Prove integral inequality $\int_{0}^{\pi} e^{-A\sin\theta}d\theta < \frac{\pi}{A}$

Question

Given the Jordan's inequality

$$\frac{\sin \theta}{ \theta} \geq \frac{2}{ \pi} $$ $\text{ for }0\leq\theta\leq \frac{\pi}{2}$, I have to prove that

$$\int_{0}^{\pi} e^{-A\sin\theta}d\theta < \frac{\pi}{A}$$

But I'm not sure how they are related, or how I can use the first to prove the second. Any help is appreciated.

Thank you.

Answer 1

The given inequality $\sin \theta\geq \frac{2}{ \pi}\theta$ leads to $e^{-A\sin\theta } \le e^{- \frac{2A}\pi\theta}$. Then $$\int_{0}^{\pi} e^{-A\sin\theta}d\theta =2\int_{0}^{\pi/2} e^{-A\sin\theta}d\theta <2 \int_{0}^{\pi/2} e^{- \frac{2A}{\pi}\theta} d\theta = \frac{\pi}A(1-e^{-A})<\frac{\pi}A $$