1. A function \(f(x)\) is defined as being equal to \(1 + x\) when \(x \leq 0\), to \(x\) when \(0 < x < 1\), to \(2 – x\) when \(1 \leq x \leq 2\), and to \(3x – x^{2}\) when \(x > 2\). Discuss the continuity of \(f(x)\) and the existence and continuity of \(f'(x)\) for \(x = 0\), \(x = 1\), and \(x = 2\).

2. Denoting \(a\), \(ax + b\), \(ax^{2} + 2bx + c\), … by \(u_{0}\), \(u_{1}\), \(u_{2}\), …, show that \(u_{0}^{2} u_{3} – 3u_{0} u_{1} u_{2} + 2u_{1}^{3}\) and \(u_{0} u_{4} – 4u_{1} u_{3} + 3u_{2}^{2}\) are independent of \(x\).

3. If \(a_{0}\), \(a_{1}\), …, \(a_{2n}\) are constants and \(U_{r} = (a_{0}, a_{1}, \dots, a_{r} | x, 1)^{r}\), then \[U_{0}U_{2n} – 2nU_{1}U_{2n-1} + \frac{2n(2n – 1)}{1\cdot2} U_{2}U_{2n-2} – \dots + U_{2n}U_{0}\] is independent of \(x\).

[Differentiate and use the relation \(U_{r}’ = rU_{r-1}\).]

4. The first three derivatives of the function \(\arcsin(\mu\sin x) – x\), where \(\mu > 1\), are positive when \(0 \leq x \leq \frac{1}{2} \pi\).

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 \(f_{1}\), \(f_{2}\), \(f_{3}\), \(f_{4}\) are polynomials of degree not greater than \(4\), then \[\begin{vmatrix} f_{1}& f_{2}& f_{3}& f_{4}\\ f_{1}’& f_{2}’& f_{3}’& f_{4}’\\ f_{1}”& f_{2}”& f_{3}”& f_{4}”\\ f_{1}”’& f_{2}”’& f_{3}”’& f_{4}”’ \end{vmatrix}\] 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 \(y^{3} + 3yx + 2x^{3} = 0\) then \(x^{2}(1 + x^{3})y” – \frac{3}{2}xy’ + y = 0\).

8. Verify that the differential equation \(y = \phi\{\psi(y_{1})\} + \phi\{x – \psi(y_{1})\}\), where \(y_{1}\) is the derivative of \(y\), and \(\psi\) is the function inverse to \(\phi’\), is satisfied by \(y = \phi(c) + \phi(x – c)\) or by \(y = 2\phi(\frac{1}{2}x)\).

9. Verify that the differential equation \(y = \{x/\psi(y_{1})\} \phi\{\psi(y_{1})\}\), where the notation is the same as that of Ex. 8, is satisfied by \(y = c\phi(x/c)\) or by \(y = \beta x\), where \(\beta = \phi(\alpha)/\alpha\) and \(\alpha\) is any root of the equation \(\phi(\alpha) – \alpha\phi'(\alpha) = 0\).

10. If \(ax + by + c = 0\) then \(y_{2} = 0\) (suffixes denoting differentiations with respect to \(x\)). We may express this by saying that *the general differential equation of all straight lines is \(y_{2} = 0\)*. 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.

11. Show that the general differential equations of all parabolas and of all conics are respectively \[D_{x}^{2} (y_{2}^{-2/3}) = 0,\quad D_{x}^{3} (y_{2}^{-2/3}) = 0.\]

[The equation of a conic may be put in the form \[y = ax + b \pm \sqrt{px^{2} + 2qx + r}.\] From this we deduce \[y_{2} = \pm(pr – q^{2})/(px^{2} + 2qx + r)^{3/2}.\] If the conic is a parabola then \(p = 0\).]

12. Denoting \(\dfrac{dy}{dx}\), \(\dfrac{1}{2!}\, \dfrac{d^{2}y}{dx^{2}}\), \(\dfrac{1}{3!}\, \dfrac{d^{3}y}{dx^{3}}\), \(\dfrac{1}{4!}\, \dfrac{d^{4}y}{dx^{4}}\), … by \(t\), \(a\), \(b\), \(c\), … and \(\dfrac{dx}{dy}\), \(\dfrac{1}{2!}\, \dfrac{d^{2}x}{dy^{2}}\), \(\dfrac{1}{3!}\, \dfrac{d^{3}x}{dy^{3}}\), \(\dfrac{1}{4!}\, \dfrac{d^{4}x}{dy^{4}}\), … by \(\tau\), \(\alpha\), \(\beta\), \(\gamma\), …, show that \[4ac – 5b^{2} = (4\alpha\gamma – 5\beta^{2})/\tau^{8},\quad bt – a^{2} = – (\beta\tau – \alpha^{2})/\tau^{6}.\] Establish similar formulae for the functions \(a^{2}d – 3abc – 2b^{3}\), \((1 + t^{2})b – 2a^{2}t\), \(2ct – 5ab\).

