5 Techniques of Integration

Transcription

1 Clculus II (prt ): Techniques of Integrtion (by Evn Dummit,, v..5) Contents 5 Techniques of Integrtion 5. Bsic Antiderivtives Substitution Integrtion by Prts Trigonometric Substitution Prtil Frctions The Weierstrss Substitution Improper Integrtion Techniques of Integrtion We discuss number stndrd techniques for computing integrls: substitution methods, integrtion by prts, prtil frctions, nd improper integrls. 5. Bsic Antiderivtives Here is list of common indenite integrls tht should lredy be fmilir: n d n+ + C, n n + d ln() + C e d e + C sin() d cos() + C cos() d sin() + C sec () d tn() + C sec() tn() d sec() + C d sin () + C + d tn () + C d sec () + C

2 Here re some other slightly more dicult ntiderivtives tht crop up occsionlly: ln() d ln() + C tn() d ln(cos()) + C sec() d ln(sec() + tn()) + C csc() d ln(csc() + cot()) + C cot() d ln(sin()) + C 5. Substitution The generl substitution formul sttes tht f (g()) g () d f(g()) + C. It is just the Chin Rule, written in terms of integrtion vi the Fundmentl Theorem of Clculus. We generlly don't use the formul written this wy. To do substitution, follow this procedure: Step : Choose substitution u g(). Step : Compute the dierentil du g () d. Step 3: Rewrite the originl integrl in terms of u: 3: Rewrite the integrl to peel o wht will become the new dierentil du. 3b: Write the remining portion of the integrnd in terms of u. 3c: Find the new limits of integrtion in terms of u, if the integrl is denite integrl. [If the old limits re nd b, the new ones will be u g() nd u g(b). In very concrete sense, these re the sme points.] 3d: Write down the new integrl. If the integrl is indenite, substitute bck in for the originl vrible. Substitution is best lerned by doing emples: Emple: Evlute 3 e d. Step : The eponentil hs 'complicted' rgument, so we try setting u. Step : The dierentil is du d. Step 3: We cn rerrnge the integrl s 3 e ( d). Step 3b: The remining portion of the integrnd is e, which is just e u. Step 3c: We see tht corresponds to u nd 3 corresponds to u 3. Step 3d: Putting it ll together gives 9 eu du e u 9 u e 9. Emple: Evlute e (ln()) d. Step : It might not look like ny function hs 'complicted' rgument, but if we think crefully we cn see tht the numertor is wht we get if we plug in ln() to the squring function. So we try setting u ln(). Step : The dierentil is du d. Step 3: We cn rerrnge the integrl s 3 [ln()] ( ) d.

3 Step 3b: The remining portion of the integrnd is [ln()], which is just u. Step 3c: We see tht corresponds to u ln() nd e corresponds to u ln(e). Step 3d: Putting it ll together gives u du 3 u3 u 3. Emple: Evlute 3 + d. Step : Here we see tht the squre root function hs the 'complicted' rgument 3 + so we try u 3 +. Step : The dierentil is du 6 d. Step 3: We cn rerrnge the integrl s 6 6, which is oky since it's just multipliction by. 3 + (6 d). Note tht we introduced fctor of 6 Step 3b: The remining portion of the integrnd is 3 + 6, which is just 6 u/. Step 3c: We see tht corresponds to u nd corresponds to u. Step 3d: Putting it ll together gives Emple: Evlute 3 d. Step : Try the denomintor: u. Step : The dierentil is du d. Step 3: We cn rerrnge the integrl s 3 6 u/ du 6 3 u3/ u 7 9. ( d). Step 3b: The remining portion of the integrnd is, which is just u. Step 3c: We see tht corresponds to u 3 nd 3 corresponds to u 8. Step 3d: Putting it ll together gives 8 3 u du ln(u) 8 u3 ln 8 ln 3. Remrk: It's possible to do this one without substitution, by using prtil frctions to see tht + +. This gives I [ln( + ) + ln( )] 3 ln()+ln() ln(3) ln(8) ln(3) s before. 5.3 Integrtion by Prts The integrtion by prts formul sttes f g d f g f g d. It is just the Product Rule, rerrnged nd rewritten in terms of integrls using the Fundmentl Theorem of Clculus. To perform n integrtion by prts, ll tht is required is to pick function f() nd function g() such tht the product f () g() is equl to the originl integrnd. Emples will mke everything cler. Emple: Evlute e d. The integrnd e is n obvious product, nd so we need to decide which of nd e should be f. Since gets more complicted if we tke its ntiderivtive (since we'd get ) we try f e nd g, to get f e nd g. Plugging into the formul gives e d e e d ( e e ). Emple: Evlute e [ln()] d. 3

