How to Avoid the Inverse Secant (and Even the Secant Itself)

Transcription

1 How to Avoi the Inverse Secant (an Even the Secant Itself) S A Fulling Stephen A Fulling (fulling@mathtamue) is Professor of Mathematics an of Physics at Teas A&M University (College Station, TX 7783) He has an AB from Harvar an a PhD from Princeton, both in physics He has written a book on quantum fiel theory in curve space-time an a tetbook on linear algebra with emphasis on applications to vector calculus an ifferential equations He has one research in various areas of mathematical physics an ifferential equations, an he has taken part in a variety of curriculum-reform efforts at the unergraate level An introctory polemic An irritating technicality that must be hanle somehow in a stanar calculus course is the efinition of the inverse secant function at negative arguments There is no particularly natural choice of principal branch from among the infinitely many caniates, an the erivatives of ifferent branches can iffer in sign (see the figure below) Some tetbook authors aopt the convention while others prefer π sec 1 < 3π for 1, sec 1 = 1 1, (1) π < sec 1 π for 1, sec 1 = 1 1 () Then authors an lecturers have a responsibility to warn stuents about the eistence of the other convention One might have epecte that after more than a ecae of calculus reform, the secant function an its inverse woul have been e-emphasize to the vanishing point, along with its even less useful siblings, cosecant an cotangent, an their inverses The persistence of sec 1 presumably stems from the perceive nee to provie a formula (see (17)) for the inefinite integral (3) 1 by inverting whichever of the ifferentiation formulas (1) an () one aopts More generally, it is argue that the secant an tangent functions unavoiably arise when algebraic functions are integrate by trigonometric substitutions, so stuents shoul have at least a noing acquaintance with the variety of integrals involving them There is, however, an alternative approach to integrals involving 1 or + 1, which has unaccountably fallen out of favor in recent ecaes: hyperbolic substitution VOL 36, NO 5, NOVEMBER 005 THE COLLEGE MATHEMATICS JOURNAL 381

2 The hyperbolic sine an cosine are efine by cosh u 1 (eu + e u ), sinh u 1 (eu e u ) () Their usefulness as replacements for trigonometric functions in integration by substitution stems from the ientities cosh u sinh u = 1, cosh u = sinh u, sinh u = cosh u (5) The hyperbolic functions () are important in their own right, constituting the natural basis of solutions of the ifferential equation y = y satisfying unit Dirichlet an Neumann initial ata at u = 0; in this role they are inispensable in upper-ivision courses in applie mathematics In a linear algebra course, () provies a beautiful eample of a change of basis, of uneniable practical importance, in a real vector space with no intrinsic inner proct or geometrical interpretation Most important in the present contet, hyperbolic substitutions are much simpler an nicer than trigonometric substitutions of the tangent an secant varieties For one thing, the inverse hyperbolic functions can be epresse in terms of more elementary functions (logarithms an square roots see (11) an (1)) Furthermore, the branch structure of these inverses is very simple: sinh is bijective, an cosh has a two-branche inverse (just like the square root), with no multiples of π to be memorize Nevertheless, many teachers of calculus have a strange islike for the hyperbolic functions an prefer not to cover them at all (An inepenent case in favor of the hyperbolics has been mae by Gearhart an Shultz [1]) In this article I hope to convince the reaer that there is nothing that the secant an inverse secant o in the traitional techniques of integration chapter that cannot be one better by the hyperbolic sine an cosine an their inverses It is time for sec, csc, cot, sec 1, csc 1, an cot 1 to be retire from our calculus syllabus, replace by sinh an cosh Our stuents will learn about two elegant an useful transcenental functions while being free from si complicate an boring ones The rest of the article has two parts First, we erive several alternative antierivative formulas, (13), (1), (16), with clear avantages over the traitional formulas (17) Then we ll see that hyperbolic substitution provies an easy way of evaluating or evaing (epening on contet) the ifficult integrals of powers of the secant that crop up in trigonometric substitutions The integral formerly known as sec 1 Since the integration problem (3) is the inverse secant s allege reason for eistence, let us see what a hyperbolic substitution oes to it Recall first that (3) makes no sense (in real analysis) unless 1 We assume temporarily that > 0, hence 1, an set = cosh u (u 0); = sinh u, 1 = sinh u (6) 38 c THE MATHEMATICAL ASSOCIATION OF AMERICA

