Techniques of Integration. Integration By Parts

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1 Techniques of Integrtion There is NO formul for Integrtion By Prts f(x)g(x)dx. ( It lmost never hppens tht f(x)g(x)dx = )( ) f (x)dx g(x)dx Notice tht df = f(x)+ C. We often shorten this to df = f to indicte tht the integrl nd differentil opertors cncel ech other. The Product Rule for derivtives, d dx (uv) = u d dx (v) + d dx (u)v hs no simple counterprt for ntiderivtives. It cn be restted in terms of differentils s d(uv) = udv + vdu, nd if we pply indefinite integrl signs, we get d(uv) = udv + vdu,oruv = udv + vdu. We usully use the equivlent formul udv = uv vdu. Exmple: Evlute x sin xdx Solution: Use integrtion by prts, with u = x, nd dv = sin xdx. Then du = dx, nd v = cos x, so ( ) x sin xdx = udv = uv vdu = x( cos x) ( cos x)dx = x cos x + cos xdx = xcos x + sin x + C

2 We cn lso use this technique with definite integrls: Evlute Solution: π 0 π 0 x cos x π 0 + π( ) = π x sin xdx ( x sin xdx = π 0 ) π udv = uv vdu = x( cos x) π 0 ( cos x)dx = 0 cos xdx = x cos x π 0 + sin x π 0 = π cos π ( 0 cos 0) + sin π sin 0 = Question: How do I know wht u nd dv should be? Students first encountering the technique of using the eqution udv = uv vdu hve trouble knowing wht to tke for u nd wht to tke for dv. Answer: Get lots of experience. This is n re where we lern lot from experience. The Integrtion by Prts technique is chrcterized by the need to select u from number of possibilities. Once u hs been chosen, dv is determined, nd we hope for the best. The bsic ide underlying Integrtion by Prts is tht we hope tht in going from udv to vdu we will end up with simpler integrl to work with. In the exmple we hve just seen, we were lucky. Let s try it gin, the unlucky wy: Exmple: Evlute (sin x)xdx Solution: Use integrtion by prts, with u = sin x, nd dv = xdx. Then du = cos xdx, nd v = x,so ( ) (sin x)xdx = udv = uv vdu = sin x x x cos xdx = sin x x cos xdx x which involves tougher looking integrl thn we strted with.

3 As rule of thumb, one-third of the possible choices will led to n esier integrl, one-third will led to hrder one, nd one-third will led to one of equl difficulty. It is lso possible to spin your wheels, nd go round in circles, s we shll soon see. Let s try strtegic pproch to our exmple: u hs to be selected so tht udv = x sin xdx, so we look t the possible choices for u: x, sin x, orx sin x. Once u is selected, we hve x sin xdx dv =, nd ll we hve to is find du(esy) nd v = dv (possibly very hrd or u impossible). Then we try to decide if we cn get nywhere with vdu. Thus, we cn let u be ny fctor of f(t), including, nd the corresponding dv is determined. (Of course, if we let u =, the problem of finding v is just our originl integrtion problem, so we will omit it.) If we cnnot then find v we know we hve non-vible selection of the pir u nd dv. We shll illustrte the rther inefficient technique of exmining ll the possibilities nd discrding the non-vible ones in the following exmples. Orgnizing our informtion in tble is helpful: u dv = v = dv du vdu vdu Better? x sin xdx x u x sin xdx = sin xdx = dx cos xdx cos xdx = x sin xdx cos x sin x YES! sin x x sin xdx = xdx = cos xdx x sin x x cos xdx NO! x sin x x xdx x sin xdx x sin x = dx = sin x x sin x dx x +x cos x +x cos x x(sin x + x cos x)dx NO! There re some simple integrls where little choice is vilble: knowing which of lrge number techniques to use is crucil. ln xdx obviously requires Exmple: u = ln x, dv = dx, so tht v = x nd du = dx x. ( ln xdx = ) udv = uv vdu = (ln x)x x dxx = x ln x dx = x ln x x + C 3

4 Exmple: rctn xdx lso obviously requires u = rctn x, dv = dx, so tht v = x nd du = dx + x ( ) rctn xdx = udv = uv vdu = (rctn x)x x rctn x ln( + x ) + C = x dx + x = x rctn x xdx + x = x rctn x ln + x + C Indefinite Integrtion of e kt times nother fctor We nowlook t fmily of integrls which showup in LSD, (Liner Systems Design), prticulrly in the clcultion of Lplce Trnsforms. We define G(G) = e kt G(t)dt, where G is ny function. The types of continuous function G tht rise in prcticl situtions re often sums nd products of polynomils, exponentil functions, nd sinusoidl functions. Fortuntely, we knowhowto evlute these using the technique of integrtion by prts. Exmples: () G(t) = c, constnt. Integrtion by prts is not needed here. Then G(c) = e kt G(t)dt = c k ekt + C () G(t) = t. Then G(t) = e kt tdt. Here f(t) = e kt t hs four different possible fctoriztions: u dv = f (t)dt v du vdu Vible? u t e kt dt k ekt dt k ekt dt Yes! e kt tdt t ke kt dt kt e kt dt No! e kt t dt t (t + k)e kt dt t(t + k)e kt dt No! We use the one vible fctoriztion, u = t, dv = e kt dt: t(e kt dt) = udv = uv vdu = t k ekt k ekt dt = t k ekt e kt + C ( ) k = t k e kt + C k 4

