Problems and Methods in Analysis

Problems and Methods in Analysis

VOLUME 2

by

W. KRYSICKI; L. WLODARSKI A. J. ZIELICKI; D. KONSTANT

P E R G A M O N P R E S S OXFORD · LONDON . EDINBURGH · NEW YORK

TORONTO · PARIS · BRAUNSCHWEIG

Pergamon Press Ltd., Headington Hill Hall, Oxford 4 & 5 Fitzroy Square, London W.l

Pergamon Press (Scotland) Ltd., 2 & 3 Teviot Place, Edinburgh 1 Pergamon Press Inc., 44-01 21st Street, Long Island City, New York 11101 Pergamon of Canada, Ltd., 6 Adelaide Street East, Toronto, Ontario,

Pergamon Press S.A.R.L., 24 rue des Ecoles, Paris 5e Friedr. Vieweg & Sohn Verlag, Postfach 185, 33 Braunschweig,

West Germany

Copyright © 1966 Panstwowe Wydawnictwo Naukowe First edition 1966

Library of Congress Catalog Card No. 65-29065

This book is based on an original Polish work by W. Krysicki and L. Wlodarski entitled Analiza

Matematyczna w Zadaniach published in two volumes by Panstwowe Wydawnictwo Naukowe, Warsaw, 1958.

This book is sold subject to the condition that it shall not, by way of trade, be lent, resold, hired out, or otherwise disposed

of without the publisher's consent, in any form of binding or cover

other than that in which it is published.

(2744/66)

FOREWORD

This book is based on a two volume book by W. Krysicki and L. Wtodarski published in Warsaw in 1952 under the title Analiza Matematyczna w Zadaniach. It has been our aim to adapt the work so that it is suitable for sixth form students, particularly scholarship students, and for those first year university students (not necessarily mathematical specialists) who need a systematic course of revision in the methods of calculus. To this end a considerable amount has been omitted from the original (notably a full treatment of differential equations), and various sections have been added to or completely rewritten.

The purpose of the book is to help the student solve problems in analytical calculus without allowing him to forget the theory involved or to approach the matter in too mechanical a way. Consequently each chapter begins with a statement without proof of any necessary definitions and theorems. There is then a set of fully worked examples, followed by a set of exercises. Answers to all the exercises, together with hints to the solution of some of the harder problems, are given at the end of the book.

It must be insisted that this is not a textbook, but should be used in conjunction with the textbook of the student's choice. We hope that carefully used in this way the book will be of real assistance to the student in helping him to master the niceties of analysis.

xii

FOREWORD xiii

Our thanks must go to Professor W. Krysicki and Dr L. Wiodarski, for allowing us so free a hand with their material, to Dr E. A. Maxwell for his encouragement, and to the Per-gamon Press for their care and efficiency.

D. K. and A. J. Z. Cardinal Vaughan School, 1964,

C H A P T E R 10

INDEFINITE INTEGRALS

§ 1. Introduction Given a function/(x) defined in the interval a<x<b, there

may be associated with it a function F(x), called its primitive function, defined in the same region, such that F'(x) =/(Λ:). Two functions having in this interval the same finite first derivative can differ only by a constant. Thus functions whose first derivative is 2x are x2+3, x2+5, or in general, x2 + C

The indefinite integral oif(x) with respect to x, written

J7(x) dx is F(x) + C, where F(x) is the primitive function of/(*) and C is an arbitrary constant.

That is,

(f(x)dx = F(x) + C,

where F'(x) =f(x).

§ 2. Some standard forms f xn+1

2.1 \xndx = - + C (n^-h x > 0 ) . J w + 1

If ft is a natural number the condition x > 0 is not required. If n is a negative integer then it is sufficient to ensure that

1

2 PROBLEMS AND METHODS IN ANALYSIS

x 7* 0. Three examples are given below:

(a) n = 0 dx = x + C;

(b)n = - I J - ^ = 2V% + C (xÔ);

(c) « = - 2 J ^ = - i + C (xÔ).

f dx 2.2 — = log | x | + C (x ^ 0).

Note that log z is defined only for z > 0.

2.3 f e*dx = e* + C.

2.4 |tf*dx = - ^ - + C ( α > 0 , ^ 1 ) . J log Ö

2.5 cos xdx = sin x + C.

2.6 sin xdx = —cos x + C.

2.7 sec2 xdx = tan x + C (cos x ^ 0).

2.8 cosec2 xdx = —cot x + C (sin x ^ 0).

Γ dx 2.9 —πτ; sr = arc sin x + C = —arc cos x + C*

( | χ | < ΐ ) .

-77-0 s^ =arc sin —+ C = —arc cos—YC * y/(a2— x 2 ) ■ tf a

(x2<a2).


dx = arc tan x + C = —arc cot x + C* +x*

dx J+:

2.13 I sinh xdx = cosh x + C.

2.14 f cosh xdx = sinh x + C.

2.15 | d * = t a n h x + C. coslr x dx

2.16 | -T-Tö—= - cothjc + C sinrr x

2.U J T

2.12 f „d x - = - arc tan -+C = - - arc cot - + C ' . ·

.13 f sir

.14 cosh xi

■ " / =

. « /

Γ dx 2.17 J - ^ ^ = s inh- ix + C = log[> + V(*2 + l)] + C.

[JVbte: The inverse hyperbolic functions are denoted in many books by ar sinh, ar cosh etc., where ar is an abbreviation for area.]

2-18 J V(*2+*2) =

= sinh"1 - + C = log Γ - + 1 V(*2 + a 2 ) l + C . a \ a a

2.19 j* ,/**_ = cosh-i x+C = log | x + J ( x 2 - l ) \ + C

( | x | > l ) .

* Note that arc sin z+arc cos z = -y,

π arc tan z+arc cot z = —,

and arc sec z+arc cosec z = -y ·

PROBLEMS AND METHODS IN ANALYSIS

= cosh- 1 —h C = log ^+IV(^2-«2) tf a

0 — t a n h ^ x + C = — log ■ 2 "· JA

. . . f dx 1 , , x , „ 1 2.22 -2 5- = - tanh"1 — + C = ^ J cr— χ·2 Λ α 2c

l+x 1 - x

-log

+ C

(x2 > fl2)

+ C (x 2 < l )

a+x a—x

+ C

(x2 < a2).

- /^r - c o t h - 1 x + C = y log x - 1 + C

„ „,, f dx 1 _ , χ ^, 1 , 2·24 j-x^ = -7coth-17+C = Talos

x + l

(x2 > 1) x—a

+ C

2.25 / ■ dx

= ± s e c h _ 1 x + C

x + a\ (x2 > a2).

( x 2 < l ) . x V(l -x2)

2·26 J w ( ^ ) = 4 s e c h - ^ + c (*2<*2>· 2.27 f dx

J xV(*2 + D ± c o s e c h - 1 x + C .

2 · 2 8 f Z / V L 2 = ± - cosech-i - + C . J x y ( x 2 + ö 2 ) a a

§ 3. Some general properties of indefinite integrals

3.1 The addition of integrals

f {Ax) +g(x)} ax = J 7 ( * ) d * + f S(x) dx.


3.2 Multiplication by a constant

kf(x) dx = k \ f(x) dx.

3.3 Integration by parts If u and v are functions of x with continuous first deriva

tives, then

u dv = uv — I v du.

3.4 Substitution Given a function g(x) = u, then ^'(Λ;) dx = du, and

jf{g(x)}g'(x)dx = jf(u)du.

3.5 Two important results

(a) f£Qdx = logflx) + C,

(b) iw§*d*= 2 V L* x ) 1 + c · Examples

1. Integrate J = x(x-l)(x-2)dx.

Solution. The integral can be expressed as the sum of integrals by multiplying out.

7-Jc*-3*+*)dx.


= \ x3 dx — 3 \ x2 dx + 2 \ x dx,

X1 „ X3 „ X2 _

= T - 3 ' T + 2 - y + c -Hence, / = —x4 - x3 + x2 + C.

4 2. A point moves in a straight line so that its acceleration

a at time t is given by:

a= 12i2 + 18 sin 3 f - 2 .

Find its velocity t? and its distance from the origin x at time t if at t = 0, #o = 10 and x0 = 5.

Solution

(i) v= \adt= f ( 1 2 i 2 + 1 8 s i n 3 i - 2 ) d i ,

= 4 i 3 - 6 c o s 3 i - 2 i + C.

For * = 0,

v0= - 6 + C = 10, hence C = 1 6 . Hence,

# = 4 ^ - 6 cos 3/ — 2t+16.

(ii) J C = I t ; d i = ( 4 i 3 - 6 c o s 3*-2 i+16)d i ,

= * 4 - 2 s i n 3 i - - * 2 + 16i-|-C". For t = 0,

x0 = C = 5. Hence,

x = i 4-2sin3*-i 2-}-16*-h5. 3. Integrate

r=jvx-^d;c (Λ;>0)

INDEFINITE INTEGRALS 7

Solution. Express the roots of x as fractional powers. Thus

I^j—p—ox,

= f(x-*'*-x-5/3)dx,

X-1/» X-2's

1 _^ 2 3

+ C.

3 2 Hence, J = —— — + C. 2V*2 V*

4. Integrate T f Xdx r f\\

Solution. Make the substitution x2+a2=w, then x dx=(1 /2) dw.

Hence, / = τ / ^ · There are two solutions: (i) n*l.

1 w~n+1

T 1 , e n · ' - 2 ^ T T + c · +c.

2(n-l)(x2 + a2)n-1

(ii) R = 1.

Then, / = i-log u + C,

= l l o g ( x 2 + a2) + C.

5. Integrate


Solution. Make the substitution 2x — 3=t2 (where i>0) ,

then dx = t dt. Hence,

>-ΙΨ-Ι«->+' or in terms of x

I=J(2x-3) + C. 6. Integrate

' - / * « " dx.

Solution. Make the substitution x2 = t, then x dx = -— dt.

Hence,

7 = 1 f etdi = ye*+C

and 7 = y e x 2 + C .

Note that ex2dx is not capable of such simple treatment.

7. Integrate

/ = J s i n * c o s * d * .

Solution. Three possible methods are given, (i) Put sin x = t, then cos x dx = dt, and

· ■ / ■ / = | tdt=^-t2 + C,

= -y sin2 x + C.


(ii) Write sin x cos x as — sin 2x, then

r = — sin 2x dx = — sin 2x d(2x),

= — r c o s 2 x + C . 4

(iii) Put cos x = t, then sin Λ: dx = — dt, and

f d t = - y i 2 + C", - /

= — I c o s 2 x + C",

It may appear that we have three different solutions, but in fact the three solutions differ from each other only by a constant.

1 · 2 -r- smz x - j — — cos 2x j = — s i r X+-T- cos 2x,

= — (2 sin2 ΛΤ+COS 2x),

= — (2 sin2 ;v +1 —2 sin2 x),

_1_

That is C = C'+4-. 4

Similarly it can be shown that C = C"+-=-.

8. Integrate

' dx. XQX i


Solution, Integrating by parts we put u = x, and dv = ex dx,

then du = ax and v = e*. Hence,

>'-$ XQX — ex dx, (by

: XQX— eX + C,

: e* (x - l ) + C.

9. Integrate

= log* dx ( x > 0 ) .

Solution, Integrating by parts we put u = log x and dv = dx,

then

Hence,

du = — dx, and v —x, x

i 1 Γ dx I = xlog x— x· — ,

= xlog x—x + C, = x(logx-l) + C.

10. Integrate ■ ■ / e* sin x dx.

Solution. Integrating by parts we put u = sin x, and dv = exdx,

then dw = cos x dx, and # = ex.

Hence,

/ = e * s i n * - J e * c o s * d * .


Similarly, by integrating e* cos x dx by parts we obtain

ex cos x dx = e* cos x+ e* sin x dx.

— e* sin x dx, Hence,

/ = e* sin X—QX cos x

= e* sin x—e* cos x—I, thus,

/ = —e* (sin x— cos x)+C.

E X E R C I S E 18 Integrate the following functions of x with respect to x9 stating where

necessary any limits on the value of x:

dx. 1. J (^-6x+3-4+ | )dx. 2 . / ^ 3. ϊ(χ2-χ + 1)(χ* + χ+1)άχ. 4. f (jc2+4)5*dje.

Γ JC dx Γ xdx Γ x2 dx 5* J T i l ? ' J (;c2+3)e * J ^Hh?

4 ,

8 jWWi^ 9. JfVWf^

10. J(3 + 2 ^ ) 3 d , . 11. j V ^ 2 ^ H 4 ^ 3 ) d j c

6 V*

12. 3

S^J^**' 13· JV(3*+l)d*. 14. fV(« + )d^. 15. f x dx

V(2*2-D 16. fxVd+^2)dx. 17. fj-TT^-^d*. x

V(3-5JC2)


22. Jxe- z' dx.

Jx log (l +x2) dx.35.

Jsins x cos x dx.

J sin x dx.a+b cos x

27.

25.

(where exp (z) is eZ).

Jel/xdx2 •X

30. Jsec2 x tan2 x dx.

J (log X)232. dx.

x

19. J"';(X~-6) dx.

JeXdx2ex +1·

37. J61 - z dx.

Jlog Iarc tan x I d39. 1+x2 x.

J x2dx41. y(1-x6) •

J(n - arc sin x) dx43. Y(1-x2).

Jx 4(l +X)3 dx. 46. Jx2ez dx.

Jx 4e2z dx. 49. Jx cos x dx.

Jx 2 sin 5x dx. 52. JeZ cos x dx.

Je-Z cos ; x dx. 55. J...;x log x dx.

J(log IX 1)2 dx 5 x.

21.

xy f1 - (log IX\)2] •

Jxez'(x2 +1) dx.

J dx(1 +x2) arc tan x .

J xdxx 4 +1 . 45.

Jx3ez dx. 48.

Jx 2 cos x dx. 51.

Je- 2Z sin 3x dx. 54.

J(log IX 1)3 dx. 57.

J x-Idx.

~(x+1)

J x2

dx

~(x3+1)

J dx2 cos2 (3x)·

Jx sin (2x2 +1) dx.

J cos x dy(l +sin x) x.

Jcos x exp (sin x) dx

Jx3sec2 (x4) dx.

J X2 dxcos2 (x3 + 1)·

J dx---. 34.eX+e- x

J"';(2+ ~Og Ix I) dx.

dx

56.

50.

53.

47.

40.

33.

36.

38.

31.

42.

29.

44.

28.

26.

20.

23.

24.

18.


58. f ^x(log\x\)3dx. 59. f 1°^χ] dx.

60. f ( 1 ° / * * <**. 61. Γ xs (log x)3 dx.

62. xn logJtdx, w* —1.

§ 4. The integration of rational functions

4.1 Introduction The rational function f(x) is the ratio of two polynomials,

i e f(x\ = S = ^Ρ^ + ^ Ρ - Ι ^ " ^ ··· + *o JK } ~ β(*) ~ ^ + ν ^ Ή · · · +*o '

If p^ q the numerator may be divided by the denominator so as to obtain a polynomial of order p—q (which is readily integrable, see 3.1), and a rational function R(x)/Q(x) in which the order of R(x) is less than that of Q(x).

lip < q it can be shown that/(x) can be expressed as the sum of terms like

(D 7LTTTÜ»> a n d (2) P (Ax+k)m' w (ax2 + bx + c)n '

where A, a, /?, A, /:, a, b, c are constants and m and « are positive integers. Also it will follow that ax2+bx+c cannot be further resolved into real factors, since if this were possible the terms like (2) could be expressed as the sums of terms like (1). This implies that the discriminant of the quadratic expression is less than zero, i.e. b2—4ac< 0. In general then,

m A s T r

r=l (*i* + &i)r " " r = l (Af* + kj)r

" KrX+ßr * λνχΛ-μτ

r=i (<^ι^2 + * ι* + Ci)r " " ~ χ (#λΧ2 + &λ* + ^ ) r *

Note that some of the coefficients A, J, α, β, λ, μ may be zero.


In expressing f(x) in this way we are said to resolve f(x) into its partial fractions. The examples below (4.2) will make the technique clear.

Examples

1. Integrate -4 Γ3χ + 4

J ^ τ dx.

Solution. Note that the integral does not exist at x = 1. The two polynomials are of the same order, so dividing numerator by denominator we obtain

3x + 4 „ 7 = 3+-x— 1 x— 1

Hence, / = 3 I dx + Ί — ^ - ,

= 3* + 7 1 o g | x - l | + C .

2. Integrate

J ax + b y a J

Solution. Put ax+b = t, then ax = — di.

Hence, 1 Λ At 1

■log If I + C , 1 Γ 1 ί - 1 a J t a

= — log\ax + b\+C. a '

3. Integrate

(ax + b)ndx. / ■


Solution. The values of x for which this function is integrable depend on the value of n.

(i) n a positive integer: no restriction on x; (ii) n a negative integer: ax+b?±0;

(iii) n a fraction —, q odd; no restriction on x; q

(iv) n a fraction —, q even: ax+b > 0. q

The case n — — 1 has been considered in the example above.

As before put ax-Yb — t, then ax =— at. a

Hence, 1 tn+1

~ a n+ 1

(ax + b)n+1

4.2 Integrals of the type

+ C ( n ^ - 1 ) .

/ ; mx + n

ax. ax2 + bx + c

(i) If mx+n is the derivative of ax2-\-bxJrc the integral is a logarithmic function (see 2.21), i.e.

2ax + b

! ax2 + bx + c ax = log | ax2 + bx + c | -f C.

We must note that the function will not be integrable for values of x for which the denominator vanishes. The discriminant Δ = b2—4ac should be examined to see whether real factors exist. If A < 0 the quadratic expression has no real factors and the integral exists for all values of x; if Δ ^ 0 the quadratic has real factors and there are either two values or one double value of x (when Δ = 0) for which the integral does not exist.


Examples

4. A point moves in a straight line so that its acceleration a after time / (t > 0) is given by

1 a = 6t + . t+l

Find its velocity v and its distance s from the origin at time t given that at t = 0, v = v0 and s = s0.

Solution

= 3i2 + log | i + l | +C. Hence, ^0 = 0 + 0 + C, and v = 3i2 + log \t+ 1 | + v0.

s= \vdt = (3i2 + log | i + l | +v0)dt,

= ts + (t-t-l)\og\t+l\-t + v0t + C. Hence, s0 = 0 + 0 - 0 + 0 + C , and s= tz + (t+l)log\t + l \-t + v0t + s0. (Note here that t is essentially positive and that the denominator (t + l) can therefore never vanish.)

5. Integrate

/ = - —s zr- dx. J 3x2-x + 2

Solution. We first examine the sign of the discriminant, Δ = l - 2 4 < 0 .

Hence the quadratic expression has no real factors and the function is integrable for all real values of x.

Moreover — (3x2 — x + 2) = 6x — 1,


i.e. the numerator is the exact derivative of the denominator, hence, / = log (3x2 - x + 2) + C.

6. Integrate ; - 3 J *2- 6x + 5 ax.

Solution. Here zl = 36—20 > 0, hence the quadratic expression factorizes. In fact

x*-6x + 5 = (x-5)(x-l). Thus the function is not integrable for x = 5 or x = 1.

Also ^-(JC2-6JC + 5) = 2 x - 6 = 2(x-3).

Hence, J = T J χ2_6χ + 5 dx =-log\x*-6x + 5\ + C.

[Note that if zl ^ 0 this method is useful provided only that the numerator is a multiple of the exact derivative of the denominator.]

(ii) If mx+n is not the derivative of ax2+bx + c the method of integration will depend on the sign of the discriminant Δ.

(1) Δ > 0. The quadratic can therefore be resolved into real (but not necessarily national) factors and the function split into a pair of partial fractions which can then be integrated as shown above.

Examples 7. Integrate

"-J: dx 2x2 + 5x + 3 '

Solution. Examining the sign of the discriminant we see that A = 25-24 > 0 .

By inspection 2x2 + 5* + 3 = (2* + 3)(*+l).


Hence the function is not integrable for x = — 1 or x = —3/2. Suppose

1 = A B 2x2 + 5x + 3 "~ 2x + 3 jc + 1 '

Then, l = 4 x + l ) + B(2x + 3).

Equating coefficients of the powers of x we obtain A+2B = 0, and A + 3B=l,

hence, A = — 2 and B = 1.

Therefore,

/=j{^-2^}dx' = log | J C + 1 | --log 12x + 3 | + C.

I x + 1 I Hence, 7 = l o g \=-—r \ + C. \2x + 3

8. Integrate

/ =J 3^-5x-2d x · Solution. Here zl = 25 +24 = 49 > 0. Thus the denominator

factorizes. 3 X 2 _ 5 X _ 2 = ( 3 J C + 1 ) ( X - 2 ) .

Hence the function is not integrable for x = —1/3 or x = 2. Suppose

l l x - 1 = A B 3x2 — 5x — 2 ~~ 3% + 1 x — 2 '

Then, 1 lx - 1 = A(x - 2) + B(3x + 1).

Equating coefficients we obtain A + 3B=ll and -2A + B=-1,

hence, A = 2 and i? = 3.


Therefore,

^/{^ττ+ώ}^ 2

= — log|3x + l | + 3 1 o g | x - 2 | + C.

(2) A = 0. The quadratic is therefore a perfect square and the function can be expressed in the form

A B (px + q)2 px + q

which can easily be integrated (see 4.2 Ex. 1 and 2).

Examples

9. Integrate

9 χ 2 - 1 2 χ + 4 *

Solution. A = 144 — 144 = 0. The denominator is therefore a perfect square, and

9 x 2 - 1 2 x + 4 = (3x-2) 2 . Hence, x cannot take the value 2/3.

J = J ( 3 ^ = J^-2)-2d^ (3X-2) - 1

+ C, 1

C- 3 ( 3 J C - 2 ) '

10. Integrate 9 x - 5

■ / ax. 9 x 2 - 6 x + l

Solution. A = 36—36 = 0. The denominator is therefore a perfect square and

9 x 2 - 6 x + l = ( 3 x - l ) 2 .


Hence, the function is not integrable for x = —. Suppose

9 x - 5 A B - + -( 3 x - l ) 2 " ( 3 x - l ) 2 ' 3 * - l "

Then 9x-5 = A+B(?>x-\).

Equating coefficients we obtain 35 = 9 and A-B = -59

hence, A = — 2 and B =3.

Therefore,

/ = _ 2 f dx + 3 f dX Z J (3x- l ) 2 + JJ (3JC-1) '

= - 2 ^ i ) + 1 ^ i 3 - - 1 i + c '

2 + l o g | 3 x - l | + C . 3 ( 3 x - l )

(3) Zl<0. The quadratic expression cannot then be resolved into real factors so it is rearranged in the form X2+A2 (where X is a linear function of x and A is independent of x) and integrated as an arc tan. If the numerator is a linear function of x it should be written as the sum of the derivative of the denominator (or a multiple of the derivative) and the appropriate compensating numerical term (see Ex. 12 below).

Examples

11. Integrate

/ dx

2χ2-12χ+2Ί

Solution. Δ = 144-216 < 0. Hence the quadratic has no real factors. We therefore rearrange the expression by completing

INDEFINITE INTEGRALS 21 the square and compensating the numerical term.

2x2-12x+27, = 2(χ2-6χ)+27, = 2 ( Λ ? - 6 Χ + 9 ) + 2 7 - 1 8 , = 2(x-3)2+9.

Therefore, dx

2 J ( * - ; -3)2+9/2 Put x - 3 = X, then djc = dX, and

dX - T /

1 V2 arc tan

arc tan l2L^Lfx-3) \ + c.

12. Integrate

3 V 2 . . . . . . ( ^ ( , - 3 ) ) -

. 8x + 5 ' 2x2 + 6x + 5

Solution, Δ = 36— 40 <0 . Hence the quadratic expression has no real factors and is integrable for all real x. We first express the numerator as the derivative of the denominator together with the appropriate numerical term.

-Γ-(2Χ2 + 6Χ + 5) = 4x + 6, ax and Sx + 5 = 2(4x + 6)-7 . Hence,

: + 6) -7 f 2(4JC " J 2x2- dx, 2x2 + 6x + 5

dx j2TO^5d^7J' 2χ 2 +6χ+5 '

= 2/1-7/2 . 7x=log(2x:2+6x + 5).


Rewriting the denominator in I2 by completing the square we obtain

2x2 + 6x + 5, = 2(χ2 + 3χ) + 5,

= 21

Therefore, (*4J4·

~ 2 J (x + 3/2)2 + l/4

Hence,

= —.2arc tan2(x+—j,

= arc tan (2x + 3). / = 2 log (2χ2 + 6χ-{-5)-7 arc tan (2x + 3) + C.

(iii) Alternatively any function of the type considered above may be integrated by expressing the denominator as the sum or differance of two squares, i.e. as X2±A2, where if is a linear function of x and A (possibly zero) is independent of x. In effect this simply bypasses the necessity of examining the sign of the discriminant.

Example 13. Integrate

■•h 6 dx :2+2x-15

Solution. Rearranging the denominator we obtain x 2+2x-15

= ( x + l ) 2 - l - 1 5 , = (x + l)2-16.

Hence, 6 ax J (* + 1)2-16


Put x +1 = X, then dx = dX, and άΧ , Γ dX

23

7 = 6 (X+4)(X-4)' ]x^ü>=6\ = 4/{τ^4-ΤΤ4-}αΖ' = A { i 0 g | z _ 4 | - l o g | Z + 4 | } + C,

= j l o g

= j l o g

X-A Z+4 x—3 x+5

+ C,

+ C.

4.3 Practical methods We now give some examples showing how rational functions

are in practice reduced to the sum of integrals of the types considered above.

14. Integrate 3 Λ ? - 5 Χ 2 + 8 Χ ■ - /:

dx. (x2-2x+l)(x2-l) Solution. First we factorize the denominator as completely

as possible: (χ 2 -2* + 1)(χ2-1) = (x- l ) 2 (x 2 - l ) ,

= (x-l)s(;c+l). Hence the function is not integrable for x = ±1. Then we express the function in its partial fractions. Suppose,

3Λ^-5Χ 2 + 8Χ A B C D · + ,.. , ^ + — Γ - + -(x-l)3(x + l) ~ ( x - l ) 3 (x-1)2 ' ΛΓ—1 ' x + l

Then, 3x*-5x2+Sx = A(x + l)+B(x + l)(x-l) + C(x-l)2(x+l)

+D(x-1)3,

2 4 PROBLEMS AND METHODS IN ANALYSIS

This is an identity and is therefore true for all values of x.

F o r x = l : 3 - 5 + 8 =2A, thus ,4 = 3 ;

F o r x = - 1 : - 3 - 5 - 8 = -8Z>, thus / ) = 2.

Equating coefficients,

of*3: 3 = C + A thus C = 1;

constant term: 0 = A-B + C-D, thus B = 2.

Hence, f 3d% Γ 2 d x f dx C lax 7-J (*-l)3+J (^lF + J ^ T + J TW

= - 4 ^ - - ^ + logl^-lK21ogU + l | + C.

15. Integrate / = J (* + l)(x» + l) d x ·

Solution. We separate the expression into its partial fractions. It must be remembered that the number of constants in the partial fractions must equal the power of the original denominator, consequently when there is a quadratic term in the denominator which has no real factors its corresponding partial fraction mustq ave a numerator linear in x, involving two constants. In particular cases one or other of these constants may be zero. Thus

x-l _ A Bx + C (x+l)(x2 + l) = 1 H T + x2 + l '

thus j c - l = A(x2 + l)+(Bx + C)(x + l).

This is true for all values of x (real and complex):

For x = - 1 : -2=2A9 thus A = - 1 ;

F o r * = /; i-\ =(Bi + C)(i + l) =(-B + C) + i(B + C),


equating real and imaginary parts

thus, and Therefore,

T -

-B+C= - 1 B+C = l,

C = 0 B = l.

- f ~dX , f * H

= I l og (x 2 + l ) - l o g | x + l | + C .

[Note. The coefficients of non-repeated linear factors can be easily found by the "cover-up" rule. The value of x for which the linear factor vanishes is substituted in the rest of the original fraction, the linear factor remaining "covered". Thus in the example above the coefficient of l/(x + l) is found by putting x = — 1 in the rest of the expression, namely (— 1 —1)/(1 + + 1) = -2/2 = -1.]

