The integral given is:

$\int \frac{1}{\sin^2 x \cos^2 x} \, dx$

To solve this, let's rewrite the integrand in a more manageable form.

1. We know that $\sin^2 x \cos^2 x$ can be rewritten using a trigonometric identity: $\sin^2 x \cos^2 x = \left(\frac{\sin 2x}{2}\right)^2 = \frac{\sin^2 2x}{4}$

2. Therefore, we can rewrite the integral as: $\int \frac{1}{\sin^2 x \cos^2 x} \, dx = \int \frac{4}{\sin^2 2x} \, dx$

3. Since $\frac{1}{\sin^2 2x} = \csc^2 2x$, we can rewrite this as: $4 \int \csc^2 2x \, dx$

4. Now, the integral of $\csc^2 u$ with respect to $u$ is $-\cot u$. Here, let $u = 2x$, so $du = 2 \, dx$ or $dx = \frac{du}{2}$.

5. Substitute into the integral: $4 \int \csc^2 u \cdot \frac{du}{2} = 2 \int \csc^2 u \, du = 2(-\cot u) = -2 \cot(2x)$

Thus, the solution to the integral is:

$\int \frac{1}{\sin^2 x \cos^2 x} \, dx = -2 \cot(2x) + C$

Would you like further explanation on any of these steps, or have questions about related concepts?

Here are five related questions for deeper understanding:

1. How would you integrate $\int \csc^2 x \, dx$?
2. What is the integral of $\sec^2 x$ and how is it different from $\csc^2 x$?
3. How can trigonometric identities help simplify complex integrals?
4. How would you solve $\int \frac{1}{\sin^4 x} \, dx$?
5. What is the general method for integrating functions involving products of sine and cosine?

Tip: Using trigonometric identities can often simplify complex integrals, making them easier to evaluate.