Rulelpatternl. Justification. Method. .sk/n+xtosxtstnx-to. Formula. Xswxtasxtc ) [$(&)( & ] = f (x)g(x) + f(x)g (x)

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1 Chapter 7 is concerned with all the integrals that can t be evaluated with simple antidifferentiation Chart of Integrals on Page Integration by Parts Like with the Chain Rule substitutions with u and du, we re going to look at integration by considering derivatives Let s look at the Product Rule:!!" [$(&)( & ] f (x)g(x) + f(x)g (x)!!" [$(&)((&)]*& (f (x)g(x) + f(x)g (x)) dx f(x) g(x) [$ + & ( & *& + or / $ & ( + & *& $ & ( + & *& f(x)g(x) [$ + & ( & *& ] Justification Integration by Parts:, *, *, Ex: Find & 123 & *& Let Rulelpatternl : Method Formula u x è du dx dv cos x dx è v sin x Sudv uvsvdu & 12& a & *& x sin x 345 & *& x sin x + cos x + C How can we check our answer? Xswxtasxtc ) ( sk/n+xtosxtstnxto 1

2 In general, we want to choose the expression that will simplify when differentiated to be u Ex: Find ln & *& Let: u ln x è du 8 " " * " better dv dx è v x fudv Sometimes you have to use the technique more than once Ex: Find 9 : ; < *9 uv xlnx Slnxdx xlnx fvdu Sdx f x ( Idd xlnx t ee#iimtel x + C Check : Differentiate 1 In t X xt * " 1+0 Sometimes it s not obvious that you re making any headway Ex: Find ; " 345 & *& (Hint: Use Integration by Parts twice, and then fold the equation)

3 fettdt u Eet Eet dv etdt z[ ztett feet dt Eet Setdt t du dt v et tet tet fetdt et + C ] 2 et t ( Set Ztdt Eet zfettdt

4 fexsinxdx dycos d dve dx e swx Utslnx fexcosxdx ye uwsx dy svcdu [vu ) ] fexsinxdx Iexsnx e sn e sw [e wsx Se swxdxe snx tfexsnxdx [ excosx dve dx Ve say )] fexfsnxdx exwsxtfexsnxdx ] exasx fexsnxdx + fexsinxdx #e sn dxexe wsxtc

5 fexsinxdx I [ exsinx e ws:

6 Ex: What is a Reduction Formula? Ex: ("#$) > '$ $("#$) > # ("#$) >?8 '$

7 72 Trigonometric Integrals First, some review: Trigonometricpalooza cos 2 x 1 sin 2 x sin 2 x 1 cos 2 x cos 2 x ½ (1 + cos 2x) sin 2 x ½ (1 cos 2x) sin A cos B ½ [sin (AB) + sin(a+b)] sin A sin B ½ [cos (AB) cos(a+b)] cos A cos B ½ [cos (AB) + cos(a+b)]!!"!!" /#$ 01$ 01$ /#$!!" 23#$ 40: $!!" 40$ 40$ 23#$ 23#$ '$ ln sec $ + > 40$ '$ ln sec $ + 23#$ + >

8 Ex: Find /#$ 01$ '$ Steps : AO Simple anti derivative? BO u or Integration Trig sub by Parts U SINX du wsxdx fudu +c (sy I+( Sw + ( Ex: Find /# : $ 01$ '$ U SIN X dv fidu cosxdx tc S N3X 3 % ( sin XP 3 t C + C

9 Slnx cos x dx U cos D Slnxdx du swxdx S ul du ) fudu + C cosy t C Arrrr! Question sw +c, wzs# Cz I +

10 sw +c, E os +c, Sihtx I cos ' sn + C, 10 + tco# C + C,, 6 + It C, SINZ + cos 2 I < #i '

11 S ' a Snx X osn Il do cosxdx X, 0 UsiNOO Sjudu ]jtz o } tswxwsxdx du D awsx Snxdx * o uqs0 sinxdx * U costs f uth Ie+

12 Ex: Find!"# G % '% U wsx du snxdx f v3 oh no! fwixdx fwix wsxdx cos ' X I smix S ( l sin fcosxdx ' x ) wsxdx fsnlxwsxdx Ex: Find!"# G % #() : % '% S cos ' xsinixwsxdxdu Want u SINX I U? u? I s N2X S w2xwix Hutu f ( I i ) i du ]du sn +C sn# + ( or fa SNX)SINK Gsxdx ssnixca#af:s:;x*iftfiiiy I sinasn# + (

