In complex analysis I'm very often using inequality for $A >0$ that:

$$\int\_0^\pi e^{-A\sin(t)} dt < \frac \pi A$$

But to be honest I'm not sure how to prove this inequality. Moreover, recently I faced a problem with bounding integral:

$$\int\_0^\pi e^{-A\cos(t)} dt $$

and I'm not sure how can I obtain analogous inequality. I tried to use fact that $\sin(t + \frac \pi 2) = \cos(t)$:

$$\int\_0^\pi e^{-A\cos(t)} dt = \int\_0^\pi e^{-A\sin(t + \frac \pi 2)} dt$$

now by substitution $u = t + \frac \pi 2$ we have that:

$$\int\_{\frac \pi 2}^ \frac{3\pi}{2}e^{-A\sin(u)}du$$

But now I cannot use my upper bound for sine integral, since the interval is now $[\frac \pi 2; \frac {3\pi}{2}]$ instead of $[0, \pi]$. Could you please give me a hand with deriving analogous inequality?