REDUCTION FORMULAS Example 1 The integrand in J x"e"dx depends on x and also on the parameter n which is the exponent of x, Let J xteXdx. Integrating this by parts, with x" as the first function and e* as the second function gives us text fet de ~ feat f etanar ‘Note that the integrand in the integral on the right hand side is similar to the one we started with. The only difference is that the exponent of x is n-1, Or, we can say that the exponent of x is reduced by 1. Thus, we can write Ix=x"e*-n I, 1) ‘The formula (1) is a reduction formula. Now suppose we want to evaluate 1,, that is,J x4e* dx. Using (1) wecan write I, = x4e* - 41, x*e* ~ 4 [x?e* —31,] using (1) for 1, 1 =xte® — axe + 121, = xte® — dx3e* + 12x7e* — 241, using (1) forl, = xte® — 4x3e% + 12x7e* ~ 24 xe* + 24g, Now lp = J x°e%dx = feta =e +e, Thus, the method of successive reduction gives us [xfetac = xtet — axve® — 122" + dire” +240" 46 Reduction Formulas for Jsinnx dx and Joos xdx1° = f sin’adx = f sin’! x sin xdx, if o> 1, Taking sin’-'x as the first function and sin x as the second and integrating by parts, we get 1, # = sin” xcosx ~ (n= 1) f sin®-*xcosx (~cosx) dx = - sin"! xcosx + (n-1) J sin""?x cos? x dx 4 = -sin™! xcosx + (n-1) [ J sine? x(I-sin? x) dx + =~ sin” xeosx + (n=1) uf sin™*x dx = f sin” x) dx] = - sin"! xcosx + (n-1) [I,-2-In] I, + (n= 1) 1, = sin"! x cosx + (a - t)1,-> Thatis, nl, = -sin™! x cosx + (n-1) 1... OF ~sin"! x cosx 1, = —————— + a n 2 This is the reduction formula for J sin” xdx (valid for n> 2).Let us now derive the reduction formula for J cos"xdx. 1, = J cos" xdx = f cos*"" xcosxdx, n> L Integrating this integral by parts we get ty = [cost tx sink — J (n= 1)cost-Px (sin). sin dx = cos"! xsinx+ (n=l) J cos”? x sin? xdx = cost! xsin x +(n=1) [ cos"? x(1-c0s? x)dx = cos™! xsinx + (n-1) (Iy.1 - Ty) By rearranging the terms we get cos™' xsinx Xan% 4. Ty = f cos" xdx = =Example We will now use the reduction formula for { sin" xdx (9 evaluate the definite 2 integral, J." sin *xdx. We first observe that =sin""' xcosx n=l pr? a lo 12 J sintxdx = = BoP sin"? xdx lb n bo sin? xdx, n 2 2. fs Thus, (an xd “ie sin?xdx wee w4 2 [Snags & Co0ss) . “15 lb2 [PP sintads = PEF? sit? a, n 22 a Using this formula repreatedly we get -tne3 ne 2 nal n=} aos £2 [?sinxds, if m is an odd number, n2 3. x n=2 33 fp sintxdx = 0 3 a dx, if n is an even number, n2 2. This means ss 2 if n is odd, and n 2 3 m2 . a n-2 3 f sin"xdx = a= 43 £ E. it nis even n 22 We can reverse the order of the factors, and write this as B23 B21 it nis odd, 23 n-2 n n-l 8 if n is even, n > 22 ‘Arguing similarly for f cos" xdxwe get -if n is odd, and n 2 3 wl ale 2 sin" x dx = 22 JF cos*xdx = 0 nm. if nis even, n22 4a 4 Rl Evaluate a) [cost xdx, b) fr cost xdx, using the reduction formula a o Reduction Formulas for fran xdx and fect xdxJ tan” xdx = f tan? x tan?xdx. = J tan®-2x(sec? x -1)dx =f tan"? xsec? xdx ~ f tan®“*xdx J tan™x seo? xax = Therefore, (2) give 1, = Thus the reduction formula for [ tan" x dx is J tan” xdx = 1J seo xdx (n > 2), 1, = J sec” xdx = f[ sec"? x sec? xdx = seo"? xtanx-(n=2) f seo"? x, sec x tan? xdx sec"”? x tanx—(n—2) J sec"? x tan? xdx " sec”? xtanx—(n-2) Joon? x(sec? x=1) dx = sec™™ xtanx—(n-2) (I, —Iy-2) ‘After rearranging the terms we get ec”? x tanx =x ag nag Let’s calculate i) 5 tan’ xdx and ii) j sec®x dx1 pom 2 0 cOsSx 1 1 —-In 4 (cos x) I tein tis int 4 