# How to integrate 2x sec^2 x dx?

May 11, 2018

$\int 2 x {\sec}^{2} x \mathrm{dx} = 2 x \tan x + 2 \ln | \cos x | + C$

#### Explanation:

We have $\int 2 x {\sec}^{2} x \mathrm{dx}$.

Apply Integration by Parts, making the following selections:

$u = 2 x$

$\mathrm{du} = 2 \mathrm{dx}$

$\mathrm{dv} = {\sec}^{2} x \mathrm{dx}$

$v = \int {\sec}^{2} x \mathrm{dx} = \tan x$

$u v - \int v \mathrm{du} = 2 x \tan x - 2 \int \tan x \mathrm{dx}$

$\int \tan x \mathrm{dx} = \int \sin \frac{x}{\cos} x \mathrm{dx}$

We can solve this with a simple substitution:

$u = \cos x$

$\mathrm{du} = - \sin x \mathrm{dx}$

$- \int \frac{\mathrm{du}}{u} = - \ln | u | + C$
$= - \ln | \cos x | + C$

Thus,

$\int 2 x {\sec}^{2} x \mathrm{dx} = 2 x \tan x + 2 \ln | \cos x | + C$