What is the integral of this function?

Question

$\int (\frac{e^{tan (3 x)}}{{cos}^{2} (3 x)}) d x$ $int ((e^tan(3x))/(cos^2(3x)))dx$

I'm confused about this problem

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Answer 1 · 2018-04-17T09:44:33

Is an inmediate integral. See below

You know that $d \frac{tan x}{d x} = {sec}^{2} x = \frac{1}{{cos}^{2} x}$ $d(tanx)/dx=sec^2x=1/cos^2x$

You also know that $inte^(u)u´du=e^(u)+C$

Appliying this, we have

$inte^(tan(3x))/cos^2(3x)dx=1/3e^(tan3x)+C$

Answer 2 · 2018-04-17T09:48:51

= $e^tan(3x)+C$

First
$color(blue)(1/cos^2(3x)=sec^2(3x)$

$inte^tan(3x)/cos^2(3x)dx=inte^tan(3x)sec^2(3x)dx$

$color(blue)(d/dxtan(3x)=sec^2(3x)dx$

$color(blue)(inte^udu=e^u+C$ $" "$ $color(green)(tan(3x)=u" " "in this case"$

Thus

$inte^tan(3x)sec^2(3x)$ = $e^tan(3x)+C$