Chpter 6 Techniques of Integrtion 6. Integrtion by Prts Every differentition rule hs corresponding integrtion rule. For instnce, the Substitution 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 by prts. The Product Rule stte tht if f nd g re differentible functions, then d [ ] f(x)g(x) = f(x)g (x)+g(x)f (x) dx In the nottion for indefinite integrls this eqution becomes [f(x)g (x)+g(x)f (x) ] dx = f(x)g(x) or f(x)g (x)dx+ g(x)f (x)dx = f(x)g(x) We cn rerrnge this eqution s f(x)g (x)dx = f(x)g(x) g(x)f (x)dx This formul is clled the formul for integrtion by prts. Let u = f(x) nd v = g(x). Then the differentil re du = f (x)dx nd dv = g (x)dx, so, by the Substitution Rule, the formul for integrtion by prts becomes udv = uv vdu 49
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 5 Exmple 6.. Find xcosxdx. Note. Our im in using integrtion by prts is to obtin simpler integrl thn the one we strt with. Thus, in Exmple 6. we strt with xcosxdx nd expressed it in terms of the simpler integrl sinxdx. If we hd chosen u = cosx nd dv = xdx, then du = sinxdx nd v = x 2 /2, so integrtion by prts gives xcosxdx = (cosx) x2 2 + 2 x 2 sinxdx Although this is true, x 2 sinxdx is 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(x) to be function tht becomes simpler when differentited (or t lest not more complicted) s long s dv = g (x)dx cn be redily integrted to give v. Exmple 6.2. Find cosxln(sinx)dx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 5 Exmple 6.3. Find x 2 sin2xdx. Exmple 6.4. Find e x cosxdx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 52 If we combine the formul for integrtion by prts with Prt 2 of the Fundmentl Theorem of Clculus, we cn evlute definite integrls by prts. Assuming f nd g re continuous, nd using the Fundmentl Theorem, we obtin b Tht is, if u = f(x) nd v = g(x), then ] b b f(x)g (x)dx = f(x)g(x) g(x)f (x)dx b ] b b udv = uv vdu. Exmple 6.5. Find tn xdx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 53 Exercise 6. Evlute the integrl.. xe 2x dx 2. x 2 lnxdx 3. xsin4xdx 4. x 2 e 3x dx 5. x 2 cos3xdx 6. e x sin4xdx 7. (lnx) 2 dx 8. cosxcos2xdx 9. xsec 2 xdx. cosxln(sinx)dx. cos(ln x) dx 2. cos xdx 3. sin xdx 4. xsin2xdx 5. 7.. 4 xe x dx 6. ln xdx 8. 2 2 lnx x 2 dx x 4 (lnx) 2 dx Answer to Exercise 6. 2 xe2x 4 e2x +C 2. 3 x2 lnx 9 x3 +C 3. xcos4x+ sin4x+c 4 6 4. 3 x2 e 3x 2 9 xe 3x 2 27 e 3x +C 5. 3 x2 cos3x+ 2 9 xcos3x 2 27 sin3x+c 6. 7 ex sin4x 4 7 ex cos4x+c 7. x(lnx) 2 2xlnx+2x+C 8. 2 sin2xcosx cos2xsinx+c 9. xtnx+ln cosx +C 3 3. sinxln(sinx) sinx+c. 2 x[sin(lnx)+cos(lnx)]+c 2. xcos x x 2 +C 3. 2 xcos x+2sin x+c 4. 4 sin2 2 cos2 5. 2 6. ln2 7. 2ln4 3 8. 3 e 2 2 2 2 (ln2)2 64 62 ln2+ 25 25 6.2 Trigonometric Integrls In this section we use trigonometric identities to integrte certin combintions of trigonometric functions. Our first im is to evlute integrls of the form sin m xcos n xdx, where m nd n re positive integers.