5 (3) G(t) = t. Then G(t ) = e kt t dt, sof(t) = e kt t. This hs four possible fctoriztions: u t t e kt dv = f (t)dt v du vdu Vible? u ( ) ( ) e kt t tdt k e kt t dt k k e kt dt Yes, but messy k e kt dt tdt Yes! k ekt k tekt dt k t dt 3 t3 ke kt dt 3 t3 e kt dt No! e kt t tdt t (t + k)e kt dt t (t + k)e kt dt No! e kt t dt t (kt + t)e kt dt t(kt + t)e kt dt No! We see tht if we use the fctoriztion u = t, dv = e kt dt, we get t e kt dt = udv = uv vdu = t k ekt k te kt dt = ( ) t k ekt t k k e kt = ( ) k t k t + e kt + C k k 3 (4) We notice howthe problem of evluting t e kt ws reduced to the evlution of e kt tdt, which hd just been done. We suspect the existence of reduction formul for e kt t n dt. Hving been successful in tking dv = e kt dt in the two preceding exmples, we decide to do this gin, nd we hve u = ekt t n dt = t n. We clculte v = e kt dt k ekt nd du = nt n dt, so tht: e kt t n dt = udv = uv vdu = k tn e kt n e kt t n dt. k Thus we hve G(t n ) = k tn e kt n k G(tn ). (5) G(t) = sin t. We hve f(t) = e kt sin t, so there re just three choices for u: u dv = f (t)dt v du vdu Vible? u sin t e kt dt k ekt cos tdt k ekt cos tdt Yes e kt sin tdt cos t kekt dt k ekt cos tdt Yes e kt sin t dt t (k sin t + cos t)e kt dt t(ksin t + cos t)e kt dt No! The first two choices re both vible, nd we see tht they both led to the evlution of e kt cos tdt. We will exmine both cses closely: First Choice: u = sin t, dv = e kt dt We get: G(sin t) = e kt sin tdt = u dv = u v v du = sin t k ekt k ekt cos tdt = sin t k ekt k e kt cos tdt = sin t k ekt G(cos t) k or G(sin t) = sin t k ekt G(cos t) k 5

6 In evluting e kt cos tdt we gin hve three choices, two of which re vible: u dv = f (t)dt v du vdu Vible? u cos t e kt dt k ekt sin tdt k ekt sin tdt Yes e kt cos tdt sin t kekt k dt ekt sin tdt Yes e kt cos t dt t (k cos t sin t)e kt dt t(ksin t + cos t)e kt dt No! Agin we hve two choices. First nd Best Choice: We will let U = cos t, dv = e kt dt, nd we get: G(cos t) = cos te kt dt = U dv = U V V du = cos t k ekt k ekt ( sin tdt) = k cos tekt + k e kt sin tdt = k cos tekt + G(sin t), k or G(cos t) = k cos tekt + G(sin t) k so we re bck where we strted! However, if we substitute this into the eqution G(sin t) = sin t k ekt G(cos t) k we get G(sin t) = sin t k ekt ( k k cos tekt + ) G(sin t) = sin t k k ekt k cos tekt G(sin t) k which cn be solved for G(sin t): ( + k )G(sin t) = + k G(sin t) = sin t k k ekt k cos tekt = e kt (sin t k k ekt cos t) = (k sin t cos t) so k k G(sin t) = k sin t cos t e kt + k Lst nd Worst Choice: We will now let U = e kt, dv = cos t, nd we get: G(cos t) = e kt ( sin t) sin t(ekt kdt) = ekt sin t k This time, however, if we substitute this into the eqution e kt sin tdt = ekt sin t k G(sin t) we get G(sin t) = sin t k ekt k so there is no informtion gined. G(sin t) = sin t k ekt G(cos t) k ( ekt sin t k ) G(sin t) =G(sin t) 6

7 Second Choice: u = e kt, dv = sin tdt We get: G(sin t) = e kt sin tdt = u dv = u v v du = e kt ( cos t) k ekt cos tdt = ekt cos t k G(cos t) or G(sin t) = ekt cos t k G(cos t) Using, from bove, G(cos t) = k cos tekt + G(sin t) k we get G(sin t) = ekt cos t k ( k cos tekt + ) G(sin t) =G(sin t) k so there is no newinformtion. On the other hnd, if we use we get the sme vlue s before. G(cos t) = ekt sin t k G(sin t) (6)G(cos t). This is esily derived from the clcultions of the previous exmple. G(cos t) = k cos tekt + k G(sin t) = k cos tekt + k so G(cos t) = ( k sin t cos t sin t + k cos t e kt + k + k ) e kt = sin t + k cos t e kt + k (7)G(t n sin t) = t n sin te kt dt Let u = t n sin t, dv = e kt dt, so tht du = (nt n sin t + t n cos t)dt, nd v = k ekt. Then G(t n sin t) = t n sin te kt dt = udv = uv vdu = t n sin t( k ekt ) k ekt ((nt n sin t + t n cos t)dt) = k tn sin te kt n k G(tn sin t) k G(tn cos t) or G(t n sin t) = k tn sin te kt n k G(tn sin t) k G(tn cos t) 7

8 (8)G(t n cos t) = t n cos te kt dt. Let U = t n cos t nd dv = e kt dt, so tht du = (nt n cos t t n sin t)dt nd V = k ekt. Then G(t n cos t) = t n cos te kt dt = UdV = UV Vdu = t n cos t k ekt e kt ((nt n cos t t n sin t)dt) = k tn cos te kt n k G(tn cos t) + k G(tn sin t). We cn nowsolve for G(t n sin t): G(t n sin t) = nd hence G(t n cos t) = [ ] (k sin t cos t)t n e kt n(g(t n (k sin t cos t))) + k [ ] (k cos t sin t)t n e kt + n(g(t n (k cos t sin t))) + k 8

