If `int ( 1+ cos 4x)/( cot x - tan x ) dx = k cos 4x + C`, then the value of k is

Options

A. `(1)/( 4)`
B. `- (1)/(2)`
C. `- ( 1)/( 8 )`
D. ` - (1)/( 4)`

Answer

To solve the integral

\[ \int \frac{1 + \cos 4x}{\cot x - \tan x} \, dx, \]

we will simplify the expression step by step and find the value of \( k \) in the equation

\[ \int \frac{1 + \cos 4x}{\cot x - \tan x} \, dx = k \cos 4x + C. \]

### Step 1: Simplify the numerator
We know that

\[ \cos 4x = 2 \cos^2 2x - 1. \]

Thus, we can rewrite the numerator:

\[ 1 + \cos 4x = 1 + (2 \cos^2 2x - 1) = 2 \cos^2 2x. \]

### Step 2: Simplify the denominator
The denominator \( \cot x - \tan x \) can be rewritten as:

\[ \cot x = \frac{\cos x}{\sin x}, \quad \tan x = \frac{\sin x}{\cos x}. \]

So,

\[ \cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos 2x}{\sin x \cos x}. \]

### Step 3: Rewrite the integral
Now we can substitute the simplified numerator and denominator into the integral:

\[ \int \frac{2 \cos^2 2x}{\frac{\cos 2x}{\sin x \cos x}} \, dx = \int \frac{2 \cos^2 2x \cdot \sin x \cos x}{\cos 2x} \, dx. \]

### Step 4: Cancel terms
We can cancel \( \cos 2x \) from the numerator and denominator:

\[ \int 2 \cos 2x \cdot \sin x \cos x \, dx. \]

### Step 5: Use the double angle identity
Recall that \( 2 \sin x \cos x = \sin 2x \). Thus, we can rewrite the integral as:

\[ \int \cos 2x \cdot \sin 2x \, dx. \]

### Step 6: Use the identity for sine
The expression \( \cos 2x \cdot \sin 2x \) can be rewritten using the identity:

\[ \sin 2x = 2 \sin x \cos x, \]

so we have:

\[ \int \frac{1}{2} \sin 4x \, dx. \]

### Step 7: Integrate
Now we can integrate:

\[ \frac{1}{2} \int \sin 4x \, dx = -\frac{1}{2} \cdot \frac{1}{4} \cos 4x + C = -\frac{1}{8} \cos 4x + C. \]

### Step 8: Identify \( k \)
From our integration result:

\[ -\frac{1}{8} \cos 4x + C, \]

we can see that \( k = -\frac{1}{8} \).

### Final Answer
Thus, the value of \( k \) is

\[ \boxed{-\frac{1}{8}}. \]