Evaluate the following integrals:
`int frac{sin^6x+cos^6x}{sin^2x cos^2x}dx`

Answer

To evaluate the integral

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

we will follow these steps:

### Step 1: Simplify the numerator
We can rewrite \(\sin^6 x + \cos^6 x\) using the identity for the sum of cubes:

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

Using the identity \(\sin^2 x + \cos^2 x = 1\), we have:

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

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

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

### Step 3: Split the integral
We can split the integral into three parts:

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

This simplifies to:

\[ I = \int \tan^2 x \, dx + \int \cot^2 x \, dx - \int 1 \, dx. \]

### Step 4: Evaluate each integral
Now we can evaluate each integral separately:

1. \(\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \tan x - x + C_1\).
2. \(\int \cot^2 x \, dx = \int (\csc^2 x - 1) \, dx = -\cot x - x + C_2\).
3. \(\int 1 \, dx = x + C_3\).

Combining these results, we have:

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

which simplifies to:

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

### Final Answer
Thus, the evaluated integral is:

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