If `intcos x log (tan""(x)/(2))dx=sin x log (tan""(x)/(2))+f(x)` then f(x) is equal to, (assuming c is an arbitrary real constant)

Options

A. c
B. `c-x`
C. `c+x`
D. `2x+c`

Answer

To solve the integral \( \int \cos x \log \left( \frac{\tan x}{2} \right) dx \) and find \( f(x) \) in the equation

\[ \int \cos x \log \left( \frac{\tan x}{2} \right) dx = \sin x \log \left( \frac{\tan x}{2} \right) + f(x), \]

we will use integration by parts.

### Step 1: Identify \( u \) and \( dv \)
Let:
- \( u = \log \left( \frac{\tan x}{2} \right) \)
- \( dv = \cos x \, dx \)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{\frac{\tan x}{2}} \cdot \frac{1}{2} \sec^2 x \, dx = \frac{1}{2 \tan x} \sec^2 x \, dx \]
- Integrate \( dv \):
\[ v = \int \cos x \, dx = \sin x \]

### Step 3: Apply integration by parts
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int \cos x \log \left( \frac{\tan x}{2} \right) dx = \sin x \log \left( \frac{\tan x}{2} \right) - \int \sin x \cdot \frac{1}{2 \tan x} \sec^2 x \, dx \]

### Step 4: Simplify the integral
The integral \( \int \sin x \cdot \frac{1}{2 \tan x} \sec^2 x \, dx \) can be simplified:
\[ \frac{\sin x}{\tan x} = \cos x \]
Thus, we have:
\[ \int \sin x \cdot \frac{1}{2 \tan x} \sec^2 x \, dx = \frac{1}{2} \int \cos x \, dx = \frac{1}{2} \sin x \]

### Step 5: Substitute back into the equation
Now substituting back, we have:
\[ \int \cos x \log \left( \frac{\tan x}{2} \right) dx = \sin x \log \left( \frac{\tan x}{2} \right) - \frac{1}{2} \sin x + C \]

### Step 6: Identify \( f(x) \)
From the original equation:
\[ \int \cos x \log \left( \frac{\tan x}{2} \right) dx = \sin x \log \left( \frac{\tan x}{2} \right) + f(x) \]
we can see that:
\[ f(x) = -\frac{1}{2} \sin x + C \]

### Final Answer
Thus, \( f(x) \) is equal to:
\[ f(x) = -\frac{1}{2} \sin x + C \]