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Unformatted text preview: Valenica, Daniel Homework 7 Due: Oct 17 2007, 3:00 am Inst: Cheng 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Evaluate the definite integral I = Z / 2 sin 3 x cos 2 xdx. 1. I = 4 15 2. I = 1 15 3. I = 8 15 4. I = 2 5 5. I = 2 15 correct 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, we see that I can be written as the sum I = Z / 2 sin x (1 cos 2 x )cos 2 xdx = Z / 2 sin x cos 2 xdx Z sin x cos 4 xdx, of two integrals, both of which can be eval uated using the substitution u = cos x . For then du = sin xdx, while x = 0 = u = 1 x = 2 = u = 0 . Thus I = Z 1 u 2 du + Z 1 u 4 du = h 1 3 u 3 + 1 5 u 5 i 1 . Consequently, I = 2 15 . keywords: Stewart5e, indefinite integral, powers of sin, powers of cos, trig substitu tion, 002 (part 1 of 1) 10 points Find the value of the definite integral I = Z 4 tan 4 xdx. 1. I = 4 + 2 3 2. I = 4 2 3 correct 3. I = 6 + 8 3 9 4. I = 6 8 3 9 5. I = 3 3 6. I = 3 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 . Valenica, Daniel Homework 7 Due: Oct 17 2007, 3:00 am Inst: Cheng 2 Consequently, I = Z 4 ( tan 2 x 1 ) sec 2 x + 1 / dx = 1 3 tan 3 x tan x + x 4 But tan 4 = 1 Consequently, I = 4 2 3 . keywords: trig identity, integral 003 (part 1 of 1) 10 points Determine the indefinite integral I = Z x (3 cos 2 x sin 2 x ) dx. 1. I = x 2 1 2 x cos2 x 1 4 sin2 x + C 2. I = 1 2 x 2 + x sin2 x + 1 2 cos2 x + C correct 3. I = 1 2 x 2 x cos2 x 1 2 sin2 x + C 4. I = x 2 1 2 x sin2 x + 1 2 cos2 x + C 5. I = x 2 + 1 2 x sin2 x + 1 4 cos2 x + C Explanation: Since cos 2 x = 1 2 (1 + cos2 x ) and sin 2 x = 1 2 (1 cos2 x ) , we see that I = 1 2 Z x { 3(1 + cos2 x ) 1 + cos2 x } dx = Z xdx + 2 Z x cos2 xdx = 1 2 x 2 + 2 Z x cos2 xdx. But after integration by parts, Z x cos2 xdx = 1 2 x sin2 x 1 2 Z sin2 xdx = 1 2 x sin2 x + 1 4 cos2 x + C . Consequently, I = 1 2 x 2 + x sin2 x + 1 2 cos2 x + C . keywords: trigonometric identities, integra tion by parts 004 (part 1 of 1) 10 points Evaluate the indefinite integral I = Z 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: Valenica, Daniel Homework 7 Due: Oct 17 2007, 3:00 am Inst: Cheng 3 Z 1 sin x cos x dx = Z (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 keywords: trig integral 005 (part 1 of 1) 10 points Find the volume obtained by rotating the region bounded by the curves y = sin x , x = 2 , x = , y = 0 about y = 1....
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This homework help was uploaded on 04/16/2008 for the course CALC 303L taught by Professor Cheng during the Fall '07 term at University of Texas at Austin.
 Fall '07
 CHENG

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