MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln + C d = + C ln sin d = cos + C cos d = sin + C sec tn csc cot d = + C d = + C sec tn d= sec + C csc cot d= csc + C tn d = ln sec + C cot d = ln sin + C sec d = ln sec + tn + C d = sin + C d tn C = + + d = sin C d tn C + = + + d = sec + C
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Section 6. Integrtion y Prts Every differentition rule hs corresponding integrtion rule. For instnce, the Sustitution Rule for integrtion corresponds to the Chin Rule for differentition. The rule tht corresponds to the Product Rule for differentition is clled the rule for integrtion y prts. The Product Rule sttes tht if f nd g re differentile functions, then d f g = f g + g f d ( ) ( ) ( ) ( ) ( ) ( ) In the nottion for indefinite integrls this eqution ecomes or ( ) ( ) + ( ) ( ) = ( ) ( ) f g g f d f g ( ) ( ) + ( ) ( ) = ( ) ( ) We cn rerrnge this eqution s f g d g f d f g ( ) ( ) = ( ) ( ) ( ) ( ) f g d f g g f d This formul is clled the formul for integrtion y prts. It is perhps esier to rememer in the following nottion. Let u= f ( ) nd v g( ) = ( ) nd dv g ( ) du f d prts ecomes =. Then the differentils re =, so, y the Sustitution Rule, the formul for integrtion y udv = uv vdu Emple : Find cos d.
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 3 Our im in using integrtion y prts is to otin simpler integrl thn the one we strted with. Thus in Emple we strted with cos dnd epressed it in term of the simpler integrl sin d. If we hd insted chosen u = cos nd dv = d, then du = sin d nd v= /, so integrtion y prts gives Although this is true, cos d= cos + sin d sin dis more difficult integrl thn the one we strted with. In generl, when deciding on choice for u nd dv, we usully try to choose u f ( ) = to e function tht ecomes simpler when differentited (or t lest not more complicted s long s ( ) dv = g d cn redily integrted to give v. Emple : Evlute ln d. Emple 3: Find cosπ d.
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 4 Emple 4: Evlute e cos d.
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 5 If we comine the formul for integrtion y prts with Prt of the Fundmentl Theorem of Clculus, we cn evlute definite integrls y prts. Assuming f nd g re continuous, nd using the Fundmentl Theorem, we otin If we let u= f ( ) nd v g( ) ( ) ( ) = ( ) ( ) ( ) ( ) f g d f g g f d =, then we hve = udv uv vdu Emple 5: Evlute 0 cos d
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 6 Section 6. Trigonometric Integrls In this section we use trigonometric identities to integrte certin comintions of trigonometric functions. Strtegy for Evluting sin m cos n d () If the power of cosine is odd ( n k ) cos = +, sve one cosine fctor nd use = sin to epress the remining fctors in terms of sine: Then sustituteu = sin. k ( ) ( ) m k+ m sin cos = sin cos cos () If the power of sine is odd ( n k ) sin d d m = sin sin cos d = +, sve one sine fctor nd use = cos to epress the remining fctors in terms of cosine: k ( ) ( ) k+ n n sin cos = sin cos sin d d k n = cos cos sin d Then sustituteu = cos. (Note tht if the powers of oth sine nd cosine re odd, either () or () cn e used.) (c) If the powers of oth sine nd cosine re even, use the hlf-ngle identities ( ) ( ) sin = cos cos = + cos It is sometimes helpful to use the identity sin cos = sin k
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 7 3 Emple : Evlute sin cos d. Emple : Evlute the integrl. ( π ) ( π ) ( ) sin cos 3 d ( ) 5 ( ) π 4 ()sin() c 0 sin d ( ) sin cos d d d
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 8 Strtegy for Evluting tn m sec n d () If the power of secnt is even ( n, k k ) =, sve the fctor of sec = + tn to epress the remining fctors in terms of tn : Then sustituteu = tn. k ( ) ( ) m k m tn sec d = tn sec sec d () If the power of tngent is odd ( n k ) k m = tn + tn sec d sec nd use = +, sve the fctor of sec tn nd use tn sec = to epress the remining fctors in terms of sec : Then sustituteu = sec. k ( ) ( ) k+ n n tn sec = tn sec sec tn d d k n sec sec sec tn = d Emple 3: Evlute the integrl () () π /4 0 sec tn 4 5 3 sec tn d d
