Homework for Week 13 Math 232 Spring 2002This homework will not be collected. It is your responsibility to do as many problems as necessaryto understand the material (this includes doing extra problems if you need more practice). Werecommend that you read each section before attempting any exercises. Next week’s quiz will be asubset 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 AnswersCaution: The hints and answers below are not necessarily full solutions. Many of them would notbe considered complete on a quiz or test. Answers are not provided for problems whoseanswers can be found in the reading or problems whose answers are easy to checkusing a calculator.Section 14.212. du = 3 cos 3x dx and dv = dx.Ru dv =Rsin 3x dx, whileRv du =R3x cos 3x dx. Thefirst integral is clearly easier, and the second integral can be rewritten in terms of the firstusing integration by parts.13. Part (a):ddx(x cos 2x) = cos 2x − 2x sin 2x.Part (b):R(cos 2x − 2x sin 2x) dx = x cos 2x + C.Part (c):R(−2x sin 2x) dx = x cos 2x −Rcos 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 = x2dx; (b) use substitution with u = x3.26. (a) use parts with u = ln x and dv =1xdx; (b) use subsitution with u = ln x.27. (a) use parts with u = x and dv = (x + 1)100dx; (b) use substitution with u = x + 1 (withback-substitution).34. Use parts with u = x2and dv = e3xdx to get13x2e3x−23Rxe3xdx. Then use parts again tosolve the new integral, with u = x and dv = e3xdx to get13x2e3x−2313xe3x−13Re3xdx=13x2e3x−29xe3x+227e3x+ C.35.32ex2+ C.42. Multiply out the integrand to get:R(x2−2xex+ e2x) dx =Rx2dx −2Rxexdx +Re2xdx.The middle integral requires parts with u = x and dv = exdx. After integrating we get:13x3− 2(xex− ex) +12e2x+ C.45. Use parts with u =√x and dv = ln x dx.48. Use parts with u = x and dv = csc2x dx to get −x cot x +Rcot x dx. Then use substitutionwith u = sin x (note that cot x =cos xsin x) to get −x cot x + ln |sin x| + C.49. Use parts three times, first with u = x3and dv = cos x dx; then with u = x2and dv =sin x dx; finally, with u = x and dv = cos x dx.Section 14.2 (continued)50. Use parts with u = x2and dv = x sin x2(this is the largest piece of the integrand that youknow how to integrate); you will have to use a substitution with u = x2to find v. Afterapplying the integration by parts formula you should have −12x2cos x2+Rx cos x2dx. Nowuse substitution with u = x2to get −12x2cos x2+12sin x2+ C.54. Use parts with u = tan−1x and dv = dx to get x tan−1x−Rx1+x2dx. Then use substitutionwith u = 1 + x2to get x tan−1x −12ln |1 + x2| + C.55. Use parts twice, and solve an equation, to get −15e2xcos x +25e2xsin x (see Example 6).59. Use parts with u = ln x and dv = dx to getx ln x 21−R21dx =x ln x 21−x 21=2 ln 2 − 1.60. Use parts with u = x and dv = e−xdx to get−xe−x 1−1+R1−1e−xdx =−xe−x 1−1+−e−x 1−1= −2e.65. Each of the problems below will involve