9 Trigonometric Integrls It is often necessry to evlute integrls of the form (sin x) m (cos x) n dx, where m nd n re integers. If one of the exponents, either m or n is nd odd, there is strightforwrd simplifiction. Cse : m = M + is odd. Then (sin x) m = (sin x) M+ = (sin x) M sin x = ( cos x) M sin x, so our integrl becomes (sin x) m (cos x) n dx = ( cos x) M sin x(cos x) n dx nd we my mke the substitution u = cos x with du = sin xdx to get (sin x) m (cos x) n dx = ( cos x) M sin x(cos x) n dx = ( u ) M u n du Exmple: (sin x) (cos x) 0 dx = ( u ) 5 u 0 du = [ 5u + 0(u ) 0(u ) 3 + 5(u ) 4 (u ) 5] u 0 du = u 0 5u + 0u 4 0u 6 + 5u 8 u 0 du = ( ) u 5u u5 5 0u u9 9 u + C = cos x + 5 cos3 x 3 cos5 x cos7 x 7 5 cos 9x 9 + cos x + C 9

10 Exmple: (sin x) (cos x) 0 dx = ( u ) 5 u 0 du = [ 5u + 0(u ) 0(u ) 3 + 5(u ) 4 (u ) 5] u 0 du = u 0 5u 8 + 0u 6 0u 4 + 5u u 0 du = ( ) u 9 9 5u u 5 5 0u u u + C = ( cos 9 x + 5 cos7 x cos 5 x + 0 ) cos 3 x 5 cos x cos x + C = sec9 x 5 7 sec7 x + sec 5 x 0 3 sec3 x + 5secx + cos x + C Exmple: (sin x) (cos x) 0 dx = (sin x) (cos x) 0 sin xdx = ( u ) 6 u 0 du cn be done, but requires the method of Prtil Frctions, which we shll see lter. The sitution is similr when the power of cos x is odd. (sin x) 0 (cos x) dx = (sin x) 0 (cos x) 0 cos xdx = Exmple: (sin x) 0 (cos x) 5 cos xdx = (sin x) 0 ( sin x) 5 cos xdx = ( letting u = sin x nd du = cos xdx) [ u 0 ( u ) 5 du = u 0 5u + 0(u ) 0(u ) 3 + 5(u ) 4 (u ) 5] du = u 0 5u + 0u 4 0u 6 + 5u 8 u 0 du = u 5u u5 5 0u u9 9 u + C = sin x 5 3 sin3 x + 5 sin5 x 0 7 sin7 x sin9 sin x + C 0

11 When both powers re odd, it is esiest to select the function with the highest power for substitution: Good Exmple: (sin x) (cos x) 3 dx = (sin x) (cos x) cos xdx = (sin x) ( sin x)cos xdx = ( letting u = sin x nd du = cos xdx) u ( u )du = u u 3 du = u u4 4 + C = sin x 4 sin3 x + C Bd Exmple: (sin x) (cos x) 3 dx = (sin x) 0 (cos x) 3 sin xdx = ( cos x) 5 (cos x) 3 sin xdx = ( letting u = cos x nd du = sin xdx) ( u ) 5 u 3 ( du) = [ + 5u 0(u ) + 0(u ) 3 5(u ) 4 + (u ) 5] u 3 du = [ u 3 + 5u 5 0u 7 + 0u 9 5u + u 3] du = u u6 6 0u u0 0 5u + u4 4 + C = 4 cos4 x cos6 x 5 4 cos8 x + cos 0 x 5 cos x + 4 cos4 x + C

12 When neither power is odd, we need to use double ngle formul from trigonometry: cos x cos x = sin x cn be solved for cos x nd sin x: cos x = + cos x Thus, if we wnt to integrte m = M nd n = N nd we hve: nd sin cos x x = (sin x) m (cos x) n dx, where m nd n re even integers, we write (sin x) m (cos x) n dx = (sin x) M (cos x) N dx = (sin x) M (cos x) N dx = ( ) cos x M ( ) + cos x N dx = (M+N) ( cos x) M ( + cos x) N dx

13 Exmple: + cos x cos xdx = dx = + cos xdx = (x + ) sin x + C = x + 4 sin x + C = x + x + sin x cos x sin x cos x + C = + C 4 Exmple: cos x sin xdx = dx = cos xdx = (x ) sin x + C = x 4 sin x + C = x x sin x cos x sin x cos x + C = + C 4 It should come s no surprise tht ( cos x + sin x) dx = x + C Exmple: ( )( cos x + cos x sin x cos xdx = + cos 4x dx = cos 4xdx = 8 cos xdx = (x 4 sin 4x ) + C = x 8 sin 4x 3 + C ) dx = Products of powers of secnt nd tngent functions sec m x tn n x cn be expressed s product of powers of sin x nd cos x, but it is often more convenient to use the identity tn x + = sec x nd the differentils d(tn x) = sec xdx nd d(secx) = secx tn xdx. We ssume tht m nd n re non-negtive. Cse : The power m of secx is positive nd even: then m = M nd we hve: sec m x tn n xdx = (sec x) M tn n xdx = (sec x) M tn n xsec xdx = (+tn x) M tn n xsec xd nd we cn mke the substitution u = tn x, du = sec xdx, nd we get ( ) M ( + tn x tn n xsec xdx = + u ) M u n du 3

14 Exmple: ( sec 0 x tn 5 xdx = + u ) 4 u 5 du = ( + 4u + 6u 4 + 4u 6 + u 8) u 5 du = u 5 + 4u 7 + 6u 9 + 4u + u 3 du = u u u u + u4 4 + C = 6 u6 + u u0 + 3 u + 4 u4 + C = 6 tn6 x + tn8 x tn0 x + 3 tn x + 4 tn4 x + C If the power of secx is not even, but the power of n of tn x is odd, so tht n = N +, nd m is positive, we cn mke the substitution u = secx, du = secx tn xdx, nd get sec m x tn n xdx = sec m x tn n xsecx tn xdx = ( ) ( N N u m tn x du = u m sec x ) du = ( N u m u ) du Exmple: sec 5 x tn 5 xdx = u 4 (u ) du = u 4 (u 4 u + )du = u 8 u 6 + u 4 du u9 9 u7 7 + u5 5 + C = 9 sec9 x 7 sec7 x + 5 sec5 x + C We re left with the cses where m is odd nd n is even, ll of which cn be reduced to the problem of finding the ntiderivtive of n odd power of secx. Exmple: secxdx = secx ( ) secx + tn x dx = secx + tn x ( ) sec x + secx tn x (sec x + secx tn x)dx dx = = secx + tn x secx + tn x d(secx + tn x) = ln(secx + tn x) + C secx + tn x 4

