If `int ( 1+cos 8x)/( tan 2x-cot 2x)dx = k cos 8x+ C`, then k equals

Options

A. `(1)/(8)`
B. `-(1)/(8)`
C. `(1)/(18)`
D. `- (1)/(16)`

Answer

To solve the integral

\[ \int \frac{1 + \cos 8x}{\tan 2x - \cot 2x} \, dx = k \cos 8x + C, \]

we will simplify the expression step by step.

### Step 1: Simplify the Denominator
First, we simplify the denominator \(\tan 2x - \cot 2x\).

\[ \tan 2x = \frac{\sin 2x}{\cos 2x}, \quad \cot 2x = \frac{\cos 2x}{\sin 2x} \]

Thus,

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

Using the identity \(\sin^2 \theta - \cos^2 \theta = -\cos 2\theta\), we have:

\[ \tan 2x - \cot 2x = \frac{-\cos 4x}{\sin 2x \cos 2x} \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral:

\[ \int \frac{1 + \cos 8x}{\tan 2x - \cot 2x} \, dx = \int \frac{(1 + \cos 8x) \sin 2x \cos 2x}{-\cos 4x} \, dx \]

### Step 3: Expand the Numerator
Next, we expand the numerator:

\[ 1 + \cos 8x = 1 + 2\cos^2 4x - 1 = 2\cos^2 4x \]

Thus, we can rewrite the integral as:

\[ \int \frac{2 \cos^2 4x \sin 2x \cos 2x}{-\cos 4x} \, dx = -2 \int \cos 4x \sin 2x \cos 2x \, dx \]

### Step 4: Use a Trigonometric Identity
Using the identity \(\sin 2x \cos 2x = \frac{1}{2} \sin 4x\), we have:

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

### Step 5: Integrate
Now we can integrate:

\[ -\int \cos 4x \sin 4x \, dx = -\frac{1}{2} \int \sin 8x \, dx = -\frac{1}{2} \cdot \left(-\frac{1}{8} \cos 8x\right) + C = \frac{1}{16} \cos 8x + C \]

### Step 6: Compare with Given Form
Now we compare with the given form \(k \cos 8x + C\):

\[ \frac{1}{16} \cos 8x + C = k \cos 8x + C \]

From this, we can see that:

\[ k = \frac{1}{16} \]

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

\[ \boxed{\frac{1}{16}} \]