`int(sin^(6)x + cos^(6)x)/(sin^(2)xcos^(2)x)dx`

Explanation

To solve the integral

\[ I = \int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} \, dx, \]

we can start by rewriting the numerator using the identity for the sum of cubes.

### Step 1: Rewrite the numerator
We can express \(\sin^6 x + \cos^6 x\) as follows:

\[ \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3. \]

Using the sum of cubes formula \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\), we have:

\[ \sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)\left((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2\right). \]

Since \(\sin^2 x + \cos^2 x = 1\), we can simplify this to:

\[ \sin^6 x + \cos^6 x = 1 \cdot \left(\sin^4 x + \cos^4 x - \sin^2 x \cos^2 x\right). \]

### Step 2: Substitute back into the integral
Now substituting this back into the integral, we get:

\[ I = \int \frac{\sin^4 x + \cos^4 x - \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} \, dx. \]

This can be separated into three integrals:

\[ I = \int \frac{\sin^4 x}{\sin^2 x \cos^2 x} \, dx + \int \frac{\cos^4 x}{\sin^2 x \cos^2 x} \, dx - \int \frac{\sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} \, dx. \]

### Step 3: Simplify each term
Now we simplify each integral:

1. The first integral becomes:
\[ \int \frac{\sin^4 x}{\sin^2 x \cos^2 x} \, dx = \int \frac{\sin^2 x}{\cos^2 x} \, dx = \int \tan^2 x \, dx. \]

2. The second integral becomes:
\[ \int \frac{\cos^4 x}{\sin^2 x \cos^2 x} \, dx = \int \frac{\cos^2 x}{\sin^2 x} \, dx = \int \cot^2 x \, dx. \]

3. The third integral simplifies to:
\[ \int 1 \, dx = x. \]

### Step 4: Integrate each term
Now we can integrate each term:

1. For \(\int \tan^2 x \, dx\):
\[ \tan^2 x = \sec^2 x - 1 \implies \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \tan x - x. \]

2. For \(\int \cot^2 x \, dx\):
\[ \cot^2 x = \csc^2 x - 1 \implies \int \cot^2 x \, dx = \int (\csc^2 x - 1) \, dx = -\cot x - x. \]

3. For \(\int 1 \, dx\):
\[ \int 1 \, dx = x. \]

### Step 5: Combine the results
Now, putting everything together, we have:

\[ I = \left(\tan x - x\right) + \left(-\cot x - x\right) - x. \]

Combining the terms gives:

\[ I = \tan x - \cot x - 3x + C, \]

where \(C\) is the constant of integration.

### Final Answer
Thus, the final result is:

\[ I = \tan x - \cot x - 3x + C. \]