a 2 +x 2 x a 2 -x 2 Figure 1: Triangles for Trigonometric Substitution

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1 I.B Trigonometric Substitution Uon comletion of the net two sections, we will be ble to integrte nlyticlly ll rtionl functions of (t lest in theory). We do so by converting binomils nd trinomils of the forms +,, ; nd + b + c into eressions involving single squred terms. We will ccomlish this through integrting methods known s "trigonometric substitution" nd "rtil frctions." I.B. Trigonometric Substitutions for Combining Squres + - θ θ θ - Figure : Tringles for Trigonometric Substitution The form + If we encounter n integrl involving n eression of the form +, we cn construct the rst tringle in Figure to conclude tht tn (becuse tn ). Then the eression + becomes q + tn + tn sec jsec j. The form On the other hnd, n integrl involving n eression of the form cn often be simli ed by concluding vi the center tringle in Figure tht sin (becuse sin ). Then the eression becomes q sin sin cos jcos j. The form Finlly, n integrl involving n eression of the form cn be trnsformed by relizing, s shown in the lst tringle of Figure, tht sec (becuse sec ). Then the eression becomes sec (sec ) tn jtn j. It is worth noting tht in ech of the three cses bove, we could hve swed the loctions of nd nd used the substitutions cot, csc, nd cos, resectively. The end result of wht follows would be the sme, however, nd so we will focus on the more trditionl choices of tn, sin, nd sec. In ddition, since we re mking substitutions, we will eventully wnt to be ble to chnge bck to the originl vribles. This mens tht we wnt the substitutions to be reversible, which mens tht we wnt to be ble to set tn sin sec if tn, if sin, nd if sec.

2 ecll tht the inverse trigonometric functions, in order to be functions, must hve restricted domins. Therefore, in order for our substitutions to be reversible, we must hve Substitution Inverse Domin tn tn < < sin sec sin sec 8 < : if < if < if Ech time we use reference tringle to introduce trigonometric substitution into n integrl, we must remin wre of the limittions to the ossible vlues for. Emle Given d, we set tn. Then d 4+ sec ; for < <, nd with in the rst tringle from Figure, we see tht 4 + This leds to d 4 + jsec j, or sec jsec j d sec d ln jsec + tn j + C: 4 + jsec j. (becuse sec > for < < ) Drwing n rorite sketch (like the rst tringle in Figure ), we see tht sec 4+ nd tn, so 4 + ln jsec + tn j + C ln + + C ln ln jj + C ln C, where C C ln (). Emle Given d, we set sin, so d cos d for 9. Then, with in the second tringle of Figure (sketch it!), we see tht 9 9 sin 9 cos, nd 9 cos j cos j. The integrl then becomes d 9 sin cos d 9 j cos j 9 sin d (becuse cos > for ) cos 9 d 9 sin + C 9 ( sin cos ) + C 9 sin! 9 + C 9 sin 9 + C.

3 Emle In the integrl d 5 ; we set 5 sec ; so d 5 sec tn d nd sec 5. Then 5 5 sec 5 5 sec 5 tn, nd d 5 5 sec tn d 5 tn sec tn d (becuse tn jtn j ) () jtn j sec d () ln jsec + tn j + C ln C () ln 5 ln 5 + C ln 5 + C (where C C ln 5). Note the use of the sign in the eression in (). This is due to the two ossible constructions of tringle with the necessry domin restriction (the tringle would be in either the rst or the second qudrnt - sketch them!) when we use the substitution sec. On the other hnd, consider the sign in front of the integrl in (). This ws originlly due to the rtio tn jtn j in (). If < <, the tngent is ositive nd so the sign is ositive (both in front of nd inside of the logrithm). If < <, the tngent is negtive, nd so both signs re negtive. Therefore, 8 d < ln C 5 or : ln 5 : (4) + C However, it cn be shown (s etr credit) tht the two eressions in (4) di er by constnt, nd so we nlly rrive t d 5 ln K. I.B. Trigonometric Substitution nd tionl Powers Emle 4 Evlute the integrl d by writing + ( +) s +. Then, with nd tn, we nd tht d sec d nd sec + (gin, consider the rorite tringle in Figure ). The integrl then becomes d d ( + ) + sec d sec d sec cos d sin + C + + C.

4 I.B. Comleting the Squre In some cses, we cn combine trigonometric substitution with comleting the squre to evlute integrls. Emle 5 We cn evlute by rst comleting the squre, i.e., d + + ( ). Then let u nd du d to get d du sin u + C sin ( ) + C. u I.B.4 Additionl Emles Emle 6 We cn evlute 9 d by noting tht the eression 9 resembles the eression we see in the second tringle of Figure. So we let sin, where. Then d cos d nd q sin 9 sin 9 cos jcos j ; but jcos j cos becuse. The integrl therefore becomes 9 d cos 9 sin cos d cot d csc d cot + C. An rorite tringle sketch is θ - nd we see from the tringle tht cot 9 nd sin. Therefore, 9 9 d sin + C. Emle 7 Note tht the integrl d could be done using the trigonometric substitution +4 tn. But you should lwys look for the simlest method, nd here we could use u-substitution with u + 4. Emle 8 We cn evlute d by rst noting tht (4 +9) Until now, the eressions we hve delt with hve ll hd terms with coe cients of. However, we cn

5 still use trigonometric substitution. The rorite tringle looks like () + θ nd from the tringle we see tht we cn let tn, which gives d sec d, nd leds to tn sec jsec j. Note tht jsec j sec becuse when we use the tn substitution. Also note tht when ; tn, so ; when, tn, so. Now, d (4 + 9) 7 8 tn 7 sec sec d tn sec d sin cos cos d sin cos d cos cos sin d. With u cos nd du integrl becomes sin d, (lso, when, u nd when, u ), this lst (4 + 9) d u + u u du u u u du u du [(5) ].

6 I.B.X Eercises (TrigSub) Evlute the following integrls d 4+. d 9 4. d d 8. 4 ( ) d. dy +9y d 4 d 9 ( ) d. 9 d 4d (4 ) d. + d Etr Chllenge. Show tht y ln + nd y ln di er by constnt. (Hint: Use roerties of logrithms to rewrite one of these s the log of frction, nd then rtionlize the denomintor.). Derive the integrtion formuls nd du u + tn u + C du u sin u + C.. Evlute the integrl d 9 +6 substitution.. It my involve substitution, comleting the squre, nd nother 8