How do you integrate $intsec3x$ ?

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Answer 1 · 2017-02-08T21:55:44

$intsec(3x)dx=1/3lnabs(sec(3x)+tan(3x))+C$

$intsec(3x)dx$

Use the substitution $u=3x$ , implying that $du=(3)dx$ . Then:

$=1/3intsec(3x)(3)dx=1/3intsec(u)du$

This is a common integral: $intsec(u)du=lnabs(sec(u)+tan(u))+C$

We can derive this integral by multiplying the integrand by $(sec(u)+tan(u))/(sec(u)+tan(u))$ :

$=1/3intsec(u)(sec(u)+tan(u))/(sec(u)+tan(u))du=1/3int(sec^2(u)+sec(u)tan(u))/(sec(u)+tan(u))du$

Now let $v=sec(u)+tan(u)$ . This implies that $dv=(sec(u)tan(u)+sec^2(u))du$ . This is the numerator:

$=1/3int(dv)/v=1/3lnabsv+C=1/3lnabs(sec(u)+tan(u))+C$

Finally:

$=1/3lnabs(sec(3x)+tan(3x))+C$

How do you integrate intsec3xintsec3x?