# How do you find the integral of sec^5 x dx?

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Aug 29, 2017

$\int {\sec}^{5} x \mathrm{dx} = \frac{1}{4} {\sec}^{3} x \cdot \tan x + \frac{3}{8} \sec x \tan x + \frac{3}{8} \ln \left\mid \sec x + \tan x \right\mid + C$

#### Explanation:

at first we will break $\int {\sec}^{5} x \mathrm{dx}$ into $\int {\sec}^{3} x \cdot {\sec}^{2} x \mathrm{dx}$ and by using $\int u \mathrm{dv} = u v - \int v \mathrm{du}$ (integration by parts) we will take $u = {\sec}^{3} x$ and $\mathrm{dv} = {\sec}^{2} x \mathrm{dx}$.

These show that $\mathrm{du} = 3 {\sec}^{2} c x \left(\sec x \tan x\right) = 3 {\sec}^{3} x \tan x$ and $v = \tan x$.

Then:

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

Grouping ${\tan}^{2} x$ and rewriting as ${\sec}^{2} x - 1$:

$= {\sec}^{3} x \tan x - \int 3 {\sec}^{3} x \left({\sec}^{2} x - 1\right) \mathrm{dx}$
$= {\sec}^{3} x \tan x - \int \left(3 {\sec}^{5} x - 3 {\sec}^{3} x\right) \mathrm{dx}$

So

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

And we know that$\int {\sec}^{3} x \mathrm{dx} = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln \left\mid \sec x + \tan x \right\mid$

So

$4 \int {\sec}^{5} x \mathrm{dx} = {\sec}^{3} x \cdot \tan x + 3 \left(\frac{1}{2} \sec x \tan x + \frac{1}{2} \ln \left\mid \sec x + \tan x \right\mid\right)$

$\int {\sec}^{5} x \mathrm{dx} = \frac{1}{4} {\sec}^{3} x \cdot \tan x + \frac{3}{8} \sec x \tan x + \frac{3}{8} \ln \left\mid \sec x + \tan x \right\mid + C$

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