Using integration tables

 Milo Lamb
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1 Using integrtion tbles Integrtion tbles re inclue in most mth tetbooks, n vilble on the Internet. Using them is nother wy to evlute integrls. Sometimes the use is strightforwr; sometimes it tkes severl steps, or nees to be combine with other techniques of integrtion. There re few types of formuls to el with, n few tricks one cn use to evlute complicte integrls using tbles. The only rel wy to lern ll tht is by working out emples, which is wht we shll o in this moule. Integrtion tbles Using integrtion tbles is, for the most prt, perfectly strightforwr. You tke your integrl, look it up in the tbles, n write own the nswer. Integrls of the sme kin re typiclly groupe up. Comprehensive tbles hve sections for squres n squre roots, trig functions, inverse trig function, eponentils, logs, hyperbolic functions, hn more. Most clculus tetbooks hve n ppeni with bsic integrtion tbles. More etensive tbles cn be foun in reference books, n on the Internet. Section 5 hs list of few ozen integrtion formuls tht you cn use to nswer questions in this moule. Strightforwr emples Emple : Suppose we wish to evlute the following integrl: 4 We go own the list of integrls in Section 5. We cn skip the trig functions, then we re through the squre roots until we hit formul 9. which sys: 9. = + sin + C To pply this formul, we set = n copy: = 4 + sin + C 4
2 Emple : A slightly more complicte version of the first emple might sk for the integrl There is no formul in the tbles tht looks ectly like this. Wht to o now? We nee to get on the bottom, in plce of /. The esiest solution is to multiply top n bottom of the frction by : is: = = 4 This is twice the integrl from Emple. Using our solution, the nswer = sin + c 3 Emple with recursion For those of you unfmilir with the term recursion, this is gret emple to see it use. In simple terms, we will epress n integrl in terms of nother integrl of the sme kin, but slightly esier. We keep oing tht until we rrive t something tht we know how to o. Let s see how tht works: Suppose we wish to evlute sin 4 () We go own the list from the beginning, where the trig functions re. Formuls.6. re for squres of vrious functions, then we hit number 7. which sys: 7. sin n () = n sinn () cos() + n n sin n () Checking the rest of the tble revels tht there is no formul specificlly for sin 4 (). We hve to use formul 7. We set n = 4, n copy own the result: sin 4 () = 4 sin3 () cos() + 3 sin () 4 We re not one yet, for there is n integrl on the right hn sie; tht is not something we wnt in our nswer. It is n integrl of sin (), for which we cn use formul.:. sin () = 4 sin() + C
3 We combine it with our nswer so fr, n obtin: sin 4 () = 4 sin3 () cos() sin() + c Smll note: the new constnt is not quite the sme s the ol one: c = 3 4 C. We never worry bout the reltionship between the constnts, since they re, well, constnts: in the en, we cn set them to whtever we like nywy. The integrl of sin 4 () cn be epresse in mny wys. A populr one is sin 4 () = sin() + 3 sin(4) + c Try if you cn show tht these nswers re the sme, using ouble ngle formuls: sin() = sin() cos() n cos() = cos () sin (). We cn use formul 7. to fin the integrl of ny power of sin(). The higher the power, the more steps we hve to perform. Even powers of sin() will le us to the use of formul., wheres o powers will le to the integrl of sin(), which we know. Before we move on, the use of formul. to integrte sin () is worth ponering. Wht woul hppen if we use formul 7. inste? It shoul, fter ll, pply to ny (positive) power of sin(). Let s just see. Using formul 7. with n = gives: sin () = sin() cos() + sin 0 () This time, the right hn integrl is etremely esy, since sin 0 () = : sin 0 () = = + c Putting tht into our solution, we get sin () = sin() cos() + + C This is ectly the sme s formul., if you use the ouble ngle ientity sin() = sin() cos(). We hve seen both versions of sin () before: in Integrtion by substitution, we got formul. for the integrl of sin (), wheres in Integrtion by prts, we got the nswer given by formul 7. Either one is perfectly cceptble. 3
4 4 Emple with substitution Sometimes, the tble simply oes not hve the kin of integrl we wnt. If tht hppens, we nee to simplify the integrl in some wy, before using the tble. To see n emple of this, let us integrte sin (e )e There is nothing like it in our integrtion tble; the formul is just too eotic. We nee to replce sin(e ) by the usul sin function. This clls for the substitution u = e, n u = e : sin (e )e = sin (u)u This we cn integrte using formul.: sin (u)u = u 4 sin(u) + C Now we substitute bck, n we re one: sin (e )e = e 4 sin(e ) + C 4
5 5 Tble of integrls, with few tips Double ngle formuls sin () = cos() cos () = + cos(). sin () = 4 sin() + C. cos () = + 4 sin() + C 3. tn () = tn() + C 4. cot () = cot() + C 5. sec () = tn() + C 6. csc () = cot() + C 7. sin n () = n sinn () cos() + n sin n () n 8. cos n () = n cosn () sin() + n cos n () n 9. tn n () = n tnn () tn n () 0. cot n () = n cotn () cot n (). sec n () = n secn () tn() + n sec n () n. csc n () = n cscn () cot() + n csc n () n 3. sin m () cos k+ () = sin m ()( sin ()) k cos() n substitute u = sin() 4. sin k+ () cos m () = ( cos ()) k sin() cos m () n substitute u = cos() 5. sin m () cos n () use both ouble ngle formuls For integrls involving substitute = sin(θ). For integrls involving + substitute = tn(θ). For integrls involving substitute = sec(θ) = + + ln C 7. + = 8 ( + ) ln C 5
6 + 8. = + ln C = + ln C 0. + = ln C. + = + ln C. + = ln C 3. + = + + C 4. = + sin + C 5. = 8 ( ) sin + C 6. = ln + + C 7. = sin + C 8. = sin + C 9. = + sin + C 30. = ln + + C 3. = + C 3. = ln + + C 33. = 8 ( ) 4 8 ln +C = cos + C 35. = + ln + + C 36. = ln + + C 6
7 = + ( ln + + C = ) sec + C 39. = + C A + B 40. ( + b)(c + ) = E ln + b + F ln c + + C, where E n F c re the solutions of A + B = E(c + ) + F ( + b) = tn () + C 4. sec() = ln sec() + tn() + C 43. tn() = ln sec() + C 44. csc() = ln csc() cot() + C 45. cot() = ln sin() + C 46. sin () = sin () + + C 47. cos () = cos () + C 48. tn () = tn () ln( + ) + C Some ifferentition formuls. sin () =. cos () = 3. tn () = + 4. csc () = 5. sec () = 6. cot () = + 7
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