13. Prove that, if \(y_{k}\) is the \(k\)th derivative of \(y = \sin(n\arcsin x)\), then \[(1 – x^{2})y_{k+2} – (2k + 1)xy_{k+1} + (n^{2} – k^{2})y_{k} = 0.\]

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

14. Prove the formula \[vD_{x}^{n}u = D_{x}^{n}(uv) – nD_{x}^{n-1}(uD_{x}v) + \frac{n(n – 1)}{1\cdot2} D_{x}^{n-2}(uD_{x}^{2}v) – \dots\] where \(n\) is any positive integer. [Use the method of induction.]

15. A curve is given by \[x = a(2\cos t + \cos 2t),\quad y = a(2\sin t – \sin 2t).\]

Prove (i) that the equations of the tangent and normal, at the point \(P\) whose parameter is \(t\), are \[x\sin \tfrac{1}{2} t + y\cos \tfrac{1}{2} t = a\sin \tfrac{3}{2} t,\quad x\cos \tfrac{1}{2} t – y\sin \tfrac{1}{2} t = 3a\cos \tfrac{3}{2} t;\] (ii) that the tangent at \(P\) meets the curve in the points \(Q\), \(R\) whose parameters are \(-\frac{1}{2} t\) and \(\pi – \frac{1}{2} t\); (iii) that \(QR = 4a\); (iv) that the tangents at \(Q\) and \(R\) are at right angles and intersect on the circle \(x^{2} + y^{2} = a^{2}\); (v) that the normals at \(P\), \(Q\), and \(R\) are concurrent and intersect on the circle \(x^{2} + y^{2} = 9a^{2}\); (vi) that the equation of the curve is \[(x^{2} + y^{2} + 12ax + 9a^{2})^{2} = 4a(2x + 3a)^{3}.\]

Sketch the form of the curve.

16. Show that the equations which define the curve of Ex. 15 may be replaced by \(\xi/a = 2u + (1/u^{2})\), \(\eta/a = (2/u) + u^{2}\), where \(\xi = x + yi\), \(\eta = x – yi\), \(u = \operatorname{Cis} t\). Show that the tangent and normal, at the point defined by \(u\), are \[u^{2}\xi – u\eta = a(u^{3} – 1),\quad u^{2}\xi + u\eta = 3a(u^{3} + 1),\] and deduce the properties (ii)–(v) of Ex. 15.

17. Show that the condition that \(x^{4} + 4px^{3} – 4qx – 1 = 0\) should have equal roots may be expressed in the form \((p + q)^{2/3} – (p – q)^{2/3} = 1\).

18. The roots of a cubic \(f(x) = 0\) are \(\alpha\), \(\beta\), \(\gamma\) in ascending order of magnitude. Show that if \({[\alpha, \beta]}\) and \({[\beta, \gamma]}\) are each divided into six equal sub-intervals, then a root of \(f'(x) = 0\) will fall in the fourth interval from \(\beta\) 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?

19. Investigate the maxima and minima of \(f(x)\), and the real roots of \(f(x) = 0\), \(f(x)\) being either of the functions \[x – \sin x – \tan\alpha (1 – \cos x),\quad x – \sin x – (\alpha – \sin\alpha) – \tan \tfrac{1}{2}\alpha (\cos\alpha – \cos x),\] and \(\alpha\) an angle between \(0\) and \(\pi\). Show that in the first case the condition for a double root is that \(\tan\alpha – \alpha\) should be a multiple of \(\pi\).

20. Show that by choice of the ratio \(\lambda : \mu\) we can make the roots of \(\lambda(ax^{2} + bx + c) + \mu(a’x^{2} + b’x + c’) = 0\) 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.

