# How do you find the integral of int 2secxtanx dx?

Sep 26, 2015

2secx+C

#### Explanation:

$I = \int 2 \sec x \tan x \mathrm{dx} = 2 \int \left(\sec x\right) ' \mathrm{dx} = 2 \int d \left(\sec x\right)$

$I = 2 \sec x + C$

Explanation:

$\left(\sec x\right) ' = \sec x \tan x \implies 2 \sec x \tan x = 2 \left(\sec x\right) '$

$I = \int 2 \sec x \tan x \mathrm{dx} = 2 \int \left(\sec x\right) ' \mathrm{dx}$

Differential of a function is defined as:

$d \left(f \left(x\right)\right) = f ' \left(x\right) \mathrm{dx} \implies d \left(\sec x\right) = \left(\sec x\right) ' \mathrm{dx}$

$I = 2 \int \left(\sec x\right) ' \mathrm{dx} = 2 \int d \left(\sec x\right)$

Elementary formula from integration theory:

$\int d \left(f \left(x\right)\right) = f \left(x\right) + C$

$I = 2 \int d \left(\sec x\right) = 2 \sec x + C$