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# What is the integral of sec(x)?

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VNVDVI Share
Apr 18, 2018

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

#### Explanation:

Integrating the secant requires a bit of manipulation.

Multiply $\sec x$ by $\frac{\sec x + \tan x}{\sec x + \tan x}$, which is really the same as multiplying by $1.$ Thus, we have

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

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

Now, make the following substitution:

$u = \sec x + \tan x$

$\mathrm{du} = \left(\sec x \tan x + {\sec}^{2} x\right) \mathrm{dx} = \left({\sec}^{2} x + \sec x \tan x\right) \mathrm{dx}$

We see that $\mathrm{du}$ appears in the numerator of the integral, so we may apply the substitution:

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

Rewrite in terms of $x$ to get

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

This is an integral worth memorizing.

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