Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln ln + c + tn + c sin cos + c sin + c cos sin + c sec + c Proof: Ech of these equlities is esy to verify by tking the derivtive of the right side. For emple, we hve ln ln +c since d ( ln ln + ln, nd we hve tn ln sec + c since d sec tn (ln sec tn, nd we hve sec sec ln sec +tn +c since d (ln sec +tn sec tn + sec sec. sec + tn 8. Emple: Find 4 5. Solution: By the Fundmentl Theorem of Clculus nd Linerity, we hve 4 5 4 8.3 Emple: Find π/6 3/ 5 / π/3 π/6 sin + cos 3. Solution: We hve π/3 ] π/3 sin + cos 3 cos + 3 sin 3 ] 4 5 5/ / ( 64 5 ( 5 5. π/6 ( 4 + ( 4 + 3 6.
Substitution 8.4 Theorem: (Substitution, or Chnge of Vribles Let u g( be differentible on n intervl nd let f(u be continuous on the rnge of g(. Then f(g(g ( f(u du nd b f(g(g ( g(b ug( f(u du. Proof: Let F (u be n ntiderivtive of f(u so F (u f(u nd f(u du F (u + c. Then from the Chin Rule, we hve d F (g( F (g(g ( f(g(g (, nd so f(g(g ( F (g( + c F (u + c f(u du nd b f(g(g ( ] b F (g( F (g(b F (g( ] g(b g(b F (u f(u du. ug( ug( 8.5 Nottion: For u g( we write du g (. More generlly, for f(u g( we write f (u du g (. This nottion mkes the bove theorem esy to remember nd to pply. 8.6 Emple: Find + 3. Solution: Mke the substitution u + 3 so du. Then + 3 u/ du 3 u3/ + c 3 ( + 33/ + c. (In pplying the Substitution Rule, we used u g( + 3 nd f(u u u /, but the nottion du g ( llows us to void eplicit mention of the function f(u in our solution. 8.7 Emple: Find e. Solution: Mke the substitution u so du. Then e eu du eu + c e + c.
ln 8.8 Emple: Find. Solution: Let u ln so du. Then ln u du u + c (ln + c. 8.9 Emple: Find tn. Solution: We hve tn sin. Let u cos so du sin. Then cos sin du tn cos ln u + c ln cos + c ln sec + c. u 8. Emple: Find +. Solution: Let u so u nd u du. Then u du + u + u du u +. Now let v u + do dv du. Then du + u + v dv ln v + c ln u + + c ln( + + c. 8. Emple: Find +. Solution: Let u + so du 4. Then 8. Emple: Find 9 + u + 3. 4 du u 9 Solution: Let u 3 so du 3. Then + 3 3 3 du + u 4 u / du u/ ] 9 3. ] 3 tn 3 u π 3 3 π 3. 3 3
Integrtion by Prts 8.3 Theorem: (Integrtion by Prts Let f( nd g( be differentible in n intervl. Then f(g ( f(g( g(f ( so b f(g ( f(g( ] b g(f (. Proof: By the Product Rule, we hve d f(g( f (g( + f(g ( nd so f (g( + f(g ( f(g( + c, which cn be rewritten s f(g ( f(g( g(f (. (We do not need to include the rbitrry constnt c since there is now n integrl on both sides of the eqution. 8.4 Nottion: If we write u f(, du f (, v g( nd dv g (, then the top formul in the bove theorem becomes u dv uv v du. 8.5 Note: To find the integrl of polynomil multiplied by n eponentil function or trigonometric function, try Integrting by prts with u equl to the polynomil (you my need to integrte by prts repetedly if the polynomil is of high degree. To integrte polynomil (or n lgebric function times logrithmic or inverse trigonometric function, try integrting by prts letting u be the logrithmic or inverse trigonometric function. To integrte n eponentil function times sine or cosine function, try integrting by prts twice, letting u be the eponentil function both times. 8.6 Emple: Find sin. Solution: Integrte by prts using u, du, v cos nd dv sin to get sin u dv uv v du cos + cos cos + sin + c. 4
