2 cos 2x 3 sin x
[PDF File]Techniques of Integration - Whitman College
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204 Chapter 10 Techniques of Integration EXAMPLE 10.1.2 Evaluate Z sin6 xdx. Use sin2 x = (1 − cos(2x))/2 to rewrite the function: Z sin6 xdx = Z (sin2 x)3 dx = Z (1− cos2x)3 8 dx = 1 8 Z 1−3cos2x+3cos2 2x− cos3 2xdx. Now we have four integrals to evaluate: Z 1dx = x and Z
[PDF File]Calculus II - Homework 3 Solutions
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(2 sin(2 )) + C: Now = arcsin x 3 and sin(2 ) = 2sin cos = 2 9 x p 9 x2. Thus the above result turns into 9 2 arcsin x 3 1 2 x p 9 x2 + C: (7.3.12) Z 2 0 dt p 4 + t2: Solution: Let t = 2tan . Then p 4 + t2 = 2sec and dt = 2sec2 . Also, = arctan t 2, and thus the interval of integration for is [0;ˇ=4]. We have Z 2 0 dt p 4 + t2 = Z ˇ=4 0 2sec2 ...
[PDF File]Questions - 2013-2019 Marks are indicated in brackets ...
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(ii) cos 2x. and 0 < x < (b) By expressing sin 3x as sin(2x+x), find the exact value of sin 3x. (a) Express 2cosx0—smx0 in the form k > 0, 0 < a < 360. (b) Hence, or otherwise, find (i) the minimum value of 6cosx0—3smx0 and (ii) the value of x for which it occurs where x < 360. (a) (b) sm 2x sm x cos2 x Show that 2 cos x sm 2x sm3 x , where ...
[PDF File]Formule trigonometrice a b a b c b a c - Math
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3. tg ctg = 1: 4. sin ˇ 2 = cos ; sin ... 1 2x x = arcctg x p 1 2; 0
[PDF File]TRIGONOMETRY LAWS AND IDENTITIES
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sin(xy)=sin(x)cos(y)cos(x)sin(y) cos(xy) = cos(x)cos(y)+sin(x)sin(y) tan(xy)= tan(x)tan(y) 1+tan(x)tan(y) LAW OF SINES sin(A) a = sin(B) b = sin(C) c DOUBLE-ANGLE IDENTITIES sin(2x)=2sin(x)cos(x) cos(2x) = cos2(x)sin2(x) = 2cos2(x)1 =12sin2(x) tan(2x)= 2tan(x) 1 2tan (x) HALF-ANGLE IDENTITIES sin ⇣x 2 ⌘ = ± r 1cos(x) 2 cos ⇣x 2 ⌘ = ± ...
[PDF File]Method of Undetermined Coefficients (aka: Method of ...
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yp(x) = Acos(2x) + B sin(2x) = −7cos(2x) − 4sin(2x) . This example illustrates that, typically, if g(x) is a sine or cosine function (or a linear combination of a sine and cosine function with the same frequency) then a linear combination
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2 cos (n9) 3 sin (n6) , where n is a positive integer , is 10 . State the value of n and hence find the largest solution of this equation in the interval O 360 [ W -12/ 11/Q71 Q33. i) Show that the equation 2 cos x = 3 tan x can be written as a quadratic equation in ii) Solve the equation 2 cos 2y 3 tan 2 y . for OV Sys [W -12/ Q34. Solve the ...
[PDF File]C3 Trigonometry - Trigonometric equations
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(i) Given that cos(x + 30)° = 3 cos(x – 30)°, prove that tan x° = −. 2 3. (5) (ii) (a) Prove that . θ θ sin 2 1−cos2. ≡ tan θ . (3) (b) Verify that θ = 180° is a solution of the equation sin 2θ = 2 – 2 cos 2θ. (1) (c) Using the result in part (a), or otherwise, find the other two solutions, 0 < θ < 360°, of the equation ...
[PDF File]Solution to Review Problems for Midterm III
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Solution to Review Problems for Midterm III MATH 1850: page 2 of 10 = (x2)0 jx2j p x 4 1 cot 1(2x)+csc 1(x2)( 2(2x)0 1+x4)+cos 1(2x)+xp 1 1 x2 x p x 1 cot 1(2x) csc 1(x2)(2 1+x4)+cos 1(2x) xp1 1 x2 (k) y= (x2 21) (2x+1)3x5 (x2+1)3(x+1)4 sin5(2x) ln(y) = 2ln(x2 1)+3ln(2x+1)+5ln(x) 3ln(x2+1) 4ln(x+1) 5ln(sin(2x)). This implies that y0 y = 2 2 x x2 1 +3 2x+1 +5 1
[PDF File]MATH 142 - Review #3 (4986576)
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19. Question Details SPreCalc6 7.3.009. [2708228] - Find sin 2 x, cos 2 x, and tan 2 x from the given information. sin 2 x= cos 2 x= tan 2 x= tan x = − , cos x > 0 1 3 20. Question Details SPreCalc6 7.3.017. [2709806] - Use an appropriate Half-Angle Formula to find the exact value of the expression. cos 15° 21. Question Details SPreCalc6 7.3 ...
