M408L hwk6 solutions - Pham, Quoc Homework 6 Due: Feb 27...

Info iconThis preview shows pages 1–3. 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Pham, Quoc Homework 6 Due: Feb 27 2007, 3:00 am Inst: Eric Katerman 1 This print-out should have 13 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 / 4 3 cos x- 4 sin x cos 3 x dx. 1. I = 1 2 2. I = 0 3. I = 1 correct 4. I = 3 2 5. I =- 1 2 Explanation: After division 3 cos x- 4 sin x cos 3 x = 3 sec 2 x- 4 tan x sec 2 x = (3- 4 tan x ) sec 2 x. Thus I = Z / 4 (3- 4 tan x ) sec 2 xdx. Let u = tan x ; then du = sec 2 xdx so I = Z 1 (3- 4 u ) du = 3 u- 2 u 2 / 1 . Consequently, I = 1 . keywords: substitution, trigonometric identi- ties 002 (part 1 of 1) 10 points Evaluate the indefinite integral I = Z 2cos 4 3 tdt. 1. I = 1 4 3 t + 2 3 cos 6 t + 1 12 cos 12 t + C 2. I = 1 4 3 t- 2 3 sin6 t + 1 12 sin 12 t + C 3. I = 1 4 3 t + 2 3 cos 6 t- 1 12 cos 12 t + C 4. I = 1 4 3 t + 2 3 sin 6 t + 1 12 sin12 t + C correct 5. I = 1 4 3 t- 2 3 cos 6 t + 1 12 cos 12 t + C 6. I = 1 4 3 t + 2 3 sin6 t- 1 12 sin 12 t + C Explanation: Since cos 2 = 1 2 1 + cos 2 , the integrand can be rewritten as 2cos 4 3 t = 1 2 1 + cos 6 t 2 = 1 2 1 + 2cos 6 t + cos 2 6 t . But in turn, this last expression can be rewrit- ten as 1 2 1 + 2cos 6 t + 1 2 n 1 + cos 12 t o . Thus 2cos 4 3 t = 1 2 3 2 + 2cos 6 t + 1 2 cos 12 t , Pham, Quoc Homework 6 Due: Feb 27 2007, 3:00 am Inst: Eric Katerman 2 and so I = 1 2 Z 3 2 + 2cos 6 t + 1 2 cos 12 t dt. Consequently, I = 1 4 3 t + 2 3 sin6 t + 1 12 sin 12 t + C with C an arbitrary constant. keywords: linearization of trigonometric func- tions 003 (part 1 of 1) 10 points Evaluate the definite integral I = Z / 4 sin2 ( 1- sin 2 ) d . 1. I = 7 8 2. I = 5 8 3. I = 3 8 correct 4. I = 1 2 5. I = 1 8 Explanation: Since sin 2 = 1 2 (1- cos 2 ) , we see that 1- sin 2 = 1- 1 2 (1- cos 2 ) = 1 2 + 1 2 cos 2 Thus I = 1 2 Z / 4 sin2 (1 + cos 2 ) d ....
View Full Document

This homework help was uploaded on 03/19/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.

Page1 / 8

M408L hwk6 solutions - Pham, Quoc Homework 6 Due: Feb 27...

This preview shows document pages 1 - 3. Sign up to view the full document.

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