ax + b dx +C ∫ a= a lna 1 ax + b ax + b e dx e = +C ∫ a ax + b 1 sin(ax + b).dx = − cos(ax + b) + C ∫ a 1 + = cos(ax b).dx sin(ax + b) + C ∫ a dx 1 = ∫ cos2 (ax + b) a tan(ax + b) + C dx 1 = − ∫ sin 2 (ax + b) a cot(ax + b) + C dx 1 1 = − ∫ (ax + b)n a (n − 1)(ax + b)n −1 + C 1 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN 4) Cách tìmnguyên hàm : Biến đổi tích hoặc thương, tổng, bạ bậc, khai triển lũy thừy, chia đa thức….. m m 1 n m −n x n x= ; n x= ; n x m−n Căn thức thành lũy thừa= : x x x 5) Công thức thường dùng : 1 + cos2u 2 1 − cos2u sin 2 u = 2 1 = 1 + tan 2 u 2 cos u 1 = 1 + cot 2 u 2 sin u cos 2 u = 3cosu + cos3u 4 3sinu − sin3u sin 3 u = 4 cos3 u = sin2u = 2sinu.cosu cos2u = cos 2 u − sin 2 u cos2u = 2cos 2 u − 1 cos2u = 1 − 2sin 2 u Ví dụ : TÌM HỌ NGUYÊN HÀM CỦA CÁC HÀM SỐ SAU: a/ f(x) = (2x 2 + 1)3 c/ f(x) = ; b/= f(x) (tan x + cot x) 2 ; 2x 3 − 5x + 2 ; x2 d/ f(x) = e 2x − 3e x + 2 ex − 1 ♥ Giải : a/ Ta có f(x) = 8x 6 + 12x 4 + 6x 2 + 1 , Suy ra : f(x) = 8∫ x 6 dx + 12 ∫ x 4 dx + 6 ∫ x 2 dx + ∫ 1dx = 8 7 12 5 x + x + 2x 3 + x + C 7 5 1 1  1   1  b/ Ta có f(x) + 2 2  = tan 2 x + cot 2 x += − 1 +  2 − 1= +2 2 2 cos x sin x  cos x   sin x  Suy ra: ∫ f(x)dx = c/ Ta có f(x) = 2x − Suy ra: 1 ∫ cos 2 x dx + ∫ 1 dx = tan x − cot x + C sin 2 x 5 2 + . x x2 1 ∫ f(x)dx = 2∫ xdx − 5∫ x dx + 2∫ x −2 dx = x 2 − 5ln x − 2 +C x e 2x − e x − 2(e x − 1) e x (e x − 1) − 2(e x − 1) (e x − 1)(e x − 2) d/ Ta có f(x) = = = ex − 2 = x x x e −1 e −1 e −1 Suy ra: ∫ f(x)dx = ∫ e x dx − ∫ 2dx = e x − 2x + C Đt : 0914.449.230 (Zalo – Facebook) 2 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN Bài Tập 1 : Tìm họ nguyên hàm các hàm số sau 1/ f(x) = x 5 + 3x 2 − 5 − 1 x 2/ f(x) = 3 7 9 20 + 4− 3+ 2 5 x x x x x 5 + 4x 7 − 2x + 8 − 7x 9 3/ f(x) = x2 4/ f(x) = x + 3 x + 4 4 x 5/ f(x) = ( x + 1)(x − x + 1) 6/ f(x) = e x (7 − 3e − x +  e− x  7/ = f(x) e x  2 + 2  sin x   8/ f(x) = (2 x e− x ) cos 2 x + 3x ) .22x −1 ♥ Giải : .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... Đt : 0914.449.230 (Zalo – Facebook) 3 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN Bài Tập 2 : Tìm họ nguyên hàm các hàm số sau 1/ f(x) = 2sinx − 3cosx + 7 x 4/ f(x) = 1 2 sin x.cos 2 x 8/ f(x) = 3x15 + 7x 4 − 2x + 8 − 10x 6 x3 9/ f(x) = 6 sin x.cos 2 x 2 2/ = f(x) tan 2 x − 3cot 2 x 3/= f(x) (2tanx + cotx) 2 5/ f(x) = ( x 5 − 3x ) ( x − 1) 6/ = f(x) 3sinx − 7cosx 2 7/ f(x) = 2 x − 3e x + 4sin x − 8 / x 3 10/ f(x) = e x (5 + 3e − x ) ♥ Giải : .