13 fcosdx fsinxwsxdx SINX fslntxasxdx i SINX du wsxdx siwx Suh sinx tc SINX Sl + C Check whhe derivative cosx 3sgNI cosx

14 cos 35k cos cos SNZX cos cos x ( I s/n2 x ) cos cos ' x cos 3 x

15 Ex: Find!"# : % '% (Hint: Use the halfangle formula cos 2 x ½ (1 + cos 2x) ) A SI ( 2x)dx It cos ftzioshdx ftzdx + Ex + I f cos Zxdx Ix u2 du Zdx G d + I fcosu Ex + I fcosudu ] lz + SINU + C I + sin Zxt C

16 Ex: Find ()* G % '% (Hint: Use tan 2 x sec 2 x 1, plus the fact that ()*%'% ln sec % + 4 ) f ( sedx anxtarxdx Sseixtanxdx Dtarxdx ftonxdx S Ik Feeney Ex: Evaluate #5*4% cos 5% '% Ttfroduct (Hint: Use one of the product rules: sin A cos B ½[sin(AB) + sin (A+B)] lnlseexkc ln/se [ sin (4 5,1) + Sw (4 +5 5) dx Ingle bnction 9 I [ swfx ) + sin (9x)]dx I [ cost ) 6s{a I ] + C zlcosfx tzwstx ) ) tacos * wsla# + ( 9x ) C OR

17 73 Trigonometric Substitution (Note: Triangles are optional Use them if they help, but skip them if they don t) Ex: Find 9 % : % '% 09 2 du Zxdx xdx sure ṫfnudu E) u du tz +c I } Ftc st u tc th xy +C

18 Ex: Find 9 % : '% Oh no! We can t use a substitution to solve this In our desperation, we try something ridiculous: Let x 3 sin ; ( M : ; M : ) then dx 3 cost do and 9 % : 20 's Inter FY as So the integral becomes 9 % : '% 9 9#5* : ; 3!"#; '; 9!"# : ; 3!"#; '; 3 cos ; 3!"#; '; 9!"# : ; '; 9 8 : ( 1 + cos 2;) '; R : [ ; + ½ sin 2;] + C

19 R [; + ½ * 2 sin ; cos ; ] + C : Note that x 3 sin ;, so sin ; " G so ; arcsin " G and cos ; 1 #5* : ; 1 "S R : [ arcsin " G + " R kswo+wiol 9 % : ] + C R R : arcsin " G + 8 : % 9 %: + C fed x But are we allowed to do this? Yes, as long as we limit ; so that x 3 sin ; is onetoone This is called an inverse substitution

20 if 4sno d 4coso th M6F t 42 X 1 SINO Sc NO dx#lt6txh 4tanodX 3secOdx3 or of 4 set 0 9 secotano

21 74 Integration of Rational Functions by Partial Fractions This section deals with integrals of rational expressions The first type we ll look at are those of the form R(x) P(x)/Q(x) where deg (P) > deg (Q) (improper fraction) Ex: Find G<?: <T8 '( Our first step is to divide the expression, using techniques from Math 90 t + I3 ] earthed I 3 + II So, G<?: <T8 '( [3 U <T8 ]'( s 3t 5 ln t+1 + C " s aty/sf, Ut + I do dt 5 ) due 5 In H + C

22 The second type of rational equation is one in which the denominator can be completely factored: Ex: Find " S T :"?8 :" V T G" S?:" '2 c degree 2 T degree :3 First, check to make sure that it s not improper Then, try factoring the denominator: I X( ) xik t2 ) zx tz Next, write the rational expression as a sum of simpler fractions: " S T :"?8 W + X + Y :" V T G" S?:" " :"?8 "T: or X3±#t s orz The idea is to simplify the problem so that the integrations will be easier, but finding A, B and C is going to take some work

23 One way to find A, B and C is to clear fractions in this equation: " S T :"?8 W + X + Y :" V T G" S?:" " :"?8 "T: C tfi#ytxfzxfxxxt2xr+zxiaxeiifeitxb*ytk*yixitxfxxza(2x73xd+b(x2+zx)tc(2x2x)x2 ZAx2t3Ax 2A / Ax x2 3A t ZBX Cx Zx 2A 1 2A + BTZC I 1 3A + ZB C 2A A± 1 2 (E) + Bt2C 1 I + B + 2C 1 B+2C0 2 " E f E o 5131 B st

