Integrals of the functional form [1/(a sin x + b cos x +c)]dx

Put sin x = (2 tan x/2)/(1 + tan² (x/2))

cos x = (1 - tan² (x/2))/(1 + tan² (x/2))

Replace (1 + tan² (x/2)) by sec²(x/2)

Put tan (x/2) = t

dt = 1/2 sec²(x/2)dx

The integral reduces to ∫[1/(at² +bt +c)]dt