Romanian Mathematical Olympiad - District Round

Since $\cos x > 0$ for any $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, the inequality in the statement can be written equivalently
$$(f''(x) - f(x)) \cdot \sin x + 2 \cdot f'(x) \cdot \cos x \geq \cos x, \quad \text{for any } x \in (-\frac{\pi}{2}, \frac{\pi}{2}),$$
or $g''(x) \ge 0, \forall x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, where $g: [-\frac{\pi}{2}, \frac{\pi}{2}] \to \mathbb{R}$ is the function defined by $g(x) = f(x) \cdot \sin x + \cos x, \forall x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

Hence, the function $g$ is convex, so that
$$\frac{g(x) + g(-x)}{2} \geq g(0) = 1, \quad \text{for any } x \in [-\frac{\pi}{2}, \frac{\pi}{2}].$$
Because $\int_{-a}^{a} h(x) \, dx = \int_{-a}^{a} h(-x) \, dx$ holds for any integrable function and any $a \ge 0$, we have
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} g(x) \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{g(x) + g(-x)}{2} \, dx \geq \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = \pi.$$

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \cdot \sin x \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (g(x) - \cos x) \, dx \geq \pi - 2.$$