`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:

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

Here, let \( a = \sin^2 x \) and \( b = \cos^2 x \). Thus, 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 the expression:

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

Now substituting this back into the integral, we get:

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

We can separate this into three integrals:

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

This simplifies to:

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

Now, we can rewrite the integrals:

1. The first integral becomes \(\int \tan^2 x \, dx\).
2. The second integral is simply \(-\int 1 \, dx = -x\).
3. The third integral becomes \(\int \cot^2 x \, dx\).

Using the identities:

\[ \tan^2 x = \sec^2 x - 1 \quad \text{and} \quad \cot^2 x = \csc^2 x - 1, \]

we can integrate each part:

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

Combining all these results, we have:

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

which simplifies to:

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

Thus, the final result of the integral is:

\[ \int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} \, dx = \tan x - \cot x - 3x + C. \]