( 495 ) 
If now we multiply this equation by 
we integrate between the limits 0 and a we find 
1 
pn 1 p li nva en 
= n I, (2) 1, (0) = = Cy 2 (1 — t°) ie sin (a sin p) dep 
1.3 JT dp 
0 
a, 1 ic 
od ap al cay 
na co 
0 
Putting for the further reduction 
ny 
u =x cos (a sin p) cos p dp 
0 
cos” p cos (ct sin p) : 
== Ct 
1—2 cos 2p” 
0 
we arrive at 
* cos? psin(asing) , 
rend = | ———_—————_ sing dg, 
dea 1—2 #? cos 2g Ht! 
“cos? pcos(asing) . . 1 
en a = — sin P ap 
da’ 1—2 ¢? cos 2g tt / 
0 
and because 
EN rl et cos 2p tt’ (1 - 2)? 
se? (QS eS 
. At At 
we find 
du à iL aigere ; 
re =mu— raf p cos (a sin p) dp, 
0 
1—?? 
where m= 
2t 
1 + cos 2p 
If we replace cos’ p by = ga we. can easily reduce this 
differential equation to 
au A n 7 7 
ei NEET a a 
_ sa Ho (a) + 1, (0) 
xz I, (a) 
JE 
sin (a@ sin p) dp and if 
7 
(1 — #?) | RR cos (a sin ~) cos p dp. 
