Evaluating $\lim_{r \to \infty} \int_0^\frac{\pi}{2} e^{-r\sin\theta} d\theta$ Evaluating $$\lim_{r \to \infty} \int_0^\frac{\pi}{2} e^{-r\sin\theta} d\theta$$
First for all, I need to show that $e^{-r\sin\theta}$ converges uniformly to a function $F(r)$ Then I can easily take the limit inside, since $e^{-r\sin\theta}$ is continuous.
How can I show it is uniformly continuous- my approach was:
$$|e^{-r\sin\theta}|\le |e^{-\delta \sin\theta}| \ for \ \delta \le r$$ 
Then If that $\int_0^\frac{\pi}{2}e^{-\delta \sin\theta}$ exists, in the sense that converges LHD is uniformly continuous so I can take the limit inside and arrive at the soulution. However I couldn't take the integration, any hints? 

Note that for $0 \le \theta \le \frac{\pi}{2}$, we have $\sin \theta \ge \frac{2}{\pi}\theta$, so:
$$0<\int_0^\frac{\pi}{2} e^{-r\sin\theta} d\theta \le \int_0^\frac{\pi}{2} e^{-(2r/\pi) \theta} d\theta \le \int_0^\infty e^{-(2r/\pi) \theta} d\theta=\frac{\pi}{2r}\to0$$
