`int cot^(3) x. cosec^(2) x dx `

Answer

To solve the integral \( \int \cot^3 x \csc^2 x \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \cot^3 x \csc^2 x \, dx \]

### Step 2: Use Substitution
We know that the derivative of \( \cot x \) is \( -\csc^2 x \). Therefore, we can use the substitution:
\[ t = \cot x \quad \Rightarrow \quad dt = -\csc^2 x \, dx \]
This implies:
\[ dx = -\frac{dt}{\csc^2 x} \]
Substituting \( \csc^2 x \) in terms of \( t \):
\[ \csc^2 x = 1 + \cot^2 x = 1 + t^2 \]
Thus, we can rewrite \( dx \) as:
\[ dx = -\frac{dt}{1 + t^2} \]

### Step 3: Substitute into the Integral
Now substituting \( t \) into the integral:
\[ I = \int t^3 \cdot (-dt) = -\int t^3 \, dt \]

### Step 4: Integrate
Now we can integrate \( -\int t^3 \, dt \):
\[ -\int t^3 \, dt = -\left( \frac{t^4}{4} \right) + C = -\frac{t^4}{4} + C \]

### Step 5: Substitute Back
Now we substitute back \( t = \cot x \):
\[ I = -\frac{\cot^4 x}{4} + C \]

### Final Answer
Thus, the final answer is:
\[ \int \cot^3 x \csc^2 x \, dx = -\frac{\cot^4 x}{4} + C \]

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