HWK8 - Valenica, Daniel Homework 7 Due: Oct 17 2007, 3:00...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 / 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....
View Full Document

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.

Page1 / 11

HWK8 - Valenica, Daniel Homework 7 Due: Oct 17 2007, 3:00...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online