Primitive sec x and sec³ x

1. Sabemos que:
d tan x sec x+sec 2 x tan x +sec x
ln∣sec x+tan x∣ = = sec x = sec x (1)
dx sec x+tan x sec x +tan x
isso implica
∫ sec x dx = ln∣sec x +tan x∣ + C (2)
Agora podemos escrever
∫ sec3 x dx = ∫ sec2 x sec x dx = ∫ (1+tan2 x )sec x dx = ∫ sec x dx+∫ tan 2 x sec x dx (3)
e integrando por partes obtemos
∫ sec3 x dx = ∫ sec2 x sec x dx = tan x sec x − ∫ tan2 x sec x dx (4)
Somando (3) e (4) e aplicando (2), obtemos
1 1
∫ sec3 x dx =
2
ln∣sec x tan x∣ + tan x sec x + C
2
(5)