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

Answer

To solve 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 start with the numerator \(\sin^6 x + \cos^6 x\). This can be expressed using the identity for the sum of cubes:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2). \]

Let \(a = \sin^2 x\) and \(b = \cos^2 x\). Then we have:

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

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

\[ \sin^6 x + \cos^6 x = \sin^4 x - \sin^2 x \cos^2 x + \cos^4 x. \]

### Step 2: Further Simplify the Expression
Next, we can express \(\sin^4 x + \cos^4 x\) using the identity:

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

Substituting this back, we get:

\[ \sin^6 x + \cos^6 x = (1 - 2\sin^2 x \cos^2 x) - \sin^2 x \cos^2 x = 1 - 3\sin^2 x \cos^2 x. \]

### Step 3: Substitute Back into the Integral
Now we substitute this back into our integral:

\[ I = \int \frac{1 - 3\sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} \, dx. \]

This can be separated into two integrals:

\[ I = \int \frac{1}{\sin^2 x \cos^2 x} \, dx - 3 \int dx. \]

### Step 4: Simplify Each Integral
The first integral can be rewritten using the identity \(\frac{1}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x} \cdot \frac{1}{\cos^2 x} = \csc^2 x \sec^2 x\):

\[ \int \csc^2 x \sec^2 x \, dx. \]

This integral can be evaluated using the substitution \(u = \tan x\), which gives us:

\[ \int \csc^2 x \sec^2 x \, dx = \int \frac{1}{\sin^2 x} \cdot \frac{1}{\cos^2 x} \, dx = \int \frac{1}{\sin^2 x} \, dx + \int \frac{1}{\cos^2 x} \, dx. \]

The integrals of \(\csc^2 x\) and \(\sec^2 x\) are:

\[ -\cot x + \tan x. \]

### Step 5: Combine the Results
Putting everything together, we have:

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

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

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

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