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by: Jack Magann

132

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8

# Chapters 6.1 and 6.2 A MTH 162

Jack Magann
UM
GPA 3.865
Calculus 2
Dr. Bibby

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These notes cover the technique of integration by parts (6.1) and how to solve for three cases of trig integrals (6.2A).
COURSE
Calculus 2
PROF.
Dr. Bibby
TYPE
Class Notes
PAGES
8
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 8 page Class Notes was uploaded by Jack Magann on Thursday February 12, 2015. The Class Notes belongs to MTH 162 at University of Miami taught by Dr. Bibby in Spring2015. Since its upload, it has received 132 views. For similar materials see Calculus 2 in Mathematics (M) at University of Miami.

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## Reviews for Chapters 6.1 and 6.2 A

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Date Created: 02/12/15
Calculus 2 Chapter 61 Integration By Parts This technique is used to solve such problems as fxcosxdx and flnx The idea is to take one integral and break it up into parts to solve it So recall that f du u And that if v vx and u ux Then it duv udv vdu Know this we can get the integration by parts formula fduv fudv fvdu uvfudvfvdu f udv 2 m7 f vdu formula for integration by parts Ex 1 fxcosx dx Step 1 select a u and a dv and nd du and v ux dudx and dvcosxdx vfcosxdxsinx Step 2 Plug these into the equation above fxcosx dx xsinx fsinx dx xsinx cosx C General Rule dV is usually the most complicated factor of the two that can be integrated using a formula 2 f1nx dx u lnx du dx dv dx 1 x lnxdxxlnx x l dxx1nx xC f f x 3f sin 1x dx u sin 1x du 162 dx dv dx 1 x f sin 1x dx x sin 1x xdx x sin 1x J x1 x2 dx x Now use u substitution but since we already are using a u lets do W substitution w1 x2 dw 2x x sin 1x fx1 x2 dx xsin 1x fw dw 1 3 xsin 1xw C xsin 1x1 x2 C 4 fxzsinx dx ux2 du2xdx dvsinx dx v cosx fx2 sinx dx x2 cosx f Zxcosx dx x2 cosx focosx dx Left with this integration of parts is needs to be done again so u x du dx dv cosxdx v sinx x2 cosx2xsinx fsinxdx x2 cosx2xsinx2cosxC Tabular Method Can only be used when you have fxquot sinax dx fxquot cosax dx or fxneaquot dx For the previous problem of f x2 sin x dx U V39 x2 sin x Step 1 make a quick table 2x cos x 2 sin x Step 2 differentiate u on the left and 0 COS x integrate v on the right until the u gets to zero Step 3 Now pair up the values diagonally so x2 pairs with cosx and so on Step 4 starting at the top line with positive and alternating between that and positive go down the rows as seen above Step 5 Now write out the pairings with their corresponding signs sz sinx dx x2 cosx 2x sinx 2 cosx O C x2 cosx 2x sinx 2 c0506 C More rules Need positive integer power of X The second function needs to be easy to integrate many times Despite these the Tabular method may not be the best depending on the situation Ex 1 f x339 dx Using the tabular method I u v x3 equot x369 39626quot 6xequot 66quot C 3x2 6quot 6x 6quot 6 equot 0 ex 2fx4lnx dx ulnx duidx dvx4 v x5 151 j141511151511 5C 5x nx 5x 5x nx 55x 5x nx 25x 3 fex sinx dx u sinx du cosx dx dv equot v equot equot sinx fex cosx dx u cosx du sinx dv equot v equot equot sinx ex cosx f equot sian dx We got the original function again Rather than integrating again we can use the rule QAB Q so 2QAB Therefore fex sinx dx equot cosx equot sinx fex sinx dx 2 f equot sinx dx equot cosx equot sinx fex Sinx dx ex cosx2ex sinx C 4 fsec3x dx fsecxseczxdx u secx du secxtanxdx dv seczxdx v tanx secxtanx ftanxsecxtanxdx