math31b_notes

math31b_notes - I AP CALCULUS. ' ,m Integrationbg Parts I '...

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Unformatted text preview: I AP CALCULUS. ' ,m Integrationbg Parts I ' Basic Formula: U CIV 7-" U V H I dLl' Use this method to integrate these gettems: I x“ sin x dx I X" Cos x dx IX“ exidx I x seczx oi)»: Ix csczx dx _ I in x dx . I sin‘1 x dx Icos'1 x dx Iexsinxdx, Iexcosxdx How? _ ._ 'l) on the side of the problem1 choose 1.: = and . dv = 2) Find the derivative of u and the do = ' _ g I v = antiderivative of our 3) Rewrite the original integral using the formula above: ' Iodv=uv -Iv'du ' 4) Nowintegrate T and add + C L. How to choose or and o'v: (A) For Ixnsinxdx, IxT‘cosxdx, Ixnexdx, I x sec2x dx , Ix {:3ch dx ' - _ choose u '= x” and dig = sin x dx, cos x dx, eX-dx, Seczx or _ csczx dx etc.§... Examgie 1: find I x cos x dx solutioni - I uzx- dv=cosxdx du=1clx _ v=sinx7 ' Using the formula, rewrite .' Ix cos xdx -=' x sin x — I sin x dx - __ ' =-x-sinx_—-cosx+C (B) For Iln x dx , I sin“1 x dx , I cos”I X'dx; etc... ' choose u = ln xor sin'1x, etc _ and ' dv =-dx . Examgie 2: find I x2 In-x dx - _ ' solution: _ g _ ' .. ' u=lnx clti=x2dx du=1ixcix ' t=x3i3 Using the formula, reWrite : Ix2 to x' dx I -= (X?! 3)_ln x — I X”! cix (X313) Inx — x3/9Ii+‘ C MW . integration bk Parts - (C) Special Trick N ll Examgte 3: find I e" sin x dx solution: : u=sinx du=exdx du=oosxdx v;=e" Using the formuia, rewrite l e" sin x dx = s“ sin x — le" cos x dx ' T again choose u and dv forthis integral u=cosx fdv=exdx du=~sinxdx vze" leKSinxdx = exsinx- [excosxw—lef‘sinxdx] lexsinxdx = exsinx— [exooex +lexsinxdx] lexsinxdx = exsinx - exoosx-lexeinxdx move this integral to the; other side exsinx — exoosx : 1/2(ex sinx — e“ cos x) +" C rileK sin x dx le" sin xdx (D) Tabular Method . . Use when a high power of x is present. I! H Examgl’e4: find lx5e" dx _ ' solution." Let u be the x to the high power. Keep differentiating "it" until you get 0 Keep integrating “dv”_. The answer is found like this: Starting with positive, keep aiternating signs ' and multiply diagonally as shown to; get: lx 5e”‘dx = x5 e"—5x4 e" + 20x 3 sex—60x2 e" +120x exm 120e“+.. C [is I ‘ miegration bg Trig Substitution Use this method to into rate ex ressions of the form? {"612 hugdu- . I Im2+u2du where o’s are numbers and iii? orefimeiions ofx. How? 1) Choose one o "the oliowin ' substitutions: at) Integron involving [wioz — again, use ' ii =Ia sin 0 I o) Iniegmis invoiving + uzdu. use it = a rim 6 e) Integron imiolving fair; 2 — o2 du. , use it = a sec 0 ' ' L. in :2)”. Find dx ' 'Exanegiies: o) 16 - 4x2 015: I choose: a = a sin 6 - ' - {Here a 1 4 and u = 2x) of 2x '-"— 4 sin 6 ' i ' x r- 2 sin (7‘ dx = 2 cos 6 d6 5,) choose: if = a {an {.9 (x‘ + 3)3 2 (Here a 2 V3 and ii = x) x m ion 9. dx : M53962 9 d8 b) Jexxlezx —1 mix choose: at; = o seeé} (Here a I 1 and Ii 2 ex) _ ex = see [9 (Look om} inside ihe 1/- exdx -“- SEC 0 tan 656’ tofino.’ a and ii) I 3) Repi’oee everything in ihe originoi iniegmi, including ofx, so their meg-e are no more x ’s and dx's. There are now oni’y 6’s and dfl’s : 4) i imiiz' and inte rate. Your answer 'wiil be in terms ofEi. AP CALCULUS Substitution; 5) ' Rewrite the answer back t'tz terms of x by drmttmg a right triangle asfillows. For :4 = a sin. 6, make these substitutions: t9 = sin-Jaye) draw-t this triangle _ sin 6 = u / a , _ strict: Lt : ash/:6 a, ,z . ' I Va2 —'a? A _ / LA. . cos 6 = ——a——~—-«_ I 34th :nUféL h/e _ _ a_ : UL e . u 1wth Peru/age k“ Ua‘iu’“ tan (9 = —'———~—,_ ere” _ \f a? '— u 2 ' Far a = at ttm 0, ' make these substitutions: 9 a (an "(u/a) ' ' a draw this triangt’e I; . sin 9 2 ———-—~— . L 4, '5‘: {rice- M = awed-J, x4} K‘JV/ - V a + “2 4'61 V: (1} M by - Q {EL/f U“ ' a ' ' N ' a" {3 COS = fan-“stifle :__ u. ' ' (Miffe’fiflfl-‘f (L _ ' I Ch» ' vazdru2 tart 6 =_ a / aj' etc.) - ’ECJfiJ/GJ- . I 2 '2 ' " ' . J: _ draw Eras triangle 1 UL / I 8m 6 _ _ “51‘1ch back?) 1? “Al/‘1. Him u ‘ I _ I) n I at" I L. n For u = [23:86.01 make these substitutions E} .