This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: momin (rrm497) Homework 07 cheng (58520) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Evaluate the definite integral I = integraldisplay / 2 (8 sin + sin 3 ) d . 1. I = 9 2. I = 26 3 correct 3. I = 7 4. I = 22 3 5. I = 25 3 6. I = 23 3 Explanation: Since sin 2 = 1 cos 2 we see that 8 sin + sin 3 = sin (8 + sin 2 ) = sin (8 + 1 cos 2 ) = sin (9 cos 2 ) . Thus I = integraldisplay / 2 sin (9 cos 2 ) d As the integral is now of the form sin f (cos ) , f ( x ) = 9 x 2 , the subsitution x = cos is suggested. For then dx = sin d , while = 0 = x = 1 , = 2 = x = 0 . In this case I = integraldisplay 1 (9 x 2 ) dx = integraldisplay 1 (9 x 2 ) dx . Consequently, I = bracketleftBig 9 x 1 3 x 3 bracketrightBig 1 = 26 3 . 002 10.0 points Evaluate the integral I = integraldisplay / 2 sin 3 x cos 2 x dx . 1. I = 2 5 2. I = 4 15 3. I = 8 15 4. I = 2 15 correct 5. I = 1 15 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 = sin x (cos 2 x cos 4 x ) , the integrand is of the form sin xf (cos x ), sug gesting use of the substitution u = cos x . For then du = sin x dx , while x = 0 = u = 1 x = 2 = u = 0 . momin (rrm497) Homework 07 cheng (58520) 2 In this case I = integraldisplay 1 ( u 2 u 4 ) du . Consequently, I = bracketleftBig 1 3 u 3 + 1 5 u 5 bracketrightBig 1 = 2 15 . keywords: Stewart5e, indefinite integral, powers of sin, powers of cos, trig substitu tion, 003 10.0 points Evaluate the indefinite integral I = integraldisplay 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 sin x ) + C 5. I = ln(1 + cos x ) + C Explanation: integraldisplay 1 sin x cos x dx = integraldisplay (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 004 10.0 points Evaluate the definite integral I = integraldisplay / 3 sec x tan x 4 + sec x dx . 1. I = ln 6 5 2. I = ln 3 4 3. I = ln 5 4 4. I = ln 6 5 correct 5. I = ln 5 4 6. I = ln 3 4 Explanation: Since d dx (sec x ) = sec x tan x , use of the substitution u = 4 + sec x is suggested. For then du = sec x tan xdx , while x = 0 = u = 5 , x = 3 = u = 6 . Thus I = integraldisplay 6 5 1 u du = bracketleftBig ln u bracketrightBig 6 5 . Consequently, I = ln 6 5 . 005 10.0 points Evaluate the definite integral I = integraldisplay / 4 2 tan 4 x dx . 1. I = 2  2 3 momin (rrm497) Homework 07 cheng (58520) 3 2. I = 2 + 2 3 3. I = 1 2  4 3 correct 4. I = + 2 3 5. I =  4 3 6. I = 1 2 + 4 3 Explanation: Since tan 2 x = sec 2 x 1 , it follows that tan 4 x = tan 2 x (sec 2 x 1) = tan 2 x sec 2 x tan 2 x ....
View
Full
Document
This note was uploaded on 12/11/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.
 Fall '08
 RAdin
 Calculus

Click to edit the document details