# How do you integrate sec^3(x)?

$\textcolor{red}{\int {\sec}^{3} x \cdot \mathrm{dx} = \frac{1}{2} \cdot \sec x \cdot \tan x + \frac{1}{2} \cdot \ln \left(\sec x + \tan x\right) + C}$

#### Explanation:

This is done using Integration by Parts

$\int u \cdot \mathrm{dv} = u v - \int v \cdot \mathrm{du}$

Let $u = \sec x$
Let $\mathrm{dv} = {\sec}^{2} x \cdot \mathrm{dx}$
Let $v = \tan x$
Let $\mathrm{du} = \sec x \cdot \tan x \cdot \mathrm{dx}$

Use the formula

$\int u \cdot \mathrm{dv} = u v - \int v \cdot \mathrm{du}$
$\int \sec x \cdot {\sec}^{2} x \cdot \mathrm{dx} = \sec x \cdot \tan x - \int \tan x \left(\sec x \cdot \tan x \cdot \mathrm{dx}\right)$

$\int {\sec}^{3} x \cdot \mathrm{dx} = \sec x \cdot \tan x - \int {\tan}^{2} x \cdot \sec x \cdot \mathrm{dx}$

Recall ${\tan}^{2} x + 1 = {\sec}^{2} x$
and ${\tan}^{2} x = {\sec}^{2} x - 1$

$\int {\sec}^{3} x \cdot \mathrm{dx} = \sec x \cdot \tan x - \int \left({\sec}^{2} x - 1\right) \sec x \cdot \mathrm{dx}$

$\int {\sec}^{3} x \cdot \mathrm{dx} = \sec x \cdot \tan x - \int \left({\sec}^{3} x - \sec x\right) \cdot \mathrm{dx}$

$\int {\sec}^{3} x \cdot \mathrm{dx} = \sec x \cdot \tan x - \int {\sec}^{3} x \cdot \mathrm{dx} + \int \sec x \cdot \mathrm{dx}$

Transpose the right $\int {\sec}^{3} x \cdot \mathrm{dx}$ to the left side of the equation

$\int {\sec}^{3} x \cdot \mathrm{dx} + \int {\sec}^{3} x \cdot \mathrm{dx} = \sec x \cdot \tan x + \int \sec \mathrm{dx}$

$2 \cdot \int {\sec}^{3} x \cdot \mathrm{dx} = \sec x \cdot \tan x + \ln \left(\sec x + \tan x\right)$

Divide both sides by $2$

$\textcolor{red}{\int {\sec}^{3} x \cdot \mathrm{dx} = \frac{1}{2} \cdot \sec x \cdot \tan x + \frac{1}{2} \cdot \ln \left(\sec x + \tan x\right) + C}$

God bless....I hope the explanation is useful.