502 MR. G. H. DARWIN ON THE PRECESSION OF A VISCOUS SPHEROID, 
Therefore 
f 1 dj 1 
A-Z 
k*-i; 
2'i _A 
2 (l+/*)\f 
— 1 
1 + 3/ A 
1 ++ / 
+ 
log 
-l+fid-jy 
I _ 
Hence the second approximation is 
N= l + n {(1 - g) +/3 (:! — P) + y ('! — f) } + i sin 2 i 
fin 
!++ G 
— 1 
1 . 1 + 3/x 
r 
+ i sill 
2 • 1^0 
/i +1 
log 
l+/dl-l)' 
( 88 ) 
It would no doubt be possible to substitute this approximate value of IV in terms of 
in the equation which gives the rate of change of obliquity, and then to find an 
approximate analytical integral of the first equation. But the integral would be very 
long and complicated, and I prefer to determine the amount of change of obliquity by 
the method of quadratures. 
In the present case it is obviously useless to try to obtain the time occupied by 
the changes, without making some hypothesis with regard to the law governing the 
variations of viscosity ; and even supposing the viscosity small but constant during 
the integration, the time would vary inversely as the coefficient of viscosity, and would 
thus be arbitrary. The only thing which can be asserted is that if the viscosity be 
small, the changes proceed more slowly than in the case winch has been already solved 
numerically. 
To return, then, to the proposed integration by quadratures : by means of the 
equation ( 88 ) we may compute four values of N (corresponding, say, to £=1, '96, *92, 
' 88 ); and since and - = — ~—, we may compute four equidistant values of all 
the terms on the right-hand side of the first of equations ( 86 ), except in as far as i is 
involved. Now i being only involved in small terms, we may take as an approximate 
final value of i that which is given by the solution of Section 15, and take as the four 
corresponding values \ 
o ■ , n {i-h) 
A+ 2 “ 
o o 
Hence four equidistant values of the right-hand side may be computed, and com- 
rzh 3 ji 
billed by the rule u x dx=—[u 0 -\-u 3 -\- 3(^ 1 +%)j, which will give the integral of the 
Jo 8 
'll % 
right-hand side from ^ to 1 ; and this is equal to log tan 2 — — log tan 2 
The integration was divided into a number of periods, just as in the solution of 
Section 15. The following were the results : 
