13.4. Integration by Parts. Introduction. Prerequisites. Learning Outcomes

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1 Integrtion by Prts 13.4 Introduction Integrtion by Prts is technique for integrting products of functions. In this Section you will lern to recognise when it is pproprite to use the technique nd hve the opportunity to prctise using it for finding both definite nd indefinite integrls. Prerequisites Before strting this Section you should... Lerning Outcomes On completion you should be ble to... understnd wht is ment by definite nd indefinite integrls be ble to use tble of integrls be ble to differentite nd integrte rnge of common functions decide when it is pproprite to use the method known s integrtion by prts pply the formul for integrtion by prts to definite nd indefinite integrls perform integrtion by prts repetedly if pproprite HELM (28): Section 13.4: Integrtion by Prts 33

2 1. Indefinite integrtion The technique known s integrtion by prts is used to integrte product of two functions, such s in these two exmples: 1 (i) e 2x sin 3x dx (ii) x 3 e 2x dx Note tht in the first exmple, the integrnd is the product of the functions e 2x nd sin 3x, nd in the second exmple the integrnd is the product of the functions x 3 nd e 2x. Note lso tht we cn chnge the order of the terms in the product if we wish nd write 1 (i) (sin 3x) e 2x dx (ii) e 2x x 3 dx Wht you must never do is integrte ech term in the product seprtely nd then multiply - the integrl of product is not the product of the seprte integrls. However, it is often possible to find integrls involving products using the method of integrtion by prts - you cn think of this s product rule for integrls. The integrtion by prts formul sttes: Key Point 5 Integrtion by Prts for Indefinite Integrls For indefinite integrls, given functions f(x) nd g(x): ( f g dx = f g dx dx ) gdx dx Alterntively, given functions u nd v: u dv dx = u.v dx v du dx dx Study the formul crefully nd note the following observtions. Firstly, to pply the formul we must be ble to differentite the function f to find, nd we must be ble to integrte the function, g. dx Secondly the formul replces one integrl, the one on the left, with different integrl, tht on the fr right. The intention is tht the ltter, whilst it my look more complicted in the formul bove, is simpler to evlute. Consider the following Exmple: 34 HELM (28): Workbook 13: Integrtion

3 Exmple 15 Find the integrl of the product of x with sin x; tht is, find x sin x dx. Compre the required integrl with the formul for integrtion by prts: we choose f = x nd g = sin x It follows tht dx = 1 nd g dx = sin x dx = cos x (When integrting g there is no need to worry bout constnt of integrtion. When you become confident with the method, you my like to think bout why this is the cse.) Applying the formul we obtin ( ) x sin x dx = f g dx dx gdx dx = x( cos x) 1( cos x) dx = x cos x + cos x dx = x cos x + sin x + c Tsk Find (5x + 1) cos 2x dx. Let f = 5x + 1 nd g = cos 2x. Now clculte dx nd g dx: Answer dx = 5 nd cos 2x dx = 1 sin 2x. 2 Substitute these results into the formul for integrtion by prts nd complete the Tsk: HELM (28): Section 13.4: Integrtion by Prts 35

4 Answer (5x + 1)( 1 2 sin 2x) 5( 1 2 sin 2x)dx = 1 2 (5x + 1) sin 2x + 5 cos 2x + c 4 Sometimes it is necessry to pply the formul more thn once, s the next Exmple shows. Exmple 16 Find 2x 2 e x dx We let f = 2x 2 nd g = e x. Then dx = 4x nd gdx = e x Using the formul for integrtion by prts we find 2x 2 e x dx = 2x 2 ( e x ) 4x( e x )dx = 2x 2 e x + 4xe x dx We now need to find 4xe x dx using integrtion by prts gin. We get 4xe x dx = 4x( e x ) 4( e x )dx = 4xe x + 4e x dx = 4xe x 4e x Altogether we hve 2x 2 e x dx = 2x 2 e x 4xe x 4e x + c = 2e x (x 2 + 2x + 2) + c Exercises In some questions below it will be necessry to pply integrtion by prts more thn once. 1. Find () x sin(2x)dx, (b) te 3t dt, (c) x cos x dx. 2. Find (x + 3) sin x dx. 3. By writing ln x s 1 ln x find ln x dx. 4. Find () tn 1 x dx, (b) 7x cos 3x dx, (c) 5x 2 e 3x dx, 5. Find () x cos kx dx, where k is constnt (b) z 2 cos kz dz, where k is constnt. 6. Find () te st dt where s is constnt, (b) Find t 2 e st dt where s is constnt. 36 HELM (28): Workbook 13: Integrtion

5 Answers 1. () 1 4 sin 2x 1 2 x cos 2x + c, (b) e3t ( 1 3 t 1 ) + c, (c) cos x + x sin x + c 9 2. (x + 3) cos x + sin x + c. 3. x ln x x + c. 4. () x tn 1 x 1 2 ln(x2 + 1) + c, (b) 7 9 cos 3x 7 3 x sin 3x + c, (c) 5 27 e3x (9x 2 6x + 2) + c, 5. () cos kx k 2 + x sin kx k + c, (b) 2z cos kz k 2 + z2 sin kz k 6. () e st (st + 1) s 2 + c, (b) e st (s 2 t 2 + 2st + 2) s 3 + c. 2 sin kz k 3 + c. 2. Definite integrtion When deling with definite integrls the relevnt formul is s follows: Key Point 6 Integrtion by Prts for Definite Integrls For definite integrls, given functions f(x) nd g(x): b ] b b ( f g dx = f g dx dx b ] b b Alterntively, given functions u nd v: uv u dv dx dx = ) gdx dx v du dx dx Exmple 17 Find 2 xe x dx. We let f = x nd g = e x. Then dx = 1 nd g dx = e x. Using integrtion by prts we obtin 2 ] 2 2 ] 2 xe x dx = xe x 1.e x dx=2e 2 e x =2e 2 e 2 1]=e 2 +1 (or to 3 d.p.) Sometimes it is necessry to pply the formul more thn once s the next Exmple shows. HELM (28): Section 13.4: Integrtion by Prts 37

6 Exmple 18 Find the definite integrl of x 2 e x from to 2; tht is, find 2 x 2 e x dx. We let f = x 2 nd g = e x. Then 2 x 2 e x dx = x 2 e x ] 2 2 dx = 2x nd 2xe x dx = 4e g dx = e x. Using integrtion by prts: xe x dx The remining integrl must be integrted by prts lso but we hve just done this in the exmple bove. So 2 x 2 e x dx = 4e 2 2e 2 + 1] = 2e 2 2 = (3 d.p.) Tsk Find π/4 (4 3x) sin x dx. Wht re your choices for f, g? Answer Tke f = 4 3x nd g = sin x. Now complete the integrl: π/4 (4 3x) sin x dx = Answer π/4 (4 3x) sin x dx = = ] π/4 (4 3x)( cos x) (4 3x)( cos x) =.716 to 3 d.p. ] π/4 3 π/4 3 sin x cos x dx ] π/4 38 HELM (28): Workbook 13: Integrtion

7 Exercises 1. Evlute the following: () 2. Find 3. Find (x + 2) sin x dx (x 2 3x + 1)e x dx 1 x cos 2x dx, (b) π/2 Answers 1. ().16, (b) π/4 =.7854, (c) x sin 2x dx, (c) 1 1 te 2t dt HELM (28): Section 13.4: Integrtion by Prts 39