Integral Calculus and Its Applications cn iii | 4. Reduction formulae. 2. Reduction formulae for f sin” x dx. f cos” x dx and evaluation of f°" sin’ x dx, | t | f"* cos” xx. 8. Reduction formula for { sin” xc0s” xdx and evaluation ot [** sin xcos” x ox. [ | 4:Reduction formulae for { tan® x dx [ cot” x dx. 5. Reduction formulae for J sec” x dx, [ cosec” xdx. | l | [x cosinx dx and [cos x sin nx dx. 8. Definite integrals. 9. Integral as the limit of a sum. 10. Areas of | | curves. 11. Lengths of curves. 12. Volumes of revolution. 13. Surface areas of revolution. 14. Objective Type of ' Questions. Bees 8. Reduction formulae tor fx” eax, [ x” (log x” ox. 7 Heduction formulae tor f x” sin mx ax, REDUCTION FORMULAE ‘The reader is already familiar with some standard methods of integrating functions of a single variable. However, there are some integrals which cannot be evaluated by the afore-said methods. In such cases, the ‘method of reduction formulae proves useful. A reduction formula connects an integral with another of the same type but of lower order. The successive application of the reduction formula enables us to evaluate the given integral. Now we shall derive some standard reduction formulae. [EEA (2) REDUCTION FORMULAE for (a) [sin® xax (6) [ cos" x dx. (@) sin" xde= [sin™* x-sin x dx Mntegrated by parts] =sint-1x-(e083)~ [ (n—1)sint-2x e0s.x cos x) dx =-sin'-txeosx+(n—D f sin"-* x —sin® x) dx wnsintVecose a (n= 1) faint? ede (n—1) J sin" de ‘Transposing, nf ata de == sinh—Vieewa x + (n= 1) fint~* ds Aa)(6) Similarly, Joos" x dx = ‘Thus we have the required reduction formulae. sin x cos"~* x ” @) To show that [ sin" x dx = {° cos" x dx _@-Din-9)(n—. (x nin —2)(n-4) x (rons vniseun) From (i), we have 2 xde ie 4 2, 2 pie 2 In=Flokg=Gh=5 fp sinxdr=3|—cosx (n= 1) (n= 3)(n-5)..2 - rin-Dn-4..3 i) Form these, we get I, = Case I. When nis even We have (n-1)(n~3)(n-5) n(n —2)(n—4) Combining (ii) and (iti), we get the required result for {°"" sin" x dx. Form these, we obtain I, = Proceeding exactly as above, we get the result for f”* cos" x dx. ‘Solution, (i) We have the reduction formula fain” xa = Putting n = 4, 2 successively, in? fsint xac=— 8% 8 sin? xe te)Join? x dx=— SPACES 4 Ef isin xP dx But flsin x! de=fde=x Join? xde=~ SRzeSe, = ‘Substituting this in (a), we get nt : [uit caro ttt, 3 sizer, 2) 2°28 Gi) We know that {°"" cos" x de = Putting n = 6, we get ait 5..1n_5n JP cos x dx = : = i [Put x =a sin 6, so that de = a cos 8 d@ 0 fy lesa [Also when x = 0, @ = 0, when x =a, @= 2 1 in? [PSS Ea cos odo=a" f° sin” 00-0" 8.4.2 185 (ii) Putting x = 20, we get (SD at cae =f? SEE nt ao 1+ cos x 1+ cos 20 = 2 [7 02 9 sin 6 cos 0 dO = 48 f°” sin* 6 do = 442-2 ° Zeos" 8 iy [ dx Put x =a tan @, so that dx = a sec” @ d0 Gd) Io Ee ep | Also whenx ~0,0~0, when x= +, 0 = W2 2 asec’ @d0_ 1 __ rit =e eg gat ee ode a Setaton: Patines asin 8 wet £ seen ri? (axing? az ez x) _fn- 100 eg (a e080) d® =a" f° sin" ode nn 2) _ @-D-3)..1 = nD) Now integrating by parts, we have Bony a Eg