Calculus II MA 241
Popular in Course
Popular in Mathematics (M)
This 8 page Class Notes was uploaded by Braeden Lind on Thursday October 15, 2015. The Class Notes belongs to MA 241 at North Carolina State University taught by Thomas Wears in Fall. Since its upload, it has received 23 views. For similar materials see /class/223698/ma-241-north-carolina-state-university in Mathematics (M) at North Carolina State University.
Reviews for Calculus II
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 10/15/15
u C Fewer ge gg 6 fx Q W3 m 1 I lt2 9W Ayn x M XYS All l A hch 3037 61 R7 1 X HV 7amp1EQWA 0 WI A2 f 94 L N h AB A Y R A l Rookiv 9 Comerqu Sir 210 Q quotPM 1 ACE 7 3 l 1 SE A 5 AL SO quot56 V eo c Arl V vs A 4 ys m A 7 A w A lt30 m 1 Ema 2 A15 39 3quot IV va M wal l vxg 34 COAVQW R 2 g a Do A In FQQZSJFRVH 2 Z K 9 9 0 QWIXSNH r H x CoA quotW No lto1nltsv lt0 1lt K A20 7 7 C BJK 1W9 3 ar mh Prom CVNquot 54quot M A law K 391 Lkd ZeaX M39 JVGWH Sam Ye SEW m J m L 7 111 f o ln gchi L 7 my 4 er Lu Wu v eomd i W at r Is L553 ii ZszW 1 th 0 i 611 rai mikzb WU K vs a s 9 39 jr LL R 44 atQV TLAC3 Z A 1A 304 0 53 x L TLM th u A m w q 1 A s A A 0 17 309 WV quotC 70 m mng k C ESQ my mothA 1 R Qty 9 3 Cx EMAX Wm r K no megsw 6 km A itch 39l3 K g 6 gq hm I quot3 MI KAJ NS 31 mm mm SMh Ad I 39j anmm czkud L mu 1 XL L lzL h w 5 am i TCAV Aquot n C v 1 nil h x N x x quot W w 3980 1 39 d J JF Mo 10 1 W U z m 1 lA 5 7 1 U xze 3amp 4 fl L 4 L4 A L QvC l Su 3K 7t a H ln 1 if U Z LJ394lT L L QVIY l L4 0 V m X r H r a AH 1 gt0 y M K T v A n V xfTX 6 L 26 339 LC W T has W s M 1M 1 5 Wm on u 7 R h DO NH 1 39 1 L 1 I k 2quot in ig i A a l 1 do 7 o W HM WA him 0 MA 241 Spring 2008 Select Trigonometric Integration 57 77Select Forms of Trigonometric Integration I fem cos dz where m7 n are integers 2 0 A Either m or n is odd i Suppose n is odldl7 with n 2k1 Thus7 we are dealing with f sz nm0052k1 dx 0 First7 leave one factor of 005m isolated7 with the goal being to use the identity s n2z 0052z 1 and then use a u substitution with u 5mm and du 005x The integral becomes smmx0052kz 005m dx oNext7 rewrite 005z 0052zk Thus7 the integral becomes sz nm 0052xkcosx dz oNow7 use the trig identity 0052 1 7 52712 s7 which gives us smm 1 7 s n2zkcosz dx 0 At this point a simple u substitution with u smx and du cosxdx is all that is needed to make the function that we are integrating a simple polynomial um17 u2k du Ex sm6z0055 dx sz n6z 0054z005 dx 6 7 2u8u10 du u u u 7 7 27 7 o 7 9 11 iszn7 25 n9 smll 0 7 9 11 ii Suppose m is odd with m 2q1 Thus we are dealing with f s n2q1cos dx 0 This case is handled almost exactly as the one immediately above We rst leave one factor of smx isolated with the goal being to use the identity s n2z 0052z 1 and then use u substitution The integral becomes sm2qzsmzcos x dx sm2qxcosnzsmz dx oNext rewrite s n2qz smzzq Thus the integral becomes sm2xqcos zsmz dx oNow use the trig identity 52712 1 7 0052 s which gives us 1 7 0052q005 x5mx dx 0 At this point a simple u substitution with u 005m and du 7sz39nzdx is all that is needed before we complete the integral 7 1 7 u2qu du Ex s n3x0058z dab sz n2zcossxsmx dab 1 7 0052z0058z5m dab 7 1 7 uz u8 du Sub with u 005x and du 7smdz 7u8 7u10 du 9 11 0 u u 3 n ug U11 77 7 o 9 11 if 0059z 00511x 9 11 0 B Both m and n are even 0 Use the Half Angle Formulas s n2z 1 7 0052 and 0052z 1 0052 s n20052x 17 00521 0052 dx i17 00522x dx 1 1 7 dx 7 700522x dx 4 4 1 1 1 7x 7 Z E 1 0054 dx angle formula again 4 1 1 52714 7 i 0 4x 4 1 sm4x i 7 32 0 11 tanmzsecnx dz where m and n are integers 2 0 A n is even with n 2k Thus we are dealing with tanm5602kx dz 0 Much like the case when we are solving f 5mm cos dz with either m or n odd7 the goal is to be able to make a u substitution and turn the function we are integrating into a polynomial We will use the identity 8662 tan2z 1 The substitution that we would like to make is u tan and7 thus7 du seczw With that in mind7 the rst step is to isolate a factor of seczw which makes the integral tanm5602kz dx tanmx5602k 2zseczz dx oNow7 rewrite the factor of 5602k 2x 8662k71 and the integral becomes tanm 8662k71 5602s dz oAt this point7 use the trig identity 8662 tan2z 17 but be sure to leave the isolated factor of 8662 alone The integral becomes tanm tan2x 11 5602s dz 3 o The integral is now set up so that a simple u substitution with u tanx and du 5602z will work EXtan7z5606z dx tan7xsec4xseczz dx tan7x 8662 seczw dx tan7x tan2x 12 5602z dx U7 u2 12 du Sub with u tan and du 5602xdx uM 2u9 u7 du U12 U10 us 7 27 7 O 12 10 8 tan z tan10 z tan8 z lt gt lt gt lt gt O 12 5 8 i 0320 I n nt L CM Pinio 9mm uh L Conchcm Mr 71 known 1V0 LR C 1 1 nl 3 L L L n H 1 Maul ns COSLCA C031K M AlJthHL r In 4 C 1 08 n on e H7554 ER 4 R1L 39 hquot a Ar q IUAh39fa or W AQKOth kI Q H k S M Nd1 No kA HAL r G va D Sr ku r 00 W 23 a M I 33 Z Eg L amp Ghuerfn LQ COM HS 95 A A
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'