resolucao de calculo de integrais, Exercícios de Cálculo Avançado

Baixe resolucao de calculo de integrais e outras Exercícios em PDF para Cálculo Avançado, somente na Docsity! 464 6.6 – EXERCÍCIOS – pg. 255 Resolver as seguintes integrais usando a técnica de integração por partes. 1. ∫ dxxsenx 5 xdxxsenvdxxsendv dxduxu 5cos 5 1 55 − ==⇒= =⇒= ∫ cxsenx x cxsenx x dxxxxI ++ − = ++ − = − − − = ∫ 5 25 1 5cos 5 5 5 1 . 5 1 5cos 5 5cos 5 1 5cos 5 1 2. ∫ − dxx)1ln( ( ) xvdxdv dx x duxu =⇒= − − =⇒−= 1 1 1ln ( ) ( ) ( ) ( ) ( ) ( ) cxxxI cxxxxI dx x xxI dx x xxxI +−−−= +−−−−=       − +−+−= − − −−= ∫ ∫ 1ln1 1ln1ln 1 1 11ln 1 1 1ln 3. ∫ dtet t4 ttt edtevdtedv dtdutu 444 4 1 ==⇒= =⇒= ∫ 465 c t e cee t dteetI t tt tt +      −= +−= −= ∫ 16 1 4 4 1 . 4 1 4 4 1 4 1 4 44 44 4. ∫ + xdxx 2cos)1( ∫ ==⇒= =⇒+= xsendxxvdxxdv dxduxu 2 2 1 2cos2cos 1 ( ) cxxsen x dxxsenxsenxI ++ + = −+= ∫ 2cos 4 1 2 2 1 2 2 1 2 2 1 1 5. ∫ dxxx 3ln ∫ ==⇒= =⇒= 2 3 3 3ln 2x dxxvxdxdv dx x duxu ( ) cx x c x x x dx x xx xI +      −= +−= −= ∫ 2 1 3ln 2 2 . 2 1 3ln 2 1 . 22 3ln 2 22 22 6. ∫ dxx 3cos ∫ ==⇒= −=⇒= xsendxxvdxxdv dxxsenxduxu coscos .cos2cos2 468 axsen a dxaxvdxaxdv dxxduxu 1 coscos 22 ==⇒= =⇒= ∫ ∫ ∫ −= −= dxaxxsen a axsen a x I dxxaxsen a axsen a xI 2 2. 11 . 2 2 ax a vdxaxsendv dxduxu cos 1− =⇒= =⇒= caxsen a ax a x axsen a x c a axsen a ax a x axsen a x dxax a ax a x a axsen a x I +−+= +−+=     − − − −= ∫ 32 2 22 2 2 2 cos 2 12 cos 2 cos 1 cos 1 . 2 11. ∫ dxxecx 2cos xgv dxxvdxxdv dxduxu cot seccosseccos 22 −= =⇒= =⇒= ∫ cxsengx dxggxI ++−= −−−= ∫ lncot. .cotcot. 12. ∫ dxxgarc 2cot xvdxdv dx x duxgarcu =⇒= + − =⇒= 241 2 2cot ´ cxxgarcx x dxx xgarcx dx x xxxgarcI +++= + += + − −= ∫ ∫ 2 2 2 41ln 4 1 2cot 41 22cot 41 2 2cot 469 13. ∫ dxbxsene ax bx b dxbxsenvdxbxsendv dxeadueu axax cos 1− ==⇒= =⇒= ∫ ∫ ∫ + − = −− − = dxbxe b a bx b e I dxeabx b bx b eI ax ax axax coscos cos 1 cos 1 bxsen b vdxbxdv dxeadueu axax 1 cos =⇒= =⇒= c ba bxseneabxeb I bxsene b a bx b e ab b I bxsene b a bx b e I b a I I b a bxsene b a bx b e I dxeabxsen b bxsen b e b a bx b e I axax ax ax ax ax ax ax axax ax + + +− =       + − + = + − =+ −+ − =     −+ − = ∫ 22 222 2 22 2 2 2 2 cos cos cos cos 11 .cos 14. ∫ + + dx bax bax )ln( ( ) ( ) ( ) 2 1 1 ln 2 1 2 1 bax a vdxbaxdv dx bax a dubaxu + =⇒+= + =⇒+= − 470 ( ) ( ) ( ) ( ) ( ) ( ) ( ) cbax a baxbax a I c bax a baxbax a I dxbaxbaxbax a I dx bax a bax a bax a baxI ++−++= + + −++= +−++= + +−++= ∫ ∫ − 4 ln 2 2 1 1 2ln 2 2ln 2 22 .ln 2 1 2 1 2 1 15. ∫ − dxxx 23 1 ( ) ( ) 2 3 1 2 1 1 2 2 3 2 1 2 2 2 x vdxxxdv dxxduxu −− =⇒−= =⇒= ( ) ( ) ( ) ( ) ( ) ( ) cxxxI c x xxI dxxxxxI +−−− − = + − −− − = − − −− − = ∫ 2 5 2 3 2 5 2 3 2 3 2 3 222 2 22 222 1 15 2 1 3 1 2 5 1 3 1 1 3 1 21 3 1 1 3 1 . 