Evaluate :
`int cot^(3)x cosec^(4)x dx`

Answer

To evaluate the integral \( \int \cot^3 x \csc^4 x \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ \int \cot^3 x \csc^4 x \, dx \]
We can express \( \csc^4 x \) as \( \csc^2 x \cdot \csc^2 x \) and rewrite the integral as:
\[ \int \cot^3 x \csc^2 x \cdot \csc^2 x \, dx \]
Using the identity \( \csc^2 x = 1 + \cot^2 x \), we can rewrite \( \csc^2 x \):
\[ \int \cot^3 x (1 + \cot^2 x) \csc^2 x \, dx \]

### Step 2: Expand the integral
Now we can expand the integral:
\[ \int \cot^3 x \csc^2 x \, dx + \int \cot^5 x \csc^2 x \, dx \]

### Step 3: Substitution
Let \( t = \cot x \). Then, the derivative \( dt = -\csc^2 x \, dx \) implies that \( dx = -\frac{dt}{\csc^2 x} \). We can substitute this into the integral:
\[ \int t^3 (-dt) + \int t^5 (-dt) \]
This simplifies to:
\[ -\int t^3 \, dt - \int t^5 \, dt \]

### Step 4: Integrate
Now we can integrate each term:
\[ -\left( \frac{t^4}{4} + \frac{t^6}{6} \right) + C \]
This gives us:
\[ -\frac{t^4}{4} - \frac{t^6}{6} + C \]

### Step 5: Substitute back
Now we substitute back \( t = \cot x \):
\[ -\frac{\cot^4 x}{4} - \frac{\cot^6 x}{6} + C \]

### Final Answer
Thus, the final answer is:
\[ -\frac{\cot^4 x}{4} - \frac{\cot^6 x}{6} + C \]

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