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. .and n = 0, 1, . . . .Several methods exist for computing the integral R(cos x, sin x) dx, one ofwhich is completely general, although not always the most efficient.a.We make the change of variable t = tan x2 . Sincecos x =dt =1 − tan2 x2,1 + tan2 x2dx,2 cos2 x2that is,sin x =2 tan x2,1 + tan2 x2dx =2 dt,1 + tan2 x25.7 Primitives319it follows thatR(cos x, sin x) dx =21 − t22tR,dt,1 + t2 1 + t2 1 + t2and the problem has been reduced to integrating a rational function.However, this way leads to a very cumbersome rational function; for that reasonone should keep in mind that in many cases there are other possibilities for rationalizing the integral.b.In the case of integrals of the form R(cos2 x, sin2 x) dx or r(tan x) dx, wherer(u) is a rational function, a convenient substitution is t = tan x, since1,1 + tan2 xdxdt =, that is,cos2 xcos2 x =tan2 x,1 + tan2 xdtdx =.1 + t2sin2 x =Carrying out this substitution, we obtain respectively dt1t2,,R cos2 x, sin2 x dx = R1 + t2 1 + t2 1 + t2dtr(tan x) dx = r(t).1 + t2c.In the case of integrals of the formR cos x, sin2 x sin x dxorR cos2 x, sin x cos x dx.One can move the functions sin x and cos x into the differential and make the substitution t = cos x or t = sin x respectively.
After these substitutions, the integralswill have the formR 1 − t 2 , t dt.− R t, 1 − t 2 dt orExample 151dx2 dt=·=2t3 + sin x3 + 1+t 2 1 + t 23205=22dt=3t 2 + 2t + 3 3d(t + 13 )(t +1 23)+892=3Differential Calculusdu√u2 + ( 2 3 2 )2=3u13t + 11= √ arctan √ + c = √ arctan √ + c =22 222 23 tan x2 + 11+ c.= √ arctan√22 2Here we have used the universal change of variable t = tan x2 .Example 16dx=(sin x + cos x)2dx=+ 1)2d tan xdt1+c====−t +1(tan x + 1)2(t + 1)2cos2 x(tan x=c−1.1 + tan xExample 17dx22 sin 3x− 3 cos2 3x+1======dx=− 3 + (1 + tan2 3x))11d tan 3xdt==33 tan2 3x − 2 33t 2 − 2& d 32 t12=3 23·2 32t − 1u − 111du+c==ln√√u2 − 1 6 6 u + 1 3 6 3t − 1 21+c=√ ln 36 62t + 1 tan 3x − 2 13 + c.√ ln6 6 tan 3x + 23cos2 3x(2 tan2 3x5.7 Primitives321Example 18cos2 x d sin x(1 − t 2 ) dtcos3 xdx===t7sin7 xsin7 x −71111=t − t −5 dt = − t −6 + t −4 + c =−+ c.644 sin4 x 6 sin6 x5.7.5 Primitives of the FormR(x, y(x)) dxLet R(x, y) be, as in Sect.
