# How do I evaluiate intsec(x)(sec(x) + tan(x)) dx?

May 1, 2018

$\int \sec x \left(\sec x + \tan x\right) \mathrm{dx} = \sec x + \tan x + c$

#### Explanation:

I don't believe that there is any trigonometric substitution involved here actually.

The integrand $\sec x \left(\sec x + \tan x\right)$ can be expanded out to get ${\sec}^{2} x + \sec x \tan x$ which are actually two simple functions to be antidifferentiated.

$\int {\sec}^{2} x \mathrm{dx} = \tan x + c$ because $\frac{d}{\mathrm{dx}} \left(\tan x\right) = {\sec}^{2} x$.

$\int \sec x \tan x \mathrm{dx} = \sec x$ which we will prove using substitution.

Expressing that in terms of sine and cosine, we get $\sec x \tan x = \sin \frac{x}{\cos} ^ 2 x$ so the integral becomes:

$\int \sin \frac{x}{\cos} ^ 2 x \mathrm{dx}$, to which we apply the substitution $u = \cos x$.

$u = \cos x$
$\therefore \frac{\mathrm{du}}{\mathrm{dx}} = - \sin x$
$\therefore - \mathrm{du} = \sin x \mathrm{dx}$

Now, using the change of variable rule, we get:

$- \int \frac{1}{u} ^ 2 \mathrm{du}$
$= \frac{1}{u} + c$
$= \sec x + c$

$\therefore \int \sec x \left(\sec x + \tan x\right) \mathrm{dx} = \sec x + \tan x + c$

And there you have it!