Homework for Week 13 Math 232 Spring 2002 This homework will not be collected It is your responsibility to do as many problems as necessary to understand the material this includes doing extra problems if you need more practice We recommend that you read each section before attempting any exercises Next week s quiz will be a subset of the problems below Section 14 2 1 3 9 11 12 13 18 19 20 24 25 26 27 28 31 34 35 42 45 48 49 50 54 55 59 60 65 Section 14 3 4 11 14 17 19 20 21 24 27 30 31 34 35 41 48 49 50 52 54ab 55ab Section 14 4 1 2 4 15 18 24 29 31 35 36 37 38 39 43 47 48 49 55 56 69 74 Selected Hints and Answers Caution The hints and answers below are not necessarily full solutions Many of them would not be considered complete on a quiz or test Answers are not provided for problems whose answers can be found in the reading or problems whose answers are easy to check using a calculator Section 14 2 R R R R 12 du 3 cos 3x dx and dv dx u dv sin 3x dx while v du 3x cos 3x dx The first integral is clearly easier and the second integral can be rewritten in terms of the first using integration by parts d 13 Part a Rdx x cos 2x cos 2x 2x sin 2x Part b R cos 2x 2x sin 2x dx x cos R 2x C Part c 2x sin 2x dx x cos 2x cos 2x dx See how the integration by parts formula comes from the product rule for differentiation 18 u ln x v x 2 20 False 24 False 25 a use parts with u ln x3 and dv x2 dx b use substitution with u x3 26 a use parts with u ln x and dv x1 dx b use subsitution with u ln x 27 a use parts with u x and dv x 1 100 dx b use substitution with u x 1 with back substitution R 34 Use parts with u x2 and dv e3x dx to get 31 x2 e3x 23 xe3x dx Then use parts R again to solve the new integral with u x and dv e3x dx to get 13 x2 e3x 23 13 xe3x 13 e3x dx 1 2 3x 2 2 3x 3x 3 x e 9 xe 27 e C 2 35 23 ex C R R R R 42 Multiply out the integrand to get x2 2xex e2x dx x2 dx 2 xex dx e2x dx The middle integral requires parts with u x and dv ex dx After integrating we get 1 3 1 2x x x 3 x 2 xe e 2 e C 45 Use parts with u x and dv ln x dx R 48 Use parts with u x and dv csc2 x dx to get x cot x cot x dx Then use substitution x with u sin x note that cot x cos sin x to get x cot x ln sin x C 49 Use parts three times first with u x3 and dv cos x dx then with u x2 and dv sin x dx finally with u x and dv cos x dx Section 14 2 continued 50 Use parts with u x2 and dv x sin x2 this is the largest piece of the integrand that you know how to integrate you will have to use a substitution with u xR2 to find v After applying the integration by parts formula you should have 12 x2 cos x2 x cos x2 dx Now use substitution with u x2 to get 21 x2 cos x2 12 sin x2 R C x 54 Use parts with u tan 1 x and dv dx to get x tan 1 x 1 x 2 dx Then use substitution 1 2 1 2 with u 1 x to get x tan x 2 ln 1 x C 55 Use parts twice and solve an equation to get 15 e2x cos x 25 e2x sin x see Example 6 2 R 2 2 2 59 Use parts with u ln x and dv dx to get x ln x 1 1 dx x ln x 1 x 1 2 ln 2 1 1 1 R1 60 Use parts with u x and dv e x dx to get xe x 1 1 e x dx xe x 1 1 e x 1 2e 65 Each of the problems below will involve using integration by parts with u x and dv cos 2x 3 dx R Part a 0 4 x cos 2x 14 3 8 1 4281 3 Part b The solutions to x cos 2x 0 in the interval 0 3 4 are x 0 x 4 and x 4 R 4 R 3 1 Here we must compute 0 x cos 2x dx 4 x cos 2x dx 8 14 2 5 8 4 1 7135 4 Part c The solutions to x cos 2x x on 0 3 4 are x 0 and x 2 However the value of f x x cos 2x is always greater than or equal to the value of g x x on this interval so we only need to calculate one definite integral The area between the two graphs is thus R 3 R 3 R 3 1 3 9 2 4 4 4 0 x cos 2x x dx 0 x cos 2x 0 x dx 4 8 32 1 34773 3 R 4 1 1 Part d 3 1 0 0 4 x cos 2x dx 3 14 3 8 3 2 0 6061 4 Section R 14 3 1 17 R21 1 cos 6x dx 12 x 12 sin 6x R R R 19 1 sin2 x 2 cos x dx cos x dx 2 sin2 x cos x dx sin4 x cos x dx sin x 23 sin3 x 1 5 x C R5 sin sin 2x 20 dx 12 ln cos 2x RC 2 R cos 2x sec 4x tan 4x sec 4x tan 4x 4x 21 sec 4x sec 4x tan 4x dx sec sec dx 14 ln sec 4x tan 4x C 4x tan 4x R R R R 24 R cot3 x csc2 x 1 R dx cot3 x csc2 x dx cot x csc2 x 1 dx cot3 x csc2 x dx cot x csc2 x dx cot x dx 41 cot4 x 12 cot2 x ln sin x C use substitution with u cot x for the first two integrals and substitution with u sin x for the integral R last 2 2 2 2 27 Use parts with u csc x and dv csc x dx to get csc x cot x 2 csc x cot2 x dx with u cot x to get csc2 x cot x 32 cot3 x C RThen use substitution R 30 R cos x 1 sin2 x sin 4 x …