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Chapter VI
Derivatives and Integrals

110. Derivatives or Differential Coefficients. Let us return to the consideration of the properties which we naturally associate with the notion of a curve. The first and most obvious property is, as we saw in the last chapter, that which gives a curve its appearance of connectedness, and which we embodied in our definition of a continuous function.

The ordinary curves which occur in elementary geometry, such as straight lines, circles and conic sections, have of course many other properties of a general character. The simplest and most noteworthy of these is perhaps that they have a definite direction at every point, or what is the same thing, that at every point of the curve we can draw a tangent to it. The reader will probably remember that in elementary geometry the tangent to a curve at P is defined to be ‘the limiting position of the chord PQ, when Q moves up towards coincidence with P’. Let us consider what is implied in the assumption of the existence of such a limiting position.

In the figure (Fig. 36) P is a fixed point on the curve, and Q a variable point; PM, QN are parallel to OY and PR to OX. We denote the coordinates of P by x, y and those of Q by x + h, y + k: h will be positive or negative according as N lies to the right or left of M.

We have assumed that there is a tangent to the curve at P, or that there is a definite ‘limiting position’ of the chord PQ. Suppose that PT, the tangent at P, makes an angle ψ with OX. Then to say that PT is the limiting position of PQ is equivalent to saying that the limit of the angle QPR is ψ, when Q approaches P along the curve from either side. We have now to distinguish two cases, a general case and an exceptional one.

pict

Fig. 36.

The general case is that in which ψ is not equal to 1 2π, so that PT is not parallel to OY . In this case RPQ tends to the limit ψ, and

RQ/PR = tan RPQ
tends to the limit tan ψ. Now
RQ/PR = (NQ − MP)/MN = {φ(x + h) − φ(x)}/h;
and so

lim h→0φ(x + h) − φ(x) h = tan ψ. (1)

The reader should be careful to note that in all these equations all lengths are regarded as affected with the proper sign, so that (e.g.) RQ is negative in the figure when Q lies to the left of P; and that the convergence to the limit is unaffected by the sign of h.

Thus the assumption that the curve which is the graph of φ(x) has a tangent at P, which is not perpendicular to the axis of x, implies that φ(x) has, for the particular value of x corresponding to P, the property that {φ(x + h) − φ(x)}/h tends to a limit when h tends to zero.

This of course implies that both of

{φ(x + h) −φ(x)}/h,{φ(x −h) −φ(x)}/(−h)
tend to limits when h →0 by positive values only, and that the two limits are equal. If these limits exist but are not equal, then the curve y = φ(x) has an angle at the particular point considered, as in Fig. 37.

Now let us suppose that the curve has (like the circle or ellipse) a tangent at every point of its length, or at any rate every portion of its length which corresponds to a certain range of variation of x. Further let us suppose this tangent never perpendicular to the axis of x: in the case of a circle this would of course restrict us to considering an arc less than a semicircle. Then an equation such as (1) holds for all values of x which fall inside this range. To each such value of x corresponds a value of tan ψ: tan ψ is a function of x, which is defined for all values of x in the range of values under consideration, and which may be calculated or derived from the original function φ(x). We shall call this function the derivative or derived function of φ(x), and we shall denote it by

φ′(x).

Another name for the derived function of φ(x) is the differential coefficient of φ(x); and the operation of calculating φ′(x) from φ(x) is generally known as differentiation. This terminology is firmly established for historical reasons: see § 115.

Before we proceed to consider the special case mentioned above, in which ψ = 1 2π, we shall illustrate our definition by some general remarks and particular illustrations.

111. Some general remarks. (1) The existence of a derived function φ′(x) for all values of x in the interval a ≦ x ≦ b implies that φ(x) is continuous at every point of this interval. For it is evident that {φ(x + h) − φ(x)}/h cannot tend to a limit unless lim φ(x + h) = φ(x), and it is this which is the property denoted by continuity.

(2) It is natural to ask whether the converse is true, i.e. whether every continuous curve has a definite tangent at every point, and

pict

Fig. 37.

every function a differential coefficient for every value of x for which it is continuous.63 The answer is obviously No: it is sufficient to consider the curve formed by two straight lines meeting to form an angle (Fig. 37). The reader will see at once that in this case {φ(x + h) − φ(x)}/h has the limit tan β when h → 0 by positive values and the limit tan α when h → 0 by negative values.

This is of course a case in which a curve might reasonably be said to have two directions at a point. But the following example, although a little more difficult, shows conclusively that there are cases in which a continuous curve cannot be said to have either one direction or several directions at one of its points. Draw the graph (Fig. 14, p. 164) of the function xsin(1/x). The function is not defined for x = 0, and so is discontinuous for x = 0. On the other hand the function defined by the equations

φ(x) = xsin(1/x)(x≠0),φ(x) = 0(x = 0)
is continuous for x = 0 (Exs. XXXVII. 14, 15), and the graph of this function is a continuous curve.

But φ(x) has no derivative for x = 0. For φ′(0) would be, by definition, lim{φ(h) −φ(0)}/h or limsin(1/h); and no such limit exists.

It has even been shown that a function of x may be continuous and yet have no derivative for any value of x, but the proof of this is much more difficult. The reader who is interested in the question may be referred to Bromwich’s Infinite Series, pp. 490–1, or Hobson’s Theory of Functions of a Real Variable, pp. 620–5.

(3) The notion of a derivative or differential coefficient was suggested to us by geometrical considerations. But there is nothing geometrical in the notion itself. The derivative φ′(x) of a function φ(x) may be defined, without any reference to any kind of geometrical representation of φ(x), by the equation

φ′(x) = lim h→0φ(x + h) − φ(x) h ;
and φ(x) has or has not a derivative, for any particular value of x, according as this limit does or does not exist. The geometry of curves is merely one of many departments of mathematics in which the idea of a derivative finds an application.

Another important application is in dynamics. Suppose that a particle is moving in a straight line in such a way that at time t its distance from a fixed point on the line is s = φ(t). Then the ‘velocity of the particle at time t’ is by definition the limit of

φ(t + h) −φ(t) h
as h →0. The notion of ‘velocity’ is in fact merely a special case of that of the derivative of a function.

Examples XXXIX. 1. If φ(x) is a constant then φ′(x) = 0. Interpret this result geometrically.

2. If φ(x) = ax + b then φ′(x) = a. Prove this (i) from the formal definition and (ii) by geometrical considerations.

3. If φ(x) = xm, where m is a positive integer, then φ′(x) = mxm−1.

[For

φ′(x) = lim (x + h)m −xm h = lim mxm−1 + m(m −1) 1 ⋅2 xm−2h + ⋯ + hm−1 .

The reader should observe that this method cannot be applied to xp/q, where p/q is a rational fraction, as we have no means of expressing (x + h)p/q as a finite series of powers of h. We shall show later on (§118) that the result of this example holds for all rational values of m. Meanwhile the reader will find it instructive to determine φ′(x) when m has some special fractional value (e.g. 1 2), by means of some special device.]

4. If φ(x) = sinx, then φ′(x) = cosx; and if φ(x) = cosx, then φ′(x) = −sinx.

[For example, if φ(x) = sinx, we have

{φ(x + h) −φ(x)}/h = {2sin 1 2hcos(x + 1 2h)}/h,
the limit of which, when h →0, is cosx, since limcos(x + 1 2h) = cosx (the cosine being a continuous function) and lim{(sin 1 2h)/1 2h}= 1 (Ex. XXXVI. 13).]

5. Equations of the tangent and normal to a curve y = φ(x). The tangent to the curve at the point (x0,y0) is the line through (x0,y0) which makes with OX an angle ψ, where tanψ = φ′(x0). Its equation is therefore

y −y0 = (x −x0)φ′(x0);
and the equation of the normal (the perpendicular to the tangent at the point of contact) is
(y −y0)φ′(x0) + x −x0 = 0.
We have assumed that the tangent is not parallel to the axis of y. In this special case it is obvious that the tangent and normal are x = x0 and y = y0 respectively.

6. Write down the equations of the tangent and normal at any point of the parabola x2 = 4ay. Show that if x0 = 2a/m, y0 = a/m2, then the tangent at (x0,y0) is x = my + (a/m).

112. We have seen that if φ(x) is not continuous for a value of x then it cannot possibly have a derivative for that value of x. Thus such functions as 1/x or sin(1/x), which are not defined for x = 0, and so necessarily discontinuous for x = 0, cannot have derivatives for x = 0. Or again the function [x], which is discontinuous for every integral value of x, has no derivative for any such value of x.

Example. Since [x] is constant between every two integral values of x, its derivative, whenever it exists, has the value zero. Thus the derivative of [x], which we may represent by [x]′, is a function equal to zero for all values of x save integral values and undefined for integral values. It is interesting to note that the function 1 −sinπx sinπx has exactly the same properties.

We saw also in Ex. XXXVII. 7 that the types of discontinuity which occur most commonly, when we are dealing with the very simplest and most obvious kinds of functions, such as polynomials or rational or trigonometrical functions, are associated with a relation of the type

φ(x) → +∞
or φ(x) →−∞. In all these cases, as in such cases as those considered above, there is no derivative for certain special values of x.

pict

Fig. 38.

In fact, as was pointed out in § 111, (1), all discontinuities of φ(x) are also discontinuities of φ′(x). But the converse is not true, as we may easily see if we return to the geometrical point of view of § 110 and consider the special case, hitherto left aside, in which the graph of φ(x) has a tangent parallel to OY . This case may be subdivided into a number of cases, of which the most typical are shown in Fig. 38. In cases (c) and (d) the function is two valued on one side of P and not defined on the other. In such cases we may consider the two sets of values of φ(x), which occur on one side of P or the other, as defining distinct functions φ1(x) and φ2(x), the upper part of the curve corresponding to φ1(x).

The reader will easily convince himself that in (a)

{φ(x + h) − φ(x)}/h → +∞,
as h → 0, and in (b)
{φ(x + h) − φ(x)}/h →−∞;
while in (c)
{φ1(x + h) − φ1(x)}/h → +∞,{φ2(x + h) − φ2(x)}/h →−∞,
and in (d)
{φ1(x + h) − φ1(x)}/h →−∞,{φ2(x + h) − φ2(x)}/h → +∞,
though of course in (c) only positive and in (d) only negative values of h can be considered, a fact which by itself would preclude the existence of a derivative.

We can obtain examples of these four cases by considering the functions defined by the equations

(a)y3 = x,(b) y3 = −x,(c) y2 = x,(d)y2 = −x,
the special value of x under consideration being x = 0.

113. Some general rules for differentiation. Throughout the theorems which follow we assume that the functions f(x) and F(x) have derivatives f′(x) and F′(x) for the values of x considered.

(1) If φ(x) = f(x) + F(x), then φ(x) has a derivative

φ′(x) = f′(x) + F′(x).

(2) If φ(x) = kf(x), where k is a constant, then φ(x) has a derivative

φ′(x) = kf′(x).

We leave it as an exercise to the reader to deduce these results from the general theorems stated in Ex. XXXV. 1.

(3) If φ(x) = f(x)F(x), then φ(x) has a derivative

φ′(x) = f(x)F′(x) + f′(x)F(x).

For

φ′(x) = lim f(x + h)F(x + h) − f(x)F(x) h = lim f(x + h)F(x + h) − F(x) h + F(x)f(x + h) − f(x) h = f(x)F′(x) + F(x)f′(x).

(4) If φ(x) = 1 f(x), then φ(x) has a derivative

φ′(x) = − f′(x) {f(x)}2.

In this theorem we of course suppose that f(x) is not equal to zero for the particular value of x under consideration. Then

φ′(x) = lim 1 h f(x) − f(x + h) f(x + h)f(x) = − f′(x) {f(x)}2.

(5) If φ(x) = f(x) F(x), then φ(x) has a derivative

φ′(x) = f′(x)F(x) − f(x)F′(x) {F(x)}2 .

This follows at once from (3) and (4).

(6) If φ(x) = F{f(x)}, then φ(x) has a derivative

φ′(x) = F′{f(x)}f′(x).

For let

f(x) = y,f(x + h) = y + k.
Then k → 0 as h → 0, and k/h → f′(x). And

φ′(x) = lim F{f(x + h)}− F{f(x)} h = lim F(y + k) − F(y) k × lim k h = F′(y)f′(x).

This theorem includes (2) and (4) as special cases, as we see on taking F(x) = kx or F(x) = 1/x. Another interesting special case is that in which f(x) = ax + b: the theorem then shows that the derivative of F(ax + b) is aF′(ax + b).

Our last theorem requires a few words of preliminary explanation. Suppose that x = ψ(y), where ψ(y) is continuous and steadily increasing (or decreasing), in the stricter sense of § 95, in a certain interval of values of y. Then we may write y = φ(x), where φ is the ‘inverse’ function (§ 109) of ψ.

(7) If y = φ(x), where φ is the inverse function of ψ, so that x = ψ(y), and ψ(y) has a derivative ψ′(y) which is not equal to zero, then φ(x) has a derivative

φ′(x) = 1 ψ′(y).

For if φ(x + h) = y + k, then k → 0 as h → 0, and

φ′(x) = lim h→0φ(x + h) − φ(x) (x + h) − x = lim k→0 (y + k) − y ψ(y + k) − ψ(y) = 1 ψ′(y).
The last function may now be expressed in terms of x by means of the relation y = φ(x), so that φ′(x) is the reciprocal of ψ′{φ(x)}. This theorem enables us to differentiate any function if we know the derivative of the inverse function.

114. Derivatives of complex functions. So far we have supposed that y = φ(x) is a purely real function of x. If y is a complex function φ(x) + iψ(x), then we define the derivative of y as being φ′(x) + iψ′(x). The reader will have no difficulty in seeing that Theorems (1)–(5) above retain their validity when φ(x) is complex. Theorems (6) and (7) have also analogues for complex functions, but these depend upon the general notion of a ‘function of a complex variable’, a notion which we have encountered at present only in a few particular cases.

115. The notation of the differential calculus. We have already explained that what we call a derivative is often called a differential coefficient. Not only a different name but a different notation is often used; the derivative of the function y = φ(x) is often denoted by one or other of the expressions

Dxy,dy dx.
Of these the last is the most usual and convenient: the reader must however be careful to remember that dy/dx does not mean ‘a certain number dy divided by another number dx’: it means ‘the result of a certain operation Dx or d/dx applied to y = φ(x)’, the operation being that of forming the quotient {φ(x + h) − φ(x)}/h and making h → 0.

Of course a notation at first sight so peculiar would not have been adopted without some reason, and the reason was as follows. The denominator h of the fraction {φ(x + h) −φ(x)}/h is the difference of the values x + h, x of the independent variable x; similarly the numerator is the difference of the corresponding values φ(x + h), φ(x) of the dependent variable y. These differences may be called the increments of x and y respectively, and denoted by δx and δy. Then the fraction is δy/δx, and it is for many purposes convenient to denote the limit of the fraction, which is the same thing as φ′(x), by dy/dx. But this notation must for the present be regarded as purely symbolical. The dy and dx which occur in it cannot be separated, and standing by themselves they would mean nothing: in particular dy and dx do not mean limδy and limδx, these limits being simply equal to zero. The reader will have to become familiar with this notation, but so long as it puzzles him he will be wise to avoid it by writing the differential coefficient in the form Dxy, or using the notation φ(x), φ′(x), as we have done in the preceding sections of this chapter.

In Ch. VII, however, we shall show how it is possible to define the symbols dx and dy in such a way that they have an independent meaning and that the derivative dy/dx is actually their quotient.

The theorems of § 113 may of course at once be translated into this notation. They may be stated as follows:

(1) if y = y1 + y2, then

dy dx = dy1 dx + dy2 dx ;

(2) if y = ky1, then

dy dx = kdy1 dx ;

(3) if y = y1y2, then

dy dx = y1dy2 dx + y2dy1 dx ;

(4) if y = 1 y1, then

dy dx = − 1 y12dy1 dx ;

(5) if y = y1 y2, then

dy dx =(y2dy1 dx − y1dy2 dx )y22;

(6) if y is a function of x, and z a function of y, then

dz dx = dz dydy dx;
(7) dy dx = 1(dx dy).