15 Exmple: I= sec 3 xdx Using Integrtion by Prts, with u = secx, dv = sec x, v = tn x, du = secx tn xdx, weget ( I= sec 3 xdx = ) udv = uv vdu = secx tn x tn xsecx tn xdx = secx tn x tn xsecxdx = secx tn x (sec x )secxdx = secx tn x sec 3 xdx + secxdx = secx tn x I+ secxdx = secx tn x + ln secx + tn x +C so I =secx tn x + ln secx + tn x +C nd sec 3 xdx = secx tn x + ln secx + tn x + C Exmple: tn xsecxdx ppers in the previous clcultion, nd is one of the simpler cses left: I=secx tn x tn xsecxdx gives us tn xsecxdx = secx tn x I = secx tn x secx tn x + ln secx + tn x secx tn x ln secx + tn x + C + C = 5

16 Integrls of products of sine nd cosine functions with different rguments The identities: cos x cos y = ( ) cos(x + y) + cos(x y) sin x sin y = ( ) cos(x y) cos(x + y) sin x cos y = ( ) sin(x y) + sin(x + y) my be used: Exmple: cos 5t cos 7tdt = (cos(5t + 7t) + cos(5t 7t)) dt = (cos t + cos( t)) dt = ( sin t + ) sin t + C = 4 sin t + sin t + C 4 6

17 Trigonometric Substitutions It is often necessry to evlute integrls contining expressions of the form x, + x, or x. For exmple, we my wish to evlute x dx or + x dx or x dx To do this, we need to mke the pproprite substitution: Expression x = dx = Expression = x sin θ cos θdθ cos θ + x tn θ sec θdθ sec θ x secθ secθ tn θdθ tn θ Exmple: x dx Letting x = sin θ, we hve dx = cos θdθ, so tht ( ) x dx = sin θ cos θdθ = ) ( sin θ cos θdθ = cos θdθ = + cos θ dθ = dθ + cos θdθ = θ + θ + rcsin x + x sin θ + C = θ + sin θ cos θ + C = 4 x sin θ cos θ + C = rcsin x + x + C ( ) x + C = 7

18 Let us use this to compute some res contined in circles of rdius : First, the re under y = x from to is tht of semicircle of rdius : ( ( x dx = rcsin x + x x ) rcsin + ) ( ( ) ( ) rcsin() + 0 rcsin( ) + 0 rcsin + = = ( ) ) = ( π π ) = π, s expected. Second, the re under y = x from 0 to is tht of qurter circle of rdius : ( x dx = 0 rcsin x + x ( rcsin + ) ( ( ) ( ) rcsin() + 0 rcsin x ) rcsin = 0 = 0 ) = π = π 4 s expected. Third, the re under y = x from x = to x = 3 is: 3 ( x dx = rcsin x + x rcsin ( 3 ) ) 3 x rcsin = + 3 rcsin rcsin = 4 [( π 3 + ) 3 8 ( π 6 + )] ( 3 π = 8 6 π ) = π ( ) = 8

19 In generl the re under y = x from x = c to x = d (c d) is: d c ( ( x dx = rcsin x + x ) d x = c rcsin d + d ) ( d rcsin c + c ) c = ( ) ( rcsin d rcsin c + d d c ) c Thus the re of the circle x + y = lying between the lines x = c nd x = d is ( rcsin d rcsin c ) ( ) + d d c c Setting c = 0, we get formul for the re A d between the y-xis nd the line x = d: A d = rcsin d + d d Exmple: + x dx We let x = tn θ, so tht + x = + tn θ = ( + tn θ) = sec θ, nd dx = sec θdθ, so we hve + x dx = ( sec θ) sec θdθ = sec 3 secθ tn θ + ln secθ + tn θ θdθ = + C Nowtn θ = x, nd secθ = x +, so our integrl in terms of x is: x + x + ln x + + x + C = x x + + ln x + + x + C = x ( ) x + + ln x + + x + C 9

20 We hve used the previously derived formul sec 3 xdx = secx tn x + ln secx + tn x + C Exmple: x dx Letting x = secθ, we hve x = sec θ = (sec θ ) = tn θ (so secθ = x nd tn θ = x ), nd dx = secθ tn θdθ,so x dx = tn θsecθ tn θdθ= secθ tn θdθ= secθ tn θ ln secθ + tn θ + C = x x ln x + x + C = x x ln x+ x + C = x x x ln + x + C We hve used the previously derived formul tn xsecxdx = secx tn x ln secx + tn x + C 0

