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

Explanation

To evaluate the integral \( \int \csc^4 x \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting \( \csc^4 x \) in terms of \( \cot x \):
\[ \csc^4 x = \frac{1}{\sin^4 x} = \frac{\cos^4 x}{\sin^4 x} = \cot^4 x + 1 \]
Thus, we can express the integral as:
\[ \int \csc^4 x \, dx = \int (\cot^4 x + 1) \, dx \]

### Step 2: Split the Integral
Now, we can split the integral into two parts:
\[ \int \csc^4 x \, dx = \int \cot^4 x \, dx + \int 1 \, dx \]

### Step 3: Integrate the Second Part
The integral of \( 1 \) is straightforward:
\[ \int 1 \, dx = x \]

### Step 4: Integrate the First Part
Now, we need to integrate \( \cot^4 x \). We can use the identity \( \cot^2 x = \csc^2 x - 1 \):
\[ \cot^4 x = (\cot^2 x)^2 = (\csc^2 x - 1)^2 \]
Expanding this gives:
\[ \cot^4 x = \csc^4 x - 2\csc^2 x + 1 \]
Thus, we can rewrite the integral:
\[ \int \cot^4 x \, dx = \int (\csc^4 x - 2\csc^2 x + 1) \, dx \]

### Step 5: Substitute and Integrate
Now we can integrate each term separately:
\[ \int \cot^4 x \, dx = \int \csc^4 x \, dx - 2\int \csc^2 x \, dx + \int 1 \, dx \]
Let \( I = \int \csc^4 x \, dx \). Then we have:
\[ I = I - 2\int \csc^2 x \, dx + x \]
Rearranging gives:
\[ 0 = -2\int \csc^2 x \, dx + x \]
Thus:
\[ 2\int \csc^2 x \, dx = x \]

### Step 6: Solve for the Integral
The integral \( \int \csc^2 x \, dx \) is known:
\[ \int \csc^2 x \, dx = -\cot x + C \]
Substituting back gives:
\[ 2(-\cot x) = x \]
So we have:
\[ I = -\frac{1}{3} \cot^3 x - \cot x + x + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \csc^4 x \, dx = -\frac{1}{3} \cot^3 x - \cot x + x + C \]

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