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HWK8 - Valenica Daniel Homework 7 Due 3:00 am Inst Cheng...

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Valenica, Daniel – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: Cheng 1 This print-out should have 21 questions. Multiple-choice 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 π/ 2 0 sin 3 x cos 2 x dx . 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 π/ 2 0 sin x (1 - cos 2 x )cos 2 x dx = Z π/ 2 0 sin x cos 2 x dx - Z sin x cos 4 x dx , of two integrals, both of which can be eval- uated using the substitution u = cos x . For then du = - sin x dx , while x = 0 = u = 1 x = π 2 = u = 0 . Thus I = - Z 0 1 u 2 du + Z 0 1 u 4 du = h - 1 3 u 3 + 1 5 u 5 i 0 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 π 4 0 tan 4 x dx . 1. I = π 4 + 2 3 2. I = π 4 - 2 3 correct 3. I = π 6 + 8 3 9 4. I = π 6 - 8 3 9 5. I = π 3 3 6. I = π 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 .
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Valenica, Daniel – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: Cheng 2 Consequently, I = Z π 4 0 £( tan 2 x - 1 ) sec 2 x + 1 / dx = 1 3 tan 3 x - tan x + x π 4 0 But tan π 4 = 1 Consequently, I = π 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 cos 2 x - 1 4 sin 2 x + C 2. I = 1 2 x 2 + x sin 2 x + 1 2 cos 2 x + C correct 3. I = 1 2 x 2 - x cos 2 x - 1 2 sin 2 x + C 4. I = x 2 - 1 2 x sin 2 x + 1 2 cos 2 x + C 5. I = x 2 + 1 2 x sin 2 x + 1 4 cos 2 x + C 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 Z x { 3 (1 + cos 2 x ) - 1 + cos 2 x } dx = Z x dx + 2 Z x cos 2 x dx = 1 2 x 2 + 2 Z x cos 2 x dx . But after integration by parts, Z x cos 2 x dx = 1 2 x sin 2 x - 1 2 Z sin 2 x dx = 1 2 x sin 2 x + 1 4 cos 2 x + C . Consequently, I = 1 2 x 2 + x sin 2 x + 1 2 cos 2 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:
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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 = π 2 , x = π , y = 0 about y = - 1. 1. π 2 2 + 4 π 2. π 2 4 + 2 π correct 3. - π 2 4 + 2 π 4. π 2 4 + π 5. - π 2 2 + 4 π Explanation: Volume = π π Z π/ 2 £ (1 + sin x ) 2 - 1 2 / dx = π π Z π/ 2 (2 sin x + sin 2 x ) dx = π - 2 cos x + x 2 - sin 2 x 4 π π/ 2 = π h‡ 2 + π 2 - 0 · - 0 + π 4 - 0 ·i = π 2 4 + 2 π keywords: 006 (part 1 of 1) 10 points Determine the indefinite integral I = Z sec 8 2 x tan 2 x dx .
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