[Consider the form of the graph of the function \((ax^{2} + bx + c)/(a’x^{2} + b’x + c’)\): cf. Exs. XLVI. 12*et seq.*]

21. Prove that \[\pi < \frac{\sin \pi x}{x(1 – x)} \leq 4\] when \(0 < x < 1\), and draw the graph of the function.

22. Draw the graph of the function \[\pi \cot\pi x – \frac{1}{x} – \frac{1}{x – 1}.\]

23. Sketch the general form of the graph of \(y\), given that \[\frac{dy}{dx} = \frac{(6x^{2} + x – 1) (x – 1)^{2} (x + 1)^{3}}{x^{2}}.\]

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 \((x^{2}/a^{2}) + (y^{2}/b^{2}) = 1\) can be cut by a concentric circle is \(\arctan\{(a^{2} – b^{2})/2ab\}\).

26. In a triangle the area \(\Delta\) and the semi-perimeter \(s\) are fixed. Show that any maximum or minimum of one of the sides is a root of the equation \(s(x – s) x^{2} + 4\Delta^{2} = 0\). Discuss the reality of the roots of this equation, and whether they correspond to maxima or minima.

[The equations \(a + b + c = 2s\), \(s(s – a)(s – b)(s – c) = \Delta^{2}\) determine \(a\) and \(b\) as functions of \(c\). Differentiate with respect to \(c\), and suppose that \(da/dc = 0\). It will be found that \(b = c\), \(s – b = s – c = \frac{1}{2} a\), from which we deduce that \(s(a – s)a^{2} + 4\Delta^{2} = 0\).This equation has three real roots if \(s^{4} > 27\Delta^{2}\), and one in the contrary case. In an equilateral triangle (the triangle of minimum perimeter for a given area) \(s^{4} = 27\Delta^{2}\); thus it is impossible that \(s^{4} < 27\Delta^{2}\). 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 \[2\Delta + \frac{a^{2} + b^{2} + c^{2}}{2\sqrt{3}},\] \(a\), \(b\), \(c\) being the sides and \(\Delta\) the area of \(ABC\).

28. If \(\Delta\), \(\Delta’\) 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 \(r = a(1 + \cos\theta)\), then \(256\Delta\Delta’ = 25a^{4}\sqrt{5}\).

29. Find the limiting values which \((x^{2} – 4y + 8)/(y^{2} – 6x + 3)\) approaches as the point \((x, y)\) on the curve \(x^{2}y – 4x^{2} – 4xy + y^{2} + 16x – 2y – 7 = 0\) approaches the position \((2, 3)\).

[If we take \((2, 3)\) as a new origin, the equation of the curve becomes \(\xi^{2} \eta – \xi^{2} + \eta^{2} = 0\), and the function given becomes \((\xi^{2} + 4\xi – 4\eta)/(\eta^{2} + 6\eta – 6\xi)\). If we put \(\eta = t\xi\), we obtain \(\xi = (1 – t^{2})/t\), \(\eta = 1 – t^{2}\). 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 \(-\frac{3}{2}\), \(-\frac{2}{3}\).]

30. If \(f(x) = \dfrac{1}{\sin x – \sin a} – \dfrac{1}{(x – a)\cos a}\), then \[\frac{d}{da}\{\lim_{x \to a} f(x)\} – \lim_{x \to a}f'(x) = \tfrac{3}{4} \sec^{3} a – \tfrac{5}{12} \sec a.\]

31. Show that if \(\phi(x) = 1/(1 + x^{2})\) then \(\phi^{n} (x) = Q_{n}(x)/(1 + x^{2})^{n+1}\), where \(Q_{n}(x)\) is a polynomial of degree \(n\). Show also that

(i) \(Q_{n+1} = (1 + x^{2}) Q_{n}’ – 2(n + 1) x Q_{n}\),

(ii) \(Q_{n+2} + 2(n + 2) x Q_{n+1} + (n + 2)(n + 1)(1 + x^{2})Q_{n} = 0\),

(iii) \((1 + x^{2}) Q_{n}” – 2nx Q_{n}’ + n(n + 1)Q_{n} = 0\),

(iv) \(Q_{n} = (-1)^{n} n!\left\{(n + 1)x^{n} – \dfrac{(n + 1)n(n – 1)}{3!} x^{n-2} + \dots\right\}\),

(v) all the roots of \(Q_{n} = 0\) are real and separated by those of \(Q_{n-1} = 0\).