16. Integrate, x5+x4 + 3x3-f-x2-2 7 ■1

- dx.

Solution. Note that the integral does not exist for x = ± 1. Since the denominator is of lower order than the numerator we divide the latter by the former and obtain

3x?+x2+x-l x + l+:

x 4 - l

Hence, / = {x + Y)dx+ 5— dx,

= / i+ / a .

Ix =-jX2+X.


The factors of the denominator of I2 are (x2 + l), (x — 1) and (x + 1), and by the same process as above we obtain

f dx C ax C x + 1 J

= log|*-l|+log|* + l | + J ^ + J ^ ,

= log | x2 — 1 | + y log (x2 +1) +arc tan x.

Thus, 7 = - y X 2 + x + l o g | x 2 - l | + -ylog(x2 + l) + arctanx + C.

17. Integrate

Ϊ 4 + 2 Χ 2 + 9 r f 2 χ 3 - χ 2 + 4 χ - 3 . / = J **-" d x ·

Solution. The denominator is a biquadratic which we resolve into two quadratic factors:

x*+2x2+9 =(x2 + 3f-4x2, = (x2 + 3-2x)(x2 + 3+2x), = (x2-2x + 3)(x2+2x + 3).

Note that neither quadratic factor will factorize since for each the discriminant is less than zero.

We now express the function in its partial fractions. Suppose,

2 x 3 - x 2 + 4 x - 3 Ax+B Cx+D (x2-2x + 3)(x2+2x + 3) X 2 -2 JC + 3 X2+2JC + 3*

Then 2x* - x2 + 4x - 3 = (Ax+B)(x2 + 2x + 3) + (Cx+D)(x2 - 2x + 3).

By equating coefficients we obtain 4 simultaneous equations in A, B, C, D with the solutions A = 1, B = -I, C = 1, Z) = 0,

INDEFINITE INTEGRALS 2 7

Hence,

/ι=τ/Ατ3^=τΐ0δ(Λ2-^+3)· \_ C 2x+2 _ Γ dx

2 ~ T J x2+2x + 3 J x2+2x + 3'

= ylog(x2+2x + 3) /2" a r c t a n "Το"^-1"1)·

Thus / = — log (x2 - 2x + 3) + y log (x2 +2x + 3) -

-—fr a r c t an -^ - (* + l) + C,

= — log (x4+2x2 +9) —-^- a r c t a n -75- (* + 1) + C.

18. Integrate dx Wo x2 + l ) n '

Solution, In this case we obtain a reduction formula, i.e. we express In as a function of 7m where m < «. We rewrite the numerator so as to obtain In_1 on the right-hand side.

- f d * _ f *2 + l - * 2 7 n " J (x2 + l)» " J (*2 + l)n

f x2

= / n - l - J (Λ2 + 1)η d*'

We integrate the second integral by parts. r> * , xdx Put M=x' d" = ö?w


Then du = dx, v = ^ τττ-α—^—7. 2(/i-l)(x2 + l) r i -1

Therefore, Γ x2 - x f dx

J (x2 + l)" X~(2«-2)(x2 + l ) ^ - 1 + J (2«-2)(x2 + l ) n - 1 ?

- x 1 (2«-2)(x2 + l ) ^ - i ^ 2 « - 2 Therefore,

T — T x * 4i - 4 - i + (2„_2)(x2 + l ) ^ - i " 2 ^ 2 n _ 1 '

_ 1 x In-3 ~ ^ 2 * (x2 + l ) n " 1 + 2 ^ 2 n - l e

Thus if « = 3, 1 x _3

"4" (x2 + l)2 + ~4 1 x _1

T x 2 + i + I

^3 — T ί^Λ , ι\2 + " Γ ^ 2 Ϊ

^2 — -Ö- ν2.ι 1 + ^" Λ 5

Γ dx A = —5—7 = arc tan x.

Therefore,

J (*: dx 1 x 3 x 3 -2 + l)3 4 (x2 + l)2 ' 8 (x2 + l)

19. Integrate

+ "F" 7~2 TV + ~F" a r c t a n x + ^·

J (*2 dx 2 - 4 x + 13)2 '

Solution. We express the denominator in the form (X2± A2)2. x 2 -4x+13 = (x-2)2 + 13-4,

= (x-2)2 + 9. Therefore,

i=[ Üi . J {(x-2)2 + 9}2

Put x—2 = 3«, then dx = 3 di/,


and Γ 3d« _ l_ C_

~ J (9a2 + 9) 2 ~ 27 J (i du

M2 + l)2

= —I2 (where In is defined as in the previous example),

= 27{T(l?TlT+2-arCtanMj+C·

Therefore, 1 j c - 2 1 A Λ : - 2 ^

/ = l 8 ' x ^ 4 x + 1 3 + 5 4 a r C t a n - T - + C

E X E R C I S E 1 9 dx 1. J(2,+ l)3d,. 2. J ^ - £ ^

J x2 — x— 6 ' J jr 2x-3 J dx.

3JC+3 x+13 , f 2x+6

f — 7 — r d x - s f 4 * - 5

J x*--Lx+4- ö · J 2*2-5* + 3

dx. 2*2 + 3*+l

d*. 2 ' 2

f Λ + 1 1ι η ^ . io. f 5

J x2 + 3x-10 J *2+3x-18 dx.

11' J x 2 + 2 * - r 12' J 6χ*-13χ+6άχ· 13· Jis£r«*· "■ J4-wd-15· J -5 + Zc-x*· 16, J 1+χΧ-χ* ■

■2

19. f ^ = - L d * . 20. f — i Z » dx. J x2-6x+9 J 4 X 2 - 4 X + 1

30

21.

23.

25.

27.

29.

31.

33.

35.

37.

39.

41 .

43.

45.

47.

49.

51.

53.

55.


2x - 1 3

5)2

f 2A:-

J ( * -j

d*.

2 A : 2 - 2 A : + 5 *

C d * J 13-6A: + A:2·

Γ 2x-1

J * 2 -Γ 2A

J *2-

2JC+5

2A : -20 8A:+25

x+6 J A : 3 - ax.

JC2 + 4 * + 1 3

4 A — 5 Γ 4;

j 6 A : + 1 0

.2

d;c.

dA:.

dx.

dA:.

5A:2+ 12

7 A : 2 + 7 A : - 1 7 6 3A:3-9A:2 + 6 A : + 5 6

3 A : 2 - 5 A : + 2 J A : 3 -

22.

24.

26.

28.

30.

32.

34.

36.

38.

40.

dA:. 42.

2 A : 2 + 3 A : - 6

A : 3 + 2 A : - 6

dA*.

Γ Α : 3 + 2 Α — J A:2-A—

J A:2 + 1

2

dA:.

dA:.

44.

46.

48.

3 A : + 1 2)2

dA:

r 3A: H J (* + ax.

J 3A:2 + 2 A : + 1 ·

3dA:

9x2-6x+2'

4x- 1

j Γ 4*

J 2 A - 2 - 2 A : + 1

2 A : - 1 0 2 A : + 1 0

3 A : + 4

dA:.

dA:.

4 A * + 8

x+6 A 2 , α dA·.

A:'2+3

1 0 A : - 4 4

Γ 3A J A:2 +

J r io.

J A:2-4A: + 20

J te

II J (

dA-.

dA-.

dA:. 2 + 3A:

2A:2+7A:+20

+ 6 A : + 2 5

* 3 - 4 A : 2 + 1

djc.

A : - 2 ) 4

2 A : + 1

dA:.

dA-. J (*2 + D 2

Γ 2 Α : 3 - 1 9 Α : 2 + 5 8 Α : - 4 2

J A : 2 - 8 A : + 1 6

f 72A:6 Λ

j 3 ? + 2 d ^

djc.

C 2 A : 4 - 1 0 A : 3 + 2 1 A : 2 - 2 0 A : + 5 f A:2 + 5A:+41 J A:2-3A:+2

d*' 5°" J (*+3)(*-1)(*-1/2) 1 7 A - 2 - A - - 2 6

D U 2 - 4 ) 2A:

C 17A-2

!

S

dx.

dx. (A:2 + 1)(JC2 + 3)

4x3-2x2 + 6x-13 A : 4 + 3 A : 2 - 4

52.

54.

A:2(6-6A:) ! d^, put 6 = / .

djc. 56.

10A:3 + 1 1 0 A : + 4 0 0 ( A : 2 - 4 A : + 2 9 ) ( ; C 2 - 2 A : + 5 )

f 1 0 A : 3 + 4 0 A : 2 + 4 0 A : + 6 J A:4+6X3+11A:2+6A:

d*.


• J I J J

• J • J • J

"■ J „ j

57,

59.

61.

63.

65

67,

69

6x3+4x+l A n — 2 — d * ·

d* x3+x2+x ' 5*3+3*2 + 12*-

58.

60.

■12 * 4 - 1 6

4 * 3 + 9 * 2 + 4 Λ Γ + 1

d*. 62.

* 4 + 3 * 3 + 3 Λ : 2 + Λ :

άχ (* 2 +*+l) 2 *

dx (JC2+4JC+8)3 '

άχ *4 + 64 *

άχ *4+6*2+25 * χ3-2χ2+5χ-$

* 4 +8* 2 + 16

άχ.

άχ.

64.

66.

68.

70.

72.

74.

/ ■ dx

άχ

ί xl+x* + \ ·

15χ2+66χ+21 (Λ;-1)(Λ:2+4Λ:+29)

άχ

άχ.

J * 3 ( Λ : - 1 ) 2 ( Λ : + 1 ) ·

f 3 ^ - m + 2 1 J

J J J J

djc. (*-2) :

*-2χ2+Ίχ+4 (x-

5x3 -1)2(*+1)2

-11JC2 + 5 * + 4

dx.

(χ-iy 9JC4-3JC3-23JC2+30JC

(x-l)4(*+3) 3*2+x-2

djc.

dx.

(*-l)3(*2 + l) dx.

§ 5. The integration of irrational functions

5.1 Integrals containing roots of linear expressions In general, if x or a function of x where the function is linear or homographic, i.e. of the form ax+b or (ax+b)/(cx+d) (ad—be ^ 0), is raised to a fractional power p/q9 where p and q are relatively prime, then we make the substitution

x = tn, or f(x) = tn

where n = q or, if different powers are involved, n is the l.c.m. of the different numbers q.


■ / dx

v*+v* (x > 0).


The substitution required is that which will make both x1/2 and x1/3 rational, hence put

x = t* (t > 0), then yjx = t3, l/x = t2, and dx = 6t5 at. Therefore,

'-J??lr*-«/7ir*-i/('-'+1-7Tr)* = 2 ί 3 -3 ί 2 +6 ί -6 log (f+ 1) + C.

Therefore, I = 2*Jx-3\lx+6\Jx-6\o%(\Jx + \) + C.

2. Integrate

J \/(3x-7)dx' (^>y)· Solution Put 3 x -

Then 3 dx and

- 7 = * 4 , = 4*3 df s

/ :

where >

■ / · ·

4

i > 0 .

4 , j

5+c. 4

Therefore, Ι = γξ V(3x - 7)5 + C.

3. Integrate

f 7*+i._L J \lx—l x+1

Solution π x+1 , t ί3 + 1 , , -6*2d/ Put - = t3, then x = -^—-, and dx = -r-=—^. x - 1 ί 3 - 1 (*3-l)2


f dt Hence, / = — 3 I ^ , and by expressing this function in terms of its partial fractions,

r C t+2 , C at

Then, by the methods shown above, we obtain

7 = y log( i 2 + i + l) + V 3 a r c t a n - ^ l o g | i - l | + C ,

where t ■ m\ E X E R C I S E 2 0

Integrate the following expressions, stating any limits there may be on the value of x:

1.

3.

f V(2* + l)dx. 2. f άχ

V(3+4x) ' r d* 4. r dx

^ ( 3 * - 4 ) V"(2JC + 1)3

5. ϊχ^/(χ-4)άχ. 6. f jt^/(3x-l)dx.

7. f *V( 2 + 3*)d·*· 8. Ϊ x yj(l-5x) ax.

9. f x^(;t-4)d;t. 10. f dx. J J ^(2^+3)

11. dx. 12. —77 —r-TT dx. ^ " ' ' 2 ' ^ 3 ^ 1 5

13. J^(2,+3)dx. 14. J^^f^·

J ^V(Ai-fl) J X — 1

17. Γ>&±!><|*. 18. f»±Vidx.


ax 19. J ( * + i W O - « ) · 20· J"V(1 + V*><*-2i. i V * _ , , „ „ f dx

23.

25.

J-^-d, 22. J .

f ^ 24. f ^ — -

Ϊ x2^/(7-2x)dx. 26. f V(*+i)+V(*+i)

2 9 . f_ ί d*. 30. r ^ - V * + l d x

^(*+l)-V(*+l) ^(*- l)

5.2 fii»cffo/w o//Ae / W e , y ( ß x 2 + ö x + c )

We consider first the two following integrals dx

/ vV-*2) where | x | < a,

a n d it a*, n » w n e r e x2 + fc =- 0. dx

(a) f dx

J V(«2-*2) Put x = a sin Θ, then dx = a cos 0 d0, and

T - f a cos Θ do

J V( a 2 - ß 2 s i n 2 e ) dö

1*1 J Λ . X

i—- arc sin —. | a | a


Thus

and

Therefore,

(b)

X for a > 0, / = arc sin —h C,

X for # < 0, 1= — arc sin —h C,

— HYO ein L r 1

— a

X = arc sin h C.

C dx x _ 1 QYC Clt! L f J V(«2-*2) M +

/ f dx J s/(x2 + k)·

The integral of this function has already been quoted (2.18), but it is interesting to notice how it may be obtained. Put t = x + <s/(x2 + k),

x

y/(x2 + k) + x

then' d'=1+7(^W*'

Therefore,

J(x2 + k) at ax

dx.

t V(*2+*)'

Therefore, / = J - ^ ^ = J * ^ = log |*| + C,

= log|x+V(x2 + /c)| + C.

The following examples show how the quadratic expression in the denominator is rearranged in one of the two forms

V(^2-Z2) and V(Z2+fc).



i dx

~J V(*2-< -6JC + 15)·

Solution. Rearranging the denominator, we obtain Χ 2 - 6 Χ + 1 5 Ξ ( Χ - 3 ) 2 + 6

Put x —3 = t, then dx = at, and

J V(i2 + 6) ' = log|i+V(i2 + 6)| + C, = log |(x-3) + V(^2-6x + 15)| + C.

5. Integrate / = f ^

J V(4-2x-x2)· Solution. Rearranging the quadratic expression, we obtain

4 _ 2 χ - χ 2 = 5-(1+χ)2 .

Put 1 +x = t, then dx = di, at and

/ v (5 -a ' = arc sin —rz + C,

• 1+* ^ = arc sin —r=- + C. V5

5.5 ftoitftou o/i te Ope j{ax2 + bxJrc)

We recall that since -^- { V[/X*)]} = 2 /lYOt)!:


then /

/ ' (*) VL/T*)]dx = 2 λ / [ / ( χ ) 1 + C ( w h e r e / W ^ 0)*

Hence in integrating a function of the above type, we first rearrange Ax+B so that it is the exact differential of the quadratic expression, compensating for the numerical term as required. The function then becomes

cc(2ax + b) ß j(ax2 + bx + c) y/(ax2 + bx + c) '

which can be integrated as described above. If a > 0, the integral of the second function will be a logarithm; if a < 0, the integral of the second function will be an arc sin.

We should note that the range of values of x for which the integral exists is determined by the discriminant of the quadratic expression. Frequently, the best way to discover the range of values is to sketch the appropriate parabola. Examples

6. Integrate 2x+l ■ -/

dx. yj(x2 + 6x-l6) Solution

X2 + 6 A. _ 1 6 = (X + 8 ) ( X _ _ 2 ) .

Hence (see Fig. 66), the function is integrable for x < — 8, and

i

-Q\y/////VA///

f ( x )

'//A

1

1 Φ

1 * 1 "

/ **

f2 *

FIG. 66


We now rearrange the numerator, 2x + 6 , f 5

- \ V(x2 + 6x-16)

I1=2y](xi+6x-\6). dx

dx J V(*2 + 6*-16) dx

■ ' / ■

V{(x + 3)2-25} ' = 5 1og|A: + 3 + V(^2+6x-16)|.

Therefore, 1 = 2 J(x2 + 6x —16) —5 log | x + 3 + <s/(x2+6x -16) |+C.

7. Integrate 2x-f-l f 2x~

J y/(2+X -3x2) dx.

I fix)

FIG. 67

Solution. We note that 2+x-3x 2 = (2+3x)(l-;c), and hence the function is integrable for — 2/3 < x < 1 (see Fig. 67). Rearranging the numerator as an exact derivative of the quadratic, we obtain

2x + l = - l ( _ 6 * + l)+i-.


Therefore, 1 Γ - 6 x + l 4 Γ dx 3 J J(2+x-3x*)dX+ 3 J V(2+^-3^)' 1 r 4 r - - y / i + j / , .

7 1 =2V(2+^-3x 2 ) .

" V 3 j

= V3"J

V(2/3+x/3-x2) ' dx

V{25/36-(x-l/6)2} ' 1 . je —1/6

= V 3 " a r C S m - 5 ^ - ' 1 . 6 x - l

=7T a r c s m -^ · Therefore,

/ = _ | ^ ( 2 + x - 3 x 2 ) + ^ - a r c s i n ^ f l + C.

5.4 Two pairs of related functions

(OV^-ând ν(Χχ2)· (ü) V(*H*) ai«l - ^ * ^ .

Both these pairs of functions can be integrated by the same technique.

(i) Write I1 = j V(*2-*2) dx, and /2 = j ^(/_χ2) dx.

Then' hss\w^^ = J V(« 2 -* 2 ) d x _ / 2 '


X Thus, Ix = a2 arc sin -— r~h · 0 )

I & I But by integrating Ιλ by parts we obtain

h= [ y/(a2-x2)dx,

Thus, Ix=x J(a2-x2) + / 2 . (2)

We now have two simultaneous equations (1) and (2) in I± and I2, and hence obtain

1 x 1 A = y tf2 arc sin -—- + — x ^J(a2-x2) + C,

1 x 1 and h=~ a2 arc sin --—- — — x V/(A2 — x2) + C. 2 | a\ 2

(ii) We obtain the integrals of {x2 +k) and x2/{yj(x2 +k)} by an exactly similar technique, recalling that

ax J: V(*2+£) Thus,

log|x + V(*2+AOI·

I and

/

V(*2+*) dx = 1 x V(*2 +*) + y * log I * + (*2+k) I + c>

dx = i- x ^ / ( x 2+^ )_ i k log | x+ V(*2+^) \ + C. J(x2+k) 2


/ = (*V(3-2x-x2)dx.


Solution. First note the range of values for x for which the function is integrable.

3 - 2 Χ - Χ 2 = (3+Χ)(1-Λ;) 5

thus, — 3 < x < 1. Rewrite the function in the form A2 — X2.

Then 3-lx-x2 = 4 - (x + l)2. Let x +1 =u9 then dx = dw, and

= JV(4-"! !) dw

= y 4 arc sin y + — Uy/(4-u2) + C.

Hence, J = 2 arc sin — (x + l ) + y 0 + 1) V(3-2x-x2) + C.

[Note that there may in fact be neater methods of solving such integrals by making use of a suitable trigonometric substitution. In the above example, for instance, by making the substitution w = 2 sin 0, we readily obtain

- J cos2 0 do,

which is easily integrated (see following § 6).] 9. Integrate

x2 + 6x + 14 f * 2 -J V(*2 dx. -2+6x + ll)

Solution. We write the numerator and denominator in terms of the appropriate perfect square.

(x+3)a+5 J V{< dx, V{(x+3)2 + 2} f (x+3)2 Γ 5 J V{(x+3)2 + 2}aX"i"J V{(x+3)2 + 2 } a x '


= l ( x + 3)V(x2 + 6x + l l ) - l o g | x + 3 + V(^2 + 6x + l l ) |

+ 51og |x + 3+V(*2 + 6 * + 1 1 ) l + C ·

Hence, 7 = 1 (x + 3) J(x2 + 6x + U)+4 log | x + 3 + V(*2 +

+ 6JC + 1 1 ) | + C .

5.5 77ze method of undetermined multipliers

Functions of the type uax2_hbx + c>), where Pn(x) is a

polynomial of order «, may be integrated as follows. The integral of the function will be of the form

C ax (1)

where Pn_x(x) is a polynomial of order (n — 1) and y4 is a constant. The coefficients of Pn_1{x) and the constant A may be found by comparing Pn(x)/{yJ(ax2+bx+c)} (which is the first derivative of the integral of the given function) with the first derivative of (1). The examples below should clarify the method.

Examples

10. Integrate 6xs-22x2+2lx-l

\ V(*2-4x + 3) ax.

Solution. First note that the denomina tor is^/Kx — 3)(x — 1)}, and that the function is only defined for x < 1 and x > 3. We now write I in the form of (1) above, i.e.

IS(a*+bx+cUV-4x+3)+AJ ^ 4 x + 3 ) ·


Differentiating the original integral and the above identity we obtain the identity

6 X 3 - 2 2 J C 2 + 2 1 X - 7 V(x2-4x + 3)

(ax2+bx + c)(2x-4) = (2ax+b)yJ(x2-4x + 3) +

+

2V(*2-4x + 3) A

V(x2-4x + 3) " Multiplying throughout by *J(x2— 4x + 3) we obtain,

6χ3-22χ2+21χ-7 = (2ax+b)(x2-4x + 3) + (ax2+bx + c)(x-2)+A.

Equating coefficients: x3: 6=2a-\-a

= 3a, hence a = 2 ; x2: - 2 2 = -Sa+b-2a+b

= - 2 0 + 2 6 , hence b = - 1 ; JC1: 21 = 6a-4b-2b+ c

= 18+c, hence c = 3;

= -9+A, hence ^ =2 .

Thus, / = (2**-* + 3 )V(* 2 -4* + 3) + 2 J ν ( , 2 ^ + 3 ) ■

Ft f dx _ f dx J V(^2-4x+3) - J V{(^-2)2-l}'

= log |x -2+V(x 2 -4x+3) | . Hence, 7=(2x2-x+3)V(x2-4x+3)+21og|x-2+V(^2-4x+3)| + C.

11. Integrate / = f (3x -2) V(*2 -2x) dx.


Solution. The function is only defined and is therefore only integrable for x < 0 and x > 2. We rearrange the function so that it appears in the required form.

(3x-2)(x2-2x) s ί

J(x2-2x)

3 x 3 - 8 x 2 + 4 x J(x2-2x)

dx

ax,

which can then be integrated by the method described, giving the solution

/ = y (x2-3x-1) J(x2-2x) — i log | x - 1 + yj{x?-2x) I + C .

12. Integrate dx J (*- ■ l ) V ( ^ 2 - 2 ) -

Solution. The function is only defined for | x \ ><J2, in which case the zero value of (x — l)4 is automatically excluded. We first transform the function by making the substitution

l / ( x - l ) = n , so that x = (1 +u)/u, and dx = —du/u2. Since x 2 >2, it follows that

l + 2 w - w 2 > 0 , and hence that u must lie between the values — G/2 — 1) and +( λ /2 + 1). Thus corresponding to the two ranges of values for x, we have

when x > Λ/2, 0 < w < ·>/2 + 1 , and x^—y/2, - ( ^ / 2 - l ) < w < 0.

In general,


Because there are two distinct ranges for which the function is defined there arise two distinct values of the integral. Multiplying numerator and denominator by ^/(w2) we must distinguish between the positive and negative roots.

When 0 < u < y/2 +1, y/(u2) = + u, and when — (^/2 — 1) < u < 0, >J(u2) = — u.

HenCe' / l = " / v ( l + 2 » - ^ ) d " (0- — V2 + D / 2=+Jv(i+l-»2)d" (-V2+1- — 0). and

We may now use the method of undetermined coefficients.

By differentiating, we obtain the identity

==(2au + b)J(l+2u-u2) V(l+2w-w2)

(au2 + bu + c)(2-2u) A 2 V ( 1 + 2 M - W 2 ) T V ( 1 + 2 M - « 2 ) '

Hence, — M3 = (2au + b)(l + 2u — u2) + {au2 + bu + c)(l — u) + A.

By equating coefficients we find the values of the constants to be a = 1/3, b = 5/6, c = 19/6, A = - 4 . Hence,

Jj = - i (2M2 + 5M + 19)V(1 + 2w -1/2) - 4 arc sin ^ ^ .

Substituting back M = 1/(Λ: — 1), we obtain

6(x~ 1)!

2 — x

h=<7Z^W*-n* + Wj\£zT?}-— 4 arc sin (V2)(x-1)·


Here x > ^2, and we take the positive value of y/[(x — l)2]. Thus,

/ = 6 ( x i 1 ) 3 (19x 2 -33x+16)V(^-2) -

+ 4 a r c s i n ( V 22

) ( - / _ 1 ) + C (* > J2).

For x< — ^/2, I = I2= —Il9 and we take the negative value of y/[(x — l)2]9 hence

6(x~l)3 / = ^ iv> ( 1 9 X 2 - 3 3 J C + 1 6 ) V ( ^ 2 - 2 ) +

+ 4 a r C S i n (V2V-D + C / ( X <^^ 2 ) · 1

[Afote. (i) Any function of the type ; can (x — κ)η^/(αχ2 +bx + c)

be transformed by the substitution l/(x—a)=w, into the function

, . η - 1

j(a'ifi + b'u + c')m

(ii) When integrating a function of this type careful attention must be paid to the distinct ranges for which the function is defined, since in general to each range there will correspond a distinct value of the integral.]

5.6 Functions of the type {axz + b)j{pxz + q)

Such functions can usually be integrated by making the substitution

y/(px2 + q) = xu,

and thus expressing x2 in terms of w,



'-Je * + 1)V(2*2 + 1)'

Solution. The function is defined and integrable for all real x. We make the substitution

Hence,

Then,

and

TVm<5

Also

hence,

and

u

du

y/(2x2 + l) = xu.

= V(2+1/*2) 1 1 (-2) 2 V(2+l/*2) x?

— ax ~ *V(2*2 + 1) '

V(2x2 + 1)~ X aU-2x2+l = xW,

x*--L-u*-2

^ + 1 = 3 ^ 0 + 1.

dx,

« 2 - l « 2 - 2 "

w2-2 (-1) w2-l w2-2 dw

dw,

1-w2 ' = tanh"1 u + C.


Thus in terms of x,

7=tanh-i{^2*; + 1>} + C,

or as a logarithmic function,

/=4log x + yj(2x2 + l) x-J(2x2 + l) + C.

E X E R C I S E Obtain the following integrals:

(8*+3)

2 1

1.

3.

5.

7.

9.

11.

13.

15.

17.

19.

21.

23.

V(4*2+3*+l)

V(2*-;*:2) *

d*.

j C _(x± J V(l-4^2)

(* + 3) djc.

f V(l-4*s)d*. r * - 5

J Λ/(5+4Λ:-Λ:2)

d* V(*2 + 3*+2)'

dx V(*2-*+m):

x+3

ax.

j

j V(*2+2*)

y/(x2-ax)

3Λ:+2 V ( ^ 2 - 4 J C + 5 )

d*.

dx, a > 0.

djc.

2.

4.

6.

8.

10.

12.

14.

16.

18.

20.

22.

24.

( 1 0 * - 1 5 )

j V ( 3 6 * 2 - 1 0 8 J C + 7 7 )

dx V ( 7 - 6 * - A : 2 ) ·

d*

d*.

V(2*r -* 2 ) , ' x

V ( l - 2 * - 3 ; t 2 ) 6JC + 5

V ( 6 + X - A : 2 )

f jc + l J V(8+2JC-X2)

2 x - 3 V ( 3 - 2 * - * 2 )

d* V(4A:2 + 3JC-1)*

d*

r > 0 .

d*.

d*.

d*.

dx.

Γ dx J ν(*-β)(*-3ο)* Λ > *

j 3*+2

j

V(*2-5x+19) 3 x - 2

V(4*2-4x + 5)

3 x - 4 V(4x2 + 5x-8)

djc.

dx.

dx.

25.

27.

29.