=- .5 {Mega/{Av cost? =a/u, etc...“ (finish (rt ml“ ____ _fl_;_..__..... If it / . u. Examgie: - has reaffirm Ha: +112de choose: ' it. =- a tan 0 X + . . '_ I _ (Here. a = 1. and It: x") ' . :6 = tart E} gfi=méew I J“ 055 I H.86026’fd6’ fijseczeae I' {seczé‘wdfl d6 ' ' ' - 0 59-636? = I (x2+1)3t2 (tan26+i)3t2 (sec: 63ft: 53036 I 5609 J O . . sin 6 + (,7 Since x = fan 6,- I Draw the triangle a \Z From the triartgfe we see that Stine: '” d” ' "" '+C _ _ _ Imeraomgéfim—q: x/x2-+1 f_ IRENE?” x/xZH ' Sgeeéei Trick: j (-x + We + 2x -— 2 -dx Rewrite as j (x 4.1)“!th +52; +1)+1.e§_- Factor and rewrite as I I(x+])1t(x+1)2 ——1-dx I Here a =x+ 1 and Choose x +-€1-= 1 {an (7' I ‘ d3: 7—9 see2 0618 j(x+1)~/(x+l)3+1-dx = Itanaxtten'zflel-seczt9-d6 = jtantQ-seeQ-seczéé-dfi crest. Ae CALCULUS rat - {sin 2 x ‘- dx Replace Sin2x with. I 1 _ C35 2x e [(3032 x ‘ dx Replace 00st with 1 + COS 2x ‘ 2 a [sin "M x- cosm‘mE x - ax Save one m and replace all other .S'ines by . costnes using SinEx = I — 00553:; a fees”; 36- Sin “WW 3: --dx Save one m and replace at! other Cannes by g"; {Air-5 Sines using e052): 2 t -— stnzx I V)"; a Isinm" x- 003%” x- 625: You wit! have to make repeated substitutions of I szx E 1—003 2x WWW coégx I 1+ 0032:: .. 2 2 _ o [tang x l dx ' Reptaee tetan with sec2x — I 0 Jr cot2 x - a’x Replace eot2x with <33ch — J t ItanM-“mmg x - dx Replace one tangx with seczxe 1 0 [SCCG‘W x - dx Save one Seezx and rept'ace at! other Secants by I tangents using sec‘2x 2 I + tan-=2); 0 [seem tan a”""”””* )9 oh: «Save one and replace at! other secants by tangentuustng 5‘8ch = 1 + tani‘ a I-secmwm‘l’“ mum x - six Save one see x tan 3: and replace at! other tangents _ by seeantfls using tangx = See2x:— I 0 fees“: )6 tan”"“”- dx first, convert at! tangents to secants aging teran = SECZJC m I 2 Use {befottowtng method of integration by parts to'fintm integrating! odd gewers at seeant: e. g. ISBCB x ’ dx ' use a = sec )6 and cat = sec2x dx get: Isecl took 2 sec x tan x. - [secxtanz x - dx (now reptace ta:an with seegx —. 1) —-' .seextanx- Jsec“ xodx +- Jsecx-dx (now j'sec3 x'dx) 2 f 36.63 x- dx r- sec x tan )5 + ln|sec x —E- tan x1 + C (finale divideby 2) o For integrafls‘ of powers of cotangent and eosecant, use .S'tmtfar patterns as the ones; used above for tangent and secant. 5 a I f at] elsefaz‘ts, try converting everything to Sines and eosines. AP CALCULUS into “rat‘on b Part'al Ftaot' Use this method to integrate rational fun-efforts." _ For example I How? 1) 2.) 3) 4’) 5) 6,: Rewrite . x2_-x+2 air or 3 2 x x —x —I—x—1 (xw2)(x+3) {fthe power ofthe numerator (top) is £3 the power of the denominatot fiottom), use long division to rewrite thefimetton. I ' 2 Factor the denominator completely ' Rewrite titefraction as a sum of Simplerfi‘actions that can be integrated in other ways. Follow the patierne betow.‘ ' Patterns: Rewrite —-——1—~—— as A + B (x~2)(x+3) x—Z x+3 _ . 1 _ .. - A I B C Rewrite a as \ + 2 + 3 (ac—2) (x#2) (xw2) (AC—2) . _ 1 + 2x Ax + B C Rewrite as + (3x2+9)(x+3) 3x2+9 x+3_ 5 x0: — 2)? (x2 + 6x +1)2 as A . B C; Dx+E Fx—I—G b + ; + . 2 + 2 ' ‘ q 2 x {it—2) (35—2) at +6x+1 (x‘+6x+1) Muttipiy eveijy'term by the'common denominator, including the. ortngnaifimction. Now you have one giant equation; “__L__'__= A + B (x.—2')(x+3) x_—-2 Muttipfi' both Sides by" (x — 2,)(xé —I— 3) to get: I x+ 3 ' ', _ 1 = Afic + 3) + B(x — 2) His is the one giant equation Repiace x by eacn root (or repiaee x by any number you want) and sotveforA, B, C, D, etc. _ In the example above, in the equation 1 = A(x + 3) + B(x — 2), First, t-‘egtace x with —- 3 andsotve to get 8 z n i/j Next, t‘egt’ace x with 2 and soive to get A 2 1/5. _ Rewrite the ortginatfitnetton and integrate the now easier integmis.E 1 12’5 ~1t5 I09, _dx=j‘ dxifx+3dxi ...
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math31b_notes - I AP CALCULUS. ' ,m Integrationbg Parts I '...

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