32. If \(f(x)\), \(\phi(x)\), \(\psi(x)\) have derivatives when \(a \leq x \leq b\), then there is a value of \(\xi\) lying between \(a\) and \(b\) and such that \[\begin{vmatrix} f(a) & \phi(a) & \psi(a)\\ f(b) & \phi(b) & \psi(b)\\ f'(\xi)& \phi'(\xi)& \psi'(\xi) \end{vmatrix} =0.\]

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

33. Deduce from Ex. 32 the formula \[\frac{f(b) – f(a)}{\phi(b) – \phi(a)} = \frac{f'(\xi)}{\phi'(\xi)}.\]

34. If \(\phi'(x) \to a\) as \(x \to \infty\), then \(\phi(x)/x \to a\). If \(\phi'(x) \to \infty\) then \(\phi(x) \to \infty\). [Use the formula \(\phi(x) – \phi(x_{0}) = (x – x_{0})\phi'(\xi)\), where \(x_{0} < \xi < x\).]

35. If \(\phi(x) \to a\) as \(x \to \infty\), then \(\phi'(x)\) cannot tend to any limit other than zero.

36. If \(\phi(x) + \phi'(x) \to a\) as \(x \to \infty\), then \(\phi(x) \to a\) and \(\phi'(x) \to 0\).

[Let \(\phi(x) = a + \psi(x)\), so that \(\psi(x) + \psi'(x) \to 0\). If \(\psi'(x)\) is of constant sign, say positive, for all sufficiently large values of \(x\), then \(\psi(x)\) steadily increases and must tend to a limit \(l\) or to \(\infty\). If \(\psi(x) \to \infty\) then \(\psi'(x) \to -\infty\), which contradicts our hypothesis. If \(\psi(x) \to l\) then \(\psi'(x) \to -l\), and this is impossible (Ex. 35) unless \(l = 0\). Similarly we may dispose of the case in which \(\psi'(x)\) is ultimately negative. If \(\psi(x)\) changes sign for values of \(x\) which surpass all limit, then these are the maxima and minima of \(\psi(x)\). If \(x\) has a large value corresponding to a maximum or minimum of \(\psi(x)\), then \(\psi(x) + \psi'(x)\) is small and \(\psi'(x) = 0\), so that \(\psi(x)\) is small.*A fortiori*are the other values of \(\psi(x)\) 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 \(\int R\left\{x, \sqrt{\frac{ax + b}{mx + n}}, \sqrt{\frac{cx + d}{mx + n}}\right\} dx\) to the integral of a rational function. [Put \(mx + n = 1/t\) and use Ex. XLIX. 13.]

38. Calculate the integrals: \[\begin{gathered} \int \frac{dx}{(1 + x^{2})^{3}},\quad \int \sqrt{\frac{x – 1}{x + 1}}\, \frac{dx}{x},\quad \int \frac{x\, dx}{\sqrt{1 + x} – \sqrt[3]{1 + x}},\\ \int \sqrt{a^{2} + \sqrt{b^{2} + \frac{c}{x}}}\, dx,\quad \int \csc^{3}x\, dx,\quad \int \frac{5\cos x + 6}{2\cos x + \sin x + 3}\, dx,\\ \int \frac{dx}{(2 – \sin^{2}x) (2 + \sin x – \sin^{2} x)},\quad \int \frac{\cos x\sin x \, dx}{\cos^{4}x + \sin^{4}x},\quad \int \csc x \sqrt{\sec 2x}\, dx,\\ \int \frac{dx}{\sqrt{(1 + \sin x) (2 + \sin x)}},\quad \int \frac{x + \sin x}{1 + \cos x}\, dx,\quad \int \operatorname{arcsec} x\, dx,\quad \int (\arcsin x)^{2}\, dx,\\ \int x\arcsin x\, dx,\quad \int \frac{x\arcsin x}{\sqrt{1 – x^{2}}}\, dx,\quad \int \frac{\arcsin x}{x^{3}}\, dx,\quad \int \frac{\arcsin x}{(1 + x)^{2}}\, dx,\\ \int \frac{\arctan x}{x^{2}}\, dx,\quad \int \frac{\arctan x}{(1 + x^{2})^{3/2}}\, dx,\quad \int \frac{\log(\alpha^{2} + \beta^{2}x^{2})}{x^{2}}\, dx,\quad \int \frac{\log(\alpha + \beta x)}{(a + bx)^{2}}\, dx.\end{gathered}\]