24 ' Bs± +2<0 ZC c st to A±,B 's,c to fxxkie#dxfetieixtefxtzfdxx+sthdxfofdxezt1n/x/ duzd wx+z dwd dfdx 'o os u + Is

25 it hh + to If few to I I In to 1h14 kµk In to lnkx to to lys A In B 2 12* In In 1 +21

26 Ex: Find " Z?:" S T["T8 " V?" S?"T8 *+ First, divide, since the degree on top is greater than on the bottom: x?xix+i Foy x * IEyIj I i Next, factor the denominator: 4x 4x x3xtx+ifkxd # it 4x_ 4x_ Grouping B ( * DC*Dl*D (xnxxdz We notice that the linear factor (x1) appears twice, which means that we ll have the partial fraction decomposition: [" W + X + Y "?8 S ("T8) ("?8) ("?8) S ("T8) i* * en nix +n ii i4 IY t II,

27 Find A, B and C: A 2 AtBxtBtCxZZCx+C atk#4xcxdyxtdax2+cx2ox2a+cobx2cx4xb2c4 At Btc 0 At OB+CO '3 Eiii 01 :# FI At CO A C B 2C4 B2Ct4 At Bt CO (e) + ( zct4)tco Ct2ct4+CO

28 Ct 2 Ct 4 + C 0 4 Ct 4 4 C 4 0 C 1 A C ) A I D 2C +4 B 2+42 A I, D 2, C 1 that

29 and then integrate: S ( + # that, )d + x + In YET, In k#

30 k# dx U 1 du dx 21*525 zfuidu 2 + C It C It + C

31 75 Strategy for Integration FourStep Strategy for Tricky Integrals: 1) Simplify the integrand, if possible, by multiplying out, or rewriting formulas: Ex: "(1 " + ") (" Sxta xtxtydx Kx±x +x }x ' )d x±ttx2+c Ex: <\>] ^_`S] (1 u tone seeo an seioao g u oe 0n" ftanotazodofsonsooncoszodo fudu Nowsada usno +C dws0d0 sn?o+c

32 2) Use a u substitution Ex: " 8?" S (" U / do Zxdx It 2 xdx SE s s tzlnly + C Ekkxytc Ex: sin 4" (" U4x du4dx df dx SINU fsinu do ytfwsu ) + C cos 4x + C

33 3) Consult one of the Four Horsemen of the Calcupocalypse a) Trigonometric Functions from 72 b) Rational Expression/Partial Fraction Decomposition 74 c) Integration by Parts 71 d) Radicals 73 (if they aren t easily dealt with using a u substitution) 4) The book helpfully suggests you try again at this point Ex: Find 9 " (" There s a cool trick to this one let u " Then u 2 x, so 2u du dx so the integral becomes 9 b 2< (< 2 9 < < (< u D du±,x d 2dux dx 2 x±dud UI Zudu dx 2udud which we can integrate by parts: 2 feoudu 2[uea Wu dyjdu dteudaeu ' 'da]2uet2eu2dee2e*tc Seo + C

34 76 Integration Using Tables and Computer Algebra Systems This section has been somewhat obviated by the appearance of online integral calculators that even show their steps, but here is the procedure anyway: Ex: Use the Table of Integrals to find " S U?[" S (" (See Reference Pages 6 10, at the back of the book) On Reference Page 7, #34 reads b S!b \ S?b S b : A : < : + \S : BCD?8 b \ + E How do we use this formula in our case? u 2x UI 4 2 du EE late a 2 tse#tsfef2o+eswt Etc ] If Exit E sin ' 2 + C ]

35 st [ x Et + E sin ' n+c] text + Is in " p t C

36 77 Approximate Integration Sometimes the exact value of an integral cannot be calculated, either because the antiderivative is difficult or impossible to find, or because a function is defined by data points and lacks a formula As long as we can evaluate f(x) at various values of x on the interval [a, b], we can use Riemann sums to approximate c \! " #" Midpoint Rule You might recall from before that we looked at left endpoint approximations and right endpoint approximations, and then discussed the fact that using the midpoints in each subinterval seemed to give us the best results:

