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Evaluate each indefinite integral. ! ! ! Integration by Substitution 8x 3 2 ! 5 4 (5 + ln (x)) 3 −5 3 dx; −16x u = −2x ( −4x+45− 1) dx; u = −4x 4 4) − 1 (5x 4 + 5) ⋅ 20x 3 dx; u = 5x 4 + 5 5) dx; u = 5 + ln 5 2) 4 3 2 3 (oftware Kuta Software Name___________________________________ −2x + 5- )Infinite x Calculus Name___________________________________ Kuta−9x Software Infinite Calculus Name___________________________________ (−3x 3 --+Infinite (3x 4 + 4) 4 dx 9) 1) dx Calculus 10) 12x ! ration by Integration Substitutionby Substitution ( Date________________ Period____ Date________________ Period____ Date________________ Period____ Evaluate each Evaluate each indefinite indefinite integral. Use Use the the provided provided substitution. substitution. ate each indefinite integral. Use the integral. provided substitution. oftware - Infinite Calculus Name___________________________________ Kuta Software Infinite Calculus Name___________________________________ Evaluate each --indefinite integral. 8x 3 Kuta Software Infinite Calculus Name___________________________________ 4 5 −5 4 5 5 3 4 4 5 −5 4 5 5 3 4 5 −5 3) − 4 5− 1)) dx; u = −3x − 1 −15x −4x 11))1 dx; uu4== −4x −− 11 u = −2x + 5 titution Integrals 1) −15x −3x −− 11 dx; u = −3x −2)1 −16x 3 (−4x 4 −2) 2) −16x −4x dx; −4x5 4 dx; (−3x 5 for Date________________ Period____ 15x −1) 1)Definite dx; u =((−3x −3x 1) −16x dx; u((= −4x−4− − (−2x + 5) Period____ 50x Substitution for Definite 3x − 5Integrals 3 4 dx; Date________________ 3 u = e 3x 4u = 5x 2 + 5 2 (e 3 by Integration Substitution ) Date________________ Period____ 3) 9) 6e 3x−9 cos − 5 dx; 4) ( ) 1) dx 10)sec (12 x (−3 xin + ss each definite integral terms of u, but do not evaluate. 5xx2 +35x) + 4 dx integral in terms of u,provided but do not evaluate. Express Evaluateeach eachdefinite indefinite integral. Use the substitution. 1 2 5 + ln8x( x)) 3 1 4 2 3 3 5 4 Kuta Software - Infinite Calculus 4+28 ln dx; u 0(=5x4x 2) 4sec −12x dx; (u4x =2)4x 1= sec (4x 4x(4x ⋅ tan−Calculus 4x1)⋅ sec dx;33 −uName_____________________________ 5Name___________________________________ 7) ) 36x3 3 (3x + 3) dx; u = 3x 4 + - Infinite Name_____________ +2x+5)1(3x2)⋅320x 3 −3 dx; u2 = Kuta 5x24 +Software 56) 3 2 4) 2 3 4 4 2 ( ) 5 1) dx; u = 4 x + 1 2) −12 x 4 x − 1 dx; u = 4 x − 1 5 