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# hw08 - padilla(tp5647 HW08 cheng(57455 1 This print-out...

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Unformatted text preview: padilla (tp5647) HW08 cheng (57455) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Evaluate the integral I = integraldisplay / 2 5 sin 3 x cos 2 x dx . 1. I = 4 3 2. I = 2 3 correct 3. I = 8 3 4. I = 2 5. I = 1 3 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 = 2 = u = 0 . In this case I =- integraldisplay 1 5( u 2- u 4 ) du . Consequently, I = bracketleftBig- 5 3 u 3 + u 5 bracketrightBig 1 = 2 3 . keywords: Stewart5e, indefinite integral, powers of sin, powers of cos, trig substitu- tion, 002 10.0 points Evaluate the definite integral I = integraldisplay / 4 4 cos x + 5 sin x cos 3 x dx . 1. I = 6 2. I = 7 3. I = 11 2 4. I = 13 2 correct 5. I = 5 Explanation: After division 4 cos x + 5 sin x cos 3 x = 4 sec 2 x + 5 tan x sec 2 x = (4 + 5 tan x ) sec 2 x . Thus I = integraldisplay / 4 (4 + 5 tan x ) sec 2 x dx . Let u = tan x ; then du = sec 2 x dx so I = integraldisplay 1 (4 + 5 u ) du = bracketleftbigg 4 u + 5 2 u 2 bracketrightbigg 1 . Consequently, I = 13 2 . padilla (tp5647) HW08 cheng (57455) 2 003 10.0 points Find the value of I = integraldisplay 4 tan 4 x dx . 1. I = 6- 8 3 9 2. I = 3 3 3. I = 4 + 2 3 4. I = 6 + 8 3 9 5. I = 3 6. I = 4- 2 3 correct Explanation: Since tan 2 x = sec 2 x- 1 , we see that tan 4 x = tan 2 x ( sec 2 x- 1 ) = tan 2 x sec 2 x- tan 2 x . Thus by trig identities yet again, tan 4 x = ( tan 2 x- 1 ) sec 2 x + 1 . In this case, I = integraldisplay 4 bracketleftbig( tan 2 x- 1 ) sec 2 x + 1 bracketrightbig dx = bracketleftbigg 1 3 tan 3 x- tan x + x bracketrightbigg 4 . On the other hand, tan 4 = 1 . Consequently, I = 4- 2 3 . 004 10.0 points Evaluate the definite integral I = integraldisplay x (4 cos 2 x + 2 sin 2 x ) dx 1. I = 3 2 2 + 1 2 2. I = 1 2 2 + 3 2 3. I = 3 2 2 correct 4. I = 2 2 5. I = 1 2 2 Explanation: Since cos 2 x = 1 2 (1 + cos 2 x ) and sin 2 x = 1 2 (1- cos 2 x ) , we see that I = 1 2 integraldisplay x { 4 (1 + cos 2 x ) + 2 (1- cos 2 x ) } dx = 3 integraldisplay x dx + integraldisplay x cos 2 x dx = bracketleftBig 3 2 x 2 bracketrightBig + integraldisplay x cos 2 x dx = 3 2 2 + integraldisplay x cos 2 x dx . padilla (tp5647) HW08 cheng (57455) 3 But after integration by parts, integraldisplay x cos 2 x dx = 1 2 bracketleftBig x sin 2 x bracketrightBig - 1 2 integraldisplay sin 2 x dx = 0 + 1 4 bracketleftBig cos 2 x bracketrightBig = 0 ....
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