ie ff 2 sin? xde= | (sin 2)-[Using (i) p. 241] when n is odd when n is even Solution, Puttiog'a wai sin 6; get B= [G28 dee f°" Gta sin? 7 a 08 0 d0 = a *1 f°" cos™* a0 aie, _2n) (2n ~2) Qn ~ (2n +1) 2n -1) 2n -3)...5.3 [+ On + 1)is always odd) Now replacing n by n ~ 1, we get a gin-1 R= 2) Bn — 2 toe 88 to T,.20"-1 a pyeegy © ager which is the second desired result. GERI (1) REDUCTION FORMULAE for J sin” x cos" x dx Jin x cos" xdx= f sin™=* x » cos" x-sin x dx (Integrate by parts) vain te (=) fem natin on ( oo) a n+l n¥t x cos" '* i. n+l m+ a - sin Eee MEE sin ® zens ede ™—1 f sin” x cost x de n+l nel n+1 ‘Transposing the last term to the left and dividing hy 1 + (m —1)n 41), i, (m+ n)lin + 1), we obtain the reduction formula Jisint-® (2 sin? x) cost x dx sin™~1 x cos**! x m+n Joint = cos sce = ee ein? x coc! x de At) (2) To show that pt rao (m=1)(m-3)..x(n-Din-B— (® oaiy JF sin x 60s" 105 ee (§: oniy if both mana n are even)From (i), we have sin™-? x os"*? x |"? ia sin x cost xa =| — men ie. m-6 m+n— 4, Im = 4, na, n= 2 2p. Finally Spent sin x cos" x dx a2 |_ soot a Wi) n+3| n+l (n+3in+0) = From these, we obtain L = (m—-1)(m~ 3) (m5)... 4: mn (m+n) (m+n—2)(m+n—4)...(n+3)(n+ 1) Case I, When m is even Wehave, Ig.gg= Gee m-5 3 Nan" Td Solution, (@) Taking n = 2, in (i) of page 241, we have the reduction formula : 1 x 006? Jin” x os? x dx = SOARES 4 OF [sin-? x cos? xde Putting m = 4, 2 successively, 2 sin x cos?x dx = sink xo0e' 58 fit x cost ede of) ~ = _ Sin x cost x 1 2 Jin? x cos? x dx = See eG Jonteds But Joost xde=% f+ ens 20) dx i(x+}sin2x)Jsin® xeos? xe = mans e+ sin20) Substituting this in (1), we get 6 2 Jsint x 00s? x dx = + far +sin2} sin’ x cos’ x oj[ meets 4 (i) Putting ¢= tan ®, so that fee [07 28 cect odo= [°" sin® @ cos? odo = 0 (1 +t") e lo (iii) Putting x = tan 0, so that f # dea [07 S88 8 sect odo [°” sin® Gees? odo = 5:B-1X5-3-1 on Be 7 a+ ey to sec™@ Example 6.6. Evaluate: (i) f°" cos" 30 sin™ 60.40 (WT, 2003 8) io) fie ~ 298 de Gin) ig x (ax =) de, (V.P.U., 2010) 26 16 Solution, @) [east 38sin’ 68.40 ~ [°° cos! 36(2 sin 98 cos 989° 8 x16 =8["" sin® 30 cos" 30.48 vetirda fll ose Also when @ = 0, x = 0; 5h sin’ x cos? x dx when 0 = 16, x = 1/2, 8, 2x6-4-2 1 “3 10-8-6-4-2 15 i) fit 2 Put x = sin ¢ so that dx = cos ¢ dt Gi) [stems When =0,¢=0; when = 1,t=n2 xi? a = JP Hint # (c00t #9 -coe te de — [7 cit tone! # de _3-1x3-1 x_ 3m “8.6.4.2 2 256° Gai) JP ax) ax 2 Put x = 2a sin? @ = aaa ae 2. dx = 4a sin 6 cos 6 d0 snl = [[) @asin® 9? Vea) eos ¥- 4a sin 9 cos 8 a8 5-3-1x1 x _bxat = 28a! ("” sin® 6cos? 0d0 = 32 a4 = Ba sin® 6 cos? 6 d6 = 32 a’ Geese | Evaluate x 6 1 1 of? ii) [~" sin? 30 40 ; ats 1. [? xa w fF Cire Gi) [8 stort ede ! 