4 We cn write [ln()] ln() ln(), but this doesn't help unless we remember the ntiderivtive of ln(). Insted we write the integrnd s [ln()], so s to tke f nd g [ln()]. Then we get f nd g ln() by the Chin Rule. So plugging in will yields e [ln()] d ln() e Emple: Evlute ( + )e 3 d. e e e ln() [ e ln e e [e (e )] e. ln() d e ] d (IBP gin) We wnt g to be the thing which gets simpler when we dierentite. The polynomil + gets much simpler if we dierentite it, while the eponentil e 3 stys bsiclly the sme. So we should tke g + nd f e 3, so tht g nd f 3 e3. Then integrting by prts yields n epression which we cn't evlute directly we hve to integrte by prts gin: ( + )e 3 d ( + ) 3 e3 ( 3 e3 ) ( ) 3 3 e3 d ( 3 e3 ) [ ( ) 3 9 e3 ( 3 e3 ) [ [ e3 ]] 7 Emple: Evlute 3 sin() d. 7 e ( ) 3 e3 d (IBP once) ] 9 e3 d (IBP gin) Here we just need to integrte by prts repetedly. We get 3 sin() d 3 cos() + 3 cos() d (IBP once) [ ] 3 cos() + 3 sin() 6 sin() d (IBP gin) [ ] 3 cos() + 3 sin() 6 cos() + 6 cos() d (IBP gin) Emple: Find 3 sin( ) d. 3 cos() + 3 sin() + 6 cos() + 6 sin() + C First we look for substitution. We try the rgument of the sine: u. The dierentil is du d, nd we cn rerrnge the integrl s sin( ) ( d) u sin(u) du. Now we integrte by prts, to get u cos(u) + cos(u) du [ u cos(u) + sin(u)] + C. [ Finlly substitute bck for to get cos( ) + sin( ) ] + C.

5 5. Trigonometric Substitution Some kinds of integrls require more clever sort of substitution to evlute, one tht's sort of 'bckwrds' from the usul wy we try to do substitutions: insted of u f() we try f(u) for some pproprite (trigonometric) function f. The ide is to use one of the Pythgoren reltions (e.g., sin () + cos () ) to simplify something more complicted. As lwys, emples mke everything cler. Emple: Evlute d. Trditionl substitution long the lines of u doesn't work like we'd hope. Insted we try sin(u); then d cos(u) du. So we get sin (u) cos(u) du cos (u) cos(u) du cos (u) du. Remembering cos (u) + cos(u) Finlly substitute bck for u sin () to obtin simplify this to the equivlent form Emple: Evlute ( + ) d. we cn evlute the integrl to get u + sin(u) sin () sin () + + sin( sin ()) + C. + C. + C. If desired, we cn This time we think of the rctngent ntiderivtive nd try tn(u). Then d sec (u) du. We obtin ( + tn (u)) sec (u) du sec (u) sec (u) du cos (u) du. Remembering the identity cos (u) + cos(u) Finlly substitute bck for u tn () to obtin simplify this to the equivlent form Emple: Evlute d. tn () + we cn evlute the integrl to get u + sin(u) tn () ( + ) + C. + sin( tn (u)) + C. + C. If desired, we cn We'd like to do the sin(u) substitution gin but it doesn't quite work, since then isn't nice. Insted we hve to try sin(u), since then sin (u) cos(u), nd we're much hppier. Then we hve d cos(u) du, so we get sin(u) cos(u) du [ sin(u)] du u+ cos(u)+ cos(u) C. Substituting bck yields sin () + cos(sin ()) + C, or equivlently, sin () + + C. 5.5 Prtil Frctions Generlly, it is dicult to integrte rtionl functions without rerrnging them in some wy rst. nding the ntiderivtive of directly.) There is generl technique for evluting such integrls, clled prtil frction decomposition (PFD): the ide is to brek down rtionl functions into simpler prts which we know how to integrte. To nd prtil frction decompositions, follow these steps: (Try 5

6 Step : Fctor the denomintor. Step : Find the form of the PFD: For ech term ( + ) n the PFD hs terms C + + C ( + ) + + C n ( + ) n. For ech non-fctorble term ( + + b) n, the PFD hs terms + C n + D n ( + + b) n. Step 3: Solve for the coecients C, C,..., C n. C + D + + b + C + D ( + + b) + The best wy is to cler ll denomintors nd then substitute 'intelligent' vlues for (e.g., the roots of the liner fctors). Sometimes plugging in other vlues of is necessry, in order to nd the coecients of the higher terms. One cn lso employ other more clever methods, such s tking derivtives. Step : Evlute the integrl. C Terms of the form cn be integrted directly using the Power Rule. ( + ) n C + D Terms of the form ( + + b) should be seprted further s E( + ) n ( + + b) n + F ( + + b). n The rst term cn then be integrted by substituting u + + b, nd the second [ term cn be ( ) + c integrted by completing the squre s ( + /) + (b /) ( + c) + d d + ], d nd then substituting tn(t) + c d. As ever, emples mke the procedure cler. Emple: Evlute + 3 d. We fctor nd get ( + 3) so we wnt to write ( + 3) A + B + 3. Cler denomintors so tht A( + 3) + B(). Set nd 3 to see tht 3A nd 3B. Then + 3 d [ /3 /3 ] d ln() ln( + 3) + C. 3 Emple: Evlute 3 + d. We fctor nd get ( + ) so we wnt to write ( + ) A + B + D +. Cler denomintors so tht A( + ) + B( + ) + D. Set to get B. Set to get D. Set to get A + B + D so tht A. Then 3 + d Emple: Evlute We wnt [ ( + )( + ) d. ( + )( + ) A + + B + D +. ] d ln() + ln( + ) + C. 6

