Math 120 Answers for Homework 14

Size: px

Start display at page:

Download "Math 120 Answers for Homework 14"

Gillian Nichols
6 years ago
Views:

1 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

2 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

3 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 =

4 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

5 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

6 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 d = d = d = 6

7 = 5 u = + du = d d = du d = du = 5 u + du = arctanu + C = arctan 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

8 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

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

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

3 2x. 3x 2. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Math B Intgration Rviw (Solutions) Do ths intgrals. Solutions ar postd at th wbsit blow. If you hav troubl with thm, sk hlp immdiatly! () 8 d () 5 d () d () sin d (5) d (6) cos d (7) d www.clas.ucsb.du/staff/vinc