8.7 Emple: Find ( + e. Solution: Integrte by prts using u +, du, v e nd dv e to get ( + e u dv uv v du ( + e e. To find e we integrte by prts gin, this time using u, du, v e nd dv e to get ( + e ( + e ( + e ( + e ( e e e ( e 4 e + c 8.8 Emple: Find ln. 4 ( + 3 e + c Solution: Integrte by prts using u ln, du, v nd dv to get ln ln ln + c. 8.9 Emple: Find 4 ln. Solution: Integrte by prts using u ln, du, v 3 3/ nd dv / to get 4 ] 4 ln 3 3/ ln 3 / ( ( 6 3 3 ln 4 9 3 ln 4 9 8. Emple: Find e sin Solution: Write I 6 3 3 3/ ln 4 9 3/ ] 4 ln 4 8 9. e sin. Integrte by prts twice, first using u e, du e, v cos nd dv sin, nd net using u e, du e, v sin nd dv cos to get I e cos + e cos ( e cos + e sin e. sin e cos + e sin I Thus I e cos + e sin + c nd so I (sin cos e + d. 5
8. Emple: Let n be n integer. Find formul for sin n in terms of sin n, nd hence find sin nd sin 4. Solution: Let I sin n sin n sin. Integrte by prts using u sin n, du (n sin n cos, v cos nd dv sin to get I sin n cos + (n sin n cos sin n cos + (n sin n ( sin sin n cos + (n sin n (n I. Add (n I to both sides to get ni sin n cos + (n sin n n sinn cos + n n In prticulr, when n we get sin sin cos + sin n. sin cos + + c sin n, tht is nd when n 4 we get sin 4 4 sin3 cos + 3 4 sin 4 sin3 cos 3 8 sin cos + 3 8 + c. 8. Emple: Let n be n integer. Find formul for sec n in terms of sec n, nd hence find sec 3. Solution: Let I sec n sec n sec. Using Integrte by Prts with u sec n, du (n sec n tn, v tn nd dv tn, we obtin I sec n tn (n sec n tn sec n tn (n sec n (sec sec n tn (n I + (n sec n Add (n I to both sides to get (n I sec n tn + (n In prticulr, when n 3 we get sec 3 sec tn + sec n n secn tn + n n sec n. sec n, tht is sec sec tn + ln sec + tn + c 6
Trigonometric Integrls 8.3 Note: To find f(sin cos n+, write cos n+ ( sin n cos then try the substitution u sin, du cos. To find f(cos sin n+, write sin n+ ( cos n sin then try the substitutionu cos, du sin. To find sin m cos n, try using the trigonometric identities sin θ cos θ nd cos θ + cos θ. Alterntively, write cosn ( sin n nd use the formul sin n n sinn cos + n sin n. n To find f(tn sec n+, write sec n+ ( + tn n sec nd try the substitution u tn, du sec. To find f(sec tn n+, write tn n+ (sec n sec tn nd sec try the substitution u sec, du sec tn. To find sec n+ tn n, write tn n (sec n nd use the formul 8.4 Emple: Find sec n n tn secn + n n π/3 sin 3 cos. sec n. Solution: Mke the substitution u cos so du sin. Then π/3 sin 3 cos 8.5 Emple: Find π/3 ] / u + u ( cos sin cos sin 6. / ( + ( +. ( u du u / u + du Solution: We could use the method of emple., but we choose insted to use the hlf-ngle formuls. We hve π/4 sin 6 π/4 π/4 π/4 ( cos 3 π/4 8 3 8 cos + 3 8 cos 8 cos3 8 3 8 cos + ( 3 8 + cos 4 ( 8 sin cos 5 6 cos + 3 6 cos 4 + 8 sin cos 5 5 4 sin + 3 64 sin 4 + 48 sin3 5π 64 4 + 48 5π 64 48. 7 ] π/4
