Answer Key for Gury No-Z ; - & integrals Evaluate and simplify the following ''°9 7 1 4) J fos*(nt) dt Solution: a ie HI (ir stant)) ye 2+ (abel ( os (2nt}at i Gs*(nt) 4 —s alle 2], Recep cinco) ‘| 2 4n +} i ° x au Fosttntdat = 4 ef + or =; ant + Sin(ent) ° 4n when nis a (integer or (-) integer, [Fos (abide on 0 2 1 2) | Sint (nt dt a . : os(ant a ii sistawdat = J (ep . 5 | ascent 2, Sin ant)” = @. Sin@an®) an f Sint(nt)dt = + ged i sint(ntyat = 20% - Sinfent) g 4n when nis a (4) integer or C) integer T | sin (nt)dt A("eet J, teat 3) Solution; oT wT J tetdt = tet o ° [by in tegration by parts, Let, ust dvzetdt duzdt y=et i 4) Betta , Solution : J ett © Uaat, eh 2e* 5 2 (e*- 0°) at ot(at-2la-))-2) [vetaes & 46 + ie a(sft Pbs( Be)dt Solution: . Si #bs(Bt)dt = Gy AS sin( By rae Let: ust dy = loc(B-t)at du =2tdt V=& sina. #) ji bs a (arta = ( at “sina. 9). ]- seminal gf bs (Qe )at = am se°sin(an,) . 282 (nes(nt) ): 7 ae (a mal) Jivtes (2a)st = a Sin (OR) 4 2 tas (B8.) - 2g sin(at.) +6 [Fe tos (2-t)at = 4 (a Sin(nit) + 20nT fos (a) ~ ea Sn(at)) ° n a eo (ets(2thse : sf6"r- 20°) sin (2) $200 (ye (| po . - | b5in(M-A)at By integration by parts, ie dve sin(aus)dt dusdt vek, oe 5) +a oo (e(aeat]6) (Tat ) ( gttein (nt) dt Solution ei at sin (nt) tp ettsin oe | -2.| es (nt) dt Tate. / J sin (nt)dt = — . p lo — [ay infegration by parts, [By inayat By pas Let: at Let: he Sintnt) ave © FL | we stot) dv: dt du=zntosintdt v= i duz-nsin(nt) V= 5 [sin (ot)at ado Wee, . * i o**sin(nt) dt ° [Pomona =(@2ae00.) pot fi a a a lo ‘ [sin ot = EAsinlo®) nn fe*Meston) -1 sal) a 2 T, (: + =) 'sin(ut) dt = aet"sin (nt) - n@" s(n) +9 = (es2°) (Fe silonat z £(EMasinla- niles (xf) *) ar / jo a a7, sin (nit) sint)) +0 fresin(otla : E(asinbon) 2 > aren r cyinteger s n is a (+) integer & at nt} (Petsnlodaes EOeT : when at n™ ‘iets (oat en (ets) at en>— — T) Compute and simplify tre Maclaurin series at 6% followings 3) $4) = bos Lt) Solution: fit) = lost) Reese) eneeaor ¥it)=- Sin lt) £'(t) = - s(t) $Ml4)= Sin lt) £! ta) = s(t) £%y « -Sin lt) 6) = bolt) fs) 2/ 4! 9.) £4) = Sin (4) Solution: £ (4): Sint) (3) When b=? ¢ (0) =Sinle) =D $'(o) = (oslo) =! Mo) = -Sinlo)= b | ££) = Sin £1) = lost) £") =-Sinlt) $A) sles Lt) $e) = - lool p(y = Sin lt) go) = Sinlo) = 0 £4)(4) = sl) Ye) = osl0) = | ¢£)(k) = -sinlt) | #%) = -Sinl)=? £'(e) 2-Sinle) = © (0) = ~ 5(0)=+1 §"@)= sinle)=0 $%%o) = (05(0)21 $e) = ~Sinlo)=0 £60) =~ losle) #1 Xp RK 24. F2D degree of the xeBE Cemapute the Furi a ° os i ay gl: tif -Tee¢ od ; Fer Fourier Serits: fsLt)= het 8 fs + & bnSin(nt) ner Solution Selving ving * des i cia is en 3 ¥ 1°33, yA; On zt ttes(nt)dt = 1 1 leap nds) [By lndegnstion by pacts] By integration by parts By integration by parts, | [ME avelesinddt | iy =Sinlnt)at | | duettdt v= Sint) | fes(nt) n ° n - A 4 talndal] | den in (nt) 7] ) wa 7 ))) ‘sin(nt) sistem) )) Ly 1 ore oa 7 lt my By an by parts | | Iver: Metis — avesindntdat | wet? duz2tdt ys~lslnt) a iD) +2 [esa Tal “stn n A ih nag boned 5 (renee + os et I bn = 2(2 (5a leon pr) #lt)=3t if -1S45 7 oo . Solution: For Fourier Series: f5lt)= lot antl) € bn Sin(nt) v3) Solving for Qo, An and oa i : 2 : is 3 [ste e3 “an aft aa wr fm) 20 eae My =1 f" a4 los(nt)at = 1 [o(eaeee [* - it ‘sin (nt)at Td T no den Ody By infegration by pacts, thst dv=Cos(at)dt duedt — ve Sin(nt) n tn! fest Fant) A sts) ° a 2 (2,(ese toe) = as (: sista 2 (Bstnot | a ( oe (2 Goslntt) + wistots)) a sel wT n ) len “ Onso bn 3 (7 se Sin nt) tt ai w -0 dv= Sin(nt}dt y= ~foslnt) n