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{.v + 1 + x®}}.* (36) Let the integral be 5d x

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(1 + 20°) assuming x + (1 + x')} = x", the transformed integral becomes

nsdz 20-12 {«+ (1 + x°)}}".

n

n

m

m

(37) If du =

(1 + x) dx

we have by assuming (1 – v*) (1 + x^)!'

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Dividing both sides of this equation by

1-2
= (1 - x)!,

– ,

1 1 + V

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(10) By the same assumption we find that
da r °

1 1(1+x')? + 2
log

sin (1 – 2*)(1+x')!

V?

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1

-1

")"

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1

assume

dx
(41) If du

(1 – 2”) (2x? – 1)!'
x = = (2x - 1)], when it becomes

dz
du

- 1

; and therefore
(22? – 1)?
U = log

tan-1
(2.x" – 1)' + x)

(2.x® – 1)?

1

2

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U = tan

{(1 + x^)! – 1937 (43) In like manner by assuming x = z{(1 + 2°)} – 2*}", we find

dr
ra

{(1 + x2") } – 13"}! These transformations are taken from Euler, Calc. Int Vol. iv. Sup. I.

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-1 = tan

CHAPTER II.

INTEGRATION BY SUCCESSIVE REDUCTION.

The method of integration by successive reduction is applicable to a great number of functions, and is the process which in practice is generally the most convenient. I shall here only give the principal formulæ of reduction with a few examples of each, taken chiefly from those integrals which more commonly occur in analysis. The reader who wishes for more numerous examples of the formula is referred to the Integral Tables compiled by Meyer Hirsch, from which work a great number of the examples in this and the preceding Chapter have been taken.

Ex. (1) Let the function to be integrated be

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s

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+

a

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n

n

(do – °) The formula of reduction is

da v" 22-' (a’ – x?)? n - 1 down - 2
(a– 2*)

(a– x*)!
By this the integral is reduced to
dc

when
(a' - x)

xdæ
and to

(a– xo)? when n is odd.
(ax?
Let n = 3;
rdx (a’ – x?)

(a” + 2a).
(a-x)!

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= sin-1

is even,

a

3

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2

-1 = vers

a

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By means of this the integral is made to depend on

dx

(2ax – 2*) Let n =

2; x2 da

3 a

3a2
· )?

+ vers
(2ax
X-!

2 2
Let n = 5;
de

9

9.7 ax! +

a2 x2 5.4

5.4.3

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-1

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a

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(2ax – 29); (

+

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1

(3) Let the function be

(a + x*)*
The formula of reduction is
dir
1

2n-3 1

+
(a’ + x^)" 2n - 2 (a+x*)n-1 2n
By this the integral is reduced to

dx 1

tan a"

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-1

+ x2

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rdxc

1 1 +

+ (a + x) 4 (a + *) 4.2 a’ (a2 + 2)

4.2 a 3 Let n = 4, m = 2; X* dx

23

X - a tan 2 (a' + @*)

2

+

%. (5) Let the function be (a* – 27), n being odd ; a– dx +2***, s(@

).

n+1

n + 1

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(1) مہک

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n

Q-1

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1 (6) Let the function be

" (v? – 1)
dx
1 (x2 – 1)? n - 2

do

+ x" (v - 1)

1
n-

(w? – 1)! By this means the integral is reduced to

do

= sec-1x when n is odd,
X (ir- 1)
dc

1) and to

when n is even. x? (x® – 1)?

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