Page images
PDF
EPUB
[merged small][merged small][ocr errors][merged small][merged small][ocr errors][merged small]

1

by (6),
(cos x)
} (tan a)?
1

+ log tan
(cosa) 2 (cos x)
(sin x)'

+ (sin x)2} + log (sec x) by (3).

[ocr errors]

+

dx (sin x)

1

COS C

[ocr errors]

s da

dx (cos x)" (cos x) 3

+ cos x + log (tan

sin a

3

[ocr errors]
[ocr errors]

:-1}

1

[ocr errors]

s do

[ocr errors]

8

dx (sin x)3

= COS X + sec X.
(cos x)'
dw (cos x)' 1

{(cos x) – 3 cos }

3 cos x} - x.
(sin x)

2 sin 2
dx (cos x)' 1 (cos x)

- 2 log (sin x).
(sin x)) (sin x) 2
dx (sin x)

{(sin x) – *}
(cos x)'

(cos x)
dx (sin x) 1

{(sin x) - 3}.
(cos x) 5 (cos x)?
dx

1

+ log (tan x).
2 (cos x)
du

1

cot 2 x.

s
(sin x)' (cos x)' 3 sin æ (cos x)
dc
8 cos 2 ir

1

2
(sin x)* (cos x)*

3
dx
1

+ log (tan).
(sin v)cos x 4 (sin x)" 2 (sin x)2

(17) If the function be (tan x)" the formula of reduction is

x sd « (tan v)" =

- sdx

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

1

1

1

1

If the function be

the formula of reduction is

(tan x)"
dc
s

sda
(tan x)" (n − 1) (tan x)n-1 (tan x)»-%"
Sdx (tan r)' = }(tan x)?

}(tan x)' – tan x + x. sdx (tan x) = (tan x)" - (tan x) + ! (tan x)2 + log (cos x).

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

sdx x" cos x = a" sin x + nu-l cos x n (n − 1) sd x æ»–2 cos x. Sdx xé cos x = x* sin x + 2.0 cos x - 2 sin x. sdx x3 cos x = x2 sin x + 3.c cos x 6x sin x 6 cos x.

In the same way we find sdx x sin x =- X cos X + sin a'. sdæ æ* sina=-x* cos x + 4 xsin x +12xocos & 24 x sin x – 24 cos18.

(19) If the function bee" (cos x)" the formula of reduction is

sd ex eo * (cos.')"
(cos x)"-'(
n-'(a cos x +n sin x) n (n − 1)

sda e* (cos x)"-2; a + na

a + n° a similar formula exists for “(sin x)".

(a cos a + 2 sin x) sdx e" (cos x)' = "" cos x

a + 4

a (a+ 1) (a sin r- 3 cosa) 66' (a sin x-cosx)

(a* + 1) (a’ +9)

[ocr errors]
[ocr errors]

261

+

Sdxe** (sin a)* = {** (sin a) = (a sin 2–3 cos x)

+

a” + 9

61

a sin 7x – 7 cos 7 & sd x 6" * (sin x)" (cos a')

61

a + 49 3(a sin 5.2-5 cos 5.0) a sin 32-3 cos 3x 5 (a sin x-cos +

+ a“ + 25

u" + 9

a? + 1

{

[ocr errors]

(20) If the function be

1
(a + b cos x)"

the formula of

reduction is

dx

- b sin x

Sta + 6 cos x)= = (n − 1) (c* = ) (a + b cos 2)* - 1 (n-1)(a-69%(a+b cosa)*-*_(n-1)02–65/(a+bcos a )*

(2n-3) a

dx

(n-2)

dx

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

1

[a

a sin x
+ b cOS

[ocr errors]
[ocr errors]

tan 1

a' - 62

[blocks in formation]

In order that a differential function of two variables of the first order, such as

Pdx + Qdy, should be the differential of a function u, it is necessary that the condition

dP dQ

dx

dy

[blocks in formation]

The application of these formulæ may be generally facilitated by observing that in the second term of the former it is only necessary to integrate the terms in Q which involve x only, and in the latter those terms of P which involve y only.

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

therefore

dP

=0

dQ

Integrating with respect to y,

[ocr errors]

(1 + x2) therefore the integral is

u = by® + ax + log C {x + (1 + x^)"}.

wy dy - y'da (2) Let

du. w* (ava + yo) Integrating with respect to y, and observing that there is no term in P involving y only, we find

(.x2 + yo)

[ocr errors]

+ C.

(3) Let

dx
(v? + y)

dy +

y

ædy
(x2 + y)

- du,

dQ

Q (20* + y)

Y (x2 + y) dP

y

dy (x2 + yo)? d.x
Since P does not contain any term independent of x,

d
Sdx (P SQdy) = const.;


therefore, integrating with respect to y,

X + (x2 + y)) u = log y + log

+ C;

U =

whence log C {x + (2x2 + y)!}. (4) Let (aʼy + xy) dx + (b3 + a+x) dy = du. The integral of this is

[blocks in formation]

(5) Let (3xy® – xo) dx – (1 + 6y 3.voy) dy = du;

dP

dQ then 6xy

« PreviousContinue »