31.

33.

35.

37.

39.

41.

43.

45.

47.

49.

51.

53.

55.

57.

59.

5 x - 4 V(2* 2 +8x- l )

5χ-Λ

άχ.

άχ.

V(*2+2*+2) 2αχ2 + 1

V(0* 2 +2x+l) 2χ2-αχ+α2

V(*2+*2) χ 3 + 2 * 2 + χ - 1

d x

ί J V(3* 2 -2x+ l )

f V(* 2 -4 )dx .

f V(* 2 -3*+2) dx

ι

j J V(*2+*+i) f * V ( 6 + * - * 2 ) d * .

r x3+5*2--3*+4 J V(*2+*+i)

* 3 + 4 J C 2 - 6 X + 3


26. f V(2*+*2)dx.

dx9 a > 1.

28. f V ( 3 - 2 x - x 2 ) d x .

30. f V(3*2 + 10x+9)dx

, J 2 J C 2 + 3 X + 1 3 6 · « V(* 2 +O d "

V ( * 2 + 2 * - l ) 3*3+2

d x

38.

40

Γ x*_ J V(*!

j-

x 3 - x + l ■2+2x+2)

*3

άχ.

V(* 2 -4x+3)

42. f x 2 V(4*-* 2 )dx

dx.

dx.

44.

46 j V(5*2+4) 5*2-2* + 10

dx.

dx.

dx. J V(5+6*-x 2 )

J ( 2 J C - 5 ) V ( 2 + 3 X - A : 2 ) dx.

V(3*2-5*+8)

48. f * V ( 8 + * - x 2 ) d x .

j J r*/7v«

Je

dx. V(2*2+3) d*

xV(10x-x 2 ) ' djc

(*+2)V(4-* 2 ) dx

xy/(x2 —2x — 1) dx

50,

52

54

d x V(2*2+3) JE4

V ( 3 + 2 J C + X 2 )

dx

dx.

* + l ) V ( l + 2 * - 3 x 2 )

j· * J (x+i)V(*2-D '

56 Γ — — ^ — _ * J XyJ(x*+X-l) *

f dx • J ( 2 * - l ) V ( * 2 - D '

u 58,

60 (3 -2* )V(* 2 -4x+3)


dx ^ C dx f dx C ax 61m J Xy/(x2+X + l)' 62' J XyJ(x2-l) '

C dx fi Γ dx

6S f d x fifi f <**

J *2 V(4-^2) ' J (x-l)2V(10*-x2) * fl f d * fi8 f dx

J ^ V ^ + l ) ' J x3V(2*2+2x+l) * 69. Γ d * - f 6X

J (Χ-1) 3 Λ/(3-2Χ 2 ) * 7°* J x2 2 y/(l-4x+x2) ' ax _ f dx f « f dx

' J x3V(l+*2 ) ' JA:4 V ( 3 - 2 X + X 2 ) * f dx

* J (x-2)4V(l-4*+x2)*

§ 6. The integration of trigonometric functions

6.1 Some standard integrals We note first of all the integrals of the derivatives of the

6 trigonometric functions.

(i) j *cos*d* = sin* + C;

(ii) sin x dx = — cos x + C;

(iii) sec2 x dx = tan x + C;

(iv) cosec2 x dx = — cot x + C;

(v) sec x tan x dx = sec x + C;

(vi) cosec x cot x dx = — cosec x + C.


6.2 Integrals of linear functions In this and in the following sections the methods used are

indicated in the examples.

Examples

1. Integrate

Solution

2. Integrate

Solution

3. Integrate

= tan x ax.

C sin x Λ = dx, J COS X

= — d (cos x), J COS X

= — log |cos x\+C. Therefore, / = log | sec x \ + C.

= cot x dx.

C cos x , = — dx,

J sin x = — d (sin x),

J smx = log | sin x | + C.

■ ■ / cosec x dx.


Solution

J sin. ax.

-I 2 sin x/2 cos x/2

(1/2) sec2 x/2

dx,

tan x/2

1

dx,

tan x/2

= log | tan x/21 + C.

d (tan x/2),

4. Integrate

■ / sec x dx.

Solution. We give three methods of solution, (i) Writing cos x = (sin π/2+χ) and using the previous

result we immediately obtain

/ = log | tan (π/4 +π/2) | + C.

(ii) Writing cos x in terms of the half angle we obtain

7 J cos2 x/2 - s i n 2 x/2 d * '

_ sec2 x/2 ~ 1-tan2 x/2

Let tan x/2 = t, then sec2 x/2 dx = 2 at, and

1+t = log

= log

+c, \-t 1 + tan x/2 1 - tan x/2 + C.


(iii) We multiply numerator and denominator of the integrand by sec x+tanx, and obtain

sec2 x+ sec x tan x l sec x + tan x ax.

The numerator is now the exact differential of the denominator, therefore / = log | sec x+tan x | + C.

We can note that a precisely similar technique can be used to integrate cosec x.

The three results obtained are equivalent since

t+tan x/2 / l 1 \ tanjr/4+ti \4t 2 ) l - t a n ^ / 4

__ 1 + tan x\2

tan; M , ,... - t a n x / 2 '

1-tan x/2 '

1 + tan2 x/2 2 tan x/2 and sec x+tan x = s—-τχ- + 1 - tan2 x/2 1 - tan2 x/2 '

(1 + tan x/2)2

1-tan2 x/2 '

1 + tan x/2 1 — tan x/2

6.3 Integrals of squared functions


Ix = cos2 x dx, I2 = sin2 x ax.

Solution. We give two methods of solution.


(i) Writing cos2 x = — (1 +cos 2x), we obtain at once

Ι± = — (1 +cos 2x) ax,

= y ( x + y s i n 2 x j + C.

Similarly, we write sin2 x = —- (1 —cos 2x) and obtain

M (1 — cos 2x) dx,

CO

= 2"(*-■2 s in2« l + C'.

A+^2 = (cos2 x+sin2 x) dx,

^Jdx, = X,

I±—12 = (cos2 x— sin2 x) dx,

= i c o s 2 x d x ' 1 · „ = -y sin 2JC.

Solving the simultaneous equations in I± and 72 and including appropriate constants,

I1=^-(x+-jsm2x ) + C.

1 / 1 . "2 X "T /2 = —( X-—sin2x | + C".


6. Integrate

Solution

= tan2 x dx.

tan2 x = sec2 x— 1,

therefore, 7 = (sec2 x - 1 ) dx,

tan x — x + C. 7. Integrate

Solution /

cot2 x dx.

7 = (cosec2 x— 1) dx,

= — cot x — x + C.

6.4 Integrals of products Examples

8. Integrate I = sin #x cos bx dx, a ^ 0, i ^ O .

Solution. Ιΐ a = ±b, the integral becomes

7 = — sin 2#x dx,

= — — cos2ax + C.

Otherwise we express the product as a sum and obtain

7 = — sin ax dx -[- — sin /?x dx,


where α = α+£ , and ß = a — b, 1 / l therefore, / = — 4r I — cos <xx + -~ cos βχ ) 2 ^α β J + C.

9. Integrate

Solution

; cosec x dx. I— sec x i

/= f-; dx, J sin x cos x

dx, J sin 2x

= log I tan x | + C (see Ex. 3 above). 10. Integrate

Solution

J sinz x (

f sin2 x J sin2;

cos* x

2 v + cos2x

dx.

dx, ■ x cos"1 x

= (sec2 x + cosec2 x) dx,

= tan x — cot x + C. 11. Integrate

f 1 J sin2 x cos3 x

dx.

Solution C sin2 x + cos2 x ,

/ = -T-2 Ö — d x , J sinz x cosd x

= s e c 3 x d x + sec x cosec2 x dx,

= / i + /».


The first integral can be evaluated by applying a reduction formula (see below 6.5), but we here use an alternative method.

h ■ / sin2 x -f- cos2 x

dx,

f sin2 x Λ Γ 1 Λ = —=— dx-f- dx, J COS3 X J COS X

= — sin x Ö— d (cos x) -f dx. J COS0 X J COS X

Integrating the first function by parts,

L = sin x · s -z 5— cos x dx + dx, 2 cos2 x J 2 cos2 x J cos x

1 . 1 f 1 Λ = — tan x sec x + — dx,

2 2 J cos x

= — ) sec x tan x + log tan \-τπ+ι^χ ) > ·

fsin2 x + cos2 x , J« = —r-ö dx ,

J s i r x cos x

f f cos x = sec x dx + —r-ö— dx,

J J sm2x = sec x dx + —r-ö— d (sin x),

J J sin2 x = log | tan (TT/4 + x/2) | — cosec x.

Thus,

r 1 3 f 7 = — sec x tan x — cosec x + — log

4* ·4&τ

tan (VT*) +c.


6.5 Reduction formulae

In the examples which follow the reduction formulae given below are proved.

(i) If Im>n= sin™ x cosn x dx,

— 1 m — 1 then, Im n = sinm_1 x cos n + 1 x-\ Im_2 n, m>n m + n m+n m 2 ' n

1 n-\ and 7m>n = — — sinm + 1 x cos"-1 * + — — An, n-2 ·

00 if / . - /* . -„* 1 · , m — \ _ / m = —— s m ™ - i x c o s x + - — — 7 m _ 2 . m m

(iii) If In = I cosn x dx,

1 , η — \ Ύ then, i« = — cosn _ 1 x sm xH A . « .

(iv) If 4 = f tann x dx,

then, 7n = — tan71-1 x—In_2-

Examples

12. Find the reduction formula for

Im,n— s i n m x c o s n x d*.

Solution. We rewrite the integrand and then integrate by parts,

sjnm-i χ s j n χ c o sn χ ^ • " " J 1


Let u = sin™""1 x, then du = (m — 1) sinm~2 x cos x dx, Λ ^ C O S n + 1 X

dv = cos71 x sin x dx, then v = -Λ— . n + 1

Thus, s i n m- l χ c o s n + l χ m — \ Im, n— —— ~ΓΓ ~"^ Γ S^nm~2 X c ° S n + 2 X dx,

s i n ^ x c o s ^ x , m-1 f . _ a n ίΛ « + 1 n + l J

— sin2 x) dx, sinm"1xcosn+1x ra —1 . .

m+w __ sinm~1xcosn+1x m —1 therefore, —rrImtn= — 1 + _ _ j m _ 2 ) n >

sin771-1 x cosn+1 x m — 1 and Im>n- —— + _ _ / T O . l f n .

Similarly, by reducing the power of cos x, __ sinm+1 x cos71"*1 x n — \

Im>n~ ^rr^Vn + ^ + ^ " m'n"2 ,

13. From the reduction formula obtained for Im n deduce the reduction formulae for

Im = sinm x dx, and In = cos71 x dx.

Solution. Putting n =0 in the first formula obtained and m =0 in the second, we immediately obtain the required results.

(i) sin771 x dx hmx' _ sinm_1 x cos x m — \


(ii) cosn x dx

__ cos71""1 x sin x n — \

We have made the tacit assumption here that m > 0 and n > 0. We now consider the appropriate formulae for negative indices. Write the formula for Im so as to make Jm_2 the subject,

_ sin™-1 x cos x m

Suppose now that m—2= -k, (k > 1), s in~ f e + 1 xcosx k — 2

then /_ f t = -■ £—-j + ^TJy-fc+2>

or, in terms of the reciprocal function, C h , cosecfe~1cosx k—2C b 9 ,

cosecftxdx = r — '""TZT c o s e c * dx.

Similarly putting « — 2 = — A, (A > 1) in the reduction formula for cosn x dx, we obtain

f h A sec*1"1 x sin x h - 2 f . sec71 x dx = T— h -7—- sec" * x dx.

14. Find the reduction formula for

In = tann x dx,

and integrate the function for n = 6.

7n = tann~2 x tan2 x dx,

= tann~2 x (sec2 x — 1) dx,

= tann~2 x d (tan x) — In_2.


tan71

Hence,

Note that

and

When n = 6:

4 = η__χ—4-2> (" > 2).

J2 = tan2 x dx,

= f(p*X-l)äx, = tan x—x,

7X = tan x dx,

= log | sec x |, (s(

76 = -y tan 5 x - / 4 ,

74= - t a n 3 x - / 2 .

(see Ex. 1 above).

Thus, J6 = — tan5 x — - tan3 x + tan x — x + C.

15. Integrate

/4 6 = sin4 x cos6 x dx.

Solution. From the formula obtained above, reducing the power of sin x first,

__ sin3 x cos7 x 3 / 4 ' 6 = ϊδ + T ö 2'6'

_ sin x cos7 x 1 4,6 — ö l·" ~o~ 4,6 >

4,6 = Z ^~~ζ4,4>

_ w . ^ 3 4,4 — T I" ~T 4,2 >

sin x cos x - i t 4,2 = Ö + ^r I dx. 4J·'


Thus, sin4 x cos6 x ax sin4 x cos6 x (

1 3 1 ■ — sin3 x cos7 x — —- sin x cos7 x + y^- sin x cos5 x 10 oU 160

1 . , 3 . 3 + T ^ s i n * c o s - + ^ΤΖ8111 x c o s -^+^^Ζ^ + ^-

12o 256 25ο 16. Integrate

Jsir 78 3 = I sin8 x cos3 x dx.

Solution. When one of the powers is an odd number, begin to reduce by this power. This shortens the work considerably.

sin9 x cos2 x 2 h,s

h,l

—

=

=

=

11

sin8

sin8

- s i n 9

— +n / « x cos x ax,

x d (sin x),

X.

1 2 Therefore, I8Z = — sin9 x cos2 x + — sin9 x + C.

17. Integrate

/5 = cos5 x dx.

Solution. We may evaluate the integral by using the formula, but it is instructive to use an alternative method which may always be used when n is odd.

/ . = cos4 x cos x dx, = COS4 X


= 1(1— sin2 x)2 d (sin x),

= (1 — 2 sin2 x + sin4x)d(sinx).

Thus, I5 = sin x- (2 sin3 x)/3 + (sin5 x)/5.

18. Integrate

sec4 x dx.

Solution. From the formula obtained in Ex. 13 above, sec3 x sin x 2

63

Jsec**d* = 3 " + T s e c 2*d*>

= — (sec3 x sin x + 2 tan x) + C.

19. Integrate

/ cosec5 x dx.

Solution. From the formula obtained in Ex. 13 above,

J ~™ 4 ' 4 - t cosec4 x cos x 3 f „ cosec5 x dx = h -r cosec3 x dx, 4 4 J

f « , cosec2 x cos x 1 f cosec* x dx = - h -=- cosec x dx,

/ cosec x dx = log tan—x

Γ 1 3 Thus cosec5 x dx = — — cosec4 x cos x —- cosec2 x cos x

J 4 8 + y l o g 1 tan—x + C.


6.6 The transformation of general trigonometric functions into rational functions

(i) Functions of the type /(sin x, cos x) which are odd with respect to either sin x or cos x, or which are even with respect to both sin x and cos x we may transform as follows.

(a) /(sin x, cos x) = —/(—sin x, cos x), (i.e. odd with respect to sin x). Here we make the substitution

cos x = t, (0 < x < 7r)

then, sin x = ->/(! — t2), and dx : V(i-'2)'


■ ■ / r sin x ax. + cos2 x

Solution. The integrand is odd with respect to sin x, hence we make the substitution cos x = t,

V(i - ί2) ( -d0 then, 14 - /■

+*2 vc 1 - ' 2 ) ' di

l + t2 ' = arc cot t+C, = arc cot (cos x) + C

(b) /(sin x, cos x) = —/(sin x, —cos x), (i.e. odd with respect to cos x). Here we make the substitution

sin x = t, (—π/2 < x < π/2)

then, cos x = >/(l — t2), and dx = —ητ—-^- .


_ f 2 + sinx 7 = ^-Ö— Ax. J {


Solution. The integrand is odd with respect to cos x9 hence we make the substitution sin x = t9

then, / = J_^±J_ d f . Expressing the integrand in its partial fractions we obtain

3 / = J | ί 2 + ί + 2 ( l - i ) + 2 ( l + 0 j d ' '

= log

2(1 - 1 ) ' 2(1

t(i+tyi2\

j + l o g | i | — j l o g ( l - i ) + 2 " log(l+/) + C,

(i-ty 3/2 -+c. Thus, / = log

s i n x ( l -fsin x)112

-2 cosecx + C ( l - s i n x ) 3 ' 2

(c) /(sin x, cos x) = / ( —sin x, —cos x), (i.e. even with respect to both sin x and cos x). Here we make the substitution, tan x = t,

t 1 then, sin x =

V0 + ' 2 ) ' COS X =


= ( 1

J sin x (sin x

νο+τ

dx.

dx ■■ at 1+t-2 '

X —COS X)

Solution. The function is even with respect to sin x and cos x, hence we make the substitution

tan x = t,

then, J t(t- 1) at,

1 1 r Λ - [at, t-\ t r

log t-\ + C.

Thus, 7 = log | l - c o t x | + C


(ii) We transform all other functions of sin x, cos x and tan

x by expressing the ratios in terms of tan — x, and making the

substitution

Then, sin x

and

Example

23. Integrate

It 1-K2

tan — x = t.

cos x =

dx =

1 + i 2 '

2 at 1+i2

tan x 2f

l-t2

T C 2 + sinx Λ I - — -pr- r- dX. J smx( l+cosx)

Solution. The function is neither odd nor even with respect to sin x and cos x, so we make the substitution

1

Then,

tan — x = t.

= f(f + l + l/0df,

= ^-t2 + t+log\t\ + C.

1 1 Thus, / = y t a n ^ x + tan^-x + log tan—x 2 + C.


E X E R C I S E 2 2

Integrate the following:

1.

3.

5.

7.

9.

11.

13.

15.

17.

19.

21.

23.

25.

27.

29.

31.

33.

cos 5x COS 7X dx. 2. sin 3x cos 2x ax.

cos 2x cos 3x ax. 4. sin x cos 3x ax.

cos 2x sin 4x ax. 6. sin 2x sin 5* dx.

cos x cos 3x dx. 8. sin 3x sin x dx.

sin 5* sin 2x dx. 10. sin3 x dx.

sin4 * dx. 12. cos4 x dx.

cos5 x dx. 14. sin5 * dx.

tan5 x dx. 16. cot4 x dx.

cot6 x dx. 18. sec7 * dx.

cosec6 x dx. 20. sin3 x cos4 x dx.

sin7 x cos6 x dx. 22. sin5 x cos2 x dx.

sin2 x cos2 x dx. 24. sin3 x cos3 x dx.

sin4 x cos5 x dx. 26. cosec8 x cos x dx.

sin x tan x dx. 28. cos x j dx. 3

V(sin2x) s i n x A„ _ Γ sin2x dx. 30. f / n

S i n 2 * dx. V(l + 2cosx) J V(l+cos2x) sin 2x , f sin 2x , , , · 2 dx. 32. - r-T—dx. l+ sm z x J 1—sm4x

cos3 x Λ C sin3 x+cos3 x dx. 34. '· J; . g ax. M. : ■ ,-

s n r x J cosx—sxnx cos x+cos2

68

35.

37.

39.

4 1 .

43.

45.

47.

49.

5 1 .

53 .

55.

57.

sec5 x dx.


36. cosec7 x ax.

s

ax.

sec x cosec3 x ax.

sec 3 * cosec5 x ax.

' sin4 x , —=— ax. cos 3 x

' cos5 x sin3 x cos 2x ,

~— d*. c o s 3 *

d* 1+s in * '

sin x cos * sin4 * + c o s 4 * cos * + s i n x (sin* —cos*)2

sin * cos x , - Γ - Ί — = - Ί Ϊ — ax. 1 + s i n 4 *

sin2 x cos2 x sin8 * + c o s 8 *

ax

ax.

dx.

dx.

38.

40.

42.

44.

46.

48.

50.

52.

54.

56.

dx.

sin·' *

sec3 x cosec * d*.

sec4 x cosec2 x dx.

C sin4 x J cos x

I s J s i n *

j J si

COS8 X

dx 5 + 4 cos **

dx + COS X '

3+sin 2 x 2 cos2 x—cos4 x sin2 JC—cos2 x 1

d*.

sin4 * + c o s 4 * dx

J si

+ 3 cos 2*) 2

dx sin4 * + c o s 4 x *

-sin**

§ 7. The integration of inverse circular functions

We illustrate the methods by a few examples.

Examples

1. Integrate

/ = arc sin x dx.

Solution. We note that — 1 < x < 1. Integrating by parts,

let u = arc sin x, and dv = dx,

then

Thus,


uu- VO-* 2 ) ' v~ — X.

69

/ = x arc sin x — —r-. ^- dx, J y/(l-X2)

= x arc sin x + ^/(l — x2) -f C.

We can integrate arc cos x dx by the same method or by

recalling that

arc sin x -f arc cos x = —

and using the result just obtained.

2. Integrate

/ = I x arc tan x dx.

Solution. Let w = arc tan x, and dv = x dx, dx Λ 1 -

then du = - ^, and v = -y x .

Thus, I = — x2 arc tan x — — - % &x>

= —x2 arc tan x Λ 2 2 ί/{'-ώ}-

3. Integrate

— x2 arc tan x —— x - f y arc tan x + C,

y (x2 + l) arc tan x — x + C.

/ = (arc sin x)2 dx, (| x | < 1).


Solution. First make the substitution

x — sin t, then ax = cos t at.

Then, / = t2 cos t at, (| f | < π/2).

Now let u = i2, and dv = cos ί di, then du = 2/ di, and Ü = sin *.

Thus, I = t2smt-2 tsintdt.

Similarly, t sin t dt = —t cos t+ cos ί di.

Thus, I=t2 sin f+2* cos t-2 sin i + C,

and in terms of x, I = x(arc sin x)2-\-2 y/(l —x2) arc sin x—2x + C.

E X E R C I S E 2 3 Integrate the following functions:

v2 f ; r arc sin x , f arc sin x , • J να-χ») άχ· 2· Jo^)^dx·

C x2 arc tan x C ax * J ϊ+χ* J (l+9;t2)V(arctan3;t) '

3

dx _ j" dx j1 (arc tan x)'5

J (l+4x2)(arccot2x)2 ' 6# J ^ + Ί

7. f d* 8. f dx

J V(l —x2) arc cos2 JC ' J V(l ~*2) a r c sin x C x arc tan x , f x arc sin * ,

9· J (i+^y d " 10· J d-*2) d"· Γ J Γ * arc tan x ,

11. x arc sin x dx. 12. — 2__i\2—^χ. - . ^ P s . J * , f arc tan e*'2 , 13. x2 arc tan * dx. 14. —^77-—-r- dx.

f arc sin x , _ _ f arc sin ex , 15. J — ^ - d * . 16. J — ^ — dx.


17. Γ x3 arc tan x dx. 18. f (2x+3) arc cos (2x-3) dx.

<Λ f Λ: arc tan x , __ f ,,_ „. . , 19. —77——=— dx. 20. V(l —x) arc sin x dx.

J V ( l + ^ ) J

21. x(l+x2) arc tan x dx. 22. arc sin( Λ_ jd*.

§ 8. The integration of logarithmic and exponential functions


= JcothI x ax, x 7* 0.

Solution. Rewriting the function as an exponential function we obtain

. = f e* + l J e * - l dx.

Let e* = i, then dx = — df.

Then,

J (ί-ΐ)ί

Thus

2.

5

Integrate

/ =

/ =

= 1

= 21og | f -= 2 log | e*

dx

- i | -- 1 | -

-log* +C. - x + C.

(a > 0, a ?t 1).


Solution. Let

ax = t, then dx = —; di. ί log a

r A* Then, (i + l)ilogtf '

loga \J t J f + 1 y

= ^ i l o g i - l o g i i + 1 ) > + c ' = j ^ {log a* - l o g (a* +1)} + C.

Therefore, / = x_i^!±H + C.

3. Integrate

/=||i^Yd*. Solution. Let

w = (log x)2, and dv = —% dx,

2 log x Λ Λ 1 then, dw = dx, and v = . x Χ

Thus, /=-^g^+2fi^dx.

Similarly, J i ^ i . dx = - 1-ψ- + j ^ dx

log x 1

Therefore, / = {(log x)2 + 2 log x + 2} + C.


E X E R C I S E 2 4

Integrate the following functions:

1. f (e3* + Ve*)dx.

'· J^r· 5. f V(e* + l)dx.

r dx J V(3+2e*)

9· iw^y«*-

2.

4.

6.

8.

10.

tanh x ax.

j* dx J e^+e-* *

iirxax-f ex^(l+ex)dx.

f (e*+e-*)2dx.

73

n Γ e* Λ 1^ f 4e* + 6e-U · J ^T5- d " 12· J9e^4e^

4e* + 6e-* J dx.

(e*+a)n

e* „_ f dx V(3-5e2*)

13. J ^ . 14. J ^ ^ d x .

15. f , „ e \ ..x dx

17. f x3e~x dx.

19. f log(x2 + l )dx.

21. f log{x+V(* 2 + l)}dx.

2 3 · L c w u i r 24· J*-** 1*1*· 25. f (4 + 3x)2 log | x | dx. 26. f x3 log (x2 + 3) dx.

27. xa*dx, Ö > 1.

16.

18

20.

22.

J V(e2*+4e* + l) Γ dx J x log x

J(log|x|)2dx.

f log|2 + 5x|dx.

CHAPTER 11

DEFINITE INTEGRALS

§ 1. General remarks

LI Definition Given a function/(x) defined in the closed interval [a, b], we

consider any dissection Dr containing n sub-intervals (where n depends on r), defined by

a ~ χ0 < χλ < x2 < . . . < xn_1 < xn = b. We define the length of the /th subinterval [χ^_ι, x j , where i = 1, 2, . . . , n, as Δχ{ = (Λ^— Χ { _Ι ) . Let the maximum length of zlxj for the dissection Dr be <5r. For any other dissection Ds the corresponding maximum length of subinterval will be ds (in general different from <5r). We call the sequence of dissections {Dr} normal if lim <5r = 0.

Let q be any point belonging to the /th subinterval. Then we define

n(r)

If the sequence {Sr} is convergent to the same limit for any normal dissection Dr, and independently of the choice of point ci9 then/(x) is integrable in the interval [a, b], and the limit of Sr we call the definite integral of the function f(x) between the limits a and b, which is written as

P/(*) dx. Ja

74

DEFINITE INTEGRALS 75

A simple way of constructing a normal dissection is to halve the intervals successively, so that

/!(r) = 2", 4 = * ^ = βΓ.

It can be proved that a function continuous in the closed interval is integrable; and also that a function bounded in the closed interval and continuous except at a finite number of points within the interval is integrable.

1.2 Geometric interpretation of a definite integral

If in the interval [a, b] the function f(x) has the same sign throughout, then the area bounded by the curve y = f(x), the x-axis and the lines x =a,x = b (b > a), is equal to the definite integral

f f(x) ax.

*\

~

I

C

\7f(x)dxX

] t > X

FIG. 68

Notice the following:

(i) If f(x) ^ 0, the area will be positive in sign, (ii) If f{x) ^ 0, the area will be negative in sign.


(iii) f'/(*) dx=- Γ/(Χ) ax. Ja Jb

(iv) fV(x)dx = 0. Ja

§ 2. Properties of definite integrals

2.1 The law of addition

Γ/(χ) dx = P / ( x ) dx + Γ/(χ) ax. Ja Ja Jb

2.2 Multiplication by a constant

[ kf{x) ax = k f f(x) ax. Ja Ja

2.3 Sum of integrals

P{/(*>+*(*)} dx = f VW dx+ (\(χ) dx. Ja Ja Ja

2.4 A mean value theorem

If f(x) is integrable in [a, b] there exists a definite number K such that

f(x)dx=K(b-a), f •>a

where Γ lies between the lower and upper bounds of the function in the interval [a, b]9 i.e. m^K=^M. Since/(x) is a continuous function, it assumes all values between m and M in the given interval, hence there exists a definite number c,a^c^b,


such that/(c) = K. Thus

2.5 A function defined as a definite integral If the function/(O is continuous in [a, b], then the function

Kx) = f 7(0 dt, (*) = f is continuous and differentiable with respect to x in the interval [a, b] and for every point of the interval

2.5 77ie relationship between definite and indefinite integrals If F(x) is the primitive function off(x) in the interval [a9 b]9

i.e. if F'(x) =/(*), then

f */(*) d* = Ja

F(b)-F(a)

(independent of an arbitrary constant). The right hand side is written in the form

M 2.7 Integration by parts

The formula already obtained for indefinite integrals becomes

J u dv = luv — v du. a l_ _}a Ja


2.8 Integration by substitution

If g'(x) is continuous in [a, b], and f(u) is continuous in feO), g(b)l where u = g(x), then

f{g(x)}g\x)dx= f(u)du. Ja Jg(a)

2.9 Dummy symbols

In evaluating definite integrals the variable in the integrand does not appear in the solution, and any symbol may be used to designate it. Thus,

fb fb f{x) ax = f{i) at.

