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HW 06 solution

# HW 06 solution - gutierrez(ig3472 HW06 Neitzke(56585 This...

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gutierrez (ig3472) – HW06 – Neitzke – (56585) 1 This print-out should have 17 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 1 1 + 9( x - 1) 2 dx . 1. I = tan - 1 3( x - 1) + C 2. I = 3 sin - 1 parenleftBig x - 1 3 parenrightBig + C 3. I = sin - 1 3( x - 1) + C 4. I = 1 3 tan - 1 3( x - 1) + C correct 5. I = 1 3 sin - 1 3( x - 1) + C 6. I = 3 tan - 1 parenleftBig x - 1 3 parenrightBig + C Explanation: Since d dx tan - 1 x = 1 1 + x 2 , the substitution u = 3( x - 1) is suggested. For then du = 3 dx , in which case I = 1 3 integraldisplay 1 1 + u 2 du = 1 3 tan - 1 u + C , with C an arbitrary constant. Consequently, I = 1 3 tan - 1 3( x - 1) + C . keywords: 002 10.0 points Evaluate the definite integral I = integraldisplay 1 8 0 1 1 - 16 x 2 dx . Correct answer: 0 . 1309. Explanation: Since integraldisplay 1 1 - x 2 dx = sin - 1 x + C , a change of variable x is needed to reduce I to this form. Set u = 4 x . Then du = 4 dx , and x = 0 = u = 0 , while x = 1 8 = u = 1 2 . In this case I = 1 4 integraldisplay 1 2 0 1 1 - u 2 du = bracketleftbigg 1 4 sin - 1 u bracketrightbigg 1 2 0 . Consequently, I = 1 4 arcsin parenleftbigg 1 2 parenrightbigg = 0 . 1309 . 003 10.0 points Determine the indefinite integral I = integraldisplay ( 1 - x 2 ) - 1 / 2 1 + 2 sin - 1 x dx . 1. I = - 1 2 ( 1 + 2 sin - 1 x ) 2 + C 2. I = 1 4 ( 1 + 2 sin - 1 x ) 2 + C 3. I = 1 4 ln vextendsingle vextendsingle 1 + 2 sin - 1 x vextendsingle vextendsingle + C 4. I = 1 2 ln vextendsingle vextendsingle 1 + 2 sin - 1 x vextendsingle vextendsingle + C correct 5. I = - 1 2 ln vextendsingle vextendsingle 1 + 2 sin - 1 x vextendsingle vextendsingle + C 6. I = - 1 4 ( 1 + 2 sin - 1 x ) 2 + C

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gutierrez (ig3472) – HW06 – Neitzke – (56585) 2 Explanation: Set u = 1 + 2 sin - 1 x . Then du = 2 1 - x 2 dx = 2 ( 1 - x 2 ) - 1 / 2 dx , so I = 1 2 integraldisplay 1 u du = 1 2 ln vextendsingle vextendsingle 1 + 2 sin - 1 x vextendsingle vextendsingle + C . Consequently, I = 1 2 ln vextendsingle vextendsingle 1 + 2 sin - 1 x vextendsingle vextendsingle + C . 004 10.0 points Evaluate the definite integral I = integraldisplay π 0 2 sin θ 1 + cos 2 θ dθ . 1. I = π correct 2. I = 5 4 π 3. I = 3 2 π 4. I = 3 4 π 5. I = 7 4 π Explanation: Since d cos θ = - sin θ , the substitution u = cos θ is suggested. For then du = - sin θ dθ , while θ = 0 = u = 1 , θ = π = u = - 1 , so that I = - 2 integraldisplay - 1 1 1 1 + u 2 du = 2 integraldisplay 1 - 1 1 1 + u 2 du , which can now be integrated using the fact that d du tan - 1 u = 1 1 + u 2 . Consequently, I = 2 bracketleftBig tan - 1 u bracketrightBig 1 - 1 = π since tan - 1 ( - 1) = - π 4 , tan - 1 1 = π 4 . 005 10.0 points Determine the integral I = integraldisplay 1 0 4 - x 1 + x 2 dx .
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