HW7 Solution - Kim Jin – Homework 7 – Due 3:00 am –...

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Unformatted text preview: Kim, Jin – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: Diane Radin 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 sin 2 x cos 3 xdx. 1. I = 4 15 2. I = 1 15 3. I = 2 15 correct 4. I = 8 15 5. I = 2 5 Explanation: Since sin 2 x cos 3 x = sin 2 x cos 2 x cos x = sin 2 x (1- sin 2 x )cos x, we see that I can be written as the sum I = Z π/ 2 sin 2 x (1- sin 2 x )cos xdx = Z π/ 2 sin 2 x cos xdx- Z π/ 2 sin 4 x cos xdx, of two integrals, both of which can be eval- uated using the substitution u = sin x . For then du = cos xdx, while x = 0 = ⇒ u = 0 x = π 2 = ⇒ u = 1 . 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 π 6 cos 4 xdx. 1. I = 9 π 32- 1 4 2. I = 3 π 16 3. I = π 8 4. I = π 4 5. I = π 16 + 9 √ 3 64 correct 6. I = 3 π 32 + 1 64 Explanation: Since cos 2 x = 1 2 (1 + cos2 x ) , we see that cos 4 x = 1 4 (1 + cos2 x ) 2 = 1 4 ( 1 + 2cos2 x + cos 2 2 x ) . Thus by trig identities yet again, cos 4 x = 1 8 (3 + 4cos2 x + cos4 x ) . Kim, Jin – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: Diane Radin 2 Consequently, I = Z π 6 1 8 (3 + 4cos2 x + cos4 x ) dx = • 3 8 x + 1 4 sin2 x + 1 32 sin4 x ‚ π 6 But sin π 3 = √ 3 2 , sin 2 π 3 = √ 3 2 Consequently, I = π 16 + 9 √ 3 64 . 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 sin2 x + 1 2 cos2 x + C 2. I = x 2 + 1 2 x sin2 x + 1 4 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 cos2 x- 1 4 sin2 x + C 5. I = 1 2 x 2 + x sin2 x + 1 2 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 = 2 Z xdx + Z x cos2 xdx = x 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 = x 2 + 1 2 x sin2 x + 1 4 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 + sin x ) + C correct 3. I =- ln(1 + cos x ) + C 4. I = ln(1 + cos x ) + C 5. I = ln(1- sin x ) + C Explanation: Kim, Jin – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: Diane Radin 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...
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This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas.

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HW7 Solution - Kim Jin – Homework 7 – Due 3:00 am –...

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