21 It is often necessry to complete squres before using trig substitution: dx Exmple: x + 4x + 5 = dx x + 4x = dx (x + ) + = du u + where we hve let u = x +. Nowwe let u = tn θ nd hve u + = sec θ nd du = sec θdθ, so we hve du sec u + = θdθ sec θ = dθ = θ + C = rctn u + C = rctn x + + C Exmple:#8, p.454 of Brown Stewrt: e t 9dt Note tht e t 9 = ( e t) 3,soifweletx = e t,weget e t 9 = ( e t) 3 = x = 3 nd we cn use the substitution x = 3secθ. We must be very creful with the differentils: we hve dx = e t dt,sodt = e t dx = e t dx = dx x Thus we hve e t 9dt = x 3 dx x Nowwe use x = 3secθ, dx = 3secθ tn θdθ, x 3 = 3 tn θ (so secθ = x 3 nd tn θ = x 3 )toget 3 x 3 dx ( ) 3secθ tn θdθ x = 3 tn θ = 3 tn θdθ = 3 (sec θ )dθ = 3(tn θ ( 3secθ x 3 θ) + C = 3 rcsec x ) + C = 3 3 x 3 3rcsec x ( ) e t 3 + C = e t 9 3rcsec + C 3

22 Prtil Frctions This is technique used to ntidifferentite proper rtionl functions: quotients of polynomils whose denomintor hs degree greter thn the numertor. Using long division, ny rtionl function cn be written s the sum of polynomil nd proper rtionl function. Theorem: Any polynomil with rel coefficients cn be written s the product of powers of liner nd qudrtic fctors. We will usully only look t prefctored exmples, or those where the fctoriztion is esy. Exmple: u du = (u )(u + ) du The bsic ide is tht we wish to write A u + B u +, (u )(u + ) where A nd B re to be determined so tht the eqution (u )(u + ) = A u + B u + is true for ll vlues of u except nd. s the sum of the form The ordinry wy of finding A nd B is to multiply by u nd to equte the coefficients of the resulting polynomils: = A(u + ) + B(u ) = (A + B)u + (A B) gives us two equtions in two unknowns: A + B = 0 nd A B = which when solved give us A = nd B =

23 A quicker wy is to substitute the illegl vlues u = nd u = into the eqution = A(u + ) + B(u ) u = : = A( + ) + B( ) = B so B = u = : = A( + ) + B( ) = A so A = We then hve (u )(u + ) = u u +,so (u )(u + ) du = u du u + du = ln u ln u + +C = ln u u + + C Definition: A prtil frction is n expression of the form A (cx + d) m or Bx ++C (x + bx + c) n. At this point we (theoreticlly) hve lerned how to ntidifferentite prtil frctions, nd the next skill to lern is howto rewrite proper rtionl function P(x) s the sum of prtil Q(x) frctions. First Cse: Q(x) = (c x +d )(c x +d ) (c k x +d k ) is the product of distinct liner fctors. Then we cn write P(x) Q(x) = P(x) (c x + d )(c x + d ) ( k x + d k ) = A c x + d + Multiplying both sides of the eqution by Q(x), we get: A c x + d + + A k c k x + d k P(x) = A (c x + d ) (c k x + d k ) + A (c x + d )(c 3 x + d 3 ) (c k x + d k ) + + A k (c x + d ) (c k x + d k ) which is esily solved by substituting the vlues x i = d i c i, i =,, 3,...,k. 3

24 Exmple, p456 of Brown Stewrt: x + x x 3 + 3x x dx x + x x 3 + 3x x = x + x x(x + )(x ) = A x + A x + + A 3 leds to x x + x = A (x + )(x ) + A x(x ) + A 3 x(x + ) nd we then substitute x = 0,, nd : x = 0: 0 + (0) = = A (0 + )((0) ) + A (0)((0) ) + A 3 (0)((0) + ) = A,soA = x = : ( ) + ( ) = = A ( + )(( ) ) + A ( )(( ) ) + A 3 ( )(( ) + ) = 0A,soA = 0 x = : ( ) + ( ) = 4 = A ( + )(( ) ) + A ( )(( ) ) + A 3( )(( ) + ) = 5 4 A 3,soA 3 = 5 Thus x + x x 3 + 3x x = nd therefore 0 x + x x x + x x 3 + 3x x dx = x + 0 x dx = x x dx 0 x + dx + 5 x dx = ln x ln x + + ln x +C 0 0 4

25 Exmple: du = u secxdx = u = ln sin x + ln sin x + C = (sin sin x + x + ln sin x + C = ln sin x ln (sin x+) + C = ln (sin x+) + C = ln sin x cos x cos x cos x dx = cos x dx = u u + + C = ln )( ) sin x + + C = sin x + + C = sin x+ cos x ln sin x cos x + cos x + C = ln tn x + secx +C cos xdx sin x = u + u + C = Second Cse: Q(x) = (c x + d ) m (c x + d ) m (c k x + d x ) m k We then hve to write P(x) Q(x) = A,m (c x + d ) m + A,m (c x + d ) m + + A, c x + d + A,m (c x + d ) m + A,m (c x + d ) m + + A, c x + d +. A k,mk (c k x + d k ) m k + A k,mk (c k x + d k ) m k + + A k, c k x + d k 5

26 Exmple: ( x ) dx = We write (x ) (x + ) dx (x ) (x + ) = A, (x ) + A, x + A (x + ) + A, x +, nd multiply by (x ) (x + ) to get = A, (x + ) + A, (x )(x + ) + A, (x ) + A, (x ) (x + ) Substituting x = nd x = gets us two constnts quickly: x = : = A, ( +) +A, ( )( +) +A, ( ) +A, ( ) ( +) = 4A, so A, = 4 x = : = A, ( + ) + A, ( )( + ) + A, ( ) + A, ( ) ( + ) = 4A, so A, = 4 The quickest wy to get the two remining coefficients is to insert the coefficients just obtined into the originl eqution: = A, (x + ) + A, (x )(x + ) + A, (x ) + A, (x ) (x + ) becomes = 4 (x + ) + A, (x )(x + ) + 4 (x ) + A, (x ) (x + ) or [ ] = (x + ) 4 + A,(x ) [ ] + (x ) 4 + A,(x + ) Differentiting, we get [ ] 0 = (x + ) 4 + A,(x ) [ ] + (x + ) A, + (x ) 4 + A,(x + ) + (x ) A, 6

