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Unformatted text preview: Rehman (aar638) HW07 sachse (56620) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Evaluate the definite integral I = integraldisplay / 2 (5 sin + 2 sin 3 ) d . 1. I = 3 2. I = 19 3 correct 3. I = 17 3 4. I = 11 3 5. I = 7 6. I = 13 3 Explanation: Since sin 2 = 1 cos 2 we see that 5 sin + 2 sin 3 = sin (5 + 2 sin 2 ) = sin [5 + 2 (1 cos 2 )] = sin (7 2 cos 2 ) . Thus I = integraldisplay / 2 sin (7 2 cos 2 ) d As the integral is now of the form sin f (cos ) , f ( x ) = 7 2 x 2 , the subsitution x = cos is suggested. For then dx = sin d , while = 0 = x = 1 , = 2 = x = 0 . In this case I = integraldisplay 1 (7 2 x 2 ) dx = integraldisplay 1 (7 2 x 2 ) dx . Consequently, I = bracketleftBig 7 x 2 3 x 3 bracketrightBig 1 = 19 3 . 002 10.0 points Evaluate the integral I = integraldisplay / 2 3 sin 3 x cos 2 x dx . 1. I = 8 5 2. I = 4 5 3. I = 2 5 correct 4. I = 6 5 5. I = 1 5 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 . Rehman (aar638) HW07 sachse (56620) 2 In this case I = integraldisplay 1 3( u 2 u 4 ) du . Consequently, I = bracketleftBig u 3 + 3 5 u 5 bracketrightBig 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 / 4 5 cos x + 4 sin x cos 3 x dx . 1. I = 11 2 2. I = 7 correct 3. I = 6 4. I = 5 5. I = 13 2 Explanation: After division 5 cos x + 4 sin x cos 3 x = 5 sec 2 x + 4 tan x sec 2 x = (5 + 4 tan x ) sec 2 x . Thus I = integraldisplay / 4 (5 + 4 tan x ) sec 2 x dx . Let u = tan x ; then du = sec 2 x dx so I = integraldisplay 1 (5 + 4 u ) du = bracketleftbig 5 u + 2 u 2 bracketrightbig 1 ....
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 Fall '09
 GOGOLEV
 Calculus

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