Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[graphic]
[ocr errors]

bave fa X=fa X'i=qX_1

fa Å'; again let X'=X":, and la

1 we shall have fa '=Q

X"

fa X", &c. Hence

" la la

1 fa X:=x

Las X' +za a X"_&c. till we arrive at an integral fa' ¿, which at least will be the most simple transcendental integral of its kind, if it be not susceptible of an exact integration.

82. We may observe that if e is the number whose logarithm =1, we shall have

se Xire X-X'+eX"_EX"+eX"-&c.
For example, let Xsx", and we shall get
X' = n, X" =n(1-1), X"=n(n-1) (-2)**, &e.
Therefore,

n (n-1)
+
(lajs

Class and consequently also

not hi-n )

[ocr errors]

as the preceding rules de

&c} 83. To find the integral of content not apply, I reduce « into a series, and obtain one (1+zla+**+ +04+&c.)= tola+*r +&c. Therefore

C++xa+4. **+ +&c. and
SiC++s+++++&c.

[ocr errors]

+

.

Let e=, and we shall have

the fluxion of a transcen

tal quantity which is equal to the infinite series C+1. lz+32 +33

+3*+&c.

Aur lz
Since = fx -=xUlz-Sillz, it is clear that site

.lz

+

2.3

2.3.4

z llz re an integral which likewise depends upon that of the transcendental quantity s

84. When the preceding rules will not apply to the integration of an exponential quantity, we must reduce it to a series by the formula

? Pa, ac3 Ba , t d=1+zla+

+ +&c. and it will be easily integrated.

Thus let y=si; by the above series į== {1 + mx Ls + m* x* x m3 3 13 +&e}=i+mw dle

+ &c, of which the fluent may be found by that of wow !" * (79), and we obtain from i=x (1–

ms

+&c.)

[ocr errors]

+

2.3

2

[ocr errors]

m

+

22

[ocr errors][ocr errors][ocr errors]

+ m x* lt (

+

&c.)

[ocr errors]
[ocr errors][ocr errors][merged small][ocr errors][merged small][merged small][merged small]
[ocr errors]

which in the particular case of x=1, becomes the converging serine

mo ms

+ 22' 38 These examples are sufficient to enable us to integrate such sorts of quantities by means of series.

+&c. 44

[ocr errors]

Examples on Logarithmic and Exponential Fluxions. 1. Required the fluent of its

? 3.

to Lx? 4.

(+*+2x+1):1st 5

Q .qcp 6.

- ez a

[graphic]
[merged small][merged small][ocr errors]

ON THE INTEGRATION OF FLUXIONS

CONTAINING SINES, COSINES, &c.

85. Since è cos x=flux sin x, and i sin r= flux cos , it is evident that si cos x= sin x, and së sin r=-cos r; as also that cos ny = Sny cos ny =

ny,

and similarly that fy sin ny cos ny. Again it is clear that cos z (sin z)"=(sin z)" flur

[ocr errors]

sin

n

[ocr errors]

1

n

1

1

(cos 2)"+1.

sin z. (sin z)*+1 , and thats (i sin 2) cos" z=n+1

n+1 Similarly if we desire to integrate y sin y cos ay, we must make sin y cos ay=sin(a+1) y-sin (a-1) y, and the integral 1

1 will become

cos (a +1)y+ cos (a1) y. 2 (a +1)

2 (a--1) 86. A similar process will serve for s sin x sin ax, or é cos x cos ax, &c. With equal facility we may integrate i sin r sin ax cos bx, &c. if we first reduce these products to simple sines or cosines, by means of the values of sin a cos b, sin a sin b, &c. given in the article trigonometry. The same treatment will serve to integrate i sin 'x, * sin 3x, * cos ?x, &c.; but it is more simple to integrate them in the following manner.

