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 \(\sin^6 x + \cos^6 x\)

We can use 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,

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

Since \(\sin^2 x + \cos^2 x = 1\), we have:

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

Now, we can express \((\sin^2 x)^2 + (\cos^2 x)^2\) as:

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

Thus,

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

### Step 2: Substitute back into the integral

Now substituting back into the integral, we get:

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

This can be split into two separate integrals:

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

### Step 3: Simplify the first integral

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

\[ I = \int \sec^2 x \csc^2 x \, dx - 3x. \]

### Step 4: Change of variables

Let \(t = \tan x\), then \(dt = \sec^2 x \, dx\). The integral becomes:

\[ I = \int \frac{1 + t^2}{t^2} \, dt - 3x. \]

This can be split into:

\[ I = \int \left(\frac{1}{t^2} + 1\right) dt - 3x. \]

### Step 5: Evaluate the integrals

Now we can evaluate each integral:

1. \(\int \frac{1}{t^2} \, dt = -\frac{1}{t} + C_1\)
2. \(\int 1 \, dt = t + C_2\)

Thus,

\[ I = -\frac{1}{t} + t - 3x + C. \]

### Step 6: Substitute back \(t = \tan x\)

Substituting back \(t = \tan x\):

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

### Final Answer

The final answer for the integral is:

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