How do you integrate #int sec^3xtanx#?

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Answer 1 · 2016-12-24T00:24:03

#intsec^3(x)tan(x)dx=sec^3(x)/3+C#

Let #u = sec(x) => du = sec(x)tan(x)dx#

then

#intsec^3(x)tan(x)dx = intsec^2(x)(sec(x)tan(x))dx#

#=intu^2du#

#=u^3/3+C#

#=sec^3(x)/3+C#

Answer 2 · 2016-12-24T00:25:59

See below.

I would use a simple #u# substitution.

#intsec^3xtanxdx#

#u=secx#

#du=secxtanxdx#

#=>intu^2du#

#=>1/3u^3+C#

#=>(sec^3x)/3 +C#