HW7 - momin (rrm497) Homework 07 cheng (58520) 1 This...

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Unformatted text preview: momin (rrm497) Homework 07 cheng (58520) 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 (8 sin + sin 3 ) d . 1. I = 9 2. I = 26 3 correct 3. I = 7 4. I = 22 3 5. I = 25 3 6. I = 23 3 Explanation: Since sin 2 = 1- cos 2 we see that 8 sin + sin 3 = sin (8 + sin 2 ) = sin (8 + 1- cos 2 ) = sin (9- cos 2 ) . Thus I = integraldisplay / 2 sin (9- cos 2 ) d As the integral is now of the form sin f (cos ) , f ( x ) = 9- 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 (9- x 2 ) dx = integraldisplay 1 (9- x 2 ) dx . Consequently, I = bracketleftBig 9 x- 1 3 x 3 bracketrightBig 1 = 26 3 . 002 10.0 points Evaluate the integral I = integraldisplay / 2 sin 3 x cos 2 x dx . 1. I = 2 5 2. I = 4 15 3. I = 8 15 4. I = 2 15 correct 5. I = 1 15 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 . momin (rrm497) Homework 07 cheng (58520) 2 In this case I =- integraldisplay 1 ( u 2- u 4 ) du . Consequently, I = bracketleftBig- 1 3 u 3 + 1 5 u 5 bracketrightBig 1 = 2 15 . 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 + sin x ) + C correct 3. I =- ln(1- cos x ) + C 4. I = ln(1- sin x ) + C 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 2. I =- ln 3 4 3. I =- ln 5 4 4. I = ln 6 5 correct 5. I = ln 5 4 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 2 tan 4 x dx . 1. I = 2 - 2 3 momin (rrm497) Homework 07 cheng (58520) 3 2. I = 2 + 2 3 3. I = 1 2 - 4 3 correct 4. I = + 2 3 5. I = - 4 3 6. I = 1 2 + 4 3 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 12/11/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.

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HW7 - momin (rrm497) Homework 07 cheng (58520) 1 This...

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