Prove that the sequence $\{\int_0^{\pi/2} \sin(t^n)\, dt\}$ converges to 0. Prove that the sequence 
$$
\left\{\int_0^{\pi/2} \sin(t^n)\, dt :n\in\mathbb N\right\}
$$
converges to 0.

First split the integral at $t=1$.  Estimating the first half is simple; since $0 \leq \sin x \leq x$ for $0 \leq x \leq 1$ we have
$$
0 \leq \int_0^1 \sin(t^n)\,dt \leq \int_0^1 t^n \,dt.
$$
For the second half, make the change of variables $t^n = (\pi/2)^n u$ to get
$$
\int_1^{\pi/2} \sin(t^n)\,dt = \frac{\pi}{2n} \int_{(2/\pi)^{\Large n}}^1 \sin\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-1}\,du.
$$
Integrating by parts yields
$$
\begin{align}
&\frac{\pi}{2n} \int_{(2/\pi)^{\Large n}}^1 \sin\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-1}\,du \\
&\qquad = \left.- \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-1}\right|_{(2/\pi)^{\Large n}}^1 \\
&\qquad\qquad + \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-2}\, du \\
&\qquad = - \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \cos\left[\left(\frac{\pi}{2}\right)^n\right] + \frac{\cos(1)}{n} \\
&\qquad\qquad + \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-2}\, du \tag{$*$}
\end{align}
$$
Finally we use the fact that $|\cos(x)| \leq 1$ to estimate the remaining integral;
$$
\begin{align}
\left| \left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-2}\, du \right| &\leq -\left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 u^{1/n-2}\,du \\
&= \left(\frac{\pi}{2}\right)^{n-1} - 1 \\
&< \left(\frac{\pi}{2}\right)^{n-1}.
\end{align}
$$
It follows that
$$
\left| \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-2}\, du \right| < \frac{1}{n}.
$$
Each term in $(*)$ tends to zero as $n \to \infty$, so the result follows.