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... Đt : 0914.449.230 (Zalo – Facebook) 4 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN .................................................................................................................................................................................... .................................................................................................................................................................................... Bài Tập 3 : Tìm họ nguyên hàm các hàm số sau x x cos 2 2 1/ f(x) = x 3 − 3x 2 + 4x + 3 ; 2/= f(x) 2x(x 2 + 3x) 2 3/ f(x) = 4sin 4/ f(x) = 2sin x + 3cos x + 5e x 5/= f(x) tan x 2 − 3 1 6/ f(x)= (2 − ) 2 x 8/ f(x) = 22x +1.33x + 2 9/ f(x) = (3x − 2) 2 7/ f(x) = ( x − 2)3 x ♥ Giải : .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... Đt : 0914.449.230 (Zalo – Facebook) 5 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN Ví dụ 2 : TÌM HỌ NGUYÊN HÀM CỦA CÁC HÀM SỐ SAU: a/ f(x) = (2x + 1)3 b/ = f(x) cos ( 3x − 2 ) ; ; d/ f(x) = e − x c/ f(x) = 2 7x + 1 e/ f(x)= (7 − 3x)10 Giải : a/ sử dụng công thức ) dx ∫ ( ax + b= α 1 (ax + b)α +1 +C a α +1 1 (2x + 1) 4 3 = + = +C f(x)dx (2x 1) dx . ∫ ∫ 2 4 b/ sử dụng công thức = ∫ cos(ax + b).dx 1 .sin ( 3x − 2 ) + C 3 = ∫ cos ( 3x − 2= )dx ∫ f(x)dx c/ sử dụng công thức 2 dx ∫ ax = +b dx dx 2 ∫ = = ∫ = ∫ f(x)dx 7x + 1 7x + 1 d/ sử dụng công thức ∫ f(x)dx =∫ e −x 1 sin(ax + b) + C a dx ∫e= ax + b 1 ln ax + b + C a 2 .ln 3x − 2 + C 7 1 ax + b e +C a 1 dx = e − x + C = −e − x + C −1 ( chú ý hệ số a trong bài này là -1 ) 1 (7 − 3x)11 e/ giống bài a/ ∫ f(x)dx = . +C ∫ (7 − 3x) dx = −3 11 10 Điền vào ô trống a/ ∫ (7 − 4x)5dx = dx c/ ∫ ( 4 x + 1) e/ ∫e −x 5 = dx = Đt : 0914.449.230 (Zalo – Facebook) dx b/ ∫ 2x + 7 = d/ ∫e f/ ∫ cos ( −π x ) = 8x + 7 dx = dx 2 6 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN Bài tập 4 : Tìm họ nguyên hàm các hàm số sau 1/ f(x) = sin 2 x ; 2 / f(x) = sin 2 7x 5/ f(x) = sin 4 2 x ; 6/ f(x) = 7 sin 2 x .cos 2 x 8/ f(x) = sin 4 x .sin 6x ; 10 / f(x) cosx . ( 3 + cosx ) = ; 3/ f(x) = cos 2 4 x ; ; 4/ f(x) = cos 4 x 7 / f(x) = sin 2 x .cos x 9 / f(x) = cos 6 x .cos 2 x ; 11 / f(x) cosx . ( sin 3x + sinx ) ; = ♥ Giải : .