Examples XL. 1. If y = y1y2y3 then

dy dx = y2y3dy1 dx + y3y1dy2 dx + y1y2dy3 dx ,
and if y = y1y2…yn then
dy dx = ∑ r=1ny 1y2…yr−1yr+1…yndyr dx .
In particular, if y = zn, then dy/dx = nzn−1(dz/dx); and if y = xn, then dy/dx = nxn−1, as was proved otherwise in Ex. XXXIX. 3.

2. If y = y1y2…yn then

1 ydy dx = 1 y1dy1 dx + 1 y2dy2 dx + ⋯ + 1 yndyn dx .
In particular, if y = zn, then 1 ydy dx = n zdz dx.

116. Standard forms. We shall now investigate more systematically the forms of the derivatives of a few of the the simplest types of functions.

A. Polynomials. If φ(x) = a0xn + a 1xn−1 + ⋯ + a n, then

φ′(x) = na0xn−1 + (n − 1)a 1xn−2 + ⋯ + a n−1.
It is sometimes more convenient to use for the standard form of a polynomial of degree n in x what is known as the binomial form, viz.
a0xn + n 1a1xn−1 + n 2a2xn−2 + ⋯ + a n.
In this case
φ′(x) = n a0xn−1 + n − 1 1 a1xn−2 + n − 1 2 a2xn−3 + ⋯ + a n−1 .

The binomial form of φ(x) is often written symbolically as

(a0,a1,…,an )( x, 1)n;
and then
φ′(x) = n(a0,a1,…,an−1 )( x, 1)n−1.

We shall see later that φ(x) can always be expressed as the product of n factors in the form

φ(x) = a0(x − α1)(x − α2)…(x − αn),
where the α’s are real or complex numbers. Then
φ′(x) = a0∑ (x − α2)(x − α3)…(x − αn),
the notation implying that we form all possible products of n − 1 factors, and add them all together. This form of the result holds even if several of the numbers α are equal; but of course then some of the terms on the right-hand side are repeated. The reader will easily verify that if
φ(x) = a0(x − α1)m1 (x − α2)m2 …(x − αν)mν ,
then
φ′(x) = a0∑ m1(x − α1)m1−1(x − α 2)m2 …(x − αν)mν .

Examples XLI. 1. Show that if φ(x) is a polynomial then φ′(x) is the coefficient of h in the expansion of φ(x + h) in powers of h.

2. If φ(x) is divisible by (x −α)2, then φ′(x) is divisible by x −α: and generally, if φ(x) is divisible by (x −α)m, then φ′(x) is divisible by (x −α)m−1.

3. Conversely, if φ(x) and φ′(x) are both divisible by x −α, then φ(x) is divisible by (x −α)2; and if φ(x) is divisible by x −α and φ′(x) by (x −α)m−1, then φ(x) is divisible by (x −α)m.

4. Show how to determine as completely as possible the multiple roots of P(x) = 0, where P(x) is a polynomial, with their degrees of multiplicity, by means of the elementary algebraical operations.

[If H1 is the highest common factor of P and P′, H2 the highest common factor of H1 and P′′, H3 that of H2 and P′′′, and so on, then the roots of H1H3/H22 = 0 are the double roots of P = 0, the roots of H2H4/H32 = 0 the treble roots, and so on. But it may not be possible to complete the solution of H1H3/H22 = 0, H2H4/H32 = 0, …. Thus if P(x) = (x −1)3(x5 −x −7)2 then H1H3/H22 = x5 −x −7 and H2H4/H32 = x −1; and we cannot solve the first equation.]

5. Find all the roots, with their degrees of multiplicity, of

x4 + 3x3 −3x2 −11x −6 = 0,x6 + 2x5 −8x4 −14x3 + 11x2 + 28x + 12 = 0.

6. If ax2 + 2bx + c has a double root, i.e. is of the form a(x −α)2, then 2(ax + b) must be divisible by x −α, so that α = −b/a. This value of x must satisfy ax2 + 2bx + c = 0. Verify that the condition thus arrived at is ac −b2 = 0.

7. The equation 1/(x −a) + 1/(x −b) + 1/(x −c) = 0 can have a pair of equal roots only if a = b = c.

(Math. Trip. 1905.)

8. Show that

ax3 + 3bx2 + 3cx + d = 0
has a double root if G2 + 4H3 = 0, where H = ac −b2, G = a2d −3abc + 2b3.

[Put ax + b = y, when the equation reduces to y3 + 3Hy + G = 0. This must have a root in common with y2 + H = 0.]

9. The reader may verify that if α, β, γ, δ are the roots of

ax4 + 4bx3 + 6cx2 + 4dx + e = 0,
then the equation whose roots are
1 12a{(α −β)(γ −δ) −(γ −α)(β −δ)},
and two similar expressions formed by permuting α, β, γ cyclically, is
4θ3 −g 2θ −g3 = 0,
where
g2 = ae −4bd + 3c2,g 3 = ace + 2bcd −ad2 −eb2 −c3.
It is clear that if two of α, β, γ, δ are equal then two of the roots of this cubic will be equal. Using the result of Ex. 8 we deduce that g23 −27g32 = 0.

10. Rolle’s Theorem for polynomials. If φ(x) is any polynomial, then between any pair of roots of φ(x) = 0 lies a root of φ′(x) = 0.

A general proof of this theorem, applying not only to polynomials but to other classes of functions, will be given later. The following is an algebraical proof valid for polynomials only. We suppose that α, β are two successive roots, repeated respectively m and n times, so that

φ(x) = (x −α)m(x −β)nθ(x),
where θ(x) is a polynomial which has the same sign, say the positive sign, for α ≦x ≦β. Then

φ′(x) = (x − α)m(x − β)nθ′(x) + {m(x − α)m−1(x − β)n + n(x − α)m(x − β)n−1}θ(x) = (x − α)m−1(x − β)n−1[(x − α)(x − β)θ′(x) + {m(x − β) + n(x − α)}θ(x)] = (x − α)m−1(x − β)n−1F(x),

say. Now F(α) = m(α −β)θ(α) and F(β) = n(β −α)θ(β), which have opposite signs. Hence F(x), and so φ′(x), vanishes for some value of x between α and β.

117. B. Rational Functions. If

R(x) = P(x) Q(x),
where P and Q are polynomials, it follows at once from § 113, (5) that
R′(x) = P′(x)Q(x) − P(x)Q′(x) {Q(x)}2 ,
and this formula enables us to write down the derivative of any rational function. The form in which we obtain it, however, may or may not be the simplest possible. It will be the simplest possible if Q(x) and Q′(x) have no common factor, i.e. if Q(x) has no repeated factor. But if Q(x) has a repeated factor then the expression which we obtain for R′(x) will be capable of further reduction.

It is very often convenient, in differentiating a rational function, to employ the method of partial fractions. We shall suppose that Q(x), as in § 116, is expressed in the form

a0(x − α1)m1 (x − α2)m2 …(x − αν)mν .
Then it is proved in treatises on Algebra64 that R(x) can be expressed in the form

Π(x) + A1,1 x − α1 + A1,2 (x − α1)2 + ⋯ + A1,m1 (x − α1)m1 + A2,1 x − α2 + A2,2 (x − α2)2 + ⋯ + A2,m2 (x − α2)m2 + …,

where Π(x) is a polynomial; i.e. as the sum of a polynomial and the sum of a number of terms of the type

A (x − α)p,
where α is a root of Q(x) = 0. We know already how to find the derivative of the polynomial: and it follows at once from Theorem (4) of § 113, or, if α is complex, from its extension indicated in § 114, that the derivative of the rational function last written is
−pA(x − α)p−1 (x − α)2p = − pA (x − α)p+1.

We are now able to write down the derivative of the general rational function R(x), in the form

Π′(x) − A1,1 (x − α1)2 − 2A1,2 (x − α1)3 −⋯ − A2,1 (x − α2)2 − 2A2,2 (x − α2)3 −….
Incidentally we have proved that the derivative of xm is mxm−1, for all integral values of m positive or negative.

The method explained in this section is particularly useful when we have to differentiate a rational function several times (see Exs. XLV).

Examples XLII. 1. Prove that

d dx x 1 + x2 = 1 −x2 (1 + x2)2, d dx 1 −x2 1 + x2 = − 4x (1 + x2)2.

2. Prove that

d dx ax2 + 2bx + c Ax2 + 2Bx + C = (ax + b)(Bx + C) −(bx + c)(Ax + B) (Ax2 + 2Bx + C)2 .

3. If Q has a factor (x −α)m then the denominator of R′(when R′ is reduced to its lowest terms) is divisible by (x −α)m+1 but by no higher power of x −α.

4. In no case can the denominator of R′have a simple factor x −α. Hence no rational function (such as 1/x) whose denominator contains any simple factor can be the derivative of another rational function.

118. C. Algebraical Functions. The results of the preceding sections, together with Theorem (6) of § 113, enable us to obtain the derivative of any explicit algebraical function whatsoever.

The most important such function is xm, where m is a rational number. We have seen already (§ 117) that the derivative of this function is mxm−1 when m is an integer positive or negative; and we shall now prove that this result is true for all rational values of m. Suppose that y = xm = xp/q, where p and q are integers and q positive; and let z = x1/q, so that x = zq and y = zp. Then

dy dx =(dy dz)(dx dz) = p qzp−q = mxm−1.

This result may also be deduced as a corollary from Ex. XXXVI. 3. For, if φ(x) = xm, we have

φ′(x) = lim h→0(x + h)m − xm h = lim ξ→xξm − xm ξ − x = mxm−1.

It is clear that the more general formula

d dx(ax + b)m = ma(ax + b)m−1
holds also for all rational values of m.

The differentiation of implicit algebraical functions involves certain theoretical difficulties to which we shall return in Ch. VII. But there is no practical difficulty in the actual calculation of the derivative of such a function: the method to be adopted will be illustrated sufficiently by an example. Suppose that y is given by the equation

x3 + y3 − 3axy = 0.
Differentiating with respect to x we find
x2 + y2 dy dx − a y + xdy dx = 0
and so
dy dx = −x2 − ay y2 − ax.

Examples XLIII. 1. Find the derivatives of

1 + x 1 −x,ax + b cx + d, ax2 + 2bx + c Ax2 + 2Bx + C,(ax + b)m(cx + d)n.

2. Prove that

d dx x a2 + x2 = a2 (a2 + x2)(3/2), d dx x a2 −x2 = a2 (a2 −x2)3/2.

3. Find the differential coefficient of y when

(i) ax2 + 2hxy + by2 + 2gx + 2fy + c = 0,(ii) x5 + y5 −5ax2y2 = 0.

119. D. Transcendental Functions. We have already proved (Ex. XXXIX. 4) that

Dx sin x = cos x,Dx cos x = − sin x.

By means of Theorems (4) and (5) of § 113, the reader will easily verify that

Dx tan x = sec 2x, D x cot x = − cosec 2x, Dx sec x = tan x sec x,Dx cosec x = − cot x cosec x.

And by means of Theorem (7) we can determine the derivatives of the ordinary inverse trigonometrical functions. The reader should verify the following formulae:

Dx arcsin x = ±1/1 − x2, D x arccos x = ∓1/1 − x2, Dx arctan x = 1/(1 + x2), D x arccot x = −1/(1 + x2), Dx arcsec x = ±1/{xx2 − 1},D x arccosec x = ∓1/{xx2 − 1}.

In the case of the inverse sine and cosecant the ambiguous sign is the same as that of  cos(arcsin x), in the case of the inverse cosine and secant the same as that of  sin(arccos x).

The more general formulae

Dx arcsin(x/a) = ±1/a2 − x2,D x arctan(x/a) = a/(x2 + a2), which are also easily derived from Theorem (7) of § 113, are also of considerable importance. In the first of them the ambiguous sign is the same as that of a cos{arcsin(x/a)}, since
a1 − (x2/a2) = ±a2 − x2
according as a is positive or negative.

Finally, by means of Theorem (6) of § 113, we are enabled to differentiate composite functions involving symbols both of algebraical and trigonometrical functionality, and so to write down the derivative of any such function as occurs in the following examples.

Examples XLIV.65 1. Find the derivatives of

cosmx,sinmx,cosxm,sinxm,cos(sinx),sin(cosx), a2 cos2x + b2 sin2x, cosxsinx a2 cos2x + b2 sin2x, xarcsinx + 1 −x2,(1 + x)arctanx −x.

2. Verify by differentiation that arcsinx + arccosx is constant for all values of x between 0 and 1, and arctanx + arccotx for all positive values of x.

3. Find the derivatives of

arcsin1 −x2,arcsin{2x1 −x2},arctan a + x 1 −ax. How do you explain the simplicity of the results?

4. Differentiate

1 ac −b2 arctan ax + b ac −b2, − 1 −aarcsin ax + b b2 −ac.

5. Show that each of the functions

2arcsinx −β α −β,2arctanx −β α −x,arcsin 2(α −x)(x −β) α −β has the derivative
1 (α −x)(x −β).

6. Prove that

d dθ arccoscos3θ cos3θ = 3 cosθcos3θ.
(Math. Trip. 1904.)

7. Show that

1 C(Ac −aC) d dx arccos C(ax2 + c) c(Ax2 + C) = 1 (Ax2 + C)ax2 + c.

8. Each of the functions

1 a2 −b2 arccos acosx + b a + bcosx, 2 a2 −b2 arctan a −b a + btan 1 2x has the derivative 1/(a + bcosx).

9. If X = a + bcosx + csinx, and

y = 1 a2 −b2 −c2 arccos aX −a2 + b2 + c2 Xb2 + c2 , then dy/dx = 1/X.

10. Prove that the derivative of F[f{φ(x)}] is F′[f{φ(x)}]f′{φ(x)}φ′(x), and extend the result to still more complicated cases.

11. If u and v are functions of x, then

Dx arctan(u/v) = (vDxu −uDxv)/(u2 + v2).

12. The derivative of y = (tanx + secx)m is mysecx.

13. The derivative of y = cosx + isinx is iy.

14. Differentiate xcosx, (sinx)/x. Show that the values of x for which the tangents to the curves y = xcosx, y = (sinx)/x are parallel to the axis of x are roots of cotx = x, tanx = x respectively.

15. It is easy to see (cf. Ex. XVII. 5) that the equation sinx = ax, where a is positive, has no real roots except x = 0 if a ≧1, and if a < 1 a finite number of roots which increases as a diminishes. Prove that the values of a for which the number of roots changes are the values of cosξ, where ξ is a positive root of the equation tanξ = ξ. [The values required are the values of a for which y = ax touches y = sinx.]

16. If φ(x) = x2 sin(1/x) when x≠0, and φ(0) = 0, then

φ′(x) = 2xsin(1/x) −cos(1/x)
when x≠0, and φ′(0) = 0. And φ′(x) is discontinuous for x = 0 (cf. §111, (2)).

17. Find the equations of the tangent and normal at the point (x0,y0) of the circle x2 + y2 = a2.

[Here y = a2 −x2, dy/dx = −x/a2 −x2, and the tangent is

y −y0 = (x −x0) −x0/a2 −x 02 ,
which may be reduced to the form xx0 + yy0 = a2. The normal is xy0 −yx0 = 0, which of course passes through the origin.]

18. Find the equations of the tangent and normal at any point of the ellipse (x/a)2 + (y/b)2 = 1 and the hyperbola (x/a)2 −(y/b)2 = 1.

19. The equations of the tangent and normal to the curve x = φ(t), y = ψ(t), at the point whose parameter is t, are

x −φ(t) φ′(t) = y −ψ(t) ψ′(t) ,{x −φ(t)}φ′(t) + {y −ψ(t)}ψ′(t) = 0.

120. Repeated differentiation. We may form a new function φ′′(x) from φ′(x) just as we formed φ′(x) from φ(x). This function is called the second derivative or second differential coefficient of φ(x). The second derivative of y = φ(x) may also be written in any of the forms

Dx2y, d dx2y,d2y dx2.