5.7.4, a rational function. Let us consider some specialintegrals of the formR x, y(x) dx,where y = y(x) is a function of x.First of all, it is clear that if one can make a change of variable x = x(t) suchthat both functions x = x(t) and y = y(x(t)) are rational functions of t, then x (t)is also a rational function andR x, y(x) dx = R x(t), y x(t) x (t) dt,that is, the problem will have been reduced to integrating a rational function.Consider the following special choices of the function y = y(x).a.If y =nax+bcx+d ,where n ∈ N, then, setting t n =x=d · tn − b,a − c · tnax+bcx+d ,we obtainy = t,and the integrand rationalizes.Example 19 3 3 &t3 + 1t +1t +13 x −1=t ·dx = t ddt −dt =33x +11−t1−t1 − t3 2t3 + 1−− 1 dt ==t·1 − t31 − t3dtt3 + 1=+t −2=t·31−t(1 − t)(1 + t + t 2 )3225Differential Calculus 12+t+dt =3(1 − t) 3(1 + t + t 2 )(t + 12 ) + 32222t+dt =ln|1−t|−=31 − t3 3(t + 12 )2 + 34%$322t11+−+ ln |1 − t| − ln t +=3241 − t3 3&2123 x −1− √ arctan √ t ++ c, where t =.2x +133=2t−21 − t3b.√Let us now consider the case when y = ax 2 + bx + c, that is, integrals of the form #R x, ax 2 + bx + c dx.By completing the square in the trinomial ax 2 + bx + c and making a suitable linearsubstitution, we reduce the general case to one of the following three simple cases: # # #R t, t 2 + 1 dt,(5.183)R t, t 2 − 1 dt,R t, 1 − t 2 dt.To rationalize these integrals it now suffices to make the following substitutions,respectively:###t 2 + 1 = tu + 1, ort 2 + 1 = tu − 1, ort 2 + 1 = t − u;###t 2 − 1 = u(t − 1), ort 2 − 1 = u(t + 1), ort 2 − 1 = t − u;###1 − t 2 = u(1 − t), or1 − t 2 = u(1 + t), or1 − t 2 = tu ± 1.These substitutions were proposed long ago by Euler (see Problem 3 at the endof this section).Let us verify, for example, that after the first substitution we will have reducedthe first integral√ to the integral of a rational function.In fact, if t 2 + 1 = tu + 1, then t 2 + 1 = t 2 u2 + 2tu + 1, from which we findt=2u1 − u2and then#t2 + 1 =1 + u2.1 − u25.7 Primitives323√Thus t and t 2 + 1 have been expressed rationally in terms of u, and consequentlythe integral has been reduced to the integral of a rational function.The integrals (5.183) can also be reduced, by means of the substitutions t =sinh ϕ, t = cosh ϕ, and t = sin ϕ (or t = cos ϕ) respectively, to the following forms:R(sinh ϕ, cosh ϕ) cosh ϕ dϕ,R(cosh ϕ, sinh ϕ) sinh ϕ dϕandR(sin ϕ, cos ϕ) cos ϕ dϕExample 20or−dxR(cos ϕ, sin ϕ) sin ϕ dϕ.dt#=.=√√22x + x + 2x + 2t − 1 + t2 + 1x + (x + 1) + 1√Setting t 2 + 1 = u − t, we have 1 = u2 − 2tu, from which it follows that t =u2 −12u .
Therefore111dt111 + 2 du =du +=√22u−12u−1ut −1+ t +11du+2u2 (u − 1) 11111− 2−du == ln |u − 1| +22u−1 uu111 u − 1 += ln |u − 1| + ln+ c.22u2u√It now remains to retrace the path of substitutions: u = t + t 2 + 1 and t = x + 1.dxc. Elliptic IntegralsAnother important class of integrals consists of those of the form #R x, P (x) dx,(5.184)where P (x) is a polynomial of degree n > 2. As Abel and Liouville showed, suchan integral cannot in general be expressed in terms of elementary functions.For n = 3 and n = 4 the integral (5.184) is called an elliptic integral, and forn > 4 it is called hyperelliptic.It can be shown that by elementary substitutions the general elliptic integral canbe reduced to the following three standard forms up to terms expressible in elemen-3245Differential Calculustary functions:##dx(1 − x 2 )(1 − k 2 x 2 )x 2 dx(1 − x 2 )(1 − k 2 x 2 ),(5.185),(5.186)dx#,(1 + hx 2 ) (1 − x 2 )(1 − k 2 x 2 )(5.187)where h and k are parameters, the parameter k lying in the interval ]0, 1[ in all threecases.By the substitution x = sin ϕ these integrals can be reduced to the followingcanonical integrals and combinations of them:dϕ#,(5.188)1 − k 2 sin2 ϕ 1 − k 2 sin2 ϕ dϕ,(5.189)dϕ#.(1 + h sin ϕ) 1 − k 2 sin2 ϕ2(5.190)The integrals (5.188), (5.189) and (5.190) are called respectively the elliptic integral of first kind, second kind, and third kind (in the Legendre form).The symbols F (k, ϕ) and E(k, ϕ) respectively denote the particular elliptic integrals (5.188) and (5.189) of first and second kind that satisfy F (k, 0) = 0 andE(k, 0) = 0.The functions F (k, ϕ) and E(k, ϕ) are frequently used, and for that reason verydetailed tables of their values have been compiled for 0 < k < 1 and 0 ≤ ϕ ≤ π/2.As Abel showed, it is natural to study elliptic integrals in the complex domain, inintimate connection with the so-called elliptic functions, which functions, which arerelated to the ellipticintegrals exactly as the function sin x, for example, is relatedto the integral √ dϕ 2 = arcsin ϕ.1−ϕ5.7.6 Problems and Exercises1.