4 (4x x+ 1) 11) ( ) −12x −4x + 2 dx 5 x − 3 0 1) 20 xsin 2) 16 x − 2dx) dx; u = 4 x − 2 (3xx ⋅−sec 12) 3) (⋅ 415x 2 (5 x2 − 3) dx; u = 1) 0 Substitution for Definite Integrals −1 (4 x + Date________________ Per Evaluate Date________________ SubstitutionPeriod____ for Definite Integrals Date____ each indefinite integral. Evaluate eachExpress indefinite integral. each definite integral terms ofintegral u, but do not evaluate. e provided substitution. Express eachindefinite in terms of u, but do not evaluate. 3 2 3 9) −9 x (−3 x 0+ 1) dx 10) 12 x 3 (3 x 4 1+ 4) 4 dx 1 4 2 x (3x 4 + 3) 0dx4 2 8 x 3 (sec 3−3x 3 (3x 5 ) dx 5) −36x 3 sec 3)4⋅8tan 6) −9sec −3x2)⋅ tan −3xx⋅2sec 33 2 + Kuta 22 −12 3 1) ( ) 3 3 2 dx; u = 4 x + 1 4 x − 1 dx; u =x 24(4xName________ 2 Calculus Software Infinite Calculus Name_______________________ ( )) 5 + ln x ( ) 8x 8x 2) 16x ⋅ sec 4x − 2 dx; u = 4x − 2 Kuta Software Infinite 8x 44 1) dx; u = 4 x + 1 2) −12 x 3− −1ln1)( x)dx; u 2 2 3 4 3 4 5) 33 44 44 dx; u = 5 + 3) − dx; u = −2x + 5 3) − dx; u = −2x + 5 3 2 ( ) 4 3 4 x + 1 2 dx; u = −2x +−155 ((5x )) 03Software 4) dx; 5xu++ 20x dx; uux== 5x 5x ++ 55 4) ⋅⋅+20x 4)x +(5x =555x 5 dx; 1) + 5) ⋅ 20x −1 (4 Kuta - Infinite 0Calculus ((−2x −2x44 ++ 55))5Integration (−2x 4 + 5) 5 by Substitution −3 2 3 Date______________ Integration by 12) Substitution Da 11) −12 x (−4 x + 2) dx (3 x 5 − 3) 5 ⋅ 15 x 4 dx ! !! ! ! ! ! !!! ! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! Substitution for Integrals Definite Evaluate each indefinite Use the integral. provided substitution. 1 integral. Evaluate each indefinite Use the provided substitution. 2 1 24x 2 2 6x3( (x 2 −4 1) )dx; u = x 2 2− 41 2 4) dx; u = 4x24 +xx4 1 5 4 4 2 each definite in terms of u, bu ( 6x 3x + 3 3) dx; u + x 4x 1Express 3= xx3x 33x dx; 3 x u 8) ( ) ) 2− 1) 2dx;4)u = 4x − 50 ( ) 6) 4sec 4x ⋅ tan 4x ⋅ sec 4x dx; = sec 6 x − 1 u = x − 1 dx;dx; u= 4 x352x+2 + 4 integral ( ) ( ) 2 2 4x + 4 3) 6e cos e − 5 dx; u = e − 5 4) u 2 4 1) ©z n2D0E1T3Y HKHug t= TaQ 8SloAfWtcw5aCr8eE JL25 KLTCZ.A h QACl04l8 erUi8gCh4t1sX Pr0eWsbe3IrnvCeVdZ.6 w BMWaYdmeV 4 3 2 2 2 0 2 ) dx; xsin (5 x −1)3) dx; u = 5(x5 x − −3 3) dx; x − 216 usec =24(x49w4nixitvh−46 UIJn−2 5− 3) 2)4x 16 x ⋅ sec (4 2) ( ) 13) −1(−2x − 4) ⋅ −32x20dx 44x x=(55+xx4)+ 20 xsin u x ⋅ Name___________________________________ sec 5 0 0 ( 14) e − 