v2 In ¥2 1 ‘4 nid 6 ii) f see®xdx = sec* xtanx 5Derive the following reduction formulas for J cot” xdx and J coseer xdx a) [ cot” xdx = cot™ x-Iyy 1 m2 b) fcosectxdx = 1, = ose xeotx , no? . n-1 I n-1 oa Evaluate 212 a a Ji cosec? xdx » sin' xdx _c) Jee? eae Integrand of the Type sin™xcos"x‘The function sin"xcos"x depends on two parameters mand n. To find a reduction formula for J sin™ xe08” dx, letus first write Tua = J sin™ xcos"xdx Since we have two parameters here, we shall take a reduction formula ty mean a formula connecting I, and, ,, where either p 1, Now, Ine = J sin™ xcos" x dx = f cos™' x (sin™ xcosx) dxIntegrating by parts we get cos' Tian = dx, if m#-1 ™ m+1 mt gin ™! _ = Oe sin BTL cin x cost”? x (10s? x) dx m+ m+1 ot y sin? os" x. Sin’ x n-1 =e Ima-2 - 1 ar aaa Umaa ~ Ine Therefore, a-1 m+n y _ cos! xsin™!x | n=l, me om+l ™ m+1? ™" m+t m+i ™"? ‘This gives us, oly inl cos™txsin™ x n-1 Ta = Tanz 8)But, surely this formula will not work ifm +n = 0.So, what do we do if m+n=0? Actually we havea simple way out. Ifm+n=0, then since, nis positive, we writem=-n. Hence 1p, = f sin™™xcos" xdx =f cot” xdx, which is easy to evaluate using the reduc- tion formula derived in Sec. 3 (See E2)). To obtain formula (3) we had started with the assumption that n> 1. Instead of this, if we assume that m > 1, we can write lan = J sin™xcos” xdx = J sin®-'x (cos" x sinx) dx. Integrating this by parts we get in™! x03"! n+1 (0, then Iya = J sin® xcos*?*!xdx = f sin” x (1-sin? x)? cosx dx 0 J e™d-t?)?dt we putt=sinx.Expanding (1-t") by binomial theorem and integrating term by term, we get mt ms + aye pretvel Ina =< = ep) 5 Ot 2) OE m+ m+5” m+2p+l — cnt’ 5 - 1) Cc C(p,l) aaa (P.2) 5 "sin ™2P41 y m+2p+l Ifmand n are positive integers, by repeated applications of formula (3) or formula (4), we keep reducing n or m by 2 at each step. Thus, eventually, we come to an integral of the form Tyo OF Im1 OF Ty, OF Io,q. In the previous section we have seen how these can be evalu- ated, This means we should be able to evaluate I,, ina finite number of steps, We shall now look at an example to see how these formulas are used. In deriving formula (4) we had assumed that m > 1. How would you evaluate, I, if =1? Formulas (3) and (4) fail when m +n = 0, We have seen how to evaluate I, ifm+n=0. and n is a positive integer. How would you evaluate it if m + n= 0 and nis a negative integer?x2 Letusevaluate ["" sin‘ xcos* x dx. Herem=4 and n=6, Since mis the lo smaller of the two, we shall employ formula (4) which reduces mat each step. . we ; Cr y nD sin? xcos” x 3 Jf sintxc0s® xax = “SE OS® += [| sin? xcos® x dx Jo 10 n 3? sin? x cos® xdx is 7, 7? 2 = 3 Jrsinxeos’x [1 (ce cant, 10 8 8 40 using formula (4) again. oa ff c08° xax 3 Sn an = — x => (from E 2)b)) = —— 80 * 96 (Som B 2)b)) S12Evaluate a2 5g 2 8 cos? xd a) J) sin’ xeos* xdx —b) [> sin’ x cos? xdx Integrand of the Type e™ sin"x . Let us denote J e™ sin" xdx by Ly ig = a Ff em sin™ xcosxdx. a a We shall now evaluate the integral on the right hand side, again by parts, with sin™!xcosx as the first function and e* as the second one. Thus, Veen {(n-1) sin"? xcos? x — sin" x} dx e*{(n-1)sin"~? x—nsin” vas] a+e b) Ly = [ e® sinxdx = inf I+a? Prove: if C, = f e* cos" xdx, then. 5 cos” 1 cost"! xsi ae™cos"x _ ne™cos™! xsinx | n(n—l scm Clem aco am asi Icy n? +a? n? +a? n+a {asinx — cosx) +c. J sinh” xdx, [ cosh” xdx,