In exactly the same way we may define the nth derivative or nth differential coefficient of y = φ(x), which may be written in any of the forms

φ(n)(x),D xny, d dxny,dny dxn.
But it is only in a few cases that it is easy to write down a general formula for the nth differential coefficient of a given function. Some of these cases will be found in the examples which follow.

Examples XLV. 1. If φ(x) = xm then

φ(n)(x) = m(m −1)…(m −n + 1)xm−n.
This result enables us to write down the nth derivative of any polynomial.

2. If φ(x) = (ax + b)m then

φ(n)(x) = m(m −1)…(m −n + 1)an(ax + b)m−n.
In these two examples m may have any rational value. If m is a positive integer, and n > m, then φ(n)(x) = 0.

3. The formula

d dxn A (x −α)p = (−1)np(p + 1)…(p + n −1)A (x −α)p+n
enables us to write down the nth derivative of any rational function expressed in the standard form as a sum of partial fractions.

4. Prove that the nth derivative of 1/(1 −x2) is

1 2(n!){(1 −x)−n−1 + (−1)n(1 + x)−n−1}.

5. Leibniz’ Theorem. If y is a product uv, and we can form the first n derivatives of u and v, then we can form the nth derivative of y by means of Leibniz’ Theorem, which gives the rule

(uv)n = unv + n 1un−1v1 + n 2un−2v2 + ⋯ + n r un−rvr + ⋯ + uvn,
where suffixes indicate differentiations, so that un, for example, denotes the nth derivative of u. To prove the theorem we observe that

(uv)1 = u1v + uv1, (uv)2 = u2v + 2u1v1 + uv2,

and so on. It is obvious that by repeating this process we arrive at a formula of the type

(uv)n = unv + an,1un−1v1 + an,2un−2v2 + ⋯ + an,run−rvr + ⋯ + uvn.

Let us assume that an,r = n r for r = 1, 2, …, n −1, and show that if this is so then an+1,r = n + 1 r for r = 1, 2, … n. It will then follow by the principle of mathematical induction that an,r = n r for all values of n and r in question.

When we form (uv)n+1 by differentiating (uv)n it is clear that the coefficient of un+1−rvr is

an,r + an,r−1 = n r + n r −1 = n + 1 r .
This establishes the theorem.

6. The nth derivative of xmf(x) is

m! (m −n)!xm−nf(x) + n m! (m −n + 1)!xm−n+1f′(x) + n(n −1) 1 ⋅2 m! (m −n + 2)!xm−n+2f′′(x) + …,

the series being continued for n + 1 terms or until it terminates.

7. Prove that Dxn cosx = cos(x + 1 2nπ), Dxn sinx = sin(x + 1 2nπ).

8. If y = Acosmx + Bsinmx then Dx2y + m2y = 0. And if

y = Acosmx + Bsinmx + Pn(x),
where Pn(x) is a polynomial of degree n, then Dxn+3y + m2Dxn+1y = 0.

9. If x2Dx2y + xDxy + y = 0 then

x2D xn+2y + (2n + 1)xD xn+1y + (n2 + 1)D xny = 0.

[Differentiate n times by Leibniz’ Theorem.]

10. If Un denotes the nth derivative of (Lx + M)/(x2 −2Bx + C), then

x2 −2Bx + C (n + 1)(n + 2)Un+2 + 2(x −B) n + 1 Un+1 + Un = 0.
(Math. Trip. 1900.)

[First obtain the equation when n = 0; then differentiate n times by Leibniz’ Theorem.]

11. The nth derivatives of a/(a2 + x2) and x/(a2 + x2). Since

a a2 + x2 = 1 2i 1 x −ai − 1 x + ai, x a2 + x2 = 1 2 1 x −ai + 1 x + ai,
we have
Dxn a a2 + x2 = (−1)nn! 2i 1 (x −ai)n+1 − 1 (x + ai)n+1 ,
and a similar formula for Dxn{x/(a2 + x2)}. If ρ = x2 + a2, and θ is the numerically smallest angle whose cosine and sine are x/ρ and a/ρ, then x + ai = ρCisθ and x −ai = ρCis(−θ), and so

Dxn{a/(a2 + x2)} = {(−1)nn!/2i}ρ−n−1[Cis{(n + 1)θ}−Cis{−(n + 1)θ}] = (−1)nn!(x2 + a2)−(n+1)/2 sin{(n + 1)arctan(a/x)}.

Similarly

Dxn{x/(a2 + x2)}= (−1)nn!(x2 + a2)−(n+1)/2 cos{(n + 1)arctan(a/x)}.

12. Prove that

Dxn{(cosx)/x} = {P n cos(x + 1 2nπ) + Qn sin(x + 1 2nπ)}/xn+1, Dxn{(sinx)/x} = {P n sin(x + 1 2nπ) −Qn cos(x + 1 2nπ)}/xn+1,

where Pn and Qn are polynomials in x of degree n and n −1 respectively.

13. Establish the formulae

dx dy = 1(dy dx),d2x dy2 = −d2y dx2(dy dx)3, d3x dy3 = −{d3y dx3 dy dx −3(d2y dx2)}(dy dx)5.

14. If yz = 1 and yr = (1/r!)Dxry, zs = (1/s!)Dxsz, then

1 z3 zz1z2 z1z2z3 z2z3z4 = 1 y2 y2y3 y3 y4 .
(Math. Trip. 1905.)

15. If

W(y,z,u) = y z u y′z′u′ y′′z′′u′′ ,
dashes denoting differentiations with respect to x, then
W(y,z,u) = y3W 1, z y, u y.

16. If

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0,
then
dy/dx = −(ax + hy + g)/(hx + by + f)
and
d2y/dx2 = (abc + 2fgh −af2 −bg2 −ch2)/(hx + by + f)3.

121. Some general theorems concerning derived functions. In all that follows we suppose that φ(x) is a function of x which has a derivative φ′(x) for all values of x in question. This assumption of course involves the continuity of φ(x).

The meaning of the sign of φ′(x). THEOREM A. If φ′(x0) > 0 then φ(x) < φ(x0) for all values of x less than x0 but sufficiently near to x0, and φ(x) > φ(x0) for all values of x greater than x0 but sufficiently near to x0.

For {φ(x0 + h) − φ(x0)}/h converges to a positive limit φ′(x0) as h → 0. This can only be the case if φ(x0 + h) − φ(x0) and h have the same sign for sufficiently small values of h, and this is precisely what the theorem states. Of course from a geometrical point of view the result is intuitive, the inequality φ′(x) > 0 expressing the fact that the tangent to the curve y = φ(x) makes a positive acute angle with the axis of x. The reader should formulate for himself the corresponding theorem for the case in which φ′(x) < 0.

An immediate deduction from Theorem A is the following important theorem, generally known as Rolle’s Theorem. In view of the great importance of this theorem it may be well to repeat that its truth depends on the assumption of the existence of the derivative φ′(x) for all values of x in question.

THEOREM B. If φ(a) = 0 and φ(b) = 0, then there must be at least one value of x which lies between a and b and for which φ′(x) = 0.

There are two possibilities: the first is that φ(x) is equal to zero throughout the whole interval [a,b]. In this case φ′(x) is also equal to zero throughout the interval. If on the other hand φ(x) is not always equal to zero, then there must be values of x for which φ(x) is positive or negative. Let us suppose, for example, that φ(x) is sometimes positive. Then, by Theorem 2 of § 102, there is a value ξ of x, not equal to a or b, and such that φ(ξ) is at least as great as the value of φ(x) at any other point in the interval. And φ′(ξ) must be equal to zero. For if it were positive then φ(x) would, by Theorem A, be greater than φ(ξ) for values of x greater than ξ but sufficiently near to ξ, so that there would certainly be values of φ(x) greater than φ(ξ). Similarly we can show that φ′(ξ) cannot be negative.

COR 1. If φ(a) = φ(b) = k, then there must be a value of x between a and b such that φ′(x) = 0.

We have only to put φ(x) − k = ψ(x) and apply Theorem B to ψ(x).

COR 2. If φ′(x) > 0 for all values of x in a certain interval, then φ(x) is an increasing function of x, in the stricter sense of § 95, throughout that interval.

Let x1 and x2 be two values of x in the interval in question, and x1 < x2. We have to show that φ(x1) < φ(x2). In the first place φ(x1) cannot be equal to φ(x2); for, if this were so, there would, by Theorem B, be a value of x between x1 and x2 for which φ′(x) = 0. Nor can φ(x1) be greater than φ(x2). For, since φ′(x1) is positive, φ(x) is, by Theorem A, greater than φ(x1) when x is greater than x1 and sufficiently near to x1. It follows that there is a value x3 of x between x1 andx2 such that φ(x3) = φ(x1); and so, by Theorem B, that there is a value of x between x1 and x3 for which φ′(x) = 0.

COR 3. The conclusion of Cor. 2 still holds if the interval [a,b] considered includes a finite number of exceptional values of x for which φ′(x) does not exist, or is not positive, provided φ(x) is continuous even for these exceptional values of x.

It is plainly sufficient to consider the case in which there is one exceptional value of x only, and that corresponding to an end of the interval, say to a. If a < x1 < x2 < b, we can choose a + δ so that a + δ < x1, and φ′(x) > 0 throughout [a + δ,b], so that φ(x1) < φ(x2), by Cor. 2. All that remains is to prove that φ(a) < φ(x1). Now φ(x1) decreases steadily, and in the stricter sense, as x1 decreases towards a, and so

φ(a) = φ(a + 0) = lim x1→a+0φ(x1) < φ(x1).

COR 4. If φ′(x) > 0 throughout the interval [a,b], and φ(a) ≧ 0, then φ(x) is positive throughout the interval [a,b].

The reader should compare the second of these corollaries very carefully with Theorem A. If, as in Theorem A, we assume only that φ′(x) is positive at a single point x = x0, then we can prove that φ(x1) < φ(x2) when x1 and x2 are sufficiently near to x0 and x1 < x0 < x2. For φ(x1) < φ(x0) and φ(x2) > φ(x0), by Theorem A. But this does not prove that there is any interval including x0 throughout which φ(x) is a steadily increasing function, for the assumption that x1 and x2 lie on opposite sides of x0 is essential to our conclusion. We shall return to this point, and illustrate it by an actual example, in a moment (§124).

122. Maxima and Minima. We shall say that the value φ(ξ) assumed by φ(x) when x = ξ is a maximum if φ(ξ) is greater than any other value assumed by φ(x) in the immediate neighbourhood of x = ξ, i.e. if we can find an interval [ξ − δ,ξ + δ] of values of x such that φ(ξ) > φ(x) when ξ − δ < x < ξ and when ξ < x < ξ + δ; and we define a minimum in a similar manner. Thus in the figure the points A correspond to maxima, the points B to minima of

pict

Fig. 39.

the function whose graph is there shown. It is to be observed that the fact that A3 corresponds to a maximum and B1 to a minimum is in no way inconsistent with the fact that the value of the function is greater at B1 than at A3.

THEOREM C. A necessary condition for a maximum or minimum value of φ(x) at x = ξ is that φ′(ξ) = 0.66

This follows at once from Theorem A. That the condition is not sufficient is evident from a glance at the point C in the figure. Thus if y = x3 then φ′(x) = 3x2, which vanishes when x = 0. But x = 0 does not give either a maximum or a minimum of x3, as is obvious from the form of the graph of x3 (Fig. 10, p. 140).

But there will certainly be a maximum at x = ξ if φ′(ξ) = 0, φ′(x) > 0 for all values of x less than but near to ξ, and φ′(x) < 0 for all values of x greater than but near to ξ: and if the signs of these two inequalities are reversed there will certainly be a minimum. For then we can (by Cor. 3 of § 121) determine an interval [ξ − δ,ξ] throughout which φ(x) increases with x, and an interval [ξ,ξ + δ] throughout which it decreases as x increases: and obviously this ensures that φ(ξ) shall be a maximum.

This result may also be stated thus. If the sign of φ′(x) changes at x = ξ from positive to negative, then x = ξ gives a maximum of φ(x): and if the sign of φ′(x) changes in the opposite sense, then x = ξ gives a minimum.

123. There is another way of stating the conditions for a maximum or minimum which is often useful. Let us assume that φ(x) has a second derivative φ′′(x): this of course does not follow from the existence of φ′(x), any more than the existence of φ′(x) follows from that of φ(x). But in such cases as we are likely to meet with at present the condition is generally satisfied.

THEOREM D. If φ′(ξ) = 0 and φ′′(ξ)≠0, then φ(x) has a maximum or minimum at x = ξ, a maximum if φ′′(ξ) < 0, a minimum if φ′′(ξ) > 0.

Suppose, e.g., that φ′′(ξ) < 0. Then, by Theorem A, φ′(x) is negative when x is less than ξ but sufficiently near to ξ, and positive when x is greater than ξ but sufficiently near to ξ. Thus x = ξ gives a maximum.

124. In what has preceded (apart from the last paragraph) we have assumed simply that φ(x) has a derivative for all values of x in the interval under consideration. If this condition is not fulfilled the theorems cease to be true. Thus Theorem B fails in the case of the function

y = 1 −x2,
where the square root is to be taken positive. The graph of this function is shown in Fig. 40. Here φ(−1) = φ(1) = 0: but φ′(x), as is evident from the figure, is equal to 1 if x is negative and to  −1 if x is positive, and never

pict

Fig. 40.

vanishes. There is no derivative for x = 0, and no tangent to the graph at P. And in this case x = 0 obviously gives a maximum of φ(x), but φ′(0), as it does not exist, cannot be equal to zero, so that the test for a maximum fails.

The bare existence of the derivative φ′(x), however, is all that we have assumed. And there is one assumption in particular that we have not made, and that is that φ′(x) itself is a continuous function. This raises a rather subtle but still a very interesting point. Can a function φ(x) have a derivative for all values of x which is not itself continuous? In other words can a curve have a tangent at every point, and yet the direction of the tangent not vary continuously? The reader, if he considers what the question means and tries to answer it in the light of common sense, will probably incline to the answer No. It is, however, not difficult to see that this answer is wrong.

Consider the function φ(x) defined, when x≠0, by the equation

φ(x) = x2 sin(1/x);
and suppose that φ(0) = 0. Then φ(x) is continuous for all values of x. If x≠0 then
φ′(x) = 2xsin(1/x) −cos(1/x);
while
φ′(0) = limh→0h2 sin(1/h) h = 0.
Thus φ′(x) exists for all values of x. But φ′(x) is discontinuous for x = 0; for 2xsin(1/x) tends to 0 as x →0, and cos(1/x) oscillates between the limits of indetermination −1 and 1, so that φ′(x) oscillates between the same limits.

What is practically the same example enables us also to illustrate the point referred to at the end of §121. Let

φ(x) = x2 sin(1/x) + ax,
where 0 < a < 1, when x≠0, and φ(0) = 0. Then φ′(0) = a > 0. Thus the conditions of Theorem A of §121 are satisfied. But if x≠0 then
φ′(x) = 2xsin(1/x) −cos(1/x) + a,
which oscillates between the limits of indetermination a −1 and a + 1 as x →0. As a −1 < 0, we can find values of x, as near to 0 as we like, for which φ′(x) < 0; and it is therefore impossible to find any interval, including x = 0, throughout which φ(x) is a steadily increasing function of x.

It is, however, impossible that φ′(x) should have what was called in Ch. V (Ex. XXXVII. 18) a ‘simple’ discontinuity; e.g. that φ′(x) →a when x →+0, φ′(x) →b when x →−0, and φ′(0) = c, unless a = b = c, in which case φ′(x) is continuous for x = 0. For a proof see §125, Ex. XLVII. 3.

Examples XLVI. 1. Verify Theorem B when φ(x) = (x −a)m(x −b)n or φ(x) = (x −a)m(x −b)n(x −c)p, where m, n, p are positive integers and a < b < c.