Ja Ja

Examples

1. Find the area (Fig. 69) bounded by the x-axis, the lines -1, x = 1 and the curve

1 x2 + l

F I G . 69

Solution. We notice that for all x, y > 0. Therefore the required area is

n άχ Γ + l + i —K r = a r C t a n x

J_^2 + l L J-i arc tan (1)—arc tan ( — 1)


=τπ-(-4π) 1

2. Find the area bounded by the parabola y2 = 2x and the line x = 8 (Fig. 70).

FIG. 70

Solution. From the symmetry of the parabola about the x-axis it is clear that the whole area is twice the area bounded by the x-axis, the line x = 8, and the curve above the axis.

Taking the curve above the x-axis, y = + >/(2x), hence

= 2 \J{2x) dx, Jo

128 ~3~'


3. Find the area bounded by the segment of the curve y = x3+x2 — 2x,

the x-axis, and the lines x = ±2 (Fig. 71).

x=-2

FIG. 71

Solution. In order to find the value of the area, we must consider whether the function is of the same sign throughout. If not we must divide the interval into subintervals within which the function is of the same sign. Solving the equation y = 0, we obtain

X-± ==: .Z, X2 z== vJ, X% == 1.

The function changes sign whenever the curve cuts the x-axis, hence we divide the interval [—2,2] into the three subintervals

[-2,0], [0,1], [1,2].

If the areas corresponding to the three intervals are Al9 A2, Az, then

J-2

-Γ J -2

ydx,

(x*-\-x2— 2x) dx,

DEFINITE INTEGRALS

= 0-(-1.16-1.8-4))

_ ^ ~ 3 '

Jo

= (i+I-)-o, - _i_

12'

: = y άχ,

-(f»4--<)-(K-1)' 3 + 12'

_ 37 ~~ 12'

Hence the numerical value of the whole area A is A=\A1\ + \At\ + \A,\.

_ £ _5_ 37_ ~ T + 12 + 12 '

- 11 6 '

Note: The value of ydx is J-2

(ίΗ-τ) _16_ 3


This differs from A since while Αλ and A3 are positive, A2 is negative. We should remember that the value of a definite integral represents the numerical value of the area only when the function is of constant sign throughout the interval.

4. Evaluate /*π/2

= x si Jo

sin x ax.

Solution. Integrating by parts we obtain Γ Ίπ/2 Λπ/2

= —x cos x -f cos x dx,

Γ Ί* /2 : 0 + sin x \ ,

= 1.

5. Evaluate , ^2 (arc sin x)3

t / = —ΤΓ« sr-dx. f

*>o V(i -*2)

Solution. Let dx _

arc sin x = t, then /n _ χ 2 \ ~ ^ί;

also when χ = 0, t = 0

and when Λ: = —, t = —π. 2 6 /•sr/6

Therefore, / = /3 df, Jo

~~ 4 I "6"


6. Prove that if f(2a-x) = -I(x), then i2a

f(x) dx = O.

Solution. We split the range of integration into two, thus

I2a f(x) dx = iaf(x) dx +1

2af(x) dx,

= iaf(x) dx-1

2af(2a-x) dx.

Let 2a-x == t, then dx.= -dt,

and i2a

f(x) dx = Laf(x) dx-1° f(t)( -dt),

= iaf(X) dx+10f(x) dx,

== o.7. Evaluate

f'!'C/ 2

o log sin x dx.

Solution. We note first of all a useful theorem,

i bf(x) dx =1b

f(a+b-x) dx.

The result follows almost immediately by putting a +b -x == t.The right-hand side becomes

Laf(t)( -dt),

= i bf(t) dt,

= left-hand side.In the given example,

f '!'C/2 f'!'C/2 ( 1 )1= 0 log sin x dx = 0 log sin 2:n:-x dx,

f'!'C/ 2

= 0 log cos x dx.


Adding the two expressions for I,

log (sin x cos x) ax,

Jo

Writing 2x = t, we obtain

f*/2 2 7 = ]

Jo Λπ/2 Λπ/2

= log sin 2x ax — log 2 ax. Jo Jo obta 1 Γ«

2 Jo 2 / = 4· I log sin t dt — — n\og 2,

K /2 1 log sin ί di —— π log 2, (*)

= 1-— Trlog 2.

Therefore, 7 = —— ττ log 2.

(*) If any function f(x) satisfies the relation f(2a—x) = f(x),

then f(x) dx = 2 /(x) dx, similar to the result proved Jo Jo

in the previous example. 8. Integrate

J 'ir/2 Λπ/2

sinm x dx, In = cosn x dx. o Jo

Solution. Applying the reduction formulae obtained above (Chapter 10, 6.5),

Γ 1 .

1 π / 2 m-\ T Im = sin™-1 x cos x H 7 m _ 2 ,

L m Jo m

m — \ = = ™ An—2 9

m

(m — l)(m — 3) m(m—2)


There are two cases to consider:

AA T ( m - l ) ( m - 3 ) . . . 4 . 2 . . A m odd: Im= , ^ — ^ | s inxdx , f»/2 sir m(m-2)... 5.3

( m - l ) ( m - 3 ) . . . 4 . 2 " m(m-2) . . .5.3

( m - l ) ( m - 3 ) . . . 3 . 1 f«/« m even: 7m = — —- ax, m(m-2)... 4.2 J0

Similarly,

(m —l)(m — 3 ) . . . 3.1 π m(m-2) . . .4.2 T "

_ (Λ2-1)(/Ζ-3)(Λ2-5). . . Λ ( # ! - 2 ) ( / Ι - 4 ) . . . '

where 0 = 1 when n is odd, and 0 = π/2 when n is even.

9. Evaluate f*/2

7 6 4 = sin6 x cos4 Λ: dx. Jo

Solution. By a precisely similar process, we can obtain a general reduction formula for an integral of this type.

_ ( m - l ) ( m - 3 ) . . . ( * - l ) ( n - 3 ) . . . m ' n ( m + n ) ( i n + n - 2 ) ( m + H - 4 ) . . . '

where 0 = 1 , unless m and w are both even, when 0 = π /2 .

T . . _ 5 3 1 3 1 π

_ 3π = 512"

10. Evaluate

ΐ2,ϋ = I s i n 1 2 x c o s S x dx-2,5 = si Jo


Solution. We reduce the odd power first, then

4 2 Jat/2

[12,5 = IT· IT 0 sin12 x cos x dx,

4 2[1. ]n/2= IT· IT IT sm13

x 0 '

421= IT·IT·IT·

This approach when one of the powers is odd will always savetime.

EX ERe I S E 25Evaluate:

1. f x 2. r dx~dx. ox2 +a2 'a:> O.

3 X -

3. f- 2 dx 4. f2 /5 dx-3 x 2 +2x+1 . -2/5 4+25x2 •

5. f 3 6. r dx-1 4x2 +4x-3 dx. 1 V(3+2x-x2) •

7. r 2x-3 8. fv'(3 /5) dx1/2 V(3+4x-4x2) dx. -y'(3/.1) y(3-5x2)·

f2 x2+1 r x9. 3 dx. 10. o V(4+x4) dx.1 -V!(x3 +3x+ 1)

11. f: 3xv(x2+4a2) dx. 12. f: V(x2-2x-l) dx.

13. flO dx 14. f: v(a2-x2) dx, a>O.1 XV (x2 +x+1) .

15. f: x2arc tan x dx. 16. f2 e2xo 1+ex dx.

17. s: xe- X dx. 18. s: x(x2+ 1)ex2 dx.

f-1 f''1 2

19. x 2e- 2x dx. 20. o e2x sin2 x dx.-2

21. f"/2 dx 22. f:12 4 dx.o 1+cos x· o 3+5 cos x

DEFINITE INTEGRALS 87 π/2 ρβτ/2 dx

o ( 2 + s i n x ) 2 * π/2 dx „^ f«/2 c o s x

dx. 3T/6 <" ;25+sin2x

23. cos3xdx. 24. J —at/2 ·/ '

p/2 dx r J jj/ssinxVCl+cosx)' " J. ί»2π/3 COS2 X Γπ/2

27. ^ - ^ d x . 28. (x + 1) cos x ax. J „i2 sin x Jo p3t/4Sin2xj ^Λ fw/2 cosx 29. — 3 - d x . 30. . -dx. Jo cos3 x Jo V( l+s inx) ί»π/4 j£ ρπ/2

31. —5—dx. 32. sin5 x cos11 x dx.

33.

ί»π/4 v ρπ/ - 4 - d j c . 32.

J o cos*2 x Jo ρπ ρπ/4

sin6 x cos6 x dx. 34. sin4x cos3 x dx. Jo Jo

35. Find the area bounded by the x-axis, the lines x = 0, x = a, and the curve y = x2. 36. Find the area between the parabolas y = x2 and y2 = x. 37. Find the area between the parabolas y2 — x and x2 = Sy. 38. Find the area between the curves y = x3 and y = 4x. 39. Find the area between the curves y = 2x3 and y2 = 4x. 40. Find the area between the curves y = x3 and y2 = x. 41. Find the area enclosed by the parabolas y = x2—x—6 and 7 = — x2 + +5x+14. 42. Find the area between the parabolas y2 = 8x and Sy = x2. 43. Find the area enclosed by the parabola y2 = 2x and the circle x2+y2_4x = 0

44. Find the area between the parabola y = x2 and the line 2x— y+3 = 0.

45. Find the area between the two parabolas y = x2 and y = — χ29 and

the line y = 3x. 2

46. Find the area between the parabola y = 2x—x2 and the straight line x+y = 0. 47. In what ratio does the parabola y2 = 2x divide the area of the circle

48. Find the area between the hyperbola xy = 4 and the line x+y = 5. 49. Find the area between the circle (x—6)2+y2 = 36 and the parabola y2 = 6x. 50. Find the area bounded by the curve y = x sin 4x, the x-axis and the

1 lines x = 0 and x = — π. o

51. Find the area bounded by the curve y = XQ~2X, the x-axis, and the lines x = 0 and x = 1/2. 52. Find the area bounded by the curve y = x cos(x/3), the x-axis, the lines x = π/2 and χ = π.


§ 3. Applications of definite integrals

3.1 The calculation of areas given parametric or polar equations of curves

(a) Given a curve y = f(x), the area bounded by the curve, the x-axis and the lines x = xx and x = x2 is

I y | dx. Ί

In calculating this area it is necessary to write | j ; | , since whenever the curve crosses the x-axis the sign of the area changes (see Ex. 3 above). If the curve is defined by the parametric equations

x = g(t\ y = Kt),

where g(t) and h(t) are continuous functions in the interval ίχ^ί^ί2 (where x1 = g(t±) andx2 = g(t2)), andg(t)is a mono-tonic function with continuous first derivative, then the area bounded by the curve, the x-axis and the lines x = xl9 x =x2

is (Fig. 72)

FIG. 72


either fh I A(0 I gV) at (for x = 8(t) monotonic Jh increasing);

J^ decreasing),

(b) If a curve is defined by the polar equation

where/(0) is a non-negative continuous function for a <s 0 =s=ß, and 0 < (ß — α) < 2ττ, then the area bounded by the arc r =/(0) and the radial lines 0 = α, 0 = /? (Fig. 73) is

1 r/V2d0. 4/ •/a

r--f(i)

FIG. 73

Examples 1. Find the area of the ellipse (Fig. 74)

x = a cos ί ] ß > 0, Z> > 0 7 = 6 sin t J 0 ^ ί ΞΞ 2π.

Solution. For 0 ί ^ττ, α cos t is monotonic decreasing, and for π ^ ί = 2π, α cos ί is monotonic increasing,


therefore, -f ί

F I G . 74

b sin t\ ( — a sin t) at

+ |&sin i | (—asini) di.

In the first interval sin t is positive, and in the second interval sin t is negative, therefore

sin2 t at+ab\ sin21 at, 0 *π

2π : ab I sin2 ί di,

Jo i = — ab\t- — sm2t\ ,

(1 - c o s 20 at,

= nab. 2. Find the area bounded by the x-axis and the cycloid

x =a(t—sin t) y = <z(l —cos t)

where 0 ^ t ^ 2π (Fig. 75).

a > 0


FIG. 75

Solution. Differentiating with respect to t we obtain

x = a{\ —cos 0·

Hence in the given interval

and thus x = a(t—sin t) is monotonic increasing in the interval.

Therefore, A = |a{\ — cos t)\a(\ —cos f) di, Jo

= fl2 f2*(l-cos/)2df, Jo

[ 3 1 Ί 2 *

-y i -2 s in i + —sin2i = 3πα2.

3. Find the area of the astroid x—a cos3 ί | a > 0 j = a sin3 ί j 0 ^ / ^ 2π (Fig. 76).

Solution. For 0 ^ ί = ^ π , a cos3/ is decreasing, and for π^ζί^2π,α cos3 f is increasing, thus we can find the required area as in Example 1. But we notice that

(i) for t = ±t0, x =x0 and y — ±y0, hence the curve is symmetrical about the x-axis;


FIG. 76

(ii) for t = t0 and (π —t{)), x = ±x0, and y — y0, hence the curve is symmetrical about the j-axis.

Therefore, the area in the first quadrant (0 =ss t =s= π/2) is a quarter of the whole area.

Thus, A = - 4

= +12a2

| a si Jo

Λπ/2 sin4

Jo

sin3 ί | ( — 3a cos2 ί sin i) df,

t cos2 ί di,

= 12a2 · — · — ·—-· — (using the reduction formula), 6 4 2 2

= - π ^ .

Note: We can eliminate t between the 2 parametric equations to obtain the cartesian equation of the astroid

x 2 /3 + j ;2 /3 = t f2 /3 #

4. Find the area of the logarithmic spiral

r = mhe, a > 0, fc > 0 ,


between the radial lines θ = 0, 0 = a, where 0 < a ^ 2π (Fig. 77).

FIG. 77

Solution

Therefore, Λ ^ ^ - Ι ) .

5. Find the area of the lemniscate

r2 = a2 cos 20, a > 0 (Fig. 78).

FIG. 78


Solution. Notice that for real r9 — π / 4 ^ θ π/4, and 3π/4 ^ θ ^Ξ 5π/4. As in Example 3 above, we can show that the curve is symmetrical about the lines 0 = 0 and θ =π/2. Thus by integrating through the interval 0 ^ θ ^Ξ π/4 we obtain as before a quarter of the total area. In this way we find the area of the shaded portion of the figure, moving along the curve in the direction shown.

Therefore, A l Γ π Μ < _ 4 " ? J . '

[ 1 Ί π

— sin 2Θ . 2 Jo

2 cos 2Θ d(9,

/4

6. Find the area enclosed by the curve

r = a sin 30, a > 0, where 0 ^ θ ^ π / 3 , 2ττ/3 ^ 0 ^ π , 4π/3 ^ θ <s 5π/3. (Fig. 79).

F I G . 79

Solution. In each of the given intervals sin 30 The curve has three lines of symmetry,

0, thus r ^ 0.

1 =τπ> 3


since for

or

or

Hence,

Therefore,

= — π±υ0>

Θ = -τττ+ θ0, ο

= ~ π±θ0

r = ÖCOS 3Θ0.

A =3 r2d0, 1 Γ 2 Jo

3 Γπ'3

— α2 sin230d0, Jo

3 Γ 1 Ί " / 3

4-α2 e - 4 s i n 6 Ö . 4 L 6 Jo

4 = — πα2. 4

EXERCISE 26 1. Find the area enclosed by the curve r = a sin 0 and the radial lines

0 = 0 and 0 = n. Sketch the curve. 2. Find the area enclosed by the curve r = a sin 20 and the radial lines

0 = 0, 0 = In. 3. Find the area enclosed by the curve r = a sin 60, where Ö > 0, and

0 ^ 0 =s In. 4. Find the area enclosed by the jc-axis and the cardioid r = a(l + cos 0),

where a > 0 and 0 <Ξ 0 <Ξ 2π. 5. Find the area enclosed by the spiral of Archimedes r = αθ, where

a > 0, and 0 ^ 0 ^ 2π. 6. Find the area of the circle r = a cos 0, a > 0, 0 =s 0 =s 2π.

7. Find the area of the ellipse — = 1 +e cos 0. r

8. Find the area of a loop of the curve x = 3t2, y = 3t—t3, where the parameter t lies in the interval — yj 3 ί^ yj3. 9. Find the area of a loop of the curve

3at y =

3at2

l+r 3


where a > 0, and the parameter t lies in the interval 0 ^ t <Ξ OO . 10. Find the area between the *-axis and the curve x = 2a sin2 Φ, y = 2a sin2 Φ tan Φ, for π/12 *£ Φ =s π/6. 11. Find the area of a loop of the curve x = t2

9 y = t+t3/3, 0 = 5 * ^ ^ / 3 . 12. Find the area of a loop of the curve x = It—t2

9 y — It2—t3. 13. Find the area of the curve (the evolute of a circle) x = 3(cos / + / sin 0 , y = 3(sin t—t cos t), 0 <Ξ t ^ 2n. 14. Find the area of a loop of the curve x = 2(2 cos t—cos It), y = 2(2 sin t — sin It).

3.2 The calculation of lengths of arc

(a) If the function y =f(x), defined in the interval a^x^b, has a continuous first derivative in the interval, then the length of arc of the curve in this interval is

and the element of arc is

dW{ i +(^)v (b) If a curve is defined parametrically by the equations

x = g(t), y = h(t), and in the interval ίλ =s= t ^ t2, the functions g(t) and h(t) have continuous first derivatives and are single-valued, then the length of arc in the interval is

and dL = J(x* + f)dt.

(c) If a curve is defined by its polar equation r =/(0), and in the interval α =Ξ= 0 β the function /(0) has a continuous first derivative and is single-valued, then


and dL= 1)^ + 1 - \ U0 . m

[Note that if β -<χ^π (or if β - α ^ 2π and r ^ 0), then r =f(ß) is certainly single-valued in the interval.]

Examples 7. Find the length of arc of the parabola y =x2 in the interval

0 < Ξ Χ ^ 2 .

Solution

ax

therefore, L= \ V C 1 * 4 * 2 ) ^ Jo

= [±x V(l + 4*2) +\ log {2x + V(l +4*2)}T,

= V ^ + | l o g ( 4 + Vl7).

(See Chapter 10, 5.4 for evaluation of the integral.) 8. Find the length of arc of the astroid

x = a cos311 y = a sin3 t j

a > 0, 0 <s t ^ 2π. j = a sm° ί J

Solution x = —3a cos2i sin f, y = 3a sin2 ί cos *.

Thus, L = V ^ 2 c o s 4 * s i n 2 '+ 9 f l 2 s i n 4 * c o s 2 0 di>

Jo

2π sin t cos ί | ^/(cos2 ί + sin2 i) df,

Λ2π • | sin 2i | Jo

3a Γ2π = y I |sin2f|df.


We have already shown that the astroid is symmetrical about the x- and j-axes, hence

la Γπ/2 L = 4 · — | s in2 id i , 3α Γ«

2 Jo = 6a\ - ^

= 6a. 9. Find the length of the arc of the cycloid

x =a(t — sin t) y =a(l —cos t)

for 0 ^ t ^ 2π.

Solution

a>09

x = a{\ —cos i)5

y = asin t. Γ2π

Hence, L = V{«2(1 - c o s 0 2 +# 2 sin2 t) at, Jo

7(1 - 2 cos ί+cos2 ί + sin21) at, o

= a ^J{2{\ - c o s 0} d/, Jo

Λ2

2(2sin24-i )M^ •2« 1

2tf | I sin— t

In this interval sin — t is positive, hence

Γ2π 1 L = 2a sin — t at, Jo 2

[ 1 Ί 2 π

- 2 cos y i

Therefore, L = Sa.


10. Find the length of the logarithmic spiral

Solution %=<**.

Hence, L = Γ y/(a2e2M+a2!e2e2M) άθ, Λ

= ay/(l+le2) f"eWdfl, Λ

^|νο+^2)ΓΗα, = -JVd+*1)(efte-l)-

E X E R C I S E 2 7

Find the lengths of arc of the following curves:

1. An arc of the semi-cubical parabola y2 = 4x3, y > 0.

2. V = 4JC3,

4. 3 / = 4x\ 0 ^ x ^ 3 . 0 ^ x = s ; l .

3 . 9y2 = 2x3, 0 ^ X ^ 2 .

5. 9y2=x\ 0 ^ s x ^ l 2 . 6. 2y2 = 3x\ O^x^l. 7.2y2 = x\ 0 ^ x = ^ 2 . 8. y = 2\lx, A(\,2),B(9,6). 9.y2 = 2x-x2, O ^ s x ^ l .

10. 2y2=x-2x2, 0 < Ξ Χ < Ξ 1 / 2 . 11. J> = 2 > / * , 0 = ^ X ^ 1 .

12. y = 2sj3x, O ^ J C ^ I . 13. ^ = y ( j c - l ) 3 / 2 , 1 < Ξ Χ ^ 4 .

14.

15.

16.

y

y

y

= log sin x9 —

= l - l o g

= c cosh

cos Λ:,

c '

■ π ^ Λ :

Ο ^ Λ

- c = ^ x

:<; -

: *

- β

~n.

1

c.

17. ^ = logjc, \ll^x*zk2y]2.


18. y = log(l-x2)9 O ^ x ^ y .

19. y = — log tanh y , 1 «s x =s= 2.

20. j> = arcsin;t + V ( l - * 2 ) , — l « s x « s l .

21. y2 = -^—, o ^ * ^ " ·

2α—Λ: 3

22. * = — y2 - — log ^, 1 < y 2.

23. >> = e*, Λ(0, 1), 5(1, e). 24. r = a(l +cos 0), 0 0 n. 25. r = a0, 0 < Ö < 1 . 26. r0 = e, ~ ^ θ ^ - ^ - .

3 4 27. r = 2α cos 0. 28. -^- = l+cos0, - y ^ 0 ^ | - .

29. χ = ί2, j = i—I/3 , 0 ^ / ^ V 3 .

30. x = a(cos f+f sin t), y = a (sin t — t cos /), 0=^ί=^π.

31. x = α cos5 —, j = asin5 —, 0=^ί^π.

32. x = 2α sin2 /, y = 2a sin2 / tan t, 0^t^ — .

33. x = 5 cos f (1 +cos 0» J> = 5 sin /(l +cos 0 , 0 =s / <= 2π.

34. x = α cos4 ί, j> = α sin4 t, Q^t^—π.

35. Given that in three dimensions the length of a curve is

find the length of the curve x = 3 cos t, y = 3 sin t, z = At, 0 t <= 2π.

5.3 Volumes of revolution and surface area

(a) If the function y =f(x) is continuous and non-negative in the closed interval [a, b], then the volume of the solid generated by rotating the arc AB and the ordinates x = a and x = b

DEFINITE INTEGRALS

about the x-axis (Fig. 80) is 101

FIG. 80

(b) The area of the curved surface of the solid so formed is

S = 2n I ydL9

-NHW* assuming that dy/dx is continuous in the given interval.

(c) Similarly, under the conditions stated in the previous paragraph, if x = g{t)9 y = h(t), tx ^ t ^ t29

and

V = π Γ2 y2x dt,

S = 2n \h yj(x2+f)dt.

Examples 11. Find the surface area and the volume of the solid formed

by revolving the parabola y2 = 4x9 between x — 0 and x = 3, about the x-axis.


Solution. We first find the element of arc length dL. ay _ 1 ax -yjx'

therefore, dL = 7(1+^)dx· Hence the surface area is

■ An I J(x + l) ax,

S = 2n P' 2Jx /( l+ -^ )dx ,

: VO-

iw , Jo

- 4 π · | - Γ ( χ + 1)3/2]3

= — π(43/2-13/2),

56

And the volume of the solid is *3

V — π I y2 ax, Jo

Jo

3 4x ax,

— 2π VI = 18π.

12. Find the surface area and volume of the solid obtained by revolving the cycloid

x — a(t —sin t), y=a(l— cost),

for 0 < t < 2π, about the x-axis.


Solution. We notice that x = a{\ —cos t) is not negative, and that consequently x is an increasing function of t. Also y is non-negative for all t.

The volume V is given by,

F = π \ a\\ - c o s i)2 · a{\ - c o s f) di, Jo

= πα3 (1—3 cos ί + 3 cos2 ί—cos3 t) at, Jo

= πα*\ ί - 3 s in i + 3 | - y i + — sin 2M

sin ί

= 5π2ο3.

The surface area is,

d(l - c o s 0 V{«2(1 - c o s i )2+ö2 sin2 r} at, o f2π

= 2π a{\ - c o s ί) · 0 yj{2(\ - c o s 0} df,

f2π 1 = 2ττ tf(l —COS t) ·2α sin — t at,

Jo 2 Γ271 l

== 8 π α 2 sin3 — t at,

Jo = 16πα2 sin3 x dx, ( where x — — / )

J'o i /2 sin3 x dx,

o 2

= 32ττα2· —· 1,

64 2 = - ^ - 7TÖ2.


13. Find the surface area and the volume of the solid generated by revolving the ellipse x2/a2 +y2jb2 = 1 about the x-axis.

Solution

then,

and

V — π \ y2 ax,

Ίν"·\'Χ"'-^ήάχ· r=2*[*-±4j,

= — nab2.

In order to find the surface area it will be simpler to express the ellipse in its parametric form,

x = a cos t, y = b sin t, 0 ^ t ^ π.

In the given interval x is a decreasing function and y is non-negative. Then,

S = 2π \ bsint ^(a2 sin2 t+b2 cos21) at. Jo

Put cos t = u, then —sin t at = dw, and the interval of integration is ( — 1, 1), then

S = 2nb P y/{a2(l-u2)+b2u2}du,

= 2nb P y/{a2-(cP-b*)tf}du.

(i) If a = b, S=4na2.

(ii) Otherwise put ^— = ε 2 ·


Then,

S=2nab I J(l-e2u2)du, J1 V(l-eV

2nab — Uy/(l —e2u2) + - y - arc sin eu

2nab } / ( l - ε2) Η— arc sin ε I .

If a > £, then

or in logarithmic form

S = 2π#6 I — + —77-5—TOT arc sin - ^ I a yj(a2-b2) a

Tf u +u · V(*2-«2) · If tf < 0, then ε = ι - ^ = IOL. a

Thus, — arc sin ε = — arc sin ια, ε ioc

= — sinhâ. a

Therefore, S = 2παΖ> J * + -77^—=r sinh"1 V(fe2~^2)) (Λ j(b2-a2) a j

)garithmic form

EXERCISE 28 1. Find the surface area and volume of revolution of the solid obtained

by revolving the curve xy2 = 1 about the jc-axis between x = a and x = b (where b > a > 0). 2. Find the volume of the solid obtained by revolving the hyperbola

xy=1 about the *-axis, for 1 x ^ 00. 3. Find the surface area and volume of the solid obtained by revolving

the sine curve y = sin x about the x-axis for 0=s= x *s π. 4. Find the volume of the solid obtained by revolving the curve

y2 (x—4) = x(x—3) about the jc-axis, 0 x =s 3. 5. Find the surface area and volume of the solid generated by rotating

the hyperbola x2— y2 = a2 about the *-axis, for a =s= x = a\Jl.


6. Find the surface area and volume of the solid generated by rotating the parabola 3y—x2 = 0, 0 =s= x ==s 1, about the *-axis. 7. Find the surface area and volume of the solid obtained by rotating

the catenary y — c cosh — about the x-axis, for — c <Ξ χ =ss c. c

8. Find the surface area of the solid obtained by revolving the curve y = *yj(2ax—x2) about the x-axis, (0 =s= x =ss 2).

9. Find the surface area and volume of the solid obtained by rotating the cissoid x3 = y2(2a—x) about the *-axis where 1 «s x = 2.

