INTEGRATION BY PARTS - Salford

1 IntegrationINTEGRATION BY PARTSG raham S McDonaldA self-contained Tutorial Module for learningthe technique of INTEGRATION by partslTable of contentslBegin Tutorialc of on using notationFull worked solutionsSection 1: Theory31. TheoryTo differentiate a product of two functions ofx, one uses the productrule:ddx(uv) =udvdx+dudxvwhereu=u(x) andv=v(x) are two functions ofx. A slightrearrangement of the product rule givesudvdx=ddx(uv) dudxvNow, integrating both sides with respect toxresults in udvdxdx=uv dudxv dxThis gives us a rule for INTEGRATION , calledINTEGRATION BYPARTS, that allows us to integrate many products of functions ofx.

2 We take one factor in this product to beu(this also appears onthe right-hand-side, along withdudx). The other factor is taken tobedvdx(on the right-hand-side onlyvappears the other factorintegrated with respect tox).TocJJIIJIBackSection 2: Usage42. UsageWe highlight here four different types of products for which integrationby PARTS can be used (as well as which factor to labeluand which oneto labeldvdx). These are:(i) xn sinbxorcosbx dx(ii) xn eaxdx udvdxudvdx(iii) xr ln (ax)dx(iv) eax sinbxorcosbx dx dvdxuudvdxwherea, bandrare given constants andnis a positive 3: Exercises53.

3 ExercisesClick onExerciselinks for full worked solutions (there are 14 exer-cises in total)Exercise 1. xcosx dxExercise 2. x2sinx dxExercise 3. xexdxExercise 4. x2e4xdxExercise 5. x2lnx dxlTheorylIntegralslFinal solutionslTipslNotationTocJJIIJIBackSect ion 3: Exercises6 Exercise 6. (x+ 1)2ln 3x dxExercise 7. e2xcosx dxExercise 8. e xsin 4x dxExercise 0xe2xdxExercise 10. /4 0xsin 2x dxlTheorylIntegralslFinal solutionslTipslNotationTocJJIIJIBackSect ion 3: Exercises7 Exercise 1/2x4ln 2x dxExercise 12. 03x2cos(x2)dxExercise 13.

4 X3exdxExercise 14. e3xcosx dxlTheorylIntegralslFinal solutionslTipslNotationTocJJIIJIBackSect ion 4: Final solutions84. Final + cosx+C,2. x2cosx+ 2xsinx+ 2 cosx+C,3.(x 1)ex+C, (8x2 4x+ 1)+C, (3 lnx 1) +C, (x+ 1)3ln 3x 19x3 12x2 x 13lnx+C, (sinx+ 2 cosx) +C,TocJJIIJIBackSection 4: Final solutions98. 117e x(4 cos 4x sin 4x) +C, , , 2 31800, ( 2 8), (x3 3x2+ 6x 6)+C, (sinx+ 3 cosx) + 5: Standard integrals105. Standard integralsf(x) f(x)dxf(x) f(x)dxxnxn+1n+1(n6= 1)[g(x)]ng (x)[g(x)]n+1n+1(n6= 1)1xln|x|g (x)g(x)ln|g(x)|exexaxaxlna(a >0)sinx cosxsinhxcoshxcosxsinxcoshxsinhxtanx ln|cosx|tanhxln coshxcosecxln tanx2 cosechxln tanhx2 secxln|secx+ tanx|sechx2 tan 1exsec2xtanxsech2xtanhxcotxln|sinx|cothx ln|sinhx|sin2xx2 sin 2x4sinh2xsinh 2x4 x2cos2xx2+sin 2x4cosh2xsinh 2x4+x2 TocJJIIJIBackSection 5.

5 Standard integrals11f(x) f(x)dxf(x) f(x)dx1a2+x21atan 1xa1a2 x212aln a+xa x (0<|x|<a)(a >0)1x2 a212aln x ax+a (|x|> a>0)1 a2 x2sin 1xa1 a2+x2ln x+ a2+x2a (a >0)( a < x < a)1 x2 a2ln x+ x2 a2a (x>a>0) a2 x2a22[sin 1(xa) a2+x2a22[sinh 1(xa)+x a2+x2a2]+x a2 x2a2] x2 a2a22[ cosh 1(xa)+x x2 a2a2]TocJJIIJIBackSection 6: Tips on using solutions126. Tips on using solutionslWhen looking at the THEORY, INTEGRALS, FINAL SOLU-TIONS, TIPS or NOTATION pages, use theBackbutton (at thebottom of the page) to return to the exerciseslUse the solutions intelligently.

6 For example, they can help you getstarted on an exercise, or they can allow you to check whether yourintermediate results are correctlTry to make less use of the full solutions as you work your waythrough the TutorialTocJJIIJIBackSection 7: Alternative notation137. Alternative notationIn this Tutorial, we express the rule for INTEGRATION by PARTS using theformula: udvdxdx=uv dudxv dxBut you may also see other forms of the formula, such as: f(x)g(x)dx=F(x)g(x) F(x)dgdxdxwheredFdx=f(x)Of course, this is simply different notation for the same rule.