27 Agin we substitute the illegl vlues of x: x = : [ ] [ ] 0 = ( +) 4 + A,( ) +( +) A, +( ) 4 + A,( + ) +( ) A, [ ] simplifies to 0 = 4 + 4A, = + 4A,,soA, = 4 4 x = : [ ] [ ] 0 = ( + ) 4 + A,( ) + ( + ) A, + ( ) 4 + A,( + ) + ( ) A, [ ] simplifies to 0 = 4 + 4A, = + 4A,,soA, = 4 4 Thus we hve (x ) (x + ) = 4 (x ) + 4 x + 4 (x + ) + 4 x + = [(x ) (x ) + (x + ) + (x + ) ] 4 Therefore: ( x ) dx = [ (x ) [ x [ (x ) (x ) + (x + ) + (x + ) ] dx = 4 ] (x + ) + ln x + + ln x + + C = + x + + ln x + ] x x + C = ( x ) + 4 ln x + x + C 7

28 Exmple: I= sec 3 xdx = cos x cos 3 x dx = cos 4 x dx = cos x (cos x) dx = cos x ( sin x) dx = ( u ) du = u ( u ) + 4 ln u + u + C = sin x ( sin x) + 4 ln sin x + sin x + C = sin x cos x + 4 ln sin x + sin x + C where we hve mde the substitution u = sin x. By trigonometric mnipultion, this cn be worked into the somewht useless formul we hve lredy derived using complicted Integrtion by Prts clcultion: sec 3 xdx = secx tn x + ln secx + tn x + C Exmple: (sin x) (cos x) 0 dx = (sin x) (cos x) 0 sin xdx = ( u ) 6 u 0 du = We write (u ) 6 (u + ) 6 u = 0 du, (where u = sin x) (u ) 6 (u + ) 6 u0 A,6 (u ) 6 + A,5 (u ) 5 + A,4 (u ) 4 + A,3 (u ) 3 + A, (u ) + A, u + A,6 (u + ) 6 + A,5 (u + ) 5 + A,4 (u + ) 4 + A,3 (u + ) 3 + A, (u + ) + A, u + + A 3,0 u + A 3,9 0 u + A 3,8 9 u + A 3,7 8 u + A 3,6 7 u + A 3,5 6 u + A 3,4 5 u + A 3,3 4 u + A 3, 3 u + A 3, u 8

29 Third Cse: Q(x) = ( x + b x + c )( x + b x + c ) ( k x + b k x + c k ) is the product of distinct irreducible qudrtic fctors. Then we cn write P(x) Q(x) = P(x) ( x + b x + c )( x + b x + c ) ( k x + b k x + c k ) = A x + B A x + B A k x + B k x + b x + c x + b x + c k x + b k x + c k Exmple: (x + )(x + 4) dx = A x + B x + dx + A x + B x + 4 dx hs to be solved for A, B, A, nd B. Multiplying by the common denomintor (x +)(x +4), we get = (A x + B )(x + 4) + (A x + B )(x + ) There re two wys of solving this for the desired constnts. One is to use the fct tht the coefficients of the two polynomils must be equl so s to get 4 equtions in 4 unknowns: = 0x 3 + 0x + 0x + = (A x + B )(x + 4) + (A x + B )(x + ) = A x 3 + B x + 4A x + 4B + A x 3 + B x + A x + B = (A + A )x 3 + (B + B )x + (4A + A )x + (4B + B ) gives us A + A = 0, B + B = 0, 4A + A = 0, 4B + B = Using B = B, we get 3B =, so B = 3 nd B = 3. Usin A = A, we get 3A = 0, so A = A = 0. Thus we hve (x + )(x + 4) dx = 3 x + dx 3 x + 4 dx = 3 rctn x 3 rctn x + C = 3 rctn x 6 rctn x + C 9

30 An lterntive methof of solving = (A x + B )(x + 4) + (A x + B )(x + ) for the desired constnts is to substitute rtfully selected vlues of x. In this cse we shll use complex numbers: x=i: Then we hve = (A i + B )(i + 4) + (A i + B )(i + ) = (A i + B )( + 4) + (A i + B )( + ) = 3(A i + B ) or + 0i = 3B + 3A i, which immeditely gives us A = 0 nd B = 3 x=i: Then we hve = (A i + B )((i) + 4) + (A i + B )((i) + ) = (A i + B )( 4 + 4) + (A i + B )( 4 + ) = 3(A + B ) or + 0i = 3B 6A i, which immeditely gives us A = 0 nd B = 3 Fourth Cse: Q(x) = ( x + b x + c ) n ( x + b x + c ) n ( k x + b k x + c k ) n k is the product of not necessrily distinct irreducible qudrtic fctors. Then we cn write P(x) ( x + b x + c ) n ( x + b x + c ) n (k x + b k x + c k ) n k = A,n x + B,n ( x + b x + c ) + A,n x + B,n n ( x + b x + c ) + + A, x + B, n ( x + b x + c ) + A,x + B, + x + b x + c A,n x + B,n ( x + b x + c ) + A,n x + B,n n ( x + b x + c ) + + A, x + B, n ( x + b x + c ) + A,x + B, + x + b x + c +. + Ak,nk x + B k,nk ( k x + b k x + c k ) n k + A k,n k x + B k,nk ( k x + b k x + c k ) n k + + A k, x + B k, ( k x + b k x + c k ) + A k,x + B k, k x + b k x + c k 30

31 Exmple: (x + ) (x + 4) dx We write (x + ) (x + 4) = A, x + B, + A,x + B, + A,x + B, + A,x + B, (x + ) x + (x + 4) x + 4 nd multiply by the common denomintor to get = (A, x + B, )(x + 4) + (A, x + B, )(x + )(x + 4) + (A, x + B, )(x + ) + (A, x + B, )(x + ) (x + 4) which will result in 8 equtions in 8 unknown constnts. Most resonble people would hve this computtion done by CAS: Computer Algebr System. 3

