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# Math HW7 - Perez(nap563 HW07 Zheng(56555 This print-out...

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Perez (nap563) – HW07 – Zheng – (56555) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the integral I = integraldisplay (5 cos θ - 2 cos 3 θ ) dθ . 1. I = 5 cos θ - 2 3 cos 3 θ + C 2. I = 5 sin θ + 2 3 sin 3 θ + C 3. I = 3 sin θ + 2 3 sin 3 θ + C correct 4. I = 3 cos θ - 2 3 cos 3 θ + C 5. I = 3 cos θ + 2 3 cos 3 θ + C 6. I = 5 sin θ - 2 3 sin 3 θ + C Explanation: Since cos 2 θ = 1 - sin 2 θ , the integrand can be rewritten as 5 cos θ - 2 cos 3 θ = cos θ (5 - 2 cos 2 θ ) = cos θ (5 - 2(1 - sin 2 θ )) = cos θ (3 + 2 sin 2 θ ) . Thus I = integraldisplay cos θ (3 + 2 sin 2 θ ) dθ . As the integrand is now of the form cos θ f (sin θ ) , f ( x ) = 3 + 2 x 2 , the substitution x = sin θ is suggested. For then dx = cos θ dθ , so that I = integraldisplay (3 + 2 x 2 ) dx = 3 x + 2 3 x 3 + C . Consequently I = 3 sin θ + 2 3 sin 3 θ + C . keywords: indefinite integral, trig function, Pythagorean identity, power cosine, 002 10.0 points Evaluate the integral I = integraldisplay π/ 2 0 sin 2 x cos 3 x dx . 1. I = 2 15 correct 2. I = 8 15 3. I = 4 15 4. I = 1 15 5. I = 2 5 Explanation: Since sin 2 x cos 3 x = (sin 2 x cos 2 x ) cos x = sin 2 x (1 - sin 2 x )cos x = (sin 2 x - sin 4 x )cos x , the integrand is of the form cos xf (sin x ), sug- gesting use of the substitution u = sin x . For then du = cos x dx , while x = 0 = u = 0 x = π 2 = u = 1 .

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Perez (nap563) – HW07 – Zheng – (56555) 2 In this case I = integraldisplay 1 0 ( u 2 - u 4 ) du . Consequently, I = bracketleftBig 1 3 u 3 - 1 5 u 5 bracketrightBig 1 0 = 2 15 . keywords: Stewart5e, indefinite integral, powers of sin, powers of cos, trig substitu- tion, 003 10.0 points The shaded region in π 2 π 3 π 2 x y is bounded by the graph of f ( x ) = 2 sin 3 x on [0 , 3 π/ 2] and the the x -axis. Find the area of this region. 1. area = 3 2. area = 4 correct 3. area = 6 π 4. area = 3 π 5. area = 6 6. area = 4 π Explanation: The area of the shaded region is given by I = integraldisplay 3 π/ 2 0 | 2 sin 3 x | dx which as the graph shows can in turn be writ- ten as I = integraldisplay π 0 2 sin 3 x dx - integraldisplay 3 π/ 2 π 2 sin 3 x dx . Since sin 2 x = 1 - cos 2 x , we thus see that I = braceleftBig integraldisplay π 0 - integraldisplay 3 π/ 2 π bracerightBig 2 sin x (1 - cos 2 x ) dx . To evaluate these integrals, set u = cos x . For then du = - sin x dx , in which case integraldisplay π 0 2 sin x (1 - cos 2 x ) dx = - 2 integraldisplay 1 1 (1 - u 2 ) du = 2 integraldisplay 1 1 (1 - u 2 ) du = 2 bracketleftBig u - 1 3 u 3 bracketrightBig 1 1 = 8 3 , while integraldisplay 3 π/ 2 π 2 sin x (1 - cos 2 x ) dx = - 2 integraldisplay 0 1 (1 - u 2 ) du = - 2 bracketleftBig u - 1 3 u 3 bracketrightBig 0 1 = - 4 3 .
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Math HW7 - Perez(nap563 HW07 Zheng(56555 This print-out...

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