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 54 Cse I: m or n Is n Odd Positive Integer If m is odd, first isolte one fctor of sinx. (You ll need this for du). Then, replce ny fctors of sin 2 x with cos 2 x nd mke the substitution u = cosx. Likewise, if n is odd, first isolte one fctor of cosx. (You ll need this for du). Then, replce ny fctors of cos 2 x with sin 2 x nd mke the substitution u = sinx. Exmple 6.6. Evlute cos 3 xdx. Exmple 6.7. Find cos 5 xsin 4 xdx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 55 Exmple 6.8. Find cos 2/3 xsin 5 xdx. Cse II: m nd n Are Both Even Positive Integers In this cse, we cn use the hlf-ngle formuls for sine nd cosine to reduce the power of in the integrnd. Hlf-ngle formuls: sin 2 θ = 2 ( cos2θ) nd cos2 θ = 2 (+cos2θ) Exmple 6.9. Find sin 4 xdx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 56 Exmple 6.. Find sin 4 xcos 2 xdx. Our next im is to devise strtegy for evluting integrls of the form tn m xsec n xdx, where m nd n re integers. Cse I: m Is n Odd Positive Integer First, isolte one fctor of secxtnx. (You ll need this for du). Then, replce ny fctors of tn 2 x with sec 2 x nd mke the substitution u = secx. Exmple 6.. Find tn 3 xsec 3/2 xdx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 57 Cse II: n Is n Even Positive Integer First, isolte one fctor of sec 2 x. (You ll need this for du). Then, replce ny remining fctors of sec 2 x with +tn 2 x nd mke the substitution u = tnx. Exmple 6.2. Find tn 6 xsec 4 xdx. For other cses, the guidelines re not s cler-cut. We my need to use identities, integrtion by prt, nd occsionlly little ingenuity. Exmple 6.3. Find sec 3 xdx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 58 Integrls of the form cot m xcsc n xdx cn be found by similr methods becuse of the identity +cot 2 x = csc 2 x. Finlly, we cn mke use of nother set of trigonometric identities: To evlute the integrls () sinmxcosnxdx, (b) sinmxsinnxdx, or (c) cosmxcosnxdx, we use the corresponding identity: [ ] sinacosb = 2 sin(a B)+sin(A+B) [ ] sinasinb = 2 cos(a B) cos(a+b) [ ] cosacosb = 2 cos(a B)+cos(A+B) Exmple 6.4. Find cos5xcos4xdx. Exmple 6.5. Find sin2xcos3xdx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 59 Exercise 6.2 Evlute the integrl.. cosxsin 4 xdx 2. cos 2 xsinxdx 3π/4 3. sin 5 xcos 3 xdx 4. cos 5 xsin 4 xdx π/2 π/2 5. sin 2 3xdx 6. cos 2 xsin 2 xdx π/4 7. ( sin2x) 2 dx 8. sin 4 xcos 2 xdx 9. sin 3 x cosxdx. cos 2 xtn 3 xdx sinx. dx 2. tn 2 xdx cosx 3. sec 4 xdx 4. tn 5 xdx π/4 5. tnxsec 3 xdx 6. tn 4 xsec 2 xdx 7. tn 3 xsecxdx 8. π/3 π/2 tn 3 xsecxdx sec 2 x 9. dx 2. cot 2 xdx cotx π/6 2. cot 3 xcsc 3 xdx 22. cot 2 ωcsc 4 ωdω sinx 23. cscxdx 24. tnx dx tn 2 x 25. dx 26. sin5xsin2xdx sec 2 x 27. cos7θcos5θdθ 28. sin4xcos5xdx Answer to Exercise 6.2. 5 sin5 x+c 2. 3 cos3 x+c 3. 384 4. 5 sin5 x 2 7 sin7 x+ 9 sin9 x+c π 5. 6. x sin4x+c 7. 3x+cos2x sin4x+c 8. (3π 4) 4 8 32 2 8 92 9. [ 2 7 cos3 x 2 3 cosx] cosx+c. 2 cos2 x ln cosx +C. ln(+sinx)+c 2. tnx x+c 3. 3 tn3 x+tnx+c 4. 4 sec4 x tn 2 x+ln secx +C 5. 3 sec3 x+c 6. 5 7. 3 sec3 x secx+c 8. 38 5 9. 2 tn2 x+c 2. 3 π 3 2. 5 csc5 x+ 3 csc3 x+c 22. 3 cot3 ω 5 cot5 ω +C