2 @ fos (a>1) (VEU, 20085) w qe sin? xeost x ds. U.N-2U, 2008),lwrecrat CALCULUS AND Irs APPLICATIONS a fa 2 4. Udy. = [°° snl x oa dx (m > 0,n> 0) show tha, = 1 Mende evalunte [sin x coe x dx Evaluate xl x2 5. (0) [win eos? x dx (Cophin, 2005), (id J” sin"® x con® x dx 6, I) aNG- Hh de (a) cont 0 sin 69 0 2a 3 ta dy ti) [x77 Qa sxe? de wi, (Madras, 2000 8) f Fee 35 ao fo ERmae (io fan te (nu, 2008) 9.101, = | x' a= de, prove that (On + 8)1, *2an 1, ,—2e"(a x) (AMfarathwodla, 2008) 2a+1 at? 5 y 10. Itn in postive integer, ahow thine [[" <"Yl2ae— 93) ee = -EL g wrtts, 2007) [GE REDUCTION FORMULAE for (a) J tan” x dx _(b) J cot” x dx (a) Let Jean xde= ftan’-* x-tan® xde= [ tan"? x. (see? x—1)dx = f tant? x see? x de f tan"? x de tan“ Thus, —~* =1,.-2 which isthe required reduction formula, (0) Let Jeot® xd = [cot** xeot? xdz = f cot"? x(cosec? x—1) dx = [ cot"? x cosec? x dx — f cot"-* x dx = ‘Thus a a n-1 which is the required reduction formula. Example 6.7.Evaluate (i) tants de (ii) [ cot! x dx. Solution. (@) Putting n = 5, 3 successively in the reduction formula for fran’ x dx, we get ie,ie, foot? de =— cottx+ beat! x —cot zx. Solution, The reduction formula for {" tan" od is 1 ale 1 ne y= | ta xf - gare ae Changing n ton + 1, we obtain 1 Iyer ey 8 MDG thd [EBGBL REDUCTION FORMULAE for (a) J sec” x dx (b) J cosec” x dx (a) Let Ty~ [occ d= [ ece"- 8x. coc" x de Integrating by parts, we have T,= sect~8x tanx— f in —2) sect 8x - sec tan x} tan x de = see" x tana —(n-2) J sect*x- tan? x de = sec'~?x tan x—(n—2) f seo“? x. (secx 1) de =sec"~?xtanx—(n-2)1,+(n-2)1, 9 ‘Transposing, we have (n= I, = sect~2 x tan x +(n 20,9 sec"? xtanx , n-2 Thus ie [= += In_awhich isthe desired reduction formula, a n (0) Let 1,2 f cosee" xdzx= f cosec"~* x-cosec” xdx Integrating by parts, we have 1,= cosec"#x, cot x)~ fin —2) cose"? x ( coseex cot 2) -(— cot x) de == cot x cosec-?x—(n-2) f cosec"~* x (cosec? x ~1) dx =~cot x cosec"~*x—(n- 2), +(n—2M,_» cot x cosee"“2x+ (n= 2,» cot x cosec"~? x | n-1 which is the required reduction formula. or [1+ (nr — 2) ‘Thus 1, Solution. (@) Putting n = 4 in the reduction formula for [ sec" x dx, we got I, = [set xae<| 0 ntan sect xtanx | 2, 3 3a? 2 4 42 sec? ede |, 3 nie 2,2 2 =3tgltmls apt D=4/3.IwTe@RaL CALCULUS AND ITS APPLICATIONS. 247 (ii) Putting n = 9 in the reduction formula for f cosec" x dz, we get — Keot x cosec r+ 31, e vusec? ee a eusee x [2 1 PF ia a zie Vis * 3 Jers as _ 19-2 x2 “tl a i 1 = q" 3. *2 eh ne 1, Evaluate (i) feast xa (W.T-U,, 2007) (i) Jest? xa 2, Show that (""" tan? x dx = (660g 2) 2 f side 6 ~ Glog ih ry t1,= [° tan” x dx, prove that (n-1) I, +1,_.)=1. (V.PU,, 2009) Hence evaluate J,. (Madras, 2000) x2 4, U1, [| cot* 0d (n> 2), prove that /, 1 _}__,. Hence evaluate I, (Marathwada, 2008) 5. Obtain the reduction formula for [feet oe, (V.TL., 2010 8) 2 6. Evatuate () [sec ade (ao PF corec® do, 7, Bvaluate f° (a? + x36"? ds, ar6 tthe [Pea eve tha oo Fates de [GGG REDUCTION FORMULAE for (a) fx" e™ ax () fx" dog x)" dx. (@) Let Jeeta or Hs fn-1 Which is the required reduction formula. (Madras, 2006) (Let a? eel ala ACORN st 1. Lee Tnya® i ~ nog az mst a a ite re log 29"? dx ot Inn == "—Imynt which is the desired reduction formula.Eo a AER TGR REDUCTION FORMULAE for which is the desired reduction formula. or (a) [x* sin mx dx (6) [ x" cos mx dx (6) J cos x sin nx dx (a)let = fx" sinmxdx Integrating by parts, we get new (=) far (cae) a mm m= =o See SATB Fx cos mx dx [Again integrate by parts} a sabi mf pan, EF y_ayyen Sn m m om “—m C08 MX, pant gin my — MAM m nm m (Madras, 2003) (Let I, = J x 008 me: dx Integrating twice by parts as above, we get x" sin mx 2 Ay 1, E108 mx — “ m (Let 1,,4= J cos” x sin nx de Integrating by parts, cos nex ” cos cos nx ™ n n Tq, == 008" x. = J moos”? 