How to show that $\lim\limits_{x\to +\infty}\int_{0}^{\frac{\pi}{2}} xe^{-x^2 \sin(t)}dt =0 $ I want to show that $\lim\limits_{x\to +\infty}\int_{0}^{\frac{\pi}{2}} xe^{-x^2 \sin(t)}dt  =0$ by other methods. I know a proof using $\sin(t)\geq \frac {2t}{\pi},\forall t\in [0,\frac{\pi}{2}]$
 A: The function $f(x) = xe^{-\sin(t)x^2}$ has its maximum at $x=\frac{1}{\sqrt[]{2\sin(t)}}$  if $t\in(0,\pi/2)$ and $x>0$. So:
\begin{align}
|xe^{-\sin(t)x^2}|\leq \frac{1}{\sqrt[]{2\sin(t)}}e^{-1/2}
\end{align}
It is clear that $$\int^{\pi/2}_0 \frac{1}{\sqrt[]{2\sin(t)}}e^{-1/2} dt < \infty $$
Since the function  behaves like $Ct^{-1/2}$ near $t=0$. Now DCT gives the required result, namely:
\begin{align}
\lim_{x\to\infty} \int_0^{\pi/2} xe^{-\sin(t)x^2}\, dt = \int_0^{\pi/2} \lim_{x\to\infty} xe^{-\sin(t)x^2}\, dt = \int^{\pi/2}_0 0\, dt =0
\end{align}
Edit. 
Since the OP asks for more methods, I'll give a method that looks a little bit as the one given in the comments below by @Tina. I use asymptotics. Note that the main contribution is near $t=0$ hence (you can make this argument more  rigourous!):
\begin{align}
\int^{\pi/2}_0 xe^{-x^2\sin(t)}\,dt \sim \int^{\pi/2}_0 xe^{-x^2t}\,dt \sim  \int^{\infty}_0 xe^{-x^2t}\,dt = \frac{1}{x}
\end{align}
as $x\to\infty$. This method does not only give you that it goes to zero, but it shows you that it goes to zero as $\frac{1}{x}$.
A: New proof: Note that
$$\int_{\pi/4}^{\pi/2}xe^{-x^2\sin t}\, dt \le \int_{\pi/4}^{\pi/2}xe^{-x^2/\sqrt 2}\, dt \to 0.$$
On $[0,\pi/4]$ we can let $t=\arcsin u$ to get
$$\int_0^{1/\sqrt 2}\frac{xe^{-x^2u}}{(1-u^2)^{1/2}}\, du \le \int_0^{1/\sqrt 2}xe^{-x^2u}\sqrt 2\, du,$$
and it's easy to show the last integral $\to 0.$ 
Previous proof: The concavity of $\sin t$ shows $\sin t \ge 2t/\pi$ on the interval $[0,\pi/2].$ Thus the integral in question is bounded above by $\int_{0}^{\frac{\pi}{2}} xe^{-2x^2t/\pi}dt.$ Let $t = u/x^2.$ Then the last integral equals
$$\frac{1}{x}\int_{0}^{x^2\pi /2} e^{-2u/\pi}\,du < \frac{1}{x}\int_{0}^{\infty} e^{-2u/\pi}\,du .$$
Since the last integral is finite, we see our integral is less than a constant times $\frac{1}{x},$ hence $\to 0.$