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 6 23. ln cscx cotx +C 24. sinx+c 25. 2 sin2x+c 26. 6 sin3x 4 sin7x+c 27. 4 sin2θ+ 24 sin2θ+c 28. 2 cosx 8 cos9x+c 6.3 Trigonometric Substitutions If n integrl contins term of the form 2 x 2, 2 +x 2 or x 2 2, for some >, you cn often evlute the integrl by mking substitution involving trig function. First, suppose tht n integrnd contins term of the form 2 x 2, for some >. If we let x = sinθ, where π θ π, then we cn eliminte the squre root, 2 2 s follows. Notice tht we now hve since for π θ π, cosθ >. 2 2 Exmple 6.6. Evlute 2 x 2 = 2 (sinθ) 2 = x 3 4 x 2 dx. 2 ( sin 2 θ) = cos 2 θ = cosθ = cosθ
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 6 Next, suppose tht n integrnd contins term of the form 2 +x 2, for some >. If we let x = tnθ, where π < θ < π, then we cn eliminte the squre root, 2 2 s follows. Notice tht in this cse, we hve since for π < θ < π, secθ >. 2 2 2 +x 2 = 2 +(tnθ) 2 = 2 (+tn 2 θ) = sec 2 θ = secθ = secθ Exmple 6.7. Evlute the integrl 9+x 2 dx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 62 Finlly, suppose tht n integrnd contins term of the form x 2 2, for some >. If we let x = secθ, where θ [ ) [ ), π 2 π, 3π 2, then we cn eliminte the squre root, s follows. Notice tht in this cse, we hve x2 2 = (secθ) 2 2 = 2 (sec 2 θ ) since for θ [ [ ), 2) π π, 3π 2, tnθ. Exmple 6.8. 9x2 Evlute the integrl dx. x = tn 2 θ = tnθ = tnθ
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 63 Exmple 6.9. Evlute x+2 x2 +4x+7 dx. We summrize the three trigonometric substitutions presented here in the following tble. Expression Substitution Identity 2 x 2 x = sinθ, π θ π 2 2 sin 2 θ = cos 2 θ 2 +x 2 x = tnθ, π 2 < θ < π 2 +tn 2 θ = sec 2 θ x2 2 x = secθ, θ [, π 2 ) [ ) π, 3π 2 sec 2 θ = tn 2 θ
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 64 Exercise 6.3 Evlute the integrl. 2 2. 2 t 3 dt 2. t 2 x2 4 dx dx 3. x 2 dx 4. 25 x2 x x 2 +3 dx x 2 4x2 5. x2 +9 dx 6. dx x2 9x2 4 7. +6dx 8. dx x x 2 x 9. dx. ( 2 x 2 ) 3/2 x2 7 dx. 3. 5. 7. 3 2/3 dx 9+x 2 2. x 3 4 9x 2 dx 4. 9x2 +6x 8 dx 6. e t 9 e 2t dt 8. x x 2 +8dx 2x x2 dx dx (x 2 +2x+2) 2 dx x2 + 2 dx Answer to Exercise 6.3 π. + 3 2. 4 x2 25 x 2 +C 3. +C 4. ( 24 8 4 2x 25x 3 )ln ( x 2 +3 3) x 5. 3 (x2 +4) 3/2 4 x 2 +4+C 6. 4 sin (2x)+ x 4x 2 2 +C 7. x x 2 2 x +4+2ln 2 + x2 +4 2 +C 8. 9x 2 4 2sec ( 3x)+C 2 +C 9. x 2 x 2 sin ( x ) +C. x2 7+C. ln(+ 2) 2. 3 (x2 +4) 3/2 +C 64 3. 4. 25 2 [sin (x )+(x ) 2x x 2 ]+C 5. ln 3 3x++ 9x2 +6x 8 +C 6. 2 [ tn (x )+ [ 7. 2 e t 9 e 2t +9sin ( et)] +C 8. ln(x+ x 3 2 + 2 )+C ] (x+) +C (x 2 +2x+2)
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 65 6.4 Integrting Rtionl Functions by Prtil Frctions In this section we show how to integrte ny rtionl function by expressing it s sum of simpler frctions, clled prtil frctions. Observe tht 3 x+2 2 x 5 = 3(x 5) 2(x+2) (x+2)(x 5) = x 9 x 2 3x. To integrte the function on the right side of this eqution, we hve ( x 9 3 x 2 3x dx = x+2 2 ) dx x 5 = 3ln x+2 2ln x 5 +C To see how the method of prtil frctions work in generl, let s consider rtionl function f(x) = P(x) (6.) Q(x) where P nd Q re polynomil. It s possible to express f s sum of simpler frctions provided the degree of P ( deg(p) ) is less thn the degree of Q ( deg(q) ). Such rtionl function is clled proper. If f is improper, tht is, deg(p) deg(q), then we must tke the preliminry step of dividing Q into P (by long division) until the reminder R(x) is obtined such tht deg(r) < deg(q). The division sttement is f(x) = P(x) Q(x) where S nd R re lso polynomil. Exmple 6.2. x 3 4x 2 +5x 2 Find dx. x 3 = S(x)+ R(x) Q(x)