[The first function vanishes for x = a and x = b. And

φ′(x) = (x −a)m−1(x −b)n−1{(m + n)x −mb −na}
vanishes for x = (mb + na)/(m + n), which lies between a and b. In the second case we have to verify that the quadratic equation
(m + n + p)x2 −{m(b + c) + n(c + a) + p(a + b)}x + mbc + nca + pab = 0
has roots between a and b and between b and c.]

2. Show that the polynomials

2x3 + 3x2 −12x + 7,3x4 + 8x3 −6x2 −24x + 19
are positive when x > 1.

3. Show that x −sinx is an increasing function throughout any interval of values of x, and that tanx −x increases as x increases from −1 2π to 1 2π. For what values of a is ax −sinx a steadily increasing or decreasing function of x?

4. Show that tanx −x also increases from x = 1 2π to x = 3 2π, from x = 3 2π to x = 5 2π, and so on, and deduce that there is one and only one root of the equation tanx = x in each of these intervals (cf. Ex. XVII. 4).

5. Deduce from Ex. 3 that sinx −x < 0 if x > 0, from this that cosx −1 + 1 2x2 > 0, and from this that sinx −x + 1 6x3 > 0. And, generally, prove that if

C2m = cosx −1 + x2 2! −⋯ −(−1)m x2m (2m)!, S2m+1 = sinx −x + x3 3! −⋯ −(−1)m x2m+1 (2m + 1)!,

and x > 0, then C2m and S2m+1 are positive or negative according as m is odd or even.

6. If f(x) and f′′(x) are continuous and have the same sign at every point of an interval [a,b], then this interval can include at most one root of either of the equations f(x) = 0, f′(x) = 0.

7. The functions u, v and their derivatives u′, v′ are continuous throughout a certain interval of values of x, and uv′−u′v never vanishes at any point of the interval. Show that between any two roots of u = 0 lies one of v = 0, and conversely. Verify the theorem when u = cosx, v = sinx.

[If v does not vanish between two roots of u = 0, say α and β, then the function u/v is continuous throughout the interval [α,β] and vanishes at its extremities. Hence (u/v)′= (u′v −uv′)/v2 must vanish between α and β, which contradicts our hypothesis.]

8. Determine the maxima and minima (if any) of (x −1)2(x + 2), x3 −3x, 2x3 −3x2 −36x + 10, 4x3 −18x2 + 27x −7, 3x4 −4x3 + 1, x5 −15x3 + 3. In each case sketch the form of the graph of the function.

[Consider the last function, for example. Here φ′(x) = 5x2(x2 −9), which vanishes for x = −3, x = 0, and x = 3. It is easy to see that x = −3 gives a maximum and x = 3 a minimum, while x = 0 gives neither, as φ′(x) is negative on both sides of x = 0.]

9. Discuss the maxima and minima of the function (x −a)m(x −b)n, where m and n are any positive integers, considering the different cases which occur according as m and n are odd or even. Sketch the graph of the function.

10. Discuss similarly the function (x −a)(x −b)2(x −c)3, distinguishing the different forms of the graph which correspond to different hypotheses as to the relative magnitudes of a, b, c.

11. Show that (ax + b)/(cx + d) has no maxima or minima, whatever values a, b, c, d may have. Draw a graph of the function.

12. Discuss the maxima and minima of the function

y = (ax2 + 2bx + c)/(Ax2 + 2Bx + C),
when the denominator has complex roots.

[We may suppose a and A positive. The derivative vanishes if

(ax + b)(Bx + C) −(Ax + B)(bx + c) = 0. (1)

This equation must have real roots. For if not the derivative would always have the same sign, and this is impossible, since y is continuous for all values of x, and y →a/A as x →+∞or x →−∞. It is easy to verify that the curve cuts the line y = a/A in one and only one point, and that it lies above this line for large positive values of x, and below it for large negative values, or vice versa, according as b/a > B/A or b/a < B/A. Thus the algebraically greater root of (1) gives a maximum if b/a > B/A, a minimum in the contrary case.]

13. The maximum and minimum values themselves are the values of λ for which ax2 + 2bx + c −λ(Ax2 + 2Bx + C) is a perfect square. [This is the condition that y = λ should touch the curve.]

14. In general the maxima and maxima of R(x) = P(x)/Q(x) are among the values of λ obtained by expressing the condition that P(x) −λQ(x) = 0 should have a pair of equal roots.

15. If Ax2 + 2Bx + C = 0 has real roots then it is convenient to proceed as follows. We have

y −(a/A) = (λx + μ)/{A(Ax2 + 2Bx + C)},
where λ = bA −aB, μ = cA −aC. Writing further ξ for λx + μ and η for (A/λ2)(Ay −a), we obtain an equation of the form
η = ξ/{(ξ −p)(ξ −q)}.

This transformation from (x,y) to (ξ,η) amounts only to a shifting of the origin, keeping the axes parallel to themselves, a change of scale along each axis, and (if λ < 0) a reversal in direction of the axis of abscissae; and so a minimum of y, considered as a function of x, corresponds to a minimum of η considered as a function of ξ, and vice versa, and similarly for a maximum.

The derivative of η with respect to ξ vanishes if

(ξ −p)(ξ −q) −ξ(ξ −p) −ξ(ξ −q) = 0,
or if ξ2 = pq. Thus there are two roots of the derivative if p and q have the same sign, none if they have opposite signs. In the latter case the form of the graph of η is as shown in Fig. 41a.

pict
Fig. 41a.
pict
Fig. 41b.
pict
Fig. 41c.

When p and q are positive the general form of the graph is as shown in Fig 41b, and it is easy to see that ξ = pq gives a maximum and ξ = −pq a minimum.67

In the particular case in which p = q the function is

η = ξ/(ξ −p)2,
and its graph is of the form shown in Fig. 41c.

The preceding discussion fails if λ = 0, i.e. if a/A = b/B. But in this case we have

y −(a/A) = μ/{A(Ax2 + 2Bx + C)} = μ/{A2(x −x 1)(x −x2)},

say, and dy/dx = 0 gives the single value x = 1 2(x1 + x2). On drawing a graph it becomes clear that this value gives a maximum or minimum according as μ is positive or negative. The graph shown in Fig. 42 corresponds to the former case.

pict

Fig. 42.

[A full discussion of the general function y = (ax2 + 2bx + c)/(Ax2 + 2Bx + C), by purely algebraical methods, will be found in Chrystal’s Algebra, vol i, pp. 464–7.]

16. Show that (x −α)(x −β)/(x −γ) assumes all real values as x varies, if γ lies between α and β, and otherwise assumes all values except those included in an interval of length 4∣α −γ∣∣β −γ∣.

17. Show that

y = x2 + 2x + c x2 + 4x + 3c
can assume any real value if 0 < c < 1, and draw a graph of the function in this case.
(Math. Trip. 1910.)

18. Determine the function of the form (ax2 + 2bx + c)/(Ax2 + 2Bx + C) which has turning values (i.e. maxima or minima) 2 and 3 when x = 1 and x = −1 respectively, and has the value 2.5 when x = 0.

(Math. Trip. 1908.)

19. The maximum and minimum of (x + a)(x + b)/(x −a)(x −b), where a and b are positive, are

−a + b a −b2, −a −b a + b2.

20. The maximum value of (x −1)2/(x + 1)3 is  2 27.

21. Discuss the maxima and minima of

x(x −1)/(x2 + 3x + 3),x4/(x −1)(x −3)3, (x −1)2(3x2 −2x −37)/(x + 5)2(3x2 −14x −1).

(Math. Trip. 1898.)

[If the last function be denoted by P(x)/Q(x), it will be found that

P′Q −PQ′= 72(x −7)(x −3)(x −1)(x + 1)(x + 2)(x + 5).]

22. Find the maxima and minima of acosx + bsinx. Verify the result by expressing the function in the form Acos(x −a).

23. Find the maxima and minima of

a2 cos2x + b2 sin2x,Acos2x + 2Hcosxsinx + Bsin2x.

24. Show that sin(x + a)/sin(x + b) has no maxima or minima. Draw a graph of the function.

25. Show that the function

sin2x sin(x + a)sin(x + b)(0 < a < b < π)
has an infinity of minima equal to 0 and of maxima equal to
−4sinasinb/sin2(a −b).
(Math. Trip. 1909.)

26. The least value of a2 sec2x + b2 cosec2x is (a + b)2.

27. Show that tan3xcot2x cannot lie between 1 9 and 3 2.

28. Show that, if the sum of the lengths of the hypothenuse and another side of a right-angled triangle is given, then the area of the triangle is a maximum when the angle between those sides is 60°.

(Math. Trip. 1909.)

29. A line is drawn through a fixed point (a,b) to meet the axes OX, OY in P and Q. Show that the minimum values of PQ, OP + OQ, and OP ⋅OQ are respectively (a2/3 + b2/3)3/2, (a + b)2, and 4ab.

30. A tangent to an ellipse meets the axes in P and Q. Show that the least value of PQ is equal to the sum of the semiaxes of the ellipse.

31. Find the lengths and directions of the axes of the conic

ax2 + 2hxy + by2 = 1.

[The length r of the semi-diameter which makes an angle θ with the axis of x is given by

1/r2 = acos2θ + 2hcosθsinθ + bsin2θ.
The condition for a maximum or minimum value of r is tan2θ = 2h/(a −b). Eliminating θ between these two equations we find
{a −(1/r2)}{b −(1/r2)}= h2.]

32. The greatest value of xmyn, where x and y are positive and x + y = k, is

mmnnkm+n/(m + n)m+n.

33. Thegreatestvalueofax + by,where x and y are positive and x2 + xy + y2 = 3κ2, is

2κa2 −ab + b2.

[If ax + by is a maximum then a + b(dy/dx) = 0. The relation between x and y gives (2x + y) + (x + 2y)(dy/dx) = 0. Equate the two values of dy/dx.]

34. If θ and φ are acute angles connected by the relation asecθ + bsecφ = c, where a, b, c are positive, then acosθ + bcosφ is a minimum when θ = φ.

125. The Mean Value Theorem. We can proceed now to the proof of another general theorem of extreme importance, a theorem commonly known as ‘The Mean Value Theorem’ or ‘The Theorem of the Mean’.

THEOREM. If φ(x) has a derivative for all values of x in the interval [a,b], then there is a value ξ of x between a and b, such that

φ(b) − φ(a) = (b − a)φ′(ξ).

Before we give a strict proof of this theorem, which is perhaps the most important theorem in the Differential Calculus, it will be well to point out its obvious geometrical meaning. This is simply (see Fig. 43) that if the curve APB has a tangent at all points of its length then there

pict

Fig. 43.

must be a point, such as P, where the tangent is parallel to AB. For φ′(ξ) is the tangent of the angle which the tangent at P makes with OX, and {φ(b) − φ(a)}/(b − a) the tangent of the angle which AB makes with OX.

It is easy to give a strict analytical proof. Consider the function

φ(b) − φ(x) −b − x b − a{φ(b) − φ(a)},
which vanishes when x = a and x = b. It follows from Theorem B of § 121 that there is a value ξ for which its derivative vanishes. But this derivative is
φ(b) − φ(a) b − a − φ′(x);
which proves the theorem. It should be observed that it has not been assumed in this proof that φ′(x) is continuous.

It is often convenient to express the Mean Value Theorem in the form

φ(b) = φ(a) + (b − a)φ′{a + θ(b − a)},
where θ is a number lying between 0 and 1. Of course a + θ(b − a) is merely another way of writing ‘some number ξ between a and b’. If we put b = a + h we obtain
φ(a + h) = φ(a) + hφ′(a + θh),
which is the form in which the theorem is most often quoted.

Examples XLVII. 1. Show that

φ(b) −φ(x) −b −x b −a{φ(b) −φ(a)}
is the difference between the ordinates of a point on the curve and the corresponding point on the chord.

2. Verify the theorem when φ(x) = x2 and when φ(x) = x3.

[In the latter case we have to prove that (b3 −a3)/(b −a) = 3ξ2, where a < ξ < b; i.e. that if 1 3(b2 + ab + a2) = ξ2 then ξ lies between a and b.]

3. Establish the theorem stated at the end of §124 by means of the Mean Value Theorem.

[Since φ′(0) = c, we can find a small positive value of x such that {φ(x) −φ(0)}/x is nearly equal to c; and therefore, by the theorem, a small positive value of ξ such that φ′(ξ) is nearly equal to c, which is inconsistent with limx→+0φ′(x) = a, unless a = c. Similarly b = c.]

4. Use the Mean Value Theorem to prove Theorem (6) of §113, assuming that the derivatives which occur are continuous.

[The derivative of F{f(x)}is by definition

lim F{f(x + h)}−F{f(x)} h .
But, by the Mean Value Theorem, f(x + h) = f(x) + hf′(ξ), where ξ is a number lying between x and x + h. And
F{f(x) + hf′(ξ)}= F{f(x)}+ hf′(ξ)F′(ξ1),
where ξ1 is a number lying between f(x) and f(x) + hf′(ξ). Hence the derivative of F{f(x)} is
limf′(ξ)F′(ξ1) = f′(x)F′{f(x)},
since ξ →x and ξ1 →f(x) as h →0.]

126. The Mean Value Theorem furnishes us with a proof of a result which is of great importance in what follows: if φ′(x) = 0, throughout a certain interval of values of x, then φ(x) is constant throughout that interval.

For, if a and b are any two values of x in the interval, then

φ(b) − φ(a) = (b − a)φ′{a + θ(b − a)} = 0.
An immediate corollary is that if φ′(x) = ψ′(x), throughout a certain interval, then the functions φ(x) and ψ(x) differ throughout that interval by a constant.

127. Integration. We have in this chapter seen how we can find the derivative of a given function φ(x) in a variety of cases, including all those of the commonest occurrence. It is natural to consider the converse question, that of determining a function whose derivative is a given function.

Suppose that ψ(x) is the given function. Then we wish to determine a function such that φ′(x) = ψ(x). A little reflection shows us that this question may really be analysed into three parts.

(1) In the first place we want to know whether such a function as φ(x) actually exists. This question must be carefully distinguished from the question as to whether (supposing that there is such a function) we can find any simple formula to express it.

(2) We want to know whether it is possible that more than one such function should exist, i.e. we want to know whether our problem is one which admits of a unique solution or not; and if not, we want to know whether there is any simple relation between the different solutions which will enable us to express all of them in terms of any particular one.

(3) If there is a solution, we want to know how to find an actual expression for it.

It will throw light on the nature of these three distinct questions if we compare them with the three corresponding questions which arise with regard to the differentiation of functions.

(1) A function φ(x) may have a derivative for all values of x, like xm, where m is a positive integer, or  sin x. It may generally, but not always have one, like x3 or  tan x or  sec x. Or again it may never have one: for example, the function considered in Ex. XXXVII. 20, which is nowhere continuous, has obviously no derivative for any value of x. Of course during this chapter we have confined ourselves to functions which are continuous except for some special values of x. The example of the function x3, however, shows that a continuous function may not have a derivative for some special value of x, in this case x = 0. Whether there are continuous functions which never have derivatives, or continuous curves which never have tangents, is a further question which is at present beyond us. Common-sense says No: but, as we have already stated in § 111, this is one of the cases in which higher mathematics has proved common-sense to be mistaken.

But at any rate it is clear enough that the question ‘has φ(x) a derivative φ′(x)?’ is one which has to be answered differently in different circumstances. And we may expect that the converse question ‘is there a function φ(x) of which ψ(x) is the derivative?’ will have different answers too. We have already seen that there are cases in which the answer is No: thus if ψ(x) is the function which is equal to a, b, or c according as x is less than, equal to, or greater than 0, then the answer is No (Ex. XLVII. 3), unless a = b = c.

This is a case in which the given function is discontinuous. In what follows, however, we shall always suppose ψ(x) continuous. And then the answer is Yes: if ψ(x) is continuous then there is always a function φ(x) such that φ′(x) = ψ(x). The proof of this will be given in Ch. VII.

