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.

$intsec^3xtanxdx$

$u=secx$

$du=secxtanxdx$

$=>intu^2du$

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

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

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