using integration by parts with u = x anddv = cos 2x dx.Part (a):R3π40x cos 2x = −14−3π8≈ −1.4281.Part (b): The solutions to x cos 2x = 0 in the interval [0,3π4] are x = 0, x =π4, and x =3π4.Here we must computeRπ40x cos 2x dx −R3π4π4x cos 2x dx = (π8−14) −(π2) =5π8−14≈ 1.7135.Part (c): The solutions to x cos 2x = −x on [0,3π4] are x = 0 and x =π2. (However, thevalue of f (x) = x cos 2x is always greater than or equal to the value of g(x) = −x on thisinterval, so we only need to calculate one definite integral.) The area between the twographs is thus:R3π40(x cos 2x − (−x)) dx =R3π40x cos 2x +R3π40x dx = (−14−3π8) +9π232≈ 1.34773.Part (d):13π4−0R3π40x cos 2x dx =43π(−14−3π8) = −13π−12≈ −0.6061.Section 14.317.12R(1 − cos 6x) dx =12x −112sin(6x).19.R(1−sin2x)2cos x dx =Rcos x dx−2Rsin2x cos x dx+Rsin4x cos x dx = sin x−23sin3x+15sin5x + C.20.Rsin 2xcos 2xdx = −12ln |cos 2x| + C.21.Rsec 4xsec 4x+tan 4xsec 4x+tan 4xdx =Rsec24x+sec 4x tan 4xsec 4x+tan 4xdx =14ln |sec 4x + tan 4x| + C.24.Rcot3x(csc2x − 1) dx =Rcot3x csc2x dx −Rcot x(csc2x − 1) dx =Rcot3x csc2x dx −Rcot x csc2x dx +Rcot x dx = −14cot4x +12cot2x + ln |sin x| + C (use substitution withu = cot x for the first two integrals, and substitution with u = sin x for the last integral).27. Use parts with u = csc2x and dv = csc2x dx to get −csc2x cot x − 2Rcsc2x cot2x dx.Then use substitution with u = cot x to get −csc2x cot x +23cot3x + C.30.Rcos x(1 − sin2x) sin4x dx =R(sin4x − sin6x) cos x dx =15sin5x −17sin7x + C.31.R1−cos 6x21+cos 6x2dx =14R(1 − cos26x) dx =14R(1 −12(1 + cos(12x))) dx =18x −196sin(12x) + C.34.Rsec7x(sec x tan x) dx =18sec8x dx.35.Rsec2x(sec2x−1)2(sec x tan x) dx =R(sec6x−2 sec4x+sec2x)(sec x tan x) dx =17sec7x−25sec5x +13sec2x + C.41.Rsin2xcos2x1sin xdx =Rsin xcos2xdx = −Ru−2du =1cos x+ C.48.R(sec2x−1)(sec x tan x) dx =Rsec2x(sec x tan x) dx−Rsec x tan x dx =13sec3x−sec x+C.Check:ddx(13sec3x−sec x) =13(3) sec2x(sec x tan x)−sec x tan x = (sec2x−1)(sec x tan x) =tan2x(sec x tan x) = sec x tan3x.Section 14.3 (continued)49. Part (a): UseR1−cos 2x21+cos 2x2dx =14R(1 − cos22x) dx =14R(1 −12(1 + cos 4x)) dx.Part (b): UseR(sin x cos x)2dx =R(12sin 2x)2dx =14R1−cos 4x2dx.50. Part (a):Rtan3x sec2x dx =Ru3du, with u = tan x.Part (b):Rsec x(sec2x −1)(sec x tan x)dx =R(sec3x −sec x)(sec x tan x)dx =R(u3−u)du,with u = sec x.52.Rπ0sin x(1 − cos2x) cos2x dx =Rπ0(cos2x − cos4x) sin x dx = −Rx=πx=0(u2− u4) du= −13cos3x −15cos5x π0=415.52. Part (b):R12(sin(−x) + sin(5x)) dx =12cos(−x) −110cos(5x) + C.55. Part (a):Rsec3x dx =12sec x tan x +12Rsec x dx =12sec x tan x + ln |sec x + tan x| + C.Rsec7x dx =16sec5x tan x +56(14sec3x tan x +34(12sec x tan x +12ln |sec x + tan x|)) + C.Section 14.44. (a)R1√x2+1dx; (b)R√x2−16xdx; (c)Rp9 − (x − 2)2dx; (d)Rxx2+1dx; (e)R1x2−1dx.29. sin u =x−53, so make a triangle with angle u, opposite side x − 5, and hypotenuse 3; thenthe remaining