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Class Note for MATH 1432 at UH


Class Note for MATH 1432 at UH

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This 5 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 21 views.

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Date Created: 02/06/15
Lecture 7Section 82 Integration by Parts Jiwen He 1 Integration by Parts 11 Undoing the Product Rule Integration by Parts Undoing the Product Rule Product Rule d d d o The product rule E uv 7 u v v u o In terms of differentials duv udv v du o Rearrange u dv duv 7 v du o Integrate udv uvivdu Integration by Parts udvuv7vdu 0 Reduction poly 1k can be reduced by Cliff 0 Cycling sin 1 cos I e7 7 sinh z cosh z retain their form after diff or int 0 Change of Form lnz7 sin 1 1 tan 1 1 completely change their form after diff Success depends on choosing u and dv so that v du is easier that u dv 1 2 Reduction Integration by Parts Reduction Integration by Parts udvuv7vdu Reduction use the fact that poly 1k can be reduced by Cliff Ezamples 1 zcoszdzzsinz7sinzdz zsinzcoszC u z dv coszdz thus du dz 1 sinz 26X vdu sinzdz is easier than udv z cos z dz 673 dz z72e 7 725 dz 72ze 74e 0 u z dv e dz thus du dz 1 72e 3 26X vdu 7 2e dz is easier than udv ze dz z2 sinzdz z27 cosz 7 7 cosz2z dz 7z200sz2zcoszdz 7z200sz2 zsinzcoszgt C u z2 dv sinzdz thus du 2zdz v 7cosz 26X vdu 7 cosz2z dz is easier than udv z2 sinzdz 1 z5 cosz3 dz gzg s1nz3 7 z2 s1nz3 dz 1 7z3sinz3 gooszSJr C 3 u zg dv z2 cos z3 dz thus du 3z2dz v sinz3 26X z2 sinz3 dz 1 1 1 3 gslnudu77gcosu4rc77gcosz C udvuv7vdu 1fF z then dz anz 7 nzn 1Fzdz Set u z dv dz du nzn 1 dz 1 Reduction Formulas 0 With e and Fz i6 z e dz lzne 7 E znile dz a a 0 With coskz and Fz isin cz7 zn coskz dz zn sinkz 7 gzn 1 sin 0 With sinkz and Fz 7 cosUcz7 zn sinkz dz 7zn coskzzni1 clt Reduction Formu s 1 n zne dz 7zne u 7 7znile udzi a a Ezamples 2 13 21 dz zgeh SIQEQTdI 13621 7 g 12621 7 162 dzgt 136202 7 g 12621 35621 7 ehd1gtgt 1 762 4z3 76z2 6z 7 3 C 13 Cycling Integration by Parts Cycling Integration by Parts udvuv7vdu Cycling use the fact that functions such as sin z7 cos z7 67 7 sinhz7 cosh z7 ftain their form after differentiation or integration zamp e ecoszdzexsinz7emsinzdz ex sinze cosz7egc coszdz u e dv coszdz7 thus du egcdz7 v sinz lex ex sinzdz is of the same form as ex coszdz 6757 cosz 7 ex7 cos z dz lex u e dv sinzdz7 thus du emdz7 v 7cosz e sinzdz is cycling after two integrations by parts Solve the equation for fey cosz dz 2excoszdz esinzcosz C 1 ex coszdz Eexsinzcosz C 14 Change of Form Integration by Parts Change of Form Integration by Parts udvuv7vdu 1 Change offozm use the fact that functions In z sin 1 z tan z completely Change their form after differentiation lnzdzzlnz7zz 1dz zlnzidzzlnz7zC Ezamples 4 u lnz dv dz thus du z ldz v z lex vdu dz is easier than udv1nzdz tan 1zdzztan 1ziLdz 1z2 ztanilzi ln11z21c u tan lz dv dz thus du g dz 1 z lex Set u 1z2 du 2z dz z 1 1 1 7d 7 7d 71 C 1I2z 2uu 2n1u1 3 z 7d 1z2 I 13 7z tan zi z2tan 1zdz zg tanilz 7 z 3 I 1z2dx 13 71 2 2 gz tan zigz 71n11z DC 1 HUD HW H 3 z u tan lz dv z2 dz thus du dz v z3 Note that 17 z 1z2 I cosilzdzzcos 1zdz x17z2 zcos 1z717z2C 1 liz z 1 1 du72zdz dz7 dui c u cos lz dv dz thus du 7 dz 1 z lex Set u 17 z2 2 Parts on De nite Integrals Parts on De nite Integrals Parts on De nite Integrals b b udvuv 27 vdu Example 5 2 2 2 zsv4712dz 7124712 7 4712721dz 0 0 0 71 64 764 7 3 5 15 u 12 d1 zx4712dz thus du Zrdz v 74 7 12 lex Set 2 o 4 u4712du721dz 4712721dz 11 du7 211 7i 0 4 5 0 5 Outline Contents 1 Integration by Parts 1 lil Undoing the Product Rule i i i i i i i i i i i i i i i i i i i i i i 1 12 Reduction i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 1 13 Cycling i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 3 14 Change of Form i i i i i i i i i i i i i i i i i i i i i i i i i i i i 4 2 Parts on De nite Integrals 5


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