Integration of trigonometric and hyperbolic functions
`I=int(cos^(3)x)/(sin^(6)x)dx.`

Answer

To solve the integral \( I = \int \frac{\cos^3 x}{\sin^6 x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We can express the integral in terms of sine and cosine functions:
\[ I = \int \frac{\cos^3 x}{\sin^6 x} \, dx = \int \cos^3 x \cdot \csc^6 x \, dx \]

### Step 2: Use the identity for cosine
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can rewrite \( \cos^3 x \):
\[ \cos^3 x = \cos^2 x \cdot \cos x = (1 - \sin^2 x) \cos x \]
Thus, we can rewrite the integral as:
\[ I = \int (1 - \sin^2 x) \cdot \cos x \cdot \csc^6 x \, dx \]

### Step 3: Substitute \( t = \sin x \)
Let \( t = \sin x \), then \( dt = \cos x \, dx \). The integral becomes:
\[ I = \int (1 - t^2) \cdot \frac{1}{t^6} \, dt = \int \frac{1 - t^2}{t^6} \, dt \]

### Step 4: Split the integral
We can split the integral into two parts:
\[ I = \int \frac{1}{t^6} \, dt - \int \frac{t^2}{t^6} \, dt = \int t^{-6} \, dt - \int t^{-4} \, dt \]

### Step 5: Integrate each term
Now we can integrate each term:
1. For \( \int t^{-6} \, dt \):
\[ \int t^{-6} \, dt = \frac{t^{-5}}{-5} = -\frac{1}{5t^5} \]
2. For \( \int t^{-4} \, dt \):
\[ \int t^{-4} \, dt = \frac{t^{-3}}{-3} = -\frac{1}{3t^3} \]

### Step 6: Combine the results
Combining the results, we have:
\[ I = -\frac{1}{5t^5} + \frac{1}{3t^3} + C \]

### Step 7: Substitute back \( t = \sin x \)
Now we substitute back \( t = \sin x \):
\[ I = -\frac{1}{5 \sin^5 x} + \frac{1}{3 \sin^3 x} + C \]

### Final Answer
Thus, the final answer is:
\[ I = \frac{1}{3 \sin^3 x} - \frac{1}{5 \sin^5 x} + C \]