Math 120 Answers for Homework 14

Similar documents
3 2x. 3x 2. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB


Math Calculus II Homework # Due Date Solutions

Integration by Parts

Solution: APPM 1360 Final (150 pts) Spring (60 pts total) The following parts are not related, justify your answers:

Calculus II (MAC )

Inverse Trigonometric Functions. September 5, 2018

1. Evaluate the integrals. a. (9 pts) x e x/2 dx. Solution: Using integration by parts, let u = x du = dx and dv = e x/2 dx v = 2e x/2.

Math 102 Spring 2008: Solutions: HW #3 Instructor: Fei Xu

u x v x dx u x v x v x u x dx d u x v x u x v x dx u x v x dx Integration by Parts Formula

Math3B Exam #02 Solution Fall 2017

Differentiation of Exponential Functions

Calculus II Solutions review final problems

u r du = ur+1 r + 1 du = ln u + C u sin u du = cos u + C cos u du = sin u + C sec u tan u du = sec u + C e u du = e u + C

Calculus II Practice Test Problems for Chapter 7 Page 1 of 6

SECTION where P (cos θ, sin θ) and Q(cos θ, sin θ) are polynomials in cos θ and sin θ, provided Q is never equal to zero.

Odd Answers: Chapter Eight Contemporary Calculus 1 { ( 3+2 } = lim { 1. { 2. arctan(a) 2. arctan(3) } = 2( π 2 ) 2. arctan(3)

dx equation it is called a second order differential equation.

EXAM. Practice for Second Exam. Math , Fall Nov 4, 2003 ANSWERS

Hyperbolic Functions Mixed Exercise 6

Math 230 Mock Final Exam Detailed Solution

SET 1. (1) Solve for x: (a) e 2x = 5 3x

Trigonometric substitutions (8.3).

Integration by parts (product rule backwards)

6.1 Integration by Parts and Present Value. Copyright Cengage Learning. All rights reserved.

Mathematics 104 Fall Term 2006 Solutions to Final Exam. sin(ln t) dt = e x sin(x) dx.

INTEGRATION BY PARTS

1 Exponential Functions Limit Derivative Integral... 5

Complex Powers and Logs (5A) Young Won Lim 10/17/13

b n x n + b n 1 x n b 1 x + b 0

Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Roy D. Yates and David J.

Review of Exponentials and Logarithms - Classwork

Southern Taiwan University

Math 112 Section 10 Lecture notes, 1/7/04

HOMEWORK SOLUTIONS MATH 1910 Sections 8.2, 8.3, 8.5 Fall 2016

Thomas Whitham Sixth Form

Assignment 6 Solution. Please do not copy and paste my answer. You will get similar questions but with different numbers!

MSLC Math 151 WI09 Exam 2 Review Solutions

I.e., the range of f(x) = arctan(x) is all real numbers y such that π 2 < y < π 2

NARAYANA I I T / P M T A C A D E M Y. C o m m o n P r a c t i c e T e s t 1 6 XII STD BATCHES [CF] Date: PHYSIS HEMISTRY MTHEMTIS

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. and θ is in quadrant IV. 1)

( ) Differential Equations. Unit-7. Exact Differential Equations: M d x + N d y = 0. Verify the condition

VTU NOTES QUESTION PAPERS NEWS RESULTS FORUMS

THE INVERSE TRIGONOMETRIC FUNCTIONS

Math Final Exam Review

1969 AP Calculus BC: Section I

Section 7.4 #1, 5, 6, 8, 12, 13, 44, 53; Section 7.5 #7, 10, 11, 20, 22; Section 7.7 #1, 4, 10, 15, 22, 44

For more important questions visit :

Thomas Whitham Sixth Form

8.3 Trigonometric Substitution

(Section 4.7: Inverse Trig Functions) 4.82 PART F: EVALUATING INVERSE TRIG FUNCTIONS. Think:

Friday 09/15/2017 Midterm I 50 minutes

Part I: Multiple Choice Mark the correct answer on the bubble sheet provided. n=1. a) None b) 1 c) 2 d) 3 e) 1, 2 f) 1, 3 g) 2, 3 h) 1, 2, 3

Power Series. x n. Using the ratio test. n n + 1. x n+1 n 3. = lim x. lim n + 1. = 1 < x < 1. Then r = 1 and I = ( 1, 1) ( 1) n 1 x n.

