REVIEW OF INTEGRATION

Transcription

1 REVIEW OF INTEGRATION Trig Fuctios ad Itegratio by Parts Oeriew I this ote we will reiew how to ealuate the sorts of itegrals we ecouter i ealuatig Fourier series. These will iclude itegratio of trig fuctios ad also a quick reiew of itegratio by arts. Deriaties of Trig Fuctios Let' s begi with a quick reiew of the deriaties of the si ad cos fuctios, sice kowig those will hel us erify the results of our later itegratios. We kow from first year calculus : We ca erify these statemets usig Mathematica : D@8Si@ xd, Cos@ xd<, xd 8 Cos@ xd, Si@ xd< d d Hsi HxLL = cos HxL; Hcos HxLL =- si HxL dx dx (1) Idefiite Itegrals of Trig Fuctios The results of the differetiatios aboe suggest that the itegrals of si (x) ad cos (x) are (igorig ay costats of itegratio) : We erify ia Mathematica : Itegrate@8Si@ xd, Cos@ xd<, xd si HxL dx = cos HxL; cos HxL dx = 1 si HxL () Cos@ xd :, Si@ xd > Ad we also erify by differetiatig the results to show they equal the origial itegrad : D@8 1 ê Cos@ xd, 1ê Si@ xd<, xd 8Si@ xd, Cos@ xd< As you hae already see, we ecouter itegrals iolig si (x) ad cos (x) ery frequetly i our studies of Fourier series.

2 itegratiofourierseries.b Itegratio by Parts Itegratio by arts is a commoly used techique of itegratio that ca be articularly useful if the itegrad is a roduct of two fuctios. Suose your itegrad is of the form : ud (3) where both u ad are fuctios of the same ariable, the the well kow statemet of itegratio by arts is : ud = u - du (4) Let' s work a few examles to show how we aly this techique. Cosider the itegral : xe x dx (5) The itegrad is a roduct, so we should iestigate whether itegratio by arts will be useful. Our first ste is to decide which term i the itegral i (5) corresods to u ad which corresods to d. Let' s work this examle through. I (5), I will set u = x ad d = e x dx. This allows me to determie exressios for du ad : u = x fl du = dx d = e x dx fl = e x (6) We get the exressio for by itegratig the exressio for d. Now that we ca write u,, du ad d, we ca use eq. (4) to sole our itegral : x Äu e x dx d = xe x É u - e x Å dxå du (7) The fial itegral o the right is elemetary, so our comlete aswer is : xe x dx = xe x - e x dx = xe x - e x = e x Hx - 1L (8) Which we quickly erify both by direct Mathematica itegratio ad also by differetiatig the result : Itegrate@x Ex@xD, xd x H 1 xl D@x Ex@xD Ex@xD, xd x x What would hae haeed if we iitially chose u = e x ad d=x dx, the we would hae: ad our itegratio by arts would yield : du = e x dx; = x xe x dx = x e x - 1 x e x dx (9)

3 itegratiofourierseries.b 3 ad our secod itegral would be more comlicated tha the first. We wat our iitial choice of u to roduce the simlest ossible itegral o the right side of our itegratio by arts. Let' s try a examle iolig trig fuctios : x si HxL dx (10) We kow that if we set u = x, the du = dx ad we ca imagie our secod itegral will be straightforward. O the other had, if we set u = si (x) ad d = x dx, the our secod itegral will be ery comlicated. This tells us to assig : u = x fl du = dx d = si HxL dx fl =- 1 cos HxL (11) Ad we roceed ia arts : Verifyig : xä u si HxL dx d = x Äu H cos HxLL - H cos HxLL dxç = -x cos HxL 1 cos HxL dx = -x cos HxL 1 si HxL du (1) Itegrate@x Si@ xd, xd x Cos@ xd Si@ xd The itegral of Ÿx cos( x) dx is easily uderstood with a selectio of u=x ad d = cos( x) dx: Verifyig : 1 x cos HxL dx = x si HxL - 1 si HxL dx = x si HxL cos HxL (13) Itegrate@x Cos@ xd, xd Cos@ xd x Si@ xd Let' s try a slightly more comlicated case : x e x dx (14) We recogize this itegrad also as a roduct, but we will see that we will hae to emloy itegratio by arts twice. Makig the obious substitios of u = x ad d = e x dx:

4 4 itegratiofourierseries.b x e x dx = x e x - xe x dx (15) Ad we realize the last itegral i (14) will require itegratio by arts. Fortuately, we hae already doe this itegral (see eq.(8) aboe), ad we ca write : x e x dx = x e x - xe x dx = x e x - Hxe x - e x L = e x Ix - x M (16) or : Itegrate@x^ Ex@xD, xd x I x x M Itegratio by Parts with Defiite Itegrals For our work with Fourier series, we are of course iterested i ealuatig these itegrals betwee limits. I show below how oe would do the itegral of eq. (10) with limits of 0 to. Not surrisigly, we just ealuate each term at these limits ad fid: Verifyig : 0 x si HxL dx = -x cos HxL 0 si HxL cos H L - 0D H L - si H0LD = - HL = (17) Itegrate@x Si@ xd, 8x, 0, π<, Assumtios Elemet@, ItegersDD H 1L π Oe Fial Examle Let' s see how we might aroach a slightly more comlex roblem. Suose we wish to itegrate : e x si HxL dx (18) Our itegrad is a roduct, but both of the roducts are o - olyomials. Let' s see how we ca attack this itegral ia arts. We set : u = e x fl du = e x dx

5 itegratiofourierseries.b 5 d = si HxL dx fl = cos HxL Usig these substitutios, our itegral (which we defie as I) becomes : I = e x si HxL dx = cos HxL ex - e x cos HxL dx Ad we are about to lose hoe whe we realize our itegratio by arts has roduced aother itegral which also ioles a roduct of trascedetal terms. But, before we desair, let' s try to itegrate the fial itegral i (19) by arts, usig i this case the substitutios : (19) u = e x fl du = e x dx; d = cos HxL dx fl = 1 si HxL With this secod substitio, we hae : I = e x si HxL dx = cos HxL ex - e x cos HxL dx = cos HxL ex 1 B 1 si HxL ex - 1 e x si HxL dxf = cos HxL ex 1 si HxL ex - 1 e x si HxL dx Ad ow this is really deressig, because we hae yet agai roduced a itegral with a roduct of trascedetal terms... but... look at this last itegral, it is simly our origial itegral I with the coefficiet of 1 ë, so we ca rewrite (0) as: I = cos HxL ex 1 si HxL ex - 1 I where I = Ÿ e x sihxl dx. Now, we just algebraically collect all terms i I: (0) 1 1 I = e x si HxL cos HxL - ad after a little algebra we sole for I : I = e x si HxL dx = ex Hsi HxL - cos HxLL 1 Ad we beseech Mathematica for a resose : (1) Itegrate@Ex@xD Si@ xd, xd x H Cos@ xd Si@ xdl 1 You may recall that we soled ery similar itegrals usig techiques from comlex umbers o the first homework set of the term.