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 9 For other cses, the guidelines re not s cler-cut. We my need to use identities, integrtion y prts, nd occsionlly little ingenuity. We will sometimes need to e le to integrte tn y using the following formul: tn d= ln sec + C We will lso need the indefinite integrl of secnt: Emple 4: Find () sec d () 3 tn d sec d= ln sec + tn + C
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 0 Integrls such s the one the preceding emple my seem vry specil ut they occur m n frequently in pplictions of integrtion. Integrls of the form cot csc d cn e found y similr methods ecuse of the identity+ cot = csc. Finlly, we cn mke use of nother set of trigonometric identities: To evlute the integrl () sin m cos n d, () sin msin n d, or (c) cos m cos n d, Use the corresponding identity: ( )sin Acos B= sin ( ) sin ( ) A B + A+ B ( )sin Asin B= cos( ) cos( ) A B A+ B ( c)cos Acos B= cos( A B) + cos( A+ B) π / Emple 5: Evlute sin 4 cos d. 0
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Section 6.3 Trigonometric Sustitution In finding the re of circle or n ellipse, n integrl of the form > 0. If it were stnds, d, the sustitution d rises, where u = would e effective ut, s it d is more difficult. If we chnge the vrile from to θ y the sustitution = sinθ, then the identity sin θ = cos θ llows us to get rid of the root sign ecuse ( ) ( ) = = = = sin θ sin θ cos θ cosθ In generl we cn mke sustitution of the form g( t) = y using the Sustitution Rule in reverse. To mke our clcultions simpler, we ssume tht g hs n inverse function; tht is, g is one-to-one. In this cse, if we replce u y nd y t in the Sustitution Rule, we otin ( ) ( ) ( ) = ( ) f d f g t g t dt This kind of sustitution is clled inverse sustitution. We cn mke the inverse sustitution = sinθ provided tht it defines one-to-one function. This cn e ccomplished y restricting θ to lie in the intervl[ π /, π /]. In the following tle we list trigonometric sustitutions tht re effective for the given rdicl epressions ecuse of the specified trigonometric identities. In ech cse the restriction on θ is imposed to ensure tht the function tht defines the sustitution is one-toone.
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi TABLE OF TRIGONOMETRIC SUBSTITUTIONS Epression Sustitution Identity π π = sin θ, θ + π π = tn θ, < θ < π 3π = sec θ, 0 θ < or π θ < sin θ = cos θ + tn θ = sec θ sec θ = tn θ Emple : Evlute 9 d
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 3 Emple : Evlute + 4 d
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 4 Emple 3: Evlute d
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 5 Emple 4: Evlute 6+ 3 d
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 6 Section 6.4 Integrting Rtionl Functions y Prtil Frctions In this section we show how to integrte ny rtionl function ( rtio of polynomils) y epressing it s sum of simpler frctions, clled prtil frctions, tht we lredy know how to integrte. To illustrte the method, oserve tht y tking the frctions ( + 3) nd ( ) to common denomintor we otin ( ) ( ) ( )( ) + 3 7 = = + 3 + 3 + 6 If we now reverse the procedure, we see how to integrte the function on the right side of this eqution: function 7 d = d = ln + 3 ln + C + 6 + 3 To see how the method of prtil frctions works in generl, let s consider rtionl f ( ) ( ) ( ) P = Q where P nd Q re polynomils. It s possile to epress f s sum of simpler frctions provided tht the degree of P, denoted s deg(p), is less thn the degree of Q, denoted s deg(q). If f is improper, tht is, deg( P) deg( Q), then we must tke the preliminry step of dividing Q into P (y long division) until reminder R() is otined such tht deg( R) < deg( Q). The division sttement is ( ) ( ) where S nd R re lso polynomils. Emple : + 6 d + 4 ( ) ( ) P R f ( ) = = S( ) + Q Q
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 7 The net step is to fctor the denomintor Q( ) s fr s possile. It cn e shown tht ny polynomil Q cn e fctored s product of liner fctors (of the form + ) nd irreducile qudrtic fctors (of the form 4 ( ) Q = 6, we could fctor it s + + c, where 4 ( ) = 6 = ( + 4)( 4) = ( + 4)( + )( ) Q The third step is to epress the proper rtionl function R( ) / ( ) prtil frctions of the form A A+ B or i ( + ) ( + + c) CASE I: The denomintor Q( ) is product of distinct liner fctors. This mens tht we cn write ( ) = ( + )( + ) ( + ) Q k k j 4c< 0). For instnce, if Q s sum of where no fctor is repeted (nd no fctor is constnt multiple of nother). In this cse the prtil frction theorem sttes tht there eist constnts A, A,..., A such tht k ( ) ( ) R A A Ak = + + + Q + + + k k () Emple : Evlute + d 3 + 3
