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Change of Variable in an Indefinite IntegralFormula (5.172) shows that in seeking a primitive for the function (f ◦ ϕ)(t) · ϕ (t)one may proceed as follows:(f ◦ ϕ)(t) · ϕ (t) dt = f ϕ(t) dϕ(t) ==f (x) dx = F (x) + c = F ϕ(t) + c,that is, first make the change of variable ϕ(t) = x in the integrand and pass to thenew variable x, then, after finding the primitive as a function of x, return to the oldvariable t by the substitution x = ϕ(t).Example 8t dt1 1d(t 2 + 1) 1dx 1= ln |x| + c = ln t 2 + 1 + c.==2222x221+t1+t3125Differential CalculusExample 9d( x2 )dx==2 sin x2 cos x2tan x2 cos2 x2d(tan u)dudv====2tan uvtan u cos u x= ln |v| + c = ln | tan u| + c = lntan + c.2dx=sin xWe have now considered several examples in which properties a, b, and c of theindefinite integral have been used individually.
Actually, in the majority of cases,these properties are used together.Example 101(sin 5x − sin x)dx =sin 2x cos 3x dx =21=sin 5x dx − sin x dx =2 1 1sin 5x d(5x) + cos x ==2 51111=sin u du + cos x = − cos u + cos x + c =102102=11cos x −cos 5x + c.210Example 11arcsin x dx = x arcsin x − x d arcsin x =x1d(1 − x 2 )dx = x arcsin x +=√√21 − x21 − x21= x arcsin x +u−1/2 du = x arcsin x + u1/2 + c =2#= x arcsin x + 1 − x 2 + c.= x arcsin x −Example 121eax cos bx dx =cos bx deax =a5.7 Primitives313=1 ax1e cos bx −aa=b1 axe cos bx +aaeax d cos bx =1b= eax cos bx + 2aaeax sin bx dx =sin bx deax =b axb1 ax= e cos bx + 2 e sin bx − 2 eax d sin bx =aaa2a cos bx + b sin bx b=− 2 eax cos bx dx.a2aFrom this result we conclude thata cos bx + b sin bx axeax cos bx dx =e + c.a 2 + b2We could have arrived at this result by using Euler’s formula and the fact that theprimitive of the function e(a+ib)x = eax cos bx + ieax sin bx isa − ib (a+ib)x1e(a+ib)x = 2e=a + iba + b2a sin x − b cos bx axa cos bx + b sin bx axe +ie .=a 2 + b2a 2 + b2It will be useful to keep this in mind in the future.
For real values of x thiscan easily be verified directly by differentiating the real and imaginary parts of the1e(a+ib)x .function a+ibIn particular, we also find from this result thata sin bx − b cos bx axe + c.eax sin bx dx =a 2 + b2Even the small set of examples we have considered suffices to show that in seeking primitives for even the elementary functions one is often obliged to resort toauxiliary transformations and clever devices, which was not at all the case in finding the derivatives of compositions of the functions whose derivatives we knew.
Itturns out that this difficulty is not accidental. For example, in contrast to differentiation, finding the primitive of an elementary function may lead to a function thatis no longer a composition of elementary functions. For that reason, one shouldnot conflate the phrase “finding a primitive” with the sometimes impossible taskof “expressing the primitive of a given elementary function in terms of elementaryfunctions”. In general, the class of elementary functions is a rather artificial object.There are very many special functions of importance in applications that have beenstudied and tabulated at least as well as, say sin x or ex .3145Differential CalculusFor example, the sine integral Si x is the primitive sinx x dx of the function sinx xthat tends to zero as x → 0. There exists such a primitive, but, like all the otherprimitives of sinx x , it is not a composition of elementary functions.Similarly, the functioncos xdx,Ci x =xspecified by the condition Ci x → 0 as x → ∞ is not elementary.
The function Ci xis called the cosine integral.The primitive lndxx of the function ln1x is also not elementary. One of the primitives of this function is denoted li x and is called the logarithmic integral. It satisfiesthe condition li x → 0 as x → +0. (More details about the functions Si x, Ci x, andli x will be given in Sect.
