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Unformatted text preview: Horstman (mdh995) HW07 Radin (58505) 1 This print-out should have 21 questions. Multiple-choice 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 - 3 sin 3 ) d . 1. I = 8 2. I = 2 3. I = 4 4. I = 7 5. I = 3 correct 6. I = 6 Explanation: Since sin 2 = 1- cos 2 we see that 5 sin - 3 sin 3 = sin (5- 3 sin 2 ) = sin [5- 3 (1- cos 2 )] = sin (2 + 3 cos 2 ) . Thus I = integraldisplay / 2 sin (2 + 3 cos 2 ) d As the integral is now of the form sin f (cos ) , f ( x ) = 2 + 3 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 (2 + 3 x 2 ) dx = integraldisplay 1 (2 + 3 x 2 ) dx . Consequently, I = bracketleftBig 2 x + x 3 bracketrightBig 1 = 3 . 002 10.0 points Evaluate the integral I = integraldisplay / 2 3 sin 3 x cos 2 x dx . 1. I = 2 5 correct 2. I = 8 5 3. I = 1 5 4. I = 4 5 5. I = 6 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 . Horstman (mdh995) HW07 Radin (58505) 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 indefinite integral I = integraldisplay 1- sin x cos x dx . 1. I =- ln(1- cos x ) + C 2. I = ln(1 + cos x ) + C 3. I = ln(1- sin x ) + C 4. I = ln(1 + sin x ) + C correct 5. I =- ln(1 + cos x ) + C Explanation: integraldisplay 1- sin x cos x dx = integraldisplay (sec x- tan x ) dx = ln | sec x + tan x | - ln | sec x | + C = ln | (sec x + tan x ) cos x | + C = ln(1 + sin x ) + C 004 10.0 points Evaluate the definite integral I = integraldisplay / 3 sec x tan x 4 + sec x dx . 1. I = ln 6 5 correct 2. I = ln 5 4 3. I =- ln 5 4 4. I = ln 3 4 5. I =- ln 6 5 6. I =- ln 3 4 Explanation: Since d dx (sec x ) = sec x tan x , use of the substitution u = 4 + sec x is suggested. For then du = sec x tan xdx , while x = 0 = u = 5 , x = 3 = u = 6 . Thus I = integraldisplay 6 5 1 u du = bracketleftBig ln u bracketrightBig 6 5 . Consequently, I = ln 6 5 . 005 10.0 points Evaluate the definite integral I = integraldisplay / 4 4 tan 4 x dx . 1. I = 4 - 4 3 Horstman (mdh995) HW07 Radin (58505) 3 2. I = 4 + 4 3 3. I = 2 - 8 3 4. I = 2 + 4 3 5. I = + 8 3 6. I = - 8 3 correct Explanation: Since tan 2 x = sec 2 x- 1 , it follows that tan 4 x = tan 2 x (sec 2 x- 1) = tan 2 x sec 2 x- tan 2 x ....
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This note was uploaded on 11/30/2008 for the course M 58510 taught by Professor Radin during the Fall '08 term at University of Texas at Austin.

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solution_pdf7 - Horstman (mdh995) HW07 Radin (58505) 1 This...

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