secxtanx ftanzxsecxdx tanzx seczx 1 secxtanx fsec2x 1 secxdx secxtanx f sec3x secxdx 2 sec x tan x f sec3 x dx f sec x dx The original question appears here so 2fsec3xdx secxtanx fsecxdx sec x tan xlnsec xtan x gt fsec3xdx C 2 5 sin In x dx u sinln x du icos n x dv dx 1 x xsinlnx fcoslnx dx u coslnx du sinln x dv dx 1 x x sinln x x cos x f sinln x dx The original question appears here so sinln x x cos x 2 2 f sinln x dx sinln x x cosx gt fsinln x dx C 6 f e dx Use rationalizing substitution from 60 t t2x dx2t fettht2fettdt now the tabular method can be used since it matches f xneaquot dx 2tet et 2 z e 26 C Calculus 2 Chapter 62 A Trig Integrals There are three types Type 1 f sinm x cosn x dx To solve for this one you either use the identity sinzx coszx 1 or on of the power reducing formulas sin2 x 1 cos 2x cos2 x 1 cos 2x To use the power reducing formulas you need to have these special cases a Only f sinm x dx where m is a positive integer b Only f cosn x dx where n is a positive integer c f sinm x cosquot x dx where n and m are positive integers If there is an odd power you need to use the trig identity to solve in terms of either cos x 0quot sin x Ex Special Cases lf sin25x dx since it is to a positive even we can use the power reducing formula 1 1 1 2 5f 1 cos10x dx 5x E sm10x C 2fsin2xcoszx f1 cost ECl cost dx if1 coszx dx 1 1 1 1 1 1 1 1 Zfdx Zf 1 cos4xdx 2x 5x 251n4x 5x 3 Zs1n4x C 3 f cos4 x dx fcos2 xcos2 xdx f1 cos2x 1 cos2x dx if cos2x2 if 2 cos 2x cos2 2xdx separate the integral to simplify if 1 2cos2x dx f 1 cos4xdx 2 ix isin2x x cos4x C Odd Integer Cases 4 f sin4 x cos3 x dx fsin4 x cos2 x cos x dx fsin2 x 1 sin2 x cos x dx f sin4 x sin6 x cos xdx use u substitution and the power rule from here u sinx du cosx dx so fsin4 x sin6 x cos xdx fu u6du u5 u7C sin5x sin7xC 5 fcos10 x sin3 x dx fcos10 x sin2 x sin x dx fcos10 x 1 cos2 x sin xdx fcos10 x cos12 x sin xdx u cosx du sinx 1 1 1 1 fu1 u12 du u11 u13 C cos11x cos13x C 11 13 11 13 Type 2 fsecm x tanquot x dx For this one either you use usecx dusecxtanxdx or utanx dusec2x dx to put the integral in terms of either one of these two trig function Make use of the identity 1 tan2 x 2 sec2 x Type 3 f cscm x cotn x dx For this one either you use u cscx du cscxcotxdx or u cotx du csc2x dx to put the integral in terms of either one of these two trig function Make use of the identity 1 cot2 x csc2 x Ex 1f tan5 x sec x dx Can t use u tan x since there is no sec2 x dx to replace for du so u secx du secxtanxdx gt ftan4 x sec x tan xdx use the identity tan2 x 2 sec2 x 1 gt fsec2 x 12secx tan xdx fsec4 x 2 sec2 x 1secx tan xdx fu4 Zu2 1du 115 u3 u C secsx sec3x secx C 2 f sec6 x dx u tan x this is u because it is easy to break du 2 sec2 x away 2 fsec4 x sec2 xdx Use the identity sec2 x tan2 x 1 ftan2 x 12sec2 xdx ftan4 x 2 tan2 x 1 sec2 xdx fu4 Zu2 1du 115 2113 u C tan5x tan3x tanx C 3 f cot3 x csc3 x dx u can t be cotx since you can t use an identity to put the Whole equation in terms of cotx when du csc2 x is taken out u csc2 x du cotxcscxdx fcot x csc3 x dx f cot2 x csc2 x cotx csc xdx Use the identity cot2 x csc2 x 1 fcsc2 x 1csc2 xcotx csc xdx fcsc4 x csc2 x cotx csc xdx fu4 u2du u5 u3C csc5x csc3xC All these notes are from a Calculus 2 class run by Dr Patrick Dibby at the University of Miami A majority of the problems used by the teacher and copied in these notes are taken from the textbook Essential Calculus Second edition By James Stewart Citation Stewart James Essential Calculus 2nd ed Np Cangage Learning 2013 Print

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