Show that $\lim_{R \to \infty} \int_0^\pi e^{-R \, \sin(t)}dt = 0$ 
Show that
$$\lim_{R \to \infty} \int_0^\pi e^{-R \, \sin(t)}dt = 0$$

I have to prove the integral from above. I tried the following,
$$
\lim_{R \to \infty} |\int_0^\pi e^{-R \, \sin(t)}dt| \leq \lim_{R \to \infty} \int_0^\pi |e^{-R \, \sin(t)}|dt,$$
and since $e^{-R\, \sin(t)} < e^{-R}$ we get
\begin{align}
\lim_{R \to \infty} \int_0^\pi |e^{-R \, \sin(t)}|dt &\leq \lim_{R \to \infty} |e^{-R}| \int_0^\pi dt\\
&=\lim_{R \to \infty} \pi \, e^{-R} = 0.&
\end{align}
From this we can conclude that $$\lim_{R \to \infty} \int_0^\pi e^{-R \, \sin(t)}dt = 0.$$
I am not sure if this is correct. I also heard that maybe I could do it with the dominated convergence theorem but am not sure how I would solve this, can someone help me with that?
 A: Observe that $\sin t \ge \frac{2t}{\pi}$ for $0 \le t \le \frac{\pi}{2} $. Thus $$\left\lvert\int_0^\pi e^{-R\sin t}\, dt\right\rvert = 2\int_0^{\pi/2}e^{-R\sin t}\, dt \le 2\int_0^{\pi/2}e^{-2Rt/\pi}\, dt = \frac{\pi(1 - e^{-R})}{R}\to 0\quad \text{as}\quad R\to \infty$$
A: Hint: let $0 < a < \frac \pi 2$ and try writing
$$\int_0^\pi e^{-R \sin t} \, dt = \int_0^a e^{-R \sin t} \, dt + \int_a^{\pi - a} e^{-R \sin t} \, dt + \int_{\pi - a}^\pi e^{-R \sin t} \, dt$$ to obtain the estimate
$$0 \le \int_0^\pi e^{-R \sin t} \, dt \le 2a + \pi e^{-R \sin a}.$$
A: Let $ R> 1 $. Using the fact that $ \left(\forall x\in\mathbb{R}\right),\ \operatorname{e}^{x}\geq 1+x $, we have : \begin{aligned}\int_{0}^{\pi}{\operatorname{e}^{-R\sin{x}}\,\mathrm{d}x}\leq\int_{0}^{\pi}{\frac{\mathrm{d}x}{1+R\sin{x}}}&=2\int_{0}^{+\infty}{\frac{\mathrm{d}y}{\left(1+y^{2}\right)\left(1+R\frac{2y}{1+y^{2}}\right)}}\\ &=2\int_{0}^{+\infty}{\frac{\mathrm{d}y}{y^{2}+2Ry+1}}\\ &=2\int_{0}^{+\infty}{\frac{\mathrm{d}y}{\left(y+R\right)^{2}+1-R^{2}}}\\ &=\frac{1}{\sqrt{R^{2}-1}}\int_{0}^{+\infty}{\left(\frac{1}{y+R-\sqrt{R^{2}-1}}-\frac{1}{y+R+\sqrt{R^{2}-1}}\right)\mathrm{d}y}\\ &=\frac{1}{\sqrt{R^{2}-1}}\left[\ln{\left(\frac{y+R-\sqrt{R^2-1}}{y+R+\sqrt{R^{2}-1}}\right)}\right]_{0}^{+\infty}\\ &=\frac{1}{\sqrt{R^{2}-1}}\ln{\left(\frac{R+\sqrt{R^{2}-1}}{R-\sqrt{R^{2}-1}}\right)}\\ &=\frac{2}{\sqrt{R^{2}-1}}\ln{\left(R+\sqrt{R^{2}-1}\right)}\underset{R\to +\infty}{\sim}\frac{2\ln{\left(2R\right)}}{R}\underset{R\to +\infty}{\longrightarrow}0\end{aligned}
A: For all $R\geq0$ we have $0\leq e^{-R\sin(t)}\leq 1$ (which is integrable over $(0,\pi)$) and $\lim\limits_{R\to \infty} e^{-R\sin(t)} = 0$ a.e. $t.$ So the dominated convergence theorem gives the result.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\lim_{R \to \infty}\int_{0}^{\pi}\expo{-R\sin\pars{t}}\,\,\dd t} =
\lim_{R \to \infty}\int_{-\pi/2}^{\pi/2}\expo{-R\cos\pars{t}}\,\,\dd t
\\[5mm] = & \
2\lim_{R \to \infty}\int_{0}^{\pi/2}\expo{-R\cos\pars{t}}\,\,\dd t =
2\lim_{R \to \infty}\int_{0}^{\pi/2}\expo{-R\sin\pars{t}}\,\,\dd t
\\[2mm] = & \
2\lim_{R \to \infty}
\overbrace{\int_{0}^{\infty}\expo{-Rt}\,\,\dd t}
^{\ds{Laplace's\ Method}} =
2\lim_{R \to \infty}{1 \over R} = \bbx{\color{#44f}{\large 0}}
\end{align}