Find the volumes of the solids generated by rotating the following curves about the #-axis in the given ranges:

2π 10. y = acos ~rx9 Q^x^b/4. b

11. 7 = sin7/2;c, Ο^χ^πβ.

1 2 . ^ 4 / < 3 * + 1 > , 0 < « 1 . V C X 2 - 6 A ; + 1 5 )

13. The ellipse 4x2+9y2 = 36. 14. 3y2 = 4x, O ^ x ^ l . 15. The ellipse 1 6 x 2 + V = 144. 16. y = 2x\ O ^ x ^ l . 17. y = cos7'2*, 0 <Ξ x <= π/2. 18. The hyperbola y(x-l) = 1, 2 *s x <= 4.

20. 25x2 + 4j>2 = 100. 21. The circle x2+.y2-20y + 75 = 0. 22. y = e - * V(sin x), 0 <= x <= 2π. 23. Find the volume and surface area of the solid generated by rotating the astroid x = a cos31, y = a sin31, 0 <Ξ / <Ξ π, about the x-axis. 24. Find the volume and surface area of the solid obtained by revolving the curve x = t2, y = f—/3/3, 0 <= /<= -^3, about the x axis.

3.4 Moment of inertia and centre of gravity

(a) Definitions

(i) If masses ml9 m2, . . ., mk are at points (xl9 y±), (x2, j>2)> . . ., (xk9 y^)9 then the moment of inertia of the masses about

the x-axis is

DEFINITE INTEGRALS

4 = Σ m$i>

107

and the moment of inertia of the system about the j-axis is k

Τυ=Σ mixi2· i = l

(ii) The moment of inertia of an arc AB of a curve y =f(x) (Fig. 81) about the x-axis is

4 = my2dL,

where m is the mass per unit length of the line.

y,

0

l t t 1 a

y = f(x) — ^ ι Β

b X

FIG. 81

(iii) If y = / (x) is continuous for a <s x =s= b9 then the moment of inertia about the x-axis of the area bounded by y = / (x) and the lines x = a and x = b is

l rb

4 = y my* dx, "a

where m is the mass per unit area. (iv) The moment of inertia about the x-axis of the solid

generated by revolving the curve y = / (x) about the x-axis


between x — a and x — b (Fig. 82) is

Ιχ==~2π my*dx' «a

where m is the mass per unit volume.

FIG. 82

(v) If masses ml9 m2, . . .,mfe are at points (xl9 yx), (x2, y2), . . ., (xk9 yk), then the coordinates (x, y) of the centre of gravity of the system are

x =

k

Σ mixi i=l

k Σ mi

i = l

k

Σ miyi i = l

y = - k — X mi

(vi) The coordinates of the centre of gravity of an arc of a curve of uniform mass per unit length (Fig. 81) are

i (B) xdL

J(A)

1 (B)

f(B) j d L

' Λ(Β) <LL, dL

(vii) The coordinates of the centre of gravity of the area bounded by y =f(x), x = a and x = b, of uniform mass per


unit area (Fig. 81) are

J %b rb xydx y2/2dx

t _ _a r-, _ Ja [b ' y ~ rb

ydx ydx

(viii) If y =f(x) revolves about the x-axis between x — a and x—b, the coordinates of the centre of gravity of the solid of uniform mass per unit volume so obtained are

rb xy2 dx

v — •'α 1 b y2 dx

y = 0.

(b) RoutKs rule for the calculation of moments of inertia. The moment of inertia of a certain number of uniform bodies of regular shape may be found according to the following rule (due to E. J. Routh):

The moment of inertia of the body about an axis of symmetry of the body

f sum of squares of semi-axes perpendicular to this axis =massX- 3, 4, or 5

according as the semi-axes terminate in a plane, elliptic, or ellipsoidal surface.

For example, for a circular solid cylinder of length 2/ and radius r, the moment of inertia about the central axis is

where M is the mass of the cylinder. About an axis through the centre perpendicular to the central axis, the moment of inertia is

M (r2 l2\


It should be noted, however, that the rule is applicable only to certain simple regular bodies.

(c) The theorems of Pappus. These are two theorems which concern the surface area and volume of solids of revolution.

(1) If an arc of a plane curve y =f(x) revolves about an axis in its plane which does not cross the arc, the area of the surface generated is equal to the length of the arc multiplied by the length of the path traced out by the centre of mass of the arc. Thus if the length of the arc is L and the coordinates of the centre of gravity are (x, y), then

S = 2 nyL. (2) If a plane closed curve revolves about an axis in its

plane which does not intersect the curve, the volume of the solid generated is equal to the area enclosed by the curve multiplied by the length of the path traced out by the centre of gravity of the area. Thus if the area enclosed by the curve y=f(x) is A, and the coordinates of the centre of gravity of the area are (x, y), then

V = 2nyA, where Fis the volume generated by revolving the curve about the x-axis.

Examples 15. Find the moment of inertia, about the x-axis of the arc

of the astroid x2/3 _|_ j2/3 = ύ ί2/3? 0 =£Ξ X ==S tf,

which is above the x-axis, assuming that the mass per unit length is uniform.

Solution j = (a2/3_X2/3)3/2

?

therefore, / = y (a2 / 3-*2 / 3)1 / 2 ( - y x^'A

_ (fl2/3-X2/3)1/2

~χϊ/3


Then, dL = J\l + XW \dx

IT*· Therefore, 7X = my2 dL, where m is mass per unit length,

1 ( α ^ - χ 2 / 3 ) 3 ^ / ^ - 1 / ^ ^ Jo

3 9 : ma1/3 | — a2x2l* - -j a*l*x*l* [

■I + i . a 2 / 3 ^ 2 _ | . ^ 8 / 3

But if M is the mass of the given length of arc, r(B)

M = m dL, Λ-Α)

= m«i/3 Γ x - i / 3 d x , Jo

3

Therefore, Ix = — Ma2.

16. Find the moment of inertia about the x-axis of the arc of the cycloid

x = a(t—sin /), 3; = #(1 —cos i),

defined by the range 0 <s t === In.


Solution. For the cycloid

aL = 2a sin — t at.

Thus, /•2« I

Ix = ml a\\ - cos tf 2a sin — t at, Jo 2

f Jo

*2π I = 8m«3 | sin5 — t at.

Put — t =u, then,

L· = I6ma* sin5 u dw, Jo Λπ/2

= 32m«3 sin5 u du, Jo

4 2

256 ,

If M is the mass of the given length of arc,

M = 2ma sin — t at, Jo

= Sma.

32 Therefore, Ιχ^ττ Μα* ·

x 15

17. Find the moment of inertia about the x-axis of the area bounded by the parabola y2 = Aax and the ordinates x=l and x=4.


Solution 1 f4

Ιχ = ~T m \ ys dx, where m is the mass per

unit area, thus

Ix = jm Γ (4ax)*l2dx

3 "'""[5 = -x- ma^2 I — x^2

496 I

ma?iz.

18. Find the moment of inertia about its axle of a wheel of internal radius r, external radius JR, thickness a, and mass M.

Solution. Consider an elementary portion of the wheel, consisting of a hoop of radius x, width δχ, and thickness a. If δΜ is the mass of the hoop, then the moment of inertia of the hoop about the axle is

δΜχ2. But, bM = m.lnxa δχ, where m is the mass per unit volume,

therefore, I = \ x2 άΜ = Γ x2

Jr

J γ ■ 2nma I xz ax

-jnma(RA-r*).

But, M= \ άΜ J = 2nma x dx

= nma(R2—r2),

therefore, / = \- M(R2+r2).


19. Find the moment of inertia of the solid generated by revolving the parabola y2 = Aax about the x-axis between x = 0 and x = b.

Solution

4 1 Cb

— ππι y* dx,

2 Jo

Jo

8nma2 I x2 ax,

— —nma2bz.

If M is the mass of the given volume, •5 Π

M = nm y2 dx, Jo

f ^0

b Anma I x dx,

= Inmab2. 4

Therefore, IX = -T Mab.

20. Find the centre of gravity of the arc of the astroid x2/3 +^2/3 = β2/3, which lies in the first quadrant.

Solution. The centre of gravity of an arc AB has coordinates, 1 f(B) _ 1 f(B)

— x dL, y = T\ ydL> Lj J(A) ^ J(A) r

X = ;(A) ^ ^(A)

MB) where L = I dL.

'(A)

We have already obtained the elementary arc length for an astroid (Ex. 15 above), namely


hence, L = I [ — ) dx, -m 3 2a-

Therefore, x = -=— xl — 1 dx,

2

From the symmetry of the equation of the astroid we may immediately deduce that the y coordinate of the centre of gravity is the same as the x coordinate.

2 2 Thus, x = — a, y = — a.

21. Find the centre of gravity of that part of the ellipse x2/a2+y2/b2 = 1, which lies in the first quadrant.

Solution. We give two methods of solution.

(1) jaydx=^ja^-x^dx,

a2 cos2 t at,

4 b Ca

- a s . ' a\_2

s >

nab.

' oJ(a2— x2) ax,

f-|W-X2)3'*T

>


1 fa h Ca

-Ε[Α-Τ*Ί· = Ι«ί".

Then, quoting the formula for the coordinates of the centre of gravity,

_ _ a2b/3 _ _ ab2!3 X ~ nab/4 ' y ~ nab/4 '

thatis, » = *?, j = g .

(2) Alternatively we may find the coordinates using Pappus' theorem.

The area of a quarter of the ellipse is nab/4. The volume obtained by revolving this area about the x-axis is 2nab2/3 (see Ex. 13 above). Hence, since V = 2nyA,

2 1 — nab2 = 2ny · — nab,

thus, y = - .

By symmetry we can immediately write

-_4a_ 3n '

22. Find the centre of gravity of the solid obtained by revolving an arc of the parabola y2 = 4ax about the x-axis, where 0 <Ξ X <S b.

DEFINITE INTEGRALS

Solution, Clearly y = 0.

_r xy2 ax

[b y2dx

b

£ 4ax2dx

y 4ax dx

l& x3/3

x2/2

-6.

117

E X E R C I S E 2 9 1. Find the moment of inertia of a rectangle of sides a and 6, about the

side of length a. 2. Find the moment of inertia about its axis of a cylinder of radius r and

height h. 3. Find the moment of inertia about its base of a triangle of base a and

height h. (It is enough to consider the moment of inertia of a right angled triangle.) 4. Find the moment of inertia about its axis of a cone of height h and

base radius r. 5. Find the moment of inertia of a ring of radius r, about a diameter. 6. Find the moment of inertia of a circle of radius r, about a diameter. 7. Find the moment of inertia of a sphere of radius r, about a diameter. 8. Find the moment of inertia of the solid obtained by revolving the

ellipse x2/a2+y2lb2 = 1 about the *-axis, about its major axis. 9. Find the moment of inertia of an arc of the catenary y = c cosh x/c,

for | x | «s c, about the .y-axis. 10. Find the centre of gravity (e.g.) of a semi-circular arc of radius r. 11. Find the e.g. of a semicircle of radius r. 12. Find the e.g. of a hemisphere of radius r. 13. Find the e.g. of a cone of height h and base radius r.


14. Find the c.g. of the paraboloid of revolution of height h and baseradius r. (Consider the rotation of the parabola hy 2 = r 2x about thex-axis.)15. Find the c.g. of the arc of the catenary y = c cosh xlc defined byIxl~c.16. Find the c.g. of the area bounded by the cycloid x = a(t-sin t),y = a(l-cos t), for 0 =E: t =E: 2n, and the x-axis.17. Find the c.g. of the solid generated by rotating the cardioid r == a(1 +cos ()), 0 ~ () ~ Tt, about the x-axis. (Expre~s parametrically incartesian coordinates.)18. Find the e.g. of the area bounded by the cissoid x = 2a sin2 t, y =

= 2a sin2 t tan t for n/12 =E: t ~ n16, and the x-axis.19. Find the e.g. of the area bounded by the astroid x = a cos3 t, Y =

= a sin3 t, for 0 ~ t ~ n12.

20. Find the c.g. of the area bounded by the curve y = a cos 2: x,

the x-axis and the y-axis.21. Find the e.g. of the solid generated by revolving the cycloid x == a(t-sin t), y = a (I-cos t) about the x-axis 0 ~ t ~ n.22. Find the e.g. of the area of the loop of the curve x = t 2,y = t-t 3/3.23. Find the e.g. of the arc of the curve x = t 2

, Y = t-t 3 /3, defined bythe range O:Ei t :Ei .J3.24. Find the c.g. of the area bounded by the curve x = t 2 - t, Y = t 3+ t 2,

and the x-axis.25. Find the c.g. of the area bounded by the parabolas y2 = ax andx 2 = aye26. Find the c.g. of the area of the cardioid r = a(1 +cos ()).27. Find the e.g. of the area bounded by the parabola y2 = 4ax, thex-axis and the line x = Xo.

§ 4. Irregular integrals

In this section we are concerned with the integration offunctions which are not defined or are not bounded at a finitenumber of points, and with integrals in which the interval ofintegration is infinitely large.

4.1 Integrals offunctions not defined at afinite number ofpoints

Let the function f(x) be defined and bounded in the closedinterval [a, b], except at a finite number of points Xb X2' ••• , Xk.

We define arbitrarily the value of the function f(x) at these


points, as al5 a2, . . ., aÄ. We then define a function φ(χ) as follows,

0 W = p O (***. * = 1,2, ..., ^ (α-i (* = *{, / = 1,2, . . ., k)

If </>(X) is integrable in [a, b] then •5

φ(χ) dx f is the irregular integral of f(x) in this interval. If we define f(xi),f(x2)> · · ·> / (*Ä) i n a different way we obtain another function φχ(χ). However,

{φ(χ)-φ1(χ)}άχ=0, •fa

since φ(χ) — φι(χ) is zero everywhere except at a finite number of points. Thus if φ{χ) is integrable so also is φ±(χ) and

•6 rb rb rb φ(χ) dx = ΦΙ(Χ) dx.

«a «a Consequently the irregular integral does not depend on the way in which we define/(x) at the given points. We will therefore denote the irregular integral in the same way as usual, namely

*b

fix) dx. y ·> a

For example, iff(x) = sin—, x^09 we can define x

φ(χ) = ' 4

jo, χ = θ' which is integrable in any interval since it is only discontinuous at the point x = 0. Thus

f1 1 f1

sin — dx = φ(χ) dx.


4.2 Integrals of unbounded functions

We assume that/(x) is defined in the interval [a, b], except perhaps at the point x = a where it is unbounded. If f{x) is integrable in any interval {a + ε, b), where ε > 0, a + ε < b, and

lim f(x) dx exists, m f(x -0 + Ja+e

then the limit is the irregular integral

f(x) dx. ; a f

* a

4.3 Integrals of infinite interval

Let f(x) be defined for all x > a. If in any interval a^x^b the function is integrable and if

lim f(x) dx exists, &->°° Ja

then we call this limit the irregular integral of f(x) from a to oo 5 and we write

* a f(x) dx.

' a

We say that this integral is convergent. If the limit does not exist we call the integral divergent.

4.4 Geometric interpretation

(i) If/(x) is non-negative in a^x=^b, if lim f(x)-* + °°,

and » 5 - ε

f(x) dx exists, lim Γ ε -> 0 + •'α

then we say that the area bounded by the lines x = a, the

DEFINITE INTEGRALS 121 x-axis, y —f{x) and x = b, (which is a finite area with an infinite bound) (Fig. 83), is equal to

f(x) ax. f •f a

FIG. 83

(ii) If f(x) is continuous for x > 0, and if

\f(x)\ dx exists, *a

then we say that the integral expresses the area bounded by x = a, the curve y =f(x) and the x-axis to the right of x = a (which is an unlimited bound) (Fig. 84).

FIG. 84


Examples

1. Find the value of I - r - dx. Γ3 _ L JoÂ

Solution. The integrand is not continuous at x = 0. For ε > 0 we calculate

- 2/3 as ε - 0 +. Λ3 1 f ä 1

77 dx = l i m "77 d x '

= 2V3.

Thus,

2. Find the area bounded by the lines x = 0, x = 9, the x-axis and the curve

*7)/}>)))>}?,,,/,>>>>>>ltlln> >>>T7 9 x

FIG. 85

Solution. This function is discontinuous for x = 1, a point which lies within the interval [0, 9]. For x < 1, /(x) < 0, for


x > 1, f{x) > 0. Thus the required area is

1 Λ iV(*-i)l ~"

f(x)dx + lim f(x)dx, - , _ 0 e'->0+ •'Ι+ε'

= - lim ΓΑ(χ- ΐ )2 /3Ί 1 - £+ lim Γ | (*_ ΐ)«/.Ί' ,

ε - > 0 + | _ Jo e ' - > 0 + | _ J l + e'

= - l i m + | | - ( - e ) 2 / 3 - | - ( - l ) 2 / 3 |

+ lim {6-Α(ε ' ) 2 ' 3 | ,

-11 2 "

3. Find the value of — τ άχ of Γ—Γ· Jo * V *

Solution. The integrand is continuous in the given range except at x = 0. Then provided the limit exists,

f1 1 f1 1

m~xdx =lim · ττχdx

Jo X\X c^0+ Je XVX

- lim Γ-τΐ = lim f-2 + ->\

4. Calculate I /-+ΑΛ dx. f (^


Solution. Provided the limit exists K [2 1 - + -x x' I (—f—» 1 d x = lim f

jcÜôi tf K2 3K*

19 ~3~

5. Find the area bounded by the x-axis and the curve

x y = x4 + l

(Fig. 86).

FIG. 86

Solution. The required area is

ax - f°° I x

·/ —oo I '

C° x Λ f°° x J = — -i—r- dx + —:—- dx


2 - ^ d x r°° x

CK χ = 2 l i m ΐοΓΓ άχ

= 2 lim — arc tan x2

= lim (arc tan K2)

1

6. Calculate

If the limit exists

e~* dx. Jo

Q~X dx = lim Q~X dx Jo K->O° J0

= lim — e - * Ä-x»L Jo

= lim ( - e - * + l)

= 1.

7. Calculate -^—

Solution. If the limit exists, fVs dx ,. fVe dx L 3ΏΤ = ^ 1 Κ x +T

Γ ~ 1 V 3

= lim arc tan x K->~1 _\_K

= lim J — π — arc tan ( — K) \


= τπ~(-τπ) 5

E X E R C I S E 30

Find the values of the following:

i. I JL·**. J o l -

3. f x~*l* ax.

5. F* (fix*)-1* dx.

16.

18.

20.

22.

26.

Ca dx JoV(«2-.

(•V(2/3) v 9. -TT^-r, dx.

2.

4.

6.

g

10.

Λ16

V x-^dx.

Cb dx )a\l{(x-a){b-x)} X 2)

Jo V ( 4 - 9 x V ~ ' " ' Jo V ( l - * 2 )

"· f_2(2^-(l3)1,3dx· f2 dx f3 x

1 2 · J t V Ö ? = I ) · 1 3 · )*W(x^4)dx-u- Slöä^dx- 15· JoVö^d*·

f2 dx f 1 dx J i x V ( * 2 - l ) ' ' J - 2 i V ( ^ - l ) '

J o V(4x-4x2) - J o V d -x) p x V ^ - 2 ) d x 2 1 - r . dx J2 V ( 4 - * ) J I / 2 V ( l — *2 arc sin x) par/4 pst/2

sec2 (2*) dx. 23. tan x dx. Jo Jo

24. f + ~^* 25. f+°° ( a r c t a n x ) 2 ^ Γ+°° dx Γ + J-o. 1+*2 * 5· J . 1+x*

V3 * 2 + 9 " " " J 3 ^ 2 " ' f°° dx Γ°° dx JV3 * 2 +9 * 2 7 ' J

28.

30,

32,

34

36

42.

J C 2 - 6 * + 1 3

Γ+0° άχ

Γ°° άχ

ι: Jo 7

r

DEFINITE INTEGRALS

ehr Λ Λ f00 dx I2*

άχ

χ 2 +2*+2 ' d* +JC3

d* *V(*2-1)

38. f °° XG-*2 ax.

j

o e2/x

2Ϊ2" ax

40. J^-^-d*.

o a2 cos2 x+b2 sin2 JC'

29.

31.

33.

35.

37.

39.

41.

a> 0,

J i 2* 2 +4' J.-1/2 dAT J _«*, x2+*+l * (·°° djc

Jx Λ:(Λ:+1)2· r~ 7JC+2 Jo * 3 - 5 * 2 + 12*-60

1 e~ax dx, a > 0.

e _ x sin JC d*.

J o a+ocos x

b>0.

C H A P T E R 12

MAXIMA AND MINIMA OF A FUNCTION OF TWO

VARIABLES

§ 1. Definition

Given a function z =f(x, y), we say that the function has a maximum at the point (x0, y0) if there exists a neighbourhood of this point such that for any (x0+h, y0 +k) belonging to the neighbourhood,

f(x0+A, y0 + k) < f(x0, j>0), where h and k are not both zero.

There is a minimum at the point (x0, >>0) if, under similar conditions,

/(*o + h> Jo + *0 >/i*o > Jo)·

§ 2. Theorem

Let a function of two variables, z = /(JC, j ) have in the neighbourhood of M(x09 y0) continuous partial first and second derivatives, where

/* (*o> Jo) = 0, and fy(x0, y0) = 0. We define

Δ =fxx(Xo , Jo ) - fyy(*0 > Jo ) -{fxy(*0 , Jo)}*· Then,

(i) If Δ > 0, the point M is an extreme point of the function. \ifxx{xQ, y0) < 0, it is a maximum. If fxxixo, >>o) >· 0> is a minimum.

128

MAXIMA AND MINIMA OF A FUNCTION 129 Similarly, it is a maximum or minimum according to whether fyy(x0, Jo) is negative or positive. Note thatfxx and/ w are unequal to zero and must have the same sign, otherwise A < 0.

(ii) If A < 0, the point M is not an extreme point of the function.

(iii) If A = 0, the point M may or may not be an extreme point of the function. For example, consider the two functions

z = x4 + y*, and w = xz +y3.

For both functions A = 0 at the origin, but z(0,0) is a minimum, while w(0, 0) is not an extreme point.

Examples 1. Find the extreme points of the function

f(x,y) = 3x* + 3x2y-yz-15x. Solution

fx = 9x2 + 6xy-15, fy = 3x*-3y\ fxx = 18x + 6j>, fyy = -6y,

Jxy == "X.

For the extreme points

fx = 0, and fy=09 which gives

3x2 +2xy-5=0, and x2 -y2 = 0.

Solving this pair of simultaneous equations, we obtain the four points

A(l9 1), 2»(-l, -1) , C(V5, -V5), . D ( - V W 5 ) ·

Tabulating the values of the second order derivatives and of A at these points, we obtain:


A

B

C

D

Jxx

24

- 2 4

12V5

-12>/5

Jxy

6

- 6

6V5

-6V5

Jyy

- 6

+ 6

6yj5

~6yj5

A JXXJ yy \ Txy)

- 2 4 x 6 - 6 2 < 0

- 2 4 x 6 - 6 2 < 0

7 2 x 5 - 3 6 x 5 > 0

7 2 x 5 - 3 6 x 5 > 0

Conclusion

No extreme

No extreme

Minimum (fxx > 0)

Maximum (fxx< 0)

2. Find the coordinates of the points on the parabola y2 z= 6x such that the distance from A(3, 12) is either a maximum or a minimum.

Solution. The square of the distance from A to any point P{x, y) on the parabola is

z = (x_3)2 + 0>-12)2.

We should notice that this is not strictly a function of two variables since in addition y is a function of x. In other words we can obtain z as a function of one variable. Thus

Z = ( J 2 / 6 - 3 ) 2 + ( J - 1 2 ) 2 ,

= j>4/36-24;; + 153. 1_

9 ' ~" dy2 ~ 3

■— = 0 for v = 6, for which value -j-z dy dy2

the point on the parabola for which z is a minimum is (6, 6), the minimum distance being 3^/5.

3. Find the dimensions of the rectangular open tank of volume 32 m3 such that the surface area is a minimum.

dz 1 d2z Then> φΓ = -9^~~24' and W ■*y

Therefore - ^ - = 0 for y = 6, for which value - ^ - > 0. Thus

MAXIMA AND MINIMA OF A FUNCTION 131 Solution. Let the dimensions of the tank be x, y, z (Fig. 87).

Then, xyz = 32,

FIG. 87

and the surface area S is given by

S = 2yz+2zx+xy,

which as a function of x and y is

c 64 64 S = — + —Vxy. x y

We now differentiate to obtain the partial derivatives,

dS dy

ds = dx

d2S _ 128 dx2" x* '■

64 ^ _ _ ^ 1

d2S d2S __ 128 dxdy ' dy2" f '

At the extreme points of S the first order partial derivatives are zero, and the points therefore satisfy the equations

64 -x2y = 0, and 64 -y2x = 0.

The solution of these equations is x = y, and hence

6 4 - ^ = 0 ,


giving the point x = 4, y = 4 as a possible extreme point of the function. At this point the values of the partial derivatives are as follows,

dx2~ ' dxdy~ ' ay 5 " ?

thus, A_<PS_ d2S f fl2S |2

~ä?'#y {a^j ' = 4 - 1 ^ 0 .

Thus the point (4, 4) is a minimum point of the function and the dimensions of the tank which give a minimum surface area are x=4m, y=4m, z = 2m.

E X E R C I S E 31 Find the extreme values of the function z =fix, y), where l.fix,y) = 2x2+3xy+y2-2x-y + l. 2. / O , y) = xi+yi-2x2+4xy-2y2. 3. f(x9 y) = x2-xy+2y2-x+4y-5. 4. / (* , y) = x3+y3—3axy. 5. fix, y) = x3+3x2y-6xy-3y2-15x-15y. 6. f(x, y) = x2-xy+y2 + 3x-2y+\. 7. fix, y) = x2+y2+xy-6x-4y+5. 8. fix, y) = x2-6xy+y3 + 3x + 6y. 9. fix, y) = x3 — 3axy+y3.

10. fix, y)=4xy+±-+—.

11. fix, y) = x2-xy+y2-2x+y. 12. fix, y) = x3+y2-6xy-4Sx. 13. fix, y) = ix+y)2-ix+5y +xy). U.fix,y) = \-îx2+y2). 15. fix, y) = xy\a-x-y)3. 16.fix,y) = i6~x-y)x2y3.

MAXIMA AND MINIMA OF A FUNCTION 133

17. f(x, y) = sin x+sin j>+sin (x+y). 18. f(x, y) = sin x+cos y+cos(x—y). 19. f{x, y) = sin x sin y sin (a—x—y). 20. /(JC, y) = sin JC siny sin(x +7), 0 =s= Λ: =«= π, 0 *s >> ^ π.

21. f(x, y) = x - 2 j + l o g \J(x2+y2)+3 arc tan -^-.

22. f(x, y) = x2+xy+y2-4 log | x 1-10 log |j>|. 23. / (χ , ^) = (Sx2-6xy + 3y2)e2x+3*. 24. / (* , >>) = (5x+7j;-25) exp{-(x2+xy+y2)}. 25. Find the maximum area of a triangle which can be inscribed in a circle. 26. Divide a given number n into three parts so that the product of the parts is a maximum. 27. Find the maximum area of a triangle with given semiperimeter s. 28. Find the coordinates of a point on the circle {x—6)2+0>—l)2 = 25, such that the distance of the point from P(0, 7) is a maximum or a minimum. 29. Find the coordinates of the points on the parabola 2y—x2 = 0 whose distance from P(4, 1) is a maximum or a minimum. 30. Find the dimensions of the cuboid of maximum volume which can be circumscribed by a sphere of radius r.

CHAPTER 13

TANGENTS AND NORMALS. CURVATURE

§ 1. Tangents and normals

1.1 Equation of tangent and normal (i) If a curve is given by the equation y =f(x) where f(x) is

differentiate, then the equation of the tangent at the point P(x0, y0) on the curve is

y-yo=fXxo)(x-xo), and the equation of the normal is

(x-x0) +f'(x0)(y -Jo) = 0, since the normal to the curve at the point P of the curve is the perpendicular to the tangent at F.