6.5.)Because of these difficulties in finding primitives, rather extensive tables of indefinite integrals have been compiled. However, in order to use these tables successfully and avoid having to resort to them when the problem is very simple, one mustacquire some skill in dealing with indefinite integrals.The remainder of this section is devoted to integrating some special classes offunctions whose primitives can be expressed as compositions of elementary functions.5.7.3 Primitives of Rational FunctionsP (x)Let us consider the problem of integrating R(x) dx, where R(x) = Q(x)is a ratioof polynomials.If we work in the domain of real numbers, then, without going outside this domain, we can express every such fraction, as we know from algebra (see formula(5.135) in Sect. 5.5.4) as a sum kj mj lnaj kbj k x + cj kP (x)= p(x) ++, (5.173)Q(x)(x − xj )k(x 2 + pj x + qj )kj =1 k=1j =1 k=1where p(x) is a polynomial (which arises when P (x) is divided by Q(x), but onlywhen the degree of P (x) is not less than the degree of Q(x)), aj k , bj k , and cj k areuniquely determined real numbers, and Q(x) = (x − x1 )k1 · · · (x − xl )kl (x 2 + p1 x +q1 )m1 · · · (x 2 + pn x + qn )mn .We have already discussed how to find the expansion (5.173) in Sect.
5.5. Oncethe expansion (5.173) has been constructed, integrating R(x) reduces to integratingthe individual terms.We have already integrated a polynomial in Example 1, so that it remains only toconsider the integration of fractions of the forms1(x − a)kand(x 2bx + c,+ px + q)kwhere k ∈ N.5.7 Primitives315The first of these problems can be solved immediately, since!1−k+1 + c for k = 1,1−k+1 (x − a)dx=k(x − a)ln |x − a| + cfor k = 1.With the integral(5.174)bx + cdx(x 2 + px + q)kwe proceed as follows. We represent the polynomial x 2 + px + q as (x + 12 p)2 +(q − 14 p 2 ), where q − 14 p 2 > 0, since the polynomial x 2 + px + q has no real roots.Setting x + 12 p = u and q − 14 p 2 = a 2 , we obtainαu + βbx + cdx=du,(x 2 + px + q)k(u2 + a 2 )kwhere α = b and β = c − 12 bp.Next,1ud(u2 + a 2 )du==22k2(u + a )(u2 + a 2 )k! 1(u2 + a 2 )−k+1= 2(1−k)1222 ln(u + a )and it remains only to study the integralIk =for k = 1,for k = 1,du.(u2 + a 2 )k(5.175)(5.176)Integrating by parts and making elementary transformations, we haveduuu2 duIk ==+2k=(u2 + a 2 )k(u2 + a 2 )k(u2 + a 2 )k+1u(u2 + a 2 ) − a 2u= 2+2kdu = 2+ 2kIk − 2ka 2 Ik+1 ,2k22k+1(u + a )(u + a )(u + a 2 )kfrom which we obtain the recursion relationIk+1 =u12k − 1+Ik ,222k2ka (u + a )2ka 2(5.177)which makes it possible to lower the exponent k in the integral (5.176).
But I1 iseasy to compute:d( ua )udu11I1 =(5.178)== arctan + c.au2 + a 2 a1 + ( ua )2 a3165Differential CalculusThus, by using (5.177) and (5.178), one can also compute the primitive (5.176).Thus we have proved the following proposition.P (x)Proposition 2 The primitive of any rational function R(x) = Q(x)can be expressedin terms of rational functions and the transcendental functions ln and arctan. Therational part of the primitive, when placed over a common denominator, will havea denominator containing all the factors of the polynomial Q(x) with multiplicitiesone less than they have in Q(x).Example 13 Let us calculate2x 2 + 5x + 5dx.(x 2 − 1)(x + 2)Since the integrand is a proper fraction, and the factorization of the denominatorinto the product (x − 1)(x + 1)(x + 2) is also known, we immediately seek a partialfraction expansion2x 2 + 5x + 5ABC=++.(x − 1)(x + 1)(x + 2) x − 1 x + 1 x + 2(5.179)Putting the right-hand side of Eq.