39. **Formulae of reduction.** (i) Show that \[\begin{gathered} 2(n – 1)(q – \tfrac{1}{4}p^{2}) \int \frac{dx}{(x^{2} + px + q)^{n}} \\ = \frac{x + \frac{1}{2}p}{(x^{2} + px + q)^{n-1}} + (2n – 3) \int \frac{dx}{(x^{2} + px + q)^{n-1}}.\end{gathered}\]

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 \(\int \frac{dx}{(x^{2} + px + q)^{n}}\) in terms of \(\int \frac{dx}{(x^{2} + px + q)^{n-1}}\), and so evaluate the integral for every value of \(n\) in turn.]

(ii) Show that if \(I_{p, q} = \int x^{p}(1 + x)^{q}\, dx\) then \[(p + 1) I_{p, q} = x^{p+1}(1 + x)^{q} – qI_{p+1, q-1},\] and obtain a similar formula connecting \(I_{p, q}\) with \(I_{p-1, q+1}\). Show also, by means of the substitution \(x = -y/(1 + y)\), that \[I_{p, q} = (-1)^{p+1} \int y^{p} (1 + y)^{-p-q-2}\, dy.\]

(iii) Show that if \(X = a + bx\) then \[\begin{aligned} \int xX^{-1/3}\, dx &= -3(3a – 2bx) X^{2/3}/10b^{2}, \\ \int x^{2}X^{-1/3}\, dx &= 3(9a^{2} – 6abx + 5b^{2}x^{2}) X^{2/3}/40b^{3},\\ \int xX^{-1/4}\, dx &= -4(4a – 3bx) X^{3/4}/21b^{2},\\ \int x^{2}X^{-1/4}\, dx &= 4(32a^{2} – 24abx + 21b^{2}x^{2}) X^{3/4}/231b^{3}.\end{aligned}\]

(iv) If \(I_{m, n} = \int \frac{x^{m}\, dx}{(1 + x^{2})^{n}}\) then \[2(n – 1)I_{m, n} = -x^{m-1} (1 + x^{2})^{-(n-1)} + (m – 1)I_{m-2, n-1}.\]

(v) If \(I_{n} = \int x^{n} \cos\beta x\, dx\) and \(J_{n} = \int x^{n} \sin\beta x\, dx\) then \[\beta I_{n} = x^{n} \sin\beta x – nJ_{n-1},\quad \beta J_{n} = -x^{n} \cos\beta x + nI_{n-1}.\]

(vi) If \(I_{n} = \int \cos^{n} x\, dx\) and \(J_{n} = \int \sin^{n} x\, dx\) then \[nI_{n} = \sin x\cos^{n-1} x + (n – 1) I_{n-2},\quad nJ_{n} = -\cos x\sin^{n-1} x + (n – 1) J_{n-2}.\]

(vii) If \(I_{n} = \int \tan^{n}x\, dx\) then \((n – 1)(I_{n} + I_{n-2}) = \tan^{n-1}x\).

(viii) If \(I_{m, n} = \int \cos^{m}x \sin^{n}x\, dx\) then \[\begin{aligned} {2} (m+n)I_{m, n} &= -&&\cos^{m+1}x \sin^{n-1}x + (n – 1) I_{m, n-2}\\ &= &&\cos^{m-1}x \sin^{n+1}x + (m – 1) I_{m-2, n}.\end{aligned}\]

[We have \[\begin{aligned} (m+1)I_{m, n} &= -\int \sin^{n-1}x \frac{d}{dx} (\cos^{m+1}x)\, dx\\ &= -\cos^{m+1}x \sin^{n-1}x + (n – 1)\int \cos^{m+2}x \sin^{n-2}x\, dx\\ &= -\cos^{m+1}x \sin^{n-1}x + (n – 1)(I_{m, n-2} – I_{m, n}),\end{aligned}\] which leads to the first reduction formula.](ix) Connect \(I_{m, n} = \int \sin^{m}x \sin nx\, dx\) with \(I_{m-2, n}\).

