If `int f(x) dx=log |tanx|+c` find f (x).

Answer

To find \( f(x) \) given that

\[ \int f(x) \, dx = \log |\tan x| + C, \]

we will differentiate both sides with respect to \( x \).

### Step 1: Differentiate both sides

We start by differentiating the left-hand side and the right-hand side:

\[ \frac{d}{dx} \left( \int f(x) \, dx \right) = f(x) \]

Now we differentiate the right-hand side:

\[ \frac{d}{dx} \left( \log |\tan x| + C \right) = \frac{d}{dx} \left( \log |\tan x| \right) \]

### Step 2: Apply the chain rule

Using the chain rule, we differentiate \( \log |\tan x| \):

\[ \frac{d}{dx} \left( \log |\tan x| \right) = \frac{1}{\tan x} \cdot \frac{d}{dx} (\tan x) \]

### Step 3: Differentiate \( \tan x \)

The derivative of \( \tan x \) is \( \sec^2 x \):

\[ \frac{d}{dx} (\tan x) = \sec^2 x \]

### Step 4: Substitute back

Substituting this back into our expression gives:

\[ \frac{d}{dx} \left( \log |\tan x| \right) = \frac{1}{\tan x} \cdot \sec^2 x \]

### Step 5: Simplify

Recall that \( \sec^2 x = \frac{1}{\cos^2 x} \) and \( \tan x = \frac{\sin x}{\cos x} \):

\[ \frac{1}{\tan x} \cdot \sec^2 x = \frac{1}{\frac{\sin x}{\cos x}} \cdot \frac{1}{\cos^2 x} = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} \]

### Step 6: Final expression for \( f(x) \)

Thus, we have:

\[ f(x) = \frac{1}{\sin x \cos x} \]

This can also be expressed as:

\[ f(x) = \frac{2}{\sin(2x)} \]

### Final Answer

Therefore, the function \( f(x) \) is:

\[ f(x) = \frac{1}{\sin x \cos x} \quad \text{or} \quad f(x) = \frac{2}{\sin(2x)} \]

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