π/4 8.6 Emple: Find tn 4. Solution: Note first tht tn 4 tn (sec tn sec tn tn sec sec +. To find tn sec, mke the substitution u tn θ, du sec θ dθ to get tn sec u du 3 u3 + c 3 tn3 + c. Thus we hve π/4 tn 4 8.7 Emple: Find π/4 tn sec sec + 3 tn3 tn + sec 3 sec. ] π/4 3 + π 4 π 4 3. Solution: Multiply the numertor nd denomintor by sec + to get sec 3 sec 3 sec (sec + sec 4 + sec 3 sec 4 (sec tn tn + Mke the substitution u tn, du sec to get sec 4 (tn tn + sec u + tn u du + u du u + c tn cot + c, u sec 3 tn. nd mke the substitution v sin, dv cos to get sec 3 cos dv tn sin v v + c csc + c. Thus sec 3 tn cot csc + c. sec sin( sin(b, cos( cos(b, or 8.8 Note: To find sin( cos(b, use one of the identities cos(a B cos(a+b sin A sin B, cos(a B+cos(A+B cos A cos B nd sin(a B + sin(a + B sin A cos B. 8.9 Emple: Find π/6 cos 3 cos. Solution: Since sin 3 cos cos(3 + cos(3 + cos + cos 5, we hve π/6 sin cos 3 π/6 ] π/6 (cos +cos 5 sin + sin 5 4 + 3. 8
8.3 Note: The Weirstrss substitution u tn, tn u, du + u converts sin nd cos into rtionl functions of u: indeed we hve sin u nd u cos so tht sin sin u cos u +u nd cos cos sin u 8.3 Emple: Find cos. +u. Solution: We use the Weirstrss substitution u tn, +u du, nd cos u +u to get cos +u du u +u du du ( + u ( u u u + c cot + c. 9
Inverse Trigonometric Substitution 8.3 Note: For n integrl involving + b ( + c +c, try the substitution θ tn so tht tn θ + c, sec θ + b ( + c nd sec θ dθ b. For n integrl involving b ( + c +c, try the substitution θ sin so tht sin θ + c, cos θ b ( + c nd cos θ dθ b. For n integrl involving b ( + c +c, try the substitution θ sec so tht sec θ + c, tn θ b ( + c nd sec θ tn θ dθ b. 8.33 Emple: Find (4 3 3/. Solution: Let sin θ 3 so cos θ 4 3 nd cos θ dθ 3. Then (4 3 3/ 8.34 Emple: Find π/3 3 π/6 / 3 cos θ dθ ( cos θ 3 + 3. π/3 ] π/3 4 3 sec θ dθ 4 tn θ 3 4. Solution: Let 3 tn θ so 3 sec θ + 3 nd 3 sec θ dθ, nd lso let u sin θ so du cos θ dθ. Then 3 π/4 3 sec + 3 θ dθ π/4 3 tn θ 3 sec θ sec θ π/4 π/6 3 tn θ dθ cos θ dθ π/6 3 sin θ 8.35 Emple: Find 4 / 3 u du 4. ] / 3u 3 + 3 3. / Solution: Let sec θ so tn θ 4 nd sec θ tn θ dθ. Then 4 4 π/3 tn θ sec θ dθ π/3 tn θ π/3 sec θ sec θ dθ sec θ sec θ π/3 8.36 Emple: Find sec θ cos θ dθ 3 (4 3/. ln sec θ + tn θ sin θ ] π/3 dθ ln( + 3 3. Solution: Let sin θ so cos θ 4 nd cos θ dθ. Then 3 (4 3/ π/6 π/6 6 cos 4 θ dθ 4 ( + cos θ dθ 4 + 8 cos θ + 4 cos θ dθ 4 + 8 cos θ + + cos 4θ dθ ] π/6 6θ + 4 sin θ + sin 4θ π + 3 + 3 4 π + 9 3 4.
