Cos6x cos5x sin6x sin5x
TRẮC NGHIỆM NGUYÊN HÀM TÍCH PHÂN Bài 1: NGUYÊN HÀM Mức độ ...
Câu 48: Một nguyên hàm của hàm số: y = cos5x.cosx là: A. cos6x B. sin6x C. 1 1 1 sin6x sin4x 2 6 4 §· ¨¸ ©¹ D. 1 sin6x sin4x 2 6 4 §· ¨¸ Câu 49: Nguyên hàm của hàm sốf(x) = 2sin3xcos2x A. 1 cos5x cosx C 5 B. 1 cos5x cosx C 5 C. 5cos5x cosx C D. Kết quả khác
[PDF File]Seminar 11
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(cos5x sin2x)dx c) Z sin3xsin2xdx b) Z cos2xsin3xdx d) Z cosxcos3xdx Problema 11.2. a) Z cos2 xdx; c) Z sin4 xdx b) Z sin2 3xdx d) Z cos5 xdx Problema 11.3. a) Z sin3 x 2 + cosx dx c) Z sinxdx cos2x b) Z sinx 1 + cosx+ cos2 x dx d) Z dx sinx+ sin2x Problema 11.4. a) Z cosx sinx dx c) Z dx cosxcos2x b) Z cos3 x sin4 x dx d) Z sin2xdx (2 + sinx)2 ...
[PDF File]SỞ GIÁO DỤC & ĐÀO TẠO CHƯƠNG 3 NGUYÊN HÀM –TÍCH PHÂN ...
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Trang 5/11 - Mã đề thi 134 Câu 44: Một nguyên hàm của hàm số: y = cos5x.cosx là: A. 1 1 1 sin6x sin4x 2 6 4 §· ¨¸ ©¹ B. sin6x C. cos6x D. 1 sin6x sin4x 2 6 4 §· ¨¸ ©¹ Câu 45: Nguyên hàm của hàm số f(x) = 2sin3xcos2x A. 1 cos5x cosx C 5 B.
[PDF File]Best Approach
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sin7x sin5x sin9x sin3x cos7x cos5x cos9x cos3x = tan 6x. Sol. L.H.S.= sin7x sin5x sin9x sin3x cos7x cos5x cos9x cos3x = 7x 5x 7x 5x 9x 3x 9x 3x 2sin cos 2sin cos 2 2 2 2 7x 5x 7x 5x 9x 3x 9x 3x 2cos cos 2cos cos 2 2 2 2 (Using "C, D" formulae) = 2sin6xcosx 2sin6xcos3x 2cos6xcosx 2cos6xcos3x = 2sin6xcosx cos3x 2cos6xcosx cos3x = sin6x cos6x ...
[PDF File]hoctot365.com
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sinx cosx cos2x cos3x cos4x cos5x sinx 1 2 1 sinxcosx sinxcos2x sinxcos3x sinxcos4x sinxcos5x sinx 2 sin2x sinx sin3x sin2x sin4x sin3x sin5x sin4x sin6x sinx sin5x sin6x 0 sin6x sin( 5x) k2 x x h2 ,k,h 11 ¢
[PDF File]PHÖÔNG TRÌNH LÖÔÏNG GIAÙC
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1) sin4x.sin7x = cos3x.cos6x ĐS : k 2 x 10 k 20 x, k Z DBÑH 2004 2) cosx.cos7x = cos3x.cos5x ĐS : 4 k x 2 k x , k Z 3) cos11x.cos3x = cos17x.cos9x ĐS : 6 k x 20 k x , k Z 4) sin18x.cos13x = sin9x.cos4x ĐS : 22 k 44 x 9 k x , k Z 5) x cos 2x 6 x sin 6 4cos x.sin
[PDF File]arXiv:1004.2168v1 [math.NT] 13 Apr 2010
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cos5x + cos2x +sin3x cos5x, s6(x) = cos5x +sinx cos6x + cosx+sin5x cos6x, s7(x) = cos3x +sin4x cos7x + cosx+sin6x cos7x − cos5x+sin2x cos7x. Our paper is organized as follows. In Section 2, we compute the generating function sm(x) when m is square-free, while in Section 3 we consider the case when m is not square-free.
[PDF File]Trigonometrikus egyenletek
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a) cos5x= sin9x x 1 = 28 + k ˇ 7 x 2 = ˇ 8 + k ˇ 2 b) cos3x= sin5x x 1 = ˇ 16 + k ˇ ˇ 4 x 2 = ˇ 4 + k 2 ˇ 1 c) sin3x= cos6x x 1 = 18 + k 2 ˇ 9 x 2 = ˇ 6 + k ˇ 3 d) sin6x= cos9x x 1 = ˇ 30 + k 2 ˇ 15 x 2 = ˇ 6 + k ˇ 2 3 e) cos4x= sin5x x 1 = ˇ 18 + k 2 ˇ 9 x 2 = 2 2 + k ˇ 1 f) cos6x= sin11x x 1 = ˇ 34 + k 2 ˇ 17 x 2 = 2 ˇ 10 ...
