1. Integrating by parts
Since #sec^3(x) = sec^2(x) sec(x)#,
#int sec^3(x)dx = int sec^2(x) sec(x)dx#
With #u=sec(x) <=> u'=sec(x)tan(x)#, #v'=sec^2(x) <=> v=tan(x)#, and #int uv'dx=uv-int u'vdx#, we have
#int sec^3(x)dx=tan(x)sec(x)-int sec(x)tan^2(x)dx#
Since #sec^2(x)-1=tan^2(x)#,
#int sec(x)tan^2(x)dx#
#= int sec(x)(sec^2(x)-1)dx#
#=int sec^3(x)dx-int sec(x)dx#
Hence
#int sec^3(x)dx#
#=tan(x)sec(x)-(int sec^3(x)dx-int sec(x)dx)#
#=tan(x)sec(x)-int sec^3(x)dx+int sec(x)dx#
If we add #int sec^3(x)dx# to both sides, we have
#2int sec^3(x)dx=tan(x)sec(x)+int sec(x)dx#
#int sec^3(x)dx=1/2(tan(x)sec(x)+int sec(x)dx)#
2. Integrating sec(x)
Now to figure out what #int sec(x)dx# is, you can either look it up in a formula sheet or derive it, as I will now.
Now there are a couple ways to derive this, but I will use the shortest and most common method for this.
#int sec(x)dx=int sec(x)/1dx#
Now if I multiply the numerator and denominator by #sec x+tan x#,
#int sec(x)dx=int (sec x(sec x+tan x))/(sec x + tan x)dx#
#=int (sec^2(x)+secxtanx)/(sec x+ tan x)dx#
Now let #u=sec x+tan x#, thus #du=(secxtanx+sec^2(x))dx#
#int (sec^2(x)+secxtanx)/(sec x+ tan x)dx#
#=int 1/u * (sec^2(x)+secxtanx)dx#
#=int 1/u du = lnabs(u)+C=lnabs(secx+tanx)+C#
3. Formulating the final answer
Hence, given #int sec(x)dx=lnabs(secx+tanx)+C#,
#int sec^3(x)dx=1/2(tan(x)sec(x)+int sec(x)dx)#
#=1/2(tan(x)sec(x)+lnabs(secx+tanx)) + C#