216 
SIR G. H. DARWIN ON THE 
In order to find Ai 1 we expand A in powers of k' j taw' y. Thus 
Ad = k cot 2 y r dy = k cot 2 y (1-^ /2 tan 2 y+ f/c' 4 tan 4 y ...) dy\ 
whence 
Ad = 
K 
2 sm 
7 
[[1 + |k /2 + Hk /4 ] [1 -COS 2 yn]-K tan 2 y (1 +|k /2 ) + |-k ' 4 tan 4 y}. 
The result may, of course, he obtained with higher accuracy if it he desired to do so. 
By (25) of the ‘ Pear-shaped Figure ’ 
where Ad = l-c/ 2 sin 2 y, with appropriate values of q for s = 0 and 2 respectively. 
It is clear that under the integral sign l/A may be expanded in powers of k 2 tan 2 y, 
hut this is not possible with l/Ad- 
The two cases s = 0, 2 have to be treated by different methods, and I begin with 
^3, where s = 0. 
Since 
— = —— [1 +i/c' 2 tan 2 y+ §k ' 4 tan 4 y...], 
A cos y 
it is necessary to consider the integrals with respect to y of the functions 
sin 8 y 
sin 15 y 
sin 10 y 
cos y Ad ’ cos 3 y Ad ’ cos" y Ad 
Writing x — sin y, and cf = this is seen to be equivalent to findin 
x 6 dx 
f x s dx f 
Jfl — x 2 ! 2 (a 2 —x 2 ) 2 ’ 
x 10 dx 
Now 
X 
(1-x 2 ) (a 2 -x 2 ) 2 ’ J(l— x 2 ) 2 (a 2 —x 2 ) 2 ’ J (1 —x 2 ) 3 (a 2 —x 2 ) 2 ' 
1 / 1 1 \ a 4 / 1 1 
(1 —x 2 ) (a 2 —x 2 ) 2 
5x3 = -! + 
2 (a 2 —l) 2 \l—x 1+x/ 4 (a 2 —1) \(a—x) 2 (a + x) 2 / 
+ 4 (a 2 —!) 2 \a—x + a+x) ’ 
a 3 (3a 2 - 5), 1 , 1 \ 
x 
(1-x 2 ) 2 (a 2 —x 2 ) 
3x5= 1 
7a 2 -3 / I | 1 \ 
+ 
1 
4 (a 2 —l) 3 \1 —x 1+x/ 4 (a 2 —l ) 2 \(1 —x ) 2 ( 1 +x ) 2 
1 
1 
a 5 (3a 2 -7)/ 1 
1 \ 
a 
4 (a 2 — l) 3 \a—x a + xj 4 (a 2 —l) 2 \(a—x ) 2 (a + x)-/ ’ 
x 
,10 
(1 —x 2 ) 3 (a 2 —x 2 ) 
3 (21a 4 —18« 2 +5) / 1 
Wi _ 1 + i a In* iV V l - 
1 \ 17a 2 -9 
1 
16 (a 2 — l) 4 \ 1 —x 1+x/ 16(a 2 —1) 3 \(1—x) 2 (1+x) 2 
1 / 1 , 1 \ , 3a 7 (a 2 -3) / 1 , 1 
+ 
8 (a 2 —l) 2 \(l —x ) 3 (1+x) 3 / 4 (a 2 —l) 4 \a—x a+x 
a 
1 
1 
4 (a 2 — l) 3 \(a—x ) 2 (a + x) 2 / 