32 Improper Integrls So fr, definite integrls hve been used to compute res of finite regions. It is possible to extend the notion of re to regions lying under the grph of function over infinite intervls, nd to functions which hve verticl symptotes. Definition: Improper Integrls of Type I re defined to be those of the form: T f (x)dx = lim f (x)dx T f (x)dx = lim T T f (x)dx f (x)dx = f (x)dx + f (x)dx, if both limits exist. Exmple: lim T T = T x dx = lim x dx = lim T T x T = lim x T T ( = lim T T ) = 0 = Exmple: T x dx = lim x dx = lim ln x T = lim ln T ln = T T T Exmple: x dx = lim T x dx = lim ln x T = lim ln ln T = T T If the limit defining n improper integrl exists, we sy tht the integrl converges or is convergent, otherwise we sy it diverges or is divergent. Exmple: If s is considered to be constnt, 0 T e sx dx = lim e sx dx = lim T 0 T s e sx T 0 = lim T s e s(t) s e s(0) = s lim T s e st = s 3

33 Definition: If f is defined on the intervl 0, ), nd treting s s constnt in integrting, we cn define function of F(s) of s by letting F(s) = 0 T e sx f (x)dx = lim f(x)e sx dx T 0 This function is clled the Lplce Trnsform of f. In the previous exmple we showed tht the Lplce trnsform of the constnt function is s. Most of the integrtion techniques lerned in this course will be used fr more often in prcticl pplictions of Clculus vi the Lplce trnsform thn in the computtion of res. Exmple:(closely relted to Exmple, p.489 of brown Stewrt) T xe sx dx = lim xe sx dx 0 T 0 We integrte xe sx dx by prts with u = x, dv = e sx dx, so tht du = dx nd v = s e sx : xe sx dx = udv = uv vdu = x ( s ) e sx s e sx dx = x s e sx + s e sx dx = x s e sx s e sx so T lim T 0 lim T xe sx dx = lim ( xs e sx s ) T T e sx = 0 ( 0s e sx s ) e s(0) ( Ts e st s e st ) = s 33

34 Note: Exmples 3&4 of brown Stewrt 7.9 re very importnt: dx + x = lim T 0 T dx + x + lim T T 0 dx + x = lim T rctn x 0 T + lim T rctn x T 0 = lim (rctn 0 rctn T ) + lim (rctn T rctn 0) = T T 0 lim rctn T + lim rctn T 0 = T T ( π π ) = π dx x p T = lim x p dx = lim T T x p+ p + T = lim T T p+ p + p+ p + = p + lim T p T p = p if nd only if p> The other type of improper integrl occurs when the integrnd f(x) hs verticl symptote t sy x = c. Then, ssuming <c<b, we define c f (x)dx = lim T c T f (x)dx b c b f (x)dx = lim f (x)dx T c + T nd b f (x)dx = c b f (x)dx + f (x)dx (if both integrls exist). c 34

35 Comprison Theorems: If we hve f(x) g(x) on the (possibly infinite) intervl (, b), then we know tht b f (x)dx b g(x)dx. Thus we cn conclude for proper integrls tht if b f (x)dx diverges, then so must b g(x)dx, nd conversely, if b g(x)dx converges, then so must b f (x)dx. { Exmple: The region R= (x, y) x, 0 y } is revolved bout the x-xis to form x solid object, known s Gbriel s Horn. It volume is π T V = dx = π lim x x T T dx = π lim = π lim x T T T x = π lim T T = π However, s we shll soon see, the formul for its surfce re is S = π + x x dx π 4 x =!!! Thus we now hve pttern for pint bucket whose volume is finite, but which cnnot be pinted becuse its surfce re is infinite! 35

36 Approximte Integrtion We hve lredy seen howriemnn sums re pproximtions to definite integrls. Recll tht the regulr prtition P of [, b] with n intervls is formed by letting x = b n, x j = + j x, so tht x 0 =,,x = + x,,x n = b x, x n = b. We hve P = x. The formuls for the specil Riemnn sums re: () The Left-hnd sum of f : L n (f ) = n f(x i ) x = [f(x 0 ) + f(x ) + +f(x n ] x = i= n ( f + (i ) b ) b n n i= () The Right-hnd sum of f : R n (f ) = n f(x i ) x = [f(x ) + f(x ) + +f(x n )] x = i= n ( f + i b ) b n n i= (3) The Midpoint sum : n ( xi + f(x i ) M n (f ) = f i= [ ( ) ( x0 + x x + x f + f n f i= ( + ( i ) x = ) + +f ) ) b b n n ( xn + x n )] x = 36

37 (4) The Inscribed sum of f : I n (f ) = [m + m + +m n ] x (5) The Exscribed sum of f : E n (f ) = [M + M + +M n ] x The Trpezoidl Estimte or Trpezoidl Rule T n (f ) = (f (x 0) + f(x ) + f(x ) + + f(x n ) + f(x n )) x = n ( f () + f + i b ) + f(b) b n n i= which is the sum of the res of the trpezoids pssing (x i, 0)(x i,f(x i ))(x i,f(x i ))(x i, 0) turns out to be the verge of the left nd right hnd sums: n n L n (f ) +R n (f ) = f(x i ) x + f(x i ) x = i= n ( f () + f + i b ) + f(b) b n n =T n(f ) i= i= We would like to knowhowclose these sums re to the ctul vlue of b b pproximtion to f (x)dx, we define E(A) = f (x)dx A) So fr, we only know the following theorem: b f (x)dx. IfA is n Theorem If f hs continuous derivtive f on [, b], nd f (x) <Mfor ll x in [, b], then E n (f ) I n (f ) M (b ) n,so lim n E n(f ) I n (f ) = 0 This ws used to prove prt of the Fundmentl Theorem of Clculus, but is not prticulry useful becuse of the difficulty of finding the Incribed nd Exscribed Sums. We cn mke some n use of it by noting tht for ny Riemnn sum R n (f ) = f(t i ) x = b n f(t i ) over n regulr prtition we hve I n (f ) E n (f ), so the theorem tells us tht E(R n (f )) M i= i= (b ). n In terms of the exmples just preceding the Substitution Method, we hve, letting y = f(x) = x, = 0, b =, (so tht b f (x)dx = 3 ), 37

