`int(cos4x)/(sin^(2)x)dx=`

Options

A. `-cotx-4x-2sin2x+c`
B. `cotx-4x-2sin2x+c`
C. `4x+cotx-2sin2x+c`
D. `4x+cotx+2sin2x+c`

Answer

To solve the integral \(\int \frac{\cos 4x}{\sin^2 x} \, dx\), we can follow these steps:

### Step 1: Rewrite the integral
We start by rewriting the integral:
\[ I = \int \frac{\cos 4x}{\sin^2 x} \, dx \]

### Step 2: Use the identity for \(\cos 4x\)
Using the trigonometric identity for \(\cos 4x\):
\[ \cos 4x = 1 - 2\sin^2 2x \]
we can substitute this into the integral:
\[ I = \int \frac{1 - 2\sin^2 2x}{\sin^2 x} \, dx \]

### Step 3: Split the integral
Now we can split the integral into two parts:
\[ I = \int \frac{1}{\sin^2 x} \, dx - 2 \int \frac{\sin^2 2x}{\sin^2 x} \, dx \]

### Step 4: Solve the first integral
The first integral, \(\int \frac{1}{\sin^2 x} \, dx\), is a standard integral:
\[ \int \frac{1}{\sin^2 x} \, dx = -\cot x \]

### Step 5: Solve the second integral
For the second integral, we can use the identity \(\sin 2x = 2 \sin x \cos x\):
\[ \sin^2 2x = 4 \sin^2 x \cos^2 x \]
Thus, we have:
\[ \int \frac{\sin^2 2x}{\sin^2 x} \, dx = \int \frac{4 \sin^2 x \cos^2 x}{\sin^2 x} \, dx = 4 \int \cos^2 x \, dx \]

### Step 6: Solve \(\int \cos^2 x \, dx\)
Using the identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\):
\[ \int \cos^2 x \, dx = \int \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \cos 2x \, dx \]
Calculating these integrals gives:
\[ \int 1 \, dx = x \quad \text{and} \quad \int \cos 2x \, dx = \frac{1}{2} \sin 2x \]
Thus,
\[ \int \cos^2 x \, dx = \frac{x}{2} + \frac{1}{4} \sin 2x \]

### Step 7: Combine results
Now we can substitute back into our expression for \(I\):
\[ I = -\cot x - 2 \left( 4 \left( \frac{x}{2} + \frac{1}{4} \sin 2x \right) \right) \]
This simplifies to:
\[ I = -\cot x - 4x - 2\sin 2x \]

### Step 8: Final result
Thus, the final result is:
\[ I = -\cot x - 4x - 2\sin 2x + C \]