# What is the the value of int sqrt(tanx/(sinxcosx)) dx?

Dec 17, 2016

$= \ln | \sec x + \tan x | + C$

#### Explanation:

Start by simplifying the expression within the integral.

$= \int \sqrt{\frac{\sin \frac{x}{\cos} x}{\sin x \cos x}} \mathrm{dx}$

$= \int \sqrt{\frac{\sin x}{\cos x \sin x \cos x}} \mathrm{dx}$

$= \int \sqrt{\frac{1}{\cos} ^ 2 x} \mathrm{dx}$

$= \int \left(\frac{1}{\cos} x\right) \mathrm{dx}$

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

This is a tricky integral to do. Multiply everything by $\sec x + \tan x$.

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

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

Let $u = \sec x + \tan x$. Then $\frac{\mathrm{du}}{\mathrm{dx}} = {\sec}^{2} x + \sec x \tan x$ and $\mathrm{du} = {\sec}^{2} x + \sec x \tan x \mathrm{dx}$. Substitute:

$= \int \frac{\mathrm{du}}{u}$

$= \ln | u | + C$

$= \ln | \sec x + \tan x | + C$

Hopefully this helps!