according as a is less or greater than unity.

291. Similarly from the known expansion

1- a2

1-2a cos x + α2

= 1 + 2a cos x+2a2 cos 2x + 2a3 cos 3x +..............,

=

....

where a is less than 1, we may deduce some definite integrals; thus if r is an integer

for every term that we have to integrate vanishes with the

assigned limits, except 2a" cos2 rx dx.

292. To find the value of
of [14

1

0

dx

1+x2 1-2a cos cx+ a2°

The term 1−2a cos cx+a2 may be expanded as in Art.

291; then each term may be integrated by Art. 286, and the results summed. Thus we shall obtain

This also follows from Art. 293, by differentiating with respect to c.

By expanding (1-x) ̄1, we find for the integral a series of which the type is

By integration by parts this is seen to be equal to

Expand the factor {1 + (cos x)2}1, and we find for the integral a series of which the type is

to

0

By integration by parts this may be shewn to be equal

(-1)"π

297. Let v denote e√(-1), that is, cos x + √(− 1) sin x ; then if ƒ denote any function, we have by Taylor's Theorem,

f(a + v) +ƒ(a + v ̄1)

And

1-c2

f"

= 2 {ƒ(a) + ƒ′ (a) cos x +3 (9) cos 2r+

1.2

......

1+2c cos x + 2c2 cos 2x + 2c3 cos 3x + ....

In this result it must be remembered that c is to be less than unity, and the functions f(a+v) and ƒ(a+v ̄1) must be such that Taylor's Theorem holds for their expansions.

In a similar way it may be shewn that

・ƒ (a+v) −ƒ (a + v1)

cos x + c2

π √(-1)

sin x dx =

{f(a+c) − f(a)}

с

Substitution of imaginary values for Constants.

298. Definite integrals are sometimes deduced from known integrals by substituting impossible values for some of the constants which occur. This process cannot be considered demonstrative, but will serve at least to suggest the forms which can be examined, and perhaps verified by other methods, (see De Morgan's Differential and Integral Calculus, p. 630). We will give some examples of it.

For p put a+b√(−1), and suppose r = √(a2+b2) and p=r {cos +√(-1) sin 0); thus

tan 0

b

==
α

so that

n-1

e-{a+b√(−1)} x x2-1 dx=r" {cos no - √(-1) sin n0} г (n). − T

For modes of verification see De Morgan, p. 630.

е

0

300. In the integral (*e-(*+*)* dæ, suppose y = x√/k;

Now put cos +√(-1) sin 0 for k; thus the right-hand member becomes

2

icos e

Thus [e (+) cos {(a + 2) sin e} de

е