# What is the integral of this function?

## $\int \left(\frac{{e}^{\tan} \left(3 x\right)}{{\cos}^{2} \left(3 x\right)}\right) \mathrm{dx}$ I'm confused about this problem

Apr 17, 2018

Is an inmediate integral. See below

#### Explanation:

You know that $d \frac{\tan x}{\mathrm{dx}} = {\sec}^{2} x = \frac{1}{\cos} ^ 2 x$

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

Appliying this, we have

$\int {e}^{\tan \left(3 x\right)} / {\cos}^{2} \left(3 x\right) \mathrm{dx} = \frac{1}{3} {e}^{\tan 3 x} + C$

Apr 17, 2018

=${e}^{\tan} \left(3 x\right) + C$

#### Explanation:

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

$\int {e}^{\tan} \frac{3 x}{\cos} ^ 2 \left(3 x\right) \mathrm{dx} = \int {e}^{\tan} \left(3 x\right) {\sec}^{2} \left(3 x\right) \mathrm{dx}$

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

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

Thus

$\int {e}^{\tan} \left(3 x\right) {\sec}^{2} \left(3 x\right)$=${e}^{\tan} \left(3 x\right) + C$