37 The Midpoint Rule c \! " #" M n "! " 8 +! " : + +! " > ] where " c?\ > and " f ½ (x i 1 + x i ) midpoint of [ x i1, x i ] If f (x) K for a x b, then the error, E M, in using the Midpoint Rule can be calculated as: E M g(c?\)v :[> S

38 Ex: Use the Midpoint Rule with n5 to approximate : 8 8 #" " : 8 8 (Note: #" ln 2 ln 1 ln 2 " ) n 5 " :?8 U 02 Intervals are: [1, 12], [12, 14], [14, 16], [16, 18] and [18, 2] Midpoints are: 11, 13, 15, 17, 19 Estimate G 8U 8i 8R What s the actual error in using the Midpoint Rule in this case? Error What is the predicted upper bound on the error? f(x) 8 " x1 f (x) x 2 f (x) 2 x 3 on [1, 2], f (x) 2 (1) 3 2 E M : (:?8)V : :[(U) S jkk

39 The Trapezoidal Rule Another way to estimate integrals is to average the values obtained by looking at the left and right endpoints of each interval The Trapezoidal Rule c \! " #" T n " :! " k + 2! " 8 + 2! " : + +2! " >?8 +! " > where " c?\ > and x i a + i " If f (x) K for a x b, then the error, E T, in using the Trapezoidal Rule can be calculated as: E T g(c?\)v 8:> S

40 Ex: How large should we take n in order to guarantee that the Trapezoidal Rule!" approximation for is accurate to within 00001? : 8 Remember that, from above, we calculated that f (x) < 2 on [1, 2] "

41 Simpson s Rule The third technique for approximating integrals comes from the idea of using parabolas, instead of lines, to approximate the function values (See page 512 for the derivation) The formula is as follows: Simpson s Rule c \! " #" S n " G! " k + 4! " 8 + 2! " : + 4! " G +2! " >?: + 4! " >?8 +! " > where " c?\ > If f (4) (x) K for a x b, then the error, E S, in using Simpson s Rule can be calculated as: E S g(c?\)n 8ok> Z

42 Ex: Use Simpson s Rule with n 10 to approximate n 10 " c?\ > 8 8k 01 : 8!" " S 10 k8 G!(2)! 1 + 4! ! ! ! ! 19 +

43 78 Improper Integrals There are two different categories of integrals which are considered improper Type 1 are integrals where one or both limits are infinite Ex: r 8!" " S The only way to evaluate this integral is to use a limit: r 8!" " S <!" < r 8 " S lim lim[?8 ]T < r " 1 lim[?8 + 1] < r < lim < r 1 8 < 1 Ex: r 8!" " <!" < r 8 " lim lim < r [ln "] T 1 lim < r [ln T ln 1] lim < r [ln T] So, what happened? Why did one converge and the other diverge?

44 In the case of y 1/x, the values simply don t get small enough fast enough for the area to converge Ex: r " G?r X?"Z dx In order to evaluate this integral, we need to pick some value c to break this integral into two pieces We can pick any value, so let s pick c 0 r " G?r X?"Z #" k " G?r X?"Z #" + r " G k X?"Z #" k " G?r X?"Z #" lim <?r k " G < X?"Z #" lim [?_vwz <?r [ ] 0 T lim [?8?8 ]?8 0 <?r [ [_ xz [ r " G k X?"Z #" lim < r < k " G X?"Z #" lim [?_vwz < r [ ] T 0 lim < r [?8 [_ xz?8 [ ] 8 [ r " G?r X?"Z #" 8 [ +?8 [ 0

45 r 8 8 #" " y is convergent if p>1 and divergent if p<1 Type 2 are integrals where the functions are discontinuous on the interval [a, b] Ex: Evaluate G k!" "?8 Why is this integral improper? 1/(x1) is discontinuous at x 1 G k!" "?8 8 k!" "?8 + G 8!" "?8 8 k!" "?8 lim < 8 v < k!" "?8 lim < 8 v < k!" "?8 lim < 8 v ln " 1 T 0 lim ln " 1 ln 1 < 8 v (It diverges) Comparison Theorem Suppose that f and g are continuous functions with f(x) > g(x) > 0 for x > a r a) If! " #" \ r b) If _ " #" \ r is convergent, then _ " #" \ r is divergent, then! " #" \ is convergent is divergent