2 Osi [ or cos nx sin x = sin nx cos x — sin (n —1)x =~ Leos x cos nx f cos” x (sin nx cos x — sin (n — 1x) dx Hine ee SS 5 eg, im, en ren A-iie~i which is the desired reduction formula, wrt . [- mos x (sin 2) x SB dx“cos (n — 1) x = cos nx cos.x + sin nx sin x or sin nx sin x = cos (n ~ 1) x — c0s nx cos x a "-* x [eos (n ~ 1) x ~ cos nx cos x] dx ‘Transposing and dividing by (1+ m/n). we get Ten = 2 = BL cos! x sin ne sin x de n bo 1, Wat MERAET which is the required result, marsh [xf sateen] Changing n ton 1, Putting 0 pols ie — > ee Hence S|e Solution, Let. n= fo sint ede = j S8* Integrating by parts, sexhitins, Boece ut £. 1,=sint x, —— finsin®“? veos x). de = 2 J (int! x 008 2). em dx [Again integrating by parts] not eos x-£— J in —Dsin"“? x @ x cos x. cos x + sin”~1x(~ sin x)} us| (asin x—ncosx) +4 [im ~1)sin*-? x x (1~sin?x)~sin" x] e™ de @ it a oe ‘Transposing and dividing by (1 + n*/a2), we get n(n -1) 2 (sin x—n ons) Ia et, e te i x(a sin x~n cos x) a +n which is the required reduction formula. Putting a = 1 and x =3, we get 1, ~ & sin? x(sin x—3eosx) | 3.2 a P49 r+o Ty But t= J é sinx dx =" sin tan. y= & Sin? x (sin = 10 osx) 3 oe sin (x — WA). E ».ry Hianen Enoineerins MatHemanics | PROBLEMS 6.3 1 It,» J x" ede, show that /, +n 1,_4= xe, Hence find (Madras, 2000) 2 Iu, = f x" &* de, prove that u,—(n+a)u,_,+a(n~Duy_.=0. (Madras, 2003) 8. Obtaina redvetion formula for fx” dg 2) a. Hence evaluate f 8 (log x)? dx. (SV.T.U,, 2009; Bhiltai, 2006) 1 1. If ia positive integer, show that x" y pmol 4. If nis a positive integer, show fy 2" daw re erm & Ifl,= Age shin” 1), prove that J, = "=? x12 1. Ifa, f 2" sin x dx, (x >), prove that u,+n(n—1)u,_g=n (2). Hence evaluate u,. (Madras, 2000 5) 8, t= fe sinar dz, show that af, =~ ax" cos ax +nx"-? sinax~n(n=1)I,_» (Marathwada, 2008) ora 9. Prove that J" cog"? sinnxdz = 25, n> 1 2 m mi(m-1) Ed hala pcend y 10, Ihe f de prove tht y= PD ty ax an mee 3 11, Find a reduction formula for fe cos” x dz. Hence evaluate &* cos’ x dx. 12, Obtain a reduction formola for, [° e"* sin" xdx where m > 2in the form (1 +m), = m(m—1I,._ » Hence evaluate I,. (Gorakhpur, 1999) [GEM DEFINITE INTEGRALS Property I. r £00) dx = r fo dt (ie, the value of a definite integral depends on the limits and not on the variable of integration). Let Jreras = 060; 2 ff rend = 40) ee Then froa =o; ef rood = ¥0)- Har Hence the result. Property Il. [ fi dx=- [* ft) dx (éc., the interchange of limita changes the aign of the integral). Let reas =; 2 free = #6)- a) and ~ f, pera =— | OG) [f =—[¢a)— 1 = 4b) - ga). ‘Hence the result,