(2) The second question presents no difficulties. In the case of differentiation we have a direct definition of the derivative which makes it clear from the beginning that there cannot possibly be more than one. In the case of the converse problem the answer is almost equally simple. It is that if φ(x) is one solution of the problem then φ(x) + C is another, for any value of the constant C, and that all possible solutions are comprised in the form φ(x) + C. This follows at once from § 126.

(3) The practical problem of actually finding φ′(x) is a fairly simple one in the case of any function defined by some finite combination of the ordinary functional symbols. The converse problem is much more difficult. The nature of the difficulties will appear more clearly later on.

DEFINITIONS. If ψ(x) is the derivative of φ(x), then we call φ(x) an integral or integral function of ψ(x). The operation of forming ψ(x) from φ(x) we call integration.

We shall use the notation

φ(x) = ∫ ψ(x)dx.
It is hardly necessary to point out that ∫ …dx like d/dx must, at present at any rate, be regarded purely as a symbol of operation: the  ∫ and the dx no more mean anything when taken by themselves than do the d and dx of the other operative symbol d/dx.

128. The practical problem of integration. The results of the earlier part of this chapter enable us to write down at once the integrals of some of the commonest functions. Thus

∫ xmdx = xm+1 m + 1,∫ cos xdx = sin x,∫ sin xdx = − cos x. (1)

These formulae must be understood as meaning that the function on the right-hand side is one integral of that under the sign of integration. The most general integral is of course obtained by adding to the former a constant C, known as the arbitrary constant of integration.

There is however one case of exception to the first formula, that in which m = −1. In this case the formula becomes meaningless, as is only to be expected, since we have seen already (Ex. XLII. 4) that 1/x cannot be the derivative of any polynomial or rational fraction.

That there really is a function F(x) such that DxF(x) = 1/x will be proved in the next chapter. For the present we shall be content to assume its existence. This function F(x) is certainly not a polynomial or rational function; and it can be proved that it is not an algebraical function. It can indeed be proved that F(x) is an essentially new function, independent of any of the classes of functions which we have considered yet, that is to say incapable of expression by means of any finite combination of the functional symbols corresponding to them. The proof of this is unfortunately too detailed and tedious to be inserted in this book; but some further discussion of the subject will be found in Ch. IX, where the properties of F(x) are investigated systematically.

Suppose first that x is positive. Then we shall write

∫ dx x = log x, (2)

and we shall call the function on the right-hand side of this equation the logarithmic function: it is defined so far only for positive values of x.

Next suppose x negative. Then − x is positive, and so log(−x) is defined by what precedes. Also

d dx log(−x) = − 1 − x = 1 x,
so that, when x is negative,

∫ dx x = log(−x). (3)

The formulae (2) and (3) may be united in the formulae

∫ dx x = log(±x) = log ∣x∣, (4)

where the ambiguous sign is to be chosen so that ± x is positive: these formulae hold for all real values of x other than x = 0.

The most fundamental of the properties of logx which will be proved in Ch. IX are expressed by the equations

log1 = 0,log(1/x) = −logx,logxy = logx + logy,
of which the second is an obvious deduction from the first and third. It is not really necessary, for the purposes of this chapter, to assume the truth of any of these formulae; but they sometimes enable us to write our formulae in a more compact form than would otherwise be possible.

It follows from the last of the formulae that logx2 is equal to 2logx if x > 0 and to 2log(−x) if x < 0, and in either case to 2log∣x∣. Either of the formulae (4) is therefore equivalent to the formula

∫ dx x = 1 2 logx2. (5)

The five formulae (1)–(3) are the five most fundamental standard forms of the Integral Calculus. To them should be added two more, viz.

∫ dx 1 + x2 = arctan x,∫ x 1 − x2 = ± arcsin x.68 (6)

129. Polynomials. All the general theorems of § 113 may of course also be stated as theorems in integration. Thus we have, to begin with, the formulae

∫ {f(x) + F(x)}dx = ∫ f(x)dx + ∫ F(x)dx,  (1) ∫ kf(x)dx = k∫ f(x)dx.  (2)

Here it is assumed, of course, that the arbitrary constants are adjusted properly. Thus the formula (1) asserts that the sum of any integral of f(x) and any integral of F(x) is an integral of f(x) + F(x).

These theorems enable us to write down at once the integral of any function of the form ∑ Aνfν(x), the sum of a finite number of constant multiples of functions whose integrals are known. In particular we can write down the integral of any polynomial: thus

∫ (a0xn + a 1xn−1 + ⋯ + a n)dx = a0xn+1 n + 1 + a1xn n + ⋯ + anx.

130. Rational Functions. After integrating polynomials it is natural to turn our attention next to rational functions. Let us suppose R(x) to be any rational function expressed in the standard form of § 117, viz. as the sum of a polynomial Π(x) and a number of terms of the form A/(x − α)p.

We can at once write down the integrals of the polynomial and of all the other terms except those for which p = 1, since

∫ A (x − α)pdx = − A p − 1 1 (x − α)p−1,
whether α be real or complex (§ 117).

The terms for which p = 1 present rather more difficulty. It follows immediately from Theorem (6) of § 113 that

∫ F′{f(x)}f′(x)dx = F{f(x)}. (3)

In particular, if we take f(x) = ax + b, where a and b are real, and write φ(x) for F(x) and ψ(x) for F′(x), so that φ(x) is an integral of ψ(x), we obtain

∫ ψ(ax + b)dx = 1 aφ(ax + b). (4)

Thus, for example,

∫ dx ax + b = 1 a log ∣ax + b∣,
and in particular, if α is real,
∫ dx x − α = log ∣x − α∣.
We can therefore write down the integrals of all the terms in R(x) for which p = 1 and α is real. There remain the terms for which p = 1 and α is complex.

In order to deal with these we shall introduce a restrictive hypothesis, viz. that all the coefficients in R(x) are real. Then if α = γ + δi is a root of Q(x) = 0, of multiplicity m, so is its conjugate ᾱ = γ − δi; and if a partial fraction Ap/(x − α)p occurs in the expression of R(x), so does Āp/(x −ᾱ)p, where Āp is conjugate to Ap. This follows from the nature of the algebraical processes by means of which the partial fractions can be found, and which are explained at length in treatises on Algebra.69

Thus, if a term (λ + μi)/(x − γ − δi) occurs in the expression of R(x) in partial fractions, so will a term (λ − μi)/(x − γ + δi); and the sum of these two terms is

2{λ(x − γ) − μδ} (x − γ)2 + δ2 .
This fraction is in reality the most general fraction of the form
Ax + B ax2 + 2bx + c,
where b2 < ac. The reader will easily verify the equivalence of the two forms, the formulae which express λ, μ, γ, δ in terms of A, B, a, b, c being
λ = A/2a,μ = −D/(2aΔ),γ = −b/a,δ = Δ/a,
where Δ = ac − b2, and D = aB − bA.

If in (3) we suppose F{f(x)} to be  log ∣f(x)∣, we obtain

∫ f′(x) f(x) dx = log ∣f(x)∣; (5)

and if we further suppose that f(x) = (x − λ)2 + μ2, we obtain

∫ 2(x − λ) (x − λ)2 + μ2dx = log{(x − λ)2 + μ2}.
And, in virtue of the equations (6) of § 128 and (4) above, we have
∫ − 2δμ (x − λ)2 + μ2dx = −2δ arctan x − λ μ .

These two formulae enable us to integrate the sum of the two terms which we have been considering in the expression of R(x); and we are thus enabled to write down the integral of any real rational function, if all the factors of its denominator can be determined. The integral of any such function is composed of the sum of a polynomial, a number of rational functions of the type

− A p − 1 1 (x − α)p−1,
a number of logarithmic functions, and a number of inverse tangents.

It only remains to add that if α is complex then the rational function just written always occurs in conjunction with another in which A and α are replaced by the complex numbers conjugate to them, and that the sum of the two functions is a real rational function.

Examples XLVIII. 1. Prove that

∫ Ax + B ax2 + 2bx + cdx = A 2alog∣X∣+ D 2a −Δlog ax + b −−Δ ax + b + −Δ
(where X = ax2 + bx + c) if Δ < 0, and
∫ Ax + B ax2 + 2bx + cdx = A 2alog∣X∣+ D 2aΔarctan ax + b Δ if Δ > 0, Δ and D having the same meanings as on p. 773.

2. In the particular case in which ac = b2 the integral is

− D a(ax + b) + A a log∣ax + b∣.

3. Show that if the roots of Q(x) = 0 are all real and distinct, and P(x) is of lower degree than Q(x), then

∫ R(x)dx = ∑ P(α) Q′(α)log∣x −α∣,
the summation applying to all the roots α of Q(x) = 0.

[The form of the fraction corresponding to α may be deduced from the facts that

Q(x) x −α →Q′(α),(x −α)R(x) → P(α) Q′(α),
as x →α.]

4. If all the roots of Q(x) are real and α is a double root, the other roots being simple roots, and P(x) is of lower degree than Q(x), then the integral is A/(x −α) + A′log∣x −α∣+ ∑ Blog∣x −β∣, where

A = −2P(α) Q′′(α),A′= 2{3P′(α)Q′′(α) −P(a)Q′′′(α)} 3{Q′′(α)}2 ,B = P(β) Q′(β),
and the summation applies to all roots β of Q(x) = 0 other than α.

5. Calculate

∫ dx {(x −1)(x2 + 1)}2.

[The expression in partial fractions is

1 4(x −1)2 − 1 2(x −1) − i 8(x −i)2 + 2 −i 8(x −i) + i 8(x + i)2 + 2 + i 8(x + i),
and the integral is
− 1 4(x −1) − 1 4(x2 + 1) −1 2 log∣x −1∣+ 1 4 log(x2 + 1) + 1 4 arctanx.]

6. Integrate

x (x −a)(x −b)(x −c), x (x −a)2(x −b), x (x −a)2(x −b)2, x (x −a)3, x (x2 + a2)(x2 + b2), x2 (x2 + a2)(x2 + b)2, x2 −a2 x2(x2 + a2), x2 −a2 x(x2 + a2)2.

7. Prove the formulae:

∫ dx 1+x4 = 1 42{ log(1+x2+x2 1−x2+x2) + 2arctan( x2 1−x2)}, ∫ x2dx 1+x4 = 1 42{−log(1+x2+x2 1−x2+x2) + 2arctan( x2 1−x2)}, ∫ dx 1+x2+x4 = 1 43{3log(1+x+x2 1−x+x2) + 2arctan( x3 1−x2)}.

131. Note on the practical integration of rational functions. The analysis of §130 gives us a general method by which we can find the integral of any real rational function R(x), provided we can solve the equation Q(x) = 0. In simple cases (as in Ex. 5 above) the application of the method is fairly simple. In more complicated cases the labour involved is sometimes prohibitive, and other devices have to be used. It is not part of the purpose of this book to go into practical problems of integration in detail. The reader who desires fuller information may be referred to Goursat’s Cours d’Analyse, second ed., vol. i, pp. 246 et seq., Bertrand’s Calcul Intégral, and Dr Bromwich’s tract Elementary Integrals (Bowes and Bowes, 1911).

If the equation Q(x) = 0 cannot be solved algebraically, then the method of partial fractions naturally fails and recourse must be had to other methods.70

132. Algebraical Functions. We naturally pass on next to the question of the integration of algebraical functions. We have to consider the problem of integrating y, where y is an algebraical function of x. It is however convenient to consider an apparently more general integral, viz.

∫ R(x,y)dx,
where R(x,y) is any rational function of x and y. The greater generality of this form is only apparent, since (Ex. XIV. 6) the function R(x,y) is itself an algebraical function of x. The choice of this form is in fact dictated simply by motives of convenience: such a function as
px + q + ax2 + 2bx + c px + q −ax2 + 2bx + c
is far more conveniently regarded as a rational function of x and the simple algebraical function ax2 + 2bx + c, than directly as itself an algebraical function of x.

133. Integration by substitution and rationalisation. It follows from equation (3) of § 130 that if ∫ ψ(x)dx = φ(x) then

∫ ψ{f(t)}f′(t)dt = φ{f(t)}. (1)

This equation supplies us with a method for determining the integral of ψ(x) in a large number of cases in which the form of the integral is not directly obvious. It may be stated as a rule as follows: put x = f(t), where f(t) is any function of a new variable t which it may be convenient to choose; multiply by f′(t), and determine ifpossible the integral of ψ{f(t)}f′(t); express the result in terms of x. It will often be found that the function of t to which we are led by the application of this rule is one whose integral can easily be calculated. This is always so, for example, if it is a rational function, and it is very often possible to choose the relation between x and t so that this shall be the case. Thus the integral of R(x), where R denotes a rational function, is reduced by the substitution x = t2 to the integral of 2tR(t2), i.e. to the integral of a rational function of t. This method of integration is called integration by rationalisation, and is of extremely wide application.

Its application to the problem immediately under consideration is obvious. If we can find a variable t such that x and y are both rational functions of t, say x = R1(t), y = R2(t), then

∫ R(x,y)dx = ∫ R{R1(t),R2(t)}R′(t)dt,
and the latter integral, being that of a rational function of t, can be calculated by the methods of § 130.

It would carry us beyond our present range to enter upon any general discussion as to when it is and when it is not possible to find an auxiliary variable t connected with x and y in the manner indicated above. We shall consider only a few simple and interesting special cases.

134. Integrals connected with conics. Let us suppose that x and y are connected by an equation of the form

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0;
in other words that the graph of y, considered as a function of x is a conic. Suppose that (ξ,η) is any point on the conic, and let x − ξ = X, y − η = Y . If the relation between x and y is expressed in terms of X and Y , it assumes the form
aX2 + 2hXY + bY 2 + 2GX + 2FY = 0,
where F = hξ + bη + f, G = aξ + hη + g. In this equation put Y = tX. It will then be found that X and Y can both be expressed as rational functions of t, and therefore x and y can be so expressed, the actual formulae being
x − ξ = − 2(G + Ft) a + 2ht + bt2,y − η = − 2t(G + Ft) a + 2ht + bt2.
Hence the process of rationalisation described in the last section can be carried out.

The reader should verify that

hx + by + f = −1 2(a + 2ht + bt2)dx dt ,
so that
∫ dx hx + by + f = −2 ∫ dt a + 2ht + bt2.

When h2 > ab it is in some ways advantageous to proceed as follows. The conic is a hyperbola whose asymptotes are parallel to the lines

ax2 + 2hxy + by2 = 0,
or
b(y − μx)(y − μ′x) = 0,
say. If we put y − μx = t, we obtain
y − μx = t,y − μ′x = −2gx + 2fy + c bt ,
and it is clear that x and y can be calculated from these equations as rational functions of t. We shall illustrate this process by an application to an important special case.

135. The integral ∫ dx ax2 + 2bx + c. Suppose in particular that y2 = ax2 + 2bx + c, where a > 0. It will be found that, if we put y + xa = t, we obtain

2dx dt = (t2 + c)a + 2bt (ta + b)2 ,2y = (t2 + c)a + 2bt ta + b ,
and so

∫ dx y = ∫ dt ta + b = 1 alog xa + y + b a. (1)

If in particular a = 1, b = 0, c = a2, or a = 1, b = 0, c = −a2, we obtain

∫ dx x2 + a2 = log{x + x2 + a2},∫ dx x2 −a2 = log∣x + x2 −a2∣, (2)

equations whose truth may be verified immediately by differentiation. With these formulae should be associated the third formula

∫ dx a2 −x2 = arcsin(x/a), (3)

which corresponds to a case of the general integral of this section in which a < 0. In (3) it is supposed that a > 0; if a < 0 then the integral is arcsin(x/∣a∣) (cf. §119). In practice we should evaluate the general integral by reducing it (as in the next section) to one or other of these standard forms.

The formula (3) appears very different from the formulae (2): the reader will hardly be in a position to appreciate the connection between them until he has read Ch. X.