Practice Differentiation Math 120 Calculus I Fall 2015

Calculus 152 Take Home Test 2 (50 points)

DSP-First, 2/e. LECTURE # CH2-3 Complex Exponentials & Complex Numbers TLH MODIFIED. Aug , JH McClellan & RW Schafer

Calculus II Practice Test 1 Problems: , 6.5, Page 1 of 10

The Matrix Exponential

Solutions to Tutorial Sheet 12 Topics: Integration by Substitution + Integration by the Method of Partial Fractions + Applications to Geometry

The answers below are not comprehensive and are meant to indicate the correct way to solve the problem. sin

6.6 Inverse Trigonometric Functions

DIFFERENTIAL EQUATION

Solution: As x approaches 3, (x 2) approaches 1, so ln(x 2) approaches ln(1) = 0. Therefore we have a limit of the form 0/0 and can apply the.

The Matrix Exponential

SAFE HANDS & IIT-ian's PACE EDT-15 (JEE) SOLUTIONS

PRACTICE PAPER 6 SOLUTIONS

nd the particular orthogonal trajectory from the family of orthogonal trajectories passing through point (0; 1).

Note: Each problem is worth 14 points except numbers 5 and 6 which are 15 points. = 3 2

MATH Section 210

10. Limits involving infinity

MATH1120 Calculus II Solution to Supplementary Exercises on Improper Integrals Alex Fok November 2, 2013

7.3 Inverse Trigonometric Functions

VANDERBILT UNIVERSITY MAT 155B, FALL 12 SOLUTIONS TO THE PRACTICE FINAL.

Math 234 Final Exam (with answers) Spring 2017

x + ye z2 + ze y2, y + xe z2 + ze x2, z and where T is the

Math 34A. Final Review

Handout 1. Research shows that students learn when they MAKE MISTAKES and not when they get it right. SO MAKE MISTAKES, because MISTAKES ARE GOOD!

3 a = 3 b c 2 = a 2 + b 2 = 2 2 = 4 c 2 = 3b 2 + b 2 = 4b 2 = 4 b 2 = 1 b = 1 a = 3b = 3. x 2 3 y2 1 = 1.

NEW APPLICATIONS OF THE ABEL-LIOUVILLE FORMULA

1973 AP Calculus AB: Section I

1997 AP Calculus AB: Section I, Part A

The graph of y = x (or y = ) consists of two branches, As x 0, y + ; as x 0, y +. x = 0 is the

If we integrate the given modulating signal, m(t), we arrive at the following FM signal:

Summary: Primer on Integral Calculus:

Math156 Review for Exam 4

Mathematics 1110H Calculus I: Limits, derivatives, and Integrals Trent University, Summer 2018 Solutions to the Actual Final Examination

Chapter 1. Chapter 10. Chapter 2. Chapter 11. Chapter 3. Chapter 12. Chapter 4. Chapter 13. Chapter 5. Chapter 14. Chapter 6. Chapter 7.

Review (2) Calculus II (201-nyb-05/05,06) Winter 2019

VII. Techniques of Integration

Exercise 1. Sketch the graph of the following function. (x 2

Math 21B - Homework Set 8

Trig Identities. or (x + y)2 = x2 + 2xy + y 2. Dr. Ken W. Smith Other examples of identities are: (x + 3)2 = x2 + 6x + 9 and

2 (x 2 + a 2 ) x 2. is easy. Do this first.

Summer Packet Greetings Future AP Calculus Scholar,

10.1 Review of Parametric Equations

Prelim 2 Math Please show your reasoning and all your work. This is a 90 minute exam. Calculators are not needed or permitted. Good luck!

Chapter 5 Analytic Trigonometry

( ) ( ) ( ) 2 6A: Special Trig Limits! Math 400

x n cos 2x dx. dx = nx n 1 and v = 1 2 sin(2x). Andreas Fring (City University London) AS1051 Lecture Autumn / 36

Transcription:

Math 0 Answrs for Homwork. Substitutions u = du = d d = du a d = du = du = u du = u + C = u = arctany du = +y dy dy = + y du b arctany arctany dy = + y du = + y + y arctany du = u du = u + C = arctan y u = lnz du = z dz dz = z du v = lnu dv = u du du = u dv c lnln z lnln z lnlnz ln u ln u z ln z dz = z ln z z du = du = ln z u du = u u dv = ln u dv = v dv = v + C = ln u + C = lnlnz It s possibl to do both of th substitutions at onc: u = lnlnz du = z lnz dz dz = z lnz du lnln z lnln z z ln z dz = z lnz du = z ln z = lnln z lnln z du = u du = u + C

