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## Calculus

by: Otilia Murray I

22

0

35

# Calculus MAT 021B

Otilia Murray I
UCD
GPA 3.88

Qinglan Xia

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COURSE
PROF.
Qinglan Xia
TYPE
Class Notes
PAGES
35
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 35 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 021B at University of California - Davis taught by Qinglan Xia in Fall. Since its upload, it has received 22 views. For similar materials see /class/187344/mat-021b-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15
84 Trigonometric Integrals Idea the general idea is to use identities to transform complicated trigono metric integrals into integrals that are easier to work with Products of powers of sines and cosines Evaluate sinm x cosquot xdx where m n are integers Key identity sin2x oos2x 1 Case 1 If m ie the power of sin x is odd then use the substitution u cos x sinm x 008quot xdx sing1 x cosquot mix sin x 008quot x sin xd k l 0082 x cosquot x sin xdx k 1 152 undu by setting to oosx So the integrand becomes a polynomial or a rational function Example 2 Cos xdx smx sinm x 008quot xdx sing1 x cosquot mix sin x 008quot x sin xd k l 0082 x cosquot x sin xdx k 1 152 undu by setting to oosx So the integrand becomes a polynomial or a rational function Example 2 Cos xdx smx 2 98 xd sum COSZLE sin2 U2 1 u sin xdx 2du by setting u cos x 1 1 1121 m d 1 u 1u u 1 1 u 1n1 u ln1u0 U 0 111 u n 2 1u 11 u n 2 1 008x 1n 1 u 1 U 1 008x 1 COS x2 1 cos2x Sl C sin LE 2 1 Cosx 1 Cosx 1 COS x LE 2 0 0 0 11 u n 2 1 008x 1n 1 u 1 U 1 008x 1 COS x2 1 cos2x Sl C sin LE 2 1 Cosx 1 Cosx 1 COS x LE 2 0 0 0 More generally sm2k1xf COS x dx stkf COS x sin xdx k l C082 x f COS x sin xd 2 k 39 1u fudubysett1ngueosx Example Sm3lt gt3COS5LEE4COSLE4CZ 1 2008 x2008x More generally sm2k1xf COS x dx stkf COS x sin xdx k l C082 x f COS x sin xd 2 k 39 1u fudubysett1ngueosx Example Sm3lt gt3COS5LEE4COSLE4CZ 1 2008 x2008x Let u cos x then 3 Boos5x 4oosx 4 sm x cm 1 2oosZ QCOSLE 3 5 4 4 l 2152 2n It becomes an integral of a rational function So we may use the method of partial fractions to solve it Evaluate sinm x cosquot xdx Case 2 If n ie the power of cos x is odd then use the substitution u sin x sinmxcosquot xdx smmxcos g1 xdx sm x 1 sm x cos xdx m 2 k u 1 u du by setting to smx Example 30s3 88m xdx Evaluate sinm x cosquot xdx Case 2 If n ie the power of cos x is odd then use the substitution u sin x sinmxcosquot xdx Si m0082kl xdx m 2 k sm x 1 sm cos xdx m 2 k u 1 u du by setting u smx Example cos3 88m mdx Let u sin x then C083 xesm xdx 1 z eudu 1 U2 8 Queudu by using integration by parts 1 u2eu2ueu 2 uC u22u 1eu0 u 12euC sinx 12esmxC Case 3 If both m and n are even that is m 2k n 2l Then we substitute 1 2 1 2 sin2x and oos2x w to reduce the integrand to one in lower powers of oos 2x sin2k x 308 xdx 1 oos2x k1oos2xl lt 2 gt lt 2 gt are Example sin4 x oos2 xdr Case 3 If both m and n are even that is m 2k n 2l Then we substitute 1 2 1 2 sin2x and oos2x w to reduce the integrand to one in lower powers of oos 2x sin2k x 308 xdx 1 oos2x k1oos2xl lt 2 gt lt 2 gt are Example sin4 x oos2 xdr OOIHOOIH sin 4 x C082 xdx 21C082 1 COS 2x 2 2 1 2 COS 2x C082 2x 1 COS 2x dx 1 cos 2x C082 2x 3083 2x dx 4 sin x C082 xdx 1 Cos2x 21C082 dx 2 2 1 2 COS 2x C082 2x 1 COS 2x dx 1 cos 2x C082 2x 3083 2x dx C082 2xdx 1 4 0208 