(ii) Given a curve F(x, y) = 0 and a point (x0, y0) lying on the curve. If the partial first derivatives Fx(x0, y0) and Fy(x0, y0) exist and at least one of these is not equal to zero, and if there exists a neighbourhood of (x0, y0) such that inside this neighbourhood both derivatives are continuous, then the curve has a tangent at (x0, y0), given by the equation

Fx(x0, yo) (x-x0)+Fy(x0, y0)(y - y0) = 0, and there also exists a normal at the point with equation

Fy(x0, yo)(x-xo)-Fx(x0, yo)(y-y0) = °-(The symbols Fx(x09 y0) and Fy(x09 y0) denote the values of the partial derivatives dF/dx, dF/dy at the point (x0, y0).)

134

TANGENTS AND NORMALS. CURVATURE 135

Examples

1. Find the equations of the tangent and normal to the curve y = x3 —x at the point P( l , 0).

Solution. First we verify that the point P lies on the curve. Right-hand side = 1 — 1 = 0 = Left-hand side. Next we find the value of the first derivative at P.

/ = 3 J C 8 - 1 ,

/ ( I ) = 3 - 1 = 2.

Then the equation of the tangent is

y=2(x-l), or y = 2x-2, and the equation of the normal is

(x-l)+2y=0, or y=~x+—.

2. Find the equations of the tangent and normal to the curve F(x, y) = x2 +f = 0 at the point P(l , - 1 ) .

Solution. First we verify that the point P lies on the curve. Left-hand side = l 2 + ( - l ) 3 = 1 -1 = 0 = Right-hand side. Next we find the values of the partial derivatives at P.

Fx = 2x, Fx(l, - 1 ) = 2 ; Fy = 3y\ Fy(l,-1)=3.

Then the equation of the tangent is

2( jc- l) + 3(y + l ) = 0 , or 2x+3y + l=0,

and the equation of the normal is 3 ( x - l ) - 2 0 + l) = 0 , or 3x-2y-5=0.

3. Find the equations of the tangent and normal to the cycloid x = a(t—sin t), y = #(1 —cos t), a > 0, at the point

1


Solution. Since all real values of t determine a point of the cycloid it is unnecessary to verify that the point lies on the curve. By diiferentiating the parametric relationships we can obtain the first derivative of y with respect to x.

ax — a{\ —cos t) at, ay = a sin t at,

hence, assuming cos t ^ 1,

ay _ sin t ax 1 —cos t '

At the point t = — π we obtain

X° = a ( T π ~~ 2 ) ' y° = ~2 a'

dy_\ V3/2 te)it„M 1-1/2

Therefore the equation of the tangent at the given point is

or

The equation of the normal is

or

χ~α[γπ--y-) + V3 (y~Ya ) = 0>

^+^ν3-^^0 = 0·


E X E R C I S E 32 1. Find the equations of the tangent and normal to the curve y = (2x3 +

+x2 + l)1/2 at the point P(l, 2). 2. Find the equations of the tangent and normal to the curve y = x3 —

- 2x+ l at the point P(l, 0). 3. Find the equations of the tangent and normal to the curve x2 —2xy+

+y2-x+l = 0 at the point P(2, 1). 4. Find the equations of the tangent and normal to the curve x2y3 —y2 —

- 4 = o at the point P(l, 2). 5. Find the equations of the tangent and normal to the curve whose

parametric form is x = 3t2, y = 3t—t3 at the point t = l/yj3. 6. Find the equation of the tangent to the curve x3—py2 = 0 (a semi-

cubical parabola) at the point (/?, p). 7. Find the equation of the tangent to the cissoid y2(2r—x)—x3 = 0 at

the point (r, r). 8. Find the equation of the tangent to the curve x3—3axy+y3 = 0 at

the point (3Ö/2, 3Ö/2). 9. Find the equation of the tangent to the curve y = ax at the point

(1. a). 10. Find the equation of the tangent to the curve x3-\-y3+2x—6 = 0 at the point of the curve whose abscissa is —3. 11. Find the equation of the tangent to the catenary y = c cosh x/c at the point of the curve whose x coordinate is c. 12. Prove that the segment of the tangent to the astroid x

2l3+y2i3 = a213

between the x~ and jy-axes is of constant length a. 13. Find the equation of the normal to the curve x3+y2+2x—6 = 0 at the point where x = 3. 14. Find the equation of the normal to the curve y = alog cos x/a at the point where x = 2πα. 15. Find the equations of the tangent and normal to the curve r = e9

(a logarithmic spiral), at the point Θ = kn, where k is an integer. (Put x = r cos 0, y = r sin Θ). 16. Find the equations of the tangent and normal to the curve x = = r (cos t + t sin t), y = r(sin /—/ cos /), r > 0, at the point t = π/4.

§ 2. Curvature and radius of curvature

2.1 Definition of curvature

The mean curvature κ(ΡΡ') of the arc PP' is defined as the ratio of the angle between the tangents to the curve at P and P' to the length of the arc PP' (Fig. 88), that is,


FIG. 88

κ(ΡΡ') = δψ/ ds, where the arc length PP' is &s, and where the angle ψ is measured in an anticlockwise sense from the positive direction of the x-axis.

The curvature κ(Ρ) at P is defined as the limit of the mean curvature of the arc PP' as P' tends to P along the curve, that is,

m = lim * ÖS-+0 OS

_ dip " ~d7*

Thus κ will be positive or negative according as ψ increases or decreases with s.

2.2 Definition of radius of curvature

If the point of intersection of the normals to the curve at P and P' tends to a point C as P' tends to P along the curve, then the length CP is defined as the radius of curvature (ρ) of the curve at the point P (Fig. 88).

1 δψ — = l im - r 1 - , ρ ÖS-»O ds

άψ = ~ds~'


Thus, κ == —. Q

A curve is called strictly tangential to another curve at a common point of the two curves if at this point each has the same centre of curvature and the same radius of curvature.

2.3 Formulae for curvature and radius of curvature

(i) Given a curve y =f(x), then the curvature κ at a point of the curve at which the second derivative exists is given by

- - Jl or κ — J v ' {l+(dj/dx)2}3/2 ' [l+{/'(x)}2]3/2 '

{l+(dj/dx)2}3/2 [1+{/'(*)}*]« and ρ=—*w—' or ρ = —röö Note:

(a) If y =f(x) has a turning point at P(x0, y0), then since / ' ( * o ) = 0 ,

κ(Ρ)=Γ(χ0).

(b) If y =f(x) has a point of inflexion at Q(xl9 Ji) , then since f"{x^) = 0,

« ( 0 = 0 .

(ii) Given a curve in parametric form x = *(*), j> = X0» a t

a point of the curve at which the second derivatives exist the curvature κ is given by

xy-xy (x2 + y2)W

and the radius of curvature ρ is correspondingly

^(χ2 + γψ2

xy-xy


2.4 Sign convention From the definition of curvature as

aw

it follows that the curvature will be positive or negative according as ψ increases or decreases with s. We always take the positive sign of {l-f(d;;/dx)2}3/2 and (i2+j>2)3/2, and consequently the sign of κ is the same as the sign of d2y/dx2 and (xy — xy). Thus κ is positive when the curve is concave upwards and negative when the curve is concave downwards. The sign of κ determines the sign of ρ.

Examples

1. Find the radius of curvature of the parabola y = ax2 +

+bx + c, (a ^ 0), at the point x — . 2a

Solution. From the equation we obtain

y' = 2ax+b, y" =2a,

t V\ P· Y P fΓί V f* L11C1 C l \JL C 5

that is,

,_ HH-mr 9 la

1 ρ=Τα·

2. Given the curve y = log x, find the least value of the curvature.

Solution. Assuming x > 0, then


and (1 + / 2 ) 3 ' 2 = (1 + 1 /x2)3'2,

_ p c 2 + l)3 /2

x3 '

therefore, x = - ( 1 + * 2 ) 3 / 2 (1)

The turning points of this function are given by the zero

values of — . ax

άκ_ ( 1+χ 2 ) 3 / 2 ( -1 ) - ( - χ )3 /2 (1+χ 2 ) ΐ / 2 · 2χ dx ~ (1 +x2)3

_ - l - x 2 + 3x2

(1 +x2)5/2 '

2 x 2 - l (1 +x2)5/2 *

d% Hence —- = 0 if 2 x 2 - l = 0, i.e. if x = ± 1/^/2. But JC > 0,

therefore x = 1/^/2 is the only possible turning point. Moreover, as x increases through the value 1/^/2 the sign of d^/dx changes from negative to positive, hence κ is a minimum for x = 1/^/2, and substituting in (1) we obtain

2 3V3'

3. Find the smallest and greatest radius of curvature of the cardioid r = a(\ +cos 0), a > 0.

Solution. Transforming into cartesian coordinates:

x = a cos 0(1 -fcos 0), j ; = a sin 0(1 +cos 0).

Since the cardioid is symmetrical about the x-axis it is sufficient to consider the interval 0 =^ 0 π.


Differentiating with respect to 0:

x' = — tf(sin 0+sin 20), / = #(cos 0+cos 20), x" = -a(cos 0 + 2 cos 20), / ' = -a(sin 0 + 2 sin 20).

Hence,

x ' 2 + / 2 = «2(sin2 0 + 2 sin 0 sin 20+sin2 20 +cos2 0 + 2 cos 0 cos 20 +cos2 20),

= a2{2+2 cos (20-0)}, = 2 A 2 ( 1 + C O S 0 ) ,

= 4a2 cos2 — .

Also,

x'y"-x"y' = a2(sin2 0 + 3 sin 0 sin 20+2 sin2 20 + cos2 0 + 3 cos 0 cos 20+2 cos2 20)

= a2{3+3 cos (20-0)}, = 3a2( l+cos0),

= 6a2 cos2 —.

But e = .^ y I ,, xy —x y 4 0

= — a cos —. 3 2

Hence the radius of curvature is minimum for 0 = π,

and gmln = 0.

The radius of curvature is maximum for 0 = 0,

and ßmax = y * ·


E X E R C I S E 33 1. Find the radius of curvature of the curve y = x3—2x at the point

Ail, - 1 ) . 2. Find the radius of curvature of the parabola y2 =px, p > 0, at the

origin. 3. Find the radius of curvature of the parabola y = x2 at the point

x = 2. 4. Find the radius of curvature of the parabola y2 = Λχ-sjl at the point

5. Find the curvature of the semi-cubical parabola y2 = x3 at the point x = 4/3. 6. Find the curvature of the ellipse x2/a2+y2/b2 = 1, a > 0, b > 0, at

the points x = 0 and x = a. 7. Find the points of the hyperbola xy = 1 at which the curvature is

greatest. Find the radii of curvature of the following curves at the points indicated: 8. 3a2y = x\ for y = \. 9. γ = χ*-4χ3-\%χ2, for x = 0.

10. The hyperbola x2/9-y2/4 = l, for x = 9. 11. y2(x—4m) = mx(x —3m), at the points of intersection of the curve with the *-axis. 12. y = tan x, at the point x = π/4.

Λ:

13. >> = a log cos —, at a general point. 14. y = 3 sin 2x9 at the point x = π/4.

x 15. >> = 4 cos —, at the point x = π/2. 16. y = 2e3x, at the point x = 0. 17. y = xerx, at the point x = — 1.

18. The catenary j> = c cosh —, at its lowest point. c

19. y = sin JC, at the point x — π/2. 20. x = a(cos t+t sin f), ^ = «(sin t—t cos /), a > 0, at the point t = tQ. 21. x = t2,y = t3, at the point (1, 1). 22. x = a{t—sin 0 , 7 — «(1 —cos 0 , at the point t = π/3. 23. Find the extreme values of the radius of curvature of the curve

, Θ r = a cos3 —, a > 0, in the interval 0 «^ 0 3π. [Hint: transform to cartesian coordinates.] 24. r = 2a cos Θ, a > 0, at the point 0 = 0O . [ifr/tf: transform to cartesian coordinates.] 25. The Archimedean spiral r = αθ, a > 0, at the point θ = θ0. 26. r2 = a2 cos 20, for 0 > 0.


27. A particle of mass m slides down the smooth curve y = x2/2. At the point A(l91/2) the velocity of the particle is v. Find the force exerted by the particle on the curve (Fig. 89). (The required force N is normal to the curve and is given by N = (ηιυ2)/ρ, where ρ is the radius of curvature.)

FIG. 89

28. The path of a moving particle is the ellipse x — a cos t, y = b sin t, where t is time, a and b are constants and a ^ b. Find the acceleration p perpendicular to the motion at time t = 0, where p = ν2/ρ, ν is the linear velocity and ρ is the radius of curvature.

§ 3. Evolute and involute

3.1 Coordinates of the centre of curvature

(i) If the curve is given by the equation y = f(x), then, assuming that y" ^ 0, the centre of curvature at the point (x, y) of the curve has coordinates (I, r\) where

1 + j ' 2

S = * - /

v=y+-

y l+y'2

y"

(ii) If the curve is given in parametric form x — g(t), y = h(t), then, assuming that xy—xy ^ 0, the centre of curva-


ture at the point / has coordinates (|, η) where

, . x2 + f I =x-y — — ^ Γ Γ , xy — xy

^. *2+f V=y+x ... ... . xy — xy

3.2 Definition

If the radius of curvature of the curve L changes continuously then the locus of the centre of curvature of L is another curve V. The second curve L' is the evolute of the curve L. The first curve L is an involute of the curve L'. Note that a curve has only one evolute, whereas it has infinitely many involutes. Referring to Fig. 90 we see that any curve "parallel" to L (in the sense that tangents at similar points are parallel) is an involute of V.

To find the evolute of y =f(x) we express y, y', y" in terms of x, substitute in the two formulae for f and η, and by eliminating x between these two equations obtain the equation of the evolute (the locus of (|, η)) in the form JF(£, ή) = 0.

If the equation is given in parametric form we eliminate the parameter t between the two equations and thus obtain the required equation.

3.3 Two properties of the evolute

(i) The normal to the given curve is tangent to the evolute. (ii) Given a curve L and its evolute L\ such that to points P

and Q on L correspond points P' and Q on L\ the arc length P'Q' is equal to the numerical difference between the radii of curvature at P and β, i.e. (Fig. 90),

arc P'Q' =\PP'-QQ'\.


FIG. 90

Examples

1. Find the equation of the circle of curvature of the cissoid

o xs Λ 2α

yz = , a > 0, at the point x = —^-. 2a—x 3

Solution. We assume that 0 < x < 2a.

Write the equation in the form

(2a-x)y*=x*. (i) Differentiate

(2a-x)2y/-y*=3x*. (ii) Differentiate again

( 2 t f - x ) 2 > y ' + ( 2 t f - x ) 2 / 2 - 2 j / - 2 j / = 6JC,

or (2a-x)yy" +(2a-x)y'2-2yy' = 3x. (in)

Substituting x = 2a/3 in equation (i) we obtain aJl aJ2

y = ~ 3 ' 0 Γ y = 3 '

From (ii) we obtain the corresponding values , 7 , 7

4V2 ' ' " 4V2 '


and from (iii) 81 „ 81

or y

But ρ =

32α yjl* * 32a sf 2

(i +y2y'2

y

hence the radius of curvature at x = 2a/3 is

9a

The coordinates of the centre of curvature are

t , 1 + / 2 ^ 1 + / 2

£ = * - / ,, , v=y+-y y therefore,

t l3Ct A , 4aJ2

£ = — j 2 " , a n d ^ = ± " ~ 3 '

and the two circles of curvature have equations

13a V / 4flfJ2\a 81^2

12 y · y-1- 3 ) 16

2. Find the circle of curvature of the parabola y — x2/2a —a, a > 0, at the point x—a, and then write the equation of the evolute of this curve.

Solution. We find the first and second derivatives.

/ x // 1 a a

Therefore the radius of curvature is ( l + / 2 ) 3 / 2

Q = —y,

(ö2+X2)3/2


and the coordinates of the centre of curvature are , 1 + / 2 x3

ξ=χ-γ

v=y+

y" a* 1 + y'2 _ 3x2

y" ~ 2a '

Hence for x = a, ρ = 2a yj2, I = — a, η = 3a\2 and the equation of the circle of curvature is

(*+*)2 + / V ~ Ö Y =8α2.

Eliminating x between the two expressions for ξ and η we obtain the relationship

27 Vs = -g- αξ2

9

hence the equation of the evolute of the given parabola is

3 2 7 2

which is a semi-cubical parabola. 3. Find the evolute of the astroid

x2/3_j_j;2/3 = # 2 / 3 #

Solution. The parametric equations of the astroid are x = a cos3 i, j = a sin3 i.

Thus, x = —3# sin f cos2 i, Je = 6ÖT sin2 ί cos t—3a cos3 f. j ; = 3a sin2 £ cos t, y = 6a sin ί cos2 t — 3a sin3 f.

Thus, x2-\-y2 = 9uf2 sin2 ί cos2 i,

and xy — xy = —9a2 sin2 ί cos2 f.


Thus, if (£, η) are the coordinates of the centre of curvature, * __ 3 3a sin2 t cos t. 9a2 sin2 ί cos2 t ξ~αο°* l -9a 2 sin2* cos2* '

= a cos3 * + 3a sin2 * cos *, similarly,

7j = a sin3 *+3a sin * cos2 *. Thus,

ξ+η = α (cos * + sin *)3, ξ—η — a (cos t—sin *)3,

therefore, (ξ+ϊ?)2'3+(!-η)2/3 = 2α2/3,

and the equation of the evolute is {x+y)2^+(x-y)2^ = 2α2/3.

Rotate the axes through π/4 so that referred to the original axes the equations of the new axes are x+y = 0 and x — y = 0 Put x+y =Ar, and x— y= — Y, then the equation of the evolute of the astroid

;c2/3 + -y2/3 = : f l 2 /3 )

is the astroid X2'* + Γ2/3 = 2a2/3. 4. Find the evolute of the curve whose parametric form is

x = a(cos t+t sin *), y = a(sin * —* cos *), a > 0. Solution. From the given relations

JC = a* cos *, x = a(cos * — * sin *), y — at sin *, j ; = a(sin t + t cos *), *2 + f = a2t2

9 xy-xy = a2t2. x2 + V2

Hence, ζ = x — y -rr.—~r- = a cos t9 xy—xy x2+y2

η =y+x—rr.—Vr- = asin *. xy — xy


Eliminating the parameter t we obtain the relationship

ξ2+η2 =α2,

hence the equation of the evolute is

x2 + y2 = a2.

Thus the given curve is an involute of a circle.

5. Find the evolute of the logarithmic spiral r = aee, a > 0.

Solution. Transforming into cartesian coordinates we obtain,

x = atB cos 0, y — ae6 sin 0. Thus,

x' = αεθ (cos Θ — sin 0), x" = —2atQ sin 0, y — ate (sin 0 +cos 0), y" = 2ate cos 0, *'a + y 2 = 2ß2e205 x y > _ ^ y = 2a

2Q2^

Therefore, I = — ate sin 0, and TJ = ate cos 0.

Therefore the evolute of the logarithmic spiral

x = #ee cos 0, y = #e0 sin 0,

is the curve I = — αοθ sin 0, η = αεθ cos 0.

Rotating the axes through π/2 and writing

Χ = ξ, Υ=-η,

we obtain the equation of the evolute in the form

X = atQ cos 0, Y = ae9 sin 0.

It follows that the evolute of a logarithmic spiral is the same spiral rotated through an angle of —π/2.


E X E R C I S E 34 1. Find the equation of the circle of curvature for the parabola y2 = 12x,

at the point P(3, - 6 ) . 2. Find the equations of the circles of curvature for the curves y —

cos x, and y = sec x, at the point x = 0. 3. Find the equation of the circle of curvature of the cardioid x =

a cos 0(1 -fcos Θ), y = a sin 0(1 +cos 0), a > 0, at the point 0 = π/2. Find the evolutes of the following curves: 4. The parabola y2 = 4ax, a > 0. 5. The ellipse x2/a2+y2/b2 = l,a>0,b>0. 6. The hyperbola x2/a2-y2/b2 = 1, a > 0, ό > 0. 7. The hyperbola Λτμ = c2. 8. * = 3/2,>> = 3/- ί3 . 9. The catenary y = c cosh x/c.

10. .y = a log cos JC/Λ. 11. The cycloid x = a(t—sin 0> 7 = e(l—cos t), a > 0. 12. The tractrix χ = a(cos ί+log tan i/2), .y = a sin /, Ö > 0. 13. The hypocycloid x = a(2 cos f+cos 20, y = #(2 sin /—sin It).

CHAPTER 14

FOURIER SERIES

§ 1. General ideas

1.1 The Euler—Fourier formulae We are concerned here with series of the form

α0-\-(αλ cos x+b1 sin x)+(ct2 cos 2x+b2 sin 2x) + +(an cos nx+bn sin nx) + . . .

oo

= £ (#ncos «χ+έη sin/zx), (1)

where the coefficients an and Z>n are constants. Assuming that this series is uniformly convergent (see Chap

ter 7,2.4) in the interval — π ==Ξ Χ =^π we denote the sum of the series as /(x). Then all the coefficients of this series can be expressed as integrals of/(x) as follows:

1 Γη a° = 2^ *^X) d x '

1 Γπ tfn = — /(x) cos nx dx

TV ]

1 Γπ όη = — /(x) sin nx dx

(2) for w = l, 2, ..

These are the Euler-Fourier formulae. 152

FOURIER SERIES 153

1.2 The Fourier series (i) If f{x) is a given integrable function in the interval — π ==Ξ

= x <=π, and if the coefficients a09 al9..., an9 and bl9 b2,..., bn are found from the given formulae (2) and substituted in (1), we obtain a series which we call the Fourier series of the function/(;c). We should note that a Fourier series so obtained is not necessarily convergent, or if convergent is not necessarily convergent to f(x). We can also note that the coefficient a0 is the mean value of the function/(x) in the interval [— π9 π\.

(ii) The Fourier series (1) can also be expressed in the form

A0+Ax sin (χ + φ^)+Α2 sin (2χ + φ2)+ . . . + + Αη$ίη(ηχ + φη) + . . .

oo

= A0+ £ Αηύη(ηχ + φώ. (3) n=l

The relationships between the coefficients an9 bn of (1) and An, φη of (3) are as follows:

α0=Αθ9 αη = Αη$ϊηφη9 bn = Ancosφn9 « = 1,2,3, . . . ,

and

An = Vfan + ^nX Φη = a r c t a n (flnlbn)> " = 1, 2 , 3 ,

1.3 Dirichlefs conditions If in the closed interval [—n9 π] the function/^) fulfills the

following two conditions then in this interval the function is the sum of the corresponding Fourier series.

(i) The closed interval [— π9π] can be divided into a finite number of sub-intervals such that at each point of these intervals f(x) is monotonic.

(ii) At every point inside the open interval ( — π, π) the value of the function/(x) is the mean value of the limit of the function from the left and the limit of the function from the right


at this point, i.e.

/ (* ) = ~2 {/(*)" +/0) + }> for -π < x <π .

At each of the extremities of the interval (x = —π, χ = π), the value of the function is equal to the arithmetic mean of the limit from the right at x = —π and the limit from the left at

χ = π, i.e. — {/(-7t)+ +f(n)~}.

1.4 Definition

A function/(x) defined in the closed interval [a, b] is called regular in its sub-intervals if the given interval can be divided into a finite number of closed sub-intervals such that in each of these intervals the first derivative of the function is a continuous function.

If the function f(x) is regular in its sub-intervals then the first of Dirichlet's conditions is automatically fulfilled.

If the function/(x) is continuous inside the interval [— π, π] then the second of Dirichlet's conditions is fulfilled since for every point of the interval f(x)~ = f(x)+ = / ( * ) .

1.5 Odd and even functions

If f(x) is an even function, i.e. if /(— x) =f(x), then in the expansion of the function as a Fourier series all the coefficients b„ are zero. yn

sin nx dx, 1 Cn

K = — fix) sir J—π

1 fo 1 r« — —- fix) s m nx dx-\— f(x) sin nx dx,

7C } 71 1

FOURIER SERIES 155

I f 0 1 Γπ = — f(—x) sin (—nx) (—dx) H— f(x) sin «x dx,

π n I

1 fO 1 Λπ = — /(*) sin «* cbM— /(#) sin nx dx, π Λ π Jo = 0.

In general, an will not be zero (since cos nx is an even function) and the Fourier series will be a series of cosines.

Similarly, if /(x) is an odd function, i.e. if/(—x)=—■/(*) the coefficients an will be zero and the Fourier series of the function will be a series of sines.

L6 Functions of period 21

If we want to find the Fourier series of g{t) defined in the interval [—/, /] , then by substituting t = Ιχ/π, we obtain the function

/(*) = g(lxfa),

where/(x) is defined in the interval [ —π, π\. Then the expansion of g(t) is

where the coefficients can be calculated by the formulae:

g(t) dt,

g(t) cos —=- dt,

, x · nnt Λ g(t) sin — dt,

} for « > 1.


7.7 Functions defined in the interval [a, bj

It may happen that we want to find the Fourier series of the continuous function f(x), regular in its sub-intervals, defined in the interval [a, b] included in the interval [— n, n]. In this case we extend the function f(x) over the whole interval. We define a function g(x) in the interval — π^χ^π, such that g(x) =f(x) for a=^x^b. Such extension can be made in different ways but we choose such a way that the function g(x) is also regular in its sub-intervals, e.g.

g(x) f(d), for — π =ΞΞ χ <s a, f(x), for a=^x=^b, f{b), for b ^ x =ss 7i.

In particular, when/(x) is defined in the half-interval Ο^χ^π, we extend the function over the interval [ — π, π] in such a way that the extended function is either odd or even. The Fourier series will then be simplified.

If f(0) = 0 we may extend the function as an odd function by defining f(x) = —f(—x) for x < 0. Then the coefficients an

are zero, and

bn= — f(x) sin nx dx,

2 Γπ

= — f(x) s i n nx d*, n > 1. 71 Jo

Alternatively we may extend the function as an even function by defining f(x) =f(—x) for x< 0. Then the coefficients bn are zero, and

1 Cn

71 Jo 2 Cn

an = — f(x) cos nx dx, n > 1. π Jo

FOURIER SERIES 157 Examples

1. Find the Fourier series for f(x) = | x | in the interval [ — π, π] Solution. From the graph of the function (Fig. 91) we can

see that the first of Dirichlet's conditions is fulfilled since the interval [—π,π] can be divided into two sub-intervals such that the function is monotonic decreasing in the one ( — π=Ξ=χ=£=0), and monotonic increasing in the other (Ο^χ^π).

FIG. 91

For the second condition it is enough to note that this function is continuous for all values of x. Also as f(—n) =/(π) = π the second Dirichlet condition is fulfilled at the extremities of the interval. Thus,

1 Cn

J — π

1 f° ■7Γ- xdx, ^ Jo

= 2π' 1 f«

an = — \x\ cos nx dx9

1 f° = — (—x) cos nx dx +

71 ]

π L Jo

1 fπ x cos nx dx,

: cos «Λ: dx.

2 Γχ . αη = — \ — si sm ηχ-\—s-cos ηχ η2

, Ö-COSWTT ^


Integrating by parts we obtain 1 — sm nx-V-n

_ 2

therefore, 0, for n even

Ö , for « odd. nrr

f(x) = \x\ is an even function of x, hence the coefficients bn

are zero. Therefore, 1 4 cos x 4 cos 3x 4 cos 5x

_ 1 4 ^ cos (2n-l) x ~ Τ π ~ ί η 4 (2«-1)2 '

The series is valid for all values of x lying in the interval —π <ΞΞ χ ΞΞΤΖ. For x = 0we obtain

Λ 1 4 / 1 1 1

which gives us π2 _ J_ 1 1 1 ΊΓ " T2_ + "32~ + "52"+ * ' * + ( 2 « - l ) 2 + * ' "

Outside the given interval [ — π, π] the sum of the series is no longer equal to | x | because the right hand side is periodic with period 2π, whereas | x | is not periodic. The graph of the periodic function is shown in Fig. 92.

2. Express as a Fourier series the function „ . f 0, for — π < x <Ξ 0

/ x, for 0 ^ x < n.