(5.179) over a common denominator,we have(A + B + C)x 2 + (3A + B)x + (2A − 2B − C)2x 2 + 5x + 5=.(x − 1)(x + 1)(x + 2)(x − 1)(x + 1)(x + 2)Equating the corresponding coefficients in the numerators, we obtain the system⎧⎨ A + B + C = 2,3A + B = 5,⎩2A − 2B − C = 5,from which we find (A, B, C) = (2, −1, 1).We remark that in this case these numbers could have been found in one’s head.Indeed, multiplying (5.179) by x − 1 and then setting x = 1 in the resulting equality,we would have A on the right-hand side, while the left-hand side would have beenthe value at x = 1 of the fraction obtained by striking out the factor x − 1 in thedenominator, that is, A = 2+5+52·3 = 2. One could proceed similarly to find B and C.Thus,2x 2 + 5x + 5dx = 2(x 2 − 1)(x + 2)dx−x −1dx+x +1dx=x+2= 2 ln |x − 1| − ln |x + 1| + ln |x + 2| + c (x − 1)2 (x + 2) + c.= lnx −15.7 Primitives317Example 14 Let us compute a primitive of the functionR(x) =x 7 − 2x 6 + 4x 5 − 5x 4 + 4x 3 − 5x 2 − x.(x − 1)2 (x 2 + 1)2We begin by remarking that this is an improper fraction, so that, removing theparentheses and finding the denominator Q(x) = x 6 − 2x 5 + 3x 4 − 4x 3 + 3x 2 −2x + 1, we divide the numerator by it, after which we obtainR(x) = x +x 5 − x 4 + x 3 − 3x 2 − 2x,(x − 1)2 (x 2 + 1)2and we then seek a partial-fraction expansion of the proper fractionx 5 − x 4 + x 3 − 3x 2 − 2xCx + DABEx + F+.
(5.180)=++ 2(x − 1)2 (x 2 + 1)2(x − 1)2 x − 1 (x 2 + 1)2x +1Of course the expansion could be obtained in the canonical way, by writing outa system of six equations in six unknowns. However, instead of doing that, we shalldemonstrate some other technical possibilities that are sometimes used.We find the coefficient A by multiplying Eq.
(5.180) by (x − 1)2 and then settingAx = 1. The result is A = −1. We then transpose the fraction (x−1)2 , in which A isnow the known quantity −1, to the left-hand side of Eq. (5.180). We then haveCx + DBEx + Fx 4 + x 3 + 2x 2 + x − 1+=+ 2x − 1 (x 2 + 1)2(x − 1)(x 2 + 1)2x +1(5.181)from which, multiplying (5.181) by x − 1 and then setting x = 1, we find B = 1.1Now, transposing the fraction x−1to the left-hand side of (5.181), we obtainx2 + x + 2Cx + DEx + F.= 2+ 2222(x + 1)(x + 1)x +1(5.182)Now, after putting the right-hand side of (5.182) over a common denominator,we equate the numeratorsx 2 + x + 2 = Ex 3 + F x 2 + (C + E)x + (D + F ),from which it follows thator (C, D, E, F ) = (1, 1, 0, 1).⎧E = 0,⎪⎪⎨F = 1,C + E = 1,⎪⎪⎩D + F = 2,3185Differential CalculusWe now know all the coefficients in (5.180). Upon integration, the first two frac1and ln |x − 1|. Thentions yield respectively x−1Cx + Dx +1dx = 2dx =22(x + 1)(x + 1)21dx−1d(x 2 + 1)=+ I2 ,+=22222(x + 1)(x + 1)2(x 2 + 1)whereI2 =xdx11=+ arctan x,(x 2 + 1)2 2 (x 2 + 1)2 2which follows from (5.177) and (5.178).Finally,1Ex + Fdx =dx = arctan x.22x +1x +1Gathering all the integrals, we finally havex131++ ln |x − 1| + arctan x + c.R(x) dx = x 2 +222x − 1 2(x + 1)2Let us now consider some frequently encountered indefinite integrals whosecomputation can be reduced to finding the primitive of a rational function.5.7.4 Primitives of the FormR(cos x, sin x) dxLet R(u, v) be a rational function in u and v, that is a quotient of polynomialsP (u,v)m nQ(u,v) , which are linear combinations of monomials u v , where m = 0, 1, 2 .
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