HW06-solutions.pdf

# HW06-solutions.pdf - young(toy68 HW06 villafuerte...

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young (toy68) – HW06 – villafuerte altu – (53615) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points To which one of the following does the inte- gral I = integraldisplay x 2 x 2 - 1 dx reduce after an appropriate trig substitution? 1. I = integraldisplay sec 3 ( θ ) correct 2. I = integraldisplay sin 2 ( θ ) sec 3 ( θ ) 3. I = integraldisplay sin 3 ( θ ) 4. I = integraldisplay tan 3 ( θ ) 5. I = integraldisplay sin 3 ( θ ) sec 2 ( θ ) 6. I = integraldisplay tan 2 ( θ ) sec 3 ( θ ) Explanation: Set x = sec( θ ). Then dx = sec( θ ) tan( θ ) dθ, radicalBig sec 2 ( θ ) - 1 = tan( θ ) . In this case, I = integraldisplay sec 2 ( θ ) tan( θ ) sec( θ ) tan( θ ) dθ . Consequently, I = integraldisplay sec 3 ( θ ) . 002 10.0points Evaluate the integral I = integraldisplay 1 0 2 4 - 3 x 2 dx . 1. I = 4 π 3 3 2. I = 2 π 3 3 correct 3. I = 1 4. I = 2 3 5. I = 2 6. I = π 3 Explanation: Set 3 x = 2 sin θ ; then 3 dx = 2 cos θ dθ and 4 - 3 x 2 = 4(1 - sin 2 θ ) = 4 cos 2 θ , while x = 0 = θ = 0 , x = 1 = θ = π 3 . In this case I = 2 3 integraldisplay π/ 3 0 2 cos θ 2 cos θ = 2 3 integraldisplay π/ 3 0 dθ . Consequently I = 2 π 3 3 . 003 10.0points Evaluate the integral I = integraldisplay 1 0 6 t 2 - t 4 dt . 1. I = 3 2 2. I = 3 π

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young (toy68) – HW06 – villafuerte altu – (53615) 2 3. I = 3 4 π correct 4. I = 3 5. I = 3 4 6. I = 3 2 π Explanation: Set t 2 = 2 sin( u ). Then 2 t dt = 2 cos( u ) du , in which case t dt = 1 2 dt . On the other hand, t = 0 = u = 0 , t = 1 = u = π 4 . Thus I = 1 2 integraldisplay π/ 4 0 6 cos( u ) 2 cos( u ) du = 1 2 bracketleftBig 6 u bracketrightBig π/ 4 0 . Consequently, I = 3 4 π . 004 10.0points Determine the integral I = integraldisplay 1 ( x 2 + 4) 3 2 dx . 1. I = x 2 + 4 4 x + C 2. I = x x 2 + 4 4 + C 3. I = 1 4 x 2 + 4 + C 4. I = x x 2 + 4 + C 5. I = x 2 + 4 x + C 6. I = x 4 x 2 + 4 + C correct Explanation: Set x = 2 tan( u ) . Then dx = 2 sec 2 ( u ) du , while ( x 2 + 4) 3 2 = ( 4(tan 2 ( u ) + 1) ) 3 2 = 8 sec 3 ( u ) . Thus I = integraldisplay 2 8 sec 2 ( u ) sec 3 ( u ) du = 1 4 integraldisplay cos( u ) du , and so I = 1 4 sin( u ) + C = 1 4 sin parenleftBig tan - 1 parenleftBig x 2 parenrightBigparenrightBig + C . But by Pythagoras u radicalbig x 2 + 4 2 x we see that sin parenleftBig tan - 1 parenleftBig x 2 parenrightBigparenrightBig = x x 2 + 4 . Consequently, I = x 4 x 2 + 4 + C with C an arbitrary constant. keywords: trig substitution 005 10.0points
young (toy68) – HW06 – villafuerte altu – (53615) 3 Determine the integral I = integraldisplay 1 + 4 x x 2 - 1 dx . 1. I = 4 ln parenleftBigvextendsingle vextendsingle vextendsingle x - radicalbig x 2 - 1 vextendsingle vextendsingle vextendsingle parenrightBig + radicalbig x 2 - 1+ C 2. I = 4 ln parenleftBigvextendsingle vextendsingle vextendsingle x + radicalbig x 2 - 1 vextendsingle vextendsingle vextendsingle parenrightBig - radicalbig x 2 - 1+ C 3. I = ln parenleftBigvextendsingle vextendsingle vextendsingle x - radicalbig x 2 - 1 vextendsingle vextendsingle vextendsingle parenrightBig +4 radicalbig x 2 - 1+ C 4. I = ln parenleftBigvextendsingle vextendsingle vextendsingle x + radicalbig x 2 - 1 vextendsingle vextendsingle vextendsingle parenrightBig + 4 radicalbig x

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• Fall '09
• GOGOLEV
• Partial fractions in complex analysis, Partial fractions in integration

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