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 66 The next step is to fctor the denomintor Q(x) s fr s possible. And the third step is to express the proper rtionl function R(x)/Q(x) s sum of prtil fctions of the form A (x+b) i or Ax+B (x 2 +bx+c) j CASE I: The denomintor Q(x) is product of distinct liner fctors. This mens tht we cn write where no fctor is repeted. Q(x) = ( x+b )( 2 x+b 2 ) ( n x+b n ) In this cse the prtil frction theorem sttes tht there exist constnts A, A 2,..., A k such tht R(x) Q(x) = = R(x) ( x+b )( 2 x+b 2 ) ( k x+b k ) A + A 2 + + A k (6.2) x+b 2 x+b 2 k x+b k These constnt cn be determined s in the following exmple. Exmple 6.2. Evlute 2x+2 x 2 6x+8 dx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 67 Exmple 6.22. Evlute 2x 2 +5x x 3 +x 2 2x dx. CASE II: Q(x) is product of distinct liner fctors, some of which re repeted. Suppose the first liner fctor ( x+b ) is repeted r times; tht is, ( x+b ) r occurs in the fctoriztion of Q(x). Then insted of the single term A /( x+b ) in Eqution (6.2), we would use For exmple, A A 2 + x+b ( x+b ) + + A r 2 ( x+b ) r. (6.3) x 2 5 x 2 (x+) = A 3 x + B x + C 2 (x+) + D (x+) 2 + E (x+) 3
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 68 Exmple 6.23. Find x x 3 +4x 2 +4x dx. CASE III: Q(x) contins irreducible qudrtic fctors, none of which is repeted. If Q(x) hs the fctor x 2 +bx+c, where b 2 4c <, then, in ddition to the prtil frctions in Equtions (6.2) nd (6.3), the expression for R(x)/Q(x) will hve term of the form Ax+B (6.4) x 2 +bx+c where A nd B re constnts to be determined.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 69 For instnce, the frction given by f(x) = x (x+2)(x 2 +)(x 2 +2) hs prtil frction decomposition of the form x (x+2)(x 2 +)(x 2 +2) = A x+2 + Bx+C x 2 + + Dx+E x 2 +2 The term given in (6.4) cn be integrte by completing the squre nd using the formul Exmple 6.24. Evlute 3x 2 7x+ (x 2)(x 2 +4) dx. x 2 + dx = ( x ) 2 tn +C
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 7 CASE IV: Q(x) contins repeted irreducible qudrtic fctors. If Q(x) hs the fctor (x 2 + bx + c) r, where b 2 4c <, then insted of the single prtil frction (6.4), the sum A x+b x 2 +bx+c + A 2x+B 2 (x 2 +bx+c) + + A r x+b r (6.5) 2 (x 2 +bx+c) r occurs in the prtil frction decomposition of R(x)/Q(x). Ech of the term of (6.5) cn be integrted by first completing the squre. Exmple 6.25. x 2x 2 x 3 Evlute dx. x(x 2 +) 2