38 L n (f ) = 3 n + 6n,soE(L n(f )) = n 6n R n (f ) = 3 + n + 6n,soE(R n(f )) = n + 6n T n (f ) = L n(f ) +R n (f ) = 3 + 6n,soE(T n(f )) = 6n M n (f ) = 3 n,soe(m n(f )) = n We stte without proof some useful fcts: If f exists nd is continuous on [, b], nd if M = mx { f (x) : x b }, then E(T n (f )) M (b ) 3 (b ) 3 nd E(M n n (f )) M 4n In our exmple, we hve f (x) =, so we tke M = nd get ( 0)3 E(T n (f )) = n 6n nd E(M (b ) 3 ( 0)3 n(f )) M = = n 4n n Exmple: Use the Trpezoidl nd Midpoint Rules to estimte 0.00 of the true vlue x dx to within Solution: We hve f(x) = (4 x ), f (x) = f (x) = so M = mx { ( 6 x ) ( 6 x ) () (x) 6 (6 x ) 3 x ( x) =, (6 x ) ( ) 6 x ( x) = 6 x + x 6 = 6 x (6 x ) 3 (6 x ) 3 } { } :0 x 3 = 6 mx :0 x 3 = 6 (6 x ) Thus E(T n (f )) 6 7 (3 0) 3 = 36 7 n 7 7n < 0.00 if n > or n> We tke n = 45 nd get L 45 (f ) = , R 45 (f ) = , so T 45 (f ) = = For the midpoint rule, we tke n = 3 nd get M 3 (f ) = The ctul vlue (to 5 deciml plces) is

39 Simpson s Rule A more sophisticted pproch is mke n be even nd to pproximte the grph of y = f(x) on the intervls [x i,x i+ ] with the prbols pssing throught the points (x i,f(x i )), (x i+,f(x i+ ), nd (x i+,f(x i+ ). The result is the pproximtion S n (f ) = [f(x 0 ) + 4f(x ) + f(x ) + 4f(x 3 ) + +f(x n ) + 4f(x n ) + f(x n )] b 3n If f exists nd is continuous on [, b], nd if M 4 = mx { f (x) : x b }, E(S n (f )) M 4 (b ) 5 80n 4 Simpson s Rule cn be expressed in terms of the Trpezoidl nd Midpoint Rules: S n (f ) = T n(f ) + M n (f ) 3 In our first exmple, we hve f (x) = 0, so we tke M 4 = 0 nd get E(S n (f )) M 4 (b ) 5 80n 4 = 0 Using S n (f ) = T n(f ) + M n (f ),wegets n (f ) = ( 6n 3 ) n = 3 3. In our second exmple, we hve, using f (x) = 6(6 x ) 3, ( f (x) = 6 3 ) (6 x ) 5 x ( x) = 48 (6 x ) 5 f (x) = 48 (6 x ) 5 () x 5 (6 x ) 3 ( x) = 48 6 x + 5x = x (6 x ) 5 (6 x ) 7 (6 x ) 7 { Thus M 4 = mx } { } x (6 x ) 7 :0 x x = 9 mx (6 x ) 7 :0 x = (3 0) 5 Thus E(S n (f )) M 4 80n = M n < n =.37 4 n 4 < 0.00 if n 4 > 37 or n>

40 We tke n = 6 nd get: L 6 (f ) =.0685, R 6 (f ) = , so T 6 (f ) = Also, M 6 (f ) = , so S 6 (f ) = T 6(f ) + S 6 (f ) = = ( ) 3 = = Is it worth ny extr effort to get better estimte of M 4? The nswer is tht it depends on difficult the expression for f (x) is. We obtined good guess for the mximum vlue of f (x) by substituting x = 3 into it. It is nowesy to sketch the grphs of functions with computers: y x The grph confirms the suspicion tht the mximum vlue of f (x) occurs t x = 3. Exmple: Evlute 0 e x dx to 5 deciml plce ccurcy. Solution: The error must be less thn =. With f(x) = e x, we hve 0, 000 f (x) = xe x, nd f (x) = e x x( x)e x = e x ( x ), [ ( f (x) = ( x)e x x ) ] + e x ( 4x) = 4e x (3x x 3 ), f (x) = 6x e x (x 3) y x 40

41 ( 0) 5 3 We hve M 4 3, so E(S n (f )) M 4 = 80n4 80n < 4 0, 000 if n 4 > 60, = 6, = 3, Then L 8 (f ) , R 8 (f ) , so T 8 (f ) = L 8(f ) +R 8 (f ) = Also M 8 (f ) = ,nd S 8 (f ) = As mtter of fct,e(s 8 (f )) of S 8 (f ). Using n = 00, we hve orn>6.6, so we tke n = , so we expect six deciml plce ccurcy 80(0) 4 L 00 (f ) = , R 00 (f ) = , T 00 (f ) = , M 00 (f ) = , S 00 (f ) = We hve E(S 00 (f )) i.e., , so our first 9 deciml plces re ccurte, 80(00) 4 e x dx = is true sttement. 4