How do you integrate $\int sec 3 x$ $intsec3x$ ?

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How do you integrate $\int sec 3 x$ $intsec3x$ ?

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Answer 1 · 2017-02-08T21:55:44

$\int sec (3 x) d x = \frac{1}{3} ln | sec (3 x) + tan (3 x) | + C$ $intsec(3x)dx=1/3lnabs(sec(3x)+tan(3x))+C$

Explanation:

$\int sec (3 x) d x$ $intsec(3x)dx$

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

$= \frac{1}{3} \int sec (3 x) (3) d x = \frac{1}{3} \int sec (u) d u$ $=1/3intsec(3x)(3)dx=1/3intsec(u)du$

This is a common integral: $\int sec (u) d u = ln | sec (u) + tan (u) | + C$ $intsec(u)du=lnabs(sec(u)+tan(u))+C$

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

$= \frac{1}{3} \int sec (u) \frac{sec (u) + tan (u)}{sec (u) + tan (u)} d u = \frac{1}{3} \int \frac{{sec}^{2} (u) + sec (u) tan (u)}{sec (u) + tan (u)} d u$ $=1/3intsec(u)(sec(u)+tan(u))/(sec(u)+tan(u))du=1/3int(sec^2(u)+sec(u)tan(u))/(sec(u)+tan(u))du$

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

$= \frac{1}{3} \int \frac{d v}{v} = \frac{1}{3} ln | v | + C = \frac{1}{3} ln | sec (u) + tan (u) | + C$ $=1/3int(dv)/v=1/3lnabsv+C=1/3lnabs(sec(u)+tan(u))+C$

Finally:

$= \frac{1}{3} ln | sec (3 x) + tan (3 x) | + C$ $=1/3lnabs(sec(3x)+tan(3x))+C$

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