Integration by substitution - Mathematics resources

1 Integrationby substitutionmc-TY-intbysub-2009-1 There are occasions when it is possible to perform an apparently difficult piece of integrationby first making asubstitution. This has the effect of changing the variable and the dealing with definite integrals, the limits of Integration can also change. In this unit wewill meet several examples of integrals where it is appropriate to make a 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: carry out Integration by making a substitution identify appropriate substitutions to make in order to evaluate an by substitutingu=ax+ f(g(x))g (x) dxby substitutingu=g(x) mathcentre 20091.

2 IntroductionThere are occasions when it is possible to perform an apparently difficult piece of integrationby first making asubstitution. This has the effect of changing the variable and the dealing with definite integrals, the limits of Integration can also change. In this unit wewill meet several examples of this type. The ability to carryout Integration by substitution is askill that develops with practice and experience. For this reason you should carry out all of thepractice exercises. Be aware that sometimes an apparently sensible substitution does not leadto an integral you will be able to evaluate. You must then be prepared to try out Integration by substitutingu=ax+bWe introduce the technique through some simple examples forwhich a linear substitution we want to find the integral (x+ 4)5dx(1)You will be familiar already with finding a similar integral u5duand know that this integral isequal tou66+c, wherecis a constant of Integration .

3 This is because you know that the rule forintegrating powers of a variable tells you to increase the power by 1 and then divide by the the integral given by Equation (1) there is still a power 5,but the integrand is more complicateddue to the presence of the termx+ 4. To tackle this problem we make asubstitution. Weletu=x+ 4. The point of doing this is to change the integrand into the much , we must take care to substitute appropriately for the terms of differentials we havedu=(dudx)dxNow, in this example, becauseu=x+ 4it follows immediately thatdudx= 1and sodu= , substituting both forx+ 4and fordxin Equation (1) we have (x+ 4)5dx= u5duThe resulting integral can be evaluated immediately to giveu66+c.

4 We can revert to an expressioninvolving the original variablexby recalling thatu=x+ 4, giving (x+ 4)5dx=(x+ 4)66+cWe have completed the Integration by mathcentre 2009 ExampleSuppose now we wish to find the integral cos(3x+ 4) dx(2)Observe that if we make a substitutionu= 3x+ 4, the integrand will then contain the muchsimpler formcosuwhich we will be able to before,du=(dudx)dxand sowithu= 3x+ 4anddudx= 3It follows thatdu=(dudx)dx= 3 dxSo, substitutingufor3x+ 4, and withdx=13duin Equation (2) we have cos(3x+ 4) dx= 13cosudu=13sinu+cWe can revert to an expression involving the original variablexby recalling thatu= 3x+ 4,giving cos(3x+ 4) dx=13sin(3x+ 4) +cWe have completed the Integration by is very easy to generalise the result of the previous example.

5 If we want to find cos(ax+b)dx,the substitutionu=ax+bleads to1a cosuduwhich equals1asinu+c, that is1asin(ax+b)+ similar argument, which you should try, shows that sin(ax+b)dx= 1acos(ax+b) + Point sin(ax+b)dx= 1acos(ax+b) +c cos(ax+b)dx=1asin(ax+b) + mathcentre 2009 ExampleSuppose we wish to find 11 make the substitutionu= 1 2xin order to simplify the integrand to1u. Recall that theintegral of1uwith respect touis the natural logarithm ofu,ln|u|. As before,du=(dudx)dxand sowithu= 1 2xanddudx= 2It follows thatdu=(dudx)dx= 2 dxThe integral becomes 1u( 12du)= 12 1udu= 12ln|u|+c= 12ln|1 2x|+cThe result of the previous example can be generalised: if we want to find 1ax+bdx, thesubstitutionu=ax+bleads to1a 1uduwhich equals1aln|ax+b|+ means, for example, that when faced with an integral such as 13x+ 7dxwe can imme-diately write down the answer as13ln|3x+ 7|+ Point 1ax+bdx=1aln|ax+b|+ mathcentre 2009A little more care must be taken with the limits of Integration when dealing with definite the following we wish to find 31(9 +x)2dxWe make the substitutionu= 9 +x.

6 As before,du=(dudx)dxand sowithu= 9 +xanddudx= 1It follows thatdu=(dudx)dx= dxThe integral becomes x=3x=1u2duwhere we have explicitly written the variable in the limits of Integration to emphasise that thoselimits were on the variablexand notu. We can write these as limits onuusing the substitutionu= 9 +x. Clearly, whenx= 1,u= 10, and whenx= 3,u= 12. So we require u=12u=10u2du=[13u3]1210=13(123 103)=7283 Note that in this example there is no need to convert the answer given in terms ofuback intoone in terms ofxbecause we had already converted the limits onxinto limits In each case use a substitution to find the integral:(a) (x 2)3dx(b) 10(x+ 5)4dx(c) (2x 1)7dx(d) 1 1(1 x) In each case use a substitution to find the integral:(a) sin(7x 3)dx(b) e3x 2dx(c) /20cos(1 x)dx(d) 17x+ mathcentre 20093.

