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)

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Prim1_secx

  • 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)