# What is the antiderivative of tan(x)?

#### Understand the Problem

The question is asking for the antiderivative (or indefinite integral) of the function tan(x). This requires applying integral calculus techniques to find a function whose derivative is tan(x).

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

The final answer is -ln|cos(x)| + C

#### Steps to Solve

1. Rewrite tan(x) using trigonometric identity

We know that $\tan(x) = \frac{\sin(x)}{\cos(x)}$. So, we can rewrite our integral as: $$\int \tan(x) , dx = \int \frac{\sin(x)}{\cos(x)} , dx$$

1. Use substitution method

Let $u = \cos(x)$. Then, $du = -\sin(x) , dx$, implying $-du = \sin(x) , dx$.

1. Integrate in terms of u

Substitute $u$ and $du$ into the integral: $$\int \frac{\sin(x)}{\cos(x)} , dx = \int \frac{-du}{u} = -\int \frac{1}{u} , du$$

1. Find the antiderivative

The antiderivative of $\frac{1}{u}$ is $\ln|u|$. Therefore, we have: $$-\int \frac{1}{u} , du = -\ln|u| + C$$

1. Substitute back in terms of x

Recall that $u = \cos(x)$. Substitute back to get the final answer: $$-\ln|\cos(x)| + C$$

The final answer is -ln|cos(x)| + C