136. The integral ∫ λx + μ ax2 + 2bx + cdx. This integral can be integrated in all cases by means of the results of the preceding sections. It is most convenient to proceed as follows. Since

λx + μ = (λ/a)(ax + b) + μ − (λb/a), ∫ ax + b ax2 + 2bx + cdx = ax2 + 2bx + c,

we have

∫ (λx + μ)dx ax2 + 2bx + c = λ aax2 + 2bx + c + μ −λb a ∫ dx ax2 + 2bx + c.

In the last integral a may be positive or negative. If a is positive we put xa + (b/a) = t, when we obtain

1 a∫ dt t2 + κ,
where κ = (ac − b2)/a. If a is negative we write A for  − a and put xA − (b/A) = t, when we obtain
1 − a∫ dt − κ − t2.

It thus appears that in any case the calculation of the integral may be made to depend on that of the integral considered in § 135, and that this integral may be reduced to one or other of the three forms

∫ dt t2 + a2,∫ dt t2 − a2,∫ dt a2 − t2.

137. The integral ∫ (λx + μ)ax2 + 2bx + cdx. In exactly the same way we find

∫ (λx + μ)ax2 + 2bx + cdx = λ 3a(ax2 + 2bx + c)3/2 + μ −λb a ∫ ax2 + 2bx + cdx;

and the last integral may be reduced to one or other of the three forms

∫ t2 + a2dt,∫ t2 −a2dt,∫ a2 −t2dt.
In order to obtain these integrals it is convenient to introduce at this point another general theorem in integration.

138. Integration by parts. The theorem of integration by parts is merely another way of stating the rule for the differentiation of a product proved in § 113. It follows at once from Theorem (3) of § 113 that

∫ f′(x)F(x)dx = f(x)F(x) −∫ f(x)F′(x)dx.
It may happen that the function which we wish to integrate is expressible in the form f′(x)F(x), and that f(x)F′(x) can be integrated. Suppose, for example, that φ(x) = xψ(x), where ψ(x) is the second derivative of a known function χ(x). Then
∫ φ(x)dx = ∫ xχ′′(x)dx = xχ′(x) −∫ χ′(x)dx = xχ′(x) − χ(x).

We can illustrate the working of this method of integration by applying it to the integrals of the last section. Taking

f(x) = ax + b,F(x) = ax2 + 2bx + c = y,
we obtain

a∫ ydx = (ax + b)y −∫ (ax + b)2 y dx = (ax + b)y −a∫ ydx + (ac −b2)∫ dx y ,

so that

∫ ydx = (ax + b)y 2a + ac −b2 2a ∫ dx y ;
and we have seen already (§135) how to determine the last integral.

Examples XLIX. 1. Prove that if a > 0 then

∫ x2 + a2dx = 1 2xx2 + a2 + 1 2a2 log{x + x2 + a2}, ∫ x2 −a2dx = 1 2xx2 −a2 −1 2a2 log∣x + x2 −a2∣, ∫ a2 −x2dx = 1 2xa2 −x2 + 1 2a2 arcsin(x/a).

2. Calculate the integrals ∫ dx a2 −x2, ∫ a2 −x2dx by means of the substitution x = asinθ, and verify that the results agree with those obtained in §135 and Ex. 1.

3. Calculate ∫ x(x + a)mdx, where m is any rational number, in three ways, viz. (i) by integration by parts, (ii) by the substitution (x + a)m = t, and (iii) by writing (x + a) −a for x; and verify that the results agree.

4. Prove, by means of the substitutions ax + b = 1/t and x = 1/u, that (in the notation of §§130 and 138)

∫ dx y3 = ax + b Δy ,∫ xdx y3 = −bx + c Δy .

5. Calculate ∫ dx (x −a)(b −x), where b > a, in three ways, viz. (i) by the methods of the preceding sections, (ii) by the substitution (b −x)/(x −a) = t2, and (iii) by the substitution x = acos2θ + bsin2θ; and verify that the results agree.

6. Integrate (x −a)(b −x) and (b −x)/(x −a).

7. Show, by means of the substitution 2x + a + b = 1 2(a −b){t2 + (1/t)2}, or by multiplying numerator and denominator by x + a −x + b, that if a > b then

∫ dx x + a + x + b = 1 2a −b t + 1 3t3 .

8. Find a substitution which will reduce ∫ dx (x + a)3/2 + (x −a)3/2 to the integral of a rational function.

(Math. Trip. 1899.)

9. Show that ∫ R{x,ax + bn}dx is reduced, by the substitution ax + b = yn, to the integral of a rational function.

10. Prove that

∫ f′′(x)F(x)dx = f′(x)F(x) −f(x)F′(x) + ∫ f(x)F′′(x)dx
and generally

∫ f(n)(x)F(x)dx = f(n−1)(x)F(x) −f(n−2)(x)F′(x) + ⋯ + (−1)n ∫ f(x)F(n)(x)dx.

11. The integral ∫ (1 + x)pxqdx, where p and q are rational, can be found in three cases, viz. (i) if p is an integer, (ii) if q is an integer, and (iii) if p + q is an integer. [In case (i) put x = us, where s is the denominator of q; in case (ii) put 1 + x = ts, where s is the denominator of p; and in case (iii) put 1 + x = xts, where s is the denominator of p.]

12. The integral ∫ xm(axn + b)qdx can be reduced to the preceding integral by the substitution axn = bt. [In practice it is often most convenient to calculate a particular integral of this kind by a ‘formula of reduction’ (cf. Misc. Ex. 39).]

13. The integral ∫ R{x,ax + b,cx + d}dx can be reduced to that of a rational function by the substitution

4x = −(b/a){t + (1/t)}2 −(d/c){t −(1/t)}2.

14. Reduce ∫ R(x,y)dx, where y2(x −y) = x2, to the integral of a rational function. [Putting y = tx we obtain x = 1/{t2(1 −t)}, y = 1/{t(1 −t)}.]

15. Reduce the integral in the same way when (a) y(x −y)2 = x, (b) (x2 + y2)2 = a2(x2 −y2). [In case (a) put x −y = t: in case (b) put x2 + y2 = t(x −y), when we obtain

x = a2t(t2 + a2)/(t4 + a4),y = a2t(t2 −a2)/(t4 + a4).]

16. If y(x −y)2 = x then

∫ dx x −3y = 1 2 log{(x −y)2 −1}.

17. If (x2 + y2)2 = 2c2(x2 −y2) then

∫ dx y(x2 + y2 + c2) = −1 c2 log x2 + y2 x −y .

139. The general integral ∫ R(x,y)dx, where y2 = ax2 + 2bx + c. The most general integral, of the type considered in §134, and associated with the special conic y2 = ax2 + 2bx + c, is

∫ R(x,X)dx, (1)

where X = y2 = ax2 + 2bx + c. We suppose that R is a real function.

The subject of integration is of the form P/Q, where P and Q are polynomials in x and X. It may therefore be reduced to the form

A + BX C + DX = (A + BX)(C −DX) C2 −D2X = E + FX,
where A, B, … are rational functions of x. The only new problem which arises is that of the integration of a function of the form FX, or, what is the same thing, G/X, where G is a rational function of x. And the integral

∫ G Xdx (2)

can always be evaluated by splitting up G into partial fractions. When we do this, integrals of three different types may arise.

(i) In the first place there may be integrals of the type

∫ xm Xdx, (3)

where m is a positive integer. The cases in which m = 0 or m = 1 have been disposed of in §136. In order to calculate the integrals corresponding to larger values of m we observe that

d dx(xm−1X) = (m −1)xm−2X + (ax + b)xm−1 X = αxm + βxm−1 + γxm−2 X ,
where α, β, γ are constants whose values may be easily calculated. It is clear that, when we integrate this equation, we obtain a relation between three successive integrals of the type (3). As we know the values of the integral for m = 0 and m = 1, we can calculate in turn its values for all other values of m.

(ii) In the second place there may be integrals of the type

∫ dx (x −p)mX, (4)

where p is real. If we make the substitution x −p = 1/t then this integral is reduced to an integral in t of the type (3).

(iii) Finally, there may be integrals corresponding to complex roots of the denominator of G. We shall confine ourselves to the simplest case, that in which all such roots are simple roots. In this case (cf. §130) a pair of conjugate complex roots of G gives rise to an integral of the type

∫ Lx + M (Ax2 + 2Bx + C)ax2 + 2bx + cdx. (5)

In order to evaluate this integral we put

x = μt + ν t + 1 ,
where μ and ν are so chosen that
aμν + b(μ + ν) + c = 0,Aμν + B(μ + ν) + C = 0;
so that μ and ν are the roots of the equation
(aB −bA)ξ2 −(cA −aC)ξ + (bC −cB) = 0.
This equation has certainly real roots, for it is the same equation as equation (1) of Ex. XLVI. 12; and it is therefore certainly possible to find real values of μ and ν fulfilling our requirements.

It will be found, on carrying out the substitution, that the integral (5) assumes the form

H∫ tdt (αt2 + β)γt2 + δ + K∫ dt (αt2 + β)γt2 + δ. (6)

The second of these integrals is rationalised by the substitution

t γt2 + δ = u,
which gives
∫ dt (αt2 + β)γt2 + δ = ∫ du β + (αδ −βγ)u2.
Finally, if we put t = 1/u in the first of the integrals (6), it is transformed into an integral of the second type, and may therefore be calculated in the manner just explained, viz. by putting u/γ + δu2 = u, i.e. 1/γt2 + δ = v.71

Examples L. 1. Evaluate

∫ dx xx2 + 2x + 3,∫ dx (x −1)x2 + 1,∫ dx (x + 1)1 + 2x −x2.

2. Prove that

∫ dx (x −p)(x −p)(x −q) = 2 q −px −q x −p.

3. If ag2 + ch2 = −ν < 0 then

∫ dx (hx + g)ax2 + c = − 1 νarctan ν(ax2 + c) ch −agx .

4. Show that ∫ dx (x −x0)y, where y2 = ax2 + 2bx + c, may be expressed in one or other of the forms

−1 y0 log axx0 + b(x + x0) + c + yy0 x −x0 , 1 z0 arctan axx0 + b(x + x0) + c yz0 , according as ax02 + 2bx0 + c is positive and equal to y02 or negative and equal to  −z02.

5. Show by means of the substitution y = ax2 + 2bx + c/(x −p) that

∫ dx (x −p)ax2 + 2bx + c = ∫ dy λy2 −μ,
where λ = ap2 + 2bp + c, μ = ac −b2. [This method of reduction is elegant but less straightforward than that explained in §139.]

6. Show that the integral

∫ dx x3x2 + 2x + 1
is rationalised by the substitution x = (1 + y2)/(3 −y2).
(Math. Trip. 1911.)

7. Calculate

∫ (x + 1)dx (x2 + 4)x2 + 9.

8. Calculate

∫ dx (5x2 + 12x + 8)5x2 + 2x −7.

[Apply the method of §139. The equation satisfied by μ and ν is ξ2 + 3ξ + 2 = 0, so that μ = −2, ν = −1, and the appropriate substitution is x = −(2t + 1)/(t + 1). This reduces the integral to

−∫ dt (4t2 + 1)9t2 −4 −∫ tdt (4t2 + 1)9t2 −4.
The first of these integrals may be rationalised by putting t/9t2 −4 = u and the second by putting 1/9t2 −4 = v.]

9. Calculate

∫ (x + 1)dx (2x2 −2x + 1)3x2 −2x + 1,∫ (x −1)dx (2x2 −6x + 5)7x2 −22x + 19.
(Math. Trip. 1911.)

10. Show that the integral ∫ R(x,y)dx, where y2 = ax2 + 2bx + c, is rationalised by the substitution t = (x −p)/(y + q), where (p,q) is any point on the conic y2 = ax2 + 2bx + c. [The integral is of course also rationalised by the substitution t = (x −p)/(y −q): cf. §134.]

140. Transcendental Functions. Owing to the immense variety of the different classes of transcendental functions, the theory of their integration is a good deal less systematic than that of the integration of rational or algebraical functions. We shall consider in order a few classes of transcendental functions whose integrals can always be found.

141. Polynomials in cosines and sines of multiples of x. We can always integrate any function which is the sum of a finite number of terms such as

A cos max sin m′ax cos nbx sin n′bx…,
where m, m′, n, n′, … are positive integers and a, b, … any real numbers whatever. For such a term can be expressed as the sum of a finite number of terms of the types
α cos{(pa + qb + …)x},β sin{(pa + qb + …)x}
and the integrals of these terms can be written down at once.

Examples LI. 1. Integrate sin3xcos22x. In this case we use the formulae

sin3x = 1 4(3sinx −sin3x),cos22x = 1 2(1 + cos4x).
Multiplying these two expressions and replacing sinxcos4x, for example, by 1 2(sin5x −sin3x), we obtain

1 16 ∫ (7sinx −5sin3x + 3sin5x −sin7x)dx = −7 16 cosx + 5 48 cos3x − 3 80 cos5x + 1 112 cos7x.

The integral may of course be obtained in different forms by different methods. For example

∫ sin3xcos22xdx = ∫ (4cos4x −4cos2x + 1)(1 −cos2x)sinxdx,
which reduces, on making the substitution cosx = t, to
∫ (4t6 −8t4 + 5t2 −1)dt = 4 7 cos7x −8 5 cos5x + 5 3 cos3x −cosx.
It may be verified that this expression and that obtained above differ only by a constant.

2. Integrate by any method cosaxcosbx, sinaxsinbx, cosaxsinbx, cos2x, sin3x, cos4x, cosxcos2xcos3x, cos32xsin23x, cos5xsin7x. [In cases of this kind it is sometimes convenient to use a formula of reduction (Misc. Ex. 39).]

142. The integrals ∫ xn cos xdx, ∫ xn sin xdx and associated integrals. The method of integration by parts enables us to generalise the preceding results. For

∫ xn cos xdx = xn sin x − n∫ xn−1 sin xdx, ∫ xn sin xdx = − xn cos x + n∫ xn−1 cos xdx,

and clearly the integrals can be calculated completely by a repetition of this process whenever n is a positive integer. It follows that we can always calculate ∫ xn cos axdx and ∫ xn sin axdx if n is a positive integer; and so, by a process similar to that of the preceding paragraph, we can calculate

∫ P(x, cos ax, sin ax, cos bx, sin bx,…)dx,
where P is any polynomial.

Examples LII. 1. Integrate xsinx, x2 cosx, x2 cos2x, x2 sin2xsin22x, xsin2xcos4x, x3 sin31 3x.

2. Find polynomials P and Q such that

∫ {(3x −1)cosx + (1 −2x)sinx}dx = Pcosx + Qsinx.

3. Prove that ∫ xn cosxdx = P n cosx + Qn sinx, where

Pn = nxn−1 −n(n −1)(n −2)xn−3 + …,Q n = xn −n(n −1)xn−2 + ….

143. Rational Functions of cos x and sin x. The integral of any rational function of cos x and sin x may be calculated by the substitution tan 1 2x = t. For

cos x = 1 − t2 1 + t2, sin x = 2t 1 + t2,dx dt = 2 1 + t2,
so that the substitution reduces the integral to that of a rational function of t.

Examples LIII. 1. Prove that

∫ secxdx = log∣secx + tanx∣,∫ cosecxdx = log∣tan 1 2x∣.

[Another form of the first integral is log∣tan(1 4π + 1 2x)∣; a third form is 1 2 log∣(1 + sinx)/(1 −sinx)∣.]

2. ∫ tanxdx = −log∣cosx∣, ∫ cotxdx = log∣sinx∣, ∫ sec2xdx = tanx, ∫ cosec2xdx = −cotx, ∫ tanxsecxdx = secx, ∫ cotxcosecxdx = −cosecx.

[These integrals are included in the general form, but there is no need to use a substitution, as the results follow at once from §119 and equation (5) of §130.]

3. Show that the integral of 1/(a + bcosx), where a + b is positive, may be expressed in one or other of the forms

2 a2 −b2 arctan ta −b a + b, 1 b2 −a2 log b + a + tb −a b + a −tb −a, where t = tan 1 2x, according as a2 > b2 or a2 < b2. If a2 = b2 then the integral reduces to a constant multiple of that of sec21 2x or cosec21 2x, and its value may at once be written down. Deduce the forms of the integral when a + b is negative.

