59] CALCULATING THE TRAJECTORIES OF SHOT. 483 



Now let 



F((f>) =f(<f>) (sec <)'" = u' (sec <f>) m+n+1 , 



then 



and 



i =/" (<) (sec $) m + 2mf (<f>) (sec <j)) m tan < 



+ mf(<f)) \_m (sec tj>) m (tan ^>) 2 + (sec ^>)" l+2 ] 

 (sec ^>)" 1 + 2rnf' (^>) (sec ^>) m tan <^> 



+ w l/"(^>) [w + 1 (sec (f>) m+ ~ m (sec <) m ]. 

 Hence, by the lemma, 



c y)" ! + (a - ^8)= (sec y)' /" (y) + 2m/( y ) tan y 



-i-| 



+ m/"(y) [m + 1 (sec y) 2 - w] 



= (a - /8) (sec y ) |/(y) + 1 (a - /3) 2 [/' (y) + 2m/(y) tan y 



_ 

 /"(y) [m + 1 (sec y) 2 TO] 



But from above 



k(l-: 



Hence, by division, 



= (secy) m jl+ (a-/8) 2 2m ^^ tan y + m [wT+T (sec y) 2 - m]l I 



It will be noticed that in this division the quantity / (y) has disappeared. 



612 