[PDF File]DOUBLE-ANGLE, POWER-REDUCING, AND HALF-ANGLE FORMULAS
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x + sin 2x + cos. 2. x = 1 + sin 2x (substitution: double-angle identity) sin. 2. x + cos. 2. x + sin 2x = 1 + sin 2x . 1 + sin 2x = 1 + sin 2x (Pythagorean identity) Therefore, 1+ sin 2x = 1 + sin 2x, is verifiable. Half-Angle Identities . The alternative form of double-angle identities are the half-angle identities. Sine • To achieve the ...
[PDF File]University of South Carolina
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X3 — 2X2 — X 2 dx cos x sin2x — 3 sin x +2 2x3 + 3x2 4 dx sec2 x tan2x + 2 tan x +2 x3 + x2 dx x4 + 2x2 1 3 + cos 0 2 — cos O 2 + 2 cos O + sin 0 1 + sin x sec x Sin X dx sin 2x 99. 101. Find the area of the surface generated by revolving the curve y = cosh x, 0 x 1, around the x-axis. 102. Find the length of the curve y = e 103.
[PDF File]Trigonometric Integrals{Solutions
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Speed Round 1. R cos(x)dx : sinx 2. R sin(x)dx: cosx 3. sin2(x)+cos2(x): 1 4. p 1 cos2(x) : sinx 5. (a+b)(a b): a2 b2 6. R sec2(x)dx: tanx 7. (1+cos(x))(1 cos(x)): sin2 x 8. cos4(x) sin4(x): (cos2 x+sin2 x)(cos2 x sin2 x) = cos2 x sin2 x = cos2x 9. (1 2x )=(1 x): 1+x 10. cos2(x)=(1 sin(x)): 1 + sinx 11.
[PDF File]Table of Fourier Transform Pairs
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Signals & Systems - Reference Tables. 3 u(t)e. t sin(0. t) 2. 2 0 0 j e. t 2. 2. 2. e. t2 /(2. 2) 2. e. 2. 2 / 2 u(t) e. t. j. 1 u(t) te. t 2 1. j. Trigonometric Fourier Series. 1 ( ) 0 cos( 0 ) sin( 0
[PDF File]Math 1A: Homework 7 Solutions
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(a;b). Note however that f0(x) = 1 + (2 3)sin(x=3)cos(x=3) = 1 + sin(2x=3) 3 >0 for all xsince sin(2x=3) 3 1=3. This shows that f0(d) cannot be zero. We therefore have a contradiction, showing that f(x) = 0 has at most one solution in the (1 ;1). Couple this with the fact that f(x) = 0 has at least one solution to
[PDF File]MATH 3321 Sample Questions for Exam 2 Second Order ...
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9. Find the general solution of y00 +4y = 2 sin 2x. Answer: y = C1 cos 2x+C2 sin 2x− 1 2 x cos 2x. 10. Find the general solution of y00 −6y0 + 8y = 2e4x +6. Answer: y = C1 e4x + C2 e2x + xe4x + 3 4. 11. A particular solution of the nonhomogeneous differential equation y00 − 2y0 −15y = 2 cos 3x+5e5x + 2 will have the form: Answer: z = A ...
[PDF File]Trigonometric integrals (Sect. 8.2) Product of sines and ...
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2 3 cos3(x) − 1 5 cos5(x)+ c. C Product of sines and cosines Example Evaluate I = Z sin6(x) dx. Solution: Since m = 6 is even, we write it as m = 2(3), I = Z sin2(x) 3 dx = Z 1 2 [1 − cos(2x)] 3 dx I = 1 8 Z 1 − 3cos(2x)+3cos2(2x) − cos3(2x) dx. The first two terms are: Z (1 − 3cos(2x))dx = x − 3 2 sin(2x). The third term can be ...
[PDF File]18.01SC Single Variable Calculus - MIT OpenCourseWare
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2 3 = 1 sin(2x) − 1 sin3(2x) + c 2 3 cos 3 (2x) dx = 1 sin(2x) − 1 sin3(2x) + C. 2 6 We can check this answer by differentiating; be sure to apply the chain rule twice when differentiating 1 6 sin 3(2x). d 1 1 1 dx 2 sin(2x) − 6 sin3(2x) + C = cos(2x) − 6 · 3 sin2(2x) cos(2x) · 2 + 0 = cos(2x) − sin2(2x) cos(2x) = cos(2x) − (1 ...
[PDF File]C3 Differentiation - Products and quotients
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(a) Differentiate with respect to x (i) x2e3x+2, (4) (ii) . 3. 3 cos(2 ) x x (4) (b) Given that x = 4 sin(2y + 6), find . x y d d in terms of x. (5) (Total 13 marks) 13. (a) Differentiate with respect to x (i) 3 sin2x + sec2x, (3) (ii) {x + ln(2x)} 3. (3) Given that y = 2 2 ( 1) 5 10 9 − − + x x x, x ≠ 1, (b) show that . x y d d = – ( 1 ...
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