4. Show that if y is defined in terms of x by means of the equation

(a + bcosx)(a −bcosy) = a2 −b2,
where a is positive and a2 > b2, then as x varies from 0 to π one value of y also varies from 0 to π. Show also that
sinx = a2 −b2 siny a −bcosy , sinx a + bcosxdx dy = siny a −bcosy;
and deduce that if 0 < x < π then
∫ dx a + bcosx = 1 a2 −b2 arccos acosx + b a + bcosx.

Show that this result agrees with that of Ex. 3.

5. Show how to integrate 1/(a + bcosx + csinx). [Express bcosx + csinx in the form b2 + c2 cos(x −α).]

6. Integrate (a + bcosx + csinx)/(α + βcosx + γsinx).

[Determine λ, μ, ν so that

a + bcosx + csinx = λ + μ(α + βcosx + γsinx) + ν(−βsinx + γcosx).
Then the integral is
μx + νlog∣α + βcosx + γsinx∣+ λ∫ dx α + βcosx + γsinx.]

7. Integrate 1/(acos2x + 2bcosxsinx + csin2x). [The subject of integration may be expressed in the form 1/(A + Bcos2x + Csin2x), where A = 1 2(a + c), B = 1 2(a −c), C = b: but the integral may be calculated more simply by putting tanx = t, when we obtain

∫ sec2xdx a + 2btanx + ctan2x = ∫ dt a + 2bt + ct2.]

144. Integrals involving arcsin x, arctan x, and log x. The integrals of the inverse sine and tangent and of the logarithm can easily be calculated by integration by parts. Thus

∫ arcsin xdx = x arcsin x −∫ xdx 1 − x2 = x arcsin x + 1 − x2, ∫ arctan xdx = x arctan x −∫ xdx 1 + x2 = x arctan x −1 2 log(1 + x2), ∫ log xdx = x log x −∫ dx = x(log x − 1).

It is easy to see that if we can find the integral of y = f(x) then we can always find that of x = φ(y), where φ is the function inverse to f. For on making the substitution y = f(x) we obtain

∫ φ(y)dy = ∫ xf′(x)dx = xf(x) −∫ f(x)dx.
The reader should evaluate the integrals of arcsin y and arctan y in this way.

Integrals of the form

∫ P(x, arcsin x)dx,∫ P(x, log x)dx, where P is a polynomial, can always be calculated. Take the first form, for example. We have to calculate a number of integrals of the type ∫ xm(arcsin x)ndx. Making the substitution x = sin y, we obtain ∫ yn sin my cos ydy, which can be found by the method of § 142. In the case of the second form we have to calculate a number of integrals of the type ∫ xm(log x)ndx. Integrating by parts we obtain
∫ xm(log x)ndx = xm+1(log x)n m + 1 − n m + 1 ∫ xm(log x)n−1dx,
and it is evident that by repeating this process often enough we shall always arrive finally at the complete value of the integral.

145. Areas of plane curves. One of the most important applications of the processes of integration which have been explained in the preceding sections is to the calculation of areas of plane curves. Suppose that P0PP′ (Fig. 44) is the graph of a continuous curve y = φ(x) which lies wholly above the axis of x, P being the point (x,y) and P′ the point (x + h,y + k), and h being either positive or negative (positive in the figure).

pict
Fig. 44.
pict
Fig. 44a.

The reader is of course familiar with the idea of an ‘area’, and in particular with that of an area such as ONPP0. This idea we shall at present take for granted. It is indeed one which needs and has received the most careful mathematical analysis: later on we shall return to it and explain precisely what is meant by ascribing an ‘area’ to such a region of space as ONPP0. For the present we shall simply assume that any such region has associated with it a definite positive number (ONPP0) which we call its area, and that these areas possess the obvious properties indicated by common sense, e.g. that

(PRP′) + (NN′RP) = (NN′P′P),(N1NPP1) < (ONPP0),
and so on.

Taking all this for granted it is obvious that the area ONPP0 is a function of x; we denote it by Φ(x). Also Φ(x) is a continuous function. For

Φ(x + h) − Φ(x) = (NN′P′P) = (NN′RP) + (PRP′) = hφ(x) + (PRP′).

As the figure is drawn, the area PRP′ is less than hk. This is not however necessarily true in general, because it is not necessarily the case (see for example Fig. 44a) that the arc PP′ should rise or fall steadily from P to P′. But the area PRP′ is always less than ∣h∣λ(h), where λ(h) is the greatest distance of any point of the arc PP′ from PR. Moreover, since φ(x) is a continuous function, λ(h) → 0 as h → 0. Thus we have

Φ(x + h) − Φ(x) = h{φ(x) + μ(h)},
where ∣μ(h)∣ < λ(h) and λ(h) → 0 as h → 0. From this it follows at once that Φ(x) is continuous. Moreover
Φ′(x) = lim h→0Φ(x + h) − Φ(x) h = lim h→0{φ(x) + μ(h)} = φ(x).
Thus the ordinate of the curve is the derivative of the area, and the area is the integral of the ordinate.

We are thus able to formulate a rule for determining the area ONPP0. Calculate Φ(x), the integral of φ(x). This involves an arbitrary constant, which we suppose so chosen that Φ(0) = 0. Then the area required is Φ(x).

If it were the area N1NPP1 which was wanted, we should of course determine the constant so that Φ(x1) = 0, where x1 is the abscissa of P1. If the curve lay below the axis of x, Φ(x) would be negative, and the area would be the absolute value of Φ(x).

146. Lengths of plane curves. The notion of the length of a curve, other than a straight line, is in reality a more difficult one even than that of an area. In fact the assumption that P0P (Fig. 44) has a definite length, which we may denote by S(x), does not suffice for our purposes, as did the corresponding assumption about areas. We cannot even prove that S(x) is continuous, i.e. that lim{S(P′) − S(P)} = 0. This looks obvious enough in the larger figure, but less so in such a case as is shown in the smaller figure. Indeed it is not possible to proceed further, with any degree of rigour, without a careful analysis of precisely what is meant by the length of a curve.

It is however easy to see what the formula must be. Let us suppose that the curve has a tangent whose direction varies continuously, so that φ′(x) is continuous. Then the assumption that the curve has a length leads to the equation

{S(x + h) − S(x)}/h = {PP′}/h = (PP′/h) × ({PP′}/PP′),
where {PP′} is the arc whose chord is PP′. Now
PP′ + PR2 + RP′2 = h1 + k2 h2,
and
k = φ(x + h) − φ(x) = hφ′(ξ),
where ξ lies between x and x + h. Hence
lim(PP′/h) = lim 1 + [φ′(ξ)]2 = 1 + [φ′(x)]2.
If also we assume that
lim{PP′}/PP′ = 1,
we obtain the result
S′(x) = lim{S(x + h) − S(x)}/h = 1 + [φ′(x)]2
and so
S(x) = ∫ 1 + [φ′(x)]2dx.

Examples LIV. 1. Calculate the area of the segment cut off from the parabola y = x2/4a by the ordinate x = ξ, and the length of the arc which bounds it.

2. Answer the same questions for the curve ay2 = x3, showing that the length of the arc is

8a 27 1 + 9ξ 4a3/2 −1.

3. Calculate the areas and lengths of the circles x2 + y2 = a2, x2 + y2 = 2ax by means of the formulae of §§145–146.

4. Show that the area of the ellipse (x2/a2) + (y2/b2) = 1 is πab.

5. Find the area bounded by the curve y = sinx and the segment of the axis of x from x = 0 to x = 2π. [Here Φ(x) = −cosx, and the difference between the values of −cosx for x = 0 and x = 2π is zero. The explanation of this is of course that between x = π and x = 2π the curve lies below the axis of x, and so the corresponding part of the area is counted negative in applying the method. The area from x = 0 to x = π is −cosπ + cos0 = 2; and the whole area required, when every part is counted positive, is twice this, i.e. is 4.]

6. Suppose that the coordinates of any point on a curve are expressed as functions of a parameter t by equations of the type x = φ(t), y = ψ(t), φ and ψ being functions of t with continuous derivatives. Prove that if x steadily increases as t varies from t0 to t1, then the area of the region bounded by the corresponding portion of the curve, the axis of x, and the two ordinates corresponding to t0 and t1, is, apart from sign, A(t1) −A(t0), where

A(t) = ∫ ψ(t)φ′(t)dt = ∫ ydx dt dt.

7. Suppose that C is a closed curve formed of a single loop and not met by any parallel to either axis in more than two points. And suppose that the coordinates of any point P on the curve can be expressed as in Ex. 6 in terms of t, and that, as t varies from t0 to t1, P moves in the same direction round the curve and returns after a single circuit to its original position. Show that the area of the loop is equal to the difference of the initial and final values of any one of the integrals

−∫ ydx dt dt,∫ xdy dtdt,1 2 ∫ xdy dt −ydx dt dt,
this difference being of course taken positively.

8. Apply the result of Ex. 7 to determine the areas of the curves given by

(i) x a = 1 −t2 1 + t2,y a = 2t 1 + t2,(ii) x = acos3t,y = bsin3t.

9. Find the area of the loop of the curve x3 + y3 = 3axy. [Putting y = tx we obtain x = 3at/(1 + t3), y = 3at2/(1 + t3). As t varies from 0 towards ∞the loop is described once. Also

1 2 ∫ ydx dt −xdy dt dt = −1 2 ∫ x2 d dt y x dt = −1 2 ∫ 9a2t2 (1+t3)2dt = 3a2 2(1+t3),
which tends to 0 as t →∞. Thus the area of the loop is 3 2a2.]

10. Find the area of the loop of the curve x5 + y5 = 5ax2y2.

11. Prove that the area of a loop of the curve x = asin2t, y = asint is 4 3a2.

(Math. Trip. 1908.)

12. The arc of the ellipse given by x = acost, y = bsint, between the points t = t1 and t = t2, is F(t2) −F(t1), where

F(t) = a∫ 1 −e2 sin2tdt,
e being the eccentricity. [This integral cannot however be evaluated in terms of such functions as are at present at our disposal.]

13. Polar coordinates. Show that the area bounded by the curve r = f(θ), where f(θ) is a one-valued function of θ, and the radii θ = θ1, θ = θ2, is F(θ2) −F(θ1), where F(θ) = 1 2 ∫ r2dθ. And the length of the corresponding arc of the curve is Φ(θ2) −Φ(θ1), where

Φ(θ) = ∫ r2 +(dr dθ)2dθ.

Hence determine (i) the area and perimeter of the circle r = 2asinθ; (ii) the area between the parabola r = 1 2lsec21 2θ and its latus rectum, and the length of the corresponding arc of the parabola; (iii) the area of the limaçon r = a + bcosθ, distinguishing the cases in which a > b, a = b, and a < b; and (iv) the areas of the ellipses 1/r2 = acos2θ + 2hcosθsinθ + bsin2θ and l/r = 1 + ecosθ. [In the last case we are led to the integral ∫ dθ (1 + ecosθ)2, which may be calculated (cf. Ex. LIII. 4) by the help of the substitution

(1 + ecosθ)(1 −ecosφ) = 1 −e2.]

14. Trace the curve 2θ = (a/r) + (r/a), and show that the area bounded by the radius vector θ = β, and the two branches which touch at the point r = a, θ = 1, is 2 3a2(β2 −1)3/2.

(Math. Trip. 1900.)

15. A curve is given by an equation p = f(r), r being the radius vector and p the perpendicular from the origin on to the tangent. Show that the calculation of the area of the region bounded by an arc of the curve and two radii vectores depends upon that of the integral 1 2 ∫ prdr r2 −p2.

Miscellaneous Examples on Chapter VI.

1. A function f(x) is defined as being equal to 1 + x when x ≦0, to x when 0 < x < 1, to 2 −x when 1 ≦x ≦2, and to 3x −x2 when x > 2. Discuss the continuity of f(x) and the existence and continuity of f′(x) for x = 0, x = 1, and x = 2.

(Math. Trip. 1908.)

2. Denoting a, ax + b, ax2 + 2bx + c, … by u0, u1, u2, …, show that u02u3 −3u0u1u2 + 2u13 and u0u4 −4u1u3 + 3u22 are independent of x.

3. If a0, a1, …, a2n are constants and Ur = (a0,a1,…,ar )(x,1)r, then

U0U2n −2nU1U2n−1 + 2n(2n −1) 1 ⋅2 U2U2n−2 −⋯ + U2nU0
is independent of x.
(Math. Trip. 1896.)

[Differentiate and use the relation U′= rUr−1.]

4. The first three derivatives of the function arcsin(μsinx) −x, where μ > 1, are positive when 0 ≦x ≦ 1 2π.

5. The constituents of a determinant are functions of x. Show that its differential coefficient is the sum of the determinants formed by differentiating the constituents of one row only, leaving the rest unaltered.

6. If f1, f2, f3, f4 are polynomials of degree not greater than 4, then

f1 f2 f3 f4 f′f′f′f′ f′′f′′f′′f′′ f′′′f′′′f′′′f′′′
is also a polynomial of degree not greater than 4. [Differentiate five times, using the result of Ex. 5, and rejecting vanishing determinants.]

7. If y3 + 3yx + 2x3 = 0 then x2(1 + x3)y′′−3 2xy′+ y = 0.

(Math. Trip. 1903.)

8. Verify that the differential equation y = φ{ψ(y1)}+ φ{x −ψ(y1)}, where y1 is the derivative of y, and ψ is the function inverse to φ′, is satisfied by y = φ(c) + φ(x −c) or by y = 2φ(1 2x).

9. Verify that the differential equation y = {x/ψ(y1)}φ{ψ(y1)}, where the notation is the same as that of Ex. 8, is satisfied by y = cφ(x/c) or by y = βx, where β = φ(α)/α and α is any root of the equation φ(α) −αφ′(α) = 0.

10. If ax + by + c = 0 then y2 = 0 (suffixes denoting differentiations with respect to x). We may express this by saying that the general differential equation of all straight lines is y2 = 0. Find the general differential equations of (i) all circles with their centres on the axis of x, (ii) all parabolas with their axes along the axis of x, (iii) all parabolas with their axes parallel to the axis of y, (iv) all circles, (v) all parabolas, (vi) all conics.

[The equations are (i) 1 + y12 + yy2 = 0, (ii) y12 + yy2 = 0, (iii) y3 = 0, (iv) (1 + y12)y3 = 3y1y22, (v) 5y32 = 3y2y4, (vi) 9y22y5 −45y2y3y4 + 40y33 = 0. In each case we have only to write down the general equation of the curves in question, and differentiate until we have enough equations to eliminate all the arbitrary constants.]

11. Show that the general differential equations of all parabolas and of all conics are respectively

Dx2(y 2−2/3) = 0,D x3(y 2−2/3) = 0.

[The equation of a conic may be put in the form

y = ax + b ±px2 + 2qx + r.
From this we deduce
y2 = ±(pr −q2)/(px2 + 2qx + r)3/2.
If the conic is a parabola then p = 0.]

12. Denoting dy dx, 1 2!d2y dx2, 1 3!d3y dx3, 1 4!d4y dx4, … by t, a, b, c, … and dx dy, 1 2!d2x dy2 , 1 3!d3x dy3 , 1 4!d4x dy4 , … by τ, α, β, γ, …, show that

4ac −5b2 = (4αγ −5β2)/τ8,bt −a2 = −(βτ −α2)/τ6.
Establish similar formulae for the functions a2d −3abc −2b3, (1 + t2)b −2a2t, 2ct −5ab.

13. Prove that, if yk is the kth derivative of y = sin(narcsinx), then
(1 −x2)y k+2 −(2k + 1)xyk+1 + (n2 −k2)y k = 0.

[Prove first when k = 0, and differentiate k times by Leibniz’ Theorem.]

14. Prove the formula

vDxnu = D xn(uv) −nD xn−1(uD xv) + n(n −1) 1 ⋅2 Dxn−2(uD x2v) −…
where n is any positive integer. [Use the method of induction.]