Prtil Frctions 8.37 Note: We cn find the integrl of rtionl function f( g( s follows: Step : use long division to find polynomils q( nd r( with deg r( < deg g( such tht f( g(q( + r( for ll, nd note tht f( r( q( + g( g( so f( g( q( + r( g(. (If deg f( < deg g( then q( nd r( f(. Step : fctor g( into liner nd irreducible qudrtic fctors. Step 3: write r( g( s sum of terms so tht for ech liner fctor ( + bk we hve the k terms A ( + b + A ( + b + + A k ( + b k nd for ech irreducible qudrtic fctor ( + b + c k we hve the k terms Writing r( g( Step 5: solve the integrl. B + C ( + b + c + B + C ( + b + c + + in this form is clled splitting r( g( B k + C k ( + b + c k. into its prtil frctions decomposition. 8.38 Emple: If g( ( 3 ( + + 3 then in step 3 we would write r( g( A + B + nd then solve for the vrious constnts. 8.39 Emple: Find 3 Solution: In order to get C ( + D ( 3 + E + F + + 3 + 7 ( ( +. 7 ( ( + A + B ( + C + A( ( + + B( + + C( 7. G + H ( + + 3. we need Equting coefficients gives A + C, A + B C nd A + B + C 7. Solving these three equtions gives A, B nd C, nd so we hve 3 7 ( ( + 3 3 A + B ( + C + ( ] 3 + ln( + ln( + (ln + ln 5 ( ln 4 ln 8 5.
3 4 3 + 8.4 Emple: Find 3. + Solution: Use long division of polynomils to show tht 4 3 + 3 + Net, note tht to get A + B + C + + + 3 + + + + 3. + we need A( + + (B + C( + +. Equting coefficients gives A + B, C nd A. Solving these three equtions gives A, B nd C. Thus 3 4 3 + 3 + 8.4 Emple: Find I Solution: To get A + B + C + D + 5 + 3 + + + + + ln ln( + + tn ] 3 ( 3 3 + ln 3 ln 4 + π 3 ( ln + π 4 3 + ln 3 + π. 5 + 4 3 5 5 ( + 5. E + F ( + 5 5 + 4 3 5 5 ( + 5 we need A( + 5 + B( + 5 + (C + D( ( + 5 + (E + F ( 5 + 4 3 5 5. Epnding the left hnd side then equting coefficients gives the 5 equtions A + C, 4A + B C + D, 4A 4B + 5C D + E A + 4B + 5D + F, 5A B 5, 5B 5 Solving these equtions gives A, B, C, D, E nd F 8, so I + + + 5 + 8 ( + 5 + + 4 6 + + 5 ( + 5 + + 5 + 4 + 5 + ( + 5 6 ( + 5 We hve ln +c nd +c. Mke the substitution u +5, du ( to get ( du + 5 u ln u + c ln( + 5 + c nd ( du ( + 5 u u + c + 5 + c.