4 ⋅ 8e dx 8x 2 1) dx; u = 4 x + 1 2 2 Date________________ Period____ −13(4 x + 1) −3 −12 x 2 (+−4lnx2 34x+)2) dx (33x 5 − 3) 51 ⋅ 1524x 4x dx u, but do not11) evaluate. (−4 12) 4x 5cos 2 1 50x 2 2 2 2 7) 4)− 8) 2 dx 2 21 3)2 56 x(dx; xdx−u 1=) 5x dx; u = x − 4) dx; u = 4 x 224 + x4 + 5 4 ( ) 3) 6 x x − 1 dx; u = x − 1 4) dx; u = 4 x 2 2 ( ) 2 x csc x − 1 5 5 2 3 4 1((sec ( ) ( )) 5 5x + 5 5 + ln x 2 3x 4 + 3 ( ) 4 x + 4 ( )) ( ) 5 + ln x 4 −1 7) 36x 3x + 3 dx; u = 0 ln (3(xx−1 4x ⋅⋅ sec 5 + ln ( x)) 5) )4x)) + 4((4x ) 6) 4sec 6) ⋅ sec 4sec ⋅ tan tanu4x sec(44x 4x dx;0 u(u4==x sec sec 5) uSoftware dx;3 uuCalculus = integral. 5+ 4 4x Evaluate 2 ((4x 6) 4sec Kuta Infinite Name___________________________________ - Infinite Calculus (4x4x) ⋅dx; ) )) dx; )− dx; 4x ⋅ tan 4x = sec dx; = 5 +each (4x(-3xindefinite −12x 1) dx;= 5u += ln 4x −) 1Name___________________________________ 2) xln 1 x x 4 13) 0 (−2 x 4 − 4) for ⋅ −32 x 3 dx x (e 4x −−34)x5 ⋅⋅tan Substitution 14) 8Date________________ e 4−3 dx n for Definite Integrals 3 4 Definite Integrals 4 Date________________ Period____ ( ) ( ) 5) −36 x sec 3 x + 3 ⋅ tan 3 x + 3 dx 6) −9sec x ⋅ sec 2 (sec −3 x) dx Period____ ate each definite integral. Evaluate each definite integral. Express each integral terms of u, but do not evaluate. definite integral terms4definite of u, but do notin evaluate. 50 x 1 8) 0x(in 50 x 3 x− 1 3x 3x 4x − 1) dx;3)u = 64x 3x 3x 8x cos (e -1−3)5) dx; ucos =16x e (e 3−x 5− 5) dx;1 u =16 4) dx; uLLC= 5 x 2 2+ 5 dx; 6 e e − 5 4) LKLTCZ.A h QACl0 l8 erUi8gCh4t1sX Pr0eWsbeIrnvCeVdZ.6 w BMWaYdmeV 9wnii2 tvh6 UIJn1fqiLnzi8tseU yCva5lCcfutlpuOsb.t e Worksheet by Kuta Software 2 2 8 x x 0 = 2x + 33 1 4x + 4 2 − dx; sec4 x(52 x+ 4+ 5) sec (5 x + 5) x(4x8+x 5) dx dx; u = 22x 2 + 3 1 6) 16) 2x +33 dx; dx 32u = 5)dx; u− 6) u =5x 2 2 2 3 2 3) 2 2 2 2 2 2 3 3 2 2 (2xdx; ) 215) ( +u3= 4x + 4 ( ) 1) dx; u = 4 x + 1 −12 x 4 x − 1 dx; u = 4 x − 1 ( ) 0(4x − 1) dx;2)u = 4x 3) 6 x x − 1 dx; u = x − 1 ( ) ( ) 4x + 1 2) −12x − 1 2 x + 3 4 x + 4 −3 0 2 2 (4 x 2 +−3x 1) ⋅ tan 1) −1 6) −1 −9sec 0 −3x) dx −3x ⋅ sec 2 (sec 0 1 each3 definite integral. 4 4Evaluate Evaluate each definite integral. 4 x 5 13) (−2 x − 4) ⋅ −32 x dx 14) (e − 4) ⋅ 8e 4 x dx 0 1 1 1 8x 2 4 0 16 x 5 4 24x 3 4 2 8 x 16 5 dx; 4 HKHugtTaQ ) ( ) 7) 36x 3 u = 3x 3 8) 4x − 1 dx; = 4x − 1 2 xintegral. 