Calculus II MATH 255
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Date Created: 10/29/15
MATH 255 Calculus II Trigonometric Substitution Section M Our objective To get zid of squaze mots of sums a2 b7 oi diffezences o2 eh in integiands by zeplacing them with equivalent tiigonometiic functions Hopefuiiy the iesuiting integiand wiii be one that is easiiy integiated bout u 39 in integiais t a t 39 t B oie we can stait discussing how to go a sin 39 ouiseives with some key concepts in geometiy and tiigonometiy Pythagorean s Theorem We know when deaiing with iight tiiangies that shoit side z shoit sidez2 hypotenuse2 oi if we abe oui hypotenuse as c and oui shoitei sides as a and i this becomes a2 h2 c7 onm this identity we get two possibie substitutions c o7 b7 01 as the guxe below illustxates chiazmg sic Witt seine Ram mic chiazmg hyacinth Witt seine Ram mic ism Ram uf Dim etseinai ism Ram etsim menus W C Figuze 1 Rewiiting Sides of the Right Iiiangie to inciude Souaie Roots of Sums oi Diffexences Trigonometric Identities hypomnug hyp sinw 2 cscs oppusiwlupp cos9 75 sec9 jig ab tan9 cot9 o 03 Pp solarium Using Trigonometric Substitution Remember we use trigonometric substitution to help us evaluate an integral with a square root of a sum or difference of perfect squares Consider using trigonometric substitution to evaluate the indefinite integral 1 dz x24 x2 Steps To Generally Follow A Speci c Example 1 Draw a right triangle 0 K 2 Identify the problematic square root term in your inte Problem term V4713 grand 1 If it has a sum of squares let the square root be the hypotenuse c Va b2 1 2 If it has a difference of squares let the square root be 9 one of the short sides a V02 7 b2 WK Then use the terms under the square root to determine the missing values on the other sides 3 There should be a single x term on one of the sides give or x is on opposite side and hypotenuse of take a constant Find the simplest trigonometric function 2 is most simple so sine relates these that incorporates the side with the at term in it and the side 31110 n with the next simplest expression 2 4 Solve the identity equation you just had for m x 2 sin0 5 Differentiate this equation so you have 126 in terms of 10 126 2cos0d0 6 If there are terms in the integral that we have not solved for We can replace x s with 2sin0 s We need to yet find a way to represent them in terms of our trigonomet represent 4 7 352 Note 0030 7V4 so ric functions V 4 i x2 2cos0 7 Make the substitutions back into the original integral DON T FORGET to sub in completely day 1 d 2 cos0 d0 x 244 7 2 2 sin6l2 2 cos0 8 Simplify the expression as much as possible 1 1 f d0 icsc20d0 4s1n 0 4 9 Complete the integration process you should end up with an answer in terms of 0 s 1 come O 4 10 You need to convert your solution to x s There are no 0 s by themselves just a cotangent 1 If there are any 0 s outside of trig functions solve your original equation relating x and trig functions for 0 2 If there are trig functions of 0 in your solution read your solution off of the triangle using the identities relating sides of the triangle to trig functions And cot0 L dj 7 x47m2 i 7 so opp m dwlv42o xgxZlixgi 4 DE MATH 255 Calculus II Derivative and Integral Formulas To Know By Heart Derivative Formulas Integral Formulas d 101 0 k dz kz C d 1 7 nzni1 n d 7 7 0 h z z 7 n 1 1 18111 1 608m cosz dz sinz C Cosml 78mm sinzdz coszC d 2 tan 1 sec m se02 z dz tanz C d 2 gleam CSC m cs02z dz 7cotzC Elsecml secmtanm secztanz dz secz C 105094 CSCQECOW csczcotzdz 7csczC kml 5m 5m dz 6m C d dm lnd dm az h az C a d 1 7 7 1 Wilma 35 dd mm i arcsin z 7 1 d 7 0 dm W W z 7 arcs1nz d 7 1 dag arccos z 7 H d t 1 1 7 are anz 7 dz 1 z2 dz arctanz C ibrccot z 1 dz 1 z2 1 7 arcsec z 7 1 dm lzlxzz 71 7 arcsec z O z z2 7 1 d 1 7arccsc z 77 dz z z2 7 1 Derivative Rules from Cale I d was m gm i M gmsz MATH 255 7 Calculus II An Introduction to Taylor Series and Taylor Polynomials 1 Deriving the Taylor Series and Polynomials for Let f be a function with derivatives of all orders throughout some interval about 0 We know from the Fundamental Theorem of Calculus that So7 solving this for x we nd that W NH 3 ds We can use integration by parts to expand fs d37 by choosing u fs d1 d3 du fusds U 87 Recall here7 that z is treated as a constant and s is our variable7 so this is a legitimate anti derivative of 13 So our equation for x is now ms M s e norat e Omlts e zgtf ltsgt ds f0xf gt7 fems ds 0 Applying integration by parts again with u f s d1 sizds du fHsds 1 and our equation becomes 1 f0 WHO 7 weens ds 7 2 N 7 m 7 2 N f0mf 0lts 