d u = du = d d = du d = du = du = u du = arcsinu + C = arcsin An altrnat solution, involving slightly mor algbra is: u = d = du d = du = d u = so = u, and = = u. du = du = u du = arcsinu + C = arcsin Anothr variation is to mak on of th sam substitutions as abov, and but us arccos instad of arcsin. E.g., in th prvious substitution w could hav don th last thr stps as u du = u du = arccosu + C = arccos Ths solutions ar all th sam i.., th prssions listd all diffr by a constant. This follows from standard trig idntits applid to arcsin or arccos. For instanc, th rlation arcsin z + arcsinz = π shows using z = that th diffrnc btwn th first two solutions is arcsin arcsin = arcsin + arcsin = π = π. u = t dt = t du t dt = = du = t dt du = u u t t du = t t du = u u u du = arcsinu + C = arcsin t du

f u = + + a du = + d = + +a d +a +a d = +a + +a du + a d = + a + a + + a du = = ln u + C = ln + + a + + a du = u du. Intgration by Parts F = / f = g = ln g = a ln d = / ln / d = / ln d = / ln 9 / F = t f = t g = t g = t F = t f = t g = t g = b t t dt = t t t t dt = t t t t t dt = t t t t t + C = t t t t + t F = f = g = g = c d = d = d = d =

An altrnat solution is to first us a substitution, and thn intgration by parts. u = d = du du =,d F = u f = u g = u g = d = du = du = u u du = u u u du = uu u + C = d ln z dz = F = z f = g = lnz g = lnz z ln z dz = z lnz z lnz z dz = z lnz ln z dz W us intgration by parts again to find th antidrivativ of ln z: ln z dz = F = z f = g = ln z g = z lnz dz = z ln z z dz = z ln z z dz = z lnz z. Substituting this back in w gt ln z dz = z lnz z ln z z + C = z lnz z ln z + z cosln d = F = f = g = cosln g = sinln cosln d = cosln sinln d = cosln + sinln d

Now w try to find th intgral by parts. sinln d sparatly, again using intgration sinln d = = sinln F = f = g = sinln g = cosln sinln d = sinln cosln d. cosln d Putting both computations togthr givs us cosln d = cosln + sinln cosln d = cosln + sinln cosln d, and now w can solv for cosln d: cosln d = cosln + sinln. Idntitis u = cos d = du sin du = sin d a sin cos cos d = sin cos sin d = cos cos sin du cos u = du = du = cos u u du = u + u + C = cos + cos Rmark: Sinc sin appard to an odd positiv powr, u = cos was a good choic for a substitution. 5

b sinθ = sinθ cosθ so sinθ cosθ = sinθ sin θcos θ dθ = sin θ dθ = 8 sin α = cosα cosθdθ = θ 8 sinθ + C = θ 8 sinθ An altrnat solution is to us ach of th half angl formulas on sin θ and cos θ sparatly, multiply out, and us th half angl formula again. sin θ = cosθ cos θ = + cosθ sin θcos θ dθ = cosθ + cosθ dθ = cos θ dθ = + cosθ dθ = 8 cosθdθ = θ 8 sinθ Sinc d dθ tanθ = cos θ c tan θ dθ = sin θ cos cos θ dθ = θ dθ = cos θ dθ = tanθ θ cos θ u = t + du = dt dt = du d t + t + t + dt = t + t + + dt = t + t + + du = u u + du = u + + C = t + + + C = t + t + + 6 + 8 d = + + 5 d = + + 5 d = 6

= 5 u = + du = d 5 5 5 d = du d = + 5 5 + + 5 + 5 du = 5 u + du = arctanu + C = arctan + 5 5 5. Anything Gos u = t du = t dt dt = tdu F = u f = u g = u g = a t dt = t t du = u u du = u u u du = u u u = t t t b = F = f = g = arctan g = + arctan d = arctan + d + arctan + + d = arctan + d = arctan + arctan 7

c z z cosz dz W want to us intgration by parts whr w diffrntiat z and intgrat z cosz. For that w ll nd to know th antidrivativ of z cosz, which will involv a sparat intgration by parts calculation. F = z f = z g = cosz g = sinz F = z f = z g = sinz g = cosz z cosz dz = z cosz z sinz dz = z cosz + z sinz dz = z cosz + z sinz z cosz dz = z cosz + z sinz z cosz dz, which w solv to gt z cosz dz = z cosz + z sinz. W will also nd th antidrivativ of z sinz: F = z f = z g = sinz g = cosz z sinz dz = z sinz z cosz dz = z sinz z cosz + z sinz = z sinz z cosz. Now w r rady to find th antidrivativ of z z cosz: F = z cosz + z sinz f = z cosz = g = z g = z z cosz dz = z z cosz + z sinz z z cosz + z z sinz z cosz + z sinz dz z cosz + sinz dz 8

= z z cosz + z z sinz z cosz + z sinz z sinz z cosz = zz cosz + z z sinz z sinz 9