Question 63
Let $f : [-\frac{\pi}{2}, \frac{\pi}{2}] \to \mathbb{R}$ be a twice differentiable function such that
$$
(f''(x) - f(x)) \cdot \operatorname{tg}(x) + 2 \cdot f'(x) \ge 1, \quad \text{for any } x \in (-\frac{\pi}{2}, \frac{\pi}{2}).
$$
Prove that
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \cdot \sin x \, dx \geq \pi - 2.
$$
$$
(f''(x) - f(x)) \cdot \operatorname{tg}(x) + 2 \cdot f'(x) \ge 1, \quad \text{for any } x \in (-\frac{\pi}{2}, \frac{\pi}{2}).
$$
Prove that
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \cdot \sin x \, dx \geq \pi - 2.
$$