CALCL HWK7 - Arwikar, Dev – Homework 7 – Due: Oct 17...

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Unformatted text preview: Arwikar, Dev – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: James Rath 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. Lucky seven The half-way mark, for sure True party time 001 (part 1 of 1) 10 points Evaluate the definite integral I = Z π/ 2 sin 3 x cos 2 xdx. 1. I = 8 15 2. I = 1 15 3. I = 2 15 correct 4. I = 4 15 5. I = 2 5 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 sin x (1- cos 2 x )cos 2 xdx = Z π/ 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 = π 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 π 3 tan 4 xdx. 1. I = π 6- 8 √ 3 9 2. I = π 6 + 8 √ 3 9 3. I = π √ 3 3 4. I = π 3 correct 5. I = π 4- 2 3 6. I = π 4 + 2 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. Arwikar, Dev – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: James Rath 2 Thus by trig identities yet again, tan 4 x = ( tan 2 x- 1 ) sec 2 x + 1 . Consequently, I = Z π 3 £( tan 2 x- 1 ) sec 2 x + 1 / dx = • 1 3 tan 3 x- tan x + x ‚ π 3 But tan π 3 = √ 3 Consequently, I = π 3 . 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 + 5 4 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 + 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 + sin x ) + C 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 + cos x ) + C correct Explanation: Arwikar, Dev – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: James Rath 3 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 02/25/2010 for the course M 54363 taught by Professor Olsen during the Fall '07 term at University of Texas.

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CALCL HWK7 - Arwikar, Dev – Homework 7 – Due: Oct 17...

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