Ostrogradskii’s35 method of separating off the rational part of the integral of aproper rational fraction.35 M.V. Ostrogradskii (1801–1861) – prominent Russian specialist in theoretical mechanics andmathematician, one of the founders of the applied area of research in the Petersburg mathematicalschool.5.7 PrimitivesLetP (x)Q(x)325be a proper rational fraction, let q(x) be the polynomial having the sameroots as Q(x), but with multiplicity 1, and let Q1 (x) =Show thatQ(x)q(x) .a) the following formula of Ostrogradskii holds:P (x)P1 (x)p(x)dx =+dx,(5.191)Q(x)Q1 (x)q(x) p(x)p(x)P1 (x)where Qandareproperrationalfractionsand(x)q(x)q(x) dx is a transcendental1function.P1 (x)(Because of this result, the fraction Qin (5.191) is called the rational part of1 (x) P (x)the integral Q(x) dx.)b) In the formulaP1 (x) p(x)P (x)=+Q(x)Q1 (x)q(x)obtained by differentiating Ostrogradskii’s formula, the sum at the right hand sidecan be given the denominator Q(x) after suitable cancellations.c) The polynomials q(x), Q1 (x), and then also the polynomials p(x), P1 (x) canbe found algebraically, without even knowing the roots of Q(x).
Thus the rationalpart of the integral (5.191) can be found completely without even computing thewhole primitive.d) Separate off the rational part of the integral (5.191) ifP (x) = 2x 6 + 3x 5 + 6x 4 + 6x 3 + 10x 2 + 3x + 2,Q(x) = x 7 + 3x 6 + 5x 5 + 7x 4 + 7x 3 + 5x 2 + 3x + 1(see Example 17 in Sect.
5.5).2. Suppose we are seeking the primitiveR(cos x, sin x) dx,where R(u, v) =Show thatP (u,v)Q(u,v)(5.192)is a rational function.a) if R(−u, v) = R(u, v), then R(u, v) has the form R1 (u2 , v);b) if R(−u, v) = −R(u, v), then R(u, v) = u · R2 (u2 , v) and the substitutiont = sin x rationalizes the integral (5.192);c) If R(−u, −v) = R(u, v), then R(u, v) = R3 ( uv , v 2 ), and the substitution t =tan x rationalizes the integral (5.192).3. Integrals of the form #R x, ax 2 + bx + c dx.(5.193)3265Differential Calculusa) Verify that the integral (5.193) can be reduced to the integral of a rationalfunction by the following Euler substitutions:√√t = ax 2 + bx + c ± ax, if a > 0,21t = x−xx−x2 if x2 and x2 are real roots of the trinomial ax + bx + c.b) Let (x0 , y0 ) be a point of the curve y 2 = ax 2 + bx + c and t the slope of theline passing through (x0 , y0 ) and intersecting this curve in the point (x, y).
Expressthe coordinates (x, y) in terms of (x0 , y0 ) and t and connect these formulas withEuler’s substitutions.c) A curve defined by an algebraic equation P (x, y) = 0 is unicursal if it admitsa parametric description x =x(t), y = y(t) in terms of rational functions x(t) andy(t). Show that the integral R(x, y(x)) dx, where R(u, v) is a rational functionand y(x) is an algebraic function satisfying the equation P (x, y) = 0 that definesthe unicursal curve, can be reduced to the integral of a rational function.d) Show that the integral (5.193) can always be reduced to computing integralsof the following three types:dxP (x)dx,√√ax 2 + bx + c(x − x0 )k · ax 2 + bx + c(Ax + B) dx.√(x 2 + px + a)m · ax 2 + bx + c4. a) Show that the integralpx m a + bx n dxwhose differential is a binomial, where m, n, and p are rational numbers, can bereduced to the integral(a + bt)p t q dt,(5.194)where p and q are rational numbers.b) The integral (5.194) can be expressed in terms of elementary functions if oneof the three numbers p, q, and p + q is an integer.
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