`int(dx)/(4cos^(3)2x-3cos2x)=`

Options

A. `1/3logabs(sec6x+tan6x)+c`
B. `1/6logabs(sec6x+tan6x)+c`
C. `logabs(sec6x+tan6x)+c`
D. `3logabs(sec6x+tan6x)+c`

Answer

To solve the integral

\[ \int \frac{dx}{4 \cos^3(2x) - 3 \cos(2x)}, \]

we can follow these steps:

### Step 1: Simplify the Denominator
We start by simplifying the denominator \(4 \cos^3(2x) - 3 \cos(2x)\). We can use the trigonometric identity for \(\cos(3x)\):

\[ \cos(3x) = 4 \cos^3(x) - 3 \cos(x). \]

In our case, we can let \(x = 2x\):

\[ \cos(3(2x)) = 4 \cos^3(2x) - 3 \cos(2x) = \cos(6x). \]

Thus, we can rewrite the denominator:

\[ 4 \cos^3(2x) - 3 \cos(2x) = \cos(6x). \]

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

\[ \int \frac{dx}{\cos(6x)}. \]

### Step 3: Change the Integral Form
We know that \(\frac{1}{\cos(6x)} = \sec(6x)\). Therefore, the integral becomes:

\[ \int \sec(6x) \, dx. \]

### Step 4: Integrate \(\sec(6x)\)
The integral of \(\sec(kx)\) is given by the formula:

\[ \int \sec(kx) \, dx = \frac{1}{k} \ln | \sec(kx) + \tan(kx) | + C. \]

In our case, \(k = 6\). Thus, we have:

\[ \int \sec(6x) \, dx = \frac{1}{6} \ln | \sec(6x) + \tan(6x) | + C. \]

### Final Answer
Putting it all together, we find:

\[ \int \frac{dx}{4 \cos^3(2x) - 3 \cos(2x)} = \frac{1}{6} \ln | \sec(6x) + \tan(6x) | + C. \]