7 To seethis, make the identifications:u=g(x) andv=F(x).TocJJIIJIBackSolutions to exercises14 Full worked solutionsExercise evaluate byintegration by PARTS : xcosx dx=x sinx (1) sinx dx, takeu=xgivingdudx= 1(by differentiation)and takedvdx= cosxgivingv= sinx(by INTEGRATION ),=xsinx sinx dx=xsinx ( cosx) +C,whereCis an arbitrary=xsinx+ cosx+Cconstant of to Exercise 1 TocJJIIJIBackSolutions to exercises15 Exercise 2. x2sinx dx=x2 ( cosx) (2x) ( cosx)dx , takeu=x2givingdudx= 2xand takedvdx= sinxgivingv= cosx,= x2cosx+ 2 xcosx dx we need to use integrationby PARTS again!

8 TocJJIIJIBackSolutions to exercises16= x2cosx+ 2{xsinx (1) sinx dx},as in question 1,= x2cosx+ 2xsinx 2 sinx dx= x2cosx+ 2xsinx 2 ( cosx) +C= x2cosx+ 2xsinx+ 2 cosx+C .Return to Exercise 2 TocJJIIJIBackSolutions to exercises17 Exercise 3. x exdx=x ex (1) exdx, takeu=xgivingdudx= 1and takedvdx=exgivingv=ex,=x ex exdx=x ex ex+C= (x 1)ex+C .Return to Exercise 3 TocJJIIJIBackSolutions to exercises18 Exercise 4. x2e4xdx=x2 14e4x 2x 14e4xdx , takeu=x2dudx= 2xdvdx=e4xv=14e4x,=14x2e4x 12 x e4xdx now use INTEGRATION by PARTS againTocJJIIJIBackSolutions to exercises19=14x2e4x 12{x 14e4x 1 14e4xdx}, this timeu=xdudx= 1dvdx=e4xv=14e4x,=14x2e4x 18xe4x 12 ( 14) e4xdx=14x2e4x 18xe4x+18 e4xdx=14x2e4x 18xe4x+18 14e4x+CTocJJIIJIBackSolutions to exercises20=832x2e4x 432xe4x+132e4x+C=132e4x(8x2 4x+ 1)+C.

9 Return to Exercise 4 TocJJIIJIBackSolutions to exercises21 Exercise 5. x2lnx dx= (lnx) (13x3) 1x (13x3)dx , lnxdudx=1xdvdx=x2v=13x3,=13x3lnx 13 x2dx=13x3lnx 13 (13x3)+C=13x3lnx 19x3+C=19x3(3 lnx 1) +C .Return to Exercise 5 TocJJIIJIBackSolutions to exercises22 Exercise 6. (x+ 1)2ln 3x dx= (ln 3x) (13)(x+ 1)3 13x (3) 13(x+ 1) ln 3xgivesdudx=13x ddx(3x) =13x (3),using the chain rule, anddvdx= (x+ 1)2givesv=11 (x+1)33,where we have used the result that ifdvdx= (ax+b)nthenv=1a(ax+b)n+1(n+1),TocJJIIJIB ackSolutions to exercises23 (x+ 1)2ln 3x dx=13(x+ 1)3ln 3x 13 (x+ 1)3xdx=13(x+ 1)3ln 3x 13 x3+ 3x2+ 3x+ 1xdx ,where we have used the binomial theorem,or just multiplied out (x+ 1)3,=13(x+ 1)3ln 3x 13 x2+ 3x+ 3 +1xdx=13(x+ 1)3ln 3x 13[x33+32x2+ 3x+ lnx]+C=13(x+ 1)3ln 3x x39 12x2 x 13lnx+C.

10 Return to Exercise 6 TocJJIIJIBackSolutions to exercises24 Exercise 7. e2xcosx dxSetu=e2xanddvdx= cosx, to givedudx= 2e2xandv= e2xcosx dx, since we will eventually getIon the right-hand-side for this type of sinx 2e2x sinx 2 e2xsinx INTEGRATION by PARTS again, withu=e2xanddvdx= sinx, givingdudx= 2e2xandv= cosxTocJJIIJIBackSolutions to 2{e2x ( cosx) 2e2x ( cosx)dx} 2{ e2xcosx+ 2 e2xcosx dx} + 2e2xcosx 4 e2xcosdx = 5I=e2xsinx+ 2e2xcosx+C1 I=15e2x(sinx+ 2 cosx) +C, whereC=15C1(another arbi-trary constant).NoteIt is customary to introduce the arbitrary constant after thelast INTEGRATION is performed, though strictly one could accommodatearbitrary constants arising from each dudx v dx(indefinite) integra-tion and these would add up to give a single arbitrary constant in thefinal to Exercise 7 TocJJIIJIBackSolutions to exercises26 Exercise 8.