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

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Aileen Simon
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1 Tchniqus of Intgration c Donald Kridr and Dwight Lahr In this sction w ar going to introduc th first approachs to valuating an indfinit intgral whos intgrand dos not hav an immdiat antidrivativ. W bgin with a list of intgrals w should rcogniz. u r du = ur+ + C, r r + du = ln u + C u sin u du = cos u + C cos u du = sin u + C sc u du = tan u + C sc u tan u du = sc u + C u du = u + C W can radily vrify an quation by diffrntiating th right hand sid and showing that w gt th intgrand on th lft hand sid. But what if an intgral is not quit in th act form that w rquir? For ampl, 5 d. What do w do thn? Is thr a systmatic mthod that can minimiz trial and rror? Th Mthod of Substitution W hav alrady usd diffrntials as an aid to intgration whn w discussd sparabl diffrntial quations. In th prsnt sction, w will s that diffrntials continu to b a vry usful tchniqu for solving intgrals. So, our first ampl will srv as a rmindr of how to calculat thm. Eampl : If y = 3, thn dy = 3 d. Or if y = sin 4, thn dy = 4 cos 4 d. Rvrsing th Chain Rul: If u = g() is a function of, and f is a function of u, thn th chain rul tlls us that (f(g())) = f (g())g () Thus, intgrating th right hand sid rvrss th chain rul and w gt f (g())g () d = f(g()) + C Now, w can rwrit th abov intgral by substituting into it u = g() and th diffrntial du = g ()d. Whn w mak ths two substitutions w gt f (u) du = f(u) + C This last formula, combind with th us of diffrntials, constituts th Mthod of Substitution. Eampl : To find 7 d by substitution, w look at th list of intgrals at th bginning of th sction and s that th targt that it appars w should aim for is u du. Thus, w lt u = 7. Thn w calculat du = 7d; thus, d = du 7. Substituting into th intgral, w gt th intgral w wr aiming for:

2 7 u d = 7 du = u 7 + C = C W hav alrady larnd to solv th abov intgral by inspction. Indd, w wr doing nothing mor than rvrsing th chain rul in an simpl cas. Th nt ampl is also on w hav larnd to do by inspction but which w can do formally by substitution. Eampl 3: sin d. Lt u = ; thn du = d. So, substitution yilds Eampl 4: sin u sin d = du = cos u + d. Lt u = + ; thn du = d. So + C = cos + C + d = u du = ln + + C Eampl 5: + d. B carful. This is not a substitution intgral. That is, it is not an intgral that rquirs substitution. W simplify th quotint to obtain two trms ach of which w can intgrat: + d = ( + ) d = + ln + C Eampl 6: d. This intgral will yild to substitution: lt u = Thn du = (3 + 3) d. Thus, w substitut to gt d = 3 u du = 3 ln u + C = 3 ln C Eampl 7: ln d. Lt u = ln ; thn du = d. Thus, ln d = u du = u + C = (ln ) Thus far, w hav not usd th Mthod of Substitution with a dfinit intgral. W will do so now. Eampl 7 (continud): + C ln This is our first ampl of a dfinit intgral rquiring substitution. Thr ar two basic ways to solv it: ithr w chang th variabl from to u and chang th limits of intgration as wll; or w lav th limits of intgration unchangd and switch back from u to. Mthod : chang th limits from to u. d u = ln

3 3 ln d = u du = u = = 3 Mthod : chang th variabl back to and rtain th original limits. Eampl 8: ln d = (ln ) = (ln ) (ln ) = = 3 π/4 tan d W rplac tan by tan = sin cos and thn us th substituiton u = cos, from which du = sin d. So, with a chang of th limits of intgration th intgral bcoms: u = cos π/4 / π/4 tan d = π/4 sin cos d = / Intgration by Parts u du = ln u / = ln W saw abov that rally th Mthod of Substitution consists of rvrsing th chain rul. Anothr tchniqu of intgration that is oftn usful involvs an undoing of th product rul. For, suppos that u and v ar functions of. Thn starting with th product rul w gt d dv (uv) = u d d + v du d u dv d = d du (uv) v d d u dv d = uv d Rwriting th last quation in trms of diffrntials yilds v du d d u dv = uv v du

4 4 This is th so-calld intgration by parts formula. In practic, w can oftn us it to transform an intgral that appars intractibl into on whos intgrand has an antidrivativ w rcogniz. Not that w first hav to choos u; thn dv is that part of th intgrand that rmains. In gnral, w must b abl to diffrntiat u and intgrat dv in ordr to us th rst of th parts formula. Also, w want an intgrand that is simplr than th on with which w startd. Ths simpl obsrvations should guid us in assigning u. Lt s considr som ampls. Eampl : Considr d. Substitution dos not appar to work. So, w try th only othr tchniqu w know, namly, parts. If w lt u =, thn dv = d. Nt, w find du and v, and us th parts formula. u = du = d d = dv = d v = d = + C Eampl : Givn ln d, w lt u = ln and thn dv = d. And w procd: u = ln du = d dv = d v = ln d = ln d = ln + C Eampl 3: To find ln d, w simply follow through in ach trm of th parts formula with th valuation at th ndpoints of th intrval. Rfrring to th prvious ampl, ln d = ln d = + = Eampl 4: Givn sin d, us parts ltting u = and dv = sin d. Thn u = du = d dv = sin d v = cos sin d = cos + cos d Now, w us parts again to valuat th nw intgral. u = du = d dv = cos d v = sin sin d = cos + ( sin + cos ) + C Eampl 5: W us parts to valuat sin d: u = du = d dv = sin d v = cos

5 5 sin d = cos + cos d Using parts again: u = du = d dv = cos d v = sin sin d = cos + sin sin d Now, w hav an quation that has th unknown intgral on both sids. Thus, w can solv for it to gt sin d = ( cos + sin ) + C Erciss: Problms Chck what you hav larnd! Vidos: Tutorial Solutions S problms workd out!

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

6.1 Integration by Parts and Present Value. Copyright Cengage Learning. All rights reserved. 6.1 Intgration by Parts and Prsnt Valu Copyright Cngag Larning. All rights rsrvd. Warm-Up: Find f () 1. F() = ln(+1). F() = 3 3. F() =. F() = ln ( 1) 5. F() = 6. F() = - Objctivs, Day #1 Studnts will b