(x) If \(I_{m, n} = \int x^{m} \csc^{n}x\, dx\) then \[\begin{gathered} (n – 1)(n – 2)I_{m, n} = (n – 2)^{2}I_{m, n-2} + m(m – 1)I_{m-2, n-2}\\ -x^{m-1} \csc^{n-1}x \{m\sin x + (n – 2) x\cos x\}.\end{gathered}\]

(xi) If \(I_{n} = \int (a + b\cos x)^{-n}\, dx\) then \[(n – 1)(a^{2} – b^{2}) I_{n} = -b\sin x (a + b\cos x)^{-(n-1)} + (2n – 3)aI_{n-1} – (n – 2)I_{n-2}.\]

(xii) If \(I_{n} = \int (a\cos^{2} x + 2h\cos x\sin x + b\sin^{2}x)^{-n}\, dx\) then \[4n(n + 1)(ab – h^{2})I_{n+2} – 2n(2n + 1)(a + b)I_{n+1} + 4n^{2}I_{n} = -\frac{d^{2} I_{n}}{dx^{2}}.\]

If \(I_{m, n} = \int x^{m}(\log x)^{n}\, dx\) then \[(m + 1)I_{m, n} = x^{m+1}(\log x)^{n} – nI_{m, n-1}.\]

40. If \(n\) is a positive integer then the value of \(\int x^{m}(\log x)^{n}\, dx\) is \[x^{m+1} \left\{\frac{(\log x)^{n}}{m + 1} – \frac{n(\log x)^{n-1}}{(m + 1)^{2}} + \frac{n(n – 1)(\log x)^{n-2}}{(m + 1)^{3}} – \dots + \frac{(-1)^{n}n!}{(m + 1)^{n+1}}\right\}.\]

41. Show that the most general function \(\phi(x)\), such that \(\phi” + a^{2}\phi = 0\) for all values of \(x\), may be expressed in either of the forms \(A\cos ax + B\sin ax\), \(\rho\cos(ax + \epsilon)\), where \(A\), \(B\), \(\rho\), \(\epsilon\) are constants. [Multiplying by \(2\phi’\) and integrating we obtain \(\phi’^{2} + a^{2}\phi^{2} = a^{2}b^{2}\), where \(b\) is a constant, from which we deduce that \(ax = \int \frac{d\phi}{\sqrt{b^{2} – \phi^{2}}}\).]

42. Determine the most general functions \(y\) and \(z\) such that \(y’ + \omega z = 0\), and \(z’ – \omega y = 0\), where \(\omega\) is a constant and dashes denote differentiation with respect to \(x\).

43. The area of the curve given by \[x = \cos\phi + \frac{\sin\alpha \sin\phi}{1 – \cos^{2}\alpha \sin^{2}\phi},\quad y = \sin\phi – \frac{\sin\alpha \cos\phi}{1 – \cos^{2}\alpha \sin^{2}\phi},\] where \(\alpha\) is a positive acute angle, is \(\frac{1}{2}\pi(1 + \sin\alpha)^{2}/\sin\alpha\).

44. The projection of a chord of a circle of radius \(a\) on a diameter is of constant length \(2a\cos\beta\); show that the locus of the middle point of the chord consists of two loops, and that the area of either is \(a^{2}(\beta – \cos\beta\sin\beta)\).

45. Show that the length of a quadrant of the curve \((x/a)^{2/3} + (y/b)^{2/3} = 1\) is \((a^{2} + ab + b^{2})/(a + b)\).

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 \(\pi(a^{2} + \frac{1}{2}b^{2})\).

47. Prove that if \((a, b, c, f, g, h | x, y, 1)^{2} = 0\) is the equation of a conic, then \[\int \frac{dx}{(lx + my + n)(hx + by + f)} = \alpha\log \frac{PT}{PT’} + \beta,\] 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 \(\alpha\), \(\beta\) are constants.

48. Show that \[\int \frac{ax^{2} + 2bx + c}{(Ax^{2} + 2Bx + C)^{2}}\, dx\] will be a rational function of \(x\) if and only if one or other of \(AC – B^{2}\) and \(aC + cA – 2bB\) is zero.^{1}

49. Show that the necessary and sufficient condition that \[\int \frac{f(x)}{\{F(x)\}^{2}}\, dx,\] 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 \(F\).

50. Show that \[\int \frac{\alpha\cos x + \beta\sin x + \gamma}{(1 – e\cos x)^{2}}\, dx\] is a rational function of \(\cos x\) and \(\sin x\) if and only if \(\alpha e + \gamma = 0\); and determine the integral when this condition is satisfied.

$\leftarrow$146. Lengths of plane curves | Main Page | Chapter VII $\rightarrow$ |