xd sin 4 C 8 OOIHOOIH 2 w NIH Q 5 C083 2xdx 2 sm3 2x 1 sin2 2 COS 2x dx 1 U2 du by setting u sin 2x 3 u C 6 C 2 6 Therefore sm4 x C082 xd 1 g 1 COS 2x C082 2 C083 2x d 1 sin2x sm4x sin2x sin32x 8 x 2 2 8 2 6 1 x sin4x sin32x 0 0 4 3 Eliminating square roots z 1 303 Eliminating square roots Example V1 COS 4xdx 0 We use the identity 1 26 C082 6 Eliminating square roots Example I v1 COS 4mm 0 We use the identity 2 1 COS 26 COS 6 2 1 v1 COS 4xdx 0 1 15 20082 2xdx COS 2x dx 0 0 2 7r 2 2 sm 2x 6I g sing sin 0 Integrals of powers of tan m and sec x Here the key identity is 2 sec x 1 tan2 x or tan2 x sec2 x 1 Example Evaluate tan3 x sec xd Integrals of powers of tan a and sec m Here the key identity is 2 sec x 1 tan2 x or tan2 x sec2 x 1 Example Evaluate tan3 x sec xdx tan3 x sec xdx tan2 x tan x sec x dx sec2 x 1 tanxsec x dx Let u sec x then du sec x tan mix Then C l w U CD 000 tan3xsecxdx X i 1ltUM1S Cgtd U2 1 du uC 3x ampmx C tan3 x sec xdx sin3 1 3 dx COS x COS x 3 sin 4 dx COS x 1 U2 4 du by settmg u cos x u u2 u4du 1 1 C u 3103 secx sec3x0 Products of Sines and Cosines To evaluate integrals sin mx sin mum sin mx cos nx and cos 77 cos nxdx we use the identities 1 sinmx sinnx 5 cosm n x cos m nx 1 sin mx cos nx 5 sin m n x sin m n x l cosmx cosnx cosm nxcosmnx Example Evaluate sin 5x cos Bxdx Products of Sines and Cosines To evaluate integrals sin mx sin mum sin mx cos nx and cos 77 cos nxdx we use the identities 1 sinmx sinnx 5 cosm n x cos m nx 1 sin mx cos nx 5 sin m n x sin m n x l cosmx cosnx cosm nxcosmnx Example Evaluate sin 5x cos Bxdx sin 5x COS Bxdx Si lt2gt31 lt8gtd 1 cos2x 208 0 5 2 8 1 1 ECOS8LE ZCOSQC 81 Basic Integration Formulas TABLE 81 Basic integration farmulas 1 duuC 13 footndulnlsinu 5 2 f rdu In C any numberk quotn lcscul C 3 dudududv H Um 15 fuquotduhu1 EC agt0al 4uquotdun C nv il l6 sinhuducushuC 14 Ham a C 5 mu c l7 fcosliudu sinth C 6 sinudu cosu C 7 cosmlu sinu c 1839 5 G C s seczudu tanu C 1939 innquot C 9 csczudu 1m c 2039 59 t C 10 seculanudusecuc 2139 S quoth C ugt0 3 c ugtugt0 12 lanudu ln lcosul c 22 coshquot 11 cscumludu cscu C 1 F a lnsccu C For more complicated integrals if possible we use techniques to rewrite them into these standard forms 1 Substitution fgxgxd fudu by setting u Example Evaluate l ez2lnz2 x Practice 2 Completing the square Example Evaluate Practise 3 Using trigonometric identities Basic trigonometric identities Example 4 Using long division to reduce an improper fraction Example de 95V 1 lln2 x dx V2 4x x2 4 de 2 x2 6x 10 singx cos2x 1 sech tatan l 2 l cos2x cos x 2 l 2 sin2 05 Cs m sin2x 2 cos x sin 05 0 l cos tdt 2x2 7x4 x2 dx 5 multiply by a form of l 0 Example Separating fractions Example Practise Example Example Example Example sec xdx x2x ld x 2xxx l i 2 2 8056 0 xsecx2 5dx HZ ln xdx x 4051112 05 85 Trigonometric Substitution If an integrand involves the terms Va2 x2 Vagm2 or x2 a2 for a gt 0 we would like to use the trigonometric substitution Case ll V02 172 Letxasin6where 363 then 1 a2 x2 a2 a2 sin2 0 a2 l sin2 6 ll IE i Va2cos20acos6 I a sin H Example Evaluate mam Example Ewiuate Example 3 Fmd ms area mclusedbyme empse Lellatenwnhrgltltgthen xa212 a2aztenz azseczsasecsasecs u a an 0 Example Evaluate 1 dx V9 052 HI Va a2 V 1 1f 4 Jig Ll Letxase060 0ltor7T 0lt37 Then A wxg a2a2se020 a2 a tan0 atanO a r 1 SEC I Example Evaluate

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