Integration by parts - Mathematics resources

1 Integrationby partsmc-TY- parts -2009-1A special rule, Integration by parts , is available for integrating products of two functions. Thisunit derives and illustrates this rule with a number of order to master the techniques explained here it is vital that you undertake plenty of practiceexercises so that they become second reading this text, and/or viewing the video tutorial on this topic, you should be able to: state the formula for Integration by parts integrate products of functions using Integration by of the formula for Integration by parts udvdxdx=u v the formula for Integration by mathcentre 20091. IntroductionFunctions often arise as products of other functions, and wemay be required to integrate theseproducts.

2 For example, we may be asked to determine xcosxdx .Here, the integrand is the product of the functionsxandcosx. A rule exists for integratingproducts of functions and in the following section we will derive Derivation of the formula for Integration by partsWe already know how to differentiate a product: ify=u vthendydx=d(uv)dx=udvdx+ this rule:udvdx=d(uv)dx integrate both sides: udvdxdx= d(uv)dxdx vdudxdx .The first term on the right simplifies since we are simply integrating what has been differentiated. udvdxdx=u v vdudxdx .This is the formula known asintegration by PointIntegration by parts udvdxdx=u v vdudxdxThe formula replaces one integral (that on the left) with another (that on the right); the intentionis that the one on the right is a simpler integral to evaluate,as we shall see in the mathcentre 20093.

3 Using the formula for Integration by partsExampleFind , we are trying to integrate the product of the functionsxandcosx. To use the integrationby parts formula we let one of the terms bedvdxand the other beu. Notice from the formula thatwhichever term we let equaluwe need to differentiate it in order to finddudx. So in this case, ifwe letuequalx, when we differentiate it we will finddudx= 1, simply a constant. Notice thatthe formula replaces one integral, the one on the left, by another, the one on the right. Carefulchoice ofuwill produce an integral which is less complicated than the cosx .With this choice, by differentiating we obtaindudx= fromdvdx= cosx, by integrating we findv= cosxdx= sinx.

4 (At this stage do not concern yourself with the constant of Integration ). Then use the formula udvdxdx=u v vdudxdx: xcosxdx=xsinx (sinx) 1 dx=xsinx+ cosx+cwherecis the constant of the next Example we will see that it is sometimes necessaryto apply the formula for integrationby parts more than mathcentre 2009 SolutionWe have to make a choice and let one of the functions in the product equaluand one a general rule we letube the function which will become simpler when we differentiate it. Inthis case it makes sense to letu=x2anddvdx= 2xandv= e3xdx= , using the formula for Integration by parts , x2e3xdx=13e3x x2 13e3x 2xdx=13x2e3x 23xe3xdx.

5 The resulting integral is still a product. It is a product of the functions23xande3x. We can usethe formula again. This time we chooseu=23xanddvdx= e3xdx= x2e3xdx=13x2e3x 23xe3xdx=13x2e3x {23x 13e3x 13e3x 23dx}=13x2e3x 29xe3x+227e3x+cwherecis the constant of Integration . So we have done Integration by parts twice to arrive atour final that to apply the formula you have to be able to integrate the function you can cause problems consider the next xln|x| mathcentre 2009 SolutionRemember the formula: udvdxdx=u v vdudxdx .It would be natural to chooseu=xso that when we differentiate it we getdudx= 1. Howeverthis choice would mean choosingdvdx= ln|x|and we would need to be able to integrate integral is not a known standard form.

6 So, in this Example we will chooseu= ln|x|anddvdx=xfrom whichdudx=1xandv= xdx= , applying the formula xln|x|dx=x22ln|x| x22 1xdx=x22ln|x| x2dx=x22ln|x| x24+cwherecis the constant of ln|x| can use the formula for Integration by parts to find this integral if we note that we can writeln|x|as1 ln|x|, a product. We choosedvdx= 1andu= ln|x|so thatv= 1 dx=xanddudx= , 1 ln|x|dx=xln|x| x 1xdx=xln|x| 1 dx=xln|x| x+cwherecis a constant of mathcentre 2009 ExampleFind terms we choose foruanddvdxit may not appear that Integration by parts is goingto produce a simpler integral. Nevertheless, let us make a choice:dvdx= sinxandu= exso thatv= sinxdx= cosxanddudx= , exsinxdx= ex cosx cosx exdx= cosx ex+ excosxdx.

7 We now integrate by parts again choosingdvdx= cosxandu= exso thatv= cosxdx= sinxanddudx= exsinxdx= cosx ex+{exsinx sinx exdx}= excosx+ exsinx exsinxdx .Notice that the integral we have ended up with is exactly the same as the one we started us call thisI. That isI= exsinx excosx Ifrom which2I= exsinx excosxandI=12(exsinx excosx).So exsinxdx=12(exsinx excosx) +cwherecis the constant of mathcentre 2009 Exercises1. Evaluate the following integrals:(a) xsinx dx(b) xcos 4x dx(c) xe xdx(d) x2cosx dx(e) 2x2exdx(f) x2ln|x|dx(g) tan 1x dx(h) sin 1x dx(i) excosx dx(j) sin3x dx(Hint: writesin3xassin2xsinx.)2. Calculate the value of each of the following:(a) 0xcos12x dx(b) 10x2exdx(c) 21x3ln|x|dx(d) /40x2sin 2x dx(e) 10xtan 1x dxAnswers1.

8 (a) xcosx+ sinx+C(b)14xsin 4x+116cos 4x+C(c) xe x e x+C(d)x2sinx+ 2xcosx 2 sinx+C(e)2x2ex 4xex+ 4ex+C(f)13x3ln|x| 19x3+C(g)xtan 1x 12ln|1 +x2|+C(h)xsin 1x+ 1 x2+C(i)12ex(cosx+ sinx) +C(j) 13(cosxsin2x+ 2 cosx) +C2.(a)2 4(b)e 2(c)4 ln 2 1516(d) 8 14(e) 4 mathcentre 2009