Homework 07 - Su Yung – Homework 7 – Due 3:00 am –...

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Unformatted text preview: Su, Yung – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: Shinko Harper 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 5 sin 3 x cos 2 xdx. 1. I = 2 2. I = 1 3 3. I = 8 3 4. I = 4 3 5. I = 2 3 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 5sin x (1- cos 2 x )cos 2 xdx = Z π/ 2 5sin x cos 2 xdx- 5 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 = π 2 = ⇒ u = 0 . Thus I =- Z 1 5 u 2 du + Z 1 5 u 4 du = h- 5 3 u 3 + u 5 i 1 . Consequently, I = 2 3 . 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 2 π 3 cos 4 xdx. 1. I = π 4- 7 √ 3 64 correct 2. I = 9 π 32- 1 4 3. I = 3 π 32 + 1 4 4. I = π 8 5. I = 3 π 16 6. I = π 16 + √ 3 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 ) . Su, Yung – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: Shinko Harper 2 Thus by trig identities yet again, cos 4 x = 1 8 (3 + 4cos2 x + cos4 x ) . Consequently, I = Z 2 π 3 1 8 (3 + 4cos2 x + cos4 x ) dx = • 3 8 x + 1 4 sin2 x + 1 32 sin4 x ‚ 2 π 3 But sin 4 π 3 =- √ 3 2 , sin 8 π 3 = 0 Consequently, I = π 4- 7 √ 3 64 . keywords: trig identity, integral 003 (part 1 of 1) 10 points Determine the indefinite integral I = Z x (9 cos 2 x- sin 2 x ) dx. 1. I = 2 x 2- 5 2 x cos2 x- 5 4 sin2 x + C 2. I = 5 2 x 2 + 2 x sin2 x + cos2 x + C 3. I = 5 2 x 2- 2 x cos2 x- sin2 x + C 4. I = 5 2 x 2- 2 x sin2 x + 5 4 cos2 x + C 5. I = 2 x 2 + 5 2 x sin2 x + 5 4 cos2 x + C cor- rect 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 { 9(1 + cos2 x )- 1 + cos2 x } dx = 4 Z xdx + 5 Z x cos2 xdx = 2 x 2 + 5 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 = 2 x 2 + 5 2 x sin2 x + 5 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- cos x sin x dx. 1. I =- ln(1 + cos x ) + C correct 2. I =- ln(1- cos x ) + C 3. I =- ln(1 + sin x ) + C 4. I = ln(1- sin x ) + C 5. I = ln(1 + sin x ) + C Su, Yung – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: Shinko Harper 3 Explanation: Z 1- cos x sin x dx = Z (csc x- cot x ) dx =- ln | csc x + cot x | - ln | sin x | + C =- ln | (csc x + cot x )sin x | + C =- ln(1 + cos x ) + C keywords: trig integral 005 (part 1 of 1) 10 points Find the volume obtained by rotating the region bounded by the curves...
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This note was uploaded on 04/15/2009 for the course M 59685 taught by Professor Harper during the Spring '09 term at University of Texas.

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Homework 07 - Su Yung – Homework 7 – Due 3:00 am –...

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