`int " cosec"^(4) 2x dx`

Answer

To solve the integral \( I = \int \csc^4(2x) \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We can express \( \csc^4(2x) \) in terms of \( \cot^2(2x) \):
\[ I = \int \csc^4(2x) \, dx = \int \csc^2(2x)(\csc^2(2x)) \, dx \]
Using the identity \( \csc^2(2x) = 1 + \cot^2(2x) \), we can rewrite the integral:
\[ I = \int \csc^2(2x) \, dx + \int \csc^2(2x) \cot^2(2x) \, dx \]

### Step 2: Solve the First Integral
The first integral is straightforward:
\[ \int \csc^2(2x) \, dx = -\frac{1}{2} \cot(2x) + C_1 \]

### Step 3: Solve the Second Integral
For the second integral, we let \( t = \cot(2x) \). Then, differentiating both sides gives:
\[ dt = -2 \csc^2(2x) \, dx \quad \Rightarrow \quad dx = -\frac{dt}{2 \csc^2(2x)} \]
Substituting this into the integral:
\[ \int \csc^2(2x) \cot^2(2x) \, dx = \int t^2 \left(-\frac{dt}{2}\right) = -\frac{1}{2} \int t^2 \, dt \]

### Step 4: Integrate \( t^2 \)
Now, we can integrate \( t^2 \):
\[ -\frac{1}{2} \int t^2 \, dt = -\frac{1}{2} \cdot \frac{t^3}{3} + C_2 = -\frac{t^3}{6} + C_2 \]

### Step 5: Substitute Back
Now we substitute \( t = \cot(2x) \) back into the equation:
\[ -\frac{t^3}{6} = -\frac{\cot^3(2x)}{6} \]

### Step 6: Combine the Results
Combining both parts of the integral, we have:
\[ I = -\frac{1}{2} \cot(2x) - \frac{\cot^3(2x)}{6} + C \]

### Final Answer
Thus, the final result for the integral is:
\[ I = -\frac{1}{2} \cot(2x) - \frac{1}{6} \cot^3(2x) + C \]