The integral in the image is:

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

Solution:

We start by rewriting the denominator: $\sin^2 x \cos^2 x = (\sin x \cos x)^2.$

Thus, the integral becomes: $\int \frac{dx}{\sin^2 x \cos^2 x} = \int \frac{dx}{(\sin x \cos x)^2}.$

Next, use the double-angle identity: $\sin(2x) = 2 \sin x \cos x \quad \implies \quad \sin x \cos x = \frac{\sin(2x)}{2}.$

Substituting this into the integral: $\int \frac{dx}{(\sin x \cos x)^2} = \int \frac{dx}{\left(\frac{\sin(2x)}{2}\right)^2}.$

Simplify the denominator: $\left(\frac{\sin(2x)}{2}\right)^2 = \frac{\sin^2(2x)}{4}.$

So the integral becomes: $\int \frac{dx}{\frac{\sin^2(2x)}{4}} = \int \frac{4 \, dx}{\sin^2(2x)}.$

The reciprocal of $\sin^2(2x)$ is $\csc^2(2x)$, so: $\int \frac{4 \, dx}{\sin^2(2x)} = 4 \int \csc^2(2x) \, dx.$

The integral of $\csc^2(kx)$ is $-\frac{1}{k} \cot(kx)$, so: $4 \int \csc^2(2x) \, dx = 4 \cdot \left(-\frac{1}{2} \cot(2x)\right) = -2 \cot(2x).$

Thus, the solution is: $\boxed{-2 \cot(2x) + C},$ where $C$ is the constant of integration.

Would you like further details or clarification?
Here are 5 related questions you can explore:

1. How is the double-angle identity for sine derived?
2. What is the integral of $\csc^2(kx)$ and why does it involve $\cot(kx)$?
3. How can trigonometric integrals be simplified using substitution?
4. What are other ways to evaluate $\int \frac{dx}{\sin^2 x \cos^2 x}$?
5. How is the constant of integration $C$ important in indefinite integrals?

Tip: Always consider trigonometric identities when simplifying integrals involving sine and cosine.