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... Đt : 0914.449.230 (Zalo – Facebook) 7 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN Bài tập 5 : Tìm họ nguyên hàm các hàm số sau 1 / f(x) = 4/ f(x) = x 3 + 3x 2 − 6x + 5 ; x +1 3 π  cos 2  2x +  4  ; 1 x +9 − x 2/ f(x) = 5 / f(x) = ; −6x + 5 2x − 5 3/ f(x) = 3x 2 − 6x + 5 2x + 1 6/= f(x) cos 4 x − sin 4 x ♥ Giải : .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... Bài tập 6 : Tìm họ nguyên hàm các hàm số sau 1/ (HV Quan Hệ Quốc Tế - 1997) f(x) = ( sin 4 x + cos4 x ) .( sin 6 x + cos6 x ) 2/ (ĐH Ngoại Thương – 1998- Khối A) f(x) = Đt : 0914.449.230 (Zalo – Facebook) x4 + x2 +1 x2 + x +1 8 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN 3/ (ĐH Ngoại Thương – 1998- Khối D) f(x) = x 4 + 2x 2 + 2 + x x2 + x +1 4/ (ĐH Ngoại Thương – 2000 - Khối D) f(x) = cos2x sinx + cosx  x −1  5 / f(x) =    x+2 2 6 / f(x) = cos 5 x .cos 2 x .sinx ♥ Giải : .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... Đt : 0914.449.230 (Zalo – Facebook) 9 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh Ví dụ 3 : a/ Tìm A, B sao cho b/ Tính I = ∫ 3x + 7 A B = + x + 4x + 3 x + 1 x + 3 ( x ≠ −1; 3 ) 2 3x + 7 dx x + 4x + 3 2 3x + 7 A B 7 A ( x + 3) + B ( x + 1) = + ⇔ 3x + = x + 4x + 3 x + 1 x + 3 Giải :a/ 2 A+B 3 = = A 2 ⇔ +B 7 = 3A= B 1 ( A + B ) .x + 3A + B ⇔  ⇔ 3x + 7 = b/= I NGUYÊN HÀM - TÍCH PHÂN ∫x 2 3x + 7 = dx + 4x + 3  2 1  dx ∫  x + 1 + x + 3 = 2 ln x + 1 + ln x + 3 + C Bài tập 7 : Tính các nguyên hàm số sau ( sử dụng pp …………….....… ) A=∫ 3x + 4 dx x + 4x − 5 ; B=∫ x+7 dx x + 8x − 9 D=∫ dx x ( x + 1) ; E=∫ x2 −1 dx ; ( x + 2 )( x − 2 )( x − 3) F=∫ −x dx x + x −6 ; 2 2 2 G=∫ 3 dx x + 7x + 12 2 C=∫ ; ; F=∫ 1 dx x −x−2 2 −8 dx x + 10x + 9 2 ♥ Giải : .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... Đt : 0914.449.230 (Zalo – Facebook) 10 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN Trắc nghiệm bổ sung TN1 : Tìm nguyên hàm của hàm số f(x) biết f ( x) = 2x + 3 x + 4x + 3 2 x 2 + 3x A. x 2 + 3x +C x2 + 4x + 3 B. − C. 1 ( ln x + 1 + 3ln x + 3 ) + C 2 D. (2 x + 3) ln x 2 + 4 x + 3 + C ( x 2 + 4 x + 3) 2 +C TN2 : Xét các mệnh đề: (I) dx cot x + C 2 ∫ sin= x (II) e3 x + 1 1 2x x ∫ e x + 1 dx= 2 e − e + x + C Khẳng định nào sau đây là đúng? A.