7 Finding f(g(x))g (x) dxby substitutingu=g(x)ExampleSuppose now we wish to find the integral 2x 1 +x2dx(3)In this example we make the substitutionu= 1 +x2, in order to simplify the square-root shall see that the rest of the integrand,2xdx, will be taken care of automatically in thesubstitution process, and that this is because2xis the derivative of that part of the integrandused in the substitution , + before,du=(dudx)dxand sowithu= 1 +x2anddudx= 2xIt follows thatdu=(dudx)dx= 2xdxSo, substitutingufor1 +x2, and with2xdx= duin Equation (3) we have 2x 1 +x2dx= udu= u1/2du=23u3/2+cWe can revert to an expression involving the original variablexby recalling thatu= 1 +x2,giving 2x 1 +x2dx=23(1 +x2)3/2+cWe have completed the Integration by us analyse this example a little further by comparing theintegrand with the general casef(g(x))g (x).

8 Suppose we writeg(x) = 1 +x2andf(u) = uThen we note that the composition1of the functionsfandgisf(g(x)) = 1 + finding the composition of functionsfandgit is the output fromgwhich is used as input tof, resultinginf(g(x)) mathcentre 2009 Further, we note that ifg(x) = 1 +x2theng (x) = 2x. So the integral 2x 1 +x2dxis of the form f(g(x))g (x) dxTo perform the Integration we used the substitutionu= 1 +x2. In the general case it will beappropriate to try substitutingu=g(x). Thendu=(dudx)dx=g (x) the substitution was made the resulting integral became udu. In the general case itwill become f(u) du. Provided that this final integral can be found the problem purposes of comparison the specific example and the general case are presented side-by-side: 2x 1 +x2dx f(g(x))g (x)dxletu= 1 +x2letu=g(x)du=(dudx)dx= 2xdxdu=(dudx)dx=g (x) dx 2x 1 +x2dx= udu f(g(x))g (x)dx= f(u) du23u3/2+c23(1 +x2)3/2+cKey PointTo evaluate f(g(x))g (x)dxsubstituteu=g(x), anddu=g (x)dxto give f(u) duIntegration is then carried out with respect tou, before reverting to the original is worth pointing out that Integration by substitution issomething of an art - and your skillat doing it will improve with practice.

9 Furthermore, a substitution which at first sight mightseem sensible, can lead nowhere. For example, if you were tryto find 1 +x2dxby lettingu= 1 +x2you would find yourself up a blind alley. Be prepared to persevere and try mathcentre 2009 ExampleSuppose we wish to evaluate 4x 2x2+ 1dxBy writing the integrand as1 2x2+ 1 4xwe note that it takes the form f(g(x))g (x)dxwheref(u) =1 u,g(x) = 2x2+ 1andg (x) = substitutionu=g(x) = 2x2+ 1transforms the integral to f(u) du= 1 uduThis is evaluated to give 1 udu= u 1/2du= 2u1/2+cFinally, usingu= 2x2+ 1to revert to the original variable gives 4x 2x2+ 1dx= 2(2x2+ 1)1/2+cor equivalently2 2x2+ 1 +cExampleSuppose we wish to find sin x the substitutionu= x.

10 Thendu=(dudx)dx=12x 1/2dx=12x1/2dx=12 xdxso that sin x xdx= 2 sinudufrom which2 sinudu= 2 cosu+c= 2 cos x+cWe can also make the following observations:the integrand can be written in the formsin x 1 mathcentre 2009 Writingf(u) = sinuandg(x) = xtheng (x) =12x 1/2=12x1/2=12 ,f(g(x)) = sin we write the given integral as2 sin x12 xdxwhich is of the form2 f(g(x))g (x) dxwithfandgas given before the substitutionu=g(x) = xproduces the integral2 f(u) du= 2 sinudufrom which2 sinudu= 2 cosu+c= 2 cos x+cExercises 21. In each case the integrand can be written asf(g(x))g (x). Identify the functionsfandganduse the general result on page 7 to complete the Integration .(a) 2xex2 5dx(b) 2xsin(1 x2)dx(c) cosx1 + In each case use the given substitution to find the integral:(a) 2xe x2dx,u= x2.