# How do you find the integral of int sin x * tan x dx?

Mar 8, 2018

The answer is $= \ln \left(| \tan x + \sec x |\right) - \sin x + C$

#### Explanation:

We need

$\tan x = \sin \frac{x}{\cos} x$

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

Therefore,

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

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

$= \int \sec x \mathrm{dx} - \int \cos x \mathrm{dx}$

$= \ln \left(| \tan x + \sec x |\right) - \sin x + C$