16. ∫ dxx2ln 3 xvdxdv dx x xduxu =⇒= =⇒= 2 2 2ln32ln 23 ∫ ∫ −= −= dxxxxI x dx xxxxI 2ln32ln 2ln.2ln 23 23 xvdxdv dx x xduxu =⇒= =⇒= 2 2 2ln22ln2 473 xx evdxedv dxxduxu =⇒= =⇒= 22 ∫−= dxxeexI xx 22 xx evduedv dxduxu =⇒= =⇒= [ ] ceexex dxeexexI xxx xxx ++−= −−= ∫ 22 2 2 2 22. ∫ dx x senarc 2 xvdxdv dx x du x senarcu =⇒= − =⇒= 4 1 2 1 2 2 cx x senarcx c xx senarcx c x x senarcx dx x xx x senarcI +−+= +      − += +       − += − −= ∫ 2 2 2 2 4 2 4 4 2 2 2 1 4 1 2 2 1 2 4 1 2 1 2 2 1 2 1 23. ∫ − dxxx 2sec)1( xtgvdxxdv dxduxu =⇒= =⇒−= 2sec 1 474 cxxtgx dxxtgxtgxI ++−= −−= ∫ |cos|ln)1( )1( 24. ∫ dxxe x 4cos3 xsenvxdxdv dxedueu xx 4 4 1 4cos 3 33 =⇒= =⇒= ∫−= dxexsenxseneI xx 33 34 4 1 4 4 1 xvxsendv dxedueu xx 4cos 4 1 4 3 33 − =⇒= =⇒=     − − − −= ∫ dxexxexsen e I xx x 33 3 34cos 4 1 4cos 4 1 . 4 3 4 4 c xexsene I xexsene II Ixexsen e I xx xx x x + + = + =+ −+= 25 4cos344 16 4cos344 16 9 16 9 4cos 16 3 4 4 33 33 3 3 25. ∫ ∈ Nndxxx n ,ln 1 1 ln 1 + =⇒= =⇒= + n x vdxxdv dx x duxu n n ( ) c n x x n x c n x n x n x dx xn x n x xI xn xn nn + + − + = + ++ − + = + − + = ++ ++ ++ ∫ 2 11 11 11 1 ln 1 11 1 ln 1 1 . 11 ln 475 26. ∫ + dxx )1ln( 2 xvdxdv dx x x duxu =⇒= + =⇒+= 1 2 )1(ln 2 2 ( ) cxtgarcxxx cxtgarcxxx dx x xx dx x x xxxI ++−+= +−−+=       + −−+= + −+= ∫ ∫ 22)1(ln 2)1(ln 1 1 12)1(ln 1 2 )1(ln 2 2 2 2 2 2 27. ( )∫ ++ dxxx 21ln ( ) ( ) xvdxdv dx xx xx duxxu =⇒= ++ ++ =⇒++= − 2 2 1 2 2 1 21 2 1 1 1ln ( ) ( ) ( ) ( ) cxxxx c x xxxI dx xx x x xxxxI ++−++= + + −++= ++ + + −++= ∫ 22 2 2 2 2 2 11ln 2 1 1 2 1 1ln 1 1 1 1ln 2 1 28. ∫ dxxtgarcx 2 1 1 2 2 x vdxxdv dx x duxtgarcu =⇒= + =⇒= 478 ( ) ( ) xvdxdv dx x xsenduxu =⇒= −=⇒= 1 lnlncos ( ) ( ) x dx xsenxxxI ∫+= lnlncos ( ) ( ) xvdxdv dx x xduxsenu =⇒= =⇒= 1 lncosln ( ) ( ) ( ) x du xxxxsenxxI ∫−+= lncoslnlncos ( ) ( ) ( )( ) cxsenxxxI xsenxxxI IxsenxxxI ++= += −+= ln)(lncos 2 1 ln)(lncos2 ln)(lncos 34. ∫ dxxarc cos xvdxdv dx x duxarcu =⇒= + − =⇒= 21 1 cos cxxarcx x dx xxarcxI +−−= − − −= ∫ 2 2 1cos 1 cos 35. ∫ dxx 3sec xtgvdxxdv dxxtgxduxu =⇒= =⇒= 2sec secsec ( ) [ ] cxtgxxtgxI cxtgxIxtgxI dxxxxtgx dxxxxtgx dxxtgxtgxI +++= +++−= +−= −−= −= ∫∫ ∫ ∫ secln.sec 2 1 secln.sec secsec.sec 1secsec.sec sec.sec 3 2 2 479 Obs. ( ) ,secln sec .secsec sec secsec sec 2 cxtgxdx xtgx xtgxx dx xtgx xtgxx dxx ++= + + = = + + = ∫ ∫∫ xxtgdu xtgu 2sec.sec sec :utilizamos onde += += 36. ∫ dxex x/1 3 1 xxx edxe x vdxe x dv dx x du x u /1/1 2 /1 2 2 11 11 ∫ −==⇒= − =⇒= cee x dx x ee x I xx xx ++−= − −−−= ∫ /1/1 2 /1/1 1 11