3 Therefore, 1 = cosh u (7) (As an asie, note that not much is gaine here by introcing the name sech u for the integran of (7)) To continue we nee cosh u = tan 1 (e u ) + C, (8) which can be either verifie by ifferentiation or iscovere through these intermeiate steps: e u + e = u e u e u + 1 = v v + 1 = tan 1 v + C Formula (8) is arguably less reconite than sec θ θ = ln sec θ + tan θ +C, (9) which inevitably plays a leaing role in traitional treatments of trigonometric substitution (An alas, there is selom time in class to present the interesting history [3] of (9)) Remark The function g u tan 1 (e u ) π, (10) which satisfies the convenient initial conition g 0 = 0, is calle the Guermannian an can be use to epress hyperbolic functions as trigonometric functions (of a ifferent variable) an vice versa [, Secs 18 an 19]; for eample, if θ = g u, then sec θ = cosh u, a formula quite pertinent to the equivalence of (1) an (16) below For more on the history an applications of g, see Robertson [] As previously remarke, one of the great charms of hyperbolic functions (as oppose to trig functions) is that their inverses can be epresse in terms of alreay familiar functions: sinh 1 = ln( + + 1) for all ; (11) cosh 1 = ln( + 1) for 1, (1) where cosh 1 enotes the positive branch Note that the right-han sie of (11) is inee an o function, though it may not look like one To prove (11) an (1), simply apply the efinitions () to their right-han sies an simplify own to Combining (7), (8), (6), an (1), one arrives at 1 = tan 1 ( + 1) + C, (13) VOL 36, NO 5, NOVEMBER 005 THE COLLEGE MATHEMATICS JOURNAL 383

4 at least for 1 It is now an elementary, though lengthy, eercise in ifferentiation to verify (13) for 1 as well However, an alternative approach leas to a neater result For negative we can let = cosh u an repeat the previous calculation to obtain 1 = tan 1 ( + 1) + C, (1) which of course agrees with (13) in the positive case The formula (1) is even in, whereas (13) is not This is not a contraiction: The constants of integration on the two isconnnecte omains, 1an 1, are inepenent At negative, (13) an (1) with the same C simply iffer by a constant (see the figure) To relate these formulas to the traitional ones base on (1) an (), we start with a known but nontrivial trigonometric ientity, ( θ sec θ + tan θ = tan + π ), (15) whose proof we omit Now let =sec θ with θ in the first quarant Then tan θ = 1, an (1) becomes, up to the arbitrary constant, 1 = tan 1 (sec θ + tan θ) ( θ = + π ) In other wors, = sec 1 + π 1 = sec 1 +C (16) is an alternative formula for the inefinite integral (for either sign of ) Similarly, after appropriate bookkeeping with quarants, one can relate (13) or (1) at negative to one s favorite efinition of sec 1 there It is to be hope that any authors an lecturers who are still unconvince of the virtues of hyperbolic substitution will at least aopt (16) in place of either of the traitional formulas, 1 = sec 1 + C, For (16) is analogous to the familiar formula 1 = sec 1 + C (17) = ln +C, (18) it is even in (unlike either of (17)), an it avois any reference to the inverse secant function at negative arguments! Our conclusions so far are summarize an mae more precise in the following theorem an figure 38 c THE MATHEMATICAL ASSOCIATION OF AMERICA

5 Theorem Define f 1 () = tan 1( + 1 ), f () = tan 1( + 1 ), f 3 () = sec 1 Further, let f () be the branch of the inverse secant etermine by (1), an f 5 () the branch etermine by () (with 0 sec 1 <π/ for 1) Then (a) f 1 an f 3 (an only they) are even functions, an f 1 = f 3 + π/ (b) On the interval [1, ), all five functions are antierivatives of 1/( 1 ), an f 1 = f,f 3 = f = f 5 (c) On the interval (, 1], the first four functions an f 5 are antierivatives of 1/( 1 ), an f = f 1 π, f = f 3 + π, f 5 = π f 3 f 3 π f () f 1 π f 1 f 5 f 3 = sec 1 π f f π π 3 π Figure 1 Graphs of the functions f 1 f 5 iscusse in the theorem (Renant names for the curves on the right are omitte) Thick curves are the traitional rival branches of the inverse secant Dashe curves are graphs of other logically possible efinitions of sec 1 Thin soli curves are not branches of sec 1, but nevertheless are useful as antierivatives of 1/ 1 VOL 36, NO 5, NOVEMBER 005 THE COLLEGE MATHEMATICS JOURNAL 385