15. A curve is given by

x = a(2cost + cos2t),y = a(2sint −sin2t).

Prove (i) that the equations of the tangent and normal, at the point P whose parameter is t, are

xsin 1 2t + ycos 1 2t = asin 3 2t,xcos 1 2t −ysin 1 2t = 3acos 3 2t;
(ii) that the tangent at P meets the curve in the points Q, R whose parameters are −1 2t and π −1 2t; (iii) that QR = 4a; (iv) that the tangents at Q and R are at right angles and intersect on the circle x2 + y2 = a2; (v) that the normals at P, Q, and R are concurrent and intersect on the circle x2 + y2 = 9a2; (vi) that the equation of the curve is
(x2 + y2 + 12ax + 9a2)2 = 4a(2x + 3a)3.

Sketch the form of the curve.

16. Show that the equations which define the curve of Ex. 15 may be replaced by ξ/a = 2u + (1/u2), η/a = (2/u) + u2, where ξ = x + yi, η = x −yi, u = Cist. Show that the tangent and normal, at the point defined by u, are

u2ξ −uη = a(u3 −1),u2ξ + uη = 3a(u3 + 1),
and deduce the properties (ii)–(v) of Ex. 15.

17. Show that the condition that x4 + 4px3 −4qx −1 = 0 should have equal roots may be expressed in the form (p + q)2/3 −(p −q)2/3 = 1.

(Math. Trip. 1898.)

18. The roots of a cubic f(x) = 0 are α, β, γ in ascending order of magnitude. Show that if [α,β] and [β,γ] are each divided into six equal sub-intervals, then a root of f′(x) = 0 will fall in the fourth interval from β on each side. What will be the nature of the cubic in the two cases when a root of f′(x) = 0 falls at a point of division?

(Math. Trip. 1907.)

19. Investigate the maxima and minima of f(x), and the real roots of f(x) = 0, f(x) being either of the functions

x −sinx −tanα(1 −cosx),x −sinx −(α −sinα) −tan 1 2α(cosα −cosx),
and α an angle between 0 and π. Show that in the first case the condition for a double root is that tanα −α should be a multiple of π.

20. Show that by choice of the ratio λ : μ we can make the roots of λ(ax2 + bx + c) + μ(a′x2 + b′x + c′) = 0 real and having a difference of any magnitude, unless the roots of the two quadratics are all real and interlace; and that in the excepted case the roots are always real, but there is a lower limit for the magnitude of their difference.

(Math. Trip. 1895.)

[Consider the form of the graph of the function (ax2 + bx + c)/(a′x2 + b′x + c′): cf. Exs. XLVI. 12 et seq.]

21. Prove that

π < sinπx x(1 −x) ≦4
when 0 < x < 1, and draw the graph of the function.

22. Draw the graph of the function

πcotπx −1 x − 1 x −1.

23. Sketch the general form of the graph of y, given that

dy dx = (6x2 + x −1)(x −1)2(x + 1)3 x2 .
(Math. Trip. 1908.)

24. A sheet of paper is folded over so that one corner just reaches the opposite side. Show how the paper must be folded to make the length of the crease a maximum.

25. The greatest acute angle at which the ellipse (x2/a2) + (y2/b2) = 1 can be cut by a concentric circle is arctan{(a2 −b2)/2ab}.
(Math. Trip. 1900.)

26. In a triangle the area Δ and the semi-perimeter s are fixed. Show that any maximum or minimum of one of the sides is a root of the equation s(x −s)x2 + 4Δ2 = 0. Discuss the reality of the roots of this equation, and whether they correspond to maxima or minima.

[The equations a + b + c = 2s, s(s −a)(s −b)(s −c) = Δ2 determine a and b as functions of c. Differentiate with respect to c, and suppose that da/dc = 0. It will be found that b = c, s −b = s −c = 1 2a, from which we deduce that s(a −s)a2 + 4Δ2 = 0.

This equation has three real roots if s4 > 27Δ2, and one in the contrary case. In an equilateral triangle (the triangle of minimum perimeter for a given area) s4 = 27Δ2; thus it is impossible that s4 < 27Δ2. Hence the equation in a has three real roots, and, since their sum is positive and their product negative, two roots are positive and the third negative. Of the two positive roots one corresponds to a maximum and one to a minimum.]

27. The area of the greatest equilateral triangle which can be drawn with its sides passing through three given points A, B, C is

2Δ + a2 + b2 + c2 23 ,
a, b, c being the sides and Δ the area of ABC.
(Math. Trip. 1899.)

28. If Δ, Δ′ are the areas of the two maximum isosceles triangles which can be described with their vertices at the origin and their base angles on the cardioid r = a(1 + cosθ), then 256ΔΔ′= 25a45.

(Math. Trip. 1907.)

29. Find the limiting values which (x2 −4y + 8)/(y2 −6x + 3) approaches as the point (x,y) on the curve x2y −4x2 −4xy + y2 + 16x −2y −7 = 0 approaches the position (2,3).

(Math. Trip. 1903.)

[If we take (2,3) as a new origin, the equation of the curve becomes ξ2η −ξ2 + η2 = 0, and the function given becomes (ξ2 + 4ξ −4η)/(η2 + 6η −6ξ). If we put η = tξ, we obtain ξ = (1 −t2)/t, η = 1 −t2. The curve has a loop branching at the origin, which corresponds to the two values t = −1 and t = 1. Expressing the given function in terms of t, and making t tend to −1 or 1, we obtain the limiting values −3 2,  −2 3.]

30. If f(x) = 1 sinx −sina − 1 (x −a)cosa, then

d da{limx→af(x)}−limx→af′(x) = 3 4 sec3a − 5 12 seca.
(Math. Trip. 1896.)

31. Show that if φ(x) = 1/(1 + x2) then φn(x) = Qn(x)/(1 + x2)n+1, where Qn(x) is a polynomial of degree n. Show also that

(i) Qn+1 = (1 + x2)Q′−2(n + 1)xQn,

(ii) Qn+2 + 2(n + 2)xQn+1 + (n + 2)(n + 1)(1 + x2)Qn = 0,

(iii) (1 + x2)Q′′−2nxQ′+ n(n + 1)Qn = 0,

(iv) Qn = (−1)nn! (n + 1)xn −(n + 1)n(n −1) 3! xn−2 + …,

(v) all the roots of Qn = 0 are real and separated by those of Qn−1 = 0.

32. If f(x), φ(x), ψ(x) have derivatives when a ≦x ≦b, then there is a value of ξ lying between a and b and such that

f(a)φ(a)ψ(a) f(b) φ(b)ψ(b) f′(ξ)φ′(ξ)ψ′(ξ) = 0.

[Consider the function formed by replacing the constituents of the third row by f(x), φ(x), ψ(x). This theorem reduces to the Mean Value Theorem (§125) when φ(x) = x and ψ(x) = 1.]

33. Deduce from Ex. 32 the formula

f(b) −f(a) φ(b) −φ(a) = f′(ξ) φ′(ξ).

34. If φ′(x) →a as x →∞, then φ(x)/x →a. If φ′(x) →∞ then φ(x) →∞. [Use the formula φ(x) −φ(x0) = (x −x0)φ′(ξ), where x0 < ξ < x.]

35. If φ(x) →a as x →∞, then φ′(x) cannot tend to any limit other than zero.

36. If φ(x) + φ′(x) →a as x →∞, then φ(x) →a and φ′(x) →0.

[Let φ(x) = a + ψ(x), so that ψ(x) + ψ′(x) →0. If ψ′(x) is of constant sign, say positive, for all sufficiently large values of x, then ψ(x) steadily increases and must tend to a limit l or to ∞. If ψ(x) →∞ then ψ′(x) →−∞, which contradicts our hypothesis. If ψ(x) →l then ψ′(x) →−l, and this is impossible (Ex. 35) unless l = 0. Similarly we may dispose of the case in which ψ′(x) is ultimately negative. If ψ(x) changes sign for values of x which surpass all limit, then these are the maxima and minima of ψ(x). If x has a large value corresponding to a maximum or minimum of ψ(x), then ψ(x) + ψ′(x) is small and ψ′(x) = 0, so that ψ(x) is small. A fortiori are the other values of ψ(x) small when x is large.

For generalisations of this theorem, and alternative lines of proof, see a paper by the author entitled “Generalisations of a limit theorem of Mr Mercer,” in volume 43 of the Quarterly Journal of Mathematics. The simple proof sketched above was suggested by Prof. E. W. Hobson.]

37. Show how to reduce ∫ R x, ax + b mx + n, cx + d mx + ndx to the integral of a rational function. [Put mx + n = 1/t and use Ex. XLIX. 13.]

38. Calculate the integrals:

∫ dx (1 + x2)3,∫ x −1 x + 1dx x ,∫ xdx 1 + x −1 + x3, ∫ a2 + b2 + c xdx,∫ cosec3xdx,∫ 5cosx + 6 2cosx + sinx + 3dx, ∫ dx (2 −sin2x)(2 + sinx −sin2x),∫ cosxsinxdx cos4x + sin4x,∫ cosecxsec2xdx, ∫ dx (1 + sinx)(2 + sinx),∫ x + sinx 1 + cosxdx,∫ arcsecxdx,∫ (arcsinx)2dx, ∫ xarcsinxdx,∫ xarcsinx 1 −x2 dx,∫ arcsinx x3 dx,∫ arcsinx (1 + x)2dx, ∫ arctanx x2 dx,∫ arctanx (1 + x2)3/2dx,∫ log(α2 + β2x2) x2 dx,∫ log(α + βx) (a + bx)2 dx.

39. Formulae of reduction. (i) Show that

2(n −1)(q −1 4p2)∫ dx (x2+px+q)n = x + 1 2p (x2 + px + q)n−1 + (2n −3)∫ dx (x2 + px + q)n−1.

[Put x + 1 2p = t, q −1 4p2 = λ: then we obtain

∫ dt (t2 + λ)n = 1 λ∫ dt (t2 + λ)n−1 −1 λ∫ t2dt (t2 + λ)n = 1 λ∫ dt (t2 + λ)n−1 + 1 2λ(n −1)∫ t d dt 1 (t2 + λ)n−1 dt,

and the result follows on integrating by parts.

A formula such as this is called a formula of reduction. It is most useful when n is a positive integer. We can then express ∫ dx (x2 + px + q)n in terms of ∫ dx (x2 + px + q)n−1, and so evaluate the integral for every value of n in turn.]

(ii) Show that if Ip,q = ∫ xp(1 + x)qdx then

(p + 1)Ip,q = xp+1(1 + x)q −qI p+1,q−1,
and obtain a similar formula connecting Ip,q with Ip−1,q+1. Show also, by means of the substitution x = −y/(1 + y), that
Ip,q = (−1)p+1 ∫ yp(1 + y)−p−q−2dy.

(iii) Show that if X = a + bx then

∫ xX−1/3dx = −3(3a −2bx)X2/3/10b2, ∫ x2X−1/3dx = 3(9a2 −6abx + 5b2x2)X2/3/40b3, ∫ xX−1/4dx = −4(4a −3bx)X3/4/21b2, ∫ x2X−1/4dx = 4(32a2 −24abx + 21b2x2)X3/4/231b3.

(iv) If Im,n = ∫ xmdx (1 + x2)n then

2(n −1)Im,n = −xm−1(1 + x2)−(n−1) + (m −1)I m−2,n−1.

(v) If In = ∫ xn cosβxdx and Jn = ∫ xn sinβxdx then

βIn = xn sinβx −nJ n−1,βJn = −xn cosβx + nI n−1.

(vi) If In = ∫ cosnxdx and Jn = ∫ sinnxdx then

nIn = sinxcosn−1x + (n −1)I n−2,nJn = −cosxsinn−1x + (n −1)J n−2.

(vii) If In = ∫ tannxdx then (n −1)(In + In−2) = tann−1x.

(viii) If Im,n = ∫ cosmxsinnxdx then

(m + n)Im,n = −cosm+1xsinn−1x + (n −1)I m,n−2 = cosm−1xsinn+1x + (m −1)I m−2,n.

[We have

(m + 1)Im,n = −∫ sinn−1x d dx(cosm+1x)dx = −cosm+1xsinn−1x + (n −1)∫ cosm+2xsinn−2xdx = −cosm+1xsinn−1x + (n −1)(I m,n−2 −Im,n),

which leads to the first reductio n formula.]

(ix) Connect Im,n = ∫ sinmxsinnxdx with Im−2,n.

(Math. Trip. 1897.)

(x) If Im,n = ∫ xm cosecnxdx then

(n −1)(n −2)Im,n = (n −2)2I m,n−2 + m(m −1)Im−2,n−2 −xm−1 cosecn−1x{msinx + (n −2)xcosx}.

(Math. Trip. 1896.)

(xi) If In = ∫ (a + bcosx)−ndx then

(n −1)(a2 −b2)I n = −bsinx(a + bcosx)−(n−1) + (2n −3)aI n−1 −(n −2)In−2.

(xii) If In = ∫ (acos2x + 2hcosxsinx + bsin2x)−ndx then

4n(n + 1)(ab −h2)I n+2 −2n(2n + 1)(a + b)In+1 + 4n2I n = −d2In dx2 .
(Math. Trip. 1898.)

(xiii) If Im,n = ∫ xm(logx)ndx then

(m + 1)Im,n = xm+1(logx)n −nI m,n−1.

40. If n is a positive integer then the value of ∫ xm(logx)ndx is

xm+1 (logx)n m + 1 −n(logx)n−1 (m + 1)2 + n(n −1)(logx)n−2 (m + 1)3 −⋯ + (−1)nn! (m + 1)n+1 .

41. Show that the most general function φ(x), such that φ′′+ a2φ = 0 for all values of x, may be expressed in either of the forms Acosax + Bsinax, ρcos(ax + ε), where A, B, ρ, ε are constants. [Multiplying by 2φ′ and integrating we obtain φ′2 + a2φ2 = a2b2, where b is a constant, from which we deduce that ax = ∫ dφ b2 −φ2.]

42. Determine the most general functions y and z such that y′+ ωz = 0, and z′−ωy = 0, where ω is a constant and dashes denote differentiation with respect to x.

43. The area of the curve given by

x = cosφ + sinαsinφ 1 −cos2αsin2φ,y = sinφ − sinαcosφ 1 −cos2αsin2φ,
where α is a positive acute angle, is 1 2π(1 + sinα)2/sinα.
(Math. Trip. 1904.)

44. The projection of a chord of a circle of radius a on a diameter is of constant length 2acosβ; show that the locus of the middle point of the chord consists of two loops, and that the area of either is a2(β −cosβsinβ).

(Math. Trip. 1903.)

45. Show that the length of a quadrant of the curve (x/a)2/3 + (y/b)2/3 = 1 is (a2 + ab + b2)/(a + b).

(Math. Trip. 1911.)

46. A point A is inside a circle of radius a, at a distance b from the centre. Show that the locus of the foot of the perpendicular drawn from A to a tangent to the circle encloses an area π(a2 + 1 2b2).

(Math. Trip. 1909.)

47. Prove that if (a,b,c,f,g,h )(x,y,1)2 = 0 is the equation of a conic, then

∫ dx (lx + my + n)(hx + by + f) = αlog PT PT′+ β,
where PT, PT′ are the perpendiculars from a point P of the conic on the tangents at the ends of the chord lx + my + n = 0, and α, β are constants.
(Math. Trip. 1902.)

48. Show that

∫ ax2 + 2bx + c (Ax2 + 2Bx + C)2dx
will be a rational function of x if and only if one or other of AC −B2 and aC + cA −2bB is zero.72

49. Show that the necessary and sufficient condition that

∫ f(x) {F(x)}2dx,
where f and F are polynomials of which the latter has no repeated factor, should be a rational function of x, is that f′F′−fF′′ should be divisible by F.
(Math. Trip. 1910.)

50. Show that

∫ αcosx + βsinx + γ (1 −ecosx)2 dx
is a rational function of cosx and sinx if and only if αe + γ = 0; and determine the integral when this condition is satisfied.
(Math. Trip. 1910.)

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