7 Cler denomintors so tht A( + ) + (B + D)( + ). Set to get A so tht A. So we wnt + + (B + D)( + ) so we see tht we need B D. Then ( + )( + ) d 5.6 The Weierstrss Substitution [ There is one dditionl substitution tht deserves discussion. ] d ln( + ) + ln( + ) + tn () + C. It is rrely covered t ny length in modern tetbooks, but it should be, since it llows one to integrte ny rtionl function in sin() nd cos(). The substitution is ( ) t tn, which is clled the Weierstrss substitution, or, occsionlly, the mgic substitution, since it llows one to evlute complicted trigonometric integrls. With this denition, one cn check tht cos() t t, sin(t), nd d + t + t + t dt. Then s we cn see, the trigonometric integrl which is rtionl function of t (though complicted). p(sin(), cos()) d becomes the integrl q(sin(), cos()) ( ) t p + t, t + t ( t q + t, t + t The mgic substitution cn lso be used to reduce ny problem involving trigonometric identities in rtionl multiples of single vrible to nite computtion with rtionl functions. (In prctice, this is not s useful s it might seem, becuse the polynomil lgebr is usully hrder thn using other techniques like the ngle-ddition formuls.) Emple: Find + cos() d. ( We set t tn ), with cos() t nd d dt, to obtin + t + t + cos() d + t + t dt + t ( + t ) + ( t ) dt 3 + t dt 3 + (t/ 3) dt In this new integrl we set u t/ 3 with du dt/ 3 to obtin u du tn (u) + C. 3 Substituting bck for t nd then yields the nswer s ( ( ) ) tn 3 tn + C. 3 ) + t 5.7 Improper Integrtion Sometimes we like to integrte to innity, by which we men, tke the limit s limit of integrtion becomes rbitrrily lrge. 7

8 In other words, we write f() d s shorthnd for lim q q f() d. Other times, we like to integrte through 'singulrity' of function tht is, through point where function is undened becuse it blows up: for emple, for the function f() /. Agin, we will write d s shorthnd for lim q + q d. Integrls with either of these two kinds of 'bd behviors' re clled improper integrls. (Perhps the terminology stems from the fct tht trying to evlute such integrls is not something done in polite compny.) Typiclly we will be interested in sking (i) if the integrl ctully converges to nite vlue, nd (ii) if it does, wht the vlue is. There re roughly two wys to solve problems like this: Method : Find the indenite integrl, nd then evlute the limit. If the limit is hrd, try rewriting the function to simplify the limit. Method : Compre the integrnd to nother function whose integrl is esier to evlute. We re ided by the following theorem: Theorem (Comprison Test for Integrls): If < f() < g() for ll in the intervl [, b] where one or both of nd b cn be innite, then if b f() d diverges, so does b g() d. If b g() d converges, then so does b f() d. The ide behind the theorem is to sy tht if < f() < g() then b f() d < b g() d, nd so if the f()-integrl goes to then so must the g()-integrl, nd conversely if the g()-integrl stys nite then so must the f()-integrl. Emple: Evlute e d. We just evlute to see e d lim q ( e ) q lim [ e q + ]. q Emple: Evlute + d. We do prtil frctions to see d ln() ln( + ) + C. + Now to ( tke the ) limit s is not so esy, unless we notice tht we cn rewrite the indenite integrl s ln. + Then we hve [ ( )] [ ( ) d lim ln q + q + lim ln ln( ] q + ) ln(). If the problem hd only sked bout convergence, we could hve observed tht + > for positive, so + <. Then the Comprison Test would hve sid tht, becuse d converges, then so does d. However, the Comprison Test doesn't help us in computing the + ctul vlue. Emple: Determine whether the integrl d converges. ( + )( + )( + 3)( + ) Doing prtil frctions is too much work (lthough it does give the correct nswer, if the trick in the previous emple is used repetedly). But it's esier to notice tht + < + 3 < + < +. 8

9 Then q ( + )( + )( + 3)( + ) d < q nite (in fct, it is ). Therefore this integrl converges. Emple: Determine whether the integrl We integrte to get d ln(q). q So s q + we see tht this diverges to. Therefore this integrl diverges to. ( + ) d ( + ) 3 q. As q this remins d converges. Well, you're t the end of my hndout. Hope it ws helpful. Copyright notice: This mteril is copyright Evn Dummit, -5. You my not reproduce or distribute this mteril without my epress permission. 9