# How do I find the integral of f(x)=sec^5(x)?

Mar 8, 2018

$\int {\sec}^{5} x \mathrm{dx} = \frac{2 {\sec}^{3} x \tan x + 3 \sec x \tan x + 3 \ln \left\mid \sec x + \tan x \right\mid}{8} + C$

#### Explanation:

Note that in general for any $n \ge 1$:

${\sec}^{2 n + 1} x = {\sec}^{2 n - 1} x {\sec}^{2} x$

and:

${\sec}^{2} x = \frac{d}{\mathrm{dx}} \tan x$

so we can integrate by parts:

$\int {\sec}^{2 n + 1} x \mathrm{dx} = \int {\sec}^{2 n - 1} d \left(\tan x\right)$

$\int {\sec}^{2 n + 1} x \mathrm{dx} = {\sec}^{2 n - 1} x \tan x - \int \tan x d \left({\sec}^{2 n - 1} x\right)$

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

Use now the trigonometric identity:

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

to get:

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

and using the linearity of the integral:

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

the integral now appears on both sides and we can solve for it:

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

$\int {\sec}^{2 n + 1} x \mathrm{dx} = \frac{{\sec}^{2 n - 1} x \tan x}{2 n} + \frac{2 n - 1}{2 n} \int {\sec}^{2 n - 1} x \mathrm{dx}$

So, for $n = 2$:

$\int {\sec}^{5} x \mathrm{dx} = \frac{{\sec}^{3} x \tan x}{4} + \frac{3}{4} \int {\sec}^{3} x \mathrm{dx}$

and for $n = 1$

$\int {\sec}^{3} x \mathrm{dx} = \frac{\sec x \tan x}{2} + \frac{1}{2} \int \sec x \mathrm{dx}$

Now:

$\int \sec x \mathrm{dx} = \int \sec x \frac{\sec x + \tan x}{\sec x + \tan x} \mathrm{dx}$

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

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

$\int \sec x \mathrm{dx} = \ln \left\mid \sec x + \tan x \right\mid + C$

Putting the partial results together:

$\int {\sec}^{5} x \mathrm{dx} = \frac{2 {\sec}^{3} x \tan x + 3 \sec x \tan x + 3 \ln \left\mid \sec x + \tan x \right\mid}{8} + C$