Mke the substitution tn θ, sec θ + 5, sec θ dθ to get 4 4 sec + 5 θ dθ ( sec θ dθ θ + c tn ( + c nd 6 6 sec ( + 5 θ dθ ( sec θ 4 dθ Thus we hve I dθ sec θ cos θ dθ θ + sin θ + c θ + sin θ cos θ + c tn ( ln + + ln( + 5 + tn + 5 ( tn + 5 ln + 5 + + 5 + ] tn ( ln 5 + 3 5 + ( tn ln 4 + 4 ln 5 8 7 + tn. + cos θ dθ + ( +5 + c. ] 3
Improper Integrtion 8.4 Definition: Suppose tht f :, b R is integrble on every closed intervl contined in, b. Then we define the improper integrl of f on, b to be b f lim t b provided the limit eists, nd we sy tht f is improperly integrble on, b, (or tht the improper integrl of f on, b converges, when the improper integrl eists nd is finite. In this definition we lso llow the cse tht b, nd then we hve t f t f lim f. t Similrly, if f : (, b] R is integrble on every closed intervl in (, b] then we define the improper integrl of f on (, b] to be b b f lim f t + t provided the limit eists, nd we sy tht f is improperly integrble on (, b] when the improper integrl is finite. In this definition we lso llow the cse tht. For function f : (, b R, which is integrble on every closed intervl in (, b, we choose point c (, b, then we define the improper integrl of f on (, b to be b f c provided tht both of the improper integrls on the right eist nd cn be dded, nd we sy tht f is improperly integrble on (, b when both of the improper integrls on the right re finite. As n eercise, you should verify tht the vlue of this integrl does not depend on the choice of c. f + 8.43 Nottion: For function F : (, b R write ] b F ( lim F ( lim F (. + b + We use similr nottion when F :, b R nd when F : (, b] R. 8.44 Note: Suppose tht f : (, b R is integrble on every closed intervl contined in (, b nd tht F is differentible with F f on (, b. Then b ] b f F ( A similr result holds for functions defined on hlf-open intervls, b nd (, b]. Proof: Choose c (, b. By the Fundmentl Theorem of Clculus we hve b f c f + b c f lim s + b c +. c lim s + ( F (c F (s + lim lim t b F (t lim s + F (s 4 s f f + lim t b t c f ( F (t F (c t b ] b F ( +.
8.45 Emple: Find nd find. Solution: We hve ] ln ( + nd 8.46 Emple: Show tht ]. + converges if nd only if p <. p Solution: The cse tht p ws delt with in the previous emple. If p > so tht p > then we hve ] p ( (p p p ( nd if p < so tht p > then we hve ] p p p 8.47 Emple: Show tht Solution: When p we hve p + + ( p ( p. converges if nd only if p >. p When p > so tht p > we hve p (p p ] ln. ] ( nd if p < so tht p > then we hve ] p p ( ( p ( p p p. 5
8.48 Emple: Find e. Solution: We hve 8.49 Emple: Find Solution: We hve e e ] (. ln. ln ln since l Hôpitl s Rule gives lim ln lim + + ] ln ( (, + lim + lim. + 8.5 Theorem: (Comprison Let f nd g be integrble on closed subintervls of (, b, nd suppose tht f( g( for ll (, b. If g is improperly integrble on (, b then so is f nd then we hve On the other hnd, if b b f diverges then f b b functions f nd g defined on hlf-open intervls. Proof: The proof is left s n eercise. 8.5 Emple: Determine whether π/ g. g diverges, too. A similr result holds for sec converges. Solution: For < π we hve cos π so sec π nd hence sec π. Since π/ π ] π/ π/ π π, which is finite, we see tht sec converges too, by comprison. 8.5 Emple: Determine whether e converges. Solution: For u we hve e u +u, so for we hve e +, so e +. ] Since + tn π, which is finite, we see tht e converges too, by comprison. 6
8.53 Theorem: (Estimtion Let f be integrble on closed subintervls of (, b. If f is improperly integrble on (, b then so is f, nd then we hve b b f f. A similr result holds for functions defined on hlf-open intervls. Proof: The proof is left s n eercise. 8.54 Emple: Show tht sin converges. Solution: We shll show tht both of the integrls sin sin nd converge. sin sin Since lim, the function f defined by f( nd f( for > is + continuous (hence integrble on, ]. By prt of the Fundmentl Theorem of Clculus, the function r f( is continuous function of r for r, ] nd so we hve sin lim r + r sin lim f( f(, r + r sin which is finite, so converges. Integrte by prts using u, du, v sin nd dv cos to get sin cos ] cos cos cos(. Since cos comprison. Thus nd cos converges, we see tht cos converges too, by lso converges by the Estimtion Theorem. 7