87. The formula i sin" :=x sin 2. sin") 1. Consequently, integrating by parts, sin" == sin) * Së sinx-;{ flux (sin+ ).

s -cos x sin3+(0–1) sé sin -2 cog-=sire-- ++ (n-1) / r sin 5-(n-1) si sın" r; and transposing, fi sin" = = cos x sin it fi sinas r. By the same

COS X

1

n

[merged small][merged small][ocr errors][merged small]

n

[ocr errors]

1 and consequently së sin" r=-- cos z sind x

1-1

cos o

n (n—2) sin-18+ (n-1) (n-3) ji sin* v=

1

cos i sin n (12—2) N--1

(n-1) (n-3) cos i sin3

cos r sin n (1-2)

*(1-2) (1-4)
(n-1) (n-3) (1-5)

cos x sin? 0—&C....-
1 (1-2) (1144) (n-6)
(11-1) (1-3) ... 2

cos x; a formula which is true only when a n (1-2) (1-4) ... 4 is an odd number, and then the integral depends simply on the quantities cos x, sin x. But when n is even, instead of the last term of the series, which would be of the form-

(n-1) (1-3) 1 cos I

2 (1–2)... O sin s we shall have + (n-1) (1—3)...1

(n-1) (n-3)...I fx sin*-* x=+ 2.4...(n-2) n

2.4....n
Consequently in this case the integral will be
1

n-1
cos x sinn-1-
cos x sina–3 1–&c.... +

(n-1) (1-3)...1 n (n–2)

2.4...n 4

4.2 Examples. få sinf x=-} cos x sino

cos x sin? 5.3

5.3 and Sc sino r=-; sin x cos x

5

5.3 sin' I cos

sin e cost 6.4

6.4.2 5.3.1 +

2.4.6

88. Make x=90°—2, and we shall have :=-2, and sin r=cos z and fż cosa z = 1 sin z cogz +

-1

sin x cos”—2 2 + n (1-2)

(n-1) (n-3) (145) sinz cosi, n(n-2) (144)

n(n-2) (114) (146) (n-1) (n-3) (1—5)...2 sin 2, if n is odd; and if even, +

-2) (1-4)...I the last term will be

(11—1) (-3) ... 1

n (1-2)...2 For example, si cos’y=sin y cobky +

4

sin y cos'y + 5.3

5.3.1

COS I;

n

(n-1)(n-3) sin 2 cog–5 % +

n

4.2 siny;

5

6.4

[ocr errors]

sine yy-9

[ocr errors]

we

sinm+1

y +

sin" +2

sin+1

1

sin"+1

cos' yt

--3

(m+1) (m+1-2)...M+2Vy sin"

and sy cos oy=sin y cos'y +

5.3
şin y cos syt sin y cos y +

6.4.2 5.3.1

y. 6.4.2

89. Let it now be required to integrate y sin y cos "y; since flux (sin py cos 'y) =p cos?+1

p+1 y

ein y

y y, have fýsin-! y cos2+1 sin py cos ry+1 cosy sinP+1

y. P

р

1 Therefore sin "y cos "y=.

1-1 y

cos m+1

m+1 cos®-y sin y. Substituting 1-cos'y in place of sin ’y, and transposing, we have sin my cos "y=.

m

y cos y +

m+n 7-1 - sin" y cos? y=

n=1

y
m+n
m+n

(m+n)(m+11-2)
y
cos"-

y

+&c.... + (m+n) (m+n2) (m+n—4) (n-1) (1-3) 2 sin

m+1

y, if n is odd; or if n is even the last (m+») (m+1-2)...m +1 term will be (n-1) (n-3) ... 1

y.

1 90. Make y=90°—2, we shall have si cosm : sin" z=

m+n

m+1 sin % cos'

(n-1) sina-5 % cos

(m+n) (m +1-2) (n-1) (1--3) com +1 z sin62

(n-1) (n-3)...2 cosm+1

&c.... (m+n) (m +16—2) (m+n-4)

(m+n)(m +1-2)...m +1 if n is odd, or to the term +(n-1) (n—3)...1 f2.com ::

if n is even. (m+n) (m +1-2)...(x+2) For example the first formula gives cos y sin' y=j sino y (cos' y +}) = sin" y (4 --sin’ y), and the second cos y sin y=-* cos 4y (sin *y+ i sin ’y+}). Hence these two results ought to be equal, or at least to differ only by a constant quantity. In the present case

, this quar:tity is a

as we shall find find by redncing the whole to sines, and comparing the two results:

sinti

y cosas y+ (n-1)(n-3) sinm+1

cosm +1

« ΠροηγούμενηΣυνέχεια »