sec x tan x

Indefinite integral of sec x tan x

∫ sec x tan x dx = ∫ (1/cos x)(sin x/cos x) dx

= ∫ sin x/cos² x dx = ∫ sin x/(1 – sin² x) dx

Now use substitution, let u=sin x  =>  du=cos x dx  and  dx= du/cos x = du/√1-u²

= ∫ u/(1 – u²)^3/2 du

Now use substituition again let a=u² =>  da =2u du  and u du = 1/2 da

= 1/2 ∫ (1 – a)^-3/2 da

=  (1-a)^-1/2 + constant(=k)

but a=u² and u=sin x  => a=sin² x

= (1 – sin² x)^-1/2 = 1/√cos² x

= 1/cos x = sec x

Pheewiit..  ∫ sec x tan x dx = sec x + constant(=k)