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 8 CASE II: Q( ) is product of liner fctors, some of which re repeted. Suppose the first liner fctor ( + ) is repeted r times; tht is, ( ) fctoriztion of Q( ). Then insted of the single term A ( ) would use r + occurs in the + in Eqution (), we A A Ar + + + + + + ( ) ( ) r () By wy of illustrtion, we could write Emple 3: Evlute 3 + 4 A B C D E = + + + + ( + ) + ( + ) ( + ) 3 3 4 3 + + 4 + d
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 9 CASE III: Q( ) contins irreducile qudrtic fctors, none of which is repeted. If Q( ) hs the fctor + + c, where 4c< 0, then, in ddition to the prtil frctions in Eqution () nd (), the epression for R( ) Q( ) will hve term of the form A + B + + c (3) where A nd B re constnts to e determined. For instnce, the function given y ( 3)( + )( + ) hs prtil frction decomposition of the form A B+ C D+ E = + + ( 3)( + )( + ) 3 + + The term given in (3) cn e integrted y completing the squre nd using the formul d = tn + C + Emple 4: Evlute + 4 3 + 4 d
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 0 CASE IV: Q( ) contins repeted irreducile qudrtic fctor. Q hs the fctor ( c) If ( ) + +, where r 4c< 0, then insted of the single prtil frction (3), the sum A + B A + B A r + Br + + + + + c + + c + + c ( ) ( ) r (4) occurs in the prtil frction decomposition of R( ) Q( ). Ech of the term in (4) cn e integrted y first completing the squre. Emple 5: Evlute + 3 d ( + )
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Section 6.5 Improper Integrls In defining definite integrl f ( ) d we delt with function f defined on finite intervl [, ] nd we ssumed tht f does not hve n infinite discontinuity. In this section we etend the concept of definite integrl to cse where the intervl is infinite nd lso to the cse where f hs n infinite discontinuity in [, ]. In either cse the integrl is clled n improper integrl. TYPE I: Infinite Intervls Definition of n improper integrl of type I t d eists for every numer t, then () If f ( ) ( ) = lim ( ) f d f d t t provided this limit eists (s finite numer). d eists for every numer t, then () If f ( ) t ( ) = lim ( ) f d f d t t provided this limit eists (s finite numer). The improper integrl f ( ) d nd f ( ) limit eists nd divergent if the limit does not eist. (c) If oth f ( ) d nd f ( ) d re clled convergent if the corresponding d re convergent, then we define ( ) = ( ) + ( ) f d f d f d In prt (c) ny rel numer cn e used.
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Emple : Determine whether the integrl d is convergent or divergent.
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 3 Emple : Evlute 0 e d.
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 4 Emple 3: Evlute d +
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 5 Emple 4: For wht vlues of p is the integrl d convergent? p
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 6 TYPE II: Discontinuous Integrnds Definition of n improper integrl of type II () If f is continuous on [, ) nd is discontinuous t, then ( ) = lim ( ) f d f d t t if this limit eists (s finite numer). () If f is continuous on (, ] nd is discontinuous t, then ( ) = lim ( ) + f d f d t if this limit eists (s finite numer). The improper integrl f ( ) divergent if the limit does not eist. t d is clled convergent if the corresponding limit eists nd (c) If f hs discontinuity t c, where c convergent, then we define c ( ) = ( ) + ( ) c < <, nd oth f ( ) d nd f ( ) f d f d f d c d re c Emple 5: Find 5 d
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 7 Emple 6: Determine whether π / sec dconverges or diverges. 0
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 8 d Emple 7: Evlute if possile 3 0
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 9 Emple 8: Evlute 0 ln d