To solve the integral

$I = \int \frac{\sin^2 x - \cos^2 x}{\sin x \cos x} \, dx,$

we start by simplifying the integrand using trigonometric identities.

Step 1: Simplify $\sin^2 x - \cos^2 x$

Recall the identity: $\sin^2 x - \cos^2 x = -\cos(2x).$

Thus, the integrand becomes: $\frac{\sin^2 x - \cos^2 x}{\sin x \cos x} = \frac{-\cos(2x)}{\sin x \cos x}.$

Step 2: Simplify further

Using the double-angle identity for sine, $\sin(2x) = 2\sin x \cos x$, we can rewrite: $\frac{-\cos(2x)}{\sin x \cos x} = -\frac{\cos(2x)}{\frac{1}{2} \sin(2x)} = -2 \cdot \frac{\cos(2x)}{\sin(2x)}.$

This simplifies to: $\frac{\sin^2 x - \cos^2 x}{\sin x \cos x} = -2 \cot(2x).$

Step 3: Integrate

The integral now becomes: $I = \int -2 \cot(2x) \, dx.$

The integral of $\cot(2x)$ is: $\int \cot(2x) \, dx = \frac{1}{2} \ln|\sin(2x)| + C.$

Thus: $I = -2 \cdot \frac{1}{2} \ln|\sin(2x)| + C = -\ln|\sin(2x)| + C.$

Final Answer:

$\int \frac{\sin^2 x - \cos^2 x}{\sin x \cos x} \, dx = -\ln|\sin(2x)| + C.$

Would you like a detailed explanation of any step? Here are some related questions:

1. What are the key trigonometric identities used in simplifying integrals?
2. How does the double-angle formula for sine simplify such expressions?
3. Why is $\cot(2x)$ related to $\ln|\sin(2x)|$ in its integral?
4. Can this integral be approached using substitution?
5. How would the integral change if $\sin^2 x + \cos^2 x$ were in the numerator instead?

Tip: Memorizing the derivatives and integrals of basic trigonometric functions like $\sin x, \cos x, \tan x,$ and $\cot x$ can greatly simplify solving such integrals.