6 Integrating powers of the secant Trigonometric substitution has an unpleasant habit of resulting in integrals like the one in (9), or, more generally, sec p θ θ with p a positive integer, that look at least as har as the original algebraic integration problem Making the natural substitution = tan θ ( π <θ< π ) yiels sec p θ θ = ( + 1) (p )/ (19) The right-han sie of (19) is elementary if p is even, but what if p is o? The avice given to the stuent by the traitional tetbook is, Make the trigonometric substitution = tan θ, which takes us straight back to the left-han sie of (19) Velleman [5](see also [3]) shows that the substitution y = sin θ turns the left-han sie into the integral of a rational function, which can be integrate by partial fractions Here we investigate what hyperbolic substitution in the right-han sie has to offer The appropriate substitution this time is = sinh u; = cosh u, + 1 = cosh u (0) It turns (19) into cosh p 1 u Let s concentrate first on the case p = 1 (that is, (9)): ( + 1) 1/ = = u + C = sinh 1 + C = ln( + + 1) + C (1) by (11) If our original interest was in integrating the algebraic function, we are one; if we really care about the secant integral for its own sake, we now use = tan θ to go from (1) to (9) in the quarants where sec θ>0 (How to hanle the other quarants is left to the reaer s taste) For p = 3wehave + 1 = cosh u () There are two ways to procee, epening on taste First, the hyperbolic functions obey ientities in close parallel to those for trigonometric functions; one can memorize, look up, or reerive the ientity eactly corresponing to the one one woul use to evaluate cos θ θ On the other han, it is again a great charm of hyperbolic functions that they can always be eliminate through () in favor of the eponential function, which obeys a much simpler an shorter, but equally powerful, list of ientities (Inee, many stuents learn, even if their calculus books never tell them, that the best way to recover trig ientities is to use the function e iθ in this same way) Thus we have cosh u= 1 (e u + e u + ) = 1 ( 1 eu 1 ) e u + u 386 c THE MATHEMATICAL ASSOCIATION OF AMERICA

7 = 1 (sinh u + u) = 1 (sinh u cosh u + u) = 1 ( sinh 1 ), hence + 1 = ln( + + 1) + C, (3) an ultimately sec 3 θ θ = 1 tan θ sec θ + 1 ln sec θ + tan θ + C () Larger o values of p can in principle be treate in the same way, although, as always in this type of problem, the compleity increases Acknowlegments I thank Philip Yasskin for comments on the manuscript, an a referee for contributing the more heuristic proof of (8) References 1 W B Gearhart an H S Shultz, Tugging a barge with hyperbolic functions, College Math J 3 (003) 9 I S Grashteyn an I M Ryzhik, Table of Integrals, Series, an Procts, Acaemic Press, V F Rickey an P M Tuchinsky, An application of geography to mathematics: History of the integral of the secant, Math Mag 53 (1980) J S Robertson, Guermann an the simple penlum, College Math J 8 (1997) D J Velleman, Partial fractions, binomial coefficients, an the integral of an o power of sec θ, Amer Math Monthly 109 (00) Where there are numbers, there is beauty Proclus = 1 + 1/3 1/5 1/7 + 1/9 + 1/11 1 1/3 + 1/5 1/7 + 1/9 1/ /3 1/5 1/7 + 1/9 + 1/11 = π/ = e 1/ 1/+1/6 1/8+1/10 Submitte by Siney Kung (sineykung@yahoocom) of Jacksonville, FL VOL 36, NO 5, NOVEMBER 005 THE COLLEGE MATHEMATICS JOURNAL 387