1. The problem is to simplify the integrand $$\frac{\csc^2 x \sin x - \sin x}{\cos x}$$ and understand why it becomes $$\frac{1 - \sin^2 x}{\sin x \cos x}$$. 2. Recall that $$\csc x = \frac{1}{\sin x}$$, so $$\csc^2 x = \frac{1}{\sin^2 x}$$. 3. Substitute $$\csc^2 x$$ in the numerator: $$\frac{\frac{1}{\sin^2 x} \cdot \sin x - \sin x}{\cos x} = \frac{\frac{\sin x}{\sin^2 x} - \sin x}{\cos x} = \frac{\frac{1}{\sin x} - \sin x}{\cos x}$$ 4. Combine the terms in the numerator over a common denominator $$\sin x$$: $$\frac{\frac{1 - \sin^2 x}{\sin x}}{\cos x}$$ 5. Dividing by $$\cos x$$ is equivalent to multiplying by $$\frac{1}{\cos x}$$, so: $$\frac{1 - \sin^2 x}{\sin x} \cdot \frac{1}{\cos x} = \frac{1 - \sin^2 x}{\sin x \cos x}$$ 6. Note that $$1 - \sin^2 x = \cos^2 x$$ by the Pythagorean identity. 7. Therefore, the integrand simplifies to: $$\frac{\cos^2 x}{\sin x \cos x} = \frac{\cancel{\cos x} \cos x}{\sin x \cancel{\cos x}} = \frac{\cos x}{\sin x}$$ 8. This is $$\cot x$$, so the integral becomes: $$\int \cot x \, dx = \ln |\sin x| + C$$ Final answer: $$\int \frac{\csc^2 x \sin x - \sin x}{\cos x} \, dx = \ln |\sin x| + C$$