Solution: Rewrite $ f(x) = \frac{\sin^2 x + 4}{\sin x} = \sin x + \frac{4}{\sin x} $. Let $ y = \sin x \in (0, 1] $. The function becomes $ f(y) = y + \frac{4}{y} $. The derivative $ f'(y) = 1 - \frac{4}{y^2} $ has critical point at $ y = 2 $, but $ y \leq 1 $. Analyze endpoints: as $ y \to 0^+ $, $ f(y) \to \infty $; at $ y = 1 $, $ f(1) = 1 + 4 = 5 $. The minimum is $ 5 $.

["# Simplifying and Minimizing a Trigonometric Expression: A Detailed Analysis", "Understanding and simplifying complex trigonometric expressions is essential for solving optimization problems in calculus. One classic example is rewriting and analyzing the function\n[\nf(x) = \frac{\sin^2 x + 4}{\sin x}, \quad x \in \mathbb{R},\ \sin x <br/>\ne 0,\n]\nto reveal its minimal value and behavior. By introducing substitution and applying basic calculus, we uncover a powerful optimization insight.", "## Rewriting the Function", "Begin with the original expression:\n[\nf(x) = \frac{\sin^2 x + 4}{\sin x}.\n]\nSince $\sin x <br/>\ne 0$, we split the fraction:\n[\nf(x) = \frac{\sin^2 x}{\sin x} + \frac{4}{\sin x} = \sin x + \frac{4}{\sin x}.\n]\nLet $ y = \sin x $. Because sine values lie in the interval $[-1, 1]$, and given the domain of the function excludes zero and focuses on $ y \in (0, 1] $, we define:\n[\nf(y) = y + \frac{4}{y}, \quad y \in (0, 1].\n]\nThis reformulation reduces a trigonometric problem into a pure algebraic minimization in one variable.", "## Applying Calculus: Finding the Minimum", "To find the minimum, compute the derivative:\n[\nf'(y) = 1 - \frac{4}{y^2}.\n]\nSet $ f'(y) = 0 $ to locate critical points:\n[\n1 - \frac{4}{y^2} = 0 \Rightarrow y^2 = 4 \Rightarrow y = 2.\n]\nHowever, $ y = 2 $ lies outside the domain $ (0, 1] $. Thus, no critical points exist within the interval. Without interior minima, the minimum must occur at an endpoint.", "Evaluate $ f(y) $ at the endpoint $ y = 1 $:\n[\nf(1) = 1 + \frac{4}{1} = 5.\n]\nAs $ y \ o 0^+ $, $ \frac{4}{y} \ o \infty $, so $ f(y) \ o \infty $. Therefore, the function attains its global minimum on $ (0, 1] $ at $ y = 1 $, with minimum value 5.", "## Interpreting the Result", "Since $ y = \sin x $, this means the minimum occurs when $ \sin x = 1 $, i.e., at $ x = \frac{\pi}{2} + 2k\pi $ for integers $ k $. At these points,\n[\nf(x) = \sin x + \frac{4}{\sin x} = 1 + 4 = 5.\n]\nThe unbounded growth near $ y = 0 $ illustrates a vertical asymptote, confirming no lower bound exists in the interval.", "## Conclusion", "Through substitution and derivative analysis, we simplified $ f(x) $ into $ f(y) = y + \frac{4}{y} $ with $ y \in (0, 1] $, identified the absence of critical points inside the interval, and confirmed the minimum value occurs at $ y = 1 $. This method demonstrates how strategic reformulation enables powerful insights in optimization—key to solving complex trigonometric expressions efficiently. Whether applied in physics, engineering, or pure mathematics, mastering such algebraic and analytic techniques is indispensable.", "---", "Key Takeaways:\n- Always simplify using substitution when $ y = \sin x \in (0, 1] $.\n- Critical points may lie outside the domain—always evaluate endpoints.\n- Asymptotic behavior governs function limits at domain boundaries.\n- Rigorous calculus ensures correct and robust conclusions.", "This approach not only solves the problem but builds a scalable framework for analyzing similar rational trigonometric functions."]