What is the integral of tan(x) dx?

The integral of ( \tan(x) , dx ) can be derived using the identity for tangent:

[

\tan(x) = \frac{\sin(x)}{\cos(x)}

]

This allows us to rewrite the integral as:

[

\int \tan(x) , dx = \int \frac{\sin(x)}{\cos(x)} , dx

]

To solve this integral, we can use the substitution method. Let ( u = \cos(x) ), which implies that ( du = -\sin(x) , dx ) or ( -du = \sin(x) , dx ). The integral then transforms to:

[

\int \frac{\sin(x)}{\cos(x)} , dx = -\int \frac{1}{u} , du

]

This integral evaluates to:

[

-\ln|u| + C = -\ln|\cos(x)| + C

]

Thus, we arrive at the integral of ( \tan(x) , dx ):

[

-\ln|\cos(x)| + C

]

This derivation confirms that the correct answer aligns with the option that states ( -\ln|