3 (3x 4 + dx; 44 + 5) − dx; u = 2 x + 3 6) dx; u = 4 xr2nvCeVdZ.6 + xw 4BMWaYdmeV 9wniidx; ©z n2D0E1T3Y 8Slou AfWtcw5aCr8eE JLKLTCZ.A h QACl0l8 erUi8gCh4t1sX tvh6 UIJn1fqiLnzi8tseU yCva5lCcfu2t 5 4 Evaluate each indefinite Evaluate each indefinite integral. 3 4 4 4) u = 4x + u =8)u2 x= x4x +(4x 3− −1 1) dx; u2 = 4x2− 16) Pr0eWsbeI u = 4x 3x +2+ 33) dx;2 u = 23x5) + 3 −8) x(4x 2− dx; ) 6x (3x + 3)7)dx;36x u =2(3x 1 dx; 2 2 2 ( ) ( ) 2 x + 3 4 x + 4 ( ) 4x + 4 −3 0 ( ) ( ) 2 x + 3 4 x + 4 0 3 −3 0 5cos (−43 + ln 4 x ) 4x ( ) 15) x 4 x + 5 dx 5 x 2 x + 3 7) − dx 8) dx 3 416) 4 3 4 4( x (sec tan( (x34 x− 1+) 3)6)dx −9sec −3 x ⋅ tan 6)−3 x−9sec x Software ⋅ −3 tanxLLC − ©Q g2c0N103Q wKbu1tuaa MSRopfHtiwLairbej eLSLaCZ.x N gAUlmlz hrkiTgvhDtPsB frDe0s5earxvge5) HwYiZtMhL yfniInUiptVeL (nC3 4aPlux cpu5) 1lVue+ sv.R 3)−36 Worksheet (Kuta ) −36 xmIpnsec ⋅ tan 3 x 3+x3)+dx3) ⋅csc ⋅ sec 2by−3 sec x Xdb.R H vMwaBdOej !! ! ! !! !! !! ! !!!! ! !! !! !! ! 0 2 dx; ! !!! ! ! ! ! ! ! !! ! ! !! ! ! ! ! ! !!! ! ! !! ! ! 1 8x !! ! ! ! ! ! ! ! ! ! ! ! ! 1 ! 28 x dx; u−= 4x Evaluate each dx; u 2= 3 x 8) u =definite 42 x 2 + 2integral. 24 x + 2 2 dx; 2 2 24x 2 ( 4x ) (4xdx;+ u2)= 4) u = x − 4) 1 0 4x 2 +0 4 (42x + 22 ) dx; u = 4 x + 4 2 2 −1 8) dx 0 (4 x + 4) 0 0 (4x + 4) 8x csc ( x 4 − 1) 5) − dx; u = 2 x 2 + 3 2 2 −3 (2 x + 3) -2-16) 5 x 2 x + ©m k2U071e3T nK5uutiaP CSTovf4tiwNaMr9ek jLwLWCH.l p GAwlpl2 erpiygDhtt3nsv 0rZews5eWrTv1ePd5.s 7 AMqaPdTef MwBistxh2 OIHngf2iqnliPtPen vCna1lGcGuBlpulsT.Q Worksheet by Kuta Softwa 15) x(4 x + 5) 0dx 13 dx 0 1 8 x 2 7) 18 x 2 (3 x 3 + 3)7) dx; 18 u =x 23(3x 3x 3++3 3) 2 dx; u = 3 x 3 + 38) − dx; u − = 4 x 2 8+x2 dx; u = 4 2 8) 2 -12 ©z n2D0E1T3Y HKHugtTaQ 8SloAfWtcw5aCr8eE eU yCva5lCcfutlpuOsb.t Worksheet by Kuta Softw 2 ( 1 JLKLTCZ.A h QACl0l8 erUi8gCh4t1sX 4 x + 2) −1 Pr0eWsbeIrnvCeVdZ.6 w BMWaYdmeV 9wniitvh6 UIJn1fqiLnzi8ts 0 ( ) 4 x + 2 −1 -1tTaQ w5a.6 Cr8w eBMWaYdmeV E JLKLTCZ.A rnvuCeVdZ tseU yCva5lCcfutlpuOsb.t 16x JLKLTCZ.A h©z QACl0n2lD0E1T 8 erUi83gYCh4HKHug t1sX Pr0 eWsb8SeloIrnAfWtc vCeVdZ 9wh QACl0 niitvh6l8UIJernUi81fgqiCh4 Lnzti18stsXePr0U yeCWsbvae5lICcf tlpuOsb.6.tw BMWaYdmeV 9wniitvh6 UIJn1fqiLnzi8-1Worksheet by0Kuta Software LLC Worksheet by Kuta Softw 2 +-13) 18x (3x 2 1) dx; u 2 3 3 2 7) 18ux=223x 3 x 3 +2 33 = 3) x 2 −−116 x x −3 1 dx; ! 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