`int (sin 4x)/(cos 2x) dx`

Answer

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

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

### Step 2: Use the Double Angle Identity
We can use the identity for sine:
\[ \sin 4x = 2 \sin 2x \cos 2x \]
Now, substituting this into the integral gives:
\[ I = \int \frac{2 \sin 2x \cos 2x}{\cos 2x} \, dx \]

### Step 3: Simplify the Integral
The \(\cos 2x\) in the numerator and denominator cancels out:
\[ I = \int 2 \sin 2x \, dx \]

### Step 4: Factor Out the Constant
We can factor out the constant 2:
\[ I = 2 \int \sin 2x \, dx \]

### Step 5: Integrate
The integral of \(\sin 2x\) is:
\[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C \]
Thus, substituting back, we have:
\[ I = 2 \left(-\frac{1}{2} \cos 2x + C\right) \]

### Step 6: Simplify the Result
This simplifies to:
\[ I = -\cos 2x + C \]

### Final Answer
Therefore, the final result of the integral is:
\[ \int \frac{\sin 4x}{\cos 2x} \, dx = -\cos 2x + C \]

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