How do you integrate #sec^3x "d"x#?

For #intsec^3x"d"x#, use integration by parts.

Let #u=secx# and #"d"u=secxtanx"d"x#

and #"d"v=sec^2x"d"x# thus #v=tanx#

Now we plug this into the formula

#intu("d"v)/("d"x)"d"x=uv-intv("d"u)/("d"x)"d"x#

So

#intsec^3x"d"x=secxtanx-intsecxtan^2x"d"x#

#=secxtanx-intsecx(sec^2x-1)"d"x#

#=secxtanx-intsec^3x"d"x+intsecx"d"x#

Thus

#2intsec^3x"d"x=secxtanx+intsecx"d"x#

So

#2intsec^3x"d"x=secxtanx+lnabs(secx+tanx)+"C"#

So

#intsec^3"d"x=1/2(secxtanx+lnabs(secx+tanx))+"c"#

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#