FOURIER SERIES

\ v

159

- 4 TT - 3 TT -ZTT ~TT 0 7Γ 2TT ZTT 47Γ X

FIG. 92

Solution. The graph of the periodic function obtained by extending the given function outside the range [-π,π] is shown in Fig. 93. Dividing the interval into the two appropriate intervals [-π, 0] and [Ο,π], we obtain

1 fo i r«

= 4"»·

1 fo 1 f r 0 η =— 0άχ-\— xcosnxdx,

-37Γ -ZTT -TT 0 27Γ 37Γ X

FIG, 93


and integrating by parts we obtain [ 0, for n even

a* = nrf

-, for n odd.

Similarly, I f 0 1 f"

bn= — \ OdxH— x sin nx dx,

i Γ * l . > = — cos /ΙΛ: + —«- sin «x

π L π Λ Jo thus,

—, for « even

H—, for « odd.

Therefore, \ / sin 2x\

■)+(-r-) 2 \ / sin2x> — cos x+sin x ] + ( =-n \ 2

+

/ 2 „ s i n 3x\ / sin 4x\

(-5Fcoe3x+^-J + (——)+··· ^ ( 2 « - 1 ) 2 π ν ' 2 n - l ^

/ sin 2nx \

At x = π the value of the function is not convergent to π9 but according to Dirichlet's second condition

/(^)=γ(0+π),

FOURIER SERIES 161

But f¥) = \*+l+^+ + (2η-1)2π + ..., 1 1 , 2 ( 1 1 1 1 )

thus, τ π = _ π + _ | Τ 2 - + _ + _ + . . . +-^=ψ+ ·. j , π^__1_ J_ J_ 1

ΟΓ 8 ~ 12 + 32 + 52 + · · · + ( 2 „ - l ) 2 + as above.

It can be shown that the series is absolutely convergent and consequently we may collect the terms and write

frx)=L„-lf cos (2/z-1)* ψ (_l)n+isjî

3. Find the amplitudes of the harmonic components of a set of oscilloscope time base potentials. The graph of the potential variation U(t) is a saw-tooth curve, where for —n ·< t < n the gradient of the curve is 1.

Solution. Since the gradient of the curve is 1, the function is of the form

U(t) = t. The graph of the function is shown as a periodic function of period 2π (Fig. 94). Since the function is an odd function the

FIG. 94


coefficients a„ are zero for all n.

" * J _ . t sin nt at,

nt cos wi + sin • T in nt \ ,

= Ö- c os n7t

>

nz

= ( - l ) n + 1 - .

The coefficient Z?n is the amplitude of the n harmonic components of the corresponding potential. The expansion of the potential function as a Fourier series is

17(0=2[ s i n f - 4 - s i n 2 i + _ + ( - l f + i i sin nt+ . 2 n

E X E R C I S E 3 5

1. Express as a Fourier series the function f(x) = x in the interval [—π, π]. What value does the series give for x = —π and x — πΐ 2. Expand as a Fourier series (as a series of sines) f(x) = a, in the in

terval 0 < x < π.

Express as Fourier series the functions:

3. ,., v f 0, 0 < J t < a , π—α<χ<π « » < : <*<π—a.

/(*) = ^ , 0 = a a, a <= α(π—Λ:)

JC ^ π — a ,

5. /(*) = \+c, 0 Τ Γ < Λ : < 0 ,

6. I 2x, — π < Λ: < 0, / ( * ) = { 0, * = 0,

( 6Λ:, 0 < χ < π .

What is the value of this series at the points x = — π and x = n ?

FOURIER SERIES 163 7. Show that in the interval — π/2 < x < π/2,

πχ _ sin * sin 3x .sin 5JC T ~~ ~Y2 Ψ~+~~Ί? ·" *·

8. Express both as a series of sines and as a series of cosines the function fix) = χ{π—χ) in the interval 0 < x < π.

9. Express as Fourier series the function f(x) = | sin x | for any value of*. 10. Express as a Fourier series the function f(x) = x2, for —π Λ: < π . 11. Express as a Fourier series the function f(x) = *3, for — π < x «STT. 12. Express as a Fourier series the function f(x) = sin a*, for — π < # < π , where a is not an integer. 13. Express as a Fourier series the function

\ sin x, Ο^χ^π.

14. Express as a Fourier series the function

I — cos x9 — π<Λ:<0, 0, * = 0,

cos JC, 0 < * < π . 15. Express as a Fourier series the function fix) = x cos x in the interval 0 < Λ: < jr. 16. Express as a Fourier series (as a series of cosines) the function fix) = (1/2) log cot JC/2 in the interval 0 < x < π. 17. Express as a series of cosines the function, fix) = log (2 cos JC/2) in the interval 0 = x < π. 18. Express as a series of cosines the function, fix) = — log (2 sin x/2) in the interval 0 < Λ: «S π. 19. Express as a Fourier series the function fix) = cos ax in the interval —π < Λ: < π, where a is any non-integral number. 20. Express as a series of cosines fix) = cos x, in the interval 0<Λ:<π. 21. A single diode valve rectifier supplies direct current according to the formula

. _ / / sin nt9 for 0 t ^ π/η, l~ \ 0 , for -njn^t^Q,

where n is the frequency of oscillation, t is time and 7 is a constant. Calculate the amplitudes of the harmonic components of the rectified current. 22. Calculate the amplitudes of the harmonic components of the magnetic induction due to a rotating magnetic field B =fix)> given that the form of the induction curve is as in Fig. 95,

* " " - ~1

v!

B

■"''""' Bo

1 / - B 0

,

J I

\

\ ir\\ X

FIG. 95

In order to express this as a Fourier series the curve must first be "rectified" as shown by the dotted line. Then the function to be considered is

/(*) - { —B0, for— π<Λ:<0, +Βθ9 ί θ Γ θ < * < π .


ANSWERS

E X E R C I S E 18, P. 11

1. / = j ^ - 3 x 2 + 3 x - 2 1 o g | x | — ! + C.

2. j = I j e « - i . x * + 2 j c * - l o g | * | + C. O 4 Z

3. 7 = y X 5 - y ^ + X + C.

4. I=^(x*+4)6 + C.

5. 7 = i - log(l+x2) + C.

6< / = 1 0 ( x 2 + 3 ) 5 + C ·

7. 7 = 1 log (e»+*») + C, xji-a.

8 . 7 = 3χ!/3 —1 χ-3/4 + C, X > 0.

9. 7 = yir *»/· - 1 | A«/« + C, x > 0.

10. 7 = 27JC + ^ *«/« + 24*»/* + —ί χ?/4 + C.

11. 7 = j *»/· - 1 χ4/3 + _*. *y(i25x»») + C.

165

166 ANSWERS

12. 1 = — ^ - + 3 i/x + C.

13. i = l(3x + iyn* + c,

14. / = ^(a+bxf + C,

15. / =- | (2x2-I)«/»+ C,

16. / = 1^/(1+x2)3 + c.

17j/=_V(3-5^) + c

1

(puta+6x = i2).

x ^ · , x * -

18. / = A(x-4)V(x+l ) 2 + C.

19. I=yJ(x2-6) + C.

20. 7 = A 5 / ( X 3 + 1)4 + C

21. 7 = -ei/* + C.

22. / = _ l e -* 2 +c.

23. / ^ - ^ t a n S x + C. 6

24. 1= --^cos(2x2 + l) + C.

25. I=-rsin« x + C. 6 26. / = 2V(l+sinjc) + C.

27. / = — -7-log \a+bcosx\ + C, a+6cosx 0

ANSWERS

28. 1= esin x+C.

29. 1= ~ tan (x4)+C, cos (x4) ~ O.

30. I = ~ tan4 x+C, cos x =P O.

131. 1= 3 tan (x3+ 1)+C, cos (x3 + 1) =p o.

32. 1= ~ (log X)3+C, x:> O.

33. I = arc tan eX + C, (put eX = t).

134. 1= 2 log (2ex+l)+C.

1+x2 x2

35.I=-2-log{l+x2)-2+C.

236. 1="3 .J(2+log Ix 1)3+C.

61- X

37.1= -log 6 +C.

38. I = arc sin (log Ix D+C.

39. I=arctan x (log larctanxl-l)+C.

140. 1= - x2ex2 +C.

2

41.1= ~ arc sin (x3)+C, -I-<x-< 1.

42. I = log Iarc tan x I+ C, x = o.43. 1= n arc sin x- ~ (arc sin X)2+C, -I -< x -< 1.

144. I = 2: arc tan (x2

) +C.

167

168 ANSWERS

45. I=jx\l+xf-^x«(l+x)2+j(jX7+jA + C.

46. I=(x2-2x+2)e+C. 47. I=(xs-3x2+6x-6)ex+C.

48. I=e^(^xi-x3 + -x2~x + j \ + C.

49. 7= xsinx+cos x + C. 50. 7 = (x2 - 2 ) sin x+2 cos x + C.

1 „ 2 2 51. 7 = — y x2cos 5 x + y y xsin 5x+—= cos 5x+C.

52. 7 = — e* (sin x+cos x) + C.

53. 7 = —-r-e-2*(2sin3x+3cos3x) + C.

9 / 2 2 2 \ 54. 7 = —TfQ~x\ - c o s y x + y sin j x J + C.

55. I=jXîlogx-j\ + C.

56. 7=x{( log |x | ) 3 -3( log |x | ) 2 +61og |x | -6} + C.

57. 7 = - y ^ ( l o g | x | 2 + 4 1 o g | x | + l) + C.

58. 7 = ^x3/2{9(log|x|)3-18(log|x|)2+241og|x|-16} + C.

5 9 . 7 = - ^ - ( l o g | x | + y ) + C.

60. I = 2s/x (log xf - 8V* log x +16 Jx+C.

61. / = j * * j(log;c)*-jlogx+-U+C, x > 0 .

62. I =*1±1(logx ^rA + C, X > 0 . «+i \ « + i '

ANSWERS

E X E R C I S E 19, P. 29

1. J = (2JC+1)4/8 + C.

2. 7 = - l /9(3x-2) s + C. 3. 7 = l o g | x - 3 | + 2 1 o g | x + 2 | + C. 4. 7 = log|x2-3;c+3| + C. 5. 7 = 31og |x -5 | - 2 1 o g | x + l | + C. 6. I = 5 1 o g | * + l / 2 | - 4 1 o g | * + l | + C. 7. 7 = 2 1 o g | x - 3 | + 4 1 o g | * - l / 2 | + C . 8. 7 = log|2x2-5;c+3|+C. 9. 7 = 3 1 o g | x - 2 | +21og |x+5 | + C.

10. J = (71og|jc-6|)/3-(31og|jc-3|)/2 + C. 11.7 = {log | (* +1 - y/2)l(x +1 + V2) |}/2 y/2 + C. 12. 7 = log |(2x-3)/(3x-2)\+C. 13. 7 = (log|10x+x2|)/2 + C. 14. 7 = (71og|4+5x2|)/10 + C. 15. 7 = {log | (x -1)/(5 -x) |}/4 + C. 16. 7 = {log|(2x + V5-l)/(-2x + V5 + l)|}/V5 + C. 17. 7 = {log | */(2 -3x) |}/2 + C. 18. 7 = (81og|x-2|)/3+(log|x + l|)/3+C. 19. 7 = 21og| jc-3 | -5 / (x-3) + C. 20. 7 = { l o g | 2 x - l | + l/(2x-l)}/4+C. 21. 7 = 2 1 o g | x - 5 | + 3 / ( x - 5 ) + C. 22. 7 = 31og|x+2|+5/(x+2) + C. 23. 7 = {arctan(2x-l)/3}/3 + C. 24. 7 = {arc tan (3x+\)l^2}j^2 + C. 25. 7 = {arc tan (x - 3)/2}/2 + C. 26. 7 = arctan(3x-l) + C. 27. 7 = (log | x2 -x+11)/2 + ,/3 arc tan (2x-1)/^/3+C. 28. 7 = log |2x 2 -2x+ l | +arc tan (2x- l ) + C.

170 ANSWERS

29. 7 = log |x 2 -2x + 5|+{arctan(x-l)/2}/2 + C. 30. I = log | x2+2x +101 - 4 arc tan (x +1)/3 + C. 31. 7 = log | x 2 -8x+25 | -4a rc t an (x -4 ) /3 + C. 32. 7 = (3 log | x2+4x+8 |)/2 - arc tan (x+2)/2 + C. 33. 7 = (log|x2-3|)/2 + V31og|(x-V3)/(* + V3)l + C. 34. 7 = (log | x2+3 |)/2+2 ^3 (arc tan x/,/3) + C. 35. 7 = 3 log |x2+4x + 13|-4arctan(x+2)/3 + C. 36. 7 = 5 log | x2-4x+201 - 6 arc tan (x-2)/4 + C. 37. 7 = 2 log |x 2 -6x + 10 |+7arctan(x-3) + C. 38. 7 = 5x/3-(101og|2+3x|)/9 + C. 39. 7 = x/5 -{2 V15 arc tan V(5/12)x}/25 + C. 40. 7 = 2x-(51og|x2 + 6x + 25|)/2-{l5arctan(x + 3)/4}/4

+ C. 41. 7= - 3 1 o g | x + 2 | + 2 1 o g | x - 4 | + 8 1 o g | x - 7 | + C. 42. 7 = -(29-30x + 6x2) /3(x-2)3+log|x-2|+C. 43. 7=(41og |x-2 | ) /7 + (171og|x2 + 3|)/14

+(arc tan x/>j3)/7 s/3 + C. 44. 7 = (arc tan x)/2+(x-2)/2(x2 + l) + C. 45. 7 = x2/2+x + 3 1og|x + l | + 2 1 o g | x - 2 | + C . 46. 7 = x 2 - 3 x - 1 4 / ( x - 4 ) + 2 1 o g | x - 4 | + C . 47. 7 = x3/3— x + arc tan x + C. 48. 7 = 9x5/5 - 2x3 +4x-{8 arc tan ^/(3/2)χ}/^/6 + C. 49. 7 = 2x3/3-2x2 + 5 x + l o g | x - 2 | + 2 1 o g | x - l | + C . 50. 7 = (51og|x + 3 |) /2+(471og|x-l | ) /2-25 1og |x-

-1/21+C. 51. 7 = (5 log | x-11 - 4 log | x +11 +10 log | x -21

- l l l o g | x + 2 | ) / 3 + C. 52. 7 = {36x/(5-6x)-121og|5/x-6|-5/x+6}/25 + C.

53. 7 = i-log(x2 + l)/(x2+3) + C.

ANSWERS 171

54. J = 21og(x2-4x+29)+31og(x2-2x+5) +6 arc tan (x-2)/5 + 8 arc tan ( x - l)/2 + C.

55. / = log |x2+4|+(51og|x + l | ) / 2 - | - l o g | x - l | 1 1

+ y arc tan — x + C.

56. / = log|x|+21og|x + l |+31og |x+2 |+41og |x+3 | + C. 57. 7 = 41og |x | +log(x2 + l ) - l / x - a r c t a n x + C . 58. / = -(log |x2-fl2|)/2a2-(log \x\)/a2 + C. 59. I = log | x | - — log | x2 +x +11 -{arc tan (2x+1)/V3}

/V3 + C. 60. 7 = {log(x2+x + l)/(x2-x + l)}/4

+{arc tan (2x + l)/V3}/2 y/3 + C.

61. 7 = 2 1 o g | x 2 - 4 | + y l o g | x 2 + 4 | + / ' 3a rc tanyx) /2 + C.

62. 7 = 31og |x - l |+61og |x 2 +4x+29 | +{42 arc tan (x+2)/5}/5 + C.

63. 7 = l/(x+l)2+log | x | + 3 log | x + 1 \+C. 64. / = l /2x 2 - l /x+2 log |x | - ( log |x + l | )/4 + C. 65. 7 = (2x+l)/3(x2+x+l)+4{arctan(2x + l)/V3}/3V3+C. 66. / = l / 2 ( x - 2 ) 2 + 5 / ( x - 2 ) + 3 1 o g | x - 2 | + C. 67. 7 = (x+2)/16(x2+4x + 8)2+3(x+2)/128(x2+4x+8)

+ i 3 arc tan — (x+2) 1/256 +C.

68. / = - 5 / 2 ( x - l ) - l o g | x - l |+3/2(x + l) +2 log 2 | x + 11 + C.

69. / = | y log (x 2 +4x + 8)/(x2-4x+8)+arctan(x+2)/2

+ arc tan (x -2)/2 ί /64 + C.

172 ANSWERS

70. 1= - l / ( x - l ) 3 + l / ( x - l ) 2 - 4 / ( x - l ) + 5 1 o g | x - l | + C. 71. 7 = {log(x2+2x+5)/(x2-2x + 5)+arctan(x+l)/2

+arc tan (x -1)/2}/40 + C. 72. 1 = - l / ( x - l ) 3 - l / ( x - l ) 2 - 5 / ( x - l ) + 7 1 o g | x - l |

+21og |x+3 |+C.

73. / = -l/2(x2 + 4) + y l o g | x 2 + 4 | - a r c t a n y X + C.

74. 7 = - l / 2 ( x - l ) 2 - 5 / 2 ( x - l ) + ( 3 1 o g | x - l | ) / 2 +(3 log |x2 + l | ) / 4 -a rc t anx+C.

E X E R C I S E 2 0, P. 33

1. / = (2x+1)3/2/3+ C.

2. l = lj(3+4x) + C.

3. / = -^(3x-4)2/3 + C.

4. 7 = 5(2x+l)2/5/4+C. 5. 7 = 3(x2-x+12)(x -4)i/3/7 + C. 6. 7 = (12x2-x-l)(3x-l)i/3/28+ C. 7. 7 = 2(2+3x)3/2{(2+3x)/5-2/3}/9 + C. 8. / = 2(1 -5x)3/2{(l -5x)/5-l/3}/25 + C. 9. 7 = 3(x2-x-12)(x -4)i/3/7 + C.

10. 7 = 2(x-2)(2x+3)3/V7 + C. 11. 7 = (5x2-12x+36)(x+2)2/3/40 + C. 12. 7 = 2(27x2 - 12x + 143)(3x + l)i/2/405 + C. 13. / = 2(10x2+3x-18)(2x+3)i/V45 + C. 14. / = [log \y(x+a)-y/a}/y(x+a)+ yfa}q/y/a + C. 15. 7 = 2{arctanN/(*/ö-l)}/Va + c'· 16. 7 = 2V*+log|(V*-l)/(V*+l)|+C. 17. 7 = 2V(l+*)+log|{V(*+l)-l}/{V(*+l) + l}l + C

ANSWERS 173

18. 7 = - ( x + 4 s / * + l o g | l - > / * l ) + C· 19. 7 = log{V(l -x)-V2}/{Va-*)+V2} + C 20. J = 4V(l+V*){(1+V*)2 /5-(l+V*)/3} + C.

21. 7 = 6 ( 4 - i 2 - i + l o g | < + l | ] + c> where ί = Λ;1/«. = 6 ( y i 2 - i + l o g | i + l | Y

22. 7 = 3( t2-t + log 12/ +11 1/2 + C, where t = x1'«.

23. 7 = {(x-5)3/2-(x-7)3/2}/3 + C. Rationalize the denominator.

24. 7 = [log|{V(x+9)-3}/{V(*+9)+3}|]/3 + C. 25. / = 3 {(7 -2x)3/10-2(7 -2x)2+49(7 -2x)/4}(7 -2x)«»/8

+ C. 26. 7 = 2(x+l)i'2-3(x + l)i/3+6(x+l)1'6

-61og|l+(jc+l)i /«| + C. 27. 7 = 2 V{(*-2)/(*-1)} + C. 28. 7 = log 11 - V ( l -*2) I - log | x \ -arc sin x + C. 29. 7 = -3(χ + 1)2/3/2-6(χ+1)5/6/5-(χ+1)-6(χ + 1)7/«/7

-3(x +1)4/3/4 -2(x +1)3/2/3 + C. 30. 7 = 3x*/3/4-6x'/e/7+x_6xs/e/5+3x2/3_2xi/2

+6xi/3_6xi/e+3 log | χΐ/"-11 +91og | x ^ + l1+ C.

E X E R C I S E 21, P. 48

1. 7=2V(4* 2 +3x+l) + C. 2. 7 = 5V(36x2-109x+77)/18 + C. 3. 7 = arcsin(jc-l) + C. 4. 7 = arc sin {(x+3)/4} + C. 5. 7 = (arcsin3x)/3 + C. 6 . 7 = arc sin (x//· — 1) + C. 7. 7 = -V( l -4x 2 ) /4+3 (arcsin2x)/2 + C.

174 8. 7 =

9. I

10. I-11. I-12. I

ANSWERS

-V( l -2x-3x 2 ) /3 -{arc sin (3x + l)/2}/3 V3 + C.

~ x V(l -4x2)+(arc sin 2x)/4 + C.

- 6 V(6+x-x2)+8{arc sin (2x-l)/5} + C. -V(5+4x-x2)-3{arcsin (x-2)/3} + C. -V(8+2x-x2)+2{arcsin(x- l ) /3} + C. 1 13. / = - (x-3) v / (6x-x 2 ) -9{arcs in(3-x) /3}/2 + C.

14. 7 = 2V(3-2x-x 2 ) -5{arc sin (x + l)/2} + C. 15. 7 = log|x+3/2+V(*2+3x+2)| + C.

16. 7 = l l o g | 2 x + 3 / 4 + V ( 4 x 2 + 3 x - l ) | + C.

17. 7 = - j l o g | x - l / 2 + v ' (x2-x+w) | + C.

18. 7 = log | χ-2α+^/{(χ-ά)(χ-3ά)}\ + C. 19. 7 = ^(x2+2x) +2 log | x + 1 +V(*2+2x) | + C. 20. 7=3V(* 2 -5x + 19) + 19{log|x-5/2+V(*2-5x + 19)|}

/2 + C. 21. 7 = V02-a*)+3a{log \χ-α/2+^/(χ2-αχ)\}/2 + α 22. 7 = 3V(4x2-4x+5)/4-{log | 2x - l+V(4x 2 -4x + 5)|}

/4 + C. 23. 7 = 3V(^2-4x + 5)+81og |x-2+V(^ 2 -4x+5) | + C. 24. 7 = 3 V(4x2 + 5x - 8)/4 - 47 {log 12x + 5/4

+V(4x2 + 5x-8)|}/16 + C. 1 25. 7 = 5V(2x2 + 8x- l ) /2-8Jlog x + 2 + /(.

/V2 + C.

x2 + 4x—-

26. 7:

27. 7:

1 ( 1 +x) V(4x+x 2 ) -y log 12+x+V(2x+x2) \ + C.

:5V(3x2-2x + l ) /3 -7{ log |x - l /3 +VO-2x/3+i /3) i} /3V 3 +c .

ANSWERS 175

28. 7 = y ( x + l )V(3-2x-x 2 )+2arcs in(x+l) /2 + C.

29. / = y x V(*2 -4 ) - 2 log | x+V(x2 -4) 1 + C.

30. J = (3x+5)V(3x2 + 10x+9)/6 +{log 13x + 5+V[3(3x2 + 10x+9)] |}/3 J3 + C.

31. / = y ( x - 3 / 2 ) V ( * 2 - 3 x + 2 ) - { l o g | x - 3 / 2

+V(*2-3x+2)|}/8 + C.

32. 7 = - — XyJ(l-x2)+—arcsmx + C.

33. / = ^ (x -3 )V(* 2 +2x+2)+ i - log |x + l+V(*2+2x+2)|

+ C. 34. According to the substitution made we obtain one of the four following solutions. Substituting y/x = iwe obtain arc sin yfx—\/(x—x^ + Q. Substituting *J(l— x) = ί we obtain arc sin >J(l —x) —y/(x—x2) + C2. Substituting ,/{x/(l—x)} = t we obtain arc tan V{*/(1—*)} — V(x ~~ χ2) + Q · Writing yj{x\{\ — x)} = x/N/(x —x2) we obtain — arc sin (2x —1) —\f(x—x2) + C4. The four constants are connected by the relationships C2 = ΟΊ+π/2, C3 = C1( C4 = ^ + π / 4 . 35. 7 = (x-3/fl)V(ö^2+2x + l)+{31og |ÖX+1

+V[a(ax2+2x + l)]|}/aVa + C 36. / = (x+3)V(x2 + l) + C. 37. / = (x-a)V(^2+a2) + C. 38. / = (2x2-5x + l)V(x2+2x+2)/6

+{5 log | x+1 +V(x2+2x+2) |}/2 + C. 39. 7 = (2x2+x+7) s/ix2+2x -1)/6

-21og |x + l + V ( * 2 + 2 x - l ) | + C. 40. 7 = (x2 + 5x+24)V(*2-4x+3)/3

+ l l l o g | x - 2 + V ( x 2 - 4 x + 3 ) | + C.

176 ANSWERS

41. 7 = (8x2-10x-l)V(*2+x + l)/8 +{531og|x + l/2+V(x2+*+l)l}/16 + C.

42. 7 = (3x3-2x2-10x-60)V(4x-^2)/12 + 10arcsin (x-2)/2 + C.

43. 7 = (8x2-2x-51)V(6+x-x2)/24 -{25 arc sin (1 -2x)/5}/16 + C.

44. 7 = (5x3-6x)V(5x2+4)/100 +{61og|xV5+V(5x2+4)|}/25V5 + C.

45. 7 = (x2/3+25x/12-163/24)V(x2+* + l) +{851og|x + l/2+V(x2+x+l)|}/16 + C.

46. 7 = (5x/6 + 17/12)V(3x2-5x+8) +{551og|3x-5/2+V[3(3x2-5x + 8)]|}/8V3 + C

47. 7 = -(x2/3+9x/2+227/6)V(5+6x-x2) -139 arc sin (3-x) /7l4 + C.

48. 7 = (x2/3-x/12-67/24)V(8+x-x2) -{33 arc sin (1 -2x)/V33}/16 + C.

49. 7 = (2x2/3-3x + l/6)V(2+3x-x2) -{17 arc sin (3-2χ)Λ/17}/4 + α

50. 7 = (x2-3)V(2x2+3)/6 + C. 51. 7 = (x4-2x2+6)V(2*2+3) + C. 52. 7 = (x3/4-7x2/12+x/3 + 5/2) V(3 +2x+x2)

+{7 log 11 +x+ N/(3+2x+x2) |}/2 + C. 53. 7= -V(10x-x2)/5x+C. 54. x < - 1 , 7 = -V{(x- l ) / (x+l )} + C,

x > 1, 7 = V{(JC-1)/(JC + 1)} + C.

55. 7 = - I v { ( 2 - x ) / ( 2 + x ) } + C.

56. x < - (V5- l ) / 2 , 7 = arcsin(2-x)/xV5 + C, x > (y/5 -1)/2, 7 = arc sin (x-2)/x J5 + C.

57. x < - Λ / 2 + Ι, 7 = arcsin(x+l)/xN/2 + C, x > V 2 + l, 7 = - a r c s i n ( x + l)/x,/2 + C.

58. x < - 1 , 7 = {arcsin(2-x)/(2x-l)}/V3+C, x > 1, 1= -{arcsin(2-x)/(2x-l)}/V3 + C

ANSWERS

59. 1= ~ arc sin 2xj(x+l)+C.

60. l<x<3/2, I={arcsinx/(3-2x)}/..j3+C,3/2<x<3,I={arcsinx/(2x-3)}/..j3+C.

61. I = log Ix/{x+2+2..j(x2+x+ I)} I+C.

62. x< -1, 1= arc sin l/x+C,x :> 1, I = -arc sin l/x+C.

177

63. 1= ..j{(a+x)/(a-x)}/a+C.

64. I = {arc sin (x+5)/2 .J2(x-2)}/..j7 +C.

65.1= -.J(x2-4x+l)/x+2Iogl{I-2x-..j(1-4x+x2)}/xl+C.

66. 1= -..j(4-x2)/4x+C.

67.1= -..j(10x-x2)/9(x-l)+{4Iog I [4x+5 +3 .J(10x-x2)]/(x-l) 1}/27 +C.

168.1= -,J(x2+ 1)/2x2 -2: log IX/{1 +V(x2+1)} I+C.

69. 1= (3x-l) .J(2x2+2x+ 1)/2x21-2: log l{x+l+.J(2x2+2x+l)}/xl+C.

