`int " cosec"^(3) " x dx "`

Answer

To solve the integral \( \int \csc^3 x \, dx \), we can follow these steps:

### Step 1: Set up the integral
Let
\[ I = \int \csc^3 x \, dx \]

### Step 2: Rewrite the integrand
We can express \( \csc^3 x \) as \( \csc^2 x \cdot \csc x \). Using the identity \( \csc^2 x = 1 + \cot^2 x \), we rewrite the integral:
\[ I = \int \csc^2 x \cdot \csc x \, dx = \int \csc^2 x \cdot \sqrt{1 + \cot^2 x} \, dx \]

### Step 3: Substitute \( \cot x \)
Let \( t = \cot x \). Then, we have:
\[ \frac{dt}{dx} = -\csc^2 x \quad \Rightarrow \quad dx = -\frac{dt}{\csc^2 x} \]
Substituting this into the integral gives:
\[ I = \int \csc^2 x \cdot \sqrt{1 + t^2} \cdot \left(-\frac{dt}{\csc^2 x}\right) = -\int \sqrt{1 + t^2} \, dt \]

### Step 4: Integrate \( \sqrt{1 + t^2} \)
The integral \( \int \sqrt{1 + t^2} \, dt \) can be solved using the formula:
\[ \int \sqrt{1 + x^2} \, dx = \frac{1}{2} x \sqrt{1 + x^2} + \frac{1}{2} \ln |x + \sqrt{1 + x^2}| + C \]
Applying this to our integral:
\[ I = -\left( \frac{1}{2} t \sqrt{1 + t^2} + \frac{1}{2} \ln |t + \sqrt{1 + t^2}| \right) + C \]

### Step 5: Substitute back for \( t \)
Now we substitute back \( t = \cot x \):
\[ I = -\left( \frac{1}{2} \cot x \sqrt{1 + \cot^2 x} + \frac{1}{2} \ln |\cot x + \sqrt{1 + \cot^2 x}| \right) + C \]
Since \( 1 + \cot^2 x = \csc^2 x \), we have:
\[ \sqrt{1 + \cot^2 x} = \csc x \]
Thus, the integral becomes:
\[ I = -\left( \frac{1}{2} \cot x \csc x + \frac{1}{2} \ln |\cot x + \csc x| \right) + C \]

### Final Result
The final answer for the integral is:
\[ \int \csc^3 x \, dx = -\frac{1}{2} \cot x \csc x - \frac{1}{2} \ln |\cot x + \csc x| + C \]