Evaluate `int tanxdx.`

Options

A. `-logsinx.`
B. `logcosx.`
C. `-logcosx.`
D. `logsecx.`

Answer

To evaluate the integral \( \int \tan x \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting \( \tan x \) in terms of sine and cosine:
\[ \tan x = \frac{\sin x}{\cos x} \]
Thus, we can express the integral as:
\[ \int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx \]

### Step 2: Use Substitution
We will use substitution to simplify the integral. Let:
\[ t = \cos x \]
Then, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = -\sin x \quad \Rightarrow \quad dt = -\sin x \, dx \]
From this, we can express \( \sin x \, dx \) in terms of \( dt \):
\[ \sin x \, dx = -dt \]

### Step 3: Substitute in the Integral
Now, we substitute \( t \) and \( dt \) into the integral:
\[ \int \frac{\sin x}{\cos x} \, dx = \int \frac{-dt}{t} \]
This simplifies to:
\[ -\int \frac{dt}{t} \]

### Step 4: Integrate
The integral of \( \frac{1}{t} \) is:
\[ -\ln |t| + C \]
Substituting back \( t = \cos x \):
\[ -\ln |\cos x| + C \]

### Step 5: Final Result
Thus, the final result of the integral is:
\[ \int \tan x \, dx = -\ln |\cos x| + C \]

### Summary
The evaluated integral is:
\[ \int \tan x \, dx = -\ln |\cos x| + C \]