(I) đúng , (II) sai B. (I) sai, (II) đúng C. Cả (I) và (II) đều đúng D. Cả (I) và (II) đều sai TN3 : Cho F(x) là một nguyên hàm của f(x) = cos 2x. π Khi đó, hiệu số F ( ) − F (0) bằng: 4 A. 15 B. 1 2 C. 2 D. 2 4 TN4 : Cho F(x) và G(x) là các nguyên hàm của hàm số f(x) trên khoảng (a,b). Khi đó (I) F(x) = G(x) + C (II) G(x) = F(x) + C Với C là một hằng số nào đó. Khẳng định nào sau đây là đúng ? A. (I) đúng, (II) sai B. (I) sai, (II) đúng C. Cả (I) và (II) đều đúng D. Cả (I) và (II) đều sai TN5 : Hàm số y = A. 2x4 + 3 có một nguyên hàm là x2 2 x3 3 − +3 3 x B. −3 x3 3 +2 x C. 2 x3 3 + +1 3 x D. x3 3 − + 2017 3 x VD4 : Tính các nguyên hàm sau ( sử dụng pp đổi biến số ) a/ A = ∫ esinx .cosxdx c/ C = ∫ ln 5 x dx x ; b/ B = ∫ ; 2x + 4 dx x + 4x − 5 d/ D = ∫ 2 ex dx ex + 1 Giải : a/ A = ∫ esinx .cosxdx ; đặt t= sinx ⇒ dt= cosxdx Vậy A = ∫ e t .dt = e t + C = esinx + C Đt : 0914.449.230 (Zalo – Facebook) 11 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN b/ B = ∫ 2x + 4 dx Đặt t = x 2 + 4x − 5 ⇒ dt = x + 4x − 5 Vậy B= ∫ 2 ( 2x + 4 ) dx dt = ln t + C= ln x 2 + 4x − 5 + C t ln 5 x dx c/ C = ∫ dx ; đặt t = ln x ⇒ dt = x x t6 ln 6 x Vậy C = ∫ t .dt = +C = +C 6 6 5 ex d/ D = ∫ x dx e +1 ∫ Vậy : D= ; đặt t = e x + 1 ⇒ dt = e x dx dt = ln t + C= ln e x + 1 + C t CÁCH ĐỔI BIẾN SỐ CẦN NHỚ Dạng Tích Phân Cách Giải + Nếu bậc tử ≥ bậc mẫu ta chia đa thức f(x) ∫ g(x) .dx + Nếu bậc tử < bậc mẫu ta xem tử có phải là đạo hàm của mẫu hay ko ? nếu có đặt t = mẫu số + Nếu ko có 2 trường hợp này ta sẽ làm theo dạng khác sẽ trình bày ở phần khác ∫ n ........dx dx ∫ f(lnx). x Đặt= t n ....... ⇒ = t n ....... sau đó lấy đạo hàm 2 vế dx x Đặt t = lnx + C ⇒ dt = ∫ f(cosx).sinxdx Đặt t = cos x + C ⇒ dt = − sin xdx ∫ f(sinx).cosxdx Đặt = t sin x + C ⇒ dt= cos xdx dx ∫ f(tanx) cos2 x Đặt = t tan x + C ⇒ dt= dx ∫ f(cotx) sin 2 x Đặt t = cot x + C ⇒ dt = − ∫ f(e x ).e x dx dx cos 2 x dx sin 2 x Đặt t = e x + C ⇒ dt = e x dx Đt : 0914.449.230 (Zalo – Facebook) 12 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh dx dx ∫ sin n x , ∫ cosn x với n chẵn ∫ sin n xdx hay ∫ cos xdx n NGUYÊN HÀM - TÍCH PHÂN Đưa về n xdx hay ∫ cos xdx 1 n−2 1 1 1 1 1 ... 