70. 1= (5 -6x) .J(3 -2x2)/2(x-l)2+ 7 log I{3 -2x-.J(3 - 2x2)}/(x-l) I+C.

171. I=.J(1+r)/2x2 - 2 1og l{l-.J(1+r)}/xl+C.

72. 1= -(1/9x3+5/54x2+ 1/54x) .J(3 -2x+x2)+2 log I{3 -x-..j[3(3 -2x+x2)]}/x 1/27v3 +C.

73. 1= {1/9(x-2)3+2/27(x-2)} .J(1-4x+x2)+C.

178 ANSWERS

E X E R C I S E 22, P. 67

1. / = (sinl2x)/24 + (sin2x)/4 + C. 2. 1= - (cos 5x)/5-(cos x)/2 + C. 3. 7 = (sin5x)/10 + (sinx)/2 + C. 4. 1 = - (cos4x)/8+(cos2x)/4 + C. 5. 1= - (cos6x) /12-(s in2x) /4 + C. 6. 7 = (sin3x)/6-(sin7x)/14 + C. 7. 7 = (sin4x)/8+(sin2x)/4 + C. 8. / = (sin 2x)/4-(sin 4x)/8 + C. 9. 7 = (sin3x)/6-(sin7x)/14 + C.

10. J = (cos 3 x) /3 -cosx + C. 11. 1 = - c o s x sin x (cos2 x + 3/2)/4 + 3x/8 + C. 12. I = cos x sin x (cos2 x + 3/2)/4 + 3x/8 + C. 13. / = (sin5 x)/5 - ( 2 sin3 x)/3 +sin x + C. 14. / = - (cos 5 x) /5 + (2cos 3 x ) /3 -cosx + C. 15. / = (tan4 x) /4-( tan2 x) /2- log cos x + C. 16. / = - ( c o t 3 x ) / 3 + c o t x + x + C. 17. / = - (co t 5 x) /5 + (cot3 x ) / 3 - c o t x - x + C. 18. / = (sec6 x sin x)/6 + (5 sec4 x sin x)/24

+ (15 sec2 x sin x)/ 48 +(15 log | sec x + t a n x |)/48. 19. / = —(cosec5 x cos x)/5 —(4 cosec3 x cos x)/15

- ( 8 cotx)/15 + C. = - cot x - ( 2 cot3 x) /3-(cot5 x)/5 + C.

20. 7 = (cos7x)/7-(cos5x)/5 + C. 21. 7 = (cos1 3x)/13-(3cos1 1x)/ l l+(cos9x)/3-(cos7x)/7 + C. 22. / = {s in 6 x-(s in 4 x) /5-(4s in 2 x) /15-8/15}(cos x)/7 + C. 23. 1 = (sin3 xcos x)/4-(s in x cos x)/8+x/8 + C. 24. / = (sin4 x)/4-(sin6 x)/6 + C. 25. I = (sin5 x)/5 - ( 2 sin7 x)/7 +(sin9 x)/9 + C. 26. 1= -(cosec7 x)/7 + C.

27. 1 = 28. / = 29. / = 30. 1 = 3 1 . . / = 32. / = 33. / = 34. / = 35. 1 =

36. / =

37. / :

38. / :

ANSWERS 179

log | tan (π/4+χ/2) | - s i n x + C. 3(sinx)i/s + C. - 3 ( l + 2 c o s x ) 2 / 3 / 4 + C. -2( l+cos2 x) 1 /2 + C. log | l + s i n 2 x | + C. arc sin (sin2 x) + C. —(1 +sin2 x)/sin x + C. sin x—cos x + C. (sin x sec4 x)/4+(3 sin x sec2 x)/8

+(3 log | tan (π/4+χ/2)|)/8 + α

—(cosec6 x)/6+(5 cosec4 x)/24+(5 cosec2 x)/16 cos x

1/16 + C. ί 5 log t a n — x j

—-y cosec2 x+log | tan x\ + C.

— tan2 x+log |tan x\ + C.

39. I = — sec2 x—cosec2 x—(cosec4 x) /4+3 log | t a n x | + C .

40. / = (sec3 x cosec x)/3 - ( 8 cot 2x)/3 + C.

41. /

42. /

43. /

44. / = (sec7 x)/7 - (sec5 x)/5 + C.

45. /

46. / = {2 arc tan [(tan x/2)/3]}/3 + C.

sin x+—sin x sec2 x+(3 log | tan (π/4—x/2) \)/2 + C.

-(s in3 x)/3 - s i n x+log | tan (π/4+χ/2) \ + C.

-=■ (sin2 x—cosec2 x) — 2 log | sin x | + C.

-— sin x sec2 x + ( 3 log |tan (π/4+χ/2) |)/2 + C.

180 ANSWERS

47. I = cot(n/4+x/2) + C. 48. / = (log | tan (π/8 +x/2) \)/y/2 + C. Use the transforma

tion sin x+cos x = sin x + sin (π/2+χ) = 2 sin (π/4+χ) cos (— π/4) = >/2 sin (π/4+χ).

49. For cos x ^ O , / = — arc tan (tan2 x) + Cl9 and

for sin x ^ 0, I——— arc tan (cot2 x) + C2.

50. 1 = 2 tan x+{arc tan (^2 tan x)}/y/2 + C. 51. J = — l/(sinjc—cos x) + C. 52. / = {log (^2 - s i n 2x)j(yj2 + sin 2x)}/2 ^ 2 + C.

53. / = -y arc tan (2 tan2 x +1) + C.

54. J = {3 arc tan (tan */V2)}/4 ^2 - t a n x/4 (tan2 x+2) + C. 55. / = {V(2+V2) arc tan [(tan 2x)/N/(4+2V2)]

- V ( 2 ~ V 2 ) arc tan [(tan 2x)/V(4-2V2)]}/4 + C. 56. 7 = {arc tan [(tan 2χ)/λ/2]}/%/2 + C.

57. / = — tan x+{arc tan (>/2 tan *)}/2 J2 + C.

E X E R C I S E 2 3, P. 70

1. I = x2/4 — - r -x>/ ( l -x 2 ) arc sin x + (arc sin x)2/4 + C.

2. I = (x arc sin x)/V(l - x 2 ) + - i log 11 -x2 | + C.

3. I = x arc tan x — — log 11 + x21 — — (arc tan x)2 + C.

4. / = {2 V(arc tan 3x)}/3 + C. 5. 7 = l / 2a r cco t2x + C. 6. 7 = (arctan;c)3/3 + C.

ANSWERS 181

7. 1 = 1/arc cos x + C. 8. I = log | arc sin x | + C. 9. / = {(x2 - 1 ) arc tan x+x}/4(l +x2) + C.

10. / = arc sin x/V(l - x 2 ) + y log | (1 - x ) / ( l + x) | +C .

11. / = y ( * 2 ~ y ) arc sin x+{xV(l -*2)}/4 + C.

12. / = l o g | { ( x - l ) / ( x + l ) } i / 8 | - i a r c t a n x j i + l / ( x 2 - - l ) |

+ C. 13. / = (x3 arc tan x)/3 - x 2 / 6 +(log 11 +x21)/6 + C.

14. 7 = x-log(l-fe3 C)-2e-x /2arctane5 C /2-( arctan—x ) +C . 2—( arc tan—x j

15. I = - ( a r c sin x ) /x - log | {l +VC1 ~χ2)}/χ\ + C. 16. 7 = x - e x a rc s in ( e* ) - l o g { l + V O +e2x)} + C.

17. 7 = ( x 7 a r c t a n x - x V 6 + x 4 / 4 - x 2 / 2 + y l o g | l + x 2 | V 7 + C .

18. / = - ( 2 x + 2 1 ) V ( - * 2 + 3x-2 ) /4 +(x2 + 3x-55/8) arc cos (2x-3 ) + C.

19. I = V(l +*2) arc tan x - l o g | x+yj(l +x2) \ + C. 20. / = - x 2 / 8 + { x V(l - * 2 ) arc sin x}/4 + (arc sin x)2/8 + C. 21. / = x/4+x 3 /12+{(l +x2)2 arc tan x}/4 + C. 22. 1= -2 sin (1 - x ) V* + 0 +*) arc sin {2 V*/(! +x)} + C.

E X E R C I S E 2 4, P. 73 1. / = e3*/3+2V(e*) + C. 2. 7==log(e*+e-*) + C = log(2coshx) + C.

3. 7 = I l o g | e 2 x - l | - x + C.

4. J = arctan(e*) + C.

182 5. 6. 7. 8. 9.

10.

11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.

27.

> 1 = 7 = I-7 = 7 =

1 =

7 = 7 = 7 = 7 = 7 = 7 = 7 = 7 = 7 = 1 = 7 = 7 = 7 = 7 = 7 = 7 =

7 =

ANSWERS

= 2V(e*+l)+log{V(ex+l)-l}/{V(e* + l) + l} + C = 2 1 o g | e x + l | - x + C. = [log {V(3+2e*)-V3}/V(3+2e*)+V3}]/V3 + C. = 2(l+ex)3'2/3 + C. = - ( e x - l ) - 1 + C .

= y (e2x-e~2x)+2x + C = 2 sinh x cosh x+2x+ C.

= log |e x +5 | + C. = (35 log |e2 x-4/9 |)/36-3x/2 + C. = l o g | l + e x | - e - x - x + C. = - l / (« - l ) ( e x +c) n - i + C. = {arcsinV(5/3)ex}/V5 + C. = x- log |2e x +l |+V(e 2 x +4e x + l) + C. = -e-x(x3+3x2+6x+6) + C. = log |logx| + C. = xlog |x2 + l | - 2 x + 2 a r c t a n x + C. = x(log |x | ) 2-2xlog |x |+2x + C. = x log |x + V(^2 + l ) | -V(^ 2 + l) + C = (2+5x)(log|2 + 5x | - l ) /5 + C. - arc tan log | x | + C. = - ( l og | x | + l)/x + C. = (4+3x)3(log|x|)/9-(641og|x|)/9-16x-6x2-x3+C. = xi (log |x2 + 3|)/4-(x4/2-3x2

+91og|x2 + 3|)/4 + C. = ax(x - 1 /log a)/log a+C.

E X E R C I S E 2 5, P. 86

1. 1 log (21/5). 2. «/4fl.

3. 1/2. 4. π/20.

ANSWERS 183

5. (3 log l/9)/8. 6. π/6.

7. l - , /3 /2- j r /6 . 8. jr/V5.

9. y (152/3-52/3). 10. -i-log (18+V10), put x2 = u.

11. (5V5-8)a3. 12. 3V7/2-V2+log(2+V2)/(3+V7)· 13. log 5(3+2 V3)/(6+Vl 1 litt. πα2/4. 15. at/12-l/6+(log 2)/6. 16. e2 - l+log2/( l+e2) . 17. -2 /e + l. 18. 2e4-e/2. 19. e2(5e2-l)/4, 20. (3e*-l)/8. 21. 1. 22. log 3. 23. 4/3. 24. (4 arc tan 3A/3)/3 , / 3 - l / 6 - ( 4 arc tan l/V3)/3 J3. 25. 1 -{log (V2 + l)/(V2-l)}/2 y/2-y/(2ß)

+{log (V2+V(3/2))/(V2-V(3/2))}/2 J2. 26. {arc tan (1/5)-arc tan (l/10)}/5. 27. -1/2+log V3. 28. π/2. 29. 2(V2-1). 30. 2(V2-1). 31. w/4+log(V2/2). 32.1/336.

33. 5π/1024, the function is even, thus I = 2 34. 9V2/560. 35. α2/3. This result was known to Archimedes. 36. 1/3. 38. 8. 40. 5/12. 42. 64/3. 44. 32/3. 46. 9/2.

37. 8/3. 39. 5/6. 41. 343/3. 43. 0-950. 45. 27/2. 47. (3π+2)/(9π -2).

184 ANSWERS

48. 15 /2 -8 log 2. 49. 6(3π + 8). 50. 1/16. 51. l / 4 - l / 2 e . 52. 3 rc (V3- l /2 ) /2+9( l -V3) /2 .

E X E R C I S E 2 6, P. 95 1. πα2/4. The curve is a circle of radius a and centre

(αβ,π/2). 2. When 0 lies in either of the intervals π/2^θ ^π, and

3ττ/2 ^ θ ζ=ζ 2π, we obtain a negative value for r. If r0 > 0, then by the point (—r0, θ0) we understand the point (r0, θ0+π). Under this convention the equation represents a four-leaved rosette. Area = πα2/2.

3. πα2/2. The curve is a two-leaved rosette. 4. 3πα2/2. The curve is symmetrical about the initial line. 5. 4π3α2/3. Draw a rough graph. 6. πα2. 7. π/?2/(1-£?2)3/2.

8. 72Λ/3/5 . 9. 3#2/2. Draw a rough graph considering the intervals 0 < t < i0, and /0 < ί < oo.

10. fl2(jr/4 + l - V 3 ) . 11. 4 ^/3/5. 12. 8/15. 13. 3(4π3 + 3π). 14. 24π.

E X E R C I S E 27, P. 99 1. 52/57. 2. 14/3. 3. 4 ( 2 V 2 - l ) / 3 . 4. 14/9. 5. 56/3. 6. 16(31 V31/8 - l ) /81 . 7. 16(13 V l 3 / 8 - l ) / 2 7 .

8. 3 Vl0-V2+y log (3+3 Vl0)/(3/2 + V2)·

9. π/2. 10. π/4. 11. ^/2+log (1 +>/2). We take ^ as the independent variable.

ANSWERS 185

12. 2+(31og3)/2. 13. 14/3. 14. (log3)/2. 15. log(V2 + l).

16. «(e-e-1). 17. 1+^-log 3/2.

18. log 3-1/2. 19. log (e^2+e-i'2). 20. a. 21. tf{2+V31og(2+V3)}. 22. 3 / 4 + 1 log 2.

23. V ( l + e 2 W 2 + l o g { V ( l + e 2 ) - l } - l - l o g ( V 2 -1)· 24.4ö. 25. 4V 2 +log( l+V 2 )} /2 . 26. 5/12 +log 3/2. 27. 2πα. 28. Pyi+\og (1+V2)}· 29. 2 V3.

30. π2α/2. 31. 5a{2 + [log(2+V3)]/V3}/8.

32. 2flJ7/2-V3 log y(V3+V7)j · 33. 40. 34. «{l+[log(V2 + l)/(V2-l)]/2V2}/2. 35. 10π.

E X E R C I S E 2 8, P. 105 1. V = π log {bid). 2. V = π. 3. Κ=π2 /2, S=27r{V2+log(V2 + l)}. 4. F=jr(15/2-81og2). 5. V = 71^(2-V2)/3,

5 = jrß2(V6-l)-a2 {log (2+V3)/(l +V2)}/4. 6. Κ=π/63, S = 2nyj2/9. 7. F=jrc3(e2+4-e-2)/4, S = 5tc2(e2+4-e-2)/2. 8. S = 4w 9. F=jr{7/3+3a+4a2+8a3 log|(2ö-2)/(2ö-l) |},

S = 2πα[(χ-4α) yj(8ax-3x2)l(x-2a) +8a {arc cos (3χ-4α)/4α}/3]|.

186 ANSWERS

10. S = a V{62 +(2πα)2}/2 + {b2 log | Inajb +ν(4π2ο2/ό2 + 1)|}/4π.

11. lftr/35. 12. F = jr{3Vl0 + 1 0 1 o g | V l 0 - 2 | - 3 V l 5 - 1 0 1 o g | V l 5 - 3 | } . 13. V=\6n. 14. S = 56n/21. 15. F = 64;r. 16. 5 = 37π V37/54. 17. Κ=16π/35. 18. Κ = 2π/3.

19. V = ^n log 9/5. 20. F = 200π/3.

21. 5 = 200π2, Γ=500π 2 . 22. Γ - π / 5 ( 1 - ε - 2 π ) . 23. V = 64nas/315, S=\6na2/35. 24. Κ = 3π/4, 5 = 3π.

EXERCISE 29, P. 117 1. Mb2/3, where M is the mass of the rectangle. 2. Mr2/2. 3. Mh2/6. 4. 3Mr2/10. 5. Mr2/2. 6. Mr2/4. 7. 2Mr2/5. 8. IMabjS. 9. mc3(e+e_1)(e2 + 10+e"2), where m is the mass per unit

length. 10. y = Irin. 11. j? = 4Γ/3ΤΓ.

12. x = 3r/8. 13. A/4 from the base. 14. x = h/3. 15. c ( e 2 +4-e - 2 ) /2 (e -e - 1 ) . 16. j? = 5tf/6. 17. x = 4a/5. 18. jc = r3(92 - 4 5 V3 - 10π)/24Α,

y = 8r3{(3V3 -7) /24+log (^3/2 cos π/12)}/Α, where ^ is the area.

19. * = 9/7, y = 5 V3/32,

ANSWERS 187

20. x = 7/5, y = V3/4. 21. χ = * ( π / 2 + 64/45π). 22. χ = 256^/315π, j? = 256α/315π. 23. je = b(n-2)/4n, y =παβ. 24. je = 83/77, j? = 9/154. 25. ic = 9a/20, y = 9a/20. 26. T = 5a/6, B = 0. 27. x = 3 x 0 / 5 , y = 3yo/5.

E X E R C I S E 3 0, P. 126

1. +°° 3. +o° 5. 4V2. 7. π/2. 9. π/12.

11. 3 1/2/S. 13. V5· 15. 3π/8. 17. -π /3 . 19. π/2. 21. log 3. not convergent. 24. si. 26. Λ/9. 28. 1. 30. 3π/8.

32. π.

34. 2π/3,/3.

2. 3/2. 4. 8. 6. 5/2. 8. π.

10. 1. 12. log(2+V3). 14. 93V4. 16. π/3. 18. π/2. 20. 7π/2. 22. ο°, the integral is 2 3 . οβ.

25. π3/12. 27. 1/3. 29. V2^/8-V2{arctanV2/2}/4. 31. n/3.

33. - y + l o g 2 .

35. π/2 7 3 — ί log 25/12.

188 ANSWERS

36. π/2. 37. I/a. 38. 1/2. 39. 1/2. 40. e - 1 . 41. n/y/(a2-b2). 42. n/2ab.

EXERCISE 31, P. 132 1. The function has no extreme values. 2. Minimum at the points ( — \J2, yjl) and (y/29 — y]2),

zm i n = — 8. The point (0, 0) is not an extreme point. 3. Minimum at (0, 1), zm i n = — 7. 4. There is an extreme at the point (a, a), where z = —a3. For

a > 0 the function is a minimum, for a < 0 the function is a maximum.

5. The first partial derivatives are zero at the points ^4(0, — 5/2), B( — 1, — 1) and C(3, — 1), but only at A is there an extreme value, when zm a x = 75/4.

6. Minimum at ( - 4 / 3 , 1/3), zm i n = - 4 / 3 . 7. Minimum at (8/3, 2/3), zm i n = - 1 1 / 3 . 8. Minimum at (27/2, 5), zm i n = -109/4. 9. Maximum at (a, a), zm a x = —a3.

10. Minimum at (^- \]29 y V 2 \ zmin = 3/V2·

11. Minimum at (1, 0), zm i n = — 1. 12. Minimum at (8, 24), zm i n = -448 . 13. Minimum at ( — 1, 3), zm i n = — 7. 14. Maximum at (0, 0), zm a x = 1. 15. Maximum at (a/6, a/3), zm a x = a6/432. 16. Maximum at (2, 3), zm a x = 108. 17. Maximum at (ττ/3, π/3), zm a x = 3 Λ/3/2.

18. Maximum at (π/3, π/6), zm a x = 3 >/3/2. 19. Maximum at (uf/3, a/3), zm a x = sin3 (a/3).

ANSWERS 189

20. Maximum at (π/3, π/3), zm a x = 3V3/8; minimum at (2π/3, 2π/3), zm i n = —3^/3/8.

21. The point (1, 1) where z = - 1 +— log 2 + 3/4, is a saddle

point, the three-dimensional equivalent of a point of inflexion.

22. Minimum at (1, 2), zm i n = 7 - 1 0 log 2. 23. Minimum at (0, 0), zm i n = 0; saddle point at ( — 1 /4, — 1 /2). 24. Maximum at (1, 3), zm a x = e"13; minimum at ( — 1/26,

-3 /26) , z m i n = -26e-i/52. 25. If the angles subtended at the centre of the circle by

the three sides are x, y and In—(x+j>), we ob ta in^ =

= — r2 {sin x + sin y+sin (2ττ—x — j )} , where r is the radius

of the circle. The area A is a maximum when the angles are equal.

26. The number is divided into three equal parts. 27. The area is a maximum when the triangle is equilateral. 28. Px(3, - 3 ) , P2(9, 5). 29. (2, 2). 30. A cube of edge 2 <j3r/3.

EXERCISE 32, P. 137 1. j> =2x , x+2y-5 =0. 2. j ; = x — 1, y+x = 1. 3. 2 j = x, j t + 2 j - 5 = 0 . 4. 2 χ + ^ - 4 = 0 , x - 2 j + 3 = 0 .

5. x->>V3 + 5/3=0> 3 X + J V 3 ~ 1 7 / 3 = 0 · 6. 3x-2>>- /?=0 , 7. 2 x - j - r = 0 . 8. x+y—3a = 0. 9. j = # l o g ax+a(l — log a).

10. 2 9 J C + 2 7 J ; - 6 = 0 . 11. (e 2 - l )x -2e j>+2c = 0.

13. 6x -5 j ;+21 = 0 , 14, χ = 2πα,

190 ANSWERS

15. k odd, y = x+eh", y = — x — ek"; k even, y = x —eft", y = — x+e*"1.

16. y = x — nr sj2/4, x+y = r y/2.

1. 3. 5. 7. 8.

10. 12. 14. 16. 18. 20. 22. e = 0 = 24. 26. 28.

V2/3. 173/2/2.

3 V3/64. Where x = ± 1 («4 + l)s/2/2a4. 1083/2/6.

53/2/4.

- 1 / 1 2 .

373/2/18.

c. fl/0.

2a.

EXERCISE 33, P. 143 2. p. 4. 8. 6. κ(0) = b/a\ κ(α) = a/ό2.

= 1/V2-9. 36.

11. 3m/8, 3m/2. 13. a sec (x/a). 15. 3V6. 17. (1 +4e2)3/2/3e. 19. 1. 21. 133/2/6. 23. R = 3a cos2 (0/3)/4. Where

: 0 or θ = 3π, x = a and y = 0, and i?max = 3a/4; where : 3π/2, x: fl.

a/3. \a\.

: 0, y = 0, and J?„ 0. 25. a(l+ θ2)3/2/(2+02).

27. j2mv2/4.

EXERCISE 34, P. 151 1. (x-15)2+0>-6)2 = 298. 2. x 2 +j 2 = l, x2+0>-2)2 = l. 3. (χ-2α/3)2+0-β/3)2 = 8α2/9. 4. 27/?j 2 = 8(x— pf, a semi-cubical parabola. 5. (ßx)2/3+(^)2/3 = (ö2-&2)2/3. 6. (tfx)2/3-(&>;)2/3= (α2+^2 ) 2 /3 .

ANSWERS 191

7. (x+y)2l*-(x-y)21* = (4α)*/». 8. x = 3(l+2t2-t*)l2,y= -4t\ 9. x = a log {y± y](y2-4a2)}l2a^y VO2-4a2)/4a.

10. x = a (arc cos t-Γ1 j(l -t2), y = a (log t-\). 11. x = a(t—unt)+na, y = a(\ — cos t)— 2a, which is the same cycloid translated.

12. x = a log tan y i, = ufcoseci =— al t a n y i + c o t y f

By making the substitution tan — t = e*/a we obtain the equa

tion of the evolute as the catenary y — a cosh (x/a). 13. x = 3<z(2 cos f—cos 2t), y = 3a(2 sin ί+sin 2t), a hypo-cycloid.

EXERCISE 35, P. 162

1. 2{s inx- ( s in2x) /2+(s in3 ;c ) /3 - . . . - f ( - l ) n + 1 (sin nx)/n

For x = ± π the series is zero. 2. 4a {sin x + (sin 3x)/3+(sin 5x)/5 + . . .}/π. 3. 4a {sin x + (sin 3x)/3+(sin 5x)/5+ . . ,}/π.¥οτχ = ±?rthe

Fourier series gives the value zero. 4. 4a {cosasinx+(cos3asin3x)/3+(cos5asin5x)/5 + ...}/7r. 5. π — 8 {cos x + (cos 3x)/32 + (cos 5x)/52 + . . .}/π

+ 8 {sin x - ( s i n 2x)/2+(sin 3*)/3 — ( - l ) n + 1 ( s i n w x ) / « + . . .}.

For x — ± π the Fourier series takes the value 2π. 6. 4a {sin a sin x + (sin 3a sin 3x)/32 + (sin 5a sin 5x)j52

+ . . .}/πα. 8. As a series of sines:

fix) = 3 {sin x + (sin 3*)/33+(sin 5x)/53+ . . .}/π. As a series of cosines:

f(x) = TC2/6-{COS 2x+(cos4x)/22+(cos 6x)/32+ . . .} .

9. 2 / π - 4 {(cos 2*)/l , 2 + ( C O S 4JC)/3 . 5-f(cos 6x)/5.7 + . . .}/π.

192 ANSWERS

10. 7i2/3-4 {cos x - ( c o s 2x)/22+(cos 3 x ) / 3 2 - . . .}. 11. 1/π-2 {(cos 2x)/l .3+(cos 4x)/3.5+(cos 6x)/5J

+ . . .}ln + (smx)/2.

12. f (-1)η(1/η*-2π21η) sin nx. n=l

13. 4{(2 sin 2*)/ l . 3 + (4 sin 4x)/3.5 + (6 sin 6x)/5.7 + 14. 2 sin απ {(sin x)/(l -a2)-(2 sin 2x) / (4-« 2 )+ . . .

+ (_l)n+l(n S i n w x ) / ( ^ 2 _ ß 2 ) + } / π

15. (sin x)/2 + £ ( - 1 ) Λ + Ι ( Λ sin ΛΛ:)/(2Λ — 3)(2/i — 1). n=2

16. cos x+(cos 3x)/3+(cos 5x)/5+ . . . . 17. cos x - ( cos 2x)/2 + (cos 3x)/3-(cos 4x)/4 + . . .. 18. cos x+(cos 2x)/2 + (cos 3x)/3+ 19. 2a sin ax{l/2a2-(cos x ) / ( a 2 - l ) + (cos 2x)/(a2-4)

+ ( - l ) n (cos nx)/(*2-n2) + . . .}/π. 20. 4{(2 sin 2x)/ l .3 +(4 sin 4x)/3.5 +(6 sin 6x)/5.7 +

21. l/rc + (sin χ)/2-(2/π) £ (cos 2nx)/(4n2 - 1 ) .

22. (450/π) Σ ( s i n (2* + ! W/(2« +1). 71 = 0

INDEX

Arc lengths 96 Areas 88

Centre of gravity 106 "Cover up" rule 25 Curvature, κ 137

centre of 144 formula 139

Definite integrals 74-127 areas 88 centre of gravity 106 definition 74 geometric interpretation 75 length of arc 96 moment of inertia 106 properties 76 relationship with indefinite inte

grals 77 surface area 100 volumes of revolution 100

DIRICHLET 153 Dummy symbols 78

Euler-Fourier formulae 152 Evolute 145 Extreme point 128

Fourier series 152-164 Function

defined as definite integral 77 primitive 1

Indefinite integrals 1-73 properties 4

Integration by parts 5, 77 by substitution 5, 78 inverse circular functions 68 irrational functions 31 logarithmic and exponential

functions 71 range of values 37 rational functions 13 trigonometric functions 51 see Definite integrals

Involute 145 Irregular integrals 118

Maxima and minima 128-132 Mean value theorem 76 Moment of inertia 106

Normal dissection 74

Pappus, theorems of 110 Partial fractions 24, 25 Primitive function 1, 77

Radius of curvature, ρ 138 formula 139 sign convention 140

Reduction formulae 27, 58, 84 Routh's rule 109

Surface area 100

Undetermined multipliers 42

Volumes of revolution 100 193

Documents

Problems and Methods in Analysis