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# solution7_pdf - rabbani(tar547 HW07 Radin(56635 This...

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rabbani (tar547) – HW07 – Radin – (56635) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the definite integral I = integraldisplay π/ 2 0 (3 sin θ + sin 3 θ ) dθ . 1. I = 11 3 correct 2. I = 2 3. I = 7 3 4. I = 4 5. I = 10 3 6. I = 8 3 Explanation: Since sin 2 θ = 1 - cos 2 θ we see that 3 sin θ + sin 3 θ = sin θ (3 + sin 2 θ ) = sin θ (3 + 1 - cos 2 θ ) = sin θ (4 - cos 2 θ ) . Thus I = integraldisplay π/ 2 0 sin θ (4 - cos 2 θ ) As the integral is now of the form sin θ f (cos θ ) , f ( x ) = 4 - x 2 , the subsitution x = cos θ is suggested. For then dx = - sin θ dθ , while θ = 0 = x = 1 , θ = π 2 = x = 0 . In this case I = - integraldisplay 0 1 (4 - x 2 ) dx = integraldisplay 1 0 (4 - x 2 ) dx . Consequently, I = bracketleftBig 4 x - 1 3 x 3 bracketrightBig 1 0 = 11 3 . 002 10.0 points Evaluate the integral I = integraldisplay π/ 2 0 3 sin 3 x cos 2 x dx . 1. I = 1 5 2. I = 8 5 3. I = 6 5 4. I = 4 5 5. I = 2 5 correct Explanation: Since sin 3 x cos 2 x = sin x (sin 2 x cos 2 x ) = sin x (1 - cos 2 x )cos 2 x = sin x (cos 2 x - cos 4 x ) , the integrand is of the form sin xf (cos x ), sug- gesting use of the substitution u = cos x . For then du = - sin x dx , while x = 0 = u = 1 x = π 2 = u = 0 .

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rabbani (tar547) – HW07 – Radin – (56635) 2 In this case I = - integraldisplay 0 1 3( u 2 - u 4 ) du . Consequently, I = bracketleftBig - u 3 + 3 5 u 5 bracketrightBig 0 1 = 2 5 . keywords: Stewart5e, indefinite integral, powers of sin, powers of cos, trig substitu- tion, 003 10.0 points Evaluate the definite integral I = integraldisplay π/ 4 0 3 cos x - 4 sin x cos 3 x dx . 1. I = 1 2 2. I = 0 3. I = 3 2 4. I = 1 correct 5. I = 2 Explanation: After division 3 cos x - 4 sin x cos 3 x = 3 sec 2 x - 4 tan x sec 2 x = (3 - 4 tan x ) sec 2 x . Thus I = integraldisplay π/ 4 0 (3 - 4 tan x ) sec 2 x dx . Let u = tan x ; then du = sec 2 x dx so I = integraldisplay 1 0 (3 - 4 u ) du = bracketleftbig 3 u - 2 u 2 bracketrightbig 1 0 . Consequently, I = 1 . 004 10.0 points A particle moves on a straight line with velocity function v ( t ) = sin ωt cos 3 ωt . Find its position function, s ( t ), if s (0) = 0. 1. s ( t ) = cos 4 ωt - sin 4 ωt 4 ω 2. s ( t ) = 1 - cos 4 ωt 4 ω correct 3. s ( t ) = 4 ω cos 4 ωt 4. s ( t ) = 1 - sin 4 ωt 4 ω 5. s ( t ) = 1 4 ω cos 4 ωt Explanation: The distance function is given in terms of velocity by s ( t ) = integraldisplay v ( t ) dt = integraldisplay sin ωt cos 3 ωt dt , where s (0) = 0. To evaluate the integral set
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