[KING] ABSORPTION PROBLEMS IN RADIOACTIVITY 99 
rc 2: 
which the plane through PA and the perpendicular makes with a fixed 
direction in the plane of the slab. We then have 
7. — Ab —h/ cose 7, — PA — 2/cosp and d2 — sing d¢ av. 
Thus (4) gives on integrating with respect to y between 0 and 27, 
m—zsec 
2a Ns 5 ? if —Kh sece | : 
js Seas e Pe SCOOUNO: Etats à (5) 
Die à 
0 
We note that the integral 
T —X secy Comes 
2e sing dg = x e du, 
wv 
o x 
by the transformation cose — —; integrating by parts, 
u 
T co —U 
2 x seco —T e du 
e sing dg = e — 2 u 
0 x 
CO 77 SL 
The last integral 2d == e du has been 
u u 