2 dx, ∫ . ... dx n−4 n−2 n−4 sin x cos x cos x cos 2 x x sin . Và Đặt = t tan x + C ⇒ dt= dx cos 2 x Dùng công thức hạ bậc = cos 2 u với n chẵn ∫ sin ∫ sin 1 + cos2u 1 − cos2u = ; sin 2 u 2 2 Tách ∫ sin n xdx = ∫ sin n −1x.sinxdx , đặt t = cosx n với n lẽ ∫ cos xdx = ∫ cos n n −1 x.cosxdx , đặt t = sinx + Nếu mẫu có 2 nghiệm x1 , x 2 , ta đưa về Ax + B ∫ a(x − x )(x − x 1 2 ) dx Sau đó dùng pp hệ số bất định Ax + B ∫ ax 2 + bx + cdx + Nếu mẫu có nghiệm kép x 0 , ta đưa về Ax + B dx 2 0) ∫ a(x − x + Nếu mẫu vô nghiệm ,đưa về Ax + B dx 2 + D2 ∫X  π π và đặt X = D.tant t ∈  − ;   2 2 1/ R(x, a − x 2 ) thì đặt x = sint 2/ R(x, a + x 2 ) thì đặt x = atant Bài tập 1 : Tính các nguyên hàm sau = A D = ∫ x(2 − x ∫ x. G=∫ 2 12 ) dx B=∫ x 2 + 1.dx e x .dx 1 + ex 8xdx x2 +1 = E I=∫ C = ∫x . 3 4 x +1 dx 3x + 1 ∫3 1 + 4sin x.cos x.dx F=∫ 1 − x.dx J=∫ 3dx 2x. 2 + ln x x 2 dx 2x 3 + e ♥ Giải : Đt : 0914.449.230 (Zalo – Facebook) 13 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... Bài tập 2 : Tính các nguyên hàm sau = K = R 5 3 ∫ x 2 − x .dx ∫ 2x .(x 7 4 − 1)5 .dx N = ∫ cos xdx 5 L=∫ O=∫ 2 + 3ln x dx x P=∫ xdx 2x + 1 M=∫ e tanx W=∫ dx cos 2 x Đt : 0914.449.230 (Zalo – Facebook) cos x dx sin 2 x S=∫ 14 xdx (2x + 1)3 1 dx x. ( 4lnx + 7 ) Q=∫ ecot x .dx sin 2 x V=∫ dx x −5 3 T = ∫ sin 3 xdx BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN ♥ Giải : .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... Đt : 0914.449.230 (Zalo – Facebook) 15 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN Bài tập 3 : Tính các nguyên hàm sau B = ∫ tanx.dx A = ∫ cot x.dx D=∫ sin2x ( 3 + cos x ) 2 4 .dx ; E=∫ ∫ ( 2 − sin x ) = C sinx − cosx .dx sinx + cosx F = 2 2 .sin2x.dx ∫ ( cos x + sin x ) .cos 2x.dx 4 4 ♥ Giải : .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... Bài tập 4 : Tính các nguyên hàm sau = G 6 6 ∫ 4 ( cos x + sin x ) .cos 2x.dx M=∫ sin 3 x .dx cos 2 x Đt : 0914.449.230 (Zalo – Facebook) K=∫ N=∫ 1 .dx x e +1 3sin 2 x 4 cos 2 x + 5sin 2 x 16 L=∫ sin 5 x .dx cos 7 x .dx BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh A = ∫ 10 NGUYÊN HÀM - TÍCH PHÂN dx x (ĐHQG Hà Nội – 1999) .dx (HV CNBCVT – 1999) B = ∫ x e − 4e − x x +1 C = ∫ 6sin 2x.cos 4 xdx (ĐH Thủy Lợi– 2001) ♥ Giải : .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... Đt : 0914.449.230 (Zalo – Facebook) 17 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN π Ví dụ 5 : a/ Tìm một nguyên hàm của hàm số f(x) = tan 2 x , biết F( ) = 0 4 π  π b/ Cho hàm số f= ( x) sin x + cos 2 x . Tìm nguyên hàm F ( x) của hàm số f ( x) biết F   = 2 2 Giải : a/ ∫ f(x)dx =  1  ∫ tan xdx= ∫  cos x − 1 dx= 2 2 tan x − x + = C F(x) π π π π π π F( ) =tan − + C =1 − + C =0 ⇔ C = − 1 ; Vậy F(x) = tan x − x + − 1 4 4 4 4 4 4 b/ ∫ ( sin x + cos 2 x )dx= π π π F( ) = ⇔ C = . 