[PDF File]AP CALCULUS ERIVATIVES TEST ADDITIONAL PROBLEMS
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AP CALCULUS DERIVATIVES TEST ADDITIONAL PROBLEMS Differentiate the following. The ONLY simplification you should do is remove all negative and fractional exponents. 1. f(x) = (2x4 x)tan(x2 41) 2. y = 4 p sin5x
[PDF File]Sample Problems - aceh.b-cdn.net
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cos5x dx Solution: We know that d dx cosx = sinx + C. We will use substitution. Let u = 5x and then du = 5dx and so du 5 = dx. Z cos5x dx = Z cosu du 5 = 1 5 Z cosu du = 1 5 sin5x+C Note: Once we have enough practice, there is no need to perform this substitution in writing. We can just simply write Z cos5x dx = 1 5 sin5x+C. 3. Z cosxsin4 x dx ...
[PDF File]Преобразования тригонометрических уравнений
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sin5x sinx cos5x cosx = sinx sin5x cosx cos5x: x = ˇ k 8; nˇ ... sin6x sinx cosx = cos6x cosx+sinx;
[PDF File]Formulaire de trigonométrie circulaire - TrigoFACILE
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Formules de trigonométrie circulaire Soient a,b,p,q,x,y ∈ R (tels que les fonctions soient bien définies) et n ∈ N. La parfaite connaissance des graphes des fonctions trigonométriques est nécessaire.
[PDF File]TRIGNOMETRY RAKESH YADAV Practice Set https://www.amazon ...
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sin7x–sin5x cos6x–cos4x – cos7x+cos5x sin6x+sin4x? (a) 2tanx (b) tanx (c) 2sinx (d) 2cosx 5. What is the value of sinx+siny sinx–siny =? cosx+cosy cosy–cosx sinx+siny sinx–siny cosx+cosy cosy–cosx dk eku D;k g\aS (a) 0 (b) 1 (c) –1 (d) 2 6.
[PDF File]Seminar 2
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(cos5x sin2x)dx c) Z sin3xsin2xdx b) Z cos2xsin3xdx d) Z cosxcos3xdx Problema 2.2. a) Z cos2 xdx; c) Z sin4 xdx b) Z sin2 3xdx d) Z cos5 xdx Problema 2.3. a) Z sin3 x 2 + cosx dx c) Z sinxdx cos2x b) Z sinx 1 + cosx+ cos2 x dx d) Z dx sinx+ sin2x Problema 2.4. a) Z cosx sinx dx c) Z dx cosxcos2x b) Z cos3 x sin4 x dx d) Z sin2xdx (2 + sinx)2 ...
[PDF File]The Generating Function for the Dirichlet Series L s
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cos5x + cos2x+sin3x cos5x; s6(x) = cos5x+sinx cos6x + cosx+sin5x cos6x; s7(x) = cos3x+sin4x cos7x + cosx+sin6x cos7x cos5x+sin2x cos7x: Our paper is organized as follows. In Section 2, we compute the generating function sm(x) when m is square-free, while in Section 3 we consider the case when m is not square-free.
[PDF File]Trigonometry solutions class 11 pdf
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NOW, Take the Municipality 2SIN4X SIN2X + 2SIN4X + SIN6X = 2SIN4X (COS2X +1) () = 2SIN4X (+1) = 2SIN4X () = RH S Question: 15 Demonstrate the following response: We know that using this, we get sin5x + sin3x = 2sin4xcosx now nultiply and divide for sin x now we know that with this our equation becomes r.h.s. Question: 16
[PDF File]HOC360.NET - TÀI LIỆU HỌC TẬP MIỄN PHÍ 1 1 25 25 541 41 41
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(sin6x + sin0) = 1 2 sin4x + 1 2 sin2x – 1 2 sin6x b) B = 8cosxsin2xsin3x = (8sin3xsin2x)cosx = 4(cosx – cos5x).cosx = 4cos2x – 4cos5x.cosx = 2(1 + cos2x) – 2(cos4x + cos6x) = 2 + 2cos2x – 2cos4x – 2cos6x c) C = sin(a + 300).cos(a – 300) = 1 2 (sin2a + sin600) = 1 2 sin2a + 3 4 d) D = sinx .sinx .cos2x 6 6
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Câu 9: Tìm (cos6x cos4x)dx là: A. 1 1 sin6x sin4x C 6 4 B. 6sin6x 5sin4x C C. 1 1 sin6x sin4x C 6 4 D. 6sin6x sin4x C Câu 10: Tính nguyên hàm 1 dx
[PDF File]Sample Problems
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cos5x dx Solution: We know that d dx cosx = sinx + C. We will use substitution. Let u = 5x and then du = 5dx and so du 5 = dx. Z cos5x dx = Z cosu du 5 = 1 5 Z cosu du = 1 5 sin5x+C Note: Once we have enough practice, there is no need to perform this substitution in writing. We can just simply write Z cos5x dx = 1 5 sin5x+C. 3. Z cosxsin4 x dx ...
[PDF File]y=− - Chino Valley Unified School District
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y=sin5x y=−sin6x y=−sin7x y=sin8x 7y=−sin9x y=−sin10x y=2sin1x y=3sin2x y=4sin3x y=5sin1 4 x y=0.6sinx 5 y=−.7sin1 6 x y=−1 8 sinx y=−9sinπx y=πsin1 π x y=−2πsinx π • y=cos1x y=cos2x y=cos3x y=−cos4x y=cos5x y=−cos6x y=−cos7x.y=cos8x y=−cos9x y=−cos10x y=2cos1x y=3cos2x y=4cos3x y=5cos1 4 x y=0.6cosx 5 y=− ...
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