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 7 Exercise 6.4 Evlute the integrl. x 5 6x. dx 2. x 2 x 2 x+2 dx x+ x+5 3. dx 4. x 2 x 6 x 3 x 2 2x dx x 3 +x+2 3x 5. dx 6. x 2 +2x 8 x 3 x dx 2x+3 x 7. dx 8. (x+2) 2 x 3 +4x 2 +4x dx x+4 x+2 9. dx. x 3 +3x 2 +2x x 3 +x dx 4x 2 3x 2 6. dx 2. x 4 x 2 x 2 dx 2x+3 x 2 +2x+ 3. dx 4. dx x 2 +2x+ x 3 +x 4x 2 +3 3x 3 + 5. dx 6. x 3 +x 2 +x x 3 x 2 +x dx Answer to Exercise 6.4. 3ln x+ 2ln x +C 2. 2ln x+ +4ln x 2 +C 3. 5 ln x+2 + 4 5 ln x 3 +C 4. 2ln x+ + 2 ln x 2 5 2 ln x +C 5. ln x+4 +2ln x 2 + 2 x2 2x+C 6. ln x+ 2ln x +ln x +C 7. 2ln x+2 (x+2) +C 8. ln x+2 3 4 2 (x+2) ln x +C 9. ln x+2 3ln x+ +2ln x +C 4. ln(x 2 +)+tn x+2ln x +C. 3 2 ln x+ + 2 ln x ln(x2 +)+tn x+c 2. 3x+ln x+ +2ln x 2 +C 3. 2ln x+ (x+) +C 4. 2tn x+ln x +C 5. 3ln x + 2 ln x2 +x+ 7 3 tn ( 2x+ 3 )+C 6. 3x+2ln x + 2 ln(x2 +) 2tn x+c
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 72 6.5 Improper Integrls In defining definite integrl b f(x)dx we delt with function f defined on finite intervl [,b] nd we ssumed tht f does not hve n infinite discontinuity. In this section we extend the concept of the definite integrl to the cse where the intervl is infinite nd lso to the cse where f hs n infinite discontinuity in [,b]. In either cse the integrl is clled n improper integrl. Type I: Infinite Intervls Definition 6.. () If (b) If t f(x)dx exists for every number t, then f(x)dx = lim t t provided this limit exists (s finite number). b t f(x)dx f(x)dx exists for every number t b, then b b f(x)dx = lim f(x)dx t t provided this limit exists (s finite number). The improper integrls t f(x)dxnd b t f(x)dxre clled convergent if the corresponding limit exists nd divergent if the limit does not exist. (c) If both f(x)dx nd f(x)dx = f(x)dx re convergent, then we define In prt (c) ny rel number cn be used. f(x)dx+ f(x)dx
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 73 Exmple 6.26. Determine whether the integrl dx is convergent or divergent. x Exmple 6.27. Evlute xe x dx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 74 Exmple 6.28. Evlute +x 2 dx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 75 Type II: Discontinuous Integrnds Definition 6.2. () If f continuous on [,b) nd is discontinuous t b, then b f(x)dx = lim t b if this limit exists (s finite number). t f(x)dx (b) If f continuous on (,b] nd is discontinuous t, then b b f(x)dx = lim f(x)dx t + t if this limit exists (s finite number). The improper integrls b f(x) dx is clled convergent if the corresponding limit exists nd divergent if the limit does not exist. (c) If f hs discontinuity t c, where < c < b, nd both b c f(x)dx re convergent, then we define b f(x)dx = c f(x)dx+ b c f(x)dx c f(x)dx nd Exmple 6.29. Find 5 2 x 2 dx.
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 76 Exmple 6.3. Determine whether the integrl π/2 sec x dx converges or diverges. Exmple 6.3. Evlute 3 dx if possible. x
MA: Prepred by Asst.Prof.Dr. Archr Pcheenburwn 77 Exercise 6.5 Determine whether ech integrl is convergent or divergent. Evlute those tht re convergent.. 3. 5. 7. 9.. 3. 5. 7. 3 2 2 2 dx 2. (3x+) 2 e x dx 4. xe x2 dx 6. cosxdx 8. lnx dx. x lnx dx 2. x2 dx 4. x2 dx 6. x4 dx 8. x 2 3 π/4 π 2 2 w dw x 3 dx (x+2)(x+3) dx xe 2x dx x +x 2 dx x dx csc 2 tdt secxdx z 2 lnzdz Answer to Exercise 6.5. 2 2. D 3. 4. D 5. 6. ln 2 3 7. D 8. e2 4 9. D. D. 2. 2 3 3. D 4. D 5. D 6. D 7. D 8. 8 3 ln2 8 9