2 2 2 1 sin 2 x − cos x + C 2 Vậy F ( x= ) 1 π sin 2 x − cos x + 2 2 Bài tập tương tự Tìm một nguyên hàm của các hàm số sau a/ f(x) = x 3 + 3x 2 + 3x − 1 biết F(1) = 1/ 3 (TN THPT – 2003) x 2 + 2x + 1 π 2 2 b/ f(x)= x + sin x biết F( ) = 4 3 d/ f(x) = c/ f(x) =e 2x −1 + cos 2x + 3 biết F(0) = 1 + 2x 2 biết F( − 1) = 3 x 3 e e/ f(x) cos x. ( 2 − 3 tan x ) biết F(π) = 1 = ♥ Giải : .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... Đt : 0914.449.230 (Zalo – Facebook) 18 BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh NGUYÊN HÀM - TÍCH PHÂN .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... CÂU HỎI TRẮC NGHIỆM PHẦN NGUYÊN HÀM Câu 01 : Họ nguyên hàm của hàm số f ( x) = A. F ( x) = 2 23 3 43 4 54 x + x + x +C 3 4 5 B. F ( x) = 2 23 3 43 4 54 x + x + x +C 3 4 5 C. F ( x) = 2 23 4 43 5 54 x + x + x +C 3 3 4 D. F ( x) = 2 32 3 43 4 54 x + x + x +C 3 4 5 x+3 x+4 x 1 x+9 − x Câu 02 : Tìm nguyên hàm của hàm số f(x) biết f ( x) = A. 2 27 ( ( x + 9) 3 ) − x3 + C Đt : 0914.449.230 (Zalo – Facebook) B. 2 27 ( 19 ( x + 9) 3 ) + x3 + C BIÊN HÒA – ĐỒNG NAI GV : ThS Nguyễn Vũ Minh 2 C. 3( ( x + 9) 3 − x ) NGUYÊN HÀM - TÍCH PHÂN D. Đáp án khác +C 3 Câu 03 : Nguyên hàm của hàm số f ( x) = tan 2 x là sin x − x cos x +C cos x tan 3 x A. +C 3 B. C. tan x − 1 + C D. Đáp án khác Câu 04 : Hàm số F ( x) = e x + tan x + C là nguyên hàm của hàm số f(x) nào A. f ( x= ) ex − 1 sin 2 x B. f ( x= ) ex +  e− x  C. f = ( x) e x 1 +  2  cos x  D. Đáp án khác Câu 05 : Nguyên hàm của hàm số f(x) = x3 − A. 1 sin 2 x 3 + 2 x là: 2 x x4 − 3ln x 2 + 2 x.ln 2 + C 4 B. x3 1 + 3 + 2x + C 3 x C. x4 3 2x + + +C 4 x ln 2 D. x4 3 + + 2 x.ln 2 + C 4 x Câu 06 : Nguyên hàm của hàm số: y = cos 2 x là: sin 2 x.cos 2 x A. tanx − cotx + C B. −tanx − cotx + C C. tanx + cotx + C D. cotx −tanx + C  e− x  Câu 07 : Nguyên hàm của hàm số: y = e  2 +  là: cos 2 x   x A. 2e x − tan x + C B. 2e x − 1 +C cos x C. 2e x + 1 +C cos x D. 2e x + tan x + C Câu 08 : Nguyên hàm của hàm số: y = cos2x.sinx là: A. 1 cos3 x + C 3 B. − cos3 x + C 1 C. - cos3 x + C 3 D. 1 3 sin x + C . 3 Câu 09 : Một nguyên hàm của hàm số: y = cos5x.cosx là: Đt : 0914.449.230 (Zalo – Facebook) 20 BIÊN HÒA – ĐỒNG NAI