2quot f Mime0 S 23 1 Ms 2 m 7 2 m flt0gtzfltogteltef ltogte 3 7 1 ads 0 Using integration by parts a third time with u fHs d1 013 du f4sds U and we see that m2 37mg m m 37mg rm fm w 7 ltmfem74 4 i Wawgt 73 m m simg ltJgffltmeA 4 i WadQ M i oifgii wws 2 3l 0 3l w MO MO We can continue on in this fashion and get what is known as the Taylor Series I H 2 m 1 f0f0f0gf 0 Note7 the Taylor Series lets us rewrite any function as some in nite polynomial where our coefficients of the polynomial are determined by the derivatives of 2 The Difference Between Series and Polynomial The Taylor Series generated by f at x 0 2 3 n W W now f 0gt f 0 f 0 Note that the Taylor Series gives an exact representation of the in nitely differentiable function f as a polynomial in terms of x We can lose the in nite quality of this representation if we look back and stop after the nth expansion by integration by parts We then have the equation M f0f 0zfquot07f 071 AEWJ W1Wd8 Hm flt0gt lt0gtz lt0gt ltngtltogt 1znltzgt where Rnx 71 for 53 f 1sds is called the remainder term and the other terms on the right hand side are called the Taylor polynomial of order n for the function 1 about the point x 0 This is still an exact representation of the function x but it is no longer an in nite representation In many cases the remainder term is small for x values near 0 and when n is a large number If we want to give up some of our exactness we can instead look at the Taylor Polynomial of order 71 Here we lose the remainder term that gave us that exactness and look only at the rst 71 terms of the Taylor Series Note also that we no longer say we have x but Pnx We call Pnx our Taylor Polynomial of order n and it has the following form The Taylor Polynomial of order n for f at 0 x2 x3 x PM W f 0m NO WW f 0g 3 Examples 1 Find the Taylor polynomial of order 5 for x em 2 Find the Taylor series for x em 3 Find the Taylor polynomial of order 7 for x sinx 4 Find the Taylor series for x sinx 4 Homework 1 Calculate by hand the Taylor polynomial of order 8 for x cos x 2 Find the Taylor series for x cos x 3 Calculate by hand the Taylor polynomials of orders 0 1 2 and 3 for x ln 1 4 Calculate by hand the Taylor polynomials of orders 0 1 2 and 3 for x x 4 5 Do the calculator worksheet on Taylor polynomials MATH 255 Calculus II Section 92 Geometric Series 1 In nite Series Partial Sums and Convergence As you know a sequence sn ih is an in nitely long list of terms numbers 317 527 537 547 Note that indexing may start with any integer but generally we use 0 or 1 We ll now introduce the mathematical concept of a series De nition 1 In nite Series A series is an in nitely long sum the form 00 Za a1a2a3a4an i1 indexing may start with any integer but generally we use 0 or 1 And we refer to an as the nth term of the series De nition 2 nM Partial Sum The nth partial sum of the series ai is denoted as 3 and we write 71 8naia1a2an i1 Thus sn is the sum of the rst n terms of the in nite series ie 81 a1 82 a1a2 83 a1a2a3 5n a1a2a3 an2il1ai De nition 3 Sequence of Partial Sums Since 3132333 are nite numbers as they are sums of a nite number of terms we can consider the sequence 81 32 33 sn as the sequence of partial sums De nition 4 Convergence of a Series Consider the series 00 Eat i1 If its corresponding sequence of partial sums 31 32 83 34 converges to some limit L then we say the in nite series converges and its sum is L ie lim sn L implies naoo 00 a1a2a3ancZanL n1 If the sequence of partial sums of the series does not conuerge we say that the series diverges 2 Geometric Series The geometric series is a particular series of the form 00 aararzarnil Ear n0 where a and r are xed numbers Examplel 3 3 3 0 1 3 Z Z n0 1 1 1 0 21ig i 7 n70 00 5 4740400747000 n0 So there are many possibilities for a and r some of which are easier to recognize as a geometric series than others Now we concern ourselves with the nth partial sum 3 Note7 if M 17 then ngtka if7 17 7 2 nili Sniaararar a 0 7 ifr71andnisodd7 ifr 71 and n is even Thus7 the sequence of partial sums diverges7 and hence the geometric series diverges If M 7 17 then 5n rsn snirsn sn17 r 3n Thus7 if M lt 17 r a as n a 007 and 3 a Else if M gt 17 then lrnl a and the series Thus If M lt 17 the geometric series a ar arz am converges to 00 a lt1 H w arnil 7 n0 If M 2 17 the series diverges NOTE These results depend on the fact that we start with n Ol Example2 3 3 1 W 1 1 3 Z3lt gt Z3lt gt n70 1 1 ltggt Zltggt n70 co 3 47404007410n2410 n0 00 4 12nzxn n70 provided that Homework pp 448 449 1 10 11 15 odd 28 31
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