Integration by Successive Reduction eee agree 4.1 Introduction The chapter deals with definite and integrals involving certain parameter or parameters, which are either positive or negative or fractional indices. Representing the integrals by I,J or U etc. with the parameter or parameters as the suffices, the integrals are reduced to a similar other integrals whose suffices are lower than that of the original integrals, by applying several use of integration. The formula which establishes a linear relation between the original integral involving parameter or parameters with some other similar type of integrals involving parameter or parameters of lower order is called a reduction formula. By repeatedly changing the suffices to its lower order in the reduction formula, the original integral is expressed to a much simpler integral which may be evaluated easily. Certain methods of finding such reduction formulae of certain integrals involving parameter or a parameters ate discussed here. 4.2 Reduction formulae for the integrals involving one parameter Lins / ae de, n € It (the set of positive integers) Integrating by parts we have he= are 2 far teae if a a near on Ina aa . aly ste nina, melt (4.1) 's the required reduction formula. 83Oo’ Advanced Calculus (Applica: 84 (Appl ications) n/2 . “ er 2. (i) In / sin"adz (ii) In = f[ sin"adz, n @ = J sin” cdr 7 foo o-sing de = -sin"“! x cosx + (n— 1) foo? w(1—sin? «)de or, nJy = sin"! 2cosa + (n— 1) [ sin Pode +. nIn = —sin"“) x cosx + (n—1)In-2; (4.2) which is the required reduction formula. (i) Jn = f 7? cin” ode By taking ae of integration from 0 to 7/2 it easily follows from (4.2) that Nn =(n—1In-2, 122 (43) which is the required reduction formula for Jn. 3.) In= fcos*ede (ii) Jn = [” cost 2s, net. Following the method as done in the previous problem we have nJn = cos"! zsingde+(n—1)Ino, n>2 (44) and Jy =(n—1)Jn2, n> 2, (48) which are the required reduction formula for In and Jy. Note By repeated application of the formulae (3.3) or (3.5) we also have a a le sin" ede = [ cos" nde 0 0 = Mol n-3 n-5 n n-2 n— n-ln-3n no n-2n-4 NIA (48) ale mle el wl according as n is even or odd.Chapter 4 Integration by Successive Reduction 85 7 — | tan® ae 4. (In i an" ede, (ii) Jy = f ee @ In = [iro sde = fax 2(sec? x — 1)de fax? rsec* dx — fia ede _ tan"! g not ie (4.7) or, (m1) Un + In~ (4.8) which is the required reduction formula. (i) Taking limit of integration from 0 to 1/4 we get from (4.7) 1 a qopo ds ie, | (n—1)(Jn + Jn—2) = 1 (4.9) which is the required reduction formula for Jn. ait © Note. If In ip cot” x dz and Jy = [ cot” x dz then it follows similarly that 0 (n= 1)(UIn — In-2) = — cot? (4.10) and (n—1)(Jn+Jn-2) = -1 (4.11) 5. In= [secteae, nelt. fo [oor asec? a dz ee GD i sec"? 2(sec? a — 1)de n>2 (4.12) Hence, (n—1)In = seo"? tanz + (n— 2)In—2, Which is the required reduction formula for In. An alternative way: 2 n olcecta—1 4 (tan a: sec” x) = sec"+? « + nsec” (sec? — 1) (n+ 1) sec"? x — nsec” #. Advanced Calculus Arvticai, : .. Integrating we get (nha) [ soot*?rde = tanasee" +n ii eo" mdr or, (n-1) [ocrade = tan sec"? x + (n— 2) [scoters or, (n= 1)Iy = tana-see”"?2 + (n—2)In-2- (by replacing n by n~2) which is the required reduction formula for Jn. @ Note. If In = / cosec" edz, then | cosec"“* cota | n—-2 | cose ore Tees n-1 na-l = ie, (n= 1p = — cose"? 2- cob a + (n= 2UIn-2 (4.3 which is the reduction formula for Jn. 6. In = J e**. cos” x dz, a being a constant. n= [er eostede _ e*- cos" & Dee aes + / e cos"! x sin x dx a a a a _ eh coste pn —= a -5 fet {(n-1eos*( = os TN ar 1 eg et co" a sina a 2 xsinx in? x) + cos" x- cos z} ug ar . -3 fe {(n-1) cos" * 2(cos* —1) + cos" x} dx eo = Fyeos"" 1 2(acosx + nsinz) n 4 [» fer cos act = (n= vf **- cos"? xda| Transposing and simplifying we get In = ga et -c0s""! a(acosa +nsina) + "=D. 7,» n? + a 24 at = hen or, (n? + a?)In = e%- cos”! g. (acosz +nsinx) + n(n —1)In—2- which is the required reduction formula for Ip. (414)hapter4 » Integration by Successive Reduction 87 za [@ + a?)"dx I= [+0 1-ae = (0? +a?) ~an [ (a? +08 1a? 4 a? — ade = 2 2)n = 2(e? 402)" —2n [(e? + a2yMde + ona? [oe+eyrae ‘Transposing and simplifying we get ae x(a? +07)" Ina? 2n+1 2n+1 which is the required reduction formula for In. Ina (4.15) @Note. Reduction formula for In = [ (a2? + be + c)"de is determined in the following way: Case (i) Let a be a positive number. Then In =a fe +p?)rdt, (4.16) b 2 4ac— where t= 2+ >> and p? = To Now using (4.15), the reduction formula for J, is determined. Case (ii) Let a be a negative number, say a = —K, K > 0. Then I, = K” | (p?——0)"dt where t, p are those given in (4.16), from which using (4.15) the required reduction formula is determined. 8 In = [@ —2?)"dxz, neélt. In = [@ =a?!" Lda = (ate fie = 2(a? —2”)"— on fio? - 2, (2n + 1)In = 2(a? — 27)" + 2na™ Ina which is the required reduction formula for Jn- n#1 2 2? —@2)de "d+ 2na? f (0? = 22)"Nde (4.17) oin= | paAdvanced Calcul 88 a“ (APDIatg, [ener | @raprte +$2(n= » [tafe $2(n—1-Ina — 2(0 ~ 1)-07F, aa ‘Tr5ansposing and simplifying we get 1 2 4 mss tt ee In = Wa2(n—1) (a? Fazyr-t © 2a(n — 1) ” (419) which is the required reduction formula for Jn. © Note 1. By substituting x = atand we get and hence using formula (4.4) the required reduction formula may also be obtained. dz = /————_ be d i @Note 2. Reduction formula for In / (ast Thapar MAY be determined in ti following way: Case (i) If a be a positive number then 1 dz b 4ac — b? 2 stt>, P= ow / a. - 4a? Case (ii) If a be a negative number, say —k, k > 0, then ol dz ee dace By using (4.18) the reduction formula for the above two integrals are determined. - nert In= / Sas Vax? + be Fe fade aun ae Cav + bo} ay Cees Vari there Va tare” nore 2. (an? suet oa Vax? + br +¢ 1 2 fy, 5g Be" Vas Fb Fe 2¢n —1) {aly + Bina + eln1}| Ins. IChapter 4 » Integration by Successive Reduction 89 ‘qransposing and then simplifying we get Seta ant yaaa (2n—1)b +br+e aooanee Ini — (4.19) which is the required reduction formula. r de 1 | Goa ia _— sing (a+ bcosx)r-t ee —1)b+acosa— (n— 2)b- cos? x dx (a+ bcos x)” A+ B(a+ cosx) + C(a + bcos)” a ted Ea al tricia A ti (e+ beosz)* , say (A,B,C are some constants) » (a? 4b and n be a positive integer greater than 1). Comparing the coefficients we have A+Ba+ Ca? =(n-1)b B-b+2Cab=a, CU? =-(n-2)b Hence, solving we get A=—(n—1)(2—8)/b, B= (2n-3)a/b, C=-(n—2)/b Thus we have nia? -B) 1 (2n—3)a, 1 ae b (et beosay* b (a+ beosx)"-? 2s (a+ boos gn? Integrating we get : : -2 P= ame P54 (Cn Ty — tne Hence, b a In = “ho n@-* (a+ beos a)" (Qn — 3)n | (4.20) ee +qen@ ey as De — b) which is the required reduction formula.Advanced Calculus (Appi 20 lication positive integer greater than 1, 12. In = i an sina da, n being @ lo Integrating by parts we have cS [roms 5 o m/e 2. ot t-test [en = (5) ae yea} n-1 7 In tn(n(n—1)In-2 =" (3) (4.21) which is the required reduction formula. 1 13. Jn = i 2” tan“! edz, n being a positive integer greater than 2. lo Integrating by parts we have n+l 1 1 cant Jn = |tan7 amy -f a era avi” i. Ae 4(n+1) nt+1Jo 142? fl antl + (n+ )n = 5 [2 iret (4.22) Replacing n by n — 2 we get from (4.22) that ed (a= 5— fo rade (423) Adding (4.22) and (4.23) we get ght] 4 gn-1 1l+2? (n+1)Jn + (n= 1)Jn_g = I alge M ! Il NIA ws an 1 Bin So SS which is'the required reduction formula.Chapter4 > Integration by Successive Reducti ion 91 43 Reduction formulae for the integrals containing two Parameters - 1. Imn = [2™(og2)"ax By integrating by parts we have I gmt n= ——. n mn = Ty (logy — fo (log 2)"“4de ai . mai (62)" — maT inmy (4.24) which is the required reduction formula, 2. () Imm = fe™(1—2)"de, Gi) Jinn = i * (1 —2)"de lo (i) Integrating by parts it follows that gmt nm Imn = pda) + wer fora 2)" de 2th on oS +h [ena-- 0-0 -a}ae mth (] 3 + —~- [Imn—1 — Im,n] ae m+1 ‘Transposing and simplifying we get aml (1— a)" n m+n+1 m+nt1 ‘which is the required reduction formula for Jm,n- (ii) Using (4.25) we also have m+1.(1—2)"]) ia | (ete m+nt+1 |, m+n+1 which is the required reduction formula for Jm,n- foc (4.25) Imn = Saige “SJmn—1 (4-26) oe ~m+n+1 © Note. By several use of (4.26) it easily follows that | “am = 2)%de 0 = [o- Which is known as the First Hulerian integral or Beta oF Baym. tn. ln aoe ee (4.27) ade = Jum = Tenet : function generally represented by BrynAdvanced Calculus (Applicay, io 92 ba)™ . 3. Inn = [ San, ae : have Integrating by parts we = (atbayn mb pat bey, . a Inn = Tatar Fad (atbo)™ mb a (Gave (429 which is the required reduction formula. dx 4. Inn = | weir Integrating by parts we have 1 n Inn = Gaal (a+ bay” on 1 oo ~~ Gaal (a+ br) 1 a Oh m1 tt = Ga at (a+ ba) Replacing n by n— 1 it follows from (4.29) that Inn ~ a(n—Da™1a + ba) which is the required reduction formula, 7 . me /2 Gi i sin” 2-cos"2dz, (ii) Jmjn = if sin™ x cos" x dx, m,n being positive integers. : On i cos"! x. (sin «cos 2)dar me = cos te Se + BO. f sin a(t — cos? 2)- cos"? xd ‘Transposing and simplifying we get sin”? +1 cos?) x = ieee na (4.30) m+n mon mn which is the required reduction formula for In. Now writing sin™ zcos" 2 = sin™-1 a ee le &= sin” 2(cos” ¢sinx) and proceeding as above it a8 sin™1 g cos™+! eee COB re ae +] — : min m+n man Which is also considered as the reduction formula for J ane oe (431)Chapter4 & Integration by Successive Reduction 93 ii) Using (4.30) or (4. Fi nea (i) (4-80) or (4.31) and taking limits of integration from 0 to ¥ we get m/2 Jinn = f sin” x cos" ede = 2—1 0 cost x da = min inno (4.32) or, J, if . m-1 7 mn = cos™ in” = es , sin a do = Inn (4.33) which are the reduction formulae for Jin. @ Note 1. Clearly, Jinn = In. @ Note 2. Using (4.32) or (4.33) a number of times we get 1.38++-(m=1)-1-3:5---(n—1) x a ce eo ee Ben) 46 (m+n) 2 (4.34) where m,n are even integers. 2-4-6---(m—1) do eee and Imm = Te Ta +8)-(n tm) coy when one of m,n, say m is considered odd integer. sin” x 6. Umm = f de, #1 cos? x Using (4.30) we have Ing = [ sin? 20s" xd et o _ sinPt¥ a-costtz | g— 1 ee (4.36) p+4a p+a Replacing q by q+ 2, it follows from (4.36) that sin?+! x cost! x qt+l ta? e ep at ere te ‘Transposing and simplifying we get inPt} x-costtiz p+q+2 sin? +1 z- cos ptat? , (4.37) ae + g+2 ager gel Again replacing p by by m and q by —n, we get from (4.37) that ae inte _ mantra (4.38) Umin =F costa on which is the required reduction formula for Um,n-a Advanced Calculus (Acton © Note. Replacing p by -m and q by —n in (4.87), the reduction formula for Ue, = fn #1 is given by ” sin™ xcos" x 1 1 oo Unn-2 n—1 sin™ 1 x- cos"? © Uma = ie / cos” «cos na da Integrating by parts it follows that Ing = Cone sine +2 fcos™4o(-sina)sinne de inn a a cos" x sinna = comtasinns | & fost x {cos(n — 1)x — cosne cos} 7 n as sinnesinx = cos(n — 1)x —cosnzcosz cos rsinnt | m = a + © Um-n-1 — Img] n n ‘Transposing and simplifying we get ie cos" xsinna m ea : m+n m+n : which is the required reduction formula for Im,n- An Alternative Reduction Formula: Integrating by parts it follows from (4.40) that. cos™asinnt om, 944 . Inn = ———— — 5 (cos! x cosna sin x) : a 7 m +3 | {om ~ 1) e08"-* 2 (cos*z ~ 1) + cos" 2} cosnede 1 =4 m +g [mn — (m= 1) Im—2,n} cos! x(nsin na cosa — m cos nar sin 2) ‘Transposing and simplifying we get Imn = 3a {nsinne cost — mcosnasin x} cos” x a _m(m ne — me In—2n which is the required reduction formula for Imm (4.39) (4.40) (4.41) (4.42) (443)Chapter4 » Integration by Successive Reduction 95 = ™ 2 sii 8. Imn / cos’ wsinna dx, m,n being positive integers. Integrating by parts we get = _SOS™rcosnt m ———s iS a mn 7 — 3 | 08" 2sin x: cosnadz cos” x cosnx =~ = m=1 pas 7 n J cos” esin na cos.x da m +2 f cos” esin(n — Vader, as cosnrsina = sinna cos — sin(n — 1) cos™ x cosnz m = n ‘Transposing and simplifying we get (m+ n)Imn = —cos™ x cosnt +m: Im—1n—1 (4.44) which is the required reduction formula for Imn- 4.4 Reduction formulae for the integrals containing three parameters a [ze + ba”)? de. Here Imny = f 2(a+ba”)Pde m+ P=. bgt. Z ae (Gt Oe) ae Os ee coe ] wf aw. (a+ ba") !de, m+1 mal = (at bo" = (m # 1 assumed) amtl(a+ be"? __npb 4.45 lage = eae | | md (448) “Tmtnnp-1 which is the required reduction formula for Im,n,p Note 1. Also Lo = emote meh fg + bo" — a)(at bx" )P de imnp = ee m+1 8 art (at ber) nD Fama oaryrae + OR | ama ter)? Ade m+i m4.5 Reduction Formulae > Example 1. [f I, = [e cosaz dz, Jn = fe sinax da then prove that (i) aI, =a" sinag —nJn_y (i) aJy = 2? cosar +n In—Chaper4 > Integration by Successive Reduction 97 Bie (asin ax + neosaz) — n(n —1)7 2 ne ~ ar Cosaz) — nn —1) J, 5 —lnsinax n,sinaz ot SNe? 8 ni, a a fe -sinac de Solution: (i) In al=2"sinag — NIn-1. ” COs at Sates fea SOE Tf sande (i) Jn = @Jn = a" cosax +n In (ii) From (i), at, = 2" sinaz —n | 2""sinaedz a. sinar—n {art Sosa Pacey 1 fer cosarde} 2 : ie @’ I, = az" sinaz + nz" 1. cos ax — n(n —1)In_2 a” (ag sinar + neosazx) — n(n —1)In—» ) From (ii), aJy = ~a" cosaz +n f 2" cosarde sinax a = a" cosar-tn {a%1, nae _ fo *sinaz ar} @ a J, = —2acosax +na".sinaz — n(n —1)Jn—2 = 2""(nsinax — axcosaz) — n(n —1)Jn_2. Proved. » Example 2. Prove that J sin? cost de = gy [sin 10x — § sin 8x + #8 sin 6x ~ 2sin dz — Tsin 2+ ua] +c Solution: Let 2 = cosa +ising Then } = cose —ising 1 af 2" =cosna +isinns, | >; = cosne —isinna. 1 Now, 2isine Roe aay) 1 2isinne 2eosna = 2" + —.ee litany, Ns) 98 ) 2 1/81 2) -2) 3 aay 1 _ sint soos? = oe (' Zz 1? aN ay (2-3) (:-2) i024 2 : Aa 1 (+39) (442) 0 +) z 1 64,1 s[(ees)*(@+8) 208) sf -8 (+ 3) -14 (2+ 3) +25] = iq [cos 102 — Gcos8x + 130s Gx ~ Bcos dar — 14.c0s 22+ 14] . [ ss8 2005? 2 de [ Fy | [costoe - Ge0s82 + 13 c0s 62 —8c0s4x — 14.cos2x + 14] +C : ig =a Faligsntoe - 4 sin 8x + — sin 6x ~2sinde — 7sin2e +142] +C Proved. m/2 > Example 3. Evaluate [ sin®/? x cos? a de. 0 Solution: m/2 a /2 ii sin®/? x cos°x de = if sin’? (1 — sin? 2). cos dx 0 1 fe 32 =[ wl oe u=sing, say. -EZ 2022. o]* 7 oO I Ble 5 Ans,chapter 4 m Integration by Successive Reduction 99 Je 4. I ln = f 2” p Exampl If In = | 2" coshadz, Jn = / Pe a () In =a" sinha — Jn (i) In = az” cosha — nI,_ Hence find I4, Ja. Solution : @) iy =a" sinha —n fo"! sinh ede = a" sinha — nJp_. (ii) Jn = 2" cosha —n [er cosh x dx = 2" cosh — nIn-1. Now, I, =a'sinha —4J3 = a‘ sinha — 4(2° cosh x — 312) = o'sinhe — 42° cosh + 12(2? sinh x — 21) = rtsinh a — 40% coshx + 122” sinh —24(xcoshx —sinhx), as Ip = sinha. = (e4 + 122? + 24) sinha — (403 + 24@) cosha +C. Ans. Also, Ja = 24-cosha — 4Ig = 2'coshx — 4(x° sinh x — 3J2) = xt cosha — 423 sinh x + 12(a? cosh — 2h) 4° sinh x + 120? cosh ¢ — 24(esinh x — cosh x) ‘4x3 + 24x) sinha + C.. Ans. = 24 coshz — = (a! + 122? + 24) cosh x — ( > Example 5. Prove that i+) tan eo 2 sce? : 4+ 30 J sec nae = zee ctang + zsccrtane 3 108 Solution: Let l= f sec’ a dx = [sock esec! adeAdvanced Calculus (Ap, cae 2 op = sect tane~8 [00 asec x tan? «dx 2 = sec? a tana = 3 f sec x(s0r aw—1)de 2 ‘ gr asec xtane +3 f secrsee ada Now, h = i seca: sec’ «dz, say 2 ‘Then, h= secrtane— f seontan adx = seca tant — [seoalsec? —1)dx =seextane — I + [ sonde tan (= 2) aa e (3+) an (745 as 3 fl 1 pee atans +7 [Feccntane + 5 log, + 2h =secrtanz + log, 1 1 ha peecxtana + 5 loge (GD tan G + 2) . Proved. 1 3 3 = poe stans + 5 seca tans + 3 OB 8 > Example 6. Prove that f Bae eg ec 3 cosa cose in> Solution: [Pie [costo d cos* © cos! g ee ‘du, cosa=u, say --/°3 m Jeary ja 1 = 2 3 costa ~ Gogg COS t+C. Proved:hapter4 » Integration by Successive Reduction 101 x/2 p Example 7. Prove that f[ sin’ ede — 16 0 7 (T n/2 sin? (F—2) dem [costae 0 n/2 a [ (1 = sin’ 2) cos de 7/2 [2 Solution : f sin’ede = [7 1 = f (l-w?)8du, u=sine say Wl 35_u]'_ 16 u-w+ lu -L] 22 pte a a7 as Proved. d Example 8. If tn = f eu prove that z (2n — 2)un — (27 — 3)tn—1 = ea Solution: Here, = So oe J +p / @apt rz —(n—1)22-a : fee = Gea ae —@ai by integrating by parts (a? +1) — = ay ey | Ga x’ cae : 4.2(n = 1)-In—1 — 2(n — Ln soca a? + 1)n-1" Gy et (Qn = 2)In — (2n = 8)In—a = Proved. ee ing a positive integer greater > Example 9. Prove ina f atten tede = 5 — 7, n being ap . 0 than 2, Solution : Integrating by parts we get for n > 2 that ntl nth 2 J, = |tan™* 2] -f eet wei n+1], a gntt eae eo oe anaes = a+)Advanced Calculus (ayy 102 antl Sy 2 nt)n= ee iva” Replacing in by n— 2 we get eo, (n-Dn-2 = r_['2 ae 7 1 gMtl 4 grt . tht (nD = 3-f Tee 37 | a” Nde aoe a Proved. 202 : sin(2n —1)r sin* nz » Example 10. If In = (ee, In al eae da prove that (i) nUInq1 — In) = sin 2ne (ii) Inga — Jn = Inst Solution: (i) i he . NInsi — In) =sin2nz. Proved. : in? a i) Ins —Jn = f Sine ite aye te sin? @ sin(2n + 1a si = [Bent de sine, sin? x = fsinQn+1)x = / sing @ =Jnu1. Proved. me /2 > Example 11. that ee ee I cos” xsin nada, m,n being positive integers, P™ 1 2 93 a Inn = Say [+3 + epee eZ]. ™mChapter > Integration by Successive Reduction 103 Solution: From (4.44) we have (m+ nim = [~ 08" x cosmalf/? + ae = Lt Mm Iman. i Finn = Bin + 3 Pty @ ol 1 = am +3 [ao aay t plmam2], by 1 1 i j =o +e tf 2m | 2(m—1) +p law 2) + yim-am-3] ae 1 1 1 Im * Bin * Boy + Imma 1 1 1 gmtg + mag + ge a 1 1 1 2m * 3m —1) * 2(m—2) * Exercises IV Fa" +n In-1 Lif hy= ij, e-*a" de. prove that In = — 2. If In = [ e*coshzde, ae = [ 2sinhade, prove that In tnJp1 = 2" sinhe and Jn +nIn—1 = 2" coshe. m/2 _ 3. Prove that i cos" a cosna de = seer 0104 bd x _ Advanced Calculus (Appi, tion, 5 _1In =1+(n— ie fe cos"? asin de, prove that a(n —1)In = 1+ (2 2)Iny ang nag f x deduce that In = ie n . If In = f (sina + cos2)"dx, show that nly = (sin + cos)": (sin — cos) + 2(n — 1)In—2. , prove that 2™ 7. -/ fax? +¢ amIm = 2") Var? + 0— (m—1)elm-2- 8 m/[2 Prove that i sin’? x cos’ xdz = 7. 0 45 If In= / ¢*. sin” ba dx then prove that _asinbr —nbcosbt on